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Open resourceIt is designed for equipment comparison, audit analysis and part-load scenario evaluation - not for comparing unmatched catalogue numbers.
1. Purpose and Engineering Decision
Specific power expresses how much electrical power is required to deliver a stated quantity of compressed air. A lower value is better only when the compared machines use the same flow reference condition, the same pressure duty and the same power boundary.
Specific Power = Package Input Power / Delivered Reference Flow
The preferred metric for the site should be configurable, for example kW per m3/min FAD or kW per 100 cfm. The engine should preserve the original source units while displaying all compared machines on one selected common basis.
Comparison mode | Primary result |
|---|---|
Single operating point | Specific power and relative efficiency |
Two or more compressors | Ranked common-basis comparison |
Part-load curve | Specific power at each matched demand point |
Annual load profile | Annual kWh and cost difference using scenario hours |
Measured audit point | Measured package kW divided by normalized measured flow |
2. Comparison Gate - What Must Match
Item | Required control |
|---|---|
Flow basis | |
Discharge pressure | Compare at the same required operating pressure or use verified performance data at each pressure |
Power boundary | Use like-for-like package input power wherever possible |
Operating state | Full load must be compared with full load; part-load comparisons require matched demand points |
Accessories included | Dryer, fan, pump, heater or ancillary loads must be included or excluded consistently |
Ambient/site condition | Flag material differences if data are not corrected or measured on the same basis |
3. Specific Power Units
SP_m3min = kW / (m3/min)
SP_100cfm = kW / (cfm / 100)
Because 1 m3/min is approximately 35.3147 cfm, the two common metrics can be converted without changing the underlying comparison.
SP_100cfm = SP_m3min x (100 / 35.3147)
SP_m3min = SP_100cfm x (35.3147 / 100)
4. Power Boundary - Package Input vs Motor or Shaft Power
A compressor can appear artificially efficient when the numerator excludes losses that another data source includes. The calculator should therefore tag every power value with its boundary.
Power label | |
|---|---|
Total package electrical input | Preferred comparison basis |
Motor electrical input | Acceptable only if ancillary package loads are separately reconciled |
Motor shaft output | Do not compare directly with package input power |
Nameplate motor kW | Not a substitute for measured or tested operating input power |
Measured plant kW | Strong audit input when synchronized with measured flow and pressure |
5. Pressure Normalization Rule
If two machines are quoted at different discharge pressures, use manufacturer/test performance at the same pressure or measured common-duty data. Do not 'correct' power or flow with a hidden generic factor.
6. Input Model
Input group | Field | Validation / note |
|---|---|---|
Machine | Name / ID | Required for comparison |
Power | Input kW | Must be > 0 |
Power | Boundary | Package / motor input / shaft / measured |
Flow | Delivered flow | Must be > 0 |
Flow | Reference basis | |
Pressure | Operating discharge pressure | |
State | Load point / demand | Full load or matched part-load point |
Annual profile | Hours at point | Optional; >= 0 |
Economics | Electricity tariff | Optional for annual cost |
Metadata | Measured / tested / estimated | Confidence classification |
7. Reference Test Case A - Full-Load Comparison
Compressor A uses 110 kW package input and delivers 20.0 m3/min FAD. Compressor B uses 100 kW package input and delivers 17.5 m3/min FAD. Both are compared at the same discharge pressure and reference-flow basis.
Machine | Package kW | Flow | Specific power |
|---|---|---|---|
A | 110 | 20.0 m3/min | 5.500 kW/(m3/min) |
B | 100 | 17.5 m3/min | 5.714 kW/(m3/min) |
At this matched operating point, Compressor A uses about 3.8% less power per unit flow than Compressor B. This conclusion applies only to the stated operating point and does not predict part-load efficiency.
8. Reference Test Case B - Power Boundary Mismatch
A data sheet lists 90 kW motor shaft power while plant measurement shows 98 kW package electrical input at the same delivered flow of 16 m3/min. Using shaft power would understate the electrical specific power.
Basis | Power | Specific power |
|---|---|---|
Shaft output | 90 kW | 5.625 kW/(m3/min) |
Package electrical input | 98 kW | 6.125 kW/(m3/min) |
The shaft-power calculation understates electrical specific power by approximately 8.2%.
9. Part-Load Specific Power
Full-load specific power is not enough for plants with variable demand. For each operating point, the calculator should divide the power actually consumed at that point by the flow actually delivered at the same point.
Demand point | Compressor A | Compressor B |
|---|---|---|
18 m3/min | 102 kW -> 5.67 kW/(m3/min) | 100 kW -> 5.56 kW/(m3/min) |
12 m3/min | 78 kW -> 6.50 kW/(m3/min) | 75 kW -> 6.25 kW/(m3/min) |
6 m3/min | 62 kW -> 10.33 kW/(m3/min) | 50 kW -> 8.33 kW/(m3/min) |
The widening part-load difference in this example shows why a compressor with similar full-load efficiency can have materially different annual energy performance.
10. Reference Test Case C - Annual Load Profile
Use the three matched demand points above for 2,000 h/y at 18 m3/min, 3,000 h/y at 12 m3/min and 1,500 h/y at 6 m3/min.
Annual Energy = Sum(Power at Point x Hours at Point)
Machine | Annual energy |
|---|---|
A | 531,000 kWh/y |
B | 500,000 kWh/y |
Difference | 31,000 kWh/y lower for B |
If an electricity tariff is entered, the calculator multiplies the annual kWh difference by the tariff. Demand charges, taxes and time-of-use tariffs should be modelled separately unless the implementation explicitly supports them.
11. Weighted Fleet / Multi-Machine Boundary
When several machines run simultaneously, total specific power is calculated from total package kW divided by total normalized delivered flow.
Fleet SP = Sum(Package kW_i) / Sum(Flow_i)
12. Data Confidence Model
Input source | Suggested confidence |
|---|---|
Synchronized measured package kW + measured flow + pressure | High |
Verified performance point with common basis | High |
Vendor curve with clearly matched pressure/reference conditions | Medium-High |
Mixed source data with reconciled boundaries | Medium |
Nameplate kW + nominal capacity | Preliminary |
Unmatched pressure or power boundary | Comparison incomplete |
13. Validation and Guardrails
Condition | Required behaviour |
|---|---|
Power <= 0 | Block calculation |
Flow <= 0 | Block calculation |
Flow bases differ | |
Operating pressures differ materially | Flag; do not rank until common-duty data are provided |
Power boundaries differ | Flag; do not issue clean ranking |
Part-load points not demand-matched | Do not compare as equivalent operating points |
Hours < 0 | Block annual profile |
Tariff < 0 | Block cost calculation |
Measured power and flow timestamps not aligned | Lower confidence / warning |
14. Output Specification
Output | Display rule |
|---|---|
Specific power | Primary result in selected metric |
Common-basis confirmation | Flow reference, pressure and power boundary visible |
Rank | Only when comparison gate passes |
Difference | Absolute and percentage specific-power difference |
Part-load curve table | SP at each matched demand point |
Annual energy | kWh/y from entered load profile |
Annual cost difference | Only when tariff is supplied |
Confidence | High / medium / preliminary / incomplete |
Warnings | Pressure mismatch, power-boundary mismatch, estimated data, unmatched load point |
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. Prices, tariffs and efficiencies shown are example values: replace them with your own. Results are not a performance guarantee or a design certificate; final selection is confirmed by an engineer against the machine data sheet and the site.
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
- Electricity tariff (₹/kWh) — enter your own: 8
- 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
- Cost of delivering the air needed: A: ₹ 34,43,478 per year
- Cost of delivering the air needed: B: ₹ 33,52,381 per year
- Difference: ₹ 91,097 per year in favour of B
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.