Chapter 4, Problem 105RQ

To determine

The volume change using compressibility factor.

The error involved between the specific volume of actual value and specific volume using compressibility chart.

Explanation of Solution

Given:

The mass of steam $\left(m\right)$ is 0.2 kg.

The initial pressure of the steam $\left(P_1\right)$ is $200\text{ kPa}$.

The initial temperature of the steam $\left(T_1\right)$ is $300°\text{C}$.

The final temperature of the steam $\left(T_2\right)$ is $150°\text{C}$.

Calculation:

Refer to Table A-1, obtain the gas constant $\left(R\right)$, critical pressure $\left(P_{cr}\right)$, and critical temperature $\left(T_{cr}\right)$ of steam.

  $R=0.4615\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$

  $T_{cr}=647.1\text{K}$

  $P_{cr}=22.06\text{MPa}$

Refer to Table A-6, obtain the specific volume $\left(v_1\right)$ at inlet condition by reading the value of $P$ and $T_1$ of 200 kPa and $300°\text{C}$.

  $v_1=1.31623{\text{m}}^{\text{3}}/\text{kg}$

Refer to Table A-6, obtain the specific volume at outlet $\left(v_2\right)$ condition by reading the value of $P$ and $T_2$ of 200 kPa and $150°\text{C}$.

  $v_2=0.95986{\text{m}}^{\text{3}}/\text{kg}$

Calculate the change in the volume of the exat value.

  $\begin{array}{c}\Delta \nu =m\left(v_1-v_2\right)\\ \Delta \nu =0.2\text{kg}\left(1.31623{\text{m}}^{\text{3}}/\text{kg}-0.95986{\text{m}}^{\text{3}}/\text{kg}\right)\\ =0.07128{\text{m}}^{\text{3}}\end{array}$

Calculate the  reduced pressure at inlet condition.

  $\begin{array}{c}{P}_{R1}=\frac{P_1}{P_{cr}}\\ P_{R1}=\frac{0.2\text{MPa}}{22.06\text{MPa}}\\ =0.0091\end{array}$

Write the equation of reduced temperature at inlet condition.

  $T_{R1}=\frac{T_1}{T_{cr}}$

  $\begin{array}{c}{T}_{R1}=\frac{300°\text{C}}{647.1\text{K}}\\ =\frac{\left(300+273\right)\text{K}}{647.1\text{K}}\\ =0.886\end{array}$

Refer to figure A-15, “The compressibility chart”, obtain the compressibility factor, $Z_1$  by reading the calculated reduced pressure and reduced temperature at inlet state of 0.0091 and 0.886.

  $Z_1=0.9956$

Write the equation of reduced pressure at outlet condition.

  $P_{R2}=\frac{P_2}{P_{cr}}$

  $\begin{array}{c}{P}_{R2}=\frac{0.2\text{MPa}}{22.06\text{MPa}}\\ =0.0091\end{array}$

Write the equation of reduced temperature at outlet condition.

  $T_{R12}=\frac{T_2}{T_{cr}}$

  $\begin{array}{c}{T}_{R2}=\frac{150°\text{C}}{647.1\text{K}}\\ =\frac{\left(150+273\right)\text{K}}{647.1\text{K}}\\ =0.65\end{array}$

Refer to figure A-15, “The compressibility chart”, obtain the compressibility factor $\left(Z_2\right)$, by reading the calculated reduced pressure and reduced temperature at inlet state of 0.0091 and 0.65.

The compressibility factor $\left(Z_2\right)$ is $0.9897$.

Write the volume of piston cylinder device at inlet state.

  ${\nu }_1=\frac{Z_{1}mR{T}_1}{P_1}$

  $\begin{array}{c}{\nu }_1=\frac{\left(0.9956\right)\left(0.2\text{kg}\right)\left(0.4615\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)\left(300°\text{C}\right)}{200\text{kPa}}\\ =\frac{\left(0.9956\right)\left(0.2\text{kg}\right)\left(0.4615\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)\left(300+273\right)\text{K}}{200\text{kPa}}\\ =0.2633{\text{m}}^{\text{3}}\end{array}$

Write the volume of piston cylinder device at outlet state.

  ${\nu }_2=\frac{Z_{2}mR{T}_2}{P_2}$

  $\begin{array}{c}{\nu }_2=\frac{\left(0.9897\right)\left(0.2\text{kg}\right)\left(0.4615\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)\left(150°\text{C}\right)}{200\text{kPa}}\\ =\frac{\left(0.9897\right)\left(0.2\text{kg}\right)\left(0.4615\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)\left(150+273\right)\text{K}}{200\text{kPa}}\\ =0.1932{\text{m}}^{\text{3}}\end{array}$

Calculate the change in the volume using compressibility factor.

  $\Delta {\nu }_{\text{chart}}={\nu }_1-{\nu }_2$

  $\begin{array}{c}\Delta {\nu }_{\text{chart}}=0.2633{\text{m}}^{\text{3}}-0.1932{\text{m}}^{\text{3}}\\ =0.07006{\text{m}}^{\text{3}}\end{array}$

Thus, the volume change using compressibility factor is $\underset{\_}{\text{0}\text{.07006}{\text{m}}^{\text{3}}}$.

Calculate the percentage of error involved.

  $\text{Error}=\frac{v_{\text{exact}}-v_{\text{chart}}}{v_{\text{chart}}}\times 100\%$        (IX)

  $\begin{array}{c}\text{Error}=\frac{0.07128{\text{m}}^{\text{3}}-0.07006{\text{m}}^{\text{3}}}{0.07006{\text{m}}^{\text{3}}}\times 100\%\\ =1.7\%\end{array}$

Thus, the error involved between the volume change of actual value and volume change using compressibility chart is $\underset{\_}{1.7\%}$.