Each of two vessels contains
of steam at a different temperature and pressure. Vessel A
is a rigid tank with a volume of 0.9 m3 and vessel B is a piston-and-cylinder device that holds 0.7 m3 of steam. |
They are connected by a pipe with a closed valve in the line. Initially, tank A contains saturated steam at 150 kPa while cylinder B
contains superheated steam at 350oC and 400
kPa. |
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When the valve is opened,
the steam in the two vessels is allowed to come to equilibrium.
a.) Determine the mass of steam in each vessel before the valve
is opened: mA1 and mB1.
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b.) If the equilibrium temperature is T2 = 240oC, calculate Q and W for the equilibration process. |
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Read : |
The key aspect of this
problem is whether ANY water remains in the cylinder, B, at equilibrium. If there is water
left in B at the final state, it will exist at T2 and P2 = PB1 because the piston would still be "floating". The other key is that this is a closed system, so the mass of water in the entire system remains constant. We can use the Steam Tables and the given initial volumes to answer part (a). In part (b), the work is done at constant pressure as the piston descends. So it is not difficult to compute. Finally, solve the 1st Law to determine Q.
This is possible because we know the initial and final
states and the work ! |
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Diagram: |
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Given: |
PA1 |
150 |
kPa |
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VB1 |
0.7 |
m3 |
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VA |
0.9 |
m3 |
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TB1 |
350 |
oC |
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xA1 |
1 |
kg vap/kg |
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PB1 |
400 |
kPa |
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T2 |
240 |
oC |
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Find: |
a.) |
mA1 |
??? |
kg |
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b.) |
Q |
??? |
kJ |
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mB1 |
??? |
kg |
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W |
??? |
kJ |
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Assumptions: |
1 - |
- The initial and final states are equilibrium states. |
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2 - |
- The process is a quasi-equilibrium process. |
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Equations
/ Data / Solve: |
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Part a.) |
We can determine mA1 because we know the volume of the tank and we can look up the specific
volume of the saturated
vapor that it contains. |
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Eqn 1 |
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NIST WebBook: |
VA1 |
1.1594 |
m3/kg |
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mA1 |
0.7762 |
kg |
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We can use the same
approach to determine mB1, but first we must determine its state. |
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At 400
kPa, Tsat = 133.52oC. Since TB1 > Tsat, tank
B initially contains superheated vapor. |
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NIST WebBook: |
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VB1 |
0.71396 |
m3/kg |
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mB1 |
0.9805 |
kg |
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Part b.) |
In part
b, there are two possibilities. At equilibrium, either B contains some water or it is completely empty. |
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Case
1 - B is not
empty: PB2 = PB1 because the piston is still
floating. |
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Case
2 - B is empty: VB2 = 0 and ALL
of the water is in tank A. |
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Let's test Case 1 first. T1 = Tsat at 150 kPa, so T1 = 111.3oC. Tsat at 400 kPa is 143.6oC. Since T2 > Tsat at 400 kPa, the water would still be superheated vapor and the specific volume would be: |
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V |
0.58314 |
m3/kg |
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Therefore the total volume occupied by this superheated vapor would be: |
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Eqn 2 |
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V2 |
1.0244 |
m3 |
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Since this volume, which the total mass of water
in the system occupies at PB1, is greater than the volume of tank
A, we can conclude
that all of the water could not fit into tank A at PB1. If P2 were less
than PB1, the water would occupy even more volume and again would not fit into tank
A. Some water must remain in the cylinder and, therefore, PB2 = PB1. |
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Therefore: |
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VB2 = V2 - VA1 = |
0.1244 |
m3 |
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Calculate the PV or boundary
work from: |
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Eqn 3 |
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But, since this
process is isobaric: |
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Eqn 4 |
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W |
-230.2 |
kJ |
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Finally, we need to
apply the 1st Law to
determine Q. Use all of the water in both vessels as the system. |
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Eqn 5 |
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Or: |
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Eqn 6 |
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Use the NIST WebBook and the ASHRAE Convention to determine all
of the specific internal energies. |
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mtot |
1.7567 |
kg |
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UA1 |
2519.2 |
kJ/kg |
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U2 |
2710.6 |
kJ/kg |
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UB1 |
2884.4 |
kJ/kg |
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DU |
-21.8 |
kJ |
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Q |
-252.0 |
kJ |
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Verify: |
None of the
assumptions can be verified from the data given in the problem statement. |
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Answers : |
a.) |
mA1 |
0.776 |
kg |
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b.) |
W |
-230 |
kJ |
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mB1 |
0.980 |
kg |
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Q |
-252 |
kJ |
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