Consider the internally reversible ammonia compressor shown below. The compression process is polytropic with
d = 1.27. |
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Determine Ws and Q in kW. |
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Read : |
The path equation given in the problem
statement tells you that the compression process follows a polytropic path with d
= 1.27. All
the properties of state 1
can be determined using the Ammonia Tables or the NIST Webbook. The polytropic path equation allows us
to determine the specific volume at state 2. This is the 2nd known intensive property at state 2
, so we can evaluate the other properties using the Ammonia Tables or the NIST Webbook. Determine the shaft
work based on the polytropic
path and then apply the 1st
Law to evaluate Q. |
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Given: |
P1 |
140 |
kPa |
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Diagram: |
See the problem
statement. |
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V1,dot |
25 |
L/s |
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x1 |
1 |
kg vap/kg |
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P2 |
750 |
kPa |
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d |
1.27 |
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Find: |
a.) |
(Ws)int rev |
??? |
kW |
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b.) |
Q |
??? |
kW |
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Assumptions: |
1 - |
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1 - |
The compressor operates at steady-state. |
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2 - |
Changes in kinetic and potential energies are negligible. |
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3 - |
The compression is internally reversible. |
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4 - |
The compression process follows a polytropic process path with d = 1.08. |
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5 - |
Shaft
work and flow work are the only forms of work that cross the system boundary. |
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Equations
/ Data / Solve: |
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Part a.) |
Work for an internally reversible, polytropic process is given by : |
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Eqn 1 |
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We can determine the mass flow rate
from the volumetric
flow rate using: |
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Eqn 2 |
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We can use the Ammonia Tables or the NIST Webbook to evaluate V1 because it is a saturated vapor at a known pressure of 140 kPa. |
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T1 |
-26.682 |
oC |
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V1 |
0.83074 |
m3/kg |
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H1 |
1409.0 |
kJ/kg |
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Next, plug values back
into Eqn 2 : |
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mdot |
0.0301 |
kg/s |
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Now, we need to
determine V2. We can make use of the
fact that the compression process follows a polytropic process path with d
= 1.08. |
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Eqn 3 |
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Solve Eqn 3 for V2 : |
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Eqn 4 |
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Now, we can plug
numbers into Eqn 4 and then Eqn 1 to complete this part of the
problem. |
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V2 |
0.22156 |
m3/kg |
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(Ws)int rev |
-7.059 |
kW |
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Part b.) |
To determine the heat transfer rate for the compressor, we must apply the 1st Law for steady-state, 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 : |
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Eqn 5 |
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We can solve Eqn 5 for the heat transfer rate: |
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Eqn 6 |
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In part
(a) we evaluated all of the unknowns on the right-hand side of
Eqn 6 except H2. So, now we need to
evaluate H2. |
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For state
2, we know the values of two intensive
properties: P2 and V2. Therfore, we can use
the Ammonia Tables or
the NIST Webbook to
evaluate any other properties of interest, in this case, H2. |
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We begin by determining
the phases present. |
At P = 750 kPa : |
Vsat liq |
0.0016228 |
m3/kg |
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Vsat vap |
0.169798 |
m3/kg |
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Since V2 > Vsat vap, state 2
is a superheated vapor. |
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T (oC) |
V (m3/kg) |
H (kJ/kg) |
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75 |
0.21661 |
1611.9 |
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T2 |
0.22156 |
H2 |
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T2 |
81.85 |
oC |
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100 |
0.23469 |
1672.2 |
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H2 |
1628.4 |
kJ/kg |
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Finally, we can plug
values back into Eqn 5 to
evaluate Q and complete
this problem: |
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Q |
-0.4586 |
kW |
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Verify: |
The assumptions made
in the solution of this problem cannot be verified with the given
information. |
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Answers : |
a.) |
(Ws)int rev |
-7.06 |
kW |
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b.) |
Q |
-0.459 |
kW |
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