What Is the Exit Temperature of Air from a Polytropic Compressor?

[Takuya925]

Homework Statement

If there's a polytropic process (n = 1.51) compressor that compresses air (Ideal Gas) as:
In: 100 kPa @ 25c
Out: 30 MPa

Given Data: Mass (m) = 105 kg, Work done by compressor (W) = 160000 kJ

a. What is the exit temperature of the air from the compressor?
b. If an air cooler is attached at the output side of the compressor, what is the heat-transfer in the air cooler?

2. The attempt at a solution

I used the energy balance and assumed it's steady state, steady flow, and ignored the potential/kinetic energy. Note Q equals heat loss/gain, and W equals the work done on system:
dE/dt = Q + W + m(h_in - h_out)

Since it's steady, state steady flow, dE/dt = 0. Also since it's polytropic which means adiabatic, Q = 0:
0 = 0 + W + m(h_in - h_out)

For Ideal Gas, change in enthalpy is influenced by temperature only so Cp(T_2 - T_1):
W = mCp(T_2 - T_1)

Sub in the known and solve. Note that Cp (specific heat for air) = 1.005, and temperature in kelvin:
160000 = 105 * 1.005 (T_2 - 298)
Part A --> T_2 = 1814.23 K

3. Question

I'm not sure how to do the part B. Based on my guess I think it's just 160000 kJ, same as the work given to the compressor. Also, I'm not sure if I did the calculation for part A properly. (Think temperature is way too high...)

 

[Andrew Mason]

2. The attempt at a solution

I used the energy balance and assumed it's steady state, steady flow, and ignored the potential/kinetic energy. Note Q equals heat loss/gain, and W equals the work done on system:
dE/dt = Q + W + m(h_in - h_out)

Since it's steady, state steady flow, dE/dt = 0. Also since it's polytropic which means adiabatic, Q = 0:
0 = 0 + W + m(h_in - h_out)

For Ideal Gas, change in enthalpy is influenced by temperature only so Cp(T_2 - T_1):
W = mCp(T_2 - T_1)

Sub in the known and solve. Note that Cp (specific heat for air) = 1.005, and temperature in kelvin:
160000 = 105 * 1.005 (T_2 - 298)
Part A --> T_2 = 1814.23 K

3. Question

I'm not sure how to do the part B. Based on my guess I think it's just 160000 kJ, same as the work given to the compressor. Also, I'm not sure if I did the calculation for part A properly. (Think temperature is way too high...)

Assuming the compression is adiabatic in part a. all you have to know is that the work done on the gas is equal to the change in internal energy of the gas (since there is no exchange of heat with surroundings). This means that:

W = \Delta U = nC_v\Delta T

where Cv is the molar specific heat at constant volume for air and n is the number of moles of air (ie. in 105 kg.)

However, the work done on the gas is not the work done by the compressor, since the compressor is not 100% efficient. You have to determine the amount of work done on the gas in compressing it from 100 KPa at 25 c to 30 MPa. I think you have to use the relationship:

PV^\gamma = K where \gamma = 1.51

AM

 

[Khamid]

Hey,

I am having some trouble understanding this question.
Is this a refridgeration cycle or a heat-pump?

Any help will be appreciated.

Q. Consider a heat pump operating with R-134a as the working fluid on an ideal cycle between the
condenser pressure of 160 psia and the evaporator pressure of 50 psia.
a) Determine the work per unit mass of working fluid (Btu/lbm) required to run the compressor.
b) Determine the power required to drive the compressor if the heat pump delivers 100,000
Btu/hr.
Consider now the same heat pump with 9.5°F subcooling of the refrigerant at the exit of the
condenser.