# Example Problem with Complete Solution

7E-5 : Power Input for an Internally Reversible, Polytropic Compressor 6 pts
Consider the internally reversible ammonia compressor shown below. The compression process is polytropic with
d = 1.27. Determine Ws and Q in kW.

Read : The path equation given in the problem statement tells you that the compression process follows a polytropic path with d = 1.27.  All the properties of state 1 can be determined using the Ammonia Tables or the NIST Webbook.  The polytropic path equation allows us to determine the specific volume at state 2.  This is the 2nd known intensive property at state 2 , so we can evaluate the other properties using the Ammonia Tables or the NIST Webbook.  Determine the shaft work based on the polytropic path and then apply the 1st Law to evaluate Q.
Given: P1 140 kPa Diagram: See the problem statement.
V1,dot 25 L/s
x1 1 kg vap/kg
P2 750 kPa
d 1.27
Find: a.) (Ws)int rev ??? kW b.) Q ??? kW
Assumptions: 1 -
1 - The compressor operates at steady-state.
2 - Changes in kinetic and potential energies are negligible.
3 - The compression is internally reversible.
4 - The compression process follows a polytropic process path with d = 1.08.
5 - Shaft work and flow work are the only forms of work that cross the system boundary.
Equations / Data / Solve:
Part a.) Work for an internally reversible, polytropic process is given by : Eqn 1
We can determine the mass flow rate
from the volumetric flow rate using: Eqn 2
We can use the Ammonia Tables or the NIST Webbook to evaluate V1 because it is a saturated vapor at a known pressure of 140 kPa.
T1 -26.682 oC V1 0.83074 m3/kg
H1 1409.0 kJ/kg
Next, plug values back into Eqn 2 : mdot 0.0301 kg/s
Now, we need to determine V2.  We can make use of the fact that the compression process follows a polytropic process path with d = 1.08. Eqn 3
Solve Eqn 3 for V2 : Eqn 4
Now, we can plug numbers into Eqn 4 and then Eqn 1 to complete this part of the problem.
V2 0.22156 m3/kg (Ws)int rev -7.059 kW
Part b.) To determine the heat transfer rate for the compressor, we must apply the 1st Law for steady-state, SISO processes.  For this compressor, changes in kinetic and potential energies are negligible and only flow work and shaft work cross the system boundaries.  The appropriate form of the 1st Law for this compressor is : Eqn 5
We can solve Eqn 5 for the heat transfer rate:  Eqn 6
In part (a) we evaluated all of the unknowns on the right-hand side of Eqn 6 except H2.  So, now we need to evaluate H2.
For state 2, we know the values of two intensive properties: P2 and V2.  Therfore, we can use the Ammonia Tables or the NIST Webbook to evaluate any other properties of interest, in this case, H2.
We begin by determining the phases present. At P = 750 kPa : Vsat liq 0.0016228 m3/kg
Vsat vap 0.169798 m3/kg
Since V2 > Vsat vap, state 2 is a superheated vapor.
T (oC) V (m3/kg) H (kJ/kg)
75 0.21661 1611.9
T2 0.22156 H2 T2 81.85 oC
100 0.23469 1672.2 H2 1628.4 kJ/kg
Finally, we can plug values back into Eqn 5 to evaluate Q and complete this problem:
Q -0.4586 kW
Verify: The assumptions made in the solution of this problem cannot be verified with the given information.
Answers : a.) (Ws)int rev -7.06 kW
b.) Q -0.459 kW 