Example Problem with Complete Solution

8B-4 : Polytropic Compression of Air 6 pts
Calculate Q & WS, in kJ/kg, when ambient air at 104 kPa and 320 K is compressed polytropically to 950 kPa.
Assume d = 1.38 for this process path and that air behaves as an ideal gas.
 
Read : The key to this problem is the fact that the process is polytropic and that the air can be assumed to be an ideal gas.  Because the process is polytropic, we can determine T2 and WS.  Because the gas is ideal, we can use the Ideal Gas Property Tables to evaluate H1 and H2.  Finish by using WS, H1 and H2 to evaluate Q.
Given: P1 104 kPa Find: WS ??? kJ/kg
T1 320 K Q ??? kJ/kg
d 1.38
P2 950 kPa
Assumptions: 1 -
The compressor operates at steady-state.
2 - Kinetic and potential energy changes are negligible.
3 - Shaft work and flow work are the only forms of work that cross the system boundary.
4 - Air is modeled as an ideal gas.
Diagram:
Equations / Data / Solve:
We can determine the shaft work for a
polytropic process on an ideal gas using:
Eqn 1
We can determine T2 using the following
PVT relationship for polytropic processes:
Eqn 2
Solve Eqn 2 for T2 :
Eqn 3
We can now either evaluate T2 or use Eqn 3 to eliminate T2 from Eqn 1.
Eqn 4
Now, we can plug values into Eqns 3 & 4 to complete the first part of this problem.
MW 29.0 g/mol T2 588.42 K
R 8.314 J/mol-K WS -279.75 kJ/kg
In order to determine the specific heat transfer 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
Solve Eqn 5 for Q :
Eqn 6
Because we know both T1 and T2 and we assumed that air behaves as an ideal gas in this process, we can use the Ideal Gas Property Table for air to evaluate H1 and H2.
T (K) Ho (kJ/kg)
590 386.57
588.42 H2 H1 107.41 kJ/kg
600 397.21 H2 384.89 kJ/kg
Now, we can plug values back into
Eqn 6 to complete this problem.		 Q -2.272 kJ/kg
Verify: Check the Ideal Gas Assumption:
V1 25.58 L/mole V2 = 5.15 L/mole
Since air can be considered to be a diatomic gas and both molar volumes are greater than 5 L/mole, it is acceptable to consider the air an ideal gas.
Answers : Part a.) WS -280 kJ/kg
Q -2.27 kJ/kg   Q > 0, the compressor loses heat to the surroundings.