# Thermodynamics Power and Entropy

1. Feb 10, 2014

### ruiwp13

1. The problem statement, all variables and given/known data
A compressor processes 1.5kg/min of air in ambient conditions (1 bar and 20ºC). The compressed air leaves at 10bar and 90ºC. It is estimated that the heat losses trough the walls of the compressor are of 25kJ/min. Calculate:

a) The power of the compressor
b) The variation of specific entropy of the air crossing the compressor
c) Verify if this process is possible, if it obeys to the 2nd Law of Thermodynamics

2. Relevant equations
1)ΔS = Q1/T1 - Q2/T2
2)ΔSuniverse=ΔSair+ΔSneighbours
3)Δs=cp*ln(T2/T1)-R*ln(P2/P1)

3. The attempt at a solution

For a I am a little bit stuck. Can I assume that the power of the compressor is all the work done by it?

In b) I think I just need to use the 3) equation.

For c) I need to use the 2) equation being ΔSneighbours=Q/Tneigh ⇔ ΔSneighbours=25kJ/min/20 and ΔSair the variation of entropy calculated in b)

Is this correct? And if it is can you help me in a)?

Best Regards,

Much appreciated!

2. Feb 11, 2014

### Sunfire

The power that goes into heating could be calculated by

$P=c_p \dot{m}\Delta T + Q$, where Q is the lost heat through the walls

3. Feb 11, 2014

### Staff: Mentor

You are probably learning about the flow form of the first law applicable to an open system operating at steady state:
$$Δh=q-w_s$$
where, per unit mass passing through the system, Δh is the change in enthalpy, q is the heat added, and ws is the shaft work done by the system on the surroundings. This will give you the result that sunfire presented.

To do part (b), you should use your equation 3. (Your process may not be reversible, so you shouldn't be using dQ/T to get the Δs).

I have to think a little more about how to answer part (c).

Chet