Consider the adiabatic air compressor
shown below. |
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a.) Calculate the minimum power input required and T2
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b.) The outlet temperature from a real, adiabatic compressor that accomplishes the same
compression is 520K. Calculate the actual power
input and the isentropic efficiency of the real compressor. |
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Read : |
Determine So(T2) for an isentropic process and then interpolate to obtain both T2S and H2S. Then, an energy balance will give you (WS)min. Use the isentropic efficiency and (WS)min to determine (WS)act. |
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Given: |
m |
9.5 |
kg/s |
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Find: |
a.) |
(WS)min |
??? |
kW |
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P1 |
110 |
kPa |
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T2S |
??? |
K |
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T1 |
310 |
K |
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b.) |
(WS)act |
??? |
kW |
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P2 |
550 |
kPa |
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hS, comp |
??? |
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T2, part (b) |
520 |
K |
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Diagram: |
The diagram in the
problem statement is adequate. |
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Assumptions: |
1 - |
The compressor operates at steady-state and there is no significant heat transfer. |
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2 - |
Kinetic and potential
energy changes are negligible. |
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3 - |
Air is modeled as an
ideal gas. |
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Equations
/ Data / Solve: |
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Part a.) |
An isentropic
compressor requires the minimum power input. |
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We can determine the isentropic work by applying the 1st Law to an isentropic compressor that takes in
the same feed and yields an effluent at the same pressure. |
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For a steady-state, single-inlet, single outlet system with negligible heat transfer, kinetic and potential energy changes, the 1st Law is: |
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Eqn 1 |
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The entropy
change for this process can determined using the 1st Gibbs Equation in terms of the Ideal Gas Entropy Function. |
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Eqn 2 |
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We can also apply Eqn 2 to our
hypothetical, isentropic compressor: |
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Eqn 3 |
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We can solve Eqn 3 for the unknown SoT2S : |
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Eqn 4 |
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We can evaluate SoT1 using the Ideal Gas Property Tables: |
So(T1) |
0.0061681 |
kJ/kg-K |
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We can get HoT1 while we are looking
in the Ideal Gas Property Tables because we will need it later when we evaluate Eqn 1. |
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H1 |
87.410 |
kJ/kg |
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Now, we can plug values into Eqn 4 : |
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R |
8.314 |
kJ/kmol-K |
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MW |
28.97 |
kg/kmol |
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So(T2S) |
0.46806 |
kJ/kg-K |
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Now, we can use SoT2S and the Ideal Gas Property Tables to
determine T2S and H2S by interpolation. |
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T (K) |
H (kJ/kg) |
So (kJ/kg-K) |
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470 |
260.49 |
0.46258 |
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T2S |
H2S |
0.46806 |
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T2S |
472.50 |
K |
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480 |
270.88 |
0.48445 |
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H2S |
263.09 |
kJ/kg |
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Now, we can plug
values back into Eqn 1 : |
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(WS)min |
-1669.0 |
kW |
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Part b.) |
We can determine the actual power input
for the compressor by
applying the 1st Law
to the real compressor, just as we did in Eqn 1 for the isentropic compressor. |
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Eqn 5 |
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We can evaluate HoT2 using the Ideal Gas Property Tables: |
T2, part (b) |
520 |
K |
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H2 |
312.65 |
kJ/kg |
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Now, we can evaluate WS,act using Eqn 5: |
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(WS)act |
-2139.78 |
kW |
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The isentropic efficiency of a compressor is defined by: |
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Eqn 6 |
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Since we determined
the isentropic work in part
(a) and the actual work in
part (b), we are ready
to plug numbers into Eqn 6 and wrap up this problem. |
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hS, comp |
78.00% |
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Verify: |
Check the Ideal Gas assumption: |
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V1
= |
23.43 |
L/mole |
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V2
= |
7.86 |
L/mole |
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Since air can be considered to be a diatomic gas and both molar volumes are greater than 5 L/mole, it is accurate to treat the air as an ideal
gas. |
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Answers : |
Part a.) |
(WS)min |
-1670 |
kW |
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Part b.) |
(WS)act |
-2140 |
kW |
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T2S |
473 |
K |
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hS, comp |
78.0% |
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