Question 3.305E: Water in a piston/cylinder, similar to Fig. P3.225, is at 21...

Water in a piston/cylinder, similar to Fig. P3.225, is at 212 F, x = 0.5 with mass 1 lbm and the piston rests on the stops. The equilibrium pressure that will float the piston is 40 psia. The water is heated to 500 F by an electrical heater. At what temperature would all the liquid be gone? Find the final (P,v), the work and heat transfer in the process.

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C.V. The 1 lbm water.

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

Process:   V = constant if  P < P _{\text {lift }}  otherwise  P = P _{ lift }  see P-v diagram.

State 1: (T,x) Table F.7.1

\begin{array}{l}v _{1}=0.01672+0.5 \times 26.7864=13.4099   ft ^{3} / lbm \\u _{1}=180.09+0.5 \times 897.51=628.845   Btu / lbm\end{array}

 

State 1a: (40 psia,  v = v _{1}> v _{ g  40  psia }=10.501   ft ^{3} / lbm)  so superheated vapor

 

Piston starts to move at state 1a,  { }_{1} W _{1 a }=0,

 

State 1b: reached before state 1a so v = v _{1}= v _{ g }  see this in F.7.1

 

T _{1 b }=250+10(13.40992-13.8247) /(11.7674-13.8247)= 2 5 2   F

 

State 2:  \left( T _{2}> T _{1 a }\right)    Table F.7.2 =>    v _{2}=14.164,    u _{2}=1180.06   Btu / lbm

 

Work is seen in the P-V diagram (when volume changes P = P _{\text {lift }} )

 

\begin{aligned}{ }_{1} W _{2} &={ }_{1 a } W _{2}= P _{2} m \left( v _{2}- v _{1}\right) \\&=40   psi \times 1   lbm (14.164-13.4099)   ft ^{3} / lbm \times 144   in ^{2} / ft ^{2} \\&=4343.5   lbf – ft = 5 . 5 8   Btu\end{aligned}

 

Heat transfer is from the energy equation

 

{ }_{1} Q _{2}=1   lbm (1180.06-628.845)   Btu / lbm +5.58   Btu = 5 5 6 . 8   Btu

 

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F.7.1
F.7.1'
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