Example Problem with Complete Solution

8B-1 : Polytropic Compression of Air 6 pts
Air is compressed from 1 bar and 310 K to 8 bar.  Calculate the specific work and heat transfer if the air follows a polytropic process path with δ = 1.32.  Assume air is an ideal gas in this process.
               
Read: Figure 1a
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 1 bar Find:
T1 310 K WS ??? kJ/kg
d 1.32 Q ??? kJ/kg
P2 8 bar
Assumptions: - The compressor operates at steady-state.
- Kinetic and potential energy changes are negligible.
- Shaft work and flow work are the only forms of work that cross the system boundary.
- Air is modeled as an ideal gas.
Diagrams: Figure 2
Equation 1
Equations / Data / Solve :
We can determine the shaft work for a polytropic process on an ideal gas using: Equation 2
Eqn 1
We can determine T2 using the folling PVT relationship for polytropic processes: Equation 3
Eqn 2
Solve Eqn 2 for T2 : Equation 4 Eqn 3
We can now either evaluate T2 or use Eqn 3 to eliminate T2 from Eqn 1.
Equation 5
Eqn 4
Now, we can plug values into Eqns 3 & 4 to complete the first part of this problem.
T2 513.21 K MW 29.0 g/mol
R 8.314 J/mol-K WS -240.31 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 potentail energies are negligible and only flowwork and shaft work cross the system boundaries.  The appropriate form of the 1st Law for this compressor is :
Equation 6
Eqn 5
Solve Eqn 5 for Q : Equation 7 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 tables to evaluate H1 and H2.
T (K) Ho (kJ/kg)
510 302.17
513.21 H2 H1 97.40 kJ/kg
520 312.65 H2 305.53 kJ/kg
Now, we can plug values back into
Eqn 6 to complete this problem.		 Q -32.18 kJ/kg
Verify : Check the IG assumption: Equation 8
V1 = 25.77 L/mole V2 = 5.33 L/mole
Since air can be considered to be a diatomic gas and all the relevant molar volumes are greater than 5 L/mole, it is accurate to consider the gas ideal.
Answers: Part a.) WS -240.3 kJ/kg
Q -32.2 kJ/kg
The compressor loses heat to the surroundings.