Calculator · Thermal Engineering
Steady-Flow Energy Equation
One stream enters a control volume at 1 and leaves at 2. Choose the per-mass form (J/kg) or the rate form (W), pick the unknown, and every term of the balance is shown. The two forms are never mixed.
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
Shaft work ws
| Term | Per mass |
|---|---|
| Heat in q / Q̇ | -5 kJ/kg |
| Shaft work out ws / Ẇs | 93.498 kJ/kg |
| Δh contribution | -100 kJ/kg |
| ΔKE contribution ΔV²/2 | 1.6 kJ/kg |
| ΔPE contribution gΔz | -98.066 J/kg |
| Residual q − ws − (Δh + ΔKE + ΔPE) | 0 J/kg |
| Property | Inlet 1 | Outlet 2 |
|---|---|---|
| Enthalpy h | 400 kJ/kg | 300 kJ/kg |
| Speed V | 20 m/s | 60 m/s |
| Elevation z | 10 m | 0 m |
Assumptions
- · steady flow
- · single inlet and outlet
- · q > 0 into the fluid
- · w_s > 0 done by the fluid
Explore in the Lab
The full control-volume state travels; the Lab adds the energy-flow schematic.
Technical reference
q − ws = (h2 − h1) + (V2² − V1²)/2 + g(z2 − z1)
Q̇ − Ẇs = ṁ[(h2 − h1) + (V2² − V1²)/2 + g(z2 − z1)]
- 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.
Solving for V2 refuses a balance that would need a negative outlet kinetic energy.
With ṁ = 0 only Q̇ or Ẇs can be solved (Q̇ = Ẇs); the stream state is undetermined.