A piston–cylinder assembly contains 2 lb of water, initially at 100 lbf/in.2 and 400°F. The water undergoes two processes in series: a constant-pressure process followed by a constant volume process. At the end of the constant-volume process, the temperature is 300°F and the water is a two-phase liquid–vapor mixture with a quality of 40%. Neglect kinetic and potential energy effects.Determine the work and heat transfer for each process, all in Btu.

Question
Asked Sep 17, 2019

A piston–cylinder assembly contains 2 lb of water, initially at 100 lbf/in.2 and 400°F. The water undergoes two processes in series: a constant-pressure process followed by a constant volume process. At the end of the constant-volume process, the temperature is 300°F and the water is a two-phase liquid–vapor mixture with a quality of 40%. Neglect kinetic and potential energy effects.

Determine the work and heat transfer for each process, all in Btu.

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Expert Answer

Step 1

Given:

Mass, m 2lbm
State 1
Temperature, T 400K
Pressure R100psi
u 1136.4Btu/ lbm
1 4.9359 ft /lbm
V9.8718 f
P (psi)
2
1
Values are taken from the thermodynamic data table
100
state 2:
P R100 psi
67.028
state 3
Temperature T = 300°C
xz = 40%= 0.4
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Mass, m 2lbm State 1 Temperature, T 400K Pressure R100psi u 1136.4Btu/ lbm 1 4.9359 ft /lbm V9.8718 f P (psi) 2 1 Values are taken from the thermodynamic data table 100 state 2: P R100 psi 67.028 state 3 Temperature T = 300°C xz = 40%= 0.4

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Step 2

Specific volume and specific energy are calculated as the

0.01745 (10.4) +6.4663(0.4)
-2.59699 ft3 /lbm
and
269.51x (10.4) +1099.8 (0.4)
601.626 Btu / lbm
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0.01745 (10.4) +6.4663(0.4) -2.59699 ft3 /lbm and 269.51x (10.4) +1099.8 (0.4) 601.626 Btu / lbm

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Step 3

Vapour quality and specific ener...

2.59699 - 0.01774
4.4327 0.01774
=0.584
298.19(1 -0.584)+1105.5(0.584)
= 769.58Btu / lbm
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2.59699 - 0.01774 4.4327 0.01774 =0.584 298.19(1 -0.584)+1105.5(0.584) = 769.58Btu / lbm

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