Calculator · Thermal Engineering
Compressor Power Calculator
Power a compressor needs, from the same steady-flow energy balance as every other device. The required power input is reported positive; internally the shaft work out of the control volume is negative. Heat transfer is stated, never assumed.
Inputs
q − ws = Δh + ΔV²/2 + gΔz (rate form × ṁ). q, Q̇ > 0 is heat into the control volume; ws, Ẇs > 0 is shaft work done by the fluid on the surroundings.
Results
Power input required −Ẇs
Specific work input −ws = 200 kJ/kg. Positive = power the compressor needs.
| Term | Per mass | Rate |
|---|---|---|
| Heat in q / Q̇ | 0 J/kg | 0 W |
| Shaft work out ws / Ẇs | -200 kJ/kg | -200 kW |
| Δh contribution | 200 kJ/kg | 200 kW |
| ΔKE contribution ΔV²/2 | 0 J/kg | 0 W |
| ΔPE contribution gΔz | 0 J/kg | 0 W |
| Residual q − ws − (Δh + ΔKE + ΔPE) | 0 J/kg | 0 W |
| Property | Inlet 1 | Outlet 2 |
|---|---|---|
| Enthalpy h | 300 kJ/kg | 500 kJ/kg |
| Speed V | 0 m/s | 0 m/s |
| Elevation z | 0 m | 0 m |
| Mass flow ṁ (steady, in = out) | 1 kg/s |
|---|
Assumptions
- · steady flow
- · single inlet and outlet
- · q > 0 into the fluid
- · w_s > 0 done by the fluid
- · adiabatic: Q̇ = 0 stated explicitly
Explore in the Lab
The full control-volume state travels; the Lab adds the energy-flow schematic.
Technical reference
−Ẇs = ṁ[(h2 − h1) + (V2² − V1²)/2 + g(z2 − z1)] − Q̇
−ws = −Ẇs / ṁ
- h
- specific enthalpy[J/kg]
- V
- stream speed[m/s]
- z
- elevation[m]
- ṁ
- mass flow rate[kg/s]
- Q̇
- heat rate into the control volume[W]
- Ẇs
- shaft power out of the control volume[W]
q − ws = Δh + ΔV²/2 + gΔz (rate form × ṁ). q, Q̇ > 0 is heat into the control volume; ws, Ẇs > 0 is shaft work done by the fluid on the surroundings.
A negative power input means the device produces work and is flagged.
A positive mass flow is required.