Calculator validation library | Reddy Compressors

Why should you trust an engineering calculator?

A calculator that shows a plausible number has not proved anything. Its units, reference conditions, boundaries and decision rules could still be wrong. For compressed-air work, small slips are costly: confusing gauge and absolute pressure, mixing cfm and m³/min, or treating free air delivery as actual flow can push a compressor selection, a pipe size or an energy saving well off target.

This library sits behind the calculators in our knowledge and calculation centre. Each calculator has a set of reference cases that anyone can repeat by hand. When a formula, unit conversion or data import changes, the cases are run again and the results recorded. The library does not certify a customer’s installation or measurement. It shows that the arithmetic and logic behave as intended.

What goes into a reference case?

Every case has a permanent identity and enough detail for someone else to reproduce the answer without the software.

A short statement of the engineering question.
All inputs with their units and reference conditions (for example, FAD at inlet conditions, pressure in bar gauge).
The method or formula, with the source it comes from.
The expected result, worked independently of the calculator.
The tolerance allowed and why.
The expected behaviour if inputs are invalid or missing.
Version numbers for the case, the formula and the calculator it was last run against.

Cases are never edited to match a new result. If a case is found to be wrong, it is replaced by a new version and the old one is kept with a note explaining why.

Types of test in the library

A small, purposeful set of tests is more useful than a large number of near-identical ones. Each calculator is covered by several kinds of case.

Test type	Purpose	Compressed-air example
Normal case	Confirms ordinary arithmetic	Specific power from measured kW and FAD
Unit check	Same answer in different units	Pipe velocity with flow entered in m³/min and in cfm
Boundary case	Behaviour exactly at a limit	A dew point exactly at +3 °C against a Class 4 requirement
Invalid input	Correct refusal or warning	Pipe outside diameter larger than hole diameter
Missing input	Clearly flagged, never assumed	No ambient temperature given for a FAD correction
Property check	Result moves the right way	Higher pressure must never reduce compressor power
Deliberately wrong	Proves the test can fail	An expected answer altered to check that the comparison detects it

Expected answers must be independent

The expected result for a case must not come from running the calculator being tested. It is worked by hand, in a spreadsheet built separately, or from a published example. Where possible two different routes are used.

Worked example: a compressor draws 110 kW at a measured FAD of 18.6 m³/min.

Route one, metric: 110 ÷ 18.6 = 5.91 kW per m³/min.
Route two, imperial: 18.6 × 35.31 = 656.8 cfm, and 110 ÷ 656.8 × 100 = 16.75 kW per 100 cfm.
Cross-check: 16.75 ÷ 100 × 35.31 = 5.91 kW per m³/min.

Both routes agree, so the expected answer is fixed at 5.91 kW per m³/min with a stated tolerance. A calculator that returns 5.91 in one unit setting and something different in the other has a unit-handling fault.

Units, reference conditions and boundaries

Most real errors in compressed-air calculations are not arithmetic. They are about what the numbers mean. The library includes cases aimed at each common trap:

Gauge against absolute pressure: 7 bar(g) is about 8 bar absolute at sea level. Ratios must use absolute values.
Free air delivery against actual flow: FAD is at inlet conditions; actual flow at pressure is much smaller.
Standard against normal reference conditions: m³ at different reference temperatures and pressures are not the same quantity.
Pressure in psig and bar: 1 bar ≈ 14.5 psi, and both appear on our pages.
Power boundary: shaft power, motor input and package input give different specific power figures.

Each case names its boundary, so that a reviewer knows whether a result is comparable with a manufacturer’s data sheet.

Tolerances without moving engineering limits

Numerical tolerance is about rounding and floating-point arithmetic. It is not permission to move an engineering limit. A result of 6,001 particles per m³ against a limit of 6,000 is a failure even though the difference is tiny.

The library therefore sets two separate things for every decision case: a numerical tolerance for comparing the calculated value with the expected one, and an exact rule for the decision itself, including what happens at equality. A dew point of exactly +3 °C meets a “+3 °C or lower” requirement; the case states that explicitly so the calculator cannot treat it as a failure through rounding.

Measurement uncertainty is a third, separate idea. It belongs to the instrument and the method, not to the software, and is handled in our measurement uncertainty worksheet.

Worked example: warmer water does not prove more cooling

A cooling-load case checks that a calculator does not draw the wrong conclusion from temperature alone.

Condition	Water flow	Temperature rise	Heat removed
Before	2.0 kg/s	8 K	4.19 × 2.0 × 8 ≈ 67.0 kW
After	1.5 kg/s	10 K	4.19 × 1.5 × 10 ≈ 62.9 kW

The outlet is warmer after the change, but less heat is being removed, because flow has fallen. A calculator that reported “improved cooling” from the higher temperature rise would fail this case.

Import, calculation and export checked separately

Many calculators read logged data, calculate and then export a report. Each step can fail on its own.

Import: check that dates, time zones, decimal separators and units are read correctly, and that gaps in logged data are shown, not filled silently.
Time meaning: a 15-minute average power value must not be added up as if it were energy without multiplying by time.
Calculation: run the reference cases.
Export: check that the report shows the same values, units and rounding as the screen, and carries the calculator version.

Versions and coverage reporting

Four things change independently and each carries its own version: the engineering method or formula, the reference case, the calculator implementation and the recorded result. Keeping them apart answers the questions a reviewer actually asks. Was this result produced by the current formula? Which cases were run against the version now on the website? Did an older case pass against an older calculator that has since changed?

Coverage is reported by behaviour rather than by an impressive total. A statement such as “40 cases passed” says little if all 40 test the same normal input. A useful coverage report shows, for each calculator, whether normal cases, unit checks, boundary cases, invalid and missing inputs, property checks and import or export checks exist, and whether each was run against the current version. Gaps are listed openly so that they can be closed, rather than hidden inside a single pass rate.

When a test fails

A failed case is diagnosed before anything is changed. The answer could be wrong, the case could be wrong, or the inputs could have been read differently.

Re-run the case by hand from the stored inputs.
Check units and reference conditions on both sides.
Compare intermediate values, not only the final result.
Record the defect, the fix and the new version number.
Run the full set again, because a fix in one place can break another.

How Reddy Compressors helps

Our calculators are there to help plant engineers, maintenance managers and drilling contractors reach better first decisions. Behind them is a team that has worked with compressors since 2002, from our office at Tarnaka, Secunderabad, with more than 20 service engineers and technicians covering Telangana and Andhra Pradesh, and supply of compressors, rentals and spares across India.

We review calculator results with you against your actual machines, nameplates and site conditions.
We carry out air audits with power and flow logging when a decision needs measured data rather than estimates.
We sell, rent and service compressors of ELGi, Atlas Copco, Ingersoll Rand, KAESER, Chicago Pneumatic and other makes as an independent multi-brand specialist.
If you find a result that looks wrong, tell us; we check it against the reference cases.

What to send us

The calculator name and the inputs you used, with units.
The result you got and the result you expected.
Any data file you imported.
Machine nameplate photos if the calculation concerns a specific compressor.

Frequently asked questions

How do I know an online compressor calculator is accurate?

Check it against a case you can work by hand. Enter simple inputs, calculate the answer yourself with the stated formula, and compare. Then change the units and confirm the answer stays the same. A trustworthy calculator states its formula, units, reference conditions and limits, and refuses impossible inputs rather than returning a number. Our calculators are checked against documented reference cases.

What is a reference case for calculator validation?

A reference case is a fixed set of inputs with a known, independently worked answer and an allowed tolerance. It also records the method, units, expected behaviour for invalid inputs and version numbers. Running the calculator against the case shows whether it gives the right answer, and running it again after any change shows that nothing has been broken.

Why do two compressor calculators give different answers?

Usually because they use different assumptions rather than different arithmetic. Common differences are gauge or absolute pressure, free air delivery or actual flow, different reference conditions for m³, and shaft power versus package input power. Check what each calculator assumes and convert to the same basis before comparing. Rounding rarely explains large differences.

Is a small rounding difference acceptable in an engineering result?

A small numerical difference between a calculated and an expected value is acceptable within a stated tolerance. That tolerance must not change an engineering decision. If a result sits exactly at or very close to a limit, the decision rule, including what happens at equality, should be explicit, and measurement uncertainty should be considered separately from software rounding.

Can calculator results replace an air audit?

No. A calculator is only as good as its inputs. Estimates of flow, power, running hours or leakage can be far from reality. Calculators are useful for screening options and understanding the effect of changes. When money or production depends on the decision, measure: log power and flow, pressure and dew point under real operating conditions.

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Calculators and worked examples

These tools give indicative figures for screening, comparison and for preparing an enquiry. They run in your browser and send nothing. No prices or tariffs are assumed: enter your own to see costs. Efficiencies and other figures shown are example values: replace them with your own. Results are not a performance guarantee or a design certificate; final selection is confirmed by a service engineer against the machine data sheet and the site.

Compressed air demand estimate

Add up the air your tools and processes use, apply duty and simultaneity, then add leakage and growth allowances to reach a capacity to enquire about.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Unit for the air each tool uses: L/s
Group A: number of tools: 6
Group A: free air each (in the unit above): 4
Group A: duty (%): 40
Group B: number of tools: 2
Group B: free air each (in the unit above): 15
Group B: duty (%): 60
Group C: number of tools: 1
Group C: free air each (in the unit above): 40
Group C: duty (%): 80
Simultaneity of all tools (%): 65
Leakage allowance (%): 15
Growth allowance (%): 10

Result

Connected demand: 94 L/s = 199 cfm
Average demand (duty-weighted): 59.6 L/s = 126 cfm
Simultaneous peak: 61.1 L/s = 129 cfm
Design basis (higher of average and peak): 61.1 L/s = 129 cfm
Design capacity with allowances: 77.29 L/s = 4.64 m³/min = 163.8 cfm

Method. Average demand = Σ(number × air per tool × duty). Peak = Σ(number × air per tool) × simultaneity. The peak can never be lower than the average, so the design basis is the higher of the two. Design capacity = design basis × (1 + leakage) × (1 + growth).

What to take from it. Leakage and growth allowances move the answer as much as any single tool. Where you can, replace catalogue figures with measured ones.

Technical content reviewed by D. Srinivas Reddy, Managing Partner and service engineer, Reddy Compressors. Last updated 7 November 2026. How our engineering content is reviewed.

Flow and reference-condition conversion

Convert a flow quoted at one set of conditions to another, and between units. Different suppliers quote at different reference conditions, so always compare like with like.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Flow as quoted: 10
Unit: m³/min
Quoted at pressure (bar a): 1.013
Quoted at temperature (°C): 20
Convert to pressure (bar a): 1.013
Convert to temperature (°C): 0

Result

Converted flow: 9.32 m³/min
155.3 L/s
559.1 m³/h
329.1 cfm

Method. Q₂ = Q₁ × (P₁ / P₂) × (T₂ / T₁), with absolute pressure and absolute temperature. Dry-air basis; humidity is ignored. Unit factors: 1 m³/min = 16.667 L/s = 60 m³/h = 35.31 cfm.

What to take from it. A flow quoted at 20 °C and the same air quoted at 0 °C differ by about 7%. Check the reference conditions before comparing two machines.

Pressure unit and gauge/absolute conversion

Convert between bar, psi, kPa and MPa and between gauge and absolute pressure. A mix-up between gauge and absolute is a classic cause of wrong sizing.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Pressure: 7
Unit: bar g

Result

bar gauge: 7
bar absolute: 8.01
psi gauge: 101.5
psi absolute: 116.2
kPa gauge: 700
MPa gauge: 0.7

Method. Absolute pressure = gauge pressure + atmospheric (1.01325 bar at sea level). 1 bar = 14.504 psi = 100 kPa = 0.1 MPa. Hyderabad is about 540 m above sea level; atmospheric pressure there is roughly 0.95 bar, so enter the local value in the flow tool if it matters.

What to take from it. A 7 bar gauge system is 8.01 bar absolute. Ratio and flow-conversion work uses absolute pressure.

Receiver storage sizing

Estimate the receiver volume needed to ride through a short burst of demand without the pressure falling below the limit.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Burst demand above compressor output (m³/min free air): 5
Burst duration (s): 20
Allowed pressure fall (bar): 0.7
Receiver volume you already have (litres; 0 to skip the hold-up check): 1520

Result

Receiver volume needed: 2,412 litres
2.41 m³
Practical size (with 25% margin): 3,016 litres
Hold-up time of the receiver you have: 12.6 s before the pressure falls by 0.7 bar

Method. V = Q × t × P_atm / ΔP, where Q is the shortfall in free air (m³/min), t the time (min), ΔP the allowed pressure fall (bar) and P_atm = 1.013 bar. The same relation gives the hold-up time of an existing receiver: t = V × ΔP / (Q × P_atm). Isothermal; for a real system add a margin.

What to take from it. A receiver buys time. It does not create capacity: once it is spent the pressure keeps falling until the compressors catch up.

Pipe pressure drop and sizing

Check the pressure drop and air velocity in a compressed-air line, or find the smallest bore that keeps the drop within a target.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Flow (m³/min free air): 12
Inlet pressure (bar g): 7
Pipe length (m): 150
Inside diameter (mm): 80
Roughness (mm): 0.046
Fittings allowance (% extra length): 30
Air temperature (°C): 30
Target pressure drop (bar): 0.1

Result

Pressure drop: 0.059 bar (0.839% of the inlet gauge pressure)
Air velocity: 5.22 m/s
Reynolds number: 2,06,066
Smallest bore for the target drop: 73 mm inside diameter
Velocity check: Within the common rule-of-thumb range for mains (roughly up to 15–20 m/s)

Method. Darcy–Weisbach: ΔP = f × (L/D) × ρv²/2, with the friction factor f from the Swamee–Jain relation and air density taken at mean line pressure. Fittings are allowed for as extra equivalent length. Typical commercial steel roughness is about 0.046 mm.

What to take from it. Pressure drop varies roughly with the fifth power of the bore at constant flow, so a larger main is a cheap way to hold pressure.

Leakage flow and cost

Estimate what a leak of a given size costs a year. A small hole at working pressure passes more air than most people expect.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Equivalent hole diameter (mm): 3
Number of such leaks: 10
System pressure (bar g): 7
Discharge coefficient: 0.7
Specific power (kW per m³/min): 7
Running hours per year: 6000

Result

Air lost: 4.66 m³/min (77.73 L/s)
Compressor power wasted: 32.64 kW
Energy wasted per year: 1,95,870 kWh
Cost per year: Enter your electricity tariff (cost per kWh) above

Method. Choked (critical) flow through an orifice: ṁ = Cd × A × P₀ × √(k / R T) × (2/(k+1))^((k+1)/(2(k−1))), converted to free air at 1.013 bar and 20 °C. Cost = flow × specific power × hours × tariff. Cd of a sharp-edged hole is about 0.6–0.8.

What to take from it. Leaks are paid for at the full cost of compressed air in every running hour, including the hours when nothing is being produced.

Specific power comparison

Compare two compressors on what matters over life: the electrical power needed per unit of air delivered, and what that costs a year.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Machine A: input power (kW): 75
Machine A: delivered flow (m³/min): 11.5
Machine B: input power (kW): 80
Machine B: delivered flow (m³/min): 12.6
Running hours per year: 6000
Air actually needed (m³/min): 11

Result

Machine A specific power: 6.52 kW per m³/min
Machine B specific power: 6.35 kW per m³/min
Energy to deliver the air needed: A: 4,30,435 kWh per year
Energy to deliver the air needed: B: 4,19,048 kWh per year
Energy difference: 11,387 kWh per year in favour of B
Cost difference: Enter your electricity tariff (cost per kWh) above

Method. Specific power = input kW ÷ delivered flow (m³/min) at the same pressure and reference conditions. Annual cost = kW × hours × tariff. Use the full-load input power on the same test basis for both machines.

What to take from it. A small gap in specific power is multiplied by thousands of running hours. Compare on the same pressure and reference conditions.

Rental versus ownership

Find the point at which owning a compressor costs less than hiring one for the same duty.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Months of need: 18

Result

Result: Enter your own prices above (purchase price and monthly hire charge are needed; no prices are assumed)

Method. Rental total = monthly hire × months + fuel/power. Ownership total = purchase price + running cost × months + service. Break-even month = (purchase) ÷ (monthly hire + monthly rental running − monthly ownership running). All prices are entered by you; none is a quotation.

What to take from it. The answer depends on how long the need really lasts. Below the break-even hire is cheaper; beyond it ownership usually is.

Moisture load at the dryer inlet

Estimate how much water a compressor puts into the air system each hour, and how much a dryer or aftercooler must remove.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Compressor flow (m³/min free air): 12
Intake air temperature (°C): 35
Intake relative humidity (%): 70
Delivery pressure (bar g): 7
Air temperature at dryer inlet (°C): 38

Result

Water entering with intake air: 19.9 kg/h
Water still held at dryer inlet: 4.1 kg/h
Condensate removed before the dryer: 15.8 kg/h (about 379 L per day)

Method. Water in = air × relative humidity × saturation vapour density at intake (Magnus formula). Water the air can hold after compression = humidity ratio at saturation at the dryer inlet temperature and pressure × dry-air mass per m³ of intake air. Condensate = the difference. Flow is free air at intake conditions; indicative.

What to take from it. This water must be removed somewhere: aftercooler separator, drains, dryer. A failed drain sends it into the pipework.

Adsorption dryer purge air

A desiccant dryer uses some of the dried air to regenerate. Put a cost on that purge.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Dryer flow (m³/min): 12
Purge fraction (%): 15
Specific power (kW per m³/min): 7
Running hours per year: 6000

Result

Purge air: 1.8 m³/min
Compressor power for purge: 12.6 kW
Purge energy per year: 75,600 kWh
Purge cost per year: Enter your electricity tariff (cost per kWh) above

Method. Purge flow = dryer flow × purge fraction. Energy = purge flow × specific power × hours × tariff. Heatless dryers are commonly in the range of 15–20% purge; heated and blower-purge types use less. Use the datasheet value for the dryer in question.

What to take from it. Purge air is generated by the compressor and paid for like any other air. Use the dryer’s measured purge rather than the nameplate where you can.

Compressor room ventilation

Estimate the cooling air a compressor room needs to keep its temperature rise within a limit.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Total compressor input power (kW): 110
Heat remaining in room air (%): 90
Allowed room temperature rise (°C): 8

Result

Heat to be removed: 99 kW
Ventilation air needed: 10.26 m³/s
36,940 m³/h
21,742 cfm

Method. Heat to room ≈ input power × fraction rejected into the room (for air-cooled units, nearly all of it unless ducted). Airflow = heat ÷ (ρ × cp × ΔT), ρ = 1.2 kg/m³, cp = 1.005 kJ/kg·K. If the cooling air is ducted out of the machine, only the radiated part stays in the room.

What to take from it. This is a screening value. The machine maker states a required ventilation flow; follow it where it is higher.

Pneumatic cylinder air consumption

Work out the free air a cylinder uses per cycle and per minute.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Bore (mm): 63
Stroke (mm): 200
Working pressure (bar g): 6
Cycles per minute: 20
Number of cylinders: 8
Double acting? (2 = yes, 1 = single): 2
Rod diameter (mm; 0 to ignore): 20

Result

Free air per cycle (each): 8.2 litres
Total consumption: 21.9 L/s
1.311 m³/min = 46.3 cfm
Theoretical force, outstroke: 1,870 N (before friction; allow 10–20% less)
Theoretical force, return stroke: 1,682 N

Method. Air per stroke (free air) = swept area × stroke × (P + 1.013) / 1.013. The outstroke uses the full bore area π/4 × D²; the return stroke of a double-acting cylinder uses the bore area less the rod area π/4 × d². Enter rod 0 to ignore it (a small over-estimate).

What to take from it. Many small cylinders add up. Reducing stroke, pressure or cycle rate lowers consumption in direct proportion.

Abrasive blasting air demand

Estimate the air a blast nozzle takes, so the compressor and hose can be chosen to hold pressure at the nozzle.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Nozzle bore (mm): 8
Pressure at the nozzle (bar g): 7
Discharge coefficient: 0.9
Number of nozzles: 1
Allowance for pot, hose and leaks (%): 10

Result

Air demand, new nozzle: 4.69 m³/min = 166 cfm
Nozzle number (bore in 1/16 inch): 5
Air demand when worn one size (1/16 in larger bore): 6.74 m³/min = 238 cfm: size the compressor for this, then replace the nozzle

Method. Nozzle air treated as orifice flow at the nozzle pressure: free air = Cd × A × choked mass flow ÷ ρ₀. Venturi nozzles with an effective Cd near 0.9 are typical; use the supplier value. Add flow for the pot and for leaks. Nozzle numbers are the bore in sixteenths of an inch (No. 6 = 3/8 in = 9.5 mm). A nozzle wears larger in use, so size the compressor for the nozzle worn one size and replace it at that point.

What to take from it. A worn nozzle passes more air for less work. Check bore wear and keep the pressure at the nozzle, not at the compressor.

Pressure budget and the cost of excess pressure

Add up the pressure that is lost between the compressor and the point of use, and see what a lower set-point would be worth.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Pressure needed at the point of use (bar g): 6
Dryer and filter loss (bar): 0.5
Piping loss (bar): 0.3
Hose, fittings and regulator loss (bar): 0.3
Control band and safety margin (bar): 0.3
Current compressor set-point (bar g): 8
Compressor input power (kW): 75
Saving per bar reduced (%): 7
Running hours per year: 6000

Result

Total pressure losses: 1.1 bar
Control band and safety margin: 0.3 bar
Compressor pressure needed: 7.4 bar g (point of use + losses + band and margin)
Possible reduction from current set-point: 0.6 bar
Indicative energy saving: 18,900 kWh per year
Indicative cost saving: Enter your electricity tariff (cost per kWh) above

Method. Required compressor pressure = point-of-use pressure + sum of losses. Savings = power × (reduction in bar) × saving per bar. A saving of roughly 6–8% of compressor power per bar of pressure is commonly quoted; it is indicative and varies with machine type and control, so edit it to suit.

What to take from it. The saving is real only if the farthest and most demanding point still works at the lower set-point. Prove it before you change the setting.

Project payback and net present value

Test whether an energy or reliability project pays back, with a discount rate for the cost of money.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Discount rate (%): 10
Life (years): 8

Result

Result: Enter your own prices above (investment and annual saving are needed; no amounts are assumed)

Method. Simple payback = investment ÷ (annual saving − upkeep). NPV = −investment + Σ (saving − upkeep) / (1 + r)ⁿ over the life. Discounted payback is the year in which the discounted savings repay the investment; IRR is the discount rate at which NPV = 0. Enter your own figures.

What to take from it. A positive NPV at your own discount rate is a stronger test than payback alone, because payback ignores what happens after it.

Temporary air fuel logistics

Plan diesel supply for hired diesel-driven compressors: litres a day, tank autonomy, refuels and the order quantity.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Number of machines: 2
Loaded hours per machine per day: 14
Fuel burn when loaded (L/h): 40
Unloaded hours per machine per day: 4
Fuel burn when unloaded (L/h): 12
Stopped hours in the working day: 2
Tank size, nominal (L): 500
Filled to (% of tank): 95
Never run below (% of tank): 10
Possible delivery delay (days): 2
Days between deliveries: 7

Result

Fuel per machine per day: 608 litres
Fuel for the site per day: 1,216 litres
Full-load burn × clock hours, for comparison: 1,600 litres a day (32% higher)
Usable fuel per tank: 425 litres
Average burn while running: 33.8 L/h
Tank autonomy: 12.6 running hours
Refuels per machine per day: 1.4 on average (plan 2 visits on a full day)
Delay reserve: 2,432 litres
Order per delivery: 10,944 litres, less fuel already on site
Fuel cost per delivery period: Enter your diesel price (cost per litre) above

Method. Fuel per machine per day = loaded hours × loaded burn + unloaded hours × unloaded burn (stopped hours use none). Usable tank = tank × (fill level − lowest level). Autonomy = usable tank ÷ average burn while running. Weekly order = days × site litres a day + delay reserve, less fuel already on site. Take burn rates from the supplier datasheet for the actual load.

What to take from it. Full-load burn multiplied by clock hours overstates fuel. Plan refuelling at shift changes, and size site storage before deciding delivery frequency.

Compressor heat recovery

Estimate the heat from a compressor that could be put to use, and what it is worth.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Compressor input power (kW): 90
Recoverable fraction (%): 70
Hours when the heat is used: 4000
Displaced heater efficiency (%): 85

Result

Recoverable heat: 63 kW
Heat recovered per year: 2,52,000 kWh
Fuel energy displaced per year: 2,96,471 kWh of fuel
Value per year: Enter the cost of the fuel you would otherwise burn (per kWh) above

Method. Recoverable heat = input power × recoverable fraction. Value = heat × hours × cost of the fuel replaced ÷ efficiency of the displaced boiler or heater. Recoverable fraction depends on the machine and the heat sink; use a measured or supplier figure.

What to take from it. Heat is only worth recovering when there is a use for it for enough hours. Confirm the sink before the equipment.

Annular area and return velocity

In drilling, the air must lift the cuttings up the annulus. Check the return velocity from the hole and rod size.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Hole diameter (mm): 115
Rod outside diameter (mm): 89
Free air supplied (m³/min): 12
Bottom-hole back pressure (bar g): 1
Target return velocity (m/s): 15

Result

Annular area: 41.7 cm²
Return velocity: 24.16 m/s = 4,757 ft/min
Free air needed for the target velocity: 7.45 m³/min = 263 cfm at 15 m/s
Assessment: Within the commonly quoted range

Method. Annular area = π/4 × (D_hole² − D_rod²). Actual flow at the back pressure = free air × (P_atm / (P_back + P_atm)). Velocity = actual flow ÷ area. A return velocity often quoted for efficient cuttings lift is around 15–25 m/s; confirm for your material.

What to take from it. Low return velocity leaves cuttings in the hole; very high velocity wears the rods and hammer. Both are warnings, not verdicts.

Lifecycle cost comparison

Compare two options over their life: purchase, energy and maintenance. The cheapest machine to buy is rarely the cheapest to own.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Option A: input power at working load (kW): 75
Option B: input power at working load (kW): 66
Running hours per year: 6000
Years: 10
Discount rate (%): 8

Result

Option A energy per year: 4,50,000 kWh
Option B energy per year: 3,96,000 kWh
Lower energy use: Option B by 54,000 kWh per year
Lifecycle cost: Enter your electricity tariff (cost per kWh) above and the purchase and maintenance amounts

Method. Lifecycle cost (present value) = purchase + Σ (annual energy + annual maintenance) / (1 + r)ⁿ. Energy = input power at working load × hours × tariff. Equivalent annual cost = present value × r / (1 − (1 + r)^−years).

What to take from it. Energy usually outweighs the purchase price over the life of a compressor. Run the comparison with your own running hours.

Load/unload cycle energy

Turn measured load and unload times (or the logged percentage of time in each state) and measured package kW into average power, annual energy, the energy spent running unloaded, and specific power based on the air actually delivered. An optional second set of times compares a change.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Times entered as: Measured cycle times (seconds)
Loaded package power (kW): 82
Unloaded package power (kW): 28
Loaded time (s per cycle, or percent): 40
Unloaded time (s per cycle, or percent): 60
Stopped time, motor off, 0 if none (s per cycle, or percent): 0
Power while stopped, controls only (kW): 0
Running hours per year with this cycle pattern: 6000
Free air delivered when loaded, FAD (m³/min): 12.5
After change, optional: loaded time, all three 0 = not used (s or percent): 0
After change, optional: unloaded time (s or percent): 0
After change, optional: stopped time (s or percent): 0

Result

Cycle time: 100 s (36 cycles per hour)
Loaded fraction: 40 %
Unloaded fraction: 60 %
Unloaded power as a share of loaded power: 34.1 %
Average package power: 49.6 kW
Annual energy: 2,97,600 kWh per year
Unload energy (buys no air): 1,00,800 kWh per year = 33.9% of the annual energy
Loaded specific power: 6.56 kW per m³/min
Average air delivered: 5 m³/min
Effective specific power over the cycle: 9.92 kW per m³/min
Energy cost per year: Enter your electricity tariff (cost per kWh) above

Method. Two input modes. Cycle timing: enter measured seconds per cycle. Fraction: enter the share of time loaded, unloaded and stopped (for example 40, 60, 0, from controller loaded/running hours); no cycle time is reported in this mode. Average kW = (P_load × t_load + P_unload × t_unload + P_stopped × t_stopped) ÷ (t_load + t_unload + t_stopped); the states must not overlap. Annual kWh = average kW × the hours this pattern runs. Unload energy = P_unload × unloaded fraction × hours. Average air delivered = loaded fraction × loaded FAD; effective specific power = average kW ÷ average air delivered. Unloaded power is never assumed to be zero: enter measured package kW (unloaded read after blowdown has settled). Scenario comparison: enter the predicted times after a change (bigger receiver, wider band, auto-stop) in the “after change” fields; the same powers, FAD and hours are used and the kWh difference is shown. An auto-stop timer is modelled by moving unload time into the stopped state (check the motor’s permitted starts per hour). For different day and night patterns, run the tool once per pattern and add the results. The rapid-cycling warning below 30 s per cycle is a screening value used by this tool. Money is shown only when you enter your own tariff.

What to take from it. A large unload-energy share, or an effective specific power well above the loaded figure, means a fixed-speed machine is running lightly loaded: look at storage, pressure band, sequencing, auto-stop or a smaller/VSD trim machine.

Part-load fleet dispatch (single demand point)

For one plant demand, list every feasible combination of up to three compressors, put the others on base load and one on trim, and rank the states by total package kW.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Plant demand to be met (m³/min free air): 10
Machine A: type: Fixed speed, load/unload
Machine A: full-load flow, FAD at duty pressure (m³/min): 6
Machine A: package power at full flow (kW): 40
Machine A, fixed speed only: unloaded power (kW): 12
Machine A, VSD only: minimum stable flow (m³/min): 0
Machine A, VSD only: power at minimum flow (kW): 0
Machine B: type: Fixed speed, load/unload
Machine B: full-load flow, FAD at duty pressure (m³/min): 6
Machine B: package power at full flow (kW): 40
Machine B, fixed speed only: unloaded power (kW): 12
Machine B, VSD only: minimum stable flow (m³/min): 0
Machine B, VSD only: power at minimum flow (kW): 0
Machine C: type: Variable speed (VSD)
Machine C: full-load flow, FAD at duty pressure (m³/min): 7.5
Machine C: package power at full flow (kW): 48
Machine C, fixed speed only: unloaded power (kW): 0
Machine C, VSD only: minimum stable flow (m³/min): 2
Machine C, VSD only: power at minimum flow (kW): 15
Running hours per year at this demand: 6000

Result

1. A + C: 67 kW, 6.7 kW per m³/min; trim C at 4 m³/min (VSD), others fully loaded
2. B + C: 67 kW, 6.7 kW per m³/min; trim C at 4 m³/min (VSD), others fully loaded
3. A + B: 70.7 kW, 7.07 kW per m³/min; trim A at 4 m³/min (67% loaded), others fully loaded
Lowest-power state: A + C trim C at 4 m³/min (VSD), others fully loaded
Specific power of that state: 6.7 kW per m³/min
Annual energy in that state: 4,02,000 kWh per year
N+1 check (lose machine C): Passes other available machines can give 12 m³/min against 10 m³/min demand
Lowest-power state that also passes N+1: Same as above
Saving against the next higher-power state (A + B): 3.67 kW = 22,000 kWh per year

Method. Single-demand-point mode. Every on/off combination of the available machines is tried, with each online machine in turn as the trim machine. Base machines run fully loaded at their FAD and full-flow kW; the trim machine takes the remainder. A state is feasible only if the remainder is above 0, no more than the trim machine’s full flow, and (for a VSD trim) not below its minimum stable flow. Fixed-speed trim on load/unload: kW = x × loaded kW + (1 − x) × unloaded kW, with x = trim flow ÷ full flow (ignores extra blowdown losses with small storage). VSD trim: kW interpolated linearly between the minimum-flow and full-flow points. Specific power = total kW ÷ demand. N+1 check: with the largest online machine lost, all other available machines together must still cover the demand (assumes standby machines can start in time). Annual kWh = kW × hours at this demand. The lowest-power state that also passes N+1 is shown when the cheapest state does not. Use package kW and FAD at the same duty pressure for every machine. Not included: the demand sweep, annual load profile and scenario modes described on the page; run the tool once per demand band and weight each result by the hours in that band (annual kWh = Σ kW × hours).

What to take from it. Run the base machines fully loaded and give the trim role to the machine with the flattest part-load curve; two fixed-speed machines both partly unloaded is usually the most expensive way to meet the same flow.

Motor input from volts, amps and power factor

Screen a compressor’s electrical input from voltage, current and measured power factor in the loaded and unloaded states, and see how far a wrongly assumed power factor overstates unloaded power.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Supply: Three-phase, balanced (line-to-line voltage)
Voltage (V): 415
Loaded: line current (A): 150
Loaded: measured power factor: 0.86
Unloaded: line current (A): 70
Unloaded: measured power factor: 0.45

Result

Loaded apparent power: 107.8 kVA
Loaded real power: 92.7 kW
Loaded reactive power: 55 kVAr
Unloaded apparent power: 50.3 kVA
Unloaded real power: 22.6 kW
Unloaded reactive power: 44.9 kVAr
Unloaded power as a share of loaded power: 24.4 %
Unloaded kW if the loaded PF were wrongly assumed: 43.3 kW, overstated by 91% against the measured value

Method. Three-phase screen (balanced supply, line-to-line voltage V_L): S (kVA) = √3 × V_L × I ÷ 1,000; P (kW) = S × PF; Q (kVAr) = √(S² − P²). Single-phase: drop the √3 and use phase voltage. The power factor must be measured in the same operating state: an induction motor’s PF falls sharply when the compressor unloads, so applying the loaded PF to unloaded amps overstates unloaded kW. Not valid on the supply side of a VSD (harmonics) or with unbalanced phases: use a true-RMS power analyser there. State the measurement boundary (motor only, or whole package) with every result.

What to take from it. Judge unloaded consumption from measured kW or a measured power factor, never from amps with the full-load power factor.

Air demand duration curve summary

Paste regular flow readings from a log and get the exceedance levels used for sizing (E1, E10, E50, E90), the data coverage, and the hours and air volume above an installed capacity.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Flow readings at a regular interval (m³/min): 6.2, 5.8, 6.0, 6.5, 9.0, 12.5, 14.8, 15.6, 16.2, 15.1, 13.4, 11.0, 12.2, 14.0, 15.4, 17.8, 18.6, 14.2, 11.6, 9.4, 7.8, 6.6
Logging interval (minutes): 60
Scheduled hours the log should cover (h): 24
Installed capacity to compare against (m³/min): 15

Result

Valid readings: 22 covering 22 h
Coverage: 91.7 % of 24 scheduled hours
Maximum: 18.6 m³/min
E1 (exceeded 1% of the time): 18.6 m³/min
E10: 16.2 m³/min
E50 (median): 12.5 m³/min
E90 (base demand): 6.2 m³/min
Minimum: 5.8 m³/min
Time-weighted mean demand: 11.8 m³/min
Time above 15 m³/min: 6 h = 27.3% of valid hours
Air needed above that capacity: 522 m³ of free air in the logged period (peak excess 3.6 m³/min)

Method. Regular-interval mode: every valid reading carries the same time weight (the logging interval), so leave out missing, negative and fault readings rather than entering them as 0 (enter 0 only when the plant was confirmed stopped). Readings are sorted from highest to lowest. Ep = the flow exceeded (reached or exceeded) for at least p% of the valid hours = the k-th highest reading with k = ⌈p × n ÷ 100⌉. Valid hours = n × interval. Coverage = valid hours ÷ scheduled hours × 100. Hours above capacity = number of readings above it × interval; air above capacity = Σ(reading − capacity) × interval. All flows must be on one basis, ideally FAD in m³/min. E1 needs at least 100 readings to differ from the maximum (with fewer, k = 1 for E1); a week at 1-minute or 5-minute logging gives plenty. The 90% coverage warning is a screening value used by this tool. Not included: timestamp import, irregular intervals and gap filling.

What to take from it. Size the base machine near E90, the trim range up to about E10, and treat the short time above that as a storage or rental question, not a reason to buy for the maximum.

Adjusted energy baseline and avoided energy

Apply a baseline model (from your regression) to the reporting period’s production and days, then compare with the metered energy to get a like-for-like avoided energy figure.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Model constant a (kWh per period): 38000
Coefficient b (kWh per unit of production): 0.85
Coefficient c (kWh per production day or hour): 120
Coefficient d, 0 if not used (kWh per degree of mean ambient temperature): 0
Reporting period: production (same units as b): 24000
Reporting period: production days or hours (same basis as c): 26
Reporting period: mean ambient temperature, used only if d is not 0 (°C): 30
Non-routine adjustment, + or − (kWh): 0
Measured reporting-period energy (kWh): 52400
Average measured baseline-period energy, for the raw comparison (kWh): 57000

Result

Production term b × production: 20,400 kWh
Days/hours term c × days: 3,120 kWh
Temperature term d × temperature: 0 kWh
Adjusted baseline energy: 61,520 kWh (incl. non-routine 0 kWh)
Measured reporting-period energy: 52,400 kWh
Avoided energy (adjusted): 9,120 kWh = 14.8% of the adjusted baseline
Raw comparison (baseline average − reporting): 4,600 kWh
Difference made by the adjustment: 4,520 kWh

Method. Adjusted baseline = a + b × reporting production + c × reporting days (or hours) + d × reporting mean ambient temperature + non-routine adjustment. Avoided energy = adjusted baseline − measured reporting-period energy. The raw comparison (average baseline period − reporting period) is shown alongside, so both what the meter said and the adjusted result stay visible. Coefficients come from a regression on baseline data (see the fit tool below or your own analysis); set d = 0 if temperature is not used. Money is a separate step with your own tariff.

What to take from it. A saving smaller than the model’s normal period-to-period scatter is not yet demonstrated; and a reporting period outside the baseline range of production is weak evidence.

Baseline regression fit (one variable)

Fit energy = a + b × production to baseline-period data, check the fit, then apply it to a reporting period to get the adjusted baseline and avoided energy.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Baseline energy, one value per period (kWh): 59030, 58120, 60810, 56840, 61440, 61420, 58210, 55120, 58980, 59090, 62150, 58780
Baseline production for the same periods, same order (units): 19800, 21200, 22600, 20400, 23100, 24500, 18600, 17900, 20800, 22100, 23700, 21500
Reporting period: production (units): 24000
Non-routine adjustment, + or − (kWh): 0
Measured reporting-period energy (kWh): 52400

Result

Periods used: 12
Constant a (base load): 39,988 kWh per period
Slope b: 0.8983 kWh per unit of production
R²: 0.77
RMSE: 1,027 kWh per period
CV(RMSE): 1.7 % (within the 20% screening guide for monthly regression baselines)
Adjusted baseline for the reporting period: 61,546 kWh
Avoided energy: 9,146 kWh = 14.9% of the adjusted baseline
Avoided energy ÷ RMSE: 8.9 more than 2 × RMSE (screening value): clearly larger than the model scatter

Method. Ordinary least squares on n baseline periods: b = Σ(x − x̄)(E − Ē) ÷ Σ(x − x̄)², a = Ē − b × x̄. R² = 1 − SS_res ÷ SS_tot. RMSE = √(SS_res ÷ (n − 2)); CV(RMSE) = RMSE ÷ Ē × 100. Screening guide used by this tool: 20% for monthly regression baselines (the figure commonly cited from ASHRAE Guideline 14; check the edition you report against). The 2 × RMSE test for the saving and the 6-period minimum are also screening values used by this tool. Adjusted baseline = a + b × reporting production + non-routine adjustment; avoided energy = adjusted baseline − measured reporting energy. One driver only: for production plus days or temperature, fit with a spreadsheet and use the adjusted-baseline tool above. Use the same metering boundary in both periods.

What to take from it. Trust the avoided energy only if the fit is good (high R², low CV(RMSE)), the reporting production is inside the baseline range, and the saving is clearly larger than the RMSE.

Compressed-air station carbon ledger

Turn station electricity and any diesel into CO₂ emissions with your own emission factors, an intensity per 1,000 m³ of delivered air, and the effect of a pressure reduction.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Compressors: average combined input power (kW): 160
Dryers: average input power (kW): 5
Other electric auxiliaries in the boundary: average power (kW): 0
Operating hours per year: 6000
Grid emission factor, illustrative default (kg CO₂ per kWh): 0.7
Diesel burnt for the air service (litres per year): 0
Diesel emission factor (kg CO₂ per litre): 2.68
Average delivered free air (m³/min): 25
Planned pressure reduction (bar): 1
Compressor power saving per bar reduced (%): 7

Result

Compressor electricity: 9,60,000 kWh per year
Dryer electricity: 30,000 kWh per year
Other auxiliaries: 0 kWh per year
Station electricity: 9,90,000 kWh per year
Electricity emissions: 693 t CO₂ per year (6,93,000 kg)
Diesel emissions: 0 t CO₂ per year
Total for the boundary: 693 t CO₂ per year
Free air delivered: 90,00,000 m³ per year
Electricity emission intensity: 77 kg CO₂ per 1,000 m³ of free air
Energy saved by the pressure reduction: 67,200 kWh per year
CO₂ avoided by the pressure reduction: 47 t CO₂ per year

Method. Electricity: kWh = average kW × hours for compressors, dryers and other electric auxiliaries inside the boundary; emissions = kWh × grid factor (kg CO₂ per kWh, numerically the same as t CO₂ per MWh). Diesel: litres × diesel factor. Intensity = electricity emissions ÷ (average free air × 60 × hours) × 1,000. Pressure reduction: kWh saved = compressor kWh × bar reduced × saving per bar (rule of thumb about 7% per bar; edit it). The default grid factor of 0.7 is illustrative only, not an official value: use the CEA CO₂ baseline database edition for your reporting year and say whether it is location- or market-based. 2.68 kg CO₂ per litre of diesel is the IPCC 2006 default CO₂-only value at 0.84 kg/L; use your company’s source. Do not add this subtotal to a site total that already includes the same meters.

What to take from it. Measured feeder kWh and a labelled emission factor make the figure traceable; the cheapest tonne of CO₂ to remove is usually a leak or excess pressure.

Energy effect versus emission-factor effect

Split a change in compressed-air emissions between two periods into the part caused by using more or less energy and the part caused by a change in the emission factor.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Earlier period: electricity (kWh): 990000
Earlier period: emission factor (kg CO₂ per kWh): 0.7
Later period: electricity (kWh): 922800
Later period: emission factor (kg CO₂ per kWh): 0.68

Result

Earlier emissions: 693 t CO₂
Later emissions: 627.5 t CO₂
Energy effect: -47 t CO₂ (reduction from using less energy)
Factor effect: -18.5 t CO₂ (from the change in emission factor)
Total change: -65.5 t CO₂ = -9.5%

Method. Energy effect = (new kWh − old kWh) × old factor. Factor effect = new kWh × (new factor − old factor). The two add up exactly to new emissions − old emissions. Only the energy effect reflects a change in the compressed-air system; report the factor effect separately. Use the same boundary and the same basis (location- or market-based) in both periods. Factors here are illustrative: enter the published values you report against.

What to take from it. If emissions fell but the energy effect is zero or positive, the system did not improve; the grid did.

Specific power uncertainty budget

Combine the uncertainties of package power and delivered flow into an expanded uncertainty for specific power, find the dominant contributor, and test whether a difference from another option is real.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Measured package power (kW): 90
Power instrument figure is: Specification limit ± (rectangular, ÷√3)
Power logger and clamps accuracy (± % of reading): 1.5
Standard deviation of the repeated power readings (kW): 0.9
Number of averaged power readings: 36
Measured delivered flow (m³/min): 14
Flow meter figure is: Specification limit ± (rectangular, ÷√3)
Flow meter accuracy (± % of reading): 3
Flow installation effect, estimated (± % of reading): 2
Coverage factor k: 2
Specific power of the option to compare, 0 to skip (kW per m³/min): 0
Expanded uncertainty of that option, used only if a comparison value is entered (± %): 0

Result

Specific power: 6.43 kW per m³/min
Power accuracy (relative standard uncertainty): 0.87 %, 15% of the variance
Power repeatability (relative standard uncertainty): 0.17 %, 1% of the variance
Flow meter accuracy (relative standard uncertainty): 1.73 %, 59% of the variance
Flow installation effect (relative standard uncertainty): 1.15 %, 26% of the variance
Combined relative standard uncertainty: 2.26 %
Expanded uncertainty (k = 2): ± 4.52 % = ± 0.29 kW per m³/min
Result: 6.43 ± 0.29 kW per m³/min (6.14 to 6.72)
Dominant contributor: Flow meter accuracy

Method. GUM method for y = P ÷ Q (a quotient, so relative uncertainties combine). For the power instrument and the flow meter, choose how the figure is stated: a ± specification limit (rectangular, u = a ÷ √3) or a calibration-certificate expanded uncertainty at k = 2 (normal, u = U ÷ 2). The installation estimate is a ± limit, rectangular. Repeatability (Type A): u = s ÷ √n, relative to P. Combined relative standard uncertainty u_c = √(Σ u_i²); expanded U = k × u_c (k = 2 gives about 95% for a near-normal result with enough readings). Comparison: the difference between the two specific powers is treated as real if it exceeds √(U₁² + U₂²) in absolute terms (independent measurements at the same k). Not covered: wrong boundary or units (errors, not uncertainty) and an unrepresentative test period.

What to take from it. Spend measurement effort on the largest contributor; in compressed-air audits that is usually the flow measurement.

Air criticality priority score

Score each air-user group by consequence and likelihood of losing air, and flag any severe consequence that hangs on a single point of failure.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Consequence level of each group (1–5): 5, 3, 2, 1
Likelihood of loss of each group’s supply path (1–5): 3, 3, 3, 3
Single point of failure in the supply path? (1 = yes, 0 = no, per group): 1, 0, 0, 0

Result

Group 1 priority score: 15 of 25 (5 × 3) – FLAGGED: severe consequence with a single point of failure
Group 2 priority score: 9 of 25 (3 × 3)
Group 3 priority score: 6 of 25 (2 × 3)
Group 4 priority score: 3 of 25 (1 × 3)
Priority order: Group 1, Group 2, Group 3, Group 4 highest score first
Flagged groups: Group 1 record the protection chosen and how it will be tested

Method. Enter one value per user group, in the same order in every list. Priority score = consequence (1–5) × likelihood of loss of the supply path (1–5). Any group with consequence 5 that depends on a single point of failure (enter 1) is flagged whatever its score. The score ranks where standby, storage, spares or rental cover are worth considering; it does not prove reliability. Default: the page example (instrument air, packaging, cleaning, workshop).

What to take from it. Act on the flagged groups first, then the highest scores, and write down the protection chosen and how it will be tested.

N+1 capacity and storage ride-through check

Check whether the plant can meet demand with its largest compressor out of service, how much load must be shed if not, and how long the receiver bridges the gap.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Capacity of each compressor at the required pressure (m³/min): 6, 6
Normal demand (m³/min): 9
Demand that can be shed during an outage (m³/min): 0
Receiver volume (m³): 3
Normal minimum pressure (bar g): 7
Lowest pressure the critical users accept (bar g): 6

Result

Total installed capacity: 12 m³/min
Capacity with the largest unit out: 6 m³/min
Shortfall with the largest unit out: 3 m³/min
N+1 on compressors: Not met: 3 m³/min short; shed load or add capacity
Usable stored free air: 2.96 m³ (3 m³ × 1 bar ÷ 1.013)
Ride-through if all air is lost: 19.7 seconds at 9 m³/min
Ride-through with the largest unit out: 59.2 seconds at the 3 m³/min shortfall

Method. Shortfall with the largest unit out = demand − (total capacity − largest unit). Usable stored free air ≈ receiver volume × allowed pressure fall ÷ 1.013 bar (isothermal). Ride-through time = stored free air ÷ the flow not being supplied. Use capacities measured at the required pressure. This checks compressors only: dryers, controls, power and headers need their own N+1 test. Default: the page example (two 6 m³/min machines, 9 m³/min demand, 3 m³ receiver, 7 → 6 bar g).

What to take from it. Two machines are not N+1 unless the remaining ones carry the critical demand. Storage buys seconds for a standby start; it does not replace a missing machine.

Spare part stocking decision

For each part line, compare the time to restore the air supply without stock with the interruption the process can tolerate, and turn that into a stocking decision and a priority order.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Part type of each line (1–4): 3, 3, 3, 4, 1
Consequence of each line (1–5): 3, 3, 5, 4, 2
Restoration time without stock (days): 7, 3, 3, 42, 3
Tolerable interruption (days): 2, 2, 0, 2, 3

Result

Line 1 (critical failure part): Stock on site, or a guaranteed supply arrangement restoration 7 d vs tolerable 2 d; exposed, 15 consequence-days
Line 2 (critical failure part): Stock on site, or a guaranteed supply arrangement restoration 3 d vs tolerable 2 d; exposed, 3 consequence-days
Line 3 (critical failure part): Stock on site, or a guaranteed supply arrangement restoration 3 d vs tolerable 0 d; exposed, 15 consequence-days
Line 4 (major assembly): No stock: plan repair or exchange route and rental cover restoration 42 d vs tolerable 2 d; exposed, 160 consequence-days
Line 5 (consumable): Service plan: keep one set ahead of the next service restoration 3 d vs tolerable 3 d; not exposed
Exposed lines, largest first: Line 4, Line 1, Line 3, Line 2

Method. Enter one value per part line, in the same order in every list. Part type: 1 = scheduled consumable, 2 = wear part, 3 = critical failure part, 4 = major assembly. A line is exposed when restoration time without stock is longer than the tolerable interruption and the consequence is 3 (significant) or more. Exposed wear and critical failure parts → stock on site or a guaranteed supply arrangement; exposed major assemblies → repair/exchange route and rental cover rather than stock; consumables → service plan, one set ahead (and one set kept on site if exposed). Exposure = consequence × (restoration − tolerable) days, used only to rank. Where a standby exists, enter as tolerable interruption how long you accept running without reserve. Default: the page example (thermostatic valve kit, temperature sensor, dryer drain valve, airend, separator element; day figures are illustrative).

What to take from it. Record the time comparison for every line, not just the score: a cheap part with a high consequence (such as a shared dryer drain) is the line a price-based list misses.

Bad-actor ranking

Rank assets or failure modes by normalised failure rate and weighted burden, and find the few items that carry most of it.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Failure events per item: 9, 4, 6, 5, 3
Running hours per item (h): 7800, 5200, 1900, 8300, 8300
Downtime per item (h): 27, 16, 6, 4, 3
Labour per item (h): 18, 20, 9, 8, 5
Consequence weight per item (1–5): 5, 3, 1, 5, 3
Repeat failures after a fix, per item: 0, 0, 0, 0, 0

Result

Item 1 score: 225 1.15 failures per 1,000 h; MTBF 867 h; 52.1% of total score
Item 2 score: 108 0.77 failures per 1,000 h; MTBF 1,300 h; 25% of total score
Item 3 score: 15 3.16 failures per 1,000 h; MTBF 317 h; 3.5% of total score
Item 4 score: 60 0.6 failures per 1,000 h; MTBF 1,660 h; 13.9% of total score
Item 5 score: 24 0.36 failures per 1,000 h; MTBF 2,767 h; 5.6% of total score
Ranking by score: Item 1, Item 2, Item 4, Item 5, Item 3 highest first
Highest failure rate: Item 3 3.16 failures per 1,000 h
Items carrying at least 80% of the burden: Item 1, Item 2, Item 4 91% of a total score of 432

Method. Enter one value per asset or failure mode, in the same order in every list. Failure rate = events ÷ running hours × 1,000 (failures per 1,000 h); MTBF = running hours ÷ events. Score = (downtime hours + labour hours) × consequence weight (for example 1 fully redundant, 3 partial standby, 5 no backup). Items are ranked by score; the Pareto line lists the top items that together make up at least 80% of the total score. Items with failures that returned after a corrective action are marked. Default: the page example over twelve months (Compressor A, B, C, Dryer 1, Dryer 2).

What to take from it. The item with the highest failure rate is often not the one to investigate first: a redundant standby can fail often and cost little.

Bid normalisation: Net flow and comparable cost

Bring up to three compressor bids to one flow unit, check net flow against the requirement, then compare specific power, energy and installed cost on a like-for-like basis.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Net flow required (m³/min): 20
Running hours per year: 6000
Bid A: offered flow: 21.5
Bid A: flow unit: m³/min
Bid A: flow stated at the required pressure?: Yes
Bid A: internal consumption (m³/min): 1.8
Bid A: total package input power (kW; 0 if not stated): 124
Bid B: offered flow: 750
Bid B: flow unit: cfm (ft³/min)
Bid B: flow stated at the required pressure?: Yes
Bid B: internal consumption (m³/min): 1
Bid B: total package input power (kW; 0 if not stated): 126
Bid C: offered flow: 22
Bid C: flow unit: m³/min
Bid C: flow stated at the required pressure?: No / unclear
Bid C: internal consumption (m³/min): 0
Bid C: total package input power (kW; 0 if not stated): 0

Result

Bid A: offered flow: 21.5 m³/min
Bid A: net flow: 19.7 m³/min – SHORT by 0.3 m³/min
Bid A: specific power on net flow: 6.29 kW per m³/min (5.77 on offered flow)
Bid A: energy per year at the required net flow: 7,55,330 for information only: this bid cannot supply the required net flow. kWh = 6.29 × 20 m³/min × 6,000 h (enter a tariff for the cost). Assumes specific power holds at this flow; use the bidder’s part-load data for a VSD or a part-load profile
Bid A: comparable installed cost: Enter your own prices above
Bid B: offered flow: 21.24 m³/min
Bid B: net flow: 20.24 m³/min – meets, margin 0.24 m³/min
Bid B: specific power on net flow: 6.23 kW per m³/min (5.93 on offered flow)
Bid B: energy per year at the required net flow: 7,47,124 kWh = 6.23 × 20 m³/min × 6,000 h (enter a tariff for the cost). Assumes specific power holds at this flow; use the bidder’s part-load data for a VSD or a part-load profile
Bid B: comparable installed cost: Enter your own prices above
Bid C: offered flow: 22 m³/min (not at the required pressure)
Bid C: net flow: Not established clarification needed: ask for data at the required pressure
Bid C: package input power: Not stated ask the bidder for total package kW at the duty point
Bid C: comparable installed cost: Enter your own prices above
Eligible bids by energy at the required net flow: B (7,47,124 kWh) only one eligible bid with stated power
Bids that pass the net-flow check: B compare cost only between these

Method. Flow is converted to m³/min (1 cfm = 0.0283168 m³/min; 1 L/s = 0.06 m³/min); this is valid only when the bids use the same reference conditions. Net flow = offered flow − internal consumption (dryer purge, drains). A bid whose data is not at the required pressure is not credited: it needs clarification, not arithmetic. Specific power = total package input kW ÷ net flow. Energy per year is compared like-for-like at the required net flow: specific power (on net flow) × required net flow × hours, so a larger machine is not charged for air the plant does not use. This assumes the specific power holds at that flow; for a VSD or a plant that runs mostly at part load, use each bidder’s supported part-load power for each hour band instead. Bids are ranked on energy only if they pass the net-flow check. Comparable installed cost = quoted price + required additions − confirmed discounts. Default: the page’s worked example B (20.0 m³/min net required at 8 bar g).

What to take from it. A low price or a larger gross flow cannot offset a failed mandatory requirement. Compare cost only between bids that pass the net-flow check.

Performance acceptance check

Compare measured flow, specific power and pressure with the agreed requirement and tolerance, and reach a disposition: accept, accept with conditions, hold or reject.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Required FAD (m³/min): 20
Measured FAD, corrected per the agreed method (m³/min): 19.5
Measured package input power (kW): 126
Declared specific power (kW per m³/min): 6.4
Agreed flow tolerance (%): 4
Agreed specific power tolerance (%): 5
Required minimum pressure at the agreed point (bar g): 7
Lowest pressure measured there while drawing the required flow (bar g): 7
Measurement uncertainty to allow beyond the tolerance (%; 0 if the tolerance already includes it, as ISO 1217 Annex C tolerances do): 0
Mandatory evidence (dew point, pressure band, safety trips, documents): All recorded and valid

Result

Flow deviation: -0.5 m³/min (-2.5% of required; tolerance −4%)
Measured FAD: 689 cfm
Specific power: 6.46 kW per m³/min
Deviation from declared specific power: +0.96 % (tolerance +5%)
Pressure margin at the agreed point: 0 bar
Disposition: Accept file the record as the baseline for future service

Method. Flow deviation = (measured FAD − required FAD) ÷ required × 100. Specific power = package input kW ÷ measured FAD; its deviation is against the declared value. Pressure margin = lowest measured pressure at the agreed point − required minimum. Disposition: mandatory evidence missing or test invalid → Hold; a pressure shortfall, or a flow or specific-power result outside tolerance by more than the measurement uncertainty → Reject; outside tolerance but within the uncertainty → Hold (shortfall not proven, retest); all within tolerance → Accept, or Accept with conditions if minor items are open. Enter the tolerance the contract states (0 if the full figure is required). For reference, the ISO 1217 Annex C bands for machines above 15 m³/min are commonly quoted as ±4% flow and ±5% specific energy; confirm from the standard. ISO 1217 Annex C tolerances already include measurement uncertainty; do not add it again when the contract uses them. The Hold-within-uncertainty rule is a default screening rule used by this tool; use the decision rule written in your contract. Default: the page example (20.0 m³/min required, 19.5 measured, 126.0 kW, declared 6.4 kW per m³/min).

What to take from it. The tool applies a default rule (outside tolerance by more than the stated uncertainty = Reject). Replace it with the rule in your contract. One favourable number does not make an acceptance if evidence is missing.

Overhaul versus replacement: Energy and whole-pathway cost

Compare an overhauled machine with a replacement on energy at your demand, and, once you enter your own quotations and tariff, on present-value cost over the same period.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Air demand at duty (m³/min): 12
Loaded hours per year: 6000
Overhauled: expected specific power (kW per m³/min): 7
Replacement: expected specific power (kW per m³/min): 6.4
Pressure reduction possible with the replacement (bar): 0
Power saving per bar reduced (%): 7
Evaluation period (years): 10
Discount rate (%): 8

Result

Overhauled: power at duty: 84 kW
Replacement: power at duty: 76.8 kW
Overhauled: energy per year: 5,04,000 kWh
Replacement: energy per year: 4,60,800 kWh
Energy difference: 43,200 kWh per year in favour of replacement
Energy cost difference: Enter your electricity tariff (cost per kWh) above
Present-value cost and break-even tariff: Enter your own prices above (both pathway costs are needed; none is assumed)

Method. Power at duty = demand × specific power. If the replacement can run at a lower pressure, its power is reduced by (bar reduction × saving per bar); about 7% per bar is a common rule of thumb for screw compressors. Annual energy = power × loaded hours. Present-value cost = pathway cost + (annual energy cost + maintenance) × PVF, where PVF = (1 − (1 + r)^−n) ÷ r (= n when r = 0), n up to 50 years. Break-even tariff = the tariff at which both present values are equal. Pathway costs should include inspection, package parts, labour, rental cover, testing and acceptance. Residual value is not included. Default: the page example (12 m³/min, 6,000 h, 7.0 vs 6.4 kW per m³/min).

What to take from it. Use measured specific power for the worn machine and verified figures for the replacement at the same boundary; brochure values against worn-machine readings bias the answer.

Leak repair persistence

For one cohort of verified repairs that has reached its follow-up date, report inspection coverage, persistence and recurrence separately, with the evidence gap and the air still being lost.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Number of eligible repairs (follow-up date passed): 120
Number of repairs re-checked: 90
Number of repairs still passing: 72
Number of repairs leaking again (recurred): 18
Number of recurrences in the most frequent component group: 11
Estimated flow of recurred leaks (m³/min): 0.9
Estimated flow of new leaks found nearby (m³/min): 0
Specific power (kW per m³/min): 7
Running hours per year: 6000

Result

Inspection coverage: 75 % (90 of 120)
Persistence rate: 80 % of re-checked repairs still pass
Recurrence rate: 20 % of re-checked repairs leak again
Other outcomes among re-checked: 0 repairs (neither passing nor recurred)
Unknown status: 30 eligible repairs not yet re-checked
Persistence range over all eligible: 60–85 % (all unknowns failed – all unknowns passed)
Most frequent component group: 61.1 % of recurrences (11 of 18)
Residual leak flow: 0.9 m³/min (32 cfm)
Compressor power for residual leaks: 6.3 kW
Energy per year: 37,800 kWh
Cost per year: Enter your electricity tariff (cost per kWh) above

Method. Coverage = re-checked ÷ eligible. Persistence = still passing ÷ re-checked; recurrence = leaking again ÷ re-checked (inspected items only). Unknown = eligible − re-checked. The persistence range assumes every unknown repair failed (low) or passed (high), each over all eligible repairs. Residual flow = estimated flow of recurred leaks + new leaks nearby; power = flow × specific power. Other outcomes = re-checked − passing − recurred (for example component removed). Default: the page example (120 eligible, 90 re-checked, 72 passing, 18 recurred, 11 of them quick couplings, 0.9 m³/min).

What to take from it. Report coverage beside persistence. A high persistence figure from a low coverage describes a small, possibly unrepresentative sample.

Section losses at the worst moment of a pressure dip

Enter the pressures read at the same moment at the four section boundaries during a low-pressure event. The tool splits the loss into treatment, distribution and point-of-use sections and points to the part of the system to test first.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Compressor-room header (bar g): 6.9
After dryer and filters (bar g): 6.8
Main header near the affected area (bar g): 6.7
Machine inlet, after its hose and coupling (bar g): 5.2
Minimum pressure the machine needs at that point (bar g): 5.8
Compressor state when the dip started: At least one compressor unloaded, off or below full speed

Result

Treatment loss: 0.1 bar
Distribution loss: 0.1 bar
Point-of-use loss: 1.5 bar
Total, compressor room to machine inlet: 1.7 bar
Machine inlet against its minimum: -0.6 bar (below the minimum)
Largest section loss: Point of use (hose, coupling, regulator, drop) 88% of the total
Where to look first: Local path: the area header is above the machine’s minimum, so the shortfall is created between the header and the machine. Check hose, coupling, regulator and drop (gauge before and after them during a cycle) before touching supply.

Method. Section loss = upstream pressure − downstream pressure, all in bar(g) at the same timestamp: treatment = compressor-room header − after dryer/filters; distribution = after dryer/filters − area header; point of use = area header − machine inlet (hose, coupling, regulator, drop). If the area header is at or above the machine minimum while the inlet is below it, the shortfall is created locally. If the compressor-room header is above the minimum but the area header is below it, the shortfall is created in treatment or distribution. Only if the compressor-room header is itself below the minimum does the compressor state at the start of the dip separate capacity (all loaded), control response or storage (a machine unloaded, off or slow) and availability (a machine tripped or isolated). A negative section loss means the two sensors disagree: compare them on a common point.

What to take from it. Raising the compressor set-point costs energy everywhere. It is only the fix when the loss is in supply, not in one machine’s hose and regulator.

Pressure event count, depth, duration and data coverage

Paste the machine-inlet pressure log at a fixed interval. The tool finds every event below the machine’s minimum and reports its depth, duration and recovery time, together with how much of the campaign the log actually covers.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Machine-inlet pressure readings in time order (bar g): 6.1, 6.1, 6.0, 6.0, 5.9, 5.7, 5.4, 5.2, 5.3, 5.6, 5.9, 6.0, 6.1, 6.1, 6.0, 6.1, 5.8, 5.6, 5.5, 5.7, 5.9, 6.0, 6.1, 6.1
Logging interval (s): 10
Minimum pressure the machine needs (bar g): 5.8
Campaign length the log should cover (min): 5

Result

Logged time: 4 min (24 readings)
Data coverage: 80 % of the campaign
Events below the minimum: 2
Time below the minimum: 80 s = 33.3% of the logged time
Lowest reading: 5.2 bar g
Deepest event: lowest pressure: 5.2 bar g, 0.6 bar below the minimum, starting at reading 6
Deepest event: duration: 50 s
Deepest event: recovery time: 30 s from the lowest reading back to the minimum
Longest event: duration: 50 s, starting at reading 6
Event rate: 30 events per hour of logged data

Method. An event starts at the first reading below the minimum and ends at the first reading back at or above it. Each reading is taken to represent one logging interval, so duration = number of readings below the minimum × interval. Depth = lowest reading in the event. Recovery time = time from the lowest reading to the first reading back at or above the minimum. Time below the minimum is duration-weighted over the logged time. Coverage = logged time ÷ campaign length; events in unlogged periods cannot be seen.

What to take from it. Few events in a log with poor coverage means insufficient data, not a minor problem. Report coverage with every event count.

Cooler approach and margin against a baseline

Compare a cooler reading now with a baseline taken at the same load, pressure and fan state. The tool splits the rise in leaving temperature into the part caused by a hotter cooling medium and the part caused by a worse cooler approach, and shows the margin to the limit.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Baseline: cooling medium entering (°C): 30
Baseline: fluid leaving the cooler (°C): 40
Now: cooling medium entering (°C): 38
Now: fluid leaving the cooler (°C): 54
Applicable limit for the leaving temperature (°C): 55
Measurement uncertainty of the comparison (K): 1

Result

Approach: baseline: 10 K
Approach: now: 16 K
Change in leaving temperature: +14 K
of which hotter cooling medium: +8 K
of which change in approach: +6 K
Margin to limit: baseline: 15 K
Margin to limit: now: 1 K
Investigation direction: Approach has risen: look at the cooler core, fan, belt and airflow (or water-side fouling), and at the thermostatic valve and oil condition.

Method. Approach = fluid leaving the cooler − cooling medium entering it (for an air-cooled oil cooler: oil out − cooling air in). Rise in leaving temperature = change in cooling-medium inlet + change in approach. Margin = applicable limit − leaving temperature. A change in approach larger than the measurement uncertainty points to the cooler core, fan or airflow; an unchanged approach with a hotter outlet points to a hotter heat sink (ventilation, recirculation). Readings must be at matched duty.

What to take from it. An unchanged approach clears the cooler but does not restore lost margin. A hot room can still trip a healthy machine.

Water-side heat duty of a water-cooled compressor

Compare two runs of the same cooling-water circuit. A larger water temperature rise does not mean more heat is being removed: heat removed depends on flow multiplied by rise.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Reference run: water flow (m³/h): 6
Reference run: water in (°C): 30
Reference run: water out (°C): 36
Later run: water flow (m³/h): 3
Later run: water in (°C): 30
Later run: water out (°C): 39

Result

Reference run: water rise: 6 K
Reference run: heat removed: 41.8 kW
Later run: water rise: 9 K
Later run: heat removed: 31.4 kW
Change in water flow: -50 %
Change in water rise: +50 %
Change in heat removed: -25 %

Method. Q (kW) = density × flow × specific heat × (water out − water in) ÷ 3,600, with flow in m³/h, density 1,000 kg/m³ and specific heat 4.18 kJ/(kg·K) for clean water.

What to take from it. If the rise goes up while the heat removed goes down, the flow has fallen. Check strainers, valves, pump and scaling before the compressor.

Particle concentration from active sampling time

Convert a particle count into a concentration using the air actually sampled. Pauses with no flow must be removed from the sampling time, or the concentration is understated.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Number of particles counted in the size band: 12
Sampling flow (L/min): 2
Recorded sampling time (min): 25
Time with no flow during the run (min): 5
Limit for this size band (particles per m³; 0 = none entered): 0

Result

Active sampling time: 20 min
Accepted sample volume: 0.04 m³
Concentration: 300 particles per m³
Concentration if the pauses were ignored: 240 particles per m³ (understated by 20%)
Effect of one particle more or less: 25 particles per m³

Method. Active time = recorded time − time with no flow. Accepted volume (m³) = sampling flow (L/min) × active time (min) ÷ 1,000. Concentration = particles counted ÷ accepted volume. One particle more or less changes the result by 1 ÷ volume.

What to take from it. Correcting the volume changes the number; it does not identify the source of the particles.

Total oil from fractions with “less than” results

Add the liquid/aerosol and vapour oil results and compare the possible total with a total-oil limit. A “less than” result is not zero: the true value can be anywhere up to the reporting limit.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Liquid and aerosol oil result (mg/m³): 0.006
Liquid and aerosol result is: a measured value
Oil vapour result (mg/m³): 0.008
Oil vapour result is: a “less than” value (below the reporting limit)
Total-oil limit at this point (mg/m³): 0.01

Result

Lowest possible total oil: 0.006 mg/m³
Highest possible total oil: 0.014 mg/m³
Limit: 0.01 mg/m³
Result: Compliance cannot be shown: the total could be above the limit.

Method. Lowest possible total = sum of the measured values (a “less than” fraction counts as 0). Highest possible total = sum of all values (a “less than” fraction counts at its reporting limit). If the highest possible total is at or below the limit, compliance is shown; if the lowest possible total is above it, the limit is exceeded; otherwise compliance cannot be shown. A fraction that was not measured leaves the highest total unknown. Stated measurement uncertainty is not included here.

What to take from it. An aerosol-only result cannot support a total-oil statement. Measure the vapour fraction, or use a more sensitive method if its reporting limit is too high.

Particle count, sample volume and ISO 8573-1 class

Turn particle-counter counts into concentrations per m³ for the three size bands, find the ISO 8573-1:2010 particle class they support, and see how firmly the sample volume supports it.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Counter sampling flow (L/min): 28.3
Active sampling time (min): 10
Number of particles counted, 0.1–0.5 µm: 3000
Number of particles counted, 0.5–1 µm: 1200
Number of particles counted, 1–5 µm: 2
Required particle class: Class 1

Result

Sample volume: 0.283 m³
Concentration 0.1–0.5 µm: 10,601 per m³ (within the Class 1 limit)
Concentration 0.5–1 µm: 4,240 per m³ (within the Class 2 limit)
Concentration 1–5 µm: 7 per m³ (within the Class 1 limit)
Particle class supported by this sample: Class 2 ISO 8573-1:2010, all three bands
Required class: Class 1 not met by this sample
Count the Class 1 limit corresponds to (1–5 µm): 2.8 particles in this volume (counting uncertainty about ±59%)
Effect of one particle more or less: 3.53 per m³

Method. Volume (m³) = counter flow (L/min) × sampling time (min) ÷ 1,000. Concentration = count ÷ volume. ISO 8573-1:2010 maximum particles per m³ (0.1–0.5 / 0.5–1 / 1–5 µm): Class 1: 20,000 / 400 / 10; Class 2: 400,000 / 6,000 / 100; Class 3: not specified / 90,000 / 1,000; Class 4: not specified / not specified / 10,000; Class 5: not specified / not specified / 100,000. The class is the lowest number whose limits are all met. Counting uncertainty at the target limit ≈ ±1/√N, where N is the count the limit corresponds to in this volume. Classes 6 and 7 are mass-based (ISO 8573-8) and Class 0 is by agreement; neither is covered here. Check the edition your requirement cites.

What to take from it. A short run at a tight class rests on a handful of particles. Sample enough volume that the limit corresponds to a meaningful count.

Acceptance coverage at the use point

Separate “sampled” from “accepted”. For each contaminant in the requirement, record what the use-point result actually shows. The tool reports whether sampling is complete and whether the requirement has been demonstrated, and lists what is still missing.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Number of planned sampling visits: 4
Number of sampling visits completed: 4
Particles at the use point: Sampled, but the sample is invalid
Water (pressure dew point) at the use point: Valid and meets the requirement
Oil (aerosol, liquid and vapour) at the use point: Not sampled

Result

Sampling: Complete 4 of 4 planned visits
Acceptance coverage: Incomplete 1 of 3 required contaminants accepted
Disposition: Requirement not yet demonstrated.
Still needed: Particles: repeat with a clean, short sampling assembly after checking flushing and volume; Oil: sample it at the use point

Method. Sampling is complete when every planned visit was made. Acceptance needs, for every contaminant in the requirement, a valid and complete result that meets the requirement at the use point (or a documented equivalent point). An invalid sample is not a failure, and an uncollected or incomplete sample is never a pass. One valid adverse result means the requirement is not met.

What to take from it. Every visit can be completed while the requirement is still not demonstrated. Report both, separately.

Pumping-test discharge stability and volume check

Screen the discharge readings for one pumping step: convert them to one unit, check they stayed within the test plan’s tolerance, look for unit mix-ups, and compare the pumped volume with the totaliser or tank.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Discharge readings during the step, in time order: 152, 150, 149, 151, 148, 150, 147, 146
Unit of the readings: litres per minute
Allowed deviation from the mean in the test plan (%): 5
Elapsed time of the step (h): 8
Time the pump was stopped within the step (min): 0
Volume from totaliser or tank for the step (m³; 0 = not available): 70

Result

Mean discharge: 8.95 m³/h = 149.1 L/min = 2.49 L/s
Lowest and highest reading: 8.76 to 9.12 m³/h
Largest deviation from the mean: 2.1 % (within the 5% tolerance)
Drift, first to last reading: -3.9 %
Hours actually pumping: 8 h
Volume from readings: 71.6 m³
Against totaliser or tank: +2.3 % (1.6 m³)
Screening result: Fit for a volume check and for drawdown screening (hydrogeologist to interpret)

Method. Each reading is converted to m³/h (1 L/min = 0.06 m³/h; 1 L/s = 3.6 m³/h; 1 imperial gallon = 4.54609 L; 1 US gallon = 3.78541 L). Mean = average of the readings. Deviation = (reading − mean) ÷ mean. Drift = (last − first) ÷ first. Volume = mean rate × hours actually pumping (elapsed time − stops). A reading that differs from the median by a factor near 3.6, 4.5, 16.7 or 60 is flagged as a possible unit mix-up. Screening result: not yet usable for drawdown analysis if the discharge is outside tolerance, a reading is zero, a unit mix-up is flagged or the pump stopped within the step; fit for a volume check when the totaliser or tank agrees within the tolerance. Screening only: aquifer properties and yield are for the hydrogeologist.

What to take from it. An average that includes a pump stop is not a pumping rate. Split the test at every interruption.

Drawdown with a measuring-point correction

Calculate drawdown from a static level and a series of pumping water levels, correcting readings taken after the measuring point changed (for example from casing top to ground level) and flagging rises during pumping.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Levels recorded in: metres
Static water level below the original measuring point (m or ft, as selected): 12.4
Pumping water levels in time order, below the measuring point in use at the time (m or ft): 15.2, 16.1, 16.6, 16.9, 16.4, 16.5, 16.6
Reading number from which levels were taken from ground level (0 = no change): 5
Stick-up: height of the original measuring point above ground (m or ft; used only if a reading number above is entered): 0.6
Rise between readings that counts as a step (m): 0.2

Result

Correction applied: +0.6 m to readings 5 to 7 (ground level converted to the original measuring point)
Final pumping water level: 17.2 m below the original measuring point
Final drawdown: 4.8 m
Maximum drawdown: 4.8 m
Drawdowns after correction: 2.8, 3.7, 4.2, 4.5, 4.6, 4.7, 4.8 m
Steps (rises) in the uncorrected data: at reading 5
Steps remaining after correction: none

Method. All levels are depths below the original measuring point (casing top). A level read from ground level is converted by adding the stick-up (height of the casing top above ground). Drawdown = pumping water level − static water level, both from the same point. A drawdown that falls by more than the step threshold between two readings is flagged: during constant-rate pumping the level should not rise, so check for a pump stop, a flow change or a misread tape. Original readings are kept; the correction is shown separately.

What to take from it. A sudden shallower reading with no change in pumping is usually a changed measuring point or a pump stop, not the aquifer. Record the correction; do not delete the reading.

Well production balance

Close the water account for one boundary and one time interval: well production, other inflows, deliveries, other known outflows and the change in storage. The residual is what is not accounted for.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Well totaliser: opening reading (m³): 12450
Well totaliser: closing reading (m³): 12546
Totaliser rollover value: the reading at which it returns to zero, e.g. 100,000 for a 5-digit m³ meter (0 = not applicable): 0
Other inflows: tanker, municipal supply, second well (m³): 12
Deliveries to use, metered (m³): 88
Other known outflows: flushing, backwash, drain-down (m³): 5
Storage at the start (m³): 30
Storage at the end (m³): 43

Result

Well production: 96 m³
Total inflow: 108 m³
Storage change: +13 m³ (accumulation, not consumption)
Residual: +2 m³
Residual as share of total inflow: +1.9 %
Statement: 2 m³ unexplained: more water entered than can be accounted for. Check for overflows, leaks, unmetered use and meter over-reading.

Method. Production = closing − opening totaliser reading (if the meter rolled over: rollover value − opening + closing). Storage change = closing volume − opening volume. Residual = production + other inflows − deliveries − other known outflows − storage change, also shown as a share of total inflow. A positive residual means more water entered than can be accounted for; a negative one means more appears to have left or been stored than entered. Either is an investigation item, not proof of a leak or meter fault.

What to take from it. Report a residual as “unexplained”, not as “leaked”, until the meters, tank levels and timings have been checked.

Supported well production: Water time, not power-on time

Work out the production a pump actually delivered in a time window, from the hours it delivered at each confirmed rate, and see how far power-on hours or the whole window overstate it.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Length of the time window (h): 10
Nominal pump rate (m³/h): 12
Confirmed delivery rate 1 (m³/h): 12
Hours delivering at rate 1 (h): 6
Confirmed delivery rate 2 (m³/h): 9
Hours delivering at rate 2 (h): 2
Hours energised with no delivery: dry running, air lock (h): 0.5
Hours stopped (h): 1.5

Result

Supported production: 90 m³
Hours delivering water: 8 h
Average rate while delivering: 11.25 m³/h
Energised hours: 8.5 h
Energised hours × nominal rate: 102 m³ +13.3% against supported production
Whole window × nominal rate: 120 m³ +33.3% against supported production
Hours not accounted for: 0 h

Method. Supported production = Σ(confirmed delivery rate × hours delivering at that rate). Energised hours = delivering hours + hours energised with no delivery (dry running, air lock). The overstated figures are energised hours × nominal rate and window × nominal rate. Unaccounted hours = window − delivering − dry − stopped.

What to take from it. Run hours are not water. Dry running also warns of a pump set too high or a falling water level.

DTH hammer duty against two compressors

Check whether each of two compressors can supply a DTH hammer’s documented air consumption at the pressure the hammer needs, after hose, string and water-column losses, and with what margin.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Hammer working pressure at the hammer (bar g): 21
Hammer air consumption at that pressure (cfm): 950
Pressure loss from compressor to hammer (bar): 1
Water column above the hammer to be lifted (m): 0
Compressor A: maximum working pressure (bar g): 17
Compressor A: data-sheet flow at the required pressure (cfm): 1150
Compressor B: maximum working pressure (bar g): 24
Compressor B: data-sheet flow at the required pressure (cfm): 1000
Site derating of delivered flow: heat, altitude, condition (%): 0

Result

Hammer working pressure: 21 bar g = 305 psig
Water column allowance: 0 bar
Required compressor discharge pressure: 22 bar g = 319 psig
Hammer consumption: 950 cfm = 26.9 m³/min
Compressor A: Fails: cannot reach 22 bar g (maximum 17 bar g); its flow does not count
Compressor B: flow available at 22 bar g: 1,000 cfm = 28.3 m³/min
Compressor B: margin: +5.3 % (covers the duty)
Result: Only compressor B covers the duty

Method. Required discharge pressure = hammer working pressure + pressure loss from compressor to hammer (hose, swivel, drill string, check valves) + water column to be lifted (0.0981 bar per metre of water). A compressor whose maximum working pressure is below the required pressure fails, whatever its flow. Available flow = data-sheet flow at the required pressure × (1 − site derating). Margin (%) = (available flow − hammer consumption) ÷ hammer consumption × 100. Use the same flow basis (FAD or scfm) for hammer and compressor. 1 bar = 14.504 psi; 1 m³/min = 35.31 cfm. Hole cleaning (annular velocity) is a separate check.

What to take from it. Compare flows at the same pressure. More cfm at a lower pressure does not run a hammer that needs higher pressure.

Air motor duty check

Translate the driven machine’s speed and torque through the gearbox to the motor shaft, then check one candidate motor’s published running and starting torque and its air use against the supply.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Required speed at the driven shaft (rpm): 300
Running torque at the driven shaft (N·m): 40
Breakaway (starting) torque at the driven shaft (N·m): 60
Gear ratio (motor speed ÷ load speed; 1 for direct drive): 5
Gearbox efficiency (%): 90
Pressure at the motor inlet while running (bar g): 6
Pressure at which the motor curve was read (bar g): 6
Candidate motor: torque at the required motor speed (N·m): 11
Candidate motor: starting torque (N·m): 16
Candidate motor: air consumption at that point (L/s free air; 0 if not yet known): 0
Number of these motors running together: 1
Spare compressor capacity at that pressure (m³/min free air; 0 if not yet checked): 0

Result

Shaft power at the driven machine: 1,257 W (1.26 kW)
Motor speed: 1,500 rpm
Motor running torque required: 8.89 N·m
Motor shaft power required (incl. gearbox loss): 1,396 W (1.4 kW)
Motor starting torque required: 13.33 N·m
Running torque margin: 23.8 % (pass)
Candidate motor output at that speed (from its curve): 1.73 kW, against 1.4 kW required at the motor shaft
Starting torque margin: 20 % (pass)
Air demand: not calculated enter the consumption from the maker’s curve at this pressure and point
Result: Hold torque checks done so far pass; still needed: air consumption

Method. Shaft power P (W) = 2 × π × n × T ÷ 60. Motor speed = load speed × gear ratio. Motor torque required = load torque ÷ (gear ratio × gearbox efficiency); the same for the breakaway (starting) torque. Margins = motor torque from the curve ÷ torque required − 1. Air demand = motor consumption × number running together. Result: Pass when running torque, starting torque and air supply are all met on data read at or below the pressure available at the motor inlet; Hold when data are missing or read at a higher pressure; Fail when any requirement is not met. For a direct drive enter ratio 1 and efficiency 100%. Peak and stall torque are not checked here: compare them with the maker’s curve separately.

What to take from it. Running power alone can choose the wrong motor. Check starting torque against the breakaway load, and read every figure at the pressure the motor actually sees while running.

Pneumatic conveying air estimate

A first estimate of conveying air from throughput and a trial solids loading ratio, with the actual velocity at pick-up and at the end of the line. For an enquiry brief, not a conveying design.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Solids throughput (t/h): 5
Conveying mode: Dilute phase (blower, suspended flow)
Trial solids loading ratio (kg solids per kg air): 10
Line pressure at pick-up (bar g; negative for vacuum systems): 1
Line pressure at the destination (bar g; negative for vacuum systems): 0
Air temperature in the line (°C): 20
Pipe internal diameter (mm): 68
Target pick-up velocity (m/s): 16

Result

Solids mass flow: 5,000 kg/h (1.389 kg/s)
Air mass flow: 500 kg/h
Free air: 416.7 m³/h = 6.94 m³/min = 245 cfm
Actual flow at pick-up: 209.7 m³/h (0.0583 m³/s)
Velocity at pick-up: 16.04 m/s
Velocity at the destination: 31.87 m/s (1.99 × pick-up)
Bore giving the target pick-up velocity: 68.1 mm internal diameter
Pick-up velocity check: Within the typical 15–25 m/s dilute-phase range typical 15–25 m/s, material-dependent

Method. Air mass flow = solids mass flow ÷ solids loading ratio (SLR). Free air = air mass flow ÷ free-air density, taken as 1.2 kg/m³ (air at about 1.01325 bar a and 20 °C). Actual flow at a line pressure p = free air × 1.01325 ÷ (p + 1.01325) × (T + 273.15) ÷ 293.15 (free air taken at 20 °C). Velocity = actual flow ÷ pipe area. The bore for a target pick-up velocity is √(4 × actual flow ÷ (π × velocity)). Bend, lift, feeder and filter losses are not calculated: the pressures are your inputs. For a vacuum (suction) system enter negative gauge pressures, pick-up nearest atmospheric. The velocity flags (15–25 m/s dilute pick-up, 30 m/s at the destination, 10 m/s dense phase) are screening values used by this tool, not material data.

What to take from it. The same free-air flow can be too little in one pipe and right in another. Air velocity rises as pressure falls towards the destination, so check both ends.

Dryer capacity correction

Apply one dryer model’s published correction factors to its rated capacity, find the margin against your design flow, and the minimum rated capacity that would meet it.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Design flow (m³/min FAD): 15
Rated capacity of the candidate dryer (m³/min): 17
Pressure correction factor (from the data sheet): 0.96
Inlet temperature correction factor: 0.72
Ambient or cooling-water correction factor: 0.84
Dew point correction factor (1 if not published): 1
Margin you want to keep (%): 10

Result

Product of factors: 0.581
Corrected capacity: 9.87 m³/min (349 cfm)
Margin against design flow: -34.2 %
Assessment: Undersized at these conditions
Minimum rated capacity (no margin): 25.83 m³/min
Minimum rated capacity with your margin: 28.42 m³/min: choose the next catalogue size at or above this
Condition reducing capacity most: Inlet temperature factor 0.72

Method. Corrected capacity = rated capacity × F(pressure) × F(inlet temperature) × F(ambient) × F(dew point). Margin (%) = (corrected capacity − design flow) ÷ design flow × 100. Required rated capacity = design flow ÷ product of factors (× (1 + your margin) if you keep one); choose the next catalogue size at or above it. Use the factors for the exact model and revision from its data sheet; enter 1 for a factor the maker does not publish. If the maker publishes a combined chart instead, enter its single factor and 1 for the rest.

What to take from it. A dryer that looks comfortably sized at rating conditions can be heavily overloaded on a summer afternoon. Recalculate after any change in pressure, inlet temperature, location or demand.

Compressor room cooling load

Split a compressor’s input power into its heat paths, find the residual heat the room must lose, and size the cooling air and cooling water for it.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Compressor package input power (kW): 100
Heat to cooling water leaving the room (kW): 0
Heat recovered and used outside the room (kW): 0
Heat leaving with the compressed air or other documented paths (kW): 5
Heat from air-cooled dryers into the room (kW): 0
Other room loads: solar gain, other equipment (kW): 0
Allowed cooling-air temperature rise (K): 8
Cooling-air density (kg/m³): 1.13
Allowed cooling-water temperature rise (K): 10

Result

Heat into the room air from the compressor: 95 kW
Room residual heat load: 95 kW
Cooling-air mass flow: 11.82 kg/s
Cooling-air volume flow: 10.46 m³/s = 37,644 m³/h = 22,156 cfm
Air flow if the allowed rise were halved: 75,287 m³/h
Cooling-water flow: none no heat to water entered (air-cooled package)

Method. Q_room = Q_input − Q_water (leaving the room) − Q_recovered (used outside the room) − Q_other (leaving the room) + Q_dryers + Q_extra (other room loads), all in kW of heat. Cooling air: mass flow = Q ÷ (cp × ΔT), cp = 1.005 kJ/kg·K; volume = mass ÷ air density. Cooling water: mass flow = Q_water ÷ (4.19 × ΔT), volume at 1,000 kg/m³. Steady state, one operating case; enter the peak case for fan and pump sizing.

What to take from it. Temperature rise is the most sensitive input. Design the room for the case with heat recovery off, and check that hot exhaust cannot recirculate to the compressor intake.

Expansion scenarios with N+1 check

Test baseline, phased growth, leak repair and an added compressor against the capacity left when the largest machine is out of service.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Existing compressors: FAD of each at duty pressure (m³/min): 8, 8, 8
Summer derating of FAD (%): 0
Baseline peak demand (m³/min): 15
Of which measured leakage (m³/min): 3
Phase 1: connected flow of new users (m³/min): 4
Phase 1: simultaneity (%): 80
Phase 2: connected flow of new users (m³/min): 6
Phase 2: simultaneity (%): 70
Leakage expected to be recovered by repairs (m³/min): 1.5
Added compressor FAD (m³/min; 0 for none): 10

Result

Installed capacity: 24 m³/min
Available with the largest unit out: 16 m³/min
Productive demand at baseline: 12 m³/min (baseline peak less measured leakage)
Leakage share of baseline: 20 %; repairs as entered free 1.5 m³/min, 1.5 m³/min of leakage remains
Baseline: 15 m³/min peak: passes N+1, margin 1 m³/min
Phase 1, no action: 18.2 m³/min peak: fails N+1, short by 2.2 m³/min
Phase 1 after leak work: 16.7 m³/min peak: fails N+1, short by 0.7 m³/min
Phase 2, no action: 22.4 m³/min peak: fails N+1, short by 6.4 m³/min
Phase 2 after leak work: 20.9 m³/min peak: fails N+1, short by 4.9 m³/min
Phase 2 after leak work with the added unit: 20.9 m³/min peak against 24 available N+1: passes, margin 3.1 m³/min
Installed capacity with the added unit: 34 m³/min: recheck dryer, receivers and header for this
Smallest single added unit for Phase 2 after leak work: 4.9 m³/min FAD, before any margin

Method. Available with the largest unit out (N+1) = Σ compressor FAD − largest FAD, after any summer derating. Phase demand = previous peak + new users’ connected flow × simultaneity factor. Productive demand = baseline peak − measured leakage. Leak work subtracts the expected recovery (which cannot exceed the measured leakage). A scenario passes when its peak demand ≤ the N+1 capacity. Minimum single added unit = Phase 2 demand − (Σ existing − largest existing); if that exceeds the largest existing unit, no single machine restores N+1.

What to take from it. Leak work is usually the cheapest capacity, but it may only delay the need for a new machine. After adding a compressor, recheck the dryer, receivers and header for the new installed capacity.

Branched network pressure drop

Solve a tree network: a trunk from the source to one junction and up to three branches, each feeding a demand. Find the pressure at each user and the critical branch.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Source pressure (bar g): 7
Air temperature (°C): 30
Pipe roughness (mm): 0.046
Fittings allowance (% extra length): 0
Trunk: length (m): 60
Trunk: internal diameter (mm): 52.5
Branch B: flow (m³/min free air): 6
Branch B: length (m): 40
Branch B: internal diameter (mm; needed if it has flow): 40.9
Branch B: minimum pressure needed (bar g; 0 if not set): 0
Branch C: flow (m³/min free air): 4
Branch C: length (m): 25
Branch C: internal diameter (mm; needed if it has flow): 27.2
Branch C: minimum pressure needed (bar g; 0 if not set): 6.5
Branch D: flow (m³/min free air; 0 if unused): 0
Branch D: length (m): 30
Branch D: internal diameter (mm; needed if it has flow): 27.2
Branch D: minimum pressure needed (bar g; 0 if not set): 0

Result

Trunk: 0.109 bar drop; 10 m³/min at 10.1 m/s
Junction pressure: 6.89 bar g
Branch B: 0.099 bar drop; 6 m³/min at 10.2 m/s
Pressure at user B: 6.79 bar g
Branch C: 0.231 bar drop; 4 m³/min at 15.4 m/s
Pressure at user C: 6.66 bar g, margin 0.16 bar over 6.5
Critical node: User C smallest margin, 0.16 bar
Largest drop on the critical path: Branch C 0.231 bar of 0.34 bar total

Method. Trunk flow = sum of branch flows. Each segment: Darcy–Weisbach ΔP = f × (L/D) × ρv²/2, Swamee–Jain friction factor, air density at the segment’s mean pressure, flow converted from free air to actual flow at line pressure and temperature. Junction pressure = source − trunk drop; user pressure = junction − branch drop. The critical node is the one with the smallest margin over its minimum pressure. Use real internal diameters; lengths may already include fittings (then enter 0% allowance). This tool solves a single-source tree only (one trunk, one junction, up to three branches). Ring mains, loops (Hardy Cross), several sources and field calibration are NOT solved here; for those, a model is built during a pressure-drop audit. An unused branch (flow 0) is ignored.

What to take from it. Fix the largest drop on the critical path first. Upsizing one short branch is usually far cheaper than raising compressor pressure for the whole plant.

Vacuum pump-down time and required pumping speed

Estimate how long a pump takes to bring a vessel from one absolute pressure to another, what the connecting line costs in speed, the lowest pressure a continuing gas load allows, and the speed needed for a target time.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Evacuated volume incl. piping and connected equipment (unit below): 2
Volume unit: m³
Start pressure (mbar a): 1013
Target pressure (mbar a): 10
Pump speed at its inlet, average over the range from the curve (unit below): 100
Line conductance, same unit as the speed (0 = no line loss): 0
Speed and conductance unit: m³/h
Continuing gas load: leaks + outgassing + vapour (0 = none): 0
Gas load unit: mbar·m³/h
Required pump-down time, for the speed check (min; 0 = skip): 0
Liquid water in the vessel or product?: No (dry)
Temperature of that water or wet product (°C): 20

Result

Evacuated volume: 2 m³ = 2,000 litres
Effective speed at the vessel: 100 m³/h = 27.78 L/s (no line loss entered)
Pressure ratio p₀ ÷ p₁: 101.3 ln = 4.618
Ideal pump-down time (no gas load): 5.54 min (333 s, 0.0924 h)

Method. Ideal pump-down: t = (V ÷ S_eff) × ln(p₀ ÷ p₁), with V the whole evacuated volume (vessel, piping, filters, condensers), absolute pressures and S_eff the average effective speed over the range. Effective speed through a line: 1/S_eff = 1/S_pump + 1/C. A continuing gas load Q (leaks, outgassing, vapour, as pressure × volume flow) sets a floor p_floor = Q ÷ S_eff; with that load the time becomes t = (V ÷ S_eff) × ln((p₀ − p_floor) ÷ (p₁ − p_floor)), and a target at or below the floor is never reached. Holding duty: the effective speed needed to hold a pressure p against Q is Q ÷ p. Required speed for a target time: S_eff = V × ln(p₀ ÷ p₁) ÷ t (solved numerically when there is a gas load), then the pump speed that gives it through the line. Units: 1 L/s = 3.6 m³/h; 1 mbar·L/s = 3.6 mbar·m³/h. Water boils when the pressure falls below its vapour pressure (Magnus formula: about 23 mbar at 20 °C, 42 mbar at 30 °C). Constant speed is assumed: real pumps lose speed towards their ultimate pressure, so treat the result as a lower bound and check the maker’s curve, or use the banded tool.

What to take from it. If the target is near the floor Q ÷ S_eff, find and cut the gas load (leaks, wet product, a condenser) before buying a bigger pump; if the line conductance is close to the pump speed, fix the line first.

Banded pump-down time from the pump curve

Split the pressure range into bands, enter the average pump speed in each band from the curve and the line conductance, and add the band times, with a continuing gas load if there is one.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Evacuated volume incl. piping and connected equipment (m³): 2
Band boundary pressures, start to target (mbar a): 1013, 500, 100, 30, 10
Average pump speed in each band, from the curve (m³/h): 100, 100, 100, 100
Line conductance: one value for all bands or one per band (m³/h; 0 = no line loss): 0
Continuing gas load (mbar·m³/h; 0 = none): 0

Result

Band 1: 1,013 → 500 mbar a: 50.8 s (0.847 min, 0.01412 h); S_eff 100 m³/h
Band 2: 500 → 100 mbar a: 115.9 s (1.931 min, 0.03219 h); S_eff 100 m³/h
Band 3: 100 → 30 mbar a: 86.7 s (1.445 min, 0.02408 h); S_eff 100 m³/h
Band 4: 30 → 10 mbar a: 79.1 s (1.318 min, 0.02197 h); S_eff 100 m³/h
Total pump-down time: 5.54 min (333 s, 0.0924 h)
Slowest band: Band 2 35% of the total time

Method. For each band from pₐ down to p_b: S_eff = 1 ÷ (1/S + 1/C), floor p_f = Q ÷ S_eff, t = (V ÷ S_eff) × ln((pₐ − p_f) ÷ (p_b − p_f)) (with Q = 0 this is the ideal V/S × ln(pₐ/p_b)). The band times are added. If a band end is at or below its floor, the pump-down stops there. Smaller bands give a better answer where the curve changes quickly. Enter one conductance for all bands, or one per band (conductance falls at lower pressure in viscous flow).

What to take from it. The band that takes longest shows where the time goes: near the target it is usually the gas load or the line, not the pump size.

Vacuum line conductance and effective speed

Work out the conductance of a round vacuum line in viscous or molecular flow, add valves and traps in series, and see how much of the pump speed actually reaches the chamber.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Pump speed at its inlet (L/s): 100
Line inside diameter (mm): 40
Line length (m): 3
Flow regime: Decide from pressure and diameter
Mean pressure in the line (mbar a): 0.001
Other series elements: valves, traps, filters (L/s; 0 = none): 0
Gas load for the required-speed check (mbar·L/s; 0 = skip): 0

Result

Mean free path of air at this pressure: 66 mm
Knudsen number λ ÷ d: 1.65 regime from this: molecular
Line conductance (molecular): 2.58 L/s = 9.29 m³/h
Total series conductance: 2.58 L/s
Effective speed at the chamber: 2.52 L/s = 9.06 m³/h, 2.5% of the pump speed
Conductance against pump speed: 0.03 times: the line, not the pump, is the limit

Method. Long round pipe, air at about 20 °C, d and L in cm: molecular flow C ≈ 12.1 × d³ ÷ L L/s; viscous (laminar) flow C ≈ 135 × d⁴ × p̄ ÷ L L/s with p̄ the mean pressure in mbar. Regime from the Knudsen number Kn = λ ÷ d, mean free path λ ≈ 0.066 mm at 1 mbar, inversely proportional to pressure: viscous below Kn 0.01, molecular above 0.5, transitional between (this tool then adds the two, an indicative figure; use maker’s data). Series elements: 1/C_total = Σ 1/Cᵢ. Effective speed: 1/S_eff = 1/S + 1/C_total. Required effective speed for a gas load: S_eff = Q ÷ p (the mean line pressure is taken as the chamber pressure for this check). 1 L/s = 3.6 m³/h. Short pipes (L less than about 20 d) need an entrance correction; near atmospheric pressure flow may be turbulent or choked and the laminar formula is optimistic.

What to take from it. Size the line so its total conductance is several times the effective speed you need, at the lowest pressure you run: a pump twice as large on a poor line changes almost nothing.

Vacuum rate-of-rise gas load

Turn an isolation test (pressure rise over time with the pump valved off) into the gas load entering the system, compare a later interval to see whether it is a leak or outgassing, and find the speed needed to hold a working pressure.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Isolated volume incl. pipework (litres): 500
Pressure at isolation (mbar a): 0.1
Pressure at the end of the interval (mbar a): 0.6
Interval (s): 100
Later interval of the same test: pressure rise (mbar; 0 = not measured): 0
Later interval: duration (s): 100
Working pressure for the pumping-speed check (mbar a; 0 = skip): 0

Result

Rate of rise: 0.005 mbar/s = 0.3 mbar/min
Gas load Q = V × Δp ÷ Δt: 2.5 mbar·L/s = 9 mbar·m³/h = 0.25 Pa·m³/s
As air at 20 °C: 10.7 g/h = 0.148 litres per minute of atmospheric air

Method. Q = V × Δp ÷ Δt, with V in litres, Δp in mbar and Δt in s, gives Q in mbar·L/s (× 3.6 = mbar·m³/h; × 0.1 = Pa·m³/s). Mass of air at 20 °C: 1 mbar·L/s ≈ 4.28 g/h. Curve shape (screening values used by this tool): a later interval rising at 80% or more of the first rate points to a real leak; 50% or less points to outgassing or vapour levelling off; between, a mixture. Required effective speed to hold a working pressure p: S = Q ÷ p. The isolated volume must include every pipe up to the closed valve; record temperature, because a warming vessel shows a rise of its own. A leak from atmosphere stays roughly constant only while the inside pressure is well below about half an atmosphere.

What to take from it. The number gives the size of the gas load, the curve shape gives its cause: test empty and loaded, and repeat after more pumping, before replacing seals.

Vacuum degassing gas and vapour load

Estimate the non-condensable gas a degasser releases, its volume at the vacuum, and how much the liquid’s own vapour adds to the flow reaching the pump.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Liquid throughput (kg/h): 2000
Dissolved gas at inlet (mg/kg): 25
Share of that gas removed (%): 100
Operating pressure (mbar a): 50
Liquid and gas temperature (°C): 20
Vapour pressure of the liquid: Water-like: from the temperature
Liquid vapour pressure at that temperature, if entered (mbar a): 23
Design air leak into vessel and line (kg/h): 0

Result

Dissolved gas released: 50 g/h = 0.05 kg/h
Non-condensable load incl. air leak: 0.05 kg/h
Specific volume of air at the vacuum: 16.83 m³/kg
Non-condensable volume at the vacuum: 0.841 m³/h
Liquid vapour pressure: 23.326 mbar a (water, Magnus formula)
Vapour share of the gas at the pump inlet: 46.7 % by volume (p_v ÷ p)
Total saturated volume flow at the vacuum: 1.577 m³/h (gas + vapour), of which vapour 0.736 m³/h
Water vapour carried with the gas: 0.0272 kg/h

Method. Gas released = throughput × dissolved content × share removed; add the design air leak. Specific volume of air v = R × T ÷ p = 287.05 × (t + 273.15) ÷ (p × 100) m³/kg (about 17 m³/kg at 50 mbar, 20 °C). Gas volume at the vacuum = mass × v. If the gas leaving is saturated with the liquid’s vapour, the vapour share by volume is p_v ÷ p and the total volume flow is the gas volume ÷ (1 − p_v ÷ p); water vapour carried = gas mass × (18.02 ÷ 28.97) × p_v ÷ (p − p_v). Water vapour pressure from the Magnus formula (about 23 mbar at 20 °C). The dissolved gas is taken as air. If the operating pressure is at or below the vapour pressure the liquid boils and the vapour flow is set by the heat input, which this estimate cannot give.

What to take from it. Vapour, not dissolved gas, usually sizes a degassing vacuum system: a condenser before the pump often removes most of it.

Vacuum dryer water removal, heat and vapour load

From the batch and moisture contents, work out the water to remove, the average and peak evaporation rate, the heat the product needs, the vapour volume at the chamber, and whether the condenser surface is cold enough.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Wet batch mass (kg): 500
Initial moisture, wet basis (%): 20
Final moisture, wet basis (%): 1
Planned cycle time (h): 8
Share of the water removed in the peak period (%): 50
Length of the peak period (h): 2
Latent heat of the liquid (kJ/kg): 2400
Chamber pressure (mbar a): 50
Vapour temperature at the chamber (°C): 30
Condenser cold-surface temperature (°C): 10

Result

Dry solids: 400 kg
Final batch mass: 404 kg
Water to remove: 96 kg
Average evaporation rate: 11.99 kg/h
Average heat to the product: 8 kW, plus heat losses
Peak evaporation rate: 23.99 kg/h (2 × the average)
Peak heat to the product: 15.99 kW, plus heat losses
Vapour specific volume at the chamber: 27.98 m³/kg (water vapour)
Peak vapour volume flow at the chamber: 671 m³/h (average 336 m³/h)
Boiling temperature of water at 50 mbar a: 32.9 °C
Note: Vapour temperature below the boiling temperature pure water vapour at this pressure is at least 32.9 °C; the product evaporates rather than boils, and the volume figure is within a few percent
Condenser surface: saturation pressure: 12.26 mbar a at 10 °C
Condenser check: Condenses 22.9 K below the boiling temperature; about 12.26 mbar of water vapour is left in the gas

Method. Wet-basis moisture: dry solids = M × (1 − w₀); final mass = dry ÷ (1 − w₁); water = M − final. Average rate = water ÷ cycle time; peak rate = water × share removed in the peak period ÷ its length. Heat (kW) = rate (kg/h) × latent heat (kJ/kg) ÷ 3,600, plus heat losses. Vapour specific volume = 461.5 × (t + 273.15) ÷ (p × 100) m³/kg (about 28 m³/kg at 50 mbar, 30 °C). Boiling temperature of water at the chamber pressure from the Magnus formula (about 33 °C at 50 mbar, 7 °C at 10 mbar). A condenser surface condenses water only if its saturation pressure is below the chamber pressure; the vapour left in the gas is about the saturation pressure at its coldest surface (over ice below 0 °C). Dry-basis or loss-on-drying figures must be converted to wet basis first.

What to take from it. Size the condenser and its coolant for the peak rate, not the average; the pump then only handles the non-condensable gas and the vapour the condenser leaves behind.

Water vapour condensation check

Compare the water vapour partial pressure in a vacuum line with the saturation pressure at a surface it meets, find the dew point, and the cold-trap temperature for a target partial pressure.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Water vapour partial pressure, or operating pressure if the gas is mostly vapour (mbar a): 20
Surface temperature: condenser, trap, pipe wall or pump casing (°C): 28
Water vapour partial pressure to leave in the gas, for cold-trap sizing (mbar a): 1

Result

Saturation pressure at the surface: 37.711 mbar a at 28 °C
At this surface: Stays vapour partial pressure 20 mbar is not above 37.711 mbar
Dew point of the vapour: 17.5 °C: surfaces colder than this collect liquid
Water vapour left in the gas after this surface: 20 mbar a (about the saturation pressure at the coldest surface)
Surface temperature needed to leave 1 mbar: -20.3 °C or colder (ice forms: plan defrosting or a second trap)

Method. Water condenses on a surface when its partial pressure is above the saturation pressure at the surface temperature. Saturation pressure (Magnus formula): over water 6.112 × exp(17.62 t ÷ (243.12 + t)) mbar; over ice, below 0 °C, 6.112 × exp(22.46 t ÷ (272.62 + t)). About 6 mbar at 0 °C, 12 at 10 °C, 23 at 20 °C, 38 at 28 °C, 42 at 30 °C; ice about 1 mbar at −20 °C. The inverse gives the dew (or frost) point: the highest surface temperature that condenses the vapour. Gas leaving a condenser carries about the saturation pressure at its coldest surface. The gauge reads total pressure: where air or other gas is present, enter the water vapour partial pressure. Water only: for solvents use their own vapour-pressure data. Range of the formula about −65 to +100 °C.

What to take from it. Check the worst-case (summer) coolant temperature: a condenser surface a few degrees too warm lets the whole vapour load through to the pump.

Vacuum forming reserve tank and pump recovery

Check whether a reserve tank pulls the forming volume down to the vacuum the part needs, how long the pump takes to recover the tank, and where the start pressure settles if it does not recover each cycle.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Reserve tank volume (litres): 500
Reserve tank pressure before the draw (mbar a): 50
Forming volume per cycle: sheet-to-tool space, passages, hoses (litres): 50
Atmospheric pressure at site (mbar a): 1013
Vacuum needed at the tool (mbar a): 100
Effective pump speed at the tank (m³/h): 100
Time available for recovery in each cycle (s): 30

Result

Equalised pressure at the tool: 137.5 mbar a (does not meet the 100 mbar a target)
Reserve volume needed for the target at this tank pressure: 913 litres (183% of the tank entered)
Recovery time back to 50 mbar a: 20 s (0.334 min, 0.00557 h) for 550 litres at 100 m³/h
Effective speed needed to recover in 30 s: 66.79 m³/h = 18.55 L/s
Recovery within the cycle: Yes 10 s to spare

Method. Equalised pressure when the valve opens (fast, isothermal transfer): p = (p_r × V_r + p₀ × V) ÷ (V_r + V). Reserve volume needed for a target p_t: V_r = V × (p₀ − p_t) ÷ (p_t − p_r). Recovery with an ideal pump of effective speed S on the whole connected volume: t = ((V_r + V) ÷ S) × ln(p ÷ p_r). If the time available is shorter, the pressure reached is p × exp(−S t ÷ (V_r + V)); repeating that each cycle, the start pressure settles at the value where one cycle’s rise equals the pump’s recovery. Ideal pump, no leakage or line losses: check against the pump curve and measured cycle data.

What to take from it. A tank that does not recover between cycles shows as detail getting worse through the shift: log the tank pressure at the start of each cycle.

Vacuum holding force

Holding force of suction cups or a vacuum table from the vacuum at the part and the sealed area, reduced by a safety factor, and checked against the part weight and acceleration.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Vacuum at the part: pressure difference below atmosphere (bar): 0.6
Sealed area entered as: Suction cups: diameter and number
Cup effective diameter (mm): 50
Number of cups sharing the load: 1
Table or pad area under vacuum (m²): 1
Share of that area covered and sealed by the workpiece (%): 100
Load case: Horizontal cup, vertical lift (safety factor 2)
Own safety factor, if chosen (times): 2
Part mass (kg; 0 = skip the check): 0
Highest acceleration incl. emergency stop (m/s²): 0

Result

Pressure difference: 600 mbar = 60 kPa (−0.6 bar g)
Effective sealed area: 19.6 cm² (1 × 19.63 cm²)
Theoretical holding force: 118 N = 0.118 kN = 12 kgf
Force per square metre of sealed area: 6.118 tonnes-force per m²
Usable force after the safety factor: 58.9 N (safety factor 2)

Method. Force F = Δp × A, with Δp the pressure difference below atmosphere in Pa (1 bar = 100,000 Pa; 1 mbar = 100 Pa) and A the effective sealed area in m² (cup: π/4 × d² × number of cups; table: area × share covered). 0.2 bar acts as about 2 tonnes-force per m². Usable force = F ÷ safety factor: at least 2 is common for a straight vertical lift, 4 or more for a cup loaded in shear on a vertical face (friction carries the load). Required force = m × (g + a), with a the highest acceleration including emergency stops. Use the vacuum measured at the part, not the pump rating; the maximum possible Δp is the local atmospheric pressure (about 1.013 bar at sea level, about 0.95 bar at Hyderabad). Use the cup maker’s data for oily, dusty, textured or curved surfaces.

What to take from it. More vacuum gives little extra force and more cup wear: add cup area or cups first, and measure the vacuum at the part on the worst workpiece.

Vacuum ejector compressed-air consumption

Average and peak compressed air used by vacuum ejectors from their rated motive-air flow, the on-time per cycle and blow-off pulses, for one gripper and for the whole line.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Ejector motive-air consumption at its feed pressure (Nl/min): 60
Cycle time (s): 10
Ejector on-time per cycle (s): 4
Blow-off flow (Nl/min; 0 = none): 100
Blow-off pulse per cycle (s): 0.2
Number of ejectors on the line: 20
Number of ejectors that can be on at the same moment: 20
Compressor specific power (kW per m³/min; 0 = skip energy): 0
Running hours per year of the line: 6000

Result

Ejector duty: 40 % of the cycle (6 cycles a minute)
Average motive air per gripper: 24 Nl/min
Blow-off per gripper: 2 Nl/min (0.333 litres per cycle)
Total per gripper: 26 Nl/min = 0.918 cfm
Average for all 20 ejectors: 520 Nl/min = 18.36 cfm = 8.67 L/s = 0.52 m³/min
Peak motive flow with 20 on together: 1,200 Nl/min = 42.4 cfm (2.3 × the average)

Method. Average motive air = rated consumption (Nl/min at the stated feed pressure) × on-time ÷ cycle time. Blow-off per cycle = blow-off flow × pulse ÷ 60 (litres); per minute = that × 60 ÷ cycle time. Total per gripper = motive + blow-off; line total = × number of ejectors. Peak = rated consumption × the number that can be on at once: it sizes the supply pipework, regulator and local receiver. 1 cfm = 28.317 l/min; normal litres are at reference conditions. Energy: line flow (m³/min) × specific power × hours. Air-saving (vacuum-hold) circuits reduce the on-time on sealed parts; use the measured on-time. Leaks are not included.

What to take from it. Average demand sizes the compressor, peak concurrency sizes the pipework and local receiver; on porous, high-duty stations an electric vacuum pump or blower usually costs less to run.

Liquid-ring seal-liquid flow and vapour-pressure margin

Seal-liquid flow needed to carry the heat out of a liquid-ring pump for an allowed temperature rise, the cooler duty, and how close the required suction pressure is to the vapour pressure of the liquid.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Total heat to remove: compression + gas + condensing vapour (kW): 50
Seal-liquid specific heat (kJ/kg·K): 4.18
Allowed temperature rise across the pump (K): 10
Seal-liquid density (kg/m³): 1000
Seal-liquid supply temperature (°C): 25
Required suction pressure (mbar a): 60
Process vapour condensing in the ring (kg/h; 0 = none): 0

Result

Seal-liquid mass flow: 1.196 kg/s = 4,306 kg/h
Seal-liquid volume flow: 4.31 m³/h = 71.8 l/min
Cooler duty in total recirculation: 50 kW
Seal-liquid outlet temperature: 35 °C
Water vapour pressure at supply temperature: 31.6 mbar a at 25 °C
Water vapour pressure at outlet temperature: 56.1 mbar a at 35 °C
Margin: suction pressure − vapour pressure at outlet: 3.9 mbar (suction is 1.07 × the vapour pressure)
Result: Too close within the 1.5 × screening margin

Method. Mass flow m = Q ÷ (c_p × ΔT) (kW, kJ/kg·K, K → kg/s); volume flow = m ÷ ρ × 3,600 m³/h. In total recirculation the cooler rejects the same heat Q. The ring liquid is hottest at the outlet (supply + ΔT); the pump cannot hold a suction pressure close to the liquid’s vapour pressure there. Water vapour pressure (Magnus formula): about 17 mbar at 15 °C, 32 at 25 °C, 56 at 35 °C, 96 at 45 °C. Screening value used by this tool: suction pressure at least 1.5 × the vapour pressure at the outlet temperature; closer than that, capacity falls and cavitation is likely. Pure water only: dissolved solvent, salt or glycol changes the vapour pressure, and the maker’s performance and cavitation curves govern. Condensing process vapour adds to the liquid inventory and must leave by overflow or purge.

What to take from it. Check the summer case: warmer supply water raises the vapour pressure and can take a working pump into cavitation without any fault in the pump.

Central vacuum plant demand and N+1 check

Add up the users of a central vacuum plant on one basis (m³/h at the plant set point), apply simultaneity, leakage and future users, then check the number of duty pumps, N+1 standby and an optional receiver for short peaks.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Group 1: number of machines: 10
Group 1: suction flow each at its machine pressure (m³/h): 20
Group 1: simultaneity (%): 60
Group 1: pressure needed at the machine (mbar a): 200
Group 2: number of machines: 0
Group 2: suction flow each at its machine pressure (m³/h): 0
Group 2: simultaneity (%): 100
Group 2: pressure needed at the machine (mbar a): 200
Group 3: number of machines: 0
Group 3: suction flow each at its machine pressure (m³/h): 0
Group 3: simultaneity (%): 100
Group 3: pressure needed at the machine (mbar a): 200
Plant set point at the receiver (mbar a): 200
Pressure loss in the header to the farthest machine (mbar): 0
Leakage measured at the plant set point (m³/h): 30
Planned future users at the plant set point (m³/h): 0
Capacity of one pump at the plant set point, from its curve (m³/h): 100
Number of pumps installed (all the same size): 3
Receiver check: short peak above pump capacity (m³/h at set point; 0 = skip): 0
Receiver check: duration of that peak (s): 10
Receiver check: pressure rise the users can accept (mbar): 20

Result

Group 1 at the set point: 120 m³/h (10 × 20 m³/h × 60%)
Users at the set point: 120 m³/h
Design suction flow incl. leakage and future: 150 m³/h at 200 mbar a
Duty pumps needed: 2 of 100 m³/h each
Capacity with all 3 pumps: 300 m³/h (100% above the design flow)
N+1 check: capacity with one pump out: 200 m³/h: passes (33% margin)
Set point check: Deep enough for the most demanding machine after the header loss

Method. Each user flow is converted to the plant set point at constant mass and temperature: q_plant = q × p_machine ÷ p_set (the same gas takes more volume at the deeper plant vacuum). Design flow = Σ(number × flow × simultaneity) + leakage + future users. Duty pumps = design flow ÷ capacity of one pump at the set point, rounded up. N+1 passes when the pumps left with one out of service still cover the design flow. The set point must be at least the header loss deeper than the most demanding machine needs; a deeper set point than that costs pump capacity. Receiver for a short peak: V = peak excess flow × duration × p_set ÷ allowed rise. All pumps are taken as the same size; use measured simultaneity and leakage where you can.

What to take from it. Set the plant at the pressure the users really need: every extra mbar of depth cuts what each pump delivers.

Aeration blower duty: Pressure budget, flow basis and power

Build the blower discharge pressure item by item from the water depth over the deepest diffuser or sparger, convert the required airflow to actual volume at the blower inlet on a hot day, and estimate the power.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Water depth above the deepest diffuser or sparger (m; at the highest level): 5
Liquid density relative to water (fresh water 1.000, sea water about 1.025): 1
Vessel head-space back-pressure (mbar g; 0 for an open basin, pond or tank): 0
Diffuser or sparger loss (mbar; fouled or end-of-life figure): 60
Pipe, header, valve, meter and non-return valve losses (mbar; worst branch at peak flow): 35
Air treatment losses (mbar; filters, cooler and dryer at end of life; 0 if none): 0
Inlet filter and silencer loss (mbar): 10
Agreed margin (mbar): 20
Site atmospheric pressure (bar a; about 1.013 at sea level, about 0.95 at Hyderabad): 1
Flow basis: Nm³/h (0 °C, 1.013 bar a)
Required airflow (in the basis chosen above): 1000
Working broth volume per vessel (m³; used only for vvm): 50
Number of vessels at peak together (used only for vvm): 1
Inlet air temperature (°C; hottest expected): 45
Machine type for the power estimate: Roots blower (no internal compression)
Overall blower and drive efficiency (%): 60

Result

Static head of the water column: 491 mbar = 0.49 bar (7.11 psi)
Blower discharge pressure required: 0.606 bar g = 606 mbar g = 8.78 psig = 60.5 kPa g
Share of the discharge pressure from static head: 81 % (the rest is losses, back-pressure and margin)
Inlet absolute pressure after the filter and silencer: 0.99 bar a
Blower differential pressure (outlet minus inlet): 616 mbar = 0.616 bar (8.93 psi)
Pressure ratio across the blower: 1.622 (absolute outlet ÷ absolute inlet)
Flow on the normal basis: 1,000 Nm³/h (0 °C, 1.013 bar a)
Actual flow at the blower inlet: 1,192 m³/h at 45 °C and 0.99 bar a = 19.87 m³/min = 702 cfm
Air mass flow: 1,292 kg/h (dry air)
Hot-intake effect: 8.5 % more inlet volume per kg of air than at a 20 °C intake
Ideal discharge temperature: 102 °C (before machine losses; the real figure is higher, read the maker’s curve)
Ideal air power: 20.38 kW (inlet flow × differential)
Estimated shaft or input power: 33.97 kW at 60% overall efficiency; select the motor with a margin

Method. Static head = depth × density × g = about 0.0981 bar per metre of fresh water (× relative density for sea water or broth). Discharge pressure (gauge) = static head + vessel back-pressure + diffuser or sparger loss + pipe, valve and meter losses + treatment losses + margin. The inlet filter and silencer loss lowers the inlet absolute pressure, so blower differential = discharge pressure + inlet loss and pressure ratio = (atmospheric + discharge) ÷ (atmospheric − inlet loss). Flow: actual inlet volume = normal volume × (T_inlet ÷ 273.15 K) × (1.01325 ÷ p_inlet absolute), dry air; vvm × working volume × vessels is taken on the normal basis (0 °C, 1.013 bar a). Power: a Roots blower has no internal compression, so its ideal air power is inlet flow × differential; a screw blower, centrifugal or compressor is taken as isentropic, P = Q₁ × p₁ × k/(k−1) × (r^((k−1)/k) − 1) with k = 1.4. Shaft or input power = ideal power ÷ overall efficiency. Ideal discharge temperature: Roots T₁ × (1 + (r − 1)(k − 1)/k); internal compression T₁ × r^((k−1)/k); real machines run hotter. Screening values; select from the maker’s curve at the hottest inlet and the fouled-diffuser pressure.

What to take from it. Size on the deepest diffuser at the highest water level, the fouled or end-of-life losses, peak flow and the hottest intake. Hot air is thinner, so the same normal flow needs more inlet volume in an Indian summer.

Blower duty points and hours-weighted energy

Build the aeration system curve (static head plus friction that rises with flow), find the pressure at each duty point, and compare one or two blowers on package input power weighted by the hours the plant actually spends at each point.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Water depth over the diffusers (m): 5
Liquid density relative to water: 1
Diffuser, pipework and valve loss at the design flow (mbar): 100
Design flow (m³/min at inlet): 100
Fouled-diffuser extra loss (mbar; 0 for the clean case): 0
Site atmospheric pressure (bar a): 1.013
Duty-point flows (m³/min at inlet): 60, 100, 120
Hours per year at each duty point (h): 3000, 4500, 1260
Blower A: package input power at each point (kW): 70, 120, 147
Blower B: package input power at each point (kW; all 0 if only one blower): 66, 124, 160

Result

Static head: 491 mbar (0.49 bar)
System pressure at the design flow: 591 mbar g (0.591 bar g, 8.56 psig)
Point 1: 60 m³/min, 3,000 h: 527 mbar g; ideal air power 45 kW; wire-to-air A 64%, B 68%
Point 2: 100 m³/min, 4,500 h: 591 mbar g; ideal air power 82.9 kW; wire-to-air A 69%, B 67%
Point 3: 120 m³/min, 1,260 h: 635 mbar g; ideal air power 105.7 kW; wire-to-air A 72%, B 66%
Blower A: annual energy: 9,35,220 kWh; hours-weighted average 106.8 kW
Blower B: annual energy: 9,57,600 kWh; hours-weighted average 109.3 kW
Lower annual energy: Blower A by 22,380 kWh a year (2.3%)
Value of the difference: Enter your electricity tariff (cost per kWh) above

Method. Static head = depth × relative density × 0.0981 bar per metre. Friction (diffusers, pipework, valves) is taken to rise with the square of flow: Δp_friction = Δp_design × (Q ÷ Q_design)². System pressure = static + friction + fouled-diffuser allowance. Ideal (isentropic) air power at each point = Q₁ × p₁ × k/(k−1) × (r^((k−1)/k) − 1), k = 1.4, inlet at site atmospheric pressure. Wire-to-air efficiency = ideal air power ÷ package input power (motor, VSD, bearings and cooling). Annual energy = Σ package kW × hours at that point; the hours-weighted average kW = annual energy ÷ total hours. The default kW figures are illustrative only: enter each bidder’s package input power at your duty points, under the same inlet conditions.

What to take from it. Compare blowers on total package input power at each operating point, weighted by the hours you spend there, not on peak efficiency. Plot every point on each maker’s map to confirm it sits clear of surge.

Blower station firm capacity, turndown and coverage

Enter each blower’s maximum flow and minimum stable flow at the station pressure, and the demand points. The tool finds the firm capacity with the largest machine out, the station turndown, any gaps between machine combinations, and which combination covers each demand point.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Each blower: maximum flow at the station pressure (m³/min): 60, 60, 60
Each blower: minimum stable flow at that pressure (m³/min): 25, 25, 25
Demand points to check (m³/min; minimum, average, peak or a profile): 20, 70, 110
Minimum air for diffuser mixing (m³/min; 0 if not set): 0

Result

Installed capacity: 180 m³/min from 3 blowers
Firm capacity (largest blower out): 120 m³/min (blower 1, 60 m³/min, out of service)
Smallest stable output: 25 m³/min
Station turndown (firm capacity ÷ smallest stable output): 4.8 : 1
Steady output range of the station: 25–180 m³/min, no gaps between combinations
Demand 1: 20 m³/min: Below range smallest stable output is 25 m³/min: over-aeration or blow-off
Demand 2: 70 m³/min: Blowers 1 + 2 range 50–120 m³/min; also met with the largest blower out
Demand 3: 110 m³/min: Blowers 1 + 2 range 50–120 m³/min; also met with the largest blower out

Method. Each running blower can deliver anywhere between its minimum stable flow (maker’s minimum speed or surge limit at that pressure) and its maximum flow, so a combination covers Σ minimum to Σ maximum. The station range is the union of all combinations; any flow outside it is a gap. Firm capacity = total maximum − largest maximum. Station turndown = firm capacity ÷ smallest stable output. A demand point below the minimum air for diffuser mixing is raised to that floor. For each demand point the tool picks the combination with the fewest machines (then the smallest total capacity) and checks whether the point can still be met with the largest machine out of service. Up to 8 blowers. Pressure: use the aeration pressure budget tool; flows here must all be at the same pressure and on the same basis.

What to take from it. Turndown is a station property. If night demand is below the smallest machine’s minimum, the station over-aerates or blows off; a smaller jockey blower or VSD on more machines usually closes the gap.

Silo and hopper aeration air estimate

Estimate the pressure needed at the aeration pads or fabric from the aerated bed head and element losses, and the simultaneous airflow from the active area of the zones that run together, plus any pulsed devices that can coincide.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Aerated bed height above the element (m): 5
Aerated bulk density (kg/m³): 1000
Element (pad or fabric) loss at design flow (bar): 0.05
Branch, valve and hose losses (bar): 0.05
Active aeration area per zone (m²): 1
Maker’s flow per unit area (m³/min free air per m²): 0.5
Number of zones installed: 8
Number of zones running at the same time: 2
Coinciding pulsed devices and air blasters (m³/min free air; 0 if none): 0
Start-up allowance (%; extra air to break a compacted bed): 0
Plant-air header pressure for comparison (bar g; 0 = no comparison): 7

Result

Bed head above the element: 0.49 bar (7.11 psi)
Pressure required at the supply point: 0.591 bar g = 8.6 psig = 591 mbar g
Simultaneous aeration flow: 1 m³/min free air = 60 m³/h = 35 cfm
of which zone air: 1 m³/min; pulsed devices 0 m³/min
If every installed zone were summed: 4 m³/min (4 × the simultaneous figure; not a sizing basis)
Likely air source: Blower (Roots, screw, side-channel or centrifugal) for continuous duty
Ideal power if fed from the 7 bar g header: 5.7 × the ideal power of compressing only to 0.59 bar g

Method. Bed head (bar) = bed height × aerated bulk density × 9.81 ÷ 100,000. Supply pressure = bed head + element (pad or fabric) loss at design flow + branch, valve and hose losses. Simultaneous flow = active area per zone × maker’s flow per unit area × zones running together × (1 + start-up allowance) + coinciding pulsed devices. The all-zones figure is shown only to show how far summing every pad oversizes the supply. Header comparison: ideal isentropic power is proportional to (r^((k−1)/k) − 1), so compressing to the plant header and throttling down costs that ratio of power against compressing only to the supply pressure. Use the element maker’s pressure–flow curve and the conveying or silo designer’s figures; a trial on the actual powder is the most reliable basis.

What to take from it. True fluidisation is a low-pressure, high-volume duty. Size on the zones that run together, not every installed pad, and do not throttle 7 bar plant air down to a fraction of a bar for continuous duty.

Combustion-air blower duty

Turn the burner’s air mass flow into actual inlet volume at your site conditions, add up the pressure losses from intake to combustion chamber for one firing case, and get the ideal air power and the low-fire turndown gap.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Air mass flow required by the burner (kg/h): 2400
Inlet absolute pressure (kPa a): 95
Inlet air temperature (°C): 35
Gas constant R (J/kg·K; 287 for dry air): 287
Element losses for this firing case (Pa; intake and filter, duct, silencer, damper, burner, chamber): 300, 760, 230, 300, 1150, 260
Blower efficiency (%; 0 = show air power only): 0
Blower stable minimum flow at the required pressure (m³/h; 0 = not checked): 1000
Low-fire actual flow needed (m³/h; 0 = not checked): 745

Result

Inlet air density: 1.074 kg/m³ at 95 kPa a and 35 °C
Actual inlet volume: 2,234 m³/h = 37.2 m³/min = 1,315 cfm
Volume at 101.325 kPa a and 20 °C: 1,993 m³/h (ρ = 1.204 kg/m³); the site case needs 12.1% more inlet volume
Normal volume (0 °C, 101.325 kPa a): 1,857 Nm³/h
Required blower pressure (sum of losses): 3,000 Pa = 30 mbar = 306 mm water gauge
Ideal air power: 1.86 kW (flow × pressure)
Low-fire turndown gap: 255 m³/h: low fire (745) is below the stable minimum (1,000)

Method. Dry-air density ρ = P ÷ (R × T), with P absolute in Pa, R = 287 J/(kg·K) and T in kelvin. Actual inlet volume = mass flow ÷ ρ. The reference volume uses 101.325 kPa and 20 °C. Required blower pressure = Σ the element losses for the case being checked (intake and filter, duct, silencer, damper or register, burner internal loss, chamber back-pressure). Enter a separate list for low, normal and peak fire, because damper losses fall as the damper opens while the others rise roughly with the square of flow. Ideal air power = (Q ÷ 3,600) × Δp; shaft power = air power ÷ blower efficiency. Turndown gap = stable minimum flow − low-fire flow. Air proving, purge and safety settings come only from the burner supplier.

What to take from it. Start from air mass, not a volume figure. A hot, lower-pressure site needs 12–16% more inlet volume for the same air mass, and low fire is often the hardest case for the blower.

Blower pulsation frequency and pipe resonance check

Work out the lobe-passing frequency band over the blower’s speed range, with harmonics, and the acoustic natural frequencies of the pipe lengths between reflection points at the discharge temperature. Lengths with a mode inside a band are flagged with the shaft speed that excites them.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Discharge pulses per revolution (2-lobe Roots 4, 3-lobe 6): 4
Minimum shaft speed (rpm): 1500
Maximum shaft speed (rpm): 3000
Highest harmonic to check: 4
Gas temperature in the pipe (°C): 100
Ratio of specific heats k (air 1.4): 1.4
Gas constant R (J/kg·K; air 287): 287
Pipe lengths between reflection points (m): 1.0, 2.5, 4.0
End condition: Half-wave: both ends open or both closed
Separation margin around each band (%): 10
Drive frequency at maximum speed (Hz; 0 if not on a VFD): 50

Result

Speed of sound: 387.2 m/s at 100 °C
Fundamental (lobe-passing) band: 100–200 Hz
Harmonic 2 band: 200–400 Hz
Harmonic 3 band: 300–600 Hz
Harmonic 4 band: 400–800 Hz
Pipe length 1 m: Resonance in band first modes 193.6, 387.2, 580.8, 774.4 Hz; excited at about 1,500, 1,940, 2,180, 2,900 rpm (drive 25, 32.3, 36.3, 48.4 Hz)
Pipe length 2.5 m: Resonance in band first modes 77.4, 154.9, 232.3, 309.8 Hz; excited at about 1,500, 1,550, 1,740, 1,940, 2,030, 2,320, 2,610, 2,710 and 2 more rpm (drive 25, 25.8, 29, 32.3, 33.9, 38.7, 43.6, 45.2 and 2 more Hz)
Pipe length 4 m: Resonance in band first modes 48.4, 96.8, 145.2, 193.6 Hz; excited at about 1,500, 1,630, 1,690, 1,820, 1,940, 2,000, 2,180, 2,360 and 6 more rpm (drive 25, 27.2, 28.2, 30.3, 32.3, 33.3, 36.3, 39.3 and 6 more Hz)

Method. Excitation f = shaft rpm ÷ 60 × discharge pulses per revolution × harmonic number (2-lobe Roots 4 pulses, 3-lobe 6, or as the maker states), taken from minimum to maximum speed. Speed of sound c = √(k × R × T), T in kelvin. Half-wave (both ends open, or both closed): f_m = m × c ÷ (2L). Quarter-wave (one end open, one closed): f_m = (2m − 1) × c ÷ (4L). A mode within a band (widened by the separation margin) is flagged with the shaft speed that would excite it; with a VFD, the drive frequency is pro-rata to speed from the frequency at maximum speed. Screening only: the acoustic and pipe-stress design for a large installation stays with the blower maker, the silencer supplier and the piping engineer.

What to take from it. Check the whole speed band and harmonics up to at least the fourth, at the hot discharge temperature, before laying out the pipe. Where a VFD must cross a resonance, set skip frequencies so it passes through quickly.

Gas compression stage screening

Convert suction and discharge to absolute pressure, find the overall and per-stage pressure ratio, the interstage pressures for an equal split, the ideal discharge temperature by the gas’s k value, and, from the mass flow, the actual suction volume, normal volume and mass transferred.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Suction pressure (in the unit below): 10
Suction pressure unit: bar a
Discharge pressure (in the unit below): 200
Discharge pressure unit: bar a
Site atmospheric pressure (bar a; for gauge entries): 1.013
Gas: Hydrogen (k 1.41, MW 2)
Other gas: k = cp/cv (used only for Other): 1.3
Other gas: molecular weight (kg/kmol; used only for Other): 17
Suction temperature (°C): 30
Number of stages (screening, equal ratio): 3
Stage ratio for the temperature check (0 = equal split): 0
Isentropic efficiency (%; 100 = ideal): 100
Maximum allowable discharge temperature (°C; 0 = not checked): 0
Mass flow (kg/h; 0 = not used): 20
Compressibility factor Z at suction (1 = ideal gas): 1
Running hours for the transferred mass (h; 0 = not used): 8
Lowest credible suction pressure (same unit as suction; equal to suction = no comparison): 10

Result

Suction, absolute: 10 bar a (8.987 bar g, 145 psia)
Discharge, absolute: 200 bar a (198.99 bar g, 2,886 psig)
Overall pressure ratio: 20 absolute discharge ÷ absolute suction
Equal stage ratio: 2.714 over 3 stages
Interstage pressures (equal split): 27.1, 73.7 bar a estimated operating pressures only, not design pressures
Exponent (k − 1)/k: 0.291 k = 1.41 for hydrogen; R = 4,124.2 J/(kg·K)
Ideal discharge temperature per stage: 132.1 °C (405.3 K) at a stage ratio of 2.714, suction 30 °C
Actual suction volume: 25 m³/h (0.417 m³/min, 14.7 acfm) at 10 bar a, Z = 1
Normal volume: 222.4 Nm³/h (0 °C, 1.013 bar a; density 0.0899 kg/Nm³)
Mass transferred: 160 kg in 8 h (a logistics check, not storage sizing)

Method. P(abs) = P(gauge) + site atmospheric pressure. Overall ratio r = P₂(abs) ÷ P₁(abs); dividing gauge figures gives a wrong ratio, worst at low suction pressure. Equal split: stage ratio = r^(1/n); interstage pressure after stage i = P₁ × (stage ratio)^i. Ideal (isentropic) discharge temperature T₂ = T₁ × (stage ratio)^((k−1)/k), kelvin, assuming intercooling back to the suction temperature before every stage; with an isentropic efficiency η, T₂ = T₁ + (T₂ideal − T₁) ÷ η. Gas constant R = 8,314.46 ÷ molecular weight. Actual suction volume = ṁ × Z × R × T₁ ÷ P₁; normal volume uses 0 °C and 1.01325 bar a with Z = 1. Transferred mass = mass flow × running hours. Rounded textbook k values near ambient; real-gas Z and the maker’s calculation for the actual composition govern. These are screening values, not equipment ratings or settings.

What to take from it. Always work in absolute pressure, and check the lowest credible suction pressure: it gives the highest ratio and discharge temperature and the largest actual suction volume for the same mass flow.

Pressure ratio for a blower or compressor duty

Turn inlet and outlet pressures, gauge or absolute, into absolute pressures, differential and pressure ratio, and see where the duty sits between blowers and compressors.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Inlet pressure (in the unit below): 0
Inlet pressure unit: bar g
Outlet pressure (in the unit below): 0.6
Outlet pressure unit: bar g
Local atmospheric pressure (bar a): 1.013
Inlet temperature (°C): 30
Ratio of specific heats k (air 1.4): 1.4

Result

Inlet, absolute: 1.013 bar a
Outlet, absolute: 1.613 bar a (0.6 bar g, 8.7 psig)
Differential pressure: 0.6 bar = 600 mbar = 8.7 psi
Pressure ratio: 1.592 absolute outlet ÷ absolute inlet
Where the duty sits: Blower territory Roots, screw, side-channel or centrifugal blower
Ideal discharge temperature, internal compression: 73 °C
Ideal discharge temperature, Roots blower: 81 °C (no internal compression; real machines run hotter still)

Method. p(abs) = p(gauge) + local atmospheric pressure (about 1.013 bar a at sea level, lower at altitude). Differential = outlet − inlet (absolute). Pressure ratio = absolute outlet ÷ absolute inlet. Ideal discharge temperature: with internal compression (screw, centrifugal, compressor) T₂ = T₁ × r^((k−1)/k); a Roots blower has no internal compression, so ideally T₂ = T₁ × (1 + (r − 1)(k − 1)/k), which is hotter. Vocabulary bands used here: up to about 1.1 fan range; up to about 2 the duty is sold as a blower (Roots, screw, side-channel, centrifugal); about 2 to 4.5 (roughly 1–3.5 bar g from atmosphere) is the grey area where screw blowers, low-pressure oil-free screw compressors and multistage centrifugals compete; plant-air compressors work at about 8–11. Select from the maker’s envelope, not a label.

What to take from it. Gauge pressure hides the duty. The same 0.6 bar differential is a ratio of about 1.6 as a pressure blower and about 2.5 as a vacuum blower, and the vacuum duty runs hotter.

Drilling booster: Pressure ratio, water head and air volume

Work out the booster’s pressure ratio from absolute pressures, the static water head above the hammer, how the actual air volume shrinks through the booster, and, if you enter the hammer data, the pressure needed at the booster outlet.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Primary compressor delivery to the booster inlet (bar g): 20
Booster discharge pressure (bar g): 34
Site atmospheric pressure (bar a): 1
Water column above the hammer (m; vertical height, not drilled depth): 100
Water density (kg/m³): 1000
Primary compressor free air delivery (cfm): 1000
Ambient temperature (°C): 40
Booster inlet temperature (°C; after the aftercooler): 40
Booster outlet temperature (°C; after its aftercooler): 40
Hammer working pressure (bar g; 0 = budget not built): 0
Hose, swivel and drill-string losses (bar): 0
Annular back-pressure from friction and cuttings (bar): 0
Margin (bar): 0

Result

Booster pressure ratio: 1.667 (35 ÷ 21 bar a)
Ratio if gauge figures were divided (wrong): 1.7 shown only to show the error
Static water head above the hammer: 9.81 bar = 9,81,000 Pa = 142 psi
Free air from the primary compressor: 28.32 m³/min (1,000 cfm); unchanged through the booster
Actual volume entering the booster: 1.348 m³/min (48 acfm)
Actual volume leaving the booster: 0.809 m³/min (29 acfm); Q₂ ÷ Q₁ = 0.6
Ideal single-stage discharge temperature: 89 °C from 40 °C inlet, before losses and cooling
Pressure needed at the booster outlet: not built enter the hammer working pressure from its maker’s data

Method. Ratio = (P₂ gauge + atmospheric) ÷ (P₁ gauge + atmospheric). Water head = ρ × g × h with g = 9.81 m/s² and h the vertical water column above the hammer (not drilled depth): about 1 bar per 10 m. Free air (mass) is set by the primary compressor and is unchanged through the booster. Actual volume entering the booster = FAD × (P_atm ÷ P₁ abs) × (T₁ ÷ T_ambient); leaving it, Q₂ = Q₁ × (P₁ ÷ P₂) × (T₂ ÷ T₁), absolute pressures and kelvin. Ideal single-stage discharge temperature T = T₁ × ratio^0.286 (air). Pressure needed at the booster outlet = hammer working pressure + string losses + annular back-pressure + water head + margin. Static planning estimates: aeration of the water column, inflow and cuttings change the real figure, and the booster’s capacity comes from its maker’s data at the stated inlet conditions.

What to take from it. A booster adds pressure, not air. The primary compressor’s free air delivery limits hole cleaning, and every hose, swivel and pipe after the booster must be rated for its maximum pressure.

Blow-tank batch air budget

Build the free-air budget of one dense-phase blow-tank batch: pressurising the vessel, conveying, line clearing and any fluidising or booster air. Get the batches per hour, the average demand and the peaks, and, if you enter the compressor delivery, the receiver needed to cover the peaks.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Free gas space in the vessel when full (m³; vessel volume less material volume): 1.4
Conveying pressure (bar g): 3
Atmospheric pressure (bar a): 1.013
Conveying air flow (m³/min free air; designer’s figure): 8
Conveying time per batch (min): 3
Line-clearing air flow (m³/min free air): 8
Line-clearing time (min): 0.5
Fluidising and booster air per batch (m³ free air; 0 if none): 0
Pressurisation time (s): 30
Cycle time (min; fill, pressurise, convey, clear and vent): 5
Usable batch mass (t; 0 = no throughput check): 0
Compressor delivery available for conveying (m³/min free air; 0 = not checked): 0
Usable receiver pressure band above conveying pressure (bar): 1

Result

Pressurisation air per batch: 4.15 m³ free air; at 8.29 m³/min over 30 s
Conveying and clearing air per batch: 28 m³ (conveying 24, clearing 4)
Total free air per batch: 32.15 m³
Batches per hour: 12
Average air demand: 6.43 m³/min = 227 cfm = 386 m³/h
Peak rate: 8.29 m³/min (pressurisation); 1.29 × the average

Method. Pressurisation air (free air) = free gas space × conveying pressure (gauge) ÷ atmospheric pressure. Conveying air = conveying flow × conveying time; clearing air = clearing flow × clearing time; fluidising and booster air is added per batch. Batches per hour = 3,600 ÷ cycle time in seconds; theoretical throughput = usable batch mass × batches per hour (a schedule check only). Average demand = air per batch × batches per hour ÷ 60. Peak rates: pressurisation air ÷ pressurisation time, and the conveying and clearing flows. Receiver: the free air the compressor cannot supply during pressurisation, conveying and clearing, Σ max(0, phase rate − compressor delivery) × phase time, is drawn from storage; V = shortfall × P_atm ÷ usable pressure band (isothermal, as in the receiver tool). The conveying pressure and air flow come from the conveying designer.

What to take from it. A compressor sized for the average starves the line; one sized for the pressurisation peak cycles heavily. A dedicated receiver between them, with the compressor set a little above conveying pressure, is the usual answer.

Vessel filling: Free air and fill time

Free air needed to raise a vessel from one pressure to another, and how long a compressor of a given FAD takes to do it.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Volume entered in: m³
Internal volume being filled (in the unit above): 1
Start pressure (bar g): 0
Target pressure (bar g): 30
Atmospheric pressure (bar a): 1.013
Allowance for cooling after the fill, hoses and leaks (%): 0
Supply flow as free air (unit below; 0 if not known): 1
Supply flow unit: m³/min
Number of compressors feeding together: 1
Fill time allowed (s; 0 if not set): 0

Result

Volume: 1,000 L = 1 m³
Pressure rise: 30 bar (435.1 psi); end pressure 31.01 bar a
Free air to add (pressure rise basis): 29.62 m³ = 29,615 L
Free air on the absolute-ratio basis: 30.62 m³ = 30,615 L (includes one volume already inside)
Free air per fill used below: 29.62 m³ = 29,615 L (pressure rise basis)
Supply flow: 1 m³/min = 1,000 L/min = 35.3 cfm
Fill time at that flow: 29.6 min = 1,777 s (minimum: ideal, isothermal)

Method. Free air to add = V × (p₂ − p₁) ÷ p₀, with p₁ and p₂ the start and end pressures (a gauge difference equals an absolute difference) and p₀ the atmospheric pressure, about 1.013 bar a. The “absolute pressure ratio” basis, V × (p₂ − p₁ + p₀) ÷ p₀, adds one volume of free air: for a volume that starts at atmospheric pressure it equals V × p₂(abs) ÷ p₀. It counts the air already inside, so it overstates what must be pumped in by one vessel volume; some procedures use it as a conservative figure. Fill time = free air ÷ supply flow (the supply flow is compressor FAD at the final pressure, or regulator flow capacity, as free air). Flow for a required time = free air ÷ time. Average demand = free air per fill × fills per hour + make-up flow. Receiver able to give one whole fill on its own = free air per fill × p₀ ÷ (receiver charge pressure − lowest pressure the user accepts). Isothermal and ideal gas: a fast fill heats the air, which then cools and loses pressure, so allow extra; real fill time through a regulator is longer than the minimum shown because flow falls as the pressure difference shrinks near the set point.

What to take from it. Halving the fill time means doubling the flow, not raising the pressure. Size the supply on free air per fill and the time allowed; a local receiver covers short, large fills cheaply.

Pipeline pigging: Air volume, run time and driving pressure

Line volume behind the pig, free air to fill it at the driving pressure, run time with the compressors on site, pig speed, and an estimate of the driving pressure needed.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Pipe internal diameter (mm): 300
Line length (m): 1000
Driving pressure behind the pig (bar g): 7
Atmospheric pressure (bar a): 1.013
Compressor FAD, each (unit below): 600
FAD unit: cfm (ft³/min)
Number of compressors running together: 1
Allowance for hoses, launcher, leakage and bypass (%): 0
Pig friction pressure, from the procedure or pig maker (bar): 0
Height of liquid column pushed uphill (m): 0
Specific gravity of that liquid (water = 1): 1
Back-pressure at the receiver (bar): 0
Maximum permitted driving pressure (bar g; 0 if not given): 0

Result

Line internal volume: 70.7 m³
Absolute pressure ratio: 7.91 (8.01 bar a ÷ 1.013)
Free air to fill the line behind the pig: 559 m³
Total compressor FAD: 16.99 m³/min = 16,990 L/min = 600 cfm
Run (fill) time: 32.9 min = 0.55 h
Average pig speed: 0.51 m/s = 30 m/min (if the line ahead vents freely)
Driving pressure needed (estimate): 0 bar = friction 0 + liquid head 0 + back-pressure 0
Note: Friction, head and back-pressure not entered enter them from the procedure to check the driving pressure

Method. Line volume V = π × D² ÷ 4 × L (internal diameter). The space behind the pig starts empty and ends full of air at the driving pressure, so free air = V × (p_g + p₀) ÷ p₀, plus your allowance for hose and launcher volume, leakage and bypass past the pig. Run time = free air ÷ total FAD (1 m³/min = 35.31 cfm). Pig speed ≈ actual air flow at driving pressure ÷ pipe area = FAD × p₀ ÷ (p_g + p₀) ÷ area. Driving pressure needed ≈ pig friction + liquid head (ρ g h: 0.0981 bar per metre of water lifted × specific gravity) + back-pressure at the receiver. Temperature and pressure gradient along the line are ignored. The maximum permitted driving pressure and any speed limit come from the pipeline owner’s procedure, never from the compressor rating.

What to take from it. Long or large-bore lines need hundreds of cubic metres of free air per run: plan several compressors or compressors plus storage, and let the procedure, not the compressor, set the pressure and speed.

Tyre top-up air, forecourt peak and receiver reserve

Free air per tyre and per vehicle, idealised transfer time, average hourly demand, the deficit when several inflation points fill at once, and whether the receiver covers it and how fast it recovers.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Tyre internal volume (L): 30
Tyre pressures entered in: psi g
Starting tyre pressure (gauge, unit above): 26
Target tyre pressure (gauge, unit above): 34
Atmospheric pressure (bar a): 1.01325
Number of tyres topped up per vehicle: 4
Effective transfer rate per inflator (L/min free air): 113.3
Vehicles per hour (average): 10
Number of inflation points filling at the same time: 2
Compressor delivery, FAD at working pressure (L/min free air): 150
Duration of the overlapping fills (min): 2
Receiver volume (L; 0 to skip): 100
Usable receiver pressure band (bar): 2
Air still being used while the receiver recovers (L/min free air): 50

Result

Pressure rise: 0.552 bar = 8 psi
Free air per tyre: 16.3 L
Idealised transfer time per tyre: 8.6 s (not a service time)
Free air per vehicle: 65.3 L
Average demand: 653 L/h = 10.9 L/min = 0.38 cfm
Coincident demand, 2 points: 226.6 L/min = 8 cfm
Deficit against the compressor: 76.6 L/min must come from the receiver
Reserve needed over 2 min: 153.2 L free air
Receiver needed for that reserve: 77.6 L at a 2 bar usable band (screening)
Usable reserve of the receiver entered: 197.4 L free air: covers the overlap
Time the receiver lasts at that deficit: 2.58 min
Recovery time after the overlap: 1.53 min at a surplus of 100 L/min

Method. Free air per tyre = tyre volume × (target − start pressure) ÷ atmospheric pressure (same pressure unit; constant tyre volume and temperature). Idealised time = free air ÷ effective transfer rate × 60; real time is longer as the pressures equalise, and handling time is extra. Average demand = free air per vehicle × vehicles per hour. Coincident demand = points × transfer rate; deficit = coincident demand − compressor delivery; reserve needed = deficit × overlap duration. Usable receiver reserve = receiver volume × usable band ÷ 1.013 (isothermal). Receiver needed = reserve × 1.013 ÷ band. Recovery time = reserve ÷ (compressor delivery − air still being used). Illustrative screening figures, not promised service times. 1 cfm = 28.3168 L/min.

What to take from it. An hourly average hides the real problem: two islands filling together. Size the receiver for the overlap deficit and check that the lowest receiver pressure still feeds the inflator after hose losses.

Valve actuator air per stroke, peak flow and reserve receiver

Free air per stroke of an on/off process-valve actuator, the flow its solenoid valve and tubing must pass for the required stroke time, the average demand, and the receiver that lets the worst simultaneous movement finish after a supply loss.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Swept volume per stroke, one direction (L): 4
Action: Spring-return or single-acting: one air stroke per cycle
Tubing inside diameter (mm; 0 if none): 0
Tubing length, valve to actuator, each port (m): 0
Other volume filled each stroke: valve passages, boosters (L): 0
Supply pressure at the actuator (bar g): 5.5
Atmospheric pressure (bar a): 1.013
Required stroke time (s; 0 to skip): 3
Operating rate in full cycles (per period below): 20
Rate period: per hour
Number of these actuators in total: 1
Number of them moving together in the worst case: 1
Steady positioner bleed, total (L/min free air): 0
Reserve receiver: normal pressure (bar g; 0 to skip): 7
Lowest pressure the actuators accept (bar g): 5.5

Result

Swept volume per stroke: 4 L
Tubing and other volume per stroke: 0 L (tubing 0 L)
Absolute pressure ratio: 6.429 (6.513 ÷ 1.013)
Free air per stroke: 25.72 L litres of free air at the site atmospheric pressure entered (of which tubing etc. 0 L)
Free air per full cycle: 25.72 L (one air stroke; the spring return uses no air)
Average flow needed during a stroke of 3 s: 8.57 L/s = 0.514 m³/min = 514 L/min = 18.2 cfm per actuator
Average demand: 8.57 L/min = 514 L/h = 0.0086 m³/min = 0.3 cfm
Free air for the worst simultaneous movement: 25.7 L (1 × 25.72 L)
Reserve receiver, 7 → 5.5 bar g: 17.4 litres minimum (isothermal; add margin and other users)

Method. Free air per stroke = (swept volume + tubing volume + other volume filled each stroke) × (p + p₀) ÷ p₀, with p the supply pressure in bar g and p₀ ≈ 1.013 bar a. Cylinder swept volume: extend = π/4 × D² × stroke; retract = π/4 × (D² − d²) × stroke. Tubing volume = π/4 × (inside diameter)² × length, filled once per stroke on the side being pressurised. Double-acting: a full cycle is two strokes (both directions). Spring-return or single-acting: one air stroke per cycle. Average flow needed during a stroke = free air per stroke ÷ stroke time; this is the flow the solenoid valve, tubing and regulator must pass, before allowing for pressure build-up time and margin (valve Kv ≈ 0.86 × Cv; select from the maker’s flow charts at an acceptable pressure drop). Average demand = free air per cycle × cycle rate × number of actuators + steady positioner bleed. Reserve receiver for the worst case = free air for the actuators that move together × p₀ ÷ (normal pressure − lowest pressure the actuators accept); isothermal, add margin.

What to take from it. Stroke time sets the flow the valve and tubing must pass; the average is usually small. Size storage for the worst simultaneous movement, not the average hour.

Rotary actuator torque margin check

Check a candidate actuator’s published torques at the lowest reliable supply pressure against the valve’s breakaway, running and seating torques with your margin, including the spring torques of a spring-return unit, and the end-of-travel energy of a swivelled load.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Actuator type: Spring-return (air opens, spring closes)
Valve breakaway torque (N·m): 100
Valve running torque (N·m): 60
Valve seating (end) torque (N·m): 90
Margin on valve torque (%): 30
Actuator air stroke start torque at minimum pressure (N·m): 210
Actuator air stroke end torque at minimum pressure (N·m): 150
Spring stroke start torque (N·m; spring-return only): 160
Spring stroke end torque (N·m; spring-return only): 125
Moment of inertia of the swivelled load (kg·m²; 0 to skip): 0
Swing angle (degrees): 90
Swing time (s): 1

Result

Required torques with 30% margin: 130 / 78 / 117 N·m breakaway / running / seating
Air stroke start vs breakaway: OK 210 N·m available against 130 N·m required (62% spare)
Air stroke end vs running: OK 150 N·m available against 78 N·m required (92% spare)
Spring start vs running: OK 160 N·m available against 78 N·m required (105% spare)
Spring end vs seating (fail direction): OK 125 N·m available against 117 N·m required (7% spare)
Result: Suitable on the torques entered confirm with the valve and actuator makers

Method. Required torque = valve torque × (1 + margin ÷ 100), with the margin the valve maker recommends for the service (often 20–50%). Double-acting: air start torque must exceed breakaway, the lower of air start and air end must exceed running, and air end must exceed seating (closing by air). Spring-return (air opens, spring closes): air start ≥ breakaway, air end ≥ running, spring start ≥ running, spring end ≥ seating, so the spring can close the valve on loss of air. Read the actuator torques from the maker’s table at the lowest pressure the actuator reliably receives, not at header pressure. Energy at end of travel = ½ × J × ω²; with ω = angle ÷ swing time this is the mean-speed value; a load that accelerates uniformly arrives at twice the mean speed and four times this energy.

What to take from it. Select at the lowest reliable pressure at the actuator, check every point of the torque envelope, and on spring-return units make sure the spring end torque closes the valve.

Robot cell air per cycle

Add up the free air of grippers and cylinders (including hose and valve dead volume), ejectors and blow-off nozzles over one robot cycle, plus re-picks and tool changes, to get the average cell flow and an indicative peak for local valve and receiver sizing.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Grippers and cylinders: swept volume per stroke, each (L): 0.05, 0.02
Hose and valve dead volume for each, same order (L): 0.03, 0.03
Strokes per robot cycle for each, same order: 2, 2
Supply pressure at the tool (bar g): 6
Ejectors and blow-off nozzles: rated consumption, each (L/min free air; 0 if none): 60, 150
On-time per robot cycle for each, same order (s): 2.5, 0.5
Robot cycle time (s): 6
Failed-pick and re-pick allowance (%): 3
Gripper close time, for the peak (s): 0.2
Free air per tool change, vent and refill (L): 0
Tool changes per hour: 0

Result

Actuators: free air per cycle: 1.8 L (pressure ratio 6.92)
Ejectors and nozzles: free air per cycle: 3.75 L
Total free air per robot cycle: 5.55 L
Average cell flow: 57.2 L/min = 0.057 m³/min = 2.02 cfm (incl. 3% re-picks)
Indicative peak flow: 376 L/min = 0.376 m³/min = 13.3 cfm (all ejectors on + largest stroke in 0.2 s)
Peak ÷ average: 6.6 times

Method. Actuator free air per stroke = (swept volume + hose and valve dead volume) × (p + 1.013) ÷ 1.013; per cycle × strokes per cycle. Ejector and nozzle air per cycle = rated consumption (L/min free air) × on-time per cycle ÷ 60. Average flow = air per cycle ÷ cycle time × (1 + re-pick allowance) + tool-change air per hour ÷ 60. Indicative peak = all ejectors and nozzles on together + the largest actuator stroke filled in the gripper close time. Illustrative defaults (the page gives no figures): replace them with your gripper and ejector data.

What to take from it. Average flow sizes the cell’s share of the compressor; the peak sizes the local regulator, valves, hoses and any small receiver. Air-saving ejectors and valves close to the tool cut both.

Machine air per cycle and per accepted item

From a metered total over a productive window and the machine’s counters, work out air per cycle, air per accepted item, average and peak flow, reject share, standby share and the spread between repeat windows.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Total free air metered in the window (m³): 54
Window duration (min): 30
Number of completed cycles in the window: 1800
Number of good (accepted) items in the window: 1764
Peak short-interval flow (m³/min; 0 if not logged): 4.5
Standby flow with air on (m³/min; 0 if not measured): 0
Air per cycle from each repeat window (L; one value is fine): 30
Meter uncertainty at this flow (± %): 3

Result

Air per cycle: 30 L = 0.03 m³
Air per accepted item: 30.61 L = 0.0306 m³ (2% more than per cycle)
Average flow: 1.8 m³/min = 1,800 L/min = 63.6 cfm = 30 L/s
Reject share: 2 % of cycles
Peak-to-average ratio: 2.5 peak 4.5 m³/min = 4,500 L/min = 159 cfm
Repeatability: needs at least two repeat windows run the window at least twice more

Method. Air per cycle = total free air ÷ completed cycles. Air per accepted item = total ÷ good items. Average flow = total ÷ window duration (1 m³/min = 35.31 cfm). Peak-to-average = peak short-interval flow (state the averaging interval) ÷ average. Reject share = 1 − good ÷ cycles. Repeatability = (highest − lowest) ÷ mean of the repeat-window results; if it exceeds the meter uncertainty, the machine is not yet stable. State the flow basis (FAD, Nm³ or scfm) and convert to the compressor’s FAD basis before sizing.

What to take from it. Judge the supply branch and local storage against the peak, the compressor energy against the average, and make standby air a separate acceptance figure.

Pneumatic cylinder force and bore selection

Theoretical push and pull force from bore, rod and the minimum dynamic pressure, the force left after exhaust back-pressure and your load ratio, the smallest standard bore for a required force, and the speed, flow and end-of-stroke energy for a travel time.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Bore (mm): 63
Rod diameter (mm): 20
Minimum dynamic pressure at the valve (bar g): 6
Back-pressure in the exhausting chamber (bar g): 0
Working direction: Push (extend)
Force required at the load (N; 0 to skip bore selection): 800
Load ratio (load ÷ theoretical force): 0.5
Stroke (mm): 200
Travel time (s; 0 to skip): 0.5
Moving mass (kg; 0 to skip): 10

Result

Piston area / annulus area: 31.17 / 28.03 cm²
Theoretical push / pull force: 1,870 / 1,682 N at 6 bar g (87 psi)
Net force, push: 1,870 N before friction
Usable force at load ratio 0.5: 935 N
Theoretical force needed: 1,600 N (800 N ÷ 0.5)
Pressure this bore needs for it: 5.13 bar g (74.4 psi): within the minimum available
Smallest standard bore for that force: 63 mm (rod 20 mm) gives 1,870 N push at 6 bar g
Mean piston speed: 0.4 m/s
Free-air flow during the stroke: 8.63 L/s = 0.518 m³/min = 518 L/min = 18.3 cfm (valve, tubing and fittings must pass this)
Kinetic energy at mean speed: 0.8 J: compare with the cushion or shock-absorber rating

Method. Push area A = π × D² ÷ 4; pull (annulus) area = π × (D² − d²) ÷ 4. Force (N) = pressure (bar g) × area (cm²) × 10, since 1 bar on 1 cm² ≈ 10 N. Net force = driving pressure × driving area − exhaust back-pressure × exhausting area (friction and load still to come off). Theoretical force needed = load ÷ load ratio (about 0.5 for dynamic duties, up to about 0.7 for slow, near-static duties). The smallest standard bore is the first in the series 12–320 mm (typical rods as in the table on the page; check the data sheet) whose net force in that direction reaches the force needed. Mean speed = stroke ÷ travel time; free-air flow during the stroke = area × speed × (p + 1.013) ÷ 1.013; kinetic energy = ½ m v² at mean speed (the real end speed can be higher; a descending vertical load adds gravitational energy).

What to take from it. Size against the lowest pressure the cylinder sees in operation, with a stated load ratio. A cylinder that looks generous on paper can stall when its exhaust side holds back-pressure.

Actuator pressure for a web tension or force

Turn a web tension (per unit width) or a known actuator force into the gauge pressure a dancer or loading cylinder needs, add a friction allowance, check regulator headroom, and give the brake torque for a roll radius.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Tension per unit width (N/m; 0 to enter the force directly): 0
Web width (m): 1
Dancer geometry factor: actuator force per N of tension: 1
Required actuator force (N; used when tension per width is 0): 600
Number of actuators sharing the force: 1
Actuator bore or effective diaphragm diameter (mm): 63
Friction and breakaway allowance (% of force): 10
Regulator inlet pressure at the worst moment (bar g): 6
Margin the regulator needs above its set point (bar): 1
Roll radius for brake torque (m; 0 to skip): 0

Result

Force per actuator: 600 N
Effective area: 31.17 cm² = 0.00312 m²
Pressure before friction: 1.92 bar g = 1,92,478 Pa = 27.9 psi
Working pressure with 10% allowance: 2.12 bar g = 30.7 psi
Regulator headroom: 3.88 bar above the working pressure: OK

Method. Web tension = tension per unit width × width (or enter the actuator force directly). Actuator force = tension × dancer geometry factor ÷ number of actuators sharing it (a dancer roll with a 180° wrap carries about twice the web tension; use the machine builder’s geometry). Effective area = π × D² ÷ 4. Pressure (Pa) = force ÷ area; 1 bar = 100,000 Pa. Working pressure = pressure × (1 + friction allowance ÷ 100). Brake torque = web tension × roll radius; it falls as the roll diameter falls. The regulator needs its inlet a margin above the set pressure.

What to take from it. Low set pressures make seal friction and regulator resolution matter most: use low-friction actuators, a precision regulator close to the actuator, and a supply that never sags near the set point.

Tool air demand: Connected, average and credible peak

For up to ten tool lines, add the connected total, the duty-weighted average and the credible peak from the number of each tool that really runs together, plus background flow (leaks, blow guns, drains), and check a compressor and receiver against the peak.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Flow unit for ratings and background: cfm
Tool lines: how many tools of each kind: 4, 2, 2
Rated consumption per tool, same order (free air): 30, 10, 15
Duty cycle of each kind, same order (%): 40, 20, 30
How many of each run in the largest credible overlap: 3, 0, 1
Background flow: leaks, blow guns, drains (unit above): 15
Margin on the peak (%): 0
Compressor FAD to check (unit above; 0 to skip): 0
Duration of the peak overlap (min): 5
Allowed pressure fall during the peak (bar): 1

Result

Connected total: 185 cfm = 5.24 m³/min (upper bound only)
Average demand: 76 cfm = 2.15 m³/min (41% of connected)
Credible peak: 120 cfm = 3.4 m³/min
Suggested capacity: 120 cfm = 3.4 m³/min (peak, no margin)
Background share of the average: 20 %

Method. Connected = Σ(number × rating) + background. Average = Σ(number × rating × duty cycle) + background. Credible peak = Σ(number running in the largest credible overlap × rating) + background. Ratings are free air at the tool’s rated pressure (usually 6.2 bar g, 90 psig); consumption rises roughly in proportion to absolute inlet pressure. Suggested capacity = peak × (1 + margin). If a compressor FAD below the peak is entered, the receiver to carry the peak for its duration = (peak − FAD) × duration × 1.013 ÷ allowed pressure fall. 1 m³/min = 35.31 cfm = 16.67 L/s.

What to take from it. Size the compressor and receiver for the sustained credible peak, let the control follow the much lower average, and cut background waste first: it is the cheapest air to save.

How many breakers or rock drills a compressor can run

Divide a compressor’s FAD by the tool’s air consumption with a margin for hose losses, leaks, wear and hot-weather derating; or, for a wanted number of tools, find the FAD and the portable class needed.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Flow unit: cfm
Compressor FAD at working pressure (unit above): 185
Air consumption per tool at its rated pressure (unit above): 85
Margin for hoses, leaks, wear and hot weather (%): 15
Number of tools you want to run together (0 to skip): 2

Result

Air per tool with margin: 97.7 cfm = 2.77 m³/min
Tools this compressor can run: 1 with 87.3 cfm left over
FAD needed for 2 tools: 195.5 cfm = 5.54 m³/min: 185–250 cfm class (this compressor is short by 10.5 cfm)
Hose guide: 25 mm main + 19 mm whips main hose to a manifold, short whips to each tool

Method. Air per tool with margin = rated consumption × (1 + margin ÷ 100); the page suggests 15–25%. Tools supported = whole number of FAD ÷ air per tool with margin. FAD needed for n tools = n × consumption × (1 + margin). Portable classes: 185–250 cfm, 250–400 cfm, 400–600 cfm (5.2–7.1, 7.1–11.3, 11.3–17 m³/min). Use the maker’s consumption at the pressure the tool really receives; consumption rises with inlet pressure. Hose guide: 19 mm (3/4 in) for one tool up to about 15–20 m; a 25 mm (1 in) or larger main hose to a manifold with short 19 mm whips for longer runs or two or more tools.

What to take from it. Plan on the FAD the compressor really delivers on a hot site, with margin, and measure pressure at the tool while it works; a weak breaker on a big compressor usually points at the hose.

Cooling heat balance and compressed-air cooling energy

Build a first heat balance (internal, wall, solar and product heat), then compare the energy of a compressed-air cooling device with an electrical alternative on the same duty and hours, and see how much air plain sensible cooling would need.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Heat generated inside the boundary (W): 300
Heat gain through walls from the surroundings (W): 100
Solar or radiant gain (W): 0
Product cooling: mass flow (kg/h; 0 if none): 0
Product specific heat (kJ/kg·K): 0.5
Product temperature drop required (K): 0
Temperature rise of the cooling air (K): 15
Cooling device air consumption, free air (unit below): 0.5
Device flow unit: m³/min
Station specific power (kW per m³/min): 6.5
Running hours per year: 4000
Alternative (fan, heat exchanger, panel AC): electrical input (kW): 0.3

Result

Product heat: 0 W
Total cooling duty: 400 W = 0.4 kW
Air needed to carry it by sensible heat: 1.32 m³/min = 1,322 L/min = 46.7 cfm at a 15 K air temperature rise (95.5 kg/h)
Device: compressor power attributable: 3.25 kW (0.5 m³/min = 500 L/min = 17.7 cfm × 6.5)
Device: energy per year: 13,000 kWh
Device: electrical kW per kW of cooling duty: 8.1
Alternative: energy per year: 1,200 kWh (0.75 kW per kW of duty)
Energy difference: 11,800 kWh per year in favour of the alternative (10.8 ×)
Cost difference per year: Enter your electricity tariff (cost per kWh) above

Method. Cooling duty (W) = internal heat + wall gain + solar/radiant gain + product heat, with product heat = mass flow (kg/h) ÷ 3,600 × specific heat (kJ/kg·K) × 1,000 × temperature drop (K). Sensible cooling by air: air mass flow = duty ÷ (1,005 J/kg·K × air temperature rise); free-air flow = mass flow ÷ 1.204 kg/m³ (20 °C, 1.013 bar a). Compressor power attributable to a device = its air consumption (free air, m³/min) × station specific power (kW per m³/min). Annual energy = kW × running hours (use thermostat-controlled hours where fitted). Illustrative defaults (the page gives no numbers): enter your own heat loads, device rating and station specific power.

What to take from it. For continuous duties an electrical alternative usually removes the same heat with a fraction of the energy; keep compressed-air cooling for small, spot, intermittent or hazardous-area duties, under thermostat or interlock control.

Pressure-decay leak rate and thermal sensitivity

Convert a pressure drop over the test time into a leak rate (cm³/s and sccm), see the error from leaving out fixture volume, compare the pressure change caused by a small temperature change with the drop at your leak limit, and estimate station air use.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Total test volume counted (cm³): 200
Extra fixture volume not yet counted (cm³; 0 if none): 50
Pressure drop measured (Pa): 50
Test (measure) time (s): 10
Reference pressure (Pa): 101325
Test pressure (bar g): 3
Atmospheric pressure (bar a): 1.013
Gas temperature (°C): 30
Temperature change during the test (K): 0.1
Leak limit (sccm; 0 to skip): 1
Test cycles per hour (0 to skip): 60

Result

Leak rate on the volume counted: 0.00987 cm³/s = 0.592 sccm
Leak rate with the extra fixture volume: 0.01234 cm³/s = 0.74 sccm (25% higher; leaving it out understates the leak)
Pressure change from the temperature change alone: 132.4 Pa (4.013 bar a, 303.15 K, 0.1 K change)
Pressure drop at the leak limit: 67.5 Pa in 10 s for 250 cm³
Measured result against the limit: Within the limit 0.74 against 1 sccm (before uncertainty)
Station air per cycle: 0.99 L free air
Station air per hour: 59.4 L = 0.00099 m³/min = 1 L/min = 0.03 cfm

Method. Leak rate Q = V × ΔP ÷ (t × P_ref) in cm³/s at the reference pressure (sccm = cm³/s × 60). V must be the total test volume: part, fixture cavities, seal spaces, tubing, valves and transducer. Thermal effect in a fixed volume: ΔP = P_abs × ΔT ÷ T_abs, with P_abs = (test pressure + atmospheric) × 100,000 Pa and T_abs = T + 273.15 K. Pressure drop at the leak limit = limit (sccm) ÷ 60 × t × P_ref ÷ V. Station air per cycle = V × (p + p₀) ÷ p₀ (filled then vented). Illustrative method only: stabilisation and test times come from trials on known good and leaking masters.

What to take from it. Count every cubic centimetre of fixture volume and control part temperature: a 0.1 K change can move the pressure more than the leak limit does.

Artificial demand after a pressure change

How the flow of pressure-sensitive users (leaks, open nozzles, unregulated cylinders) changes when the supply pressure changes, the compressor saving from the lower set point, and a combined estimate that does not double-count the two.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Present flow of pressure-sensitive users (free air, unit below): 20
Flow unit: m³/h
Present supply pressure (bar g): 7
Proposed supply pressure (bar g): 6
Atmospheric pressure (bar a): 1.013
Compressor input power (kW; 0 to skip): 0
Power saving per bar reduced (%): 7
Running hours per year: 0
Specific power (kW per m³/min; 0 to skip): 0

Result

Absolute pressures: 8.01 → 7.01 bar a
New flow of the pressure-sensitive users: 17.5 m³/h = 0.292 m³/min = 292 L/min = 10.3 cfm
Air saved: 2.5 m³/h = 12.5% of that flow
Gauge-ratio figure (do not use): 17.14 m³/h would claim 14.3%, overstating the change
Energy: enter compressor kW and specific power for the power and combined figures

Method. Pressure-sensitive users behave like a fixed opening: at choked conditions free-air flow is proportional to absolute upstream pressure, so Q_new = Q_old × (p_new + p₀) ÷ (p_old + p₀). The gauge ratio p_new ÷ p_old overstates the change and is shown only as the pitfall. Compressor saving from the lower set point = input kW × Δp × saving per bar (about 7% per bar is typical; edit it). Energy of the air no longer used = air saved (m³/min) × specific power × hours. The two overlap; the combined estimate = kW − (kW − air saved × specific power) × (1 − saving per bar × Δp). Regulated users and process users with a fixed mass flow do not respond. Teaching estimate: prove the saving by a stepped trial with logged flow and power.

What to take from it. Lower pressure in small steps against the minimum each critical user needs, and report pressure reduction, leak repair and running-time controls separately with one measured combined figure.

System leakage test: Load/unload timing, residual flow or pressure decay

Turn a non-production leakage test into a leakage flow, a share of compressor delivery, leak power and annual energy. Choose the method you used; only that method’s inputs are read. Optionally enter the result after repairs, or the measured station power before and after, to state the reduction.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Test method: Load/unload timing (fixed-speed compressor)
Compressor actual delivered air, FAD (m³/min): 10
Load/unload: loaded time per cycle, averaged (s): 30
Load/unload: unloaded time per cycle, averaged (s): 70
Residual: measured flow with production stopped (m³/min): 5
Residual: verified legitimate continuous users (m³/min): 1.2
Decay: system volume, receivers plus pipework (m³): 5
Decay: start pressure (bar g): 7
Decay: end pressure (bar g): 6
Decay: time for the pressure fall (s): 60
Atmospheric pressure (bar a): 1.013
System pressure during the test (bar g): 7.5
Normal operating pressure to restate the result at (bar g): 7.5
Leakage after repairs, same method and pressure (m³/min; 0 if not yet measured): 0
Station specific power (kW per m³/min): 6.5
Pressurised hours per year: 6000
Optional: station power in the idle test before repairs (kW; 0 to skip): 0
Optional: station power in the idle test after repairs (kW): 0
Idle (non-production) hours per year for the measured power: 2000

Result

Loaded share of the cycle: 30% T ÷ (T + t) = 30 ÷ 100 s
Estimated leakage at the test pressure: 3 m³/min free air = 50 L/s = 105.9 cfm (load/unload)
Leakage as a share of compressor delivery: 30% of 10 m³/min
Leak power at the operating pressure: 19.5 kW at 6.5 kW per m³/min
Leak energy per year: 1,17,000 kWh over 6,000 pressurised hours
Leak energy cost per year: Enter your electricity tariff (cost per kWh) above

Method. Load/unload timing (fixed-speed load/unload compressor, production stopped, timed over several cycles): leakage ≈ FAD × T ÷ (T + t), with T the loaded and t the unloaded time per cycle. Residual flow (flow meter, production stopped): unexplained residual = measured residual flow − verified legitimate users (dryer purge, instrument air, drains); it is the figure to investigate, not automatically all leakage. Pressure decay (compressors stopped and isolated, no users): leakage in free air ≈ V × (p_start − p_end) ÷ (p_atm × time), with V the receiver plus pipework volume in m³, pressures in bar and time in minutes; the pressure difference is the same in gauge or absolute terms. Leak flow through an orifice rises roughly in proportion to upstream absolute pressure, so a result taken at one pressure is restated at the normal operating pressure by × (p_op + p_atm) ÷ (p_test + p_atm). Leak power = leakage × station specific power (kW per m³/min); annual energy = power × pressurised hours. The after-repair figure must come from the same method at the same pressure. The measured-power option gives (kW before − kW after) × idle hours, which is a measured saving, not an estimate. Money is shown only when you enter your own tariff.

What to take from it. Record the method, the pressure and the plant state with every result, so the next test is comparable. A lower leak flow saves energy only if the compressors then unload, slow down or stop.

Receiver volume to limit load/unload cycling

Find the receiver volume that keeps a load/unload compressor within the cycle limit its maker allows, and check the shortest cycle with the receiver you already have.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Compressor free air delivery: 100
Unit: L/s
Shortest load/unload cycle the maker allows (s per cycle): 30
Load/unload pressure band (bar): 0.5
Atmospheric pressure (bar a): 1.013
Receiver volume you already have (litres; 0 to skip): 0

Result

Compressor delivery: 100 L/s = 211.9 cfm = 6 m³/min
Cycle limit: 0.0333 cycles per second = 120 cycles per hour
Receiver volume needed: 1,519 litres = 1.52 m³
Rough guide, 6–12 L per cfm: 1,271 to 2,543 litres, for comparison only

Method. V = 0.25 × q × p_atm ÷ (f × Δp), with V in litres, q the compressor free air delivery in L/s, f the maximum load cycles per second the maker allows (1 cycle per 30 s = 0.033 per s) and Δp the load/unload band in bar. The worst case is demand equal to half the compressor output, which is where 0.25 comes from. Rearranged, the shortest cycle with an existing receiver = V × Δp ÷ (0.25 × q × p_atm). Unit factors: 1 cfm = 0.472 L/s, 1 m³/min = 16.67 L/s. Pipework adds a little storage. The 6–12 litres per cfm range is the page’s rough guide for a single load/unload screw compressor; peak and standby-bridging duties are sized with the receiver storage tool, and the largest of the volumes is used.

What to take from it. Widening the band reduces the volume needed but raises average pressure and power (about 7% per bar). Piston compressors on start/stop are limited in motor starts per hour and often need more storage.

Receiver pump-up test: Estimated free air delivery

Time how long the compressor takes to raise the pressure of a receiver of known volume (outlet closed) between two pressures, and convert it into an approximate free air delivery to compare with the rating.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Receiver volume (litres): 1000
Connected pipe volume up to the closed outlet (litres): 0
Start pressure (bar g): 6
End pressure (bar g): 7
Measured pump-up time (s): 10
Atmospheric pressure (bar a): 1.013
Intake air temperature (°C): 30
Air temperature in the receiver at the end (°C): 30
Rated FAD to compare with (m³/min; 0 to skip): 7

Result

Volume pressurised: 1,000 litres
Free air pumped in: 987 litres of free air for 1 bar
Estimated FAD: 5.92 m³/min = 98.72 L/s = 209.2 cfm
Timing sensitivity: ±10% for a ±1 s timing error
Expected time at the rated FAD: 8.5 s
Measured delivery as a share of rating: 84.6% of 7 m³/min

Method. FAD ≈ V × (p_end − p_start) ÷ p_atm ÷ t × (T_intake ÷ T_receiver), with V in m³ (receiver plus connected pipe), pressures in bar, t in minutes and temperatures in kelvin. With equal temperatures the correction is 1 (isothermal). Expected time at the rated FAD = V × Δp ÷ (p_atm × FAD_rated) × (T_intake ÷ T_receiver). It is a screening test: warm air in the receiver, an unknown pipe volume and a 1 s timing error all move the result, so a machine is judged far off its rating only with a clear margin. For a definitive figure use a flow meter or an air audit.

What to take from it. Run the test at the working pressure band after the machine is warm, repeat it three times and use the average time. Record the receiver volume from its nameplate.

Nitrogen generator feed-air compressor sizing

Work out the compressor FAD and discharge pressure a nitrogen generator needs, from the nitrogen flow and air factor or from the maker’s feed-air figure, adding dryer purge and losses the air factor does not cover.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Feed-air basis: Nitrogen flow × air factor
Nitrogen flow per generator (in the flow unit below): 100
Air factor: feed air ÷ nitrogen, same basis: 3
Feed-air flow per generator from the data sheet (in the flow unit below): 300
Flow unit: scfm (standard reference)
Reference temperature of the scfm figure (°C): 20
Number of generators running together: 2
Losses outside the air factor, e.g. dryer purge (% of compressor delivery): 0
Design margin on flow (%): 0
Site inlet temperature (°C): 20
Site inlet pressure (bar a): 1.013
Generator minimum inlet pressure (bar g): 8
Pressure drop through filters, dryer and pipe (bar): 0.5

Result

Feed air per generator: 300 scfm = 100 × 3
Total nitrogen flow: 200 scfm from 2 generators
Total feed air before other losses: 600 scfm
Compressor FAD to enquire at site inlet: 16.99 m³/min = 600.1 cfm at 20 °C, 1.013 bar a
Minimum compressor discharge pressure: 8.5 bar g = 123 psig (8 + 0.5 bar losses)

Method. Feed air per generator = nitrogen flow × air factor (same reference basis), or the feed-air flow from the data sheet. Total = per generator × number running together. Losses outside the air factor (for example heatless desiccant dryer purge, as a share of compressor delivery): total ÷ (1 − loss). Margin: × (1 + margin). Conversion to FAD at site inlet: Q_FAD = Q_ref × (p_ref ÷ p_inlet) × (T_inlet ÷ T_ref), absolute values; Nm³/h is 0 °C and 1.01325 bar a; for scfm enter the reference temperature its quoter uses (at 1.01325 bar a). Minimum compressor discharge pressure = generator minimum inlet pressure + filter, dryer and pipe losses. The air factor must come from the selected generator’s data at your purity, pressure and temperature; it rises steeply with purity.

What to take from it. Size on the worst summer inlet temperature and check the pressure at the generator inlet during bed switching, not at the compressor. Do not add pressure “for safety”: about 7% more power per bar.

Electric vs diesel compressor running cost

Compare the monthly running cost of an electric compressor (on grid power or on a DG set) with a diesel portable at the same duty, from documented input power, fuel use at the expected load and your own prices.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Running hours per month: 200
Electric: package input power at the duty (kW): 75
Electric supply: Grid or site substation
DG set: diesel per kWh generated at its expected load (L/kWh): 0.3
Diesel compressor: fuel at the expected load (L/h): 20

Result

Electric energy per month: 15,000 kWh = 75 kW × 200 h
Electric running cost per month: Enter your electricity tariff (cost per kWh) above
Diesel compressor fuel per month: 4,000 litres = 20 L/h × 200 h
Diesel compressor running cost per month: Enter your own prices above

Method. Electric energy = package input kW at the duty × running hours. On grid power: cost = kWh × tariff + demand or fixed charges. On a DG set: DG diesel = kWh × the DG’s litres per kWh at its expected load (from the DG data; light loading raises it), cost = litres × diesel price + fixed charges. Diesel portable: fuel = litres per hour at the expected load × hours; cost = litres × diesel price + engine servicing per hour × hours. Break-even tariff = (diesel L/h × price + servicing per hour) ÷ electric kW: below it, grid electric is cheaper to run. No prices are assumed; enter your own.

What to take from it. Use input power and fuel burn at the real load, not full-load catalogue values, and compare at the same FAD and pressure. A DG brought in only for the compressor removes most of the electric advantage.

Heat recovery: Compressor supply vs hot-water demand

Compare the heat a compressor can give in the hours it is wanted with the heat a hot-water user needs, and express the useful recovery as fuel displaced.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Recoverable heat from the compressor (kW): 40
Hours per day the heat is available and wanted (overlap): 16
Hot water needed per day (litres): 5000
Temperature rise needed (°C): 30
Cold-water supply temperature (°C): 25
Displaced heater efficiency (%): 85
Fuel displaced: Diesel (litres)
Days per year the heat user runs: 300

Result

Recoverable heat per day: 640 kWh = 40 kW × 16 h
Heat the user needs per day: 174 kWh = 5,000 L × 30 °C × 1.163 kWh per 1,000 L·°C
Useful recovery per day: 174 kWh, limited by the heat user’s demand
Recoverable heat left unused: 466 kWh per day (27% of the recoverable heat is used)
Water temperature target: 55 °C
Fuel energy displaced per day: 205 kWh at 85% heater efficiency
Fuel saved per day: 20.73 litres of diesel
Fuel saved per year: 6,218 litres of diesel over 300 days
Value per year: Enter the price of the fuel you would otherwise use above

Method. Supply per day = recoverable heat (kW) × overlapping hours. Demand per day = litres × temperature rise × 4.186 kJ/(kg·K) ÷ 3,600, about 1.16 kWh per 1,000 litres per °C. Useful recovery = the lower of the two. Fuel energy displaced = useful heat ÷ efficiency of the heater it replaces; fuel quantity = that energy ÷ the fuel’s calorific value (net values used here: diesel about 9.9 kWh per litre, furnace oil about 10.7 kWh per litre, LPG about 12.8 kWh per kg). Recoverable heat = compressor input power × the share the maker states is recoverable for the model and recovery method (see the heat recovery tool).

What to take from it. The size of the heat user, not the compressor, usually sets the saving. If supply and demand hours do not overlap, add hot-water storage before counting the heat.

Combined savings of interacting energy measures

Combine the stand-alone savings of several compressed-air measures applied one after another, instead of adding them, and see how much the plain sum overstates the result.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Baseline station energy (kWh per year): 1000000
Savings entered as: kWh per year
Stand-alone saving of each measure, in order of application: 150000, 80000

Result

Measure 1: alone: 1,50,000 kWh per year (15% of baseline)
Measure 1: in sequence: 1,50,000 kWh per year, leaving 8,50,000 kWh
Measure 2: alone: 80,000 kWh per year (8% of baseline)
Measure 2: in sequence: 68,000 kWh per year, leaving 7,82,000 kWh
Combined saving: 2,18,000 kWh per year (21.8% of baseline)
Plain sum of stand-alone savings: 2,30,000 kWh per year
Overstatement of the plain sum: 12,000 kWh per year (5.5% too high)
Energy remaining after all measures: 7,82,000 kWh per year
Value of the combined saving per year: Enter your electricity tariff (cost per kWh) above

Method. Each measure’s stand-alone saving is turned into a fraction of the baseline. Applied in sequence, each measure saves its fraction of the energy left by the measures before it: saving_i = residual_(i−1) × fraction_i, residual_i = residual_(i−1) − saving_i. Combined saving = baseline − final residual. The combined total does not depend on the order, but the saving credited to each measure does, so list the measures in the order they will be done (usually leaks and demand first, then pressure, then supply matching, then replacement and heat recovery). This is a multiplicative screening model; verify each stage by measurement against the adjusted baseline.

What to take from it. Report the sequenced figure, not the sum. Value VSD, replacement and heat-recovery projects on the demand left after leak and pressure work.

VSD trim control-gap check

Check whether a VSD trim compressor’s usable flow range covers the step of the largest fixed-speed machine it works with, find the demand band where a control gap occurs, and the VSD size that would close it.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

VSD maximum flow, FAD (m³/min): 10
VSD minimum stable flow (% of maximum): 30
Largest fixed-speed machine FAD, the step (m³/min): 10
Lowest demand, e.g. nights or weekends (m³/min; 0 to skip): 0
Margin wanted on the usable range (%): 10

Result

VSD minimum flow: 3 m³/min
VSD usable range: 7 m³/min = 247 cfm
Largest fixed-speed step: 10 m³/min = 353 cfm
Control gap: Yes 3 m³/min wide
Demand band the station cannot follow smoothly: 10 to 13 m³/min
VSD maximum flow needed to cover the step with margin: 15.71 m³/min = 555 cfm

Method. VSD minimum flow = maximum × minimum %. Usable range = maximum − minimum. A control gap exists when the largest fixed-speed step is larger than the usable range. When demand rises just above the VSD maximum, the fixed machine starts and the VSD must fall to demand − step; below its minimum it cannot, so demand between the VSD maximum and (fixed step + VSD minimum) cannot be followed smoothly. Gap width = step − usable range. Minimum VSD maximum flow to close the gap = step × (1 + margin) ÷ (1 − minimum %), at the same turndown. Use the maker’s minimum flow; many VSD screws turn down to about 20–40%.

What to take from it. If demand often sits in the gap band, resize the VSD, change which fixed machine runs, or add storage. Log demand before choosing.

FMEA risk priority number with severity review

Rank compressed-air failure modes by risk priority number (severity × occurrence × detection) and list every high-severity item for separate review, whatever its RPN.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Severity of each failure mode (1–10): 8, 6, 10
Occurrence of each failure mode (1–10): 6, 5, 2
Detection of each failure mode (1–10, 10 = not detected): 3, 4, 3
Severity at or above which an item is reviewed separately (1–10): 9

Result

Item 1 RPN: 144 8 × 6 × 3, rank 1 of 3
Item 2 RPN: 120 6 × 5 × 4, rank 2 of 3
Item 3 RPN: 60 10 × 2 × 3, rank 3 of 3, severity review
Ranking by RPN: Item 1, Item 2, Item 3 highest first
Reviewed separately (severity 9 or more): Item 3 whatever their RPN

Method. Enter one score per failure mode in each list, in the same order (Item 1, Item 2 …), on the site’s own 1–10 scales. Detection: 10 = will not be detected before it acts. RPN = S × O × D (maximum 1,000). Items are ranked by RPN; equal RPNs are ordered by severity. Any item with severity at or above the review threshold is listed separately, because multiplying ordinal scores can hide a rare but serious failure behind a frequent nuisance. The RPN sorts work; it is not a probability.

What to take from it. Act on the separately reviewed items and the highest RPNs, give each action an owner and date, and re-score after the action is complete.

Compressor motor overload setting and supply unbalance

Find the overload relay setting for the starter type and relay position, the typical starting current, and the voltage and current unbalance from three measured phase readings.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Motor full-load current at the actual supply voltage (A): 100
Starting method: Star-delta
Overload relay position (star-delta): Motor winding (phase) circuit
Service factor or ambient correction from the makers (multiplier): 1
Measured phase-to-phase voltages under load (V, three values): 415, 410, 400
Measured line currents under load (A, three values): 95, 97, 100
Motor maker’s voltage unbalance limit (%): 1

Result

Overload relay setting: 57.7 A = FLC ÷ √3 (57.7% of FLC)
Typical starting current: 200 to 267 A (2–2.7 × FLC, typical values only)
Average voltage: 408.3 V; largest deviation 8.3 V
Voltage unbalance: 2.04% limit entered 1%
Average current: 97.3 A = 97% of FLC
Current unbalance: 2.74% 1.3 × the voltage unbalance

Method. Overload setting = motor nameplate full-load current (at the actual supply voltage) × the service factor or ambient correction from the relay and motor makers. On a star-delta starter with the relay in the winding (phase) circuit, the winding current is line current ÷ √3, so the setting is FLC ÷ √3 (about 58% of FLC); in the line circuit it is FLC. Typical starting current (page table, typical values only): DOL 6–8 × FLC; star-delta about one third of DOL (2–2.7 × FLC); soft starter 2–4 × FLC; VFD about 1 × FLC. Unbalance (%) = largest deviation from the average ÷ average × 100 (the common definition used in motor standards); voltages phase to phase, currents in the three lines, all under load. A small voltage unbalance produces a much larger current unbalance and extra winding heating.

What to take from it. Never raise the overload to cure trips: find the cause (pressure, cooler, voltage, phase, bearing, starts per hour). Confirm the relay position on the starter maker’s diagram.

Drilling fuel and cost per metre: Two air packages

Compare two compressor packages on the metres they actually drill: accepted metres, fuel per metre, time per metre, effective rate after downtime and cost per metre with your own prices.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Fuel entered as: Litres used in the period
Package A: hours in the period, drilling plus downtime (h): 10
Package A: downtime within those hours (h): 0
Package A: average penetration rate (m/h): 12
Package A: fuel (litres, or L/h, as selected): 300
Package B: hours in the period, drilling plus downtime (h): 10
Package B: downtime within those hours (h): 0
Package B: average penetration rate (m/h): 18
Package B: fuel (litres, or L/h, as selected): 380
Metres to drill on the project (m): 1000
Optional: MTBF, operating time ÷ failures (h; 0 to skip): 500
Optional: MTTR, repair time ÷ repairs (h): 10

Result

Package A: accepted metres: 120 m = 10 productive h × 12 m/h
Package A: fuel per metre: 2.5 L/m (300 litres)
Package A: time per metre: 5 minutes per metre
Package A: effective rate: 12 m per scheduled hour (100% of hours productive)
Package A: cost per metre: Enter your own prices above
Package B: accepted metres: 180 m = 10 productive h × 18 m/h
Package B: fuel per metre: 2.111 L/m (380 litres)
Package B: time per metre: 3.33 minutes per metre
Package B: effective rate: 18 m per scheduled hour (100% of hours productive)
Package B: cost per metre: Enter your own prices above
B vs A: fuel per metre: -0.389 L/m 15.6% less for B
B vs A: time per metre: -1.67 min/m 33.3% less for B
Project drilling hours, A / B: 83.3 / 55.6 h at the penetration rates
Project scheduled hours, A / B: 83.3 / 55.6 h including downtime
Project fuel, A / B: 2,500 / 2,111 litres; B saves 389 litres
Availability from MTBF and MTTR: 98% = 500 ÷ (500 + 10)

Method. Productive hours = hours in the period − downtime. Accepted metres = productive hours × average penetration rate. Fuel: either the litres used in the period, or litres per drilling hour × productive hours. Fuel per metre = litres ÷ metres. Time per metre = 60 ÷ penetration rate (minutes). Effective rate = metres ÷ all hours in the period (penetration × availability). Cost per metre = (fuel × diesel price + other period costs such as crew, hire or ownership, bits, hammer, servicing) ÷ accepted metres. Project: drilling hours = metres ÷ penetration rate; scheduled hours = metres ÷ effective rate; fuel = metres × fuel per metre. Availability from records = MTBF ÷ (MTBF + MTTR). Use several holes or a month, not one day; the comparison works in litres, metres and hours whatever your prices.

What to take from it. Judge the air package by litres and cost per accepted metre, not litres per hour. Downtime keeps crew and ownership costs running while adding no metres.

Pumping test: Drawdown, specific capacity and pumped volume

Work out drawdown, specific capacity and pumped volume from a pumping test, with every level measured from the same datum.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Static water level below datum (m): 30
Pumping water level below the same datum, stabilised (m): 50
Discharge: 600
Discharge unit: L/min
Pumping duration at constant discharge (h): 24

Result

Drawdown: 20 m
Discharge: 600 L/min = 10 L/s = 36 m³/h
Specific capacity: 30 L/min per m of drawdown, at 600 L/min after 24 h
1.8 m³/h per m
0.5 L/s per m
Pumped volume at constant discharge: 864 m³ = 8,64,000 litres

Method. Drawdown s = pumping water level − static water level (both as depths below the same datum, e.g. top of casing). Specific capacity = discharge Q ÷ s, quoted with the rate and the pumping duration, because it usually falls as the rate rises and as pumping continues. Pumped volume V = Q × t, valid only while the discharge is constant; for a varying discharge add the measured interval volumes. 1 L/s = 3.6 m³/h = 60 L/min.

What to take from it. Choose the operating rate from the constant-rate and recovery tests with headroom for the dry season and nearby pumping, not from the airlift figure or the maximum tested rate.

Borewell air-lift start pressure and submergence

Estimate the pressure a compressor needs to start an air-lift (development, testing or redevelopment), the submergence ratio, and the effect of hot intake air on the mass of air delivered.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Submerged length of the air line below standing water (m): 60
Lift from the water level up to the discharge point (m): 40
Hose and fitting losses (bar): 0
Compressor rated pressure, to check margin (bar g; 0 to skip): 0
Compressor intake temperature (°C): 45

Result

Air-lift start pressure: 5.89 bar g = 85 psig = 6 kg/cm² (60 m × 0.0981)
Submergence ratio: 60% of the air-line length below the discharge point
Air mass at this intake temperature: -7.9% compared with 20 °C intake air

Method. Start pressure ≈ submerged length of the air line × 0.0981 bar per metre of fresh water (10 m ≈ 0.98 bar ≈ 14.2 psi) + hose and fitting losses. The submerged length is measured from the water level at the time of starting to the air-line foot; once lifting begins the working pressure usually falls. Submergence ratio = submerged length ÷ (submerged length + lift from the water level to the discharge point); about 60% is commonly quoted as a target, and lower submergence lifts water poorly. Free air delivery is a volume at inlet conditions, so hot intake air carries less mass: mass ratio = (20 + 273.15) ÷ (intake °C + 273.15).

What to take from it. Choose a compressor with clear margin above the start pressure, and record the datum and submerged length with the result.

Borewell production forecast

Forecast the water a borewell will deliver in a period from the approved rate, scheduled hours and an availability factor, with transfer losses, an optional second interval (for example a reduced dry-season rate) and a check against demand.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Approved operating rate (m³/h): 10
Pumping hours per day: 8
Pumping days in the period: 25
Planned maintenance hours removed from the schedule (h): 4
Availability factor for unplanned stops (0 to 1): 0.95
Transfer losses, wellhead to tank (%): 3
Optional second interval: rate (m³/h; 0 if none): 0
Second interval: pumping hours per day: 0
Second interval: pumping days: 0
Second interval: availability factor (0 to 1): 0.95
Demand in the period, at the tank (m³; 0 to skip): 0

Result

Scheduled hours: 196 h = 8 × 25 − 4 maintenance
Wellhead volume, interval 1: 1,862 m³ = 10 × 196 × 0.95
Forecast wellhead volume: 1,862 m³
Water entering the tank: 1,806 m³ after 3% transfer losses

Method. Scheduled hours H = hours per day × days − planned maintenance hours removed from the schedule. Forecast wellhead volume V = Q × H × A, with A the availability factor for unplanned stops only (power cuts, faults, trips); maintenance already removed from H is not deducted again through A. For a second interval with a different rate or schedule, V = Σ(Q × H × A). Water entering the tank = V × (1 − transfer loss). Shortfall or surplus = tank volume − demand. The result is arithmetic on assumptions, not evidence that the well can sustain the rate: that comes from testing and monitoring.

What to take from it. Give a central and a low case (dry-season rate, lower availability). If the low case does not meet demand, plan storage, a second source or redevelopment now.

Borewell production variance: Rate and hours effects

Put a number on the gap between forecast and actual production and split it into a rate effect and an hours effect, so the investigation starts in the right place.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Data available: Average rates and pumping hours
Forecast average rate (m³/h): 10
Forecast pumping hours (h): 200
Actual average rate (m³/h): 9
Actual pumping hours (h): 190
Volumes only: forecast volume (m³): 2000
Volumes only: actual volume (m³): 1760

Result

Forecast volume: 2,000 m³ = 10 × 200
Actual volume: 1,710 m³ = 9 × 190
Absolute variance: -290 m³ (negative = below forecast)
Percentage variance: -14.5% of the forecast
Rate effect: -200 m³ = (9 − 10) × 200
Hours effect: -90 m³ = 9 × (190 − 200)
Check: rate + hours effect: -290 m³, equals the total variance
Larger share: Rate look first at pump, head, water level, valve setting and meter

Method. Forecast volume Vf = Qf × Hf; actual Va = Qa × Ha. Absolute variance = Va − Vf; percentage = 100 × (Va − Vf) ÷ Vf. Sign convention: negative = less than forecast (not necessarily bad, if demand fell). Rate effect = (Qa − Qf) × Hf; hours effect = Qa × (Ha − Hf); the two add up exactly to the total. Other conventions place the interaction term differently, so keep one convention month to month. Compare forecast and actual at the same boundary (wellhead, tank inflow or usable water) and period.

What to take from it. A rate effect points to the pump, head, water level, valve or meter; an hours effect points to power, trips, maintenance or tank-full standby. Check the meter record before diagnosing the well.

Borewell derating options: Rate factor, daily volume and shortfall

Compare restriction options against a stated reference duty: daily volume, rate factor, volume factor and the shortfall against daily demand.

The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.

Worked example

Inputs

Reference (baseline) rate (m³/h): 16
Reference run hours per day: 6
Option rates (m³/h, one per option): 12, 16, 12
Option run hours per day (one per option): 6, 4.5, 8
Daily demand to be met (m³ per day; 0 to skip): 96

Result

Reference daily volume: 96 m³ = 16 m³/h × 6 h
Option A: daily volume: 72 m³ = 12 × 6 h
Option A: rate factor fQ: 0.75 volume factor 0.75
Option A: shortfall against demand: 24 m³ per day
Option B: daily volume: 72 m³ = 16 × 4.5 h
Option B: rate factor fQ: 1 volume factor 0.75
Option B: shortfall against demand: 24 m³ per day
Option C: daily volume: 96 m³ = 12 × 8 h
Option C: rate factor fQ: 0.75 volume factor 1
Option C: surplus against demand: 0 m³ per day

Method. Enter one rate and one run time per option, in the same order. Daily volume = rate × run hours. Rate factor fQ = Q_restricted ÷ Q_reference (0.75 means 75% of the reference rate; it is not a 25% groundwater, energy or impact saving). Volume factor = option daily volume ÷ reference daily volume. Shortfall = demand − option daily volume. “Reduce pumping by 25%” is ambiguous: it can mean rate, hours or volume, so decide which the restriction must control.

What to take from it. A lower rate run for longer can take the same daily volume. Check any reduced rate against the pump motor’s minimum cooling flow, and record why the restriction was set and what evidence would lift it.

Calculator validation and worked example library

Why should you trust an engineering calculator?

What goes into a reference case?

Types of test in the library

Expected answers must be independent

Units, reference conditions and boundaries

Tolerances without moving engineering limits

Worked example: warmer water does not prove more cooling

Import, calculation and export checked separately

Versions and coverage reporting

When a test fails

How Reddy Compressors helps

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Frequently asked questions

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Is a small rounding difference acceptable in an engineering result?

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