Chapter 4, Problem 106RQ

(a)

To determine

The pressure of the steam using the ideal gas equation.

Explanation of Solution

Given:

The temperture of the steam is $\left(T\right)$ is $400°\text{C}$.

The specific volume of the steam $\left(v\right)$ is $0.02{\text{m}}^{\text{3}}/\text{kg}$.

Calculation:

Refer to Table A-1 to find the gas constant $\left(R\right)$, the critical pressure $\left(P_{cr}\right)$, and the critical temperature $\left(T_{cr}\right)$ of water as $0.4615\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$, $22.06\text{MPa}$, and $647.1\text{K}$.

Determine the pressure of the steam using the ideal gas equation.

  $P=\frac{RT}{v}$

  $\begin{array}{c}P=\frac{\left(0.4615\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(400°\text{C}\right)}{0.02{\text{m}}^{\text{3}}/\text{kg}}\\ =\frac{\left(0.4615\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(400°\text{C+273}\right)}{0.02{\text{m}}^{\text{3}}/\text{kg}}\\ =15,529\text{kPa}\end{array}$

Thus, the pressure of the steam using the ideal gas equation is $\underset{\_}{15,529\text{kPa}}$.

(b)

To determine

The pressure of the steam using the compressibility chart.

Explanation of Solution

Determine the reduced temperature of the steam.

  $\begin{array}{c}{T}_R=\frac{T}{T_{\text{cr}}}\\ T_R=\frac{400°\text{C}}{647.1\text{K}}\\ =\frac{400°\text{C+273}}{647.1\text{K}}\\ =1.040\end{array}$

Determine the reduced specific volume of the steam.

  $\begin{array}{c}{v}_R=\frac{v_{\text{actual}}}{R{T}_{\text{cr}}/P_{\text{cr}}}\\ v_R=\frac{\left(0.02{\text{m}}^{\text{3}}/\text{kg}\right)}{\left(0.4615\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(647.1\text{K}\right)/\left(22,060\text{kPa}\right)}\\ =\frac{\left(0.02{\text{m}}^{\text{3}}/\text{kg}\right)}{\left(0.013537{\text{m}}^{\text{3}}/\text{kg}\right)}\\ =1.477\\ \cong 1.48\end{array}$

Determine the pressure of the steam using the compressibility chart.

  $\begin{array}{c}P=P_R\times P_{\text{cr}}\\ P=0.57\times \left(22,060\text{kPa}\right)\\ =12,574\text{kPa}\end{array}$

Thus, the pressure of the steam using the compressibility chart is $\underset{\_}{12,574\text{kPa}}$.

(c)

To determine

The pressure of the steam using the steam table.

Explanation of Solution

Refer to Table A-6, “Superheated water”, obtain the below properties at the specific volume $0.02{\text{m}}^{\text{3}}/\text{kg}$ using interpolation method of two variables.

Show the pressure at $12.5\text{MPa}$ and $15\text{MPa}$ as in Table (1).

S. No	specific volume ${\text{m}}^{\text{3}}/\text{kg}$ $\left(x\right)$	Pressure, MPa $\left(y\right)$
1	0.020030	12.5
2	0.02	$y_2=?$
3	0.015671	15.0

Calculate pressure at the specific volume $0.02{\text{m}}^{\text{3}}/\text{kg}$ for liquid phase using interpolation method.

Calculate the pressure of the steam using interpolation method of two variables.

  $\begin{array}{c}{y}_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1\\ y_2=\frac{\left(0.015671{\text{m}}^{\text{3}}/\text{kg}-0.020030{\text{m}}^{\text{3}}/\text{kg}\right)\left(15.0\text{MPa}-12.5\text{MPa}\right)}{\left(0.015671{\text{m}}^{\text{3}}/\text{kg}-0.020030{\text{m}}^{\text{3}}/\text{kg}\right)}+12.5\text{MPa}\\ =12.517\text{MPa}\end{array}$

Here, the variables denote by x and y are temperature and final specific volume.

From above calculation the pressure of $0.02{\text{m}}^{\text{3}}/\text{kg}$ at specific volume is $12.517\text{MPa}$.

Thus, the pressure of the steam using the steam table is $\underset{\_}{12.517\text{MPa}}$.