Chapter 3.8, Problem 120RP

To determine

The volume change using compressibility factor.

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

Answer to Problem 120RP

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

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

Explanation of Solution

Refer to Table A-1, obtain the gas constant, critical pressure, and critical temperature 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 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 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.

$\Delta \nu =m\left(v_1-v_2\right)$ (I)

Here, mass of steam is m.

Write the equation of reduced pressure at inlet condition.

$P_{R1}=\frac{P_1}{P_{cr}}$ (II)

Here, critical pressure of steam is $P_{cr}$ and inlet pressure of steam is $P_1$.

Write the equation of reduced temperature at inlet condition.

$T_{R1}=\frac{T_1}{T_{cr}}$ (III)

Here, critical temperature of steam is $T_{cr}$ and inlet temperature of steam is $T_1$.

Write the equation of reduced pressure at outlet condition.

$P_{R2}=\frac{P_2}{P_{cr}}$ (IV)

Here, outlet pressure of steam is $P_2$.

Write the equation of reduced temperature at outlet condition.

$T_{R12}=\frac{T_2}{T_{cr}}$ (V)

Here, outlet temperature of steam is $T_2$.

Write the volume of piston cylinder device at inlet state.

${\nu }_1=\frac{Z_{1}mR{T}_1}{P_1}$ (VI)

Here, the compressibility factor at inlet state is $Z_1$ and gas constant is R.

Write the volume of piston cylinder device at outlet state.

${\nu }_2=\frac{Z_{2}mR{T}_2}{P_2}$ (VII)

Here, the compressibility factor at outlet state is $Z_2$.

Calculate the change in the volume using compressibility factor.

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

Calculate the percentage of error involved.

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

Conclusion:

Substitute 0.2 kg for m, $1.31623{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$, and $0.95986{\text{m}}^{\text{3}}/\text{kg}$ for $v_2$ in Equation (I).

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

Substitute 0.2 MPa for $P_1$ and $22.06\text{MPa}$ for $P_{cr}$ in Equation (II).

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

Substitute $300°\text{C}$ for $T_1$ and 647.1 K for $T_{cr}$ in Equation (III).

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

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.

$Z_1=0.9956$

Substitute 0.2 MPa for $P_2$ and $22.06\text{MPa}$ for $P_{cr}$ in Equation (IV).

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

Substitute $150°\text{C}$ for $T_2$ and 647.1 K for $T_{cr}$ in Equation (V).

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

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

$Z_2=0.9897$

Substitute 0.9956 for $Z_1$, 0.2 kg for $m$, $0.4615\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ for R, $300°\text{C}$ for $T_1$, and 200 kPa for $P_1$ in Equation (VI).

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

Substitute 0.9897 for $Z_2$, 0.2 kg for $m$, $0.4615\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ for R, $150°\text{C}$ for $T_2$, and 200 kPa for $P_2$ in Equation (VII).

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

Substitute $0.2633{\text{m}}^{\text{3}}$ for ${\nu }_1$ and $0.1932{\text{m}}^{\text{3}}$ for ${\nu }_2$ in Equation (VIII).

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

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

Substitute $0.07006{\text{m}}^{\text{3}}$ for $\Delta {\nu }_{\text{chart}}$ and $0.07128{\text{m}}^{\text{3}}$ for $\Delta {\nu }_{\text{exact}}$ in Equation (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 $\underset{\_}{1.7\%}$.