Chapter 4, Problem 82P

(a)

To determine

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

The error involved.

Explanation of Solution

Given:

The pressure of super heated water vapor $\left(P\right)$ is $15\text{ MPa}\text{.}$

The temperatue of superheated water vapor $\left(T\right)$ is $350°\text{C}$.

Calculation:

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

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

  $\begin{array}{c}v=\frac{RT}{P}\\ v=\frac{\left(0.4615\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(350°\text{C}\right)}{15\text{MPa}}\\ =\frac{\left(0.4615\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(350+273\right)\text{K}}{15\text{MPa}\times {\text{10}}^3\frac{\text{kPa}}{\text{MPa}}}\\ =0.01917{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Thus, the specific volume of superheated water vapour based on the ideal gas equation is $\underset{\_}{0.01917{\text{m}}^{\text{3}}/\text{kg}}$.

Refer to Table A-6, obtain the value of $v_{@15\text{MPa},350°\text{C}}$ at pressure and temperature of 15 MPa and $350°\text{C}$ as $0.011481{\text{m}}^{\text{3}}/\text{kg}$.

Calculate the percentage of error involved.

  $\begin{array}{c}\text{Error}=\frac{v_{\text{calculated}}-v_{@15\text{MPa},350°\text{C}}}{v_{@15\text{MPa},350°\text{C}}}\times 100\%\\ \text{Error}=\frac{0.01917{\text{m}}^{\text{3}}/\text{kg}-0.011481{\text{m}}^{\text{3}}/\text{kg}}{0.011481{\text{m}}^{\text{3}}/\text{kg}}\times 100\%\\ =67.0\%\end{array}$

Thus, the error involved is $\underset{\_}{67.0\%}$.

(b)

To determine

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

The error involved.

Explanation of Solution

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

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

Calculate the reduced pressure.

  $\begin{array}{c}{P}_R=\frac{P}{P_{cr}}\\ P_R=\frac{15\text{MPa}}{22.06\text{MPa}}\\ =0.679\end{array}$

Calculate the reduced temperature.

  $\begin{array}{c}{T}_R=\frac{T}{T_{cr}}\\ T_R=\frac{350°\text{C}}{647.1\text{K}}\\ =\frac{\left(350+273\right)\text{K}}{647.1\text{K}}\\ =0.962\end{array}$

Refer to figure A-15, “The compressibility chart”, obtain the compressibility factor, Z by reading the calculated reduced pressure and reduced temperature as 0.65.

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

  $\begin{array}{c}v=Z{v}_{\text{ideal}}\\ v=\left(0.65\right)\left(0.01917{\text{m}}^{\text{3}}/\text{kg}\right)\\ =0.01246{\text{m}}^{\text{3}}/\text{kg}\end{array}$        (V)

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

Thus, the specific volume of superheated water vapour based on the generalized compressibility chart is $\underset{\_}{0.01246{\text{m}}^{\text{3}}/\text{kg}}$.

Substitute $0.01246{\text{m}}^{\text{3}}/\text{kg}$ for $v{\text{calculated}}_{}$ and $0.011481{\text{m}}^{\text{3}}/\text{kg}$ for $v_{@15\text{MPa},350°\text{C}}$ in Equation (II).

  $\begin{array}{c}\text{Error}=\frac{0.01246{\text{m}}^{\text{3}}/\text{kg}-0.011481{\text{m}}^{\text{3}}/\text{kg}}{0.011481{\text{m}}^{\text{3}}/\text{kg}}\times 100\%\\ =8.5\%\end{array}$

Thus, the error involved is $\underset{\_}{8.5\%}$.

(c)

To determine

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

Explanation of Solution

Refer to Table A-6, obtain the specific volume at pressure and temperature of 15 MPa and $350°\text{C}$ as $0.01148{\text{m}}^{\text{3}}/\text{kg}$.

Thus, the specific volume of superheated water vapour based on data from steam tables is $\underset{\_}{0.01148{\text{m}}^{\text{3}}/\text{kg}}$.