Chapter 3.8, Problem 84P

Determine the specific volume of superheated water vapor at 3.5 MPa and 450°C based on (a) the ideal-gas equation, (b) the generalized compressibility chart, and (c) the steam tables. Determine the error involved in the first two cases.

To determine

The specific volume of superheated water vapour based on the ideal gas equation.

The error involved.

Answer to Problem 84P

The specific volume of superheated water vapour based on the ideal gas equation is $\underset{\_}{0.09533{\text{m}}^{\text{3}}/\text{kg}}$.

The error involved is $\underset{\_}{\text{3}\text{.7\%}}$.

Explanation of Solution

Write the equation of specific volume of superheated water using ideal gas equation of state.

$v=\frac{RT}{P}$ (I)

Here, gas constant is R, pressure and temperature of R-134a are P and T respectively.

Calculate the percentage of error involved.

$\text{Error}=\frac{v_{\text{calculated}}-v_{@3.5\text{MPa},450°\text{C}}}{v_{@3.5\text{MPa},450°\text{C}}}\times 100\%$ (II)

Here, specific volume at pressure and temperature of 3.5 MPa and $450°\text{C}$ is $v_{@3.5\text{MPa},450°\text{C}}$.

Conclusion:

Refer to Table A-1, obtain the gas constant, R of water as $0.4615\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$.

Substitute $0.4615\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for R, $450°\text{C}$ for $T$, and 3.5 MPa for P in Equation (I).

$\begin{array}{c}v=\frac{\left(0.4615\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(450°\text{C}\right)}{3.5\text{MPa}}\\ =\frac{\left(0.4615\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(450+273\right)\text{K}}{3.5\text{MPa}\times {\text{10}}^3\frac{\text{kPa}}{\text{MPa}}}\\ =0.09533{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Thus, the specific volume of superheated water vapour based on the ideal gas equation is $\underset{\_}{0.09533{\text{m}}^{\text{3}}/\text{kg}}$.

Refer to Table A-6, obtain the value of $v_{@3.5\text{MPa},450°\text{C}}$ at pressure and temperature of 3.5 MPa and $450°\text{C}$ as $0.09198{\text{m}}^{\text{3}}/\text{kg}$.

Substitute $0.09533{\text{m}}^{\text{3}}/\text{kg}$ for $v{\text{calculated}}_{}$ and $0.09198{\text{m}}^{\text{3}}/\text{kg}$ for $v_{@3.5\text{MPa},450°\text{C}}$ in Equation (II).

$\begin{array}{c}\text{Error}=\frac{0.09533{\text{m}}^{\text{3}}/\text{kg}-0.09198{\text{m}}^{\text{3}}/\text{kg}}{0.09198{\text{m}}^{\text{3}}/\text{kg}}\times 100\%\\ =3.7\%\end{array}$

Thus, the error involved is $\underset{\_}{3.7\%}$.

To determine

The specific volume of superheated water vapour based on the generalized compressibility chart.

The error involved.

Answer to Problem 84P

The specific volume of superheated water vapour based on the generalized compressibility chart is $\underset{\_}{0.09161{\text{m}}^{\text{3}}/\text{kg}}$.

The error involved is $\underset{\_}{-0.4\%}$.

Explanation of Solution

Calculate the reduced pressure.

$P_R=\frac{P}{P_{cr}}$ (III)

Here, pressure of superheated water vapour is P and critical pressure is $P_{cr}$

Calculate the reduced temperature.

$T_R=\frac{T}{T_{cr}}$ (IV)

Here, temperature of superheated water vapor is T and critical temperature is $T_{cr}$

Calculate the specific volume of superheated water vapour based on the generalized compressibility chart.

$v=Z{v}_{\text{ideal}}$ (V)

Here, the specific volume at ideal condition is $v_{\text{ideal}}$.

Conclusion:

Refer to Table A-1, obtain the critical pressure and critical temperature of water.

$\begin{array}{l}{P}_{cr}=22.06\text{MPa}\\ T_{cr}=647.1\text{K}\end{array}$

Substitute 22.06 MPa for $P_{cr}$ and 3.5 MPa for P in Equation (III).

$\begin{array}{c}{P}_R=\frac{3.5\text{MPa}}{22.06\text{MPa}}\\ =0.159\end{array}$

Substitute 647.1 K for $T_{cr}$ and $450°\text{C}$ for T in Equation (IV).

$\begin{array}{c}{T}_R=\frac{450°\text{C}}{647.1\text{K}}\\ =\frac{\left(450+273\right)\text{K}}{647.1\text{K}}\\ =1.12\end{array}$

Refer to figure A-15, “The compressibility chart”, obtain the compressibility factor, Z by reading the calculated reduced pressure and reduced temperature as 0.961.

Substitute 0.961 for Z and $0.09533{\text{m}}^{\text{3}}/\text{kg}$ for $v_{\text{ideal}}$ in Equation (V).

$\begin{array}{c}v=\left(0.961\right)\left(0.09533{\text{m}}^{\text{3}}/\text{kg}\right)\\ =0.09161{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Thus, the specific volume of superheated water vapour based on the generalized compressibility chart is $\underset{\_}{0.09161{\text{m}}^{\text{3}}/\text{kg}}$.

Substitute $0.09161{\text{m}}^{\text{3}}/\text{kg}$ for $v{\text{calculated}}_{}$ and $0.09198{\text{m}}^{\text{3}}/\text{kg}$ for $v_{@3.5\text{MPa},450°\text{C}}$ in Equation (II).

$\begin{array}{c}\text{Error}=\frac{0.09161{\text{m}}^{\text{3}}/\text{kg}-0.09198{\text{m}}^{\text{3}}/\text{kg}}{0.09198{\text{m}}^{\text{3}}/\text{kg}}\times 100\%\\ =-0.4\%\end{array}$

Thus, the error involved is $\underset{\_}{-0.4\%}$.

To determine

The specific volume of superheated water vapour based on the data from tables.

Answer to Problem 84P

The specific volume of superheated water vapour based on the data from table is $\underset{\_}{0.09196{\text{m}}^{\text{3}}/\text{kg}}$.

Explanation of Solution

Refer to Table A-6, obtain the specific volume at pressure and temperature of 3.5 MPa and $450°\text{C}$ as $0.09196{\text{m}}^{\text{3}}/\text{kg}$.

Thus, the specific volume of superheated water vapour based on data from steam tables is $\underset{\_}{0.09196{\text{m}}^{\text{3}}/\text{kg}}$.