Question 3.301E: A piston/cylinder contains 2 lbm of water at 70 F with a vol...

A piston/cylinder contains 2 lbm of water at 70 F with a volume of 0.1 ft ^{3}, shown in Fig. P3.225. Initially the piston rests on some stops with the top surface open to the atmosphere, Po, so a pressure of 40 lbf/in .^{2} is required to lift it. To what temperature should the water be heated to lift the piston? If it is heated to saturated vapor find the final temperature, volume, and the heat transfer.

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C.V. Water. This is a control mass.

m _{2}= m _{1}= m ; \quad m \left( u _{2}- u _{1}\right)={ }_{1} Q _{2}-{ }_{1} W _{2}

State 1: 20 C,  v _{1}= V / m =0.1 / 2=0.05   ft ^{3} / lbm

 

x =(0.05-0.01605) / 867.579=0.0003913

 

u _{1}=38.09+0.0003913 \times 995.64=38.13   Btu / lbm

 

To find state 2 check on state 1a:

P = 40 psia,    v = v _{1}=0.05   ft ^{3} / lbm

Table F.7.1:      v _{ f }< v < v _{ g }=10.501, x _{1 a }>0

State 2 is saturated vapor at 40 psia as state 1a is two-phase.  T _{2}=267.3  F

 

v _{2}= vg =10.501   ft ^{3} / lbm , V _{2}= m v _{2}=21.0   ft ^{3}, u _{2}= ug =1092.27   Btu / lbm

 

Pressure is constant as volume increase beyond initial volume.

 

{ }_{1} W _{2}=\int P d V = P _{ lift }\left( V _{2}- V _{1}\right)=40(21.0-0.1) \times 144 / 778=154.75   Btu

 

{ }_{1} Q _{2}= m \left( u _{2}- u _{1}\right)+{ }_{1} W _{2}=2(1092.27-38.13)+154.75=2263   Btu
1
F.7.1
F.7.1'

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