## Related questions with answers

A cylinder having an initial volume of 100 ft3 contains 0.2 lbm of water at 100 F. The water is then compressed in an isothermal quasi-equilibrium process until it has a quality of 50%. Calculate the work done in the process, assuming that water vapor is an ideal gas.

Solution

VerifiedWe are given following data for Water:

$T_{1}=100\text{ F}$

$m=0.2\text{ lbm}$

$V_1=100\text{ ft}^3$

$x_{3}=0.5$

From Saturated Water table F.7.1. corresponding to $T=100\text{ F}$ we can obtain stage 2 specific volume and pressure:

$\nu_{2}=\nu_g=349.9764\frac{\text{ ft}^3}{\text{ lbm}}$

$P_2=P_g=0.95\frac{\text{ lbf}}{\text{ in}^2}$

Initial specific volume and pressure are equal to:

$\nu_1=\dfrac{V_1}{m}=\dfrac{100}{0.2}=500\frac{\text{ ft}^3}{\text{ lbm}}$

$P_1=\dfrac{P_2\cdot \nu_2}{\nu_1}=\dfrac{0.95\cdot 349.9764}{500}=0.665\frac{\text{ lbf}}{\text{ in}^2}$

Stage 2 volume is equal to:

$V_2=m\cdot \nu_2=0.2\cdot 349.9764=70\text{ ft}^3$

We can now calculate stage 1-2 work:

$\begin{align*} W_{1-2}&=\int PdV=P_1\cdot V_1\cdot \int \dfrac{dV}{V}=P_1\cdot V_1\cdot \ln \dfrac{V_2}{V_1}\\\\ &=\dfrac{0.665\cdot 144\cdot 100}{778}\cdot \ln \dfrac{70}{100}\\\\ &=-4.39\text{ Btu} \end{align*}$

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