How to use these scenarios
Every scenario on this page uses invented but realistic numbers. None of them describes a Reddy Compressors customer or a measured result. They follow the same pattern:
- State the engineering question.
- List the inputs with units and reference conditions.
- Apply the method or formula.
- Check that the answer is physically sensible.
- Vary the inputs that matter most.
- Note what must be measured before a decision is made.
The calculators on our engineering tools pages apply the same methods. These scenarios explain the reasoning behind them.
Units that cause most mistakes
| Quantity | Common confusion | Rule to follow |
|---|---|---|
| Pressure | Gauge (bar g, psig) versus absolute (bar a) | Use absolute pressure in gas-law calculations: bar a ≈ bar g + 1.013 at sea level |
| Flow | Free air delivery versus actual volume at line pressure | Compressor ratings (FAD, cfm) are at inlet conditions; line volume is much smaller |
| Flow units | cfm, m³/min, l/s | 1 m³/min ≈ 35.3 cfm; 1 l/s ≈ 2.12 cfm |
| Power | Motor nameplate kW versus measured input kW | Use measured electrical input for energy work |
| Energy | kW versus kWh | kWh = average kW × hours |
| Dew point | Pressure dew point versus atmospheric dew point | Always state the pressure at which dew point is measured |
| Pressure conversion | bar and psi | 1 bar = 14.5 psi |
Scenario 1: sizing a compressor from demand
Question. A workshop runs pneumatic tools and machines. What compressor capacity does it need?
Inputs (illustrative). Ten tools at 20 cfm each, used about 40% of the time; four machines at 15 cfm each, running continuously; required pressure 6.5 bar (94.3 psig) at the tools.
Method. Average demand = (10 × 20 × 0.4) + (4 × 15) = 80 + 60 = 140 cfm. Add an allowance for leakage and future growth agreed with the plant, for example 20%: 140 × 1.2 = 168 cfm. Allow for pipework, dryer and filter losses when setting compressor pressure, typically 7–7.5 bar (101.5–108.8 psig).
Sensitivity. If the tools are used together at peak, short-term demand can be much higher than the average. A receiver can cover short peaks; sustained peaks need compressor capacity.
Check on site. Log actual flow and pressure over a full shift before buying. Many plants size from nameplates and end up with an oversized, inefficient compressor.
Scenario 2: the cost of extra pressure
Question. What happens to compressor power if the setpoint is raised from 7 bar to 8 bar (about 100 to 116 psig)?
Inputs (illustrative). A screw compressor drawing 75 kW at 7 bar, running 6,000 hours a year at a similar load.
Method. As a typical rule, each extra bar needs about 7% more power for the same flow. Extra power ≈ 75 × 0.07 = 5.25 kW. Extra energy ≈ 5.25 × 6,000 = 31,500 kWh a year.
Sensitivity. Leaks and unregulated users also consume more air at higher pressure, so the real increase is usually higher than the compressor effect alone.
Check on site. Find out why higher pressure is being asked for. A pressure-drop survey often shows a restriction that can be fixed instead.
Scenario 3: leakage screening by the load-time method
Question. How much air is the plant losing to leaks?
Inputs (illustrative). With production stopped, a 300 cfm load/unload compressor runs loaded for 24 seconds in every 80-second cycle.
Method. Leakage ≈ capacity × loaded time ÷ total time = 300 × 24 ÷ 80 = 90 cfm. In this example, about 90 cfm is being produced only to feed leaks.
Sensitivity. Drains, dryers and machines left on standby also use air during the test. Note which equipment was running.
Check on site. Follow up with an ultrasonic leak survey, tag and repair leaks, then repeat the test to confirm the reduction.
Scenario 4: comparing compressors by specific power
Question. Which of two compressors is more efficient?
Inputs (illustrative). Compressor A delivers 10 m³/min at 7 bar with 62 kW measured package input. Compressor B delivers 9 m³/min at 7 bar with 58 kW.
Method. Specific power = input kW ÷ flow. A = 62 ÷ 10 = 6.2 kW per m³/min. B = 58 ÷ 9 ≈ 6.4 kW per m³/min. A is slightly more efficient at this point.
Sensitivity. Compare at the same pressure and include fans, dryers and controls if they are part of the package. Part-load efficiency often matters more than full-load efficiency.
Check on site. Use measured power and verified flow wherever possible, and data sheets prepared to the same test standard where not.
Scenario 5: receiver size for a short peak
Question. What receiver volume will cover a short peak without pressure falling too far?
Inputs (illustrative). A peak of 100 cfm above compressor capacity lasting 30 seconds; allowable pressure fall from 7.5 to 6.5 bar(g) (a 1 bar drop).
Method. Volume ≈ (excess flow × time × atmospheric pressure) ÷ allowable pressure drop. Excess air = 100 cfm × 0.5 min = 50 ft³ of free air ≈ 1.42 m³. Volume ≈ 1.42 × 1.013 ÷ 1 ≈ 1.43 m³, so a receiver of about 1.5 m³ (1,500 litres) is indicated.
Sensitivity. Halving the allowable drop doubles the volume required.
Check on site. Confirm peak duration with a logger, and make sure any receiver is designed, tested and inspected to the applicable pressure-vessel rules.
Scenario 6: temporary air for site work
Question. What portable compressor suits two breakers and a blasting nozzle on a construction site?
Inputs (illustrative). Two breakers at about 100 cfm each at 7 bar (101.5 psig); one No. 5 blast nozzle needing about 195–210 cfm at 7 bar.
Method. Total ≈ 200 + 210 = 410 cfm. Add margin for hose losses and wear, which points to the 400–600 cfm class, for example a 450 cfm unit.
Sensitivity. If the breakers and blasting run at different times, a smaller unit may do. Longer hoses and worn nozzles raise demand.
Check on site. Confirm tool ratings, hose lengths and diameters, and whether the blasting needs an aftercooler and moisture separator for coating quality.
Scenario 7: borewell compressor duty
Question. What compressor class suits a borewell rig drilling a deep hole in hard rock?
Inputs (illustrative). DTH hammer requiring high pressure; bore deep enough that water column and cuttings increase back-pressure; good uphole velocity needed to lift cuttings.
Method. Borewell compressors typically range from 600–1,500 cfm at 200–400 psig (13.8–27.6 bar), with special units up to 35 bar (507.6 psig). Deeper holes and harder formations usually push towards the higher pressure bands; larger hole diameters need more flow to keep uphole velocity adequate.
Check on site. Match the compressor to the hammer maker’s pressure and flow requirement, the bit and drill-pipe size, and the expected water level.
Scenario 8: altitude and hot ambient correction
Question. Does a compressor deliver its rated free air at high altitude or in high heat?
Method. Free air delivery is stated at reference inlet conditions. At higher altitude the air is thinner, so the mass of air delivered falls even if the volume at inlet conditions is similar. In Indian summers of 40–45 °C, and above 45 °C on open sites, the main concern is cooling: high ambient temperature raises discharge temperature, which in an oil-injected screw is normally about 75–95 °C, and can lead to high-temperature shutdowns if coolers are dusty or ventilation is poor.
Check on site. Confirm the compressor’s rated maximum ambient and keep coolers clean, especially on dusty construction and drilling sites.
Common mistakes in compressed-air calculations
- Mixing gauge and absolute pressure in gas-law calculations.
- Sizing from tool nameplates without considering how often tools are used.
- Using motor nameplate power instead of measured input power for energy work.
- Comparing compressors at different pressures.
- Assuming pressure drop scales simply with flow across a whole network.
- Treating an estimate as a measured result in a report.
How Reddy Compressors helps
Reddy Compressors, at Tarnaka, Secunderabad, has served industry since 2002. We are an independent multi-brand specialist: we sell, rent and service compressors of ELGi, Atlas Copco, Ingersoll Rand, KAESER, Chicago Pneumatic and other brands, and we are not an authorised dealer of any of them. We have tie-ups with all major and reputed brands. Our engineers can take the methods on this page and apply them to your plant with measured data: demand logging, leakage tests, pressure-drop surveys and specific-power comparisons.
Our team of over 20 service engineers and technicians covers Telangana and Andhra Pradesh, and we supply compressors, rentals and spares across India. Our rental fleet covers 50–2,000 cfm at 2–30 bar (29–435.1 psig), so we can also size and supply temporary air for site work and shutdowns.
What to send us
- The question you want to answer: sizing, energy, leakage, storage or temporary air.
- Compressor nameplate photos and setpoints.
- Any logged pressure, flow or power data.
- A list of main users with their pressure and flow needs.
- Running hours and site conditions.
Frequently asked questions
How do I calculate the size of air compressor I need?
Add up the air demand of each user, multiplied by how often it is used, then add an agreed allowance for leakage and growth. Set the compressor pressure above the highest user requirement plus pipework, dryer and filter losses. For an existing plant, logging flow and pressure over a full shift is far more reliable than adding nameplate figures, which usually leads to oversizing.
How much more power does a compressor use at higher pressure?
As a typical rule, a compressor needs about 7% more power for each extra bar of discharge pressure at the same flow. Higher pressure also increases the air lost through leaks and unregulated users. Raising the setpoint to solve a low-pressure problem at one machine is therefore expensive; it is usually better to find and fix the restriction in the pipework.
How can I estimate compressed-air leakage in my plant?
With production stopped, record how long a load/unload compressor runs loaded compared with the total cycle time. Multiply the compressor capacity by that fraction to estimate leakage flow. For example, 300 cfm loaded 24 seconds out of every 80 suggests about 90 cfm of leakage. Confirm with an ultrasonic survey and retest after repairs.
What is specific power for an air compressor?
Specific power is the electrical input power divided by the air delivered, usually in kW per m³/min or kW per 100 cfm. A lower figure means a more efficient compressor. Comparisons are only fair at the same discharge pressure and with the same package items included. Measured input power is better than motor nameplate power for real comparisons.
Are these worked examples from real customer projects?
No. Every scenario on this page uses invented inputs to show a method. They are not customer projects, measured results or promised savings. Our real projects are described separately on our evidence pages, using only facts from our records. Before any decision, the inputs in these scenarios should be replaced with measured values from your own system.
Why do I need absolute pressure in compressed-air calculations?
Gas laws relate volume and pressure using absolute pressure, which is gauge pressure plus atmospheric pressure, about 1.013 bar at sea level. Using gauge pressure in storage, expansion or compression-ratio calculations gives wrong answers, sometimes by a large margin at low pressures. Compressor ratings and line pressures are usually stated as gauge, so convert before calculating.
Related pages
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.
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.