Assuming, I've got a machine, that runs with pressurized air and provides the following measurements:

  • $\dot m$ - mass-flow through the machine (same for input and output)
  • $p_{1}$ - input pressure
  • $T_1$ - input temperature
  • $p_{2}$ - output pressure
  • $T_2$ - output temperature
  • $P_{mech}$ - mechanical output power

How can I calculate the power $P_{fluid}$ the air gives to the machine? In a first step, the heat-flow through the housing of the machine can be neglected, yet its influence to the process is interesting to me.

  • $\begingroup$ Mass flow, input pressure and temperature tell you the power the air gives to the machine; the others just tell you what the machine does with it. $\endgroup$ Commented Oct 18, 2016 at 11:28
  • $\begingroup$ @Brian how can the fluid power be calculated? $\endgroup$
    – DPF
    Commented Oct 18, 2016 at 12:16

1 Answer 1


Well here goes.

  1. Given that there is no heat transfer to the surroundings, indicated by the neglection of heat to the machine, this means that we can assume the system to be adiabatic - a fancy term for no heat exchange. This is important because it makes the equations that we can use much simpler.
  2. We want to calculate power, which is work over time. $$P = \frac{W}{t}$$
  3. Work done in a thermodynamics setting is the integral of pressure multiplied by the change in volume$$W = \int_a^b PdV$$
  4. Since our system is adiabatic, we can use the equation $$PV^{\gamma} = K (constant)$$ which is derived here.
  5. Where gamma is derived from the constant pressure and constant volume specific heats for the fluid which can be looked up $$\gamma = \frac{C_p}{C_v}$$
  6. This gives us $$W = K\int_{Vi}^{Vf}\frac{dV}{V^{\gamma}}$$

  7. Integrating gives $$W = \frac{K(V_f^{1-{\gamma}}-V_i^{1-{\gamma}})}{1-{\gamma}}$$

  8. Substituting K with Pressure and volume as before gives $$W = \frac{PV^{\gamma}*(V_f^{1-{\gamma}}-V_i^{1-{\gamma}})}{1-{\gamma}}$$

  9. Rearranging the equation gives $$W = \frac{PV*V^{\gamma-1}*(V_f^{1-{\gamma}}-V_i^{1-{\gamma}})}{1-{\gamma}}$$

  10. Since volume equals mass times specific volume, which can be obtained from the fluid tables based on your temperature and pressure, $$V = m\nu $$ we can insert this into the previous equation

  11. Which gives $$W = \frac{PV^{\gamma}*(m\nu_f^{1-{\gamma}}-m\nu_i^{1-{\gamma}})}{1-{\gamma}}$$ The values of $\nu_i$ and $\nu_f$ can be found from the property tables of your fluid in question (air) based on your initial and final conditions for pressure and temperature

  12. All of a sudden, we can replace m with $\dot{m}$ which is mass / time and after pulling it out of the brackets we get $$\frac{\dot{m}PV^{\gamma}*(\nu_f^{1-{\gamma}}-\nu_i^{1-{\gamma}})}{1-{\gamma}}$$

  13. Which is the same as $$\frac{W}{t} = P$$

And there is your answer. Much of this comes from the help of hyperphysics, as I'm a bit rusty from my thermodynamics courses

Hope this helps.


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