Chapter 3.8, Problem 110RP

Complete the blank cells in the following table of properties of refrigerant-134a. In the last column, describe the condition of refrigerant-134a as compressed liquid, saturated mixture, superheated vapor, or insufficient information, and, if applicable, give the quality.

To determine

The following table for refrigerant-134a which are blank.

P, kPa	$T,°\text{C}$	$v,{\text{m}}^{\text{3}}/\text{kg}$	u, kJ/kg	x, quality	Phase description
320	-12
1000	39.37
	40	1.17794
180		0.0700
200			249

Explanation of Solution

State 1

Refer to Table A-12, obtain the value of saturated temperature at a pressure of 320 kPa as $2.46°\text{C}$.

The given temperature in state 1 is less than the saturated temperature at a pressure of 320 kPa.

$T_1<T_{sat@320\text{kPa}}$

Hence, state 1 is compressed liquid.

As wee see now there is no data for compressed liquid water in table A-7 for pressure 320 kPa, so calculate the specific internal energy and specific volume of a mixture at a saturated refrigerant-134a at a temperature of $-12°\text{C}$ fom table A-11.

$u_{@-12°\text{C}}=\underset{\_}{35.76\text{kJ/kg}}$

$v_{@30°\text{C}}=\underset{\_}{0.0007498{\text{m}}^{\text{3}}/\text{kg}}$

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 $39.37°\text{C}$.

$v_f=v_{39.37°\text{C}\simeq 40°\text{C}}=\underset{\_}{0.0008720{\text{m}}^{\text{3}}/\text{kg}}$

$u_f=u_{39.37°\text{C}\simeq \text{40}°\text{C}}=\underset{\_}{107.39\text{kJ/kg}}$

Thus, the state 2 condition is saturated liquid.

State 3

Refer to Table A-13, “Superheater refrigerant-134a”, obtain the pressure and specific internal energy at a temperature and specific volume of $40°\text{C}$ and $0.0700{\text{m}}^{\text{3}}/\text{kg}$.

$\begin{array}{l}{P}_{40°\text{C}}=0.14\text{MPa}=\underset{\_}{140\text{kPa}}\\ u_{@40°\text{C}}=\underset{\_}{263.80\text{kJ/kg}}\end{array}$

The given specific internal energy is greater than the specific internal energy at saturated vapour at a pressure of 140 kPa refer from Table A-12.

$u_3>u_g$

Thus, state 3 is a superheated steam.

State 4

Refer to Table A-12, 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 180 kPa.

$\begin{array}{l}{v}_{f@18\text{0}\text{kPa}}=0.0007485{\text{m}}^{\text{3}}/\text{kg}\\ v_{g@18\text{0}\text{kPa}}=0.11049{\text{m}}^{\text{3}}/\text{kg}\\ u_{f@18\text{0}\text{kPa}}=34.81\text{kJ}/\text{kg}\\ u_{g@18\text{0}\text{kPa}}=223.01\text{kJ}/\text{kg}\end{array}$

As we see now the given specific volume of the mixture $\left(0.0700{\text{m}}^{\text{3}}/\text{kg}\right)$ 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-12, obtain the temperature at a pressure of 180 kPa as $\underset{\_}{-12.73°\text{C}}$.

Calculate the quality at state 1.

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

Substitute $0.0007485{\text{m}}^{\text{3}}/\text{kg}$ for $v_f$, $0.0700{\text{m}}^{\text{3}}/\text{kg}$ for $v$, and $0.11049{\text{m}}^{\text{3}}/\text{kg}$ for $v_g$  in Equation (I).

$\begin{array}{c}x=\frac{0.0700{\text{m}}^{\text{3}}/\text{kg}-0.0007485{\text{m}}^{\text{3}}/\text{kg}}{0.11049{\text{m}}^{\text{3}}/\text{kg}-0.0007485{\text{m}}^{\text{3}}/\text{kg}}\\ =\underset{\_}{0.631}\end{array}$

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 $34.81\text{kJ}/\text{kg}$ for $u_f$, 0.631 for x, and $223.01\text{kJ}/\text{kg}$ for $u_g$ in Equation (II).

$\begin{array}{c}u=34.81\text{kJ}/\text{kg}+0.631\left(223.01\text{kJ}/\text{kg}-34.81\text{kJ}/\text{kg}\right)\\ =\underset{\_}{153.56\text{kJ/kg}}\end{array}$

State 5

Since $u=249\text{kJ/kg}$ which is greater than the specific internal energy at vapour phase at a pressure of 200 kPa from Table A-12 as 224.51 kJ/kg.

$u_5>u_{g@200\text{kPa}}$

Thus, the state 5 is superheated steam.

Convert the unit of pressure from kPa to MPa.

$\begin{array}{c}P=200\text{kPa}\\ =200\text{kPa}\times \frac{1}{1000\text{kPa}}\text{MPa}\\ =0.20\text{MPa}\end{array}$

Refer to Table A-13, obtain the temperature and specific volume at a pressure of 0.20 MPa and specific intenal energy of 249 kJ/kg as $\underset{\_}{22.10°\text{C}}$ and $\underset{\_}{0.1197{\text{m}}^{\text{3}}/\text{kg}}$ after interpolation method of two variables.

From the above calculations and referred from the steam table, complete the table of $\text{R-134a}$ 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
320	-12	$\underset{\_}{0.0007498{\text{m}}^{\text{3}}/\text{kg}}$	$\underset{\_}{35.76\text{kJ/kg}}$	---	compressed liquid
1000	39.37	$\underset{\_}{0.0008720{\text{m}}^{\text{3}}/\text{kg}}$	$\underset{\_}{107.39\text{kJ/kg}}$	--	saturated liquid
$\underset{\_}{140\text{kPa}}$	40	1.17794	$263.80\text{kJ/kg}$	-	superheated steam
180	$\underset{\_}{-12.73°\text{C}}$	0.0700	$\underset{\_}{153.56\text{kJ/kg}}$	$\underset{\_}{0.631}$	saturated mixture
200	$\underset{\_}{22.10°\text{C}}$	$\underset{\_}{0.1197{\text{m}}^{\text{3}}/\text{kg}}$	249	--	superheated steam