Chapter 4, Problem 84P

(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 3.5 MPa.

The temperature of super heated water vapor $\left(T\right)$ is $450°\text{C}$.

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.

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

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 $0.09198{\text{m}}^{\text{3}}/\text{kg}$.

Calculate the percentage of error involved.

  $\begin{array}{c}\text{Error}=\frac{v_{\text{calculated}}-v_{@3.5\text{MPa},450°\text{C}}}{v_{@3.5\text{MPa},450°\text{C}}}\times 100\%\\ \text{Error}=\frac{0.09533{\text{m}}^{\text{3}}/\text{kg}-0.09198{\text{m}}^{\text{3}}/\text{kg}}{0.09198{\text{m}}^{\text{3}}/\text{kg}}\times 100\%\\ =3.7\%\end{array}$

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

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

(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.

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

Calculate the reduced temperature.

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

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

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

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

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 $\underset{\_}{0.09161{\text{m}}^{\text{3}}/\text{kg}}$.

Calculate the percentage of error involved.

  $\begin{array}{c}\text{Error}=\frac{v_{\text{calculated}}-v_{@3.5\text{MPa},450°\text{C}}}{v_{@3.5\text{MPa},450°\text{C}}}\times 100\%\\ \text{Error}=\frac{0.09161{\text{m}}^{\text{3}}/\text{kg}-0.09198{\text{m}}^{\text{3}}/\text{kg}}{0.09198{\text{m}}^{\text{3}}/\text{kg}}\times 100\%\\ =-0.4\%\end{array}$

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

(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 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 $\underset{\_}{0.09196{\text{m}}^{\text{3}}/\text{kg}}$.