Chapter 3.8, Problem 109RP

Complete the blank cells in the following table of properties of steam. In the last column, describe the condition of steam as compressed liquid, saturated mixture, superheated vapor, or insufficient information, and, if applicable, give the quality.

To determine

The following table for ${\text{H}}_{\text{2}}\text{O}$ which are blank.

P, kPa	$T,°\text{C}$	$v,{\text{m}}^{\text{3}}/\text{kg}$	u, kJ/kg	x, quality	Phase description
200	30
270.3	130
	400	1.5493
300		0.500
500			3084

Explanation of Solution

State 1

Refer to Table A-5, “Saturated water-Pressure table”, obtain the value of saturated temperature at a pressure of 200 kPa as $120.21°\text{C}$.

The given temperature in state 1 is less than the saturated temperature at a pressure of 200 kPa.

$T_1<T_{sat@200\text{kPa}}$

Hence, state 1 is compressed liquid.

As wee see now there is no data for compressed liquid water in table A-7 for pressure 200 kPa, so calculate the specific internal energy and specific volume of a mixture at a saturated liquid at a temperature of $30°\text{C}$ fom table A-4.

$u_{@30°\text{C}}=\underset{\_}{125.73\text{kJ/kg}}$

$v_{@30°\text{C}}=\underset{\_}{0.001004{\text{m}}^{\text{3}}/\text{kg}}$

State 2

Refer to Table A-4, obtain the specific volume at saturated liquid and specific internal energy at saturated liquid at a temperature of $130°\text{C}$.

$v_f=v_{130°\text{C}}=\underset{\_}{0.001070{\text{m}}^{\text{3}}/\text{kg}}$

$u_f=u_{130°\text{C}}=\underset{\_}{546.10\text{kJ/kg}}$

Thus, the state 2 condition is saturated liquid.

State 3

Refer to Table A-6, “Superheater water”, obtain the pressure and specific internal energy at a temperature and specific volume of $400°\text{C}$ and $1.5493{\text{m}}^{\text{3}}/\text{kg}$.

$\begin{array}{l}{P}_{400°\text{C}}=0.20\text{MPa}=\underset{\_}{200\text{kPa}}\\ u_{@400°\text{C}}=\underset{\_}{2967.2\text{kJ/kg}}\end{array}$

The given specific internal energy is greater than the specific internal energy at saturated vapour at a pressure of 200 kPa refer from Table A-5.

$u_3>u_g$

Thus, state 3 is a superheated steam.

State 4

Refer to Table A-4, “Saturated water-Pressure table”, obtain the specific volume and specific internal energy  at saturated liquid $\left(v_f\right)$ and saturated vapour $\left(v_g\right)$ at a pressure of 300 kPa.

$\begin{array}{l}{v}_{f@3\text{00}\text{kPa}}=0.001073{\text{m}}^{\text{3}}/\text{kg}\\ v_{g@3\text{00}\text{kPa}}=0.60582{\text{m}}^{\text{3}}/\text{kg}\\ u_{f@3\text{00}\text{kPa}}=561.11\text{kJ}/\text{kg}\\ u_{g@3\text{00}\text{kPa}}=2543.2\text{kJ}/\text{kg}\end{array}$

As we see now the given specific volume of the mixture $\left(0.500{\text{m}}^{\text{3}}/\text{kg}\right)$ is greater than specific volume at saturated liquid and less than the specific volume at saturated vapour.

$v_f<v<v_g$

Hence, the state 4 is known as saturated mixture.

Refer to Table A-4, “Satuated water-pressure table”, obtain the temperature at a pressure of 200 kPa as $\underset{\_}{133.52°\text{C}}$.

Calculate the quality at state 1.

$x=\frac{v-v_f}{v_g-v_f}$ (I)

Substitute $0.001073{\text{m}}^{\text{3}}/\text{kg}$ for $v_f$, $0.500{\text{m}}^{\text{3}}/\text{kg}$ for $v$, and $0.60582{\text{m}}^{\text{3}}/\text{kg}$ for $v_g$  in Equation (I).

$\begin{array}{c}x=\frac{0.500{\text{m}}^{\text{3}}/\text{kg}-0.001073{\text{m}}^{\text{3}}/\text{kg}}{0.60582{\text{m}}^{\text{3}}/\text{kg}-0.001073{\text{m}}^{\text{3}}/\text{kg}}\\ =\underset{\_}{0.825}\end{array}$

Calculate the specific internal state.

$u=u_f+x\left(u_g-u_f\right)$ (II)

Here, specific internal energy at saturated liquid and saturated vapour is $u_f\text{and}{u}_g$ respectively.

Substitute $561.11\text{kJ}/\text{kg}$ for $u_f$, 0.825 for x, and $2543.2\text{kJ}/\text{kg}$ for $u_g$ in Equation (II).

$\begin{array}{c}u=561.11\text{kJ}/\text{kg}+0.825\left(2543.2\text{kJ}/\text{kg}-561.11\text{kJ}/\text{kg}\right)\\ =\underset{\_}{2196.663\text{kJ/kg}}\end{array}$

State 5

Since $u=3084\text{kJ/kg}$ which is greater than the specific internal energy at vapour phase at a pressure of 500 kPa from Table A-5 as 2560.7 kJ/kg.

$u_5>u_{g@500\text{kPa}}$

Thus, the state 5 is superheated steam.

Convert the unit of pressure from kPa to MPa.

$\begin{array}{c}P=500\text{kPa}\\ =500\text{kPa}\times \frac{1}{1000\text{kPa}}\text{MPa}\\ =0.50\text{MPa}\end{array}$

Refer to Table A-6, “Superheated water”, obtain the temperature and specific volume at a pressure of 0.50 MPa and specific intenal energy of 3084 kJ/kg as $\underset{\_}{473.1°\text{C}}$ and $\underset{\_}{0.6858{\text{m}}^{\text{3}}/\text{kg}}$ after interpolation method of two variables.

From the above calculations and referred from the steam table, complete the table of ${\text{H}}_{\text{2}}\text{O}$ as shown below in tabular form.

P, kPa	$T,°\text{C}$	$v,{\text{m}}^{\text{3}}/\text{kg}$	u, kJ/kg	x, quality	Phase description
200	30	$\underset{\_}{0.001004{\text{m}}^{\text{3}}/\text{kg}}$	$\underset{\_}{125.73\text{kJ/kg}}$	--	compressed liquid
270.3	130	$\underset{\_}{0.001070{\text{m}}^{\text{3}}/\text{kg}}$	$\underset{\_}{546.10\text{kJ/kg}}$	--	saturated liquid
$\underset{\_}{200\text{kPa}}$	400	1.5493	$\underset{\_}{2967.2\text{kJ/kg}}$		superheated steam
300	$\underset{\_}{133.52°\text{C}}$	0.500	$\underset{\_}{2196.663\text{kJ/kg}}$	$\underset{\_}{0.825}$	saturated mixture
500	$\underset{\_}{473.1°\text{C}}$	$\underset{\_}{0.6858{\text{m}}^{\text{3}}/\text{kg}}$	3084		superheated steam