# Example Problem with Complete Solution

8B-3 : Ideal Gas Compressor and Heat Exchanger Combination 8 pts

Assume air behaves as an ideal gas and heat losses to the surroundings are negligible.
a.) Calculate the power requirement for the compressor and the required cooling water mass flow rate for the heat exchanger.
b.) Calculate the entropy production rate for the compressor and for the heat exchanger separately.

Read : Use the ideal gas EOS and the volumetric flow rate to determine the mass flow rate.
Use an energy balance to determine the work for the compressor in kW.
To determine the water flow rate, draw the control volume enclosing the heat exchanger. This control volume has four mass flows entering or leaving but no Q or W. An energy balance on this control volume yields the water flow rate.
Very important point: the air and the water DO NOT MIX in the heat exchanger!
For the enthalpy of the cooling water, use H(T) for saturated liquid water from the Steam Tables.
Part (b) Don't forget about the cooling water when you calculate the entropy generated. Use S(T) for saturated liquid water from the Steam Tables.
Given: TA 15 oC P2 255 kPa
TB 35 oC T2 143 oC
P1 104 kPa P3 255 kPa
T1 30 oC T3 65 oC
mair 35 kg/min
Find: Part (a) WS ??? kW Part (b) Sgen, comp ??? kW/K
mcw ??? kg/s Sgen, HEX ??? kW/K
Diagram: Diagram already provided in the problem statement.
Assumptions: 1 - Both the compressor and heat exchanger operate at steady-state.
2 - Heat exchange between the equipment in this process and the surroundings is negliqible.
3 - There is no shaft work in the heat exchanger.
4 - Kinetic and potential energy changes are negligible.
5 - The air behaves as an ideal gas.
6 - The properties of the cooling water are the same as the properties of saturated liquid water at the same temperature.
Equations / Data / Solve:
Part a.) R 8.314 kJ/kmol K
MW 28.97 kg/kmol mair 0.5833 kg/s
We can determine the properties of air at all three states by interpolating on the Ideal Gas Property Tables for air because all three temperatures are given in the problem statement.
T1 303.15 K T (K) Ho (kJ/kg) So (kJ/kg-K)
T2 416.15 K 300 87.410 0.0061681
T3 338.15 K 310 97.396 0.038914 H1 90.556 kJ/kg
410 198.63 0.32178
420 208.88 0.34649 H2 204.93 kJ/kg
330 117.45 0.10159
340 127.51 0.13163 H3 125.65 kJ/kg
Now, we can plug values back into Eqn 1 to evaluate Ws: WS -66.72 kW
The mass flow rate of the cooling water can be determined by an energy balance on the heat exchanger.  For a steady-state process with negligible heat transfer, kinetic and potential energy changes and no shaft work:
Eqn 3
Eqn 3 can be simplified because no shaft work crosses the system boundary and when we use the entire HEX as the system, there is no heat transfer across the system boundary either.
Eqn 4
Now, we can solve Eqn 4 for mcw : Eqn 5
We do not know the pressure of the cooling water, so we cannot look up its properties.  Therefore, we assume that the enthalpy of the cooling water is the same as the enthalpy of saturated liquid water at the same temperature.  This assumption is accurate as long as water is nearly an incompressible liquid.  The properties of saturated liquid water were determined from NIST WebBook:
Saturated liquid water at TA: HA 62.981 kJ/kg
Saturated liquid water at TB: HB 146.63 kJ/kg
Now, we can plug values into Eqn 5 and evaluate mcw: mcw 0.5529 kg/s
Part b.) Entropy generation is defined by: Eqn 6
No heat transfer crosses the system boundary for either the compressor or the HEX, so Eqn 6 simplifes to:
Eqn 7
We can determine the entropy change for air (ideal gas) in the compressor from the 2nd Gibbs Equation:
Eqn 8
Therefore, the rate at which entropy is generated in the compressor is:
Eqn 9
We can determine the properties of air at all three states by interpolating on the Ideal Gas Property Tables for air because all three temperatures are given in the problem statement.
So(T1) 0.016483 kJ/kg-K
So(T2) 0.33698 kJ/kg-K So(T3) 0.12607 kJ/kg-K
Now, we can plug values back into Eqn 9 : Sgen, comp 0.03681 kW/K
The entropy generated in the HEX must take into account the entropy change of BOTH the air and the cooling water.
Eqn 10
Again, we assume that the entropy of the cooling water is the same as the entropy of saturated liquid water at the same temperature.  The properties of saturated liquid water were determined from NIST WebBook:
Saturated liquid water at TA: S(TA) 0.22446 kJ/kg-K
Saturated liquid water at TB: S(TB) 0.50513 kJ/kg-K
Now, we can plug value into Eqn 10 : Sgen, HEX 0.03215 kW/K
Verify: Check the Ideal Gas Assumption: V1 = 24.23 L/mole
V2 = 13.57 L/mole
V3 = 11.03 L/mole
Since air can be considered to be a diatomic gas and all three molar volumes are greater than 5 L/mole, it is accurate to treat the air as an ideal gas.
Answers : Part a.) WS -66.7 kW Part b.) Sgen, comp 0.0368 kW/K
mcw 0.553 kg/s Sgen, HEX 0.0321 kW/K