Q.4.29 - An Introduction to Thermal Physics

Question

Liquid HFC-134a at its boiling point at 12 bars pressure is throttled to 1 bar pressure. What is the final temperature? What fraction of the liquid vaporizes?

Table 4.3. Properties of the refrigerant HFC-134a under saturated conditions (at its boiling point for each pressure). All values are for $1 \mathrm{kg}$ of fluid, and are measured relative to an arbitrarily chosen reference state, the saturated liquid at $-40°$c. Excerpted from Moran and Shapiro (1995).

Step-by-Step Solution

Answer

• $\text{ The liquid vaporization is }0.465\text{. }$
• The final temperature is $-26.4°\mathrm{C}$

Step 1: To find

The final temperature and the fraction of liquid that vaporizes

Step 2: Explanation

Given:

Initial pressure, P=12 bars

Final pressure, P=1 bar

Formula: $H_f=a\left(H_{\text{liquid }}\right)+(1-a)\left(H_{gas}\right)$

$\text{ Where, }a=\text{ is the fraction of the HFC-134a, which ends up as liquid. }$

Calculation: 

Using table 4.3, the initial temperature and enthalpy of the HFC $-134 a$ at a pressure of 12.0 bar are as follows:

$$\begin{gathered} T_i=46.3°\mathrm{C} \\ {\mathrm{H}}_{\mathrm{i}}=116 \mathrm{kJ} \end{gathered}$$

At a pressure of 12.0 bar, the final temperature of HFC-134a is as follows:

$$\begin{gathered} Tf=-26.4°\mathrm{C} \\ =(-26.4+273)\mathrm{K} \\ =246.6 \mathrm{K} \end{gathered}$$

$$\begin{gathered} \text{ Therefore, the final temperature of the liquid }\mathrm{HFC}-134\mathrm{a}\text{ is } \\ -26.4°\mathrm{C}\text{ or }246.6 \mathrm{K} \end{gathered}$$

The enthalpy of the liquid phase of HFC-134a at the boiling point at a final pressure of 1.0 bar is $16 \mathrm{kJ}$, while the gas phase is$231 \mathrm{kJ}$.

$$\begin{gathered} H_{\text{liquid }}=16 \mathrm{kJ} \\ {\mathrm{H}}_{\mathrm{gas}}=231 \mathrm{kJ} \end{gathered}$$

Step 3: Further calculation

A throttling operation conserves the enthalpy. The initial enthalpy of liquid HFC-134a ranges from$16 \mathrm{kJ}$ to $231 \mathrm{kJ}$. The boiling point of

HFC-134a is $-26.4°\mathrm{C}$, which results in a mixture of liquid and gas.

$H_f=a\left(H_{\text{liquid }}\right)+(1-a)\left(H_{gas}\right)$

Substitute $H_{\text{liquid }}=$$16 \mathrm{kJ}$ 

                 $H_{gas}=$$231 \mathrm{kJ}$ 

$$\begin{gathered} H_f=a(16 \mathrm{kJ})+(1-a)(231 \mathrm{kJ}) \\ =231 \mathrm{kJ}-(215 \mathrm{kJ})a \end{gathered}$$

HFC-134a has the same initial and ultimate enthalpies as water.

$H_f=H_i$

$$\begin{gathered} \text{ Substitute} H_f=231 \mathrm{kJ}-(215 \mathrm{kJ})a \\ H_i=116 \mathrm{kJ} \end{gathered}$$

solving for a is :

$$\begin{gathered} a=\frac{231 \mathrm{kJ}-116 \mathrm{kJ}}{215 \mathrm{kJ}} \\ =0.535 \end{gathered}$$

Hence, the fraction of liquid vaporizes is as follows:

$$\begin{gathered} 1-a \\ =1-0.535 \\ = 0.465 \end{gathered}$$ 

Thus, the liquid vaporization is 0.465 .