8B4 :  Polytropic Compression of Air  6 pts 

Calculate Q & W_{S}, 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 T_{2} and W_{S}. Because the gas is ideal, we can use the Ideal Gas Property Tables to evaluate H_{1} and H_{2}. Finish by using W_{S}, H_{1} and H_{2} to evaluate Q.  
Given:  P_{1}  104  kPa  Find:  W_{S}  ???  kJ/kg  
T_{1}  320  K  Q  ???  kJ/kg  
d  1.38  
P_{2}  950  kPa  
Assumptions:  1  


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 T_{2} using the following PVT relationship for polytropic processes: 

Eqn 2  
Solve Eqn 2 for T_{2} : 

Eqn 3  
We can now either evaluate T_{2} or use Eqn 3 to eliminate T_{2} 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  T_{2}  588.42  K  
R  8.314  J/molK  W_{S}  279.75  kJ/kg  
In order to determine the specific heat transfer for the compressor, we must apply the 1st Law for steadystate, 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 T_{1} and T_{2} 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 H_{1} and H_{2}.  
T (K)  H^{o} (kJ/kg)  
590  386.57  
588.42  H_{2}  H_{1}  107.41  kJ/kg  
600  397.21  H_{2}  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: 


V_{1}  25.58  L/mole  V_{2} =  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.)  W_{S}  280  kJ/kg  
Q  2.27  kJ/kg  Q > 0, the compressor loses heat to the surroundings. 