Water in a piston/cylinder is at 150 lbf/in.^2, 900 F, as shown in Fig. P6.177. There are two stops, a lower one at which Vmin = 35 ft^3 and an upper one at Vmax = 105 ft^3. The piston is loaded with a mass and outside atmosphere such that it floats when the pressure is 75 lbf/in.^2. This setup is

Question

Water in a piston/cylinder is at 150 [latex]lbf / in .^{2}[/latex], 900 F, as shown in Fig. P6.177. There are two stops, a lower one at which [latex]V_{\min }=35 ft ^{3}[/latex] and an upper one at [latex]V_{\max }=105 ft ^{3}[/latex]. The piston is loaded with a mass and outside atmosphere such that it floats when the pressure is 75 [latex]lbf / in .^{2}[/latex]. This setup is now cooled to 210 F by rejecting heat to the surroundings at 70 F. Find the total entropy generated in the process.

Accepted Answer

C.V. Water

State 1: Table F.7.2 [latex]v _{1}=5.3529 ft ^{3} / lbm , \quad u _{1}=1330.2 btu / lbm[/latex],

[latex]s _{1}=1.8381 Btu / lbm[/latex]

[latex]m = V / v _{1}=105 / 5.353=19.615 lbm[/latex]

State 2: 210 F and on line in P-v diagram.

Notice the following: [latex]v _{ g }\left( P _{ ext {float }} ight)=5.818 ft ^{3} / lbm , \quad v _{ ext {bot }}= V _{\min } / m =1.7843[/latex]

[latex]T _{ ext {sat }}\left( P _{ ext {float }} ight)=307.6 F , \quad T _{2}< T _{ ext {sat }}\left( P _{ ext {float }} ight) \Rightarrow V _{2}= V _{\min }[/latex]

State 2: 210 F, [latex]v _{2}= v _{ bot } \Rightarrow x _{2}=(1.7843-0.0167) / 27.796=0.06359[/latex]

[latex]\begin{array}{l}u _{2}=178.1+0.06359 imes 898.9=235.26 btu / lbm , \s _{2}=0.3091+0.06359 imes 1.4507=0.4014 btu / lbm R\end{array}[/latex]

Now we can do the work and then the heat transfer from the energy equation

[latex]{ }_{1} W _{2}=\int PdV = P _{ ext {float }}\left( V _{2}- V _{1} ight)=75(35-105) \frac{144}{778}=-971.72 Btu[/latex]

[latex]{ }_{1} Q _{2}= m \left( u _{2}- u _{1} ight)+{ }_{1} W _{2}=19.615(235.26-1330.2)-971.72=-22449 Btu[/latex]

Take C.V. total out to where we have 70 F:

[latex]m \left( s _{2}- s _{1} ight)={ }_{1} Q _{2} / T _{0}+ S _{ gen } \Rightarrow[/latex]

[latex]\begin{aligned}S _{ gen } &= m \left( s _{2}- s _{1} ight)-{ }_{1} Q _{2} / T _{0}=19.615(0.4014-1.8381)+\frac{22449}{529.67} \&= 1 4 . 2 0 B t u / R \quad\left(=\Delta S _{ ext {water }}+\Delta S _{ ext {sur }} ight)\end{aligned}[/latex]

Question 6.243E: Water in a piston/cylinder is at 150 lbf/in.^2, 900 F, as sh...

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