Chapter 5.7, Problem 16P

(A)

Interpretation Introduction

Interpretation:

The efficiency of the compressor.

Concept Introduction:

The entropy balance around the reversible compressor is,

$\frac{d\left(N\widehat{S}\right)}{dt}={\dot{n}}_{in}{\underset{\_}{S}}_{in}-{\dot{n}}_{out}{\underset{\_}{S}}_{out}+\frac{\dot{Q}}{T}+{\dot{S}}_{gen}$

Here, number of moles in the system is N, specific entropy of the system is $\widehat{S}$, molar flow rates of individual streams entering and leaving the system is ${\dot{n}}_{in}$, ${\dot{n}}_{out}$, molar entropies of streams entering and leaving the system is ${\widehat{S}}_{in},{\widehat{S}}_{out}$, actual rate at which heat is added to or removed from the system is $\dot{Q}$, the temperature of the system at the boundary is $T$, and the rate at which entropy is generated within the boundaries of the system is ${\dot{S}}_{gen}$ and time is t.

The molar entropy difference across the compressor modeling the gas as ideal is,

$\begin{array}{c}d\underset{\_}{S}=C_P^{*}\ln \left(\frac{T_2}{T_1}\right)+R\ln \left(\frac{P_1}{P_2}\right)\\ =\left(C_V^{*}+R\right)\ln \left(\frac{T_2}{T_1}\right)+R\ln \left(\frac{P_1}{P_2}\right)\end{array}$

Here, heat capacity at constant pressure and constant volume for an ideal gas condition is $C_P^{*}$ and $C_V^{*}$, gas constant is R, subscripts “1” and “2” refers inlet and outlet, temperature at state 1 and 2 is $T_1$ and $T_2$, pressure at state 1 and 2 is $P_1$ and $P_2$ respectively.

The expression of the enthalpy of effluent of the reversible compressor is,

$\begin{array}{l}d\underset{\_}{H}=C_P^{*}dT\\ {\underset{\_}{H}}_{2,rev}-{\underset{\_}{H}}_1=\left(C_V^{*}+R\right)\left(T_2-T_1\right)\end{array}$

Here, change in molar enthalpy is $d\underset{\_}{H}$, molar enthalpy at state 1 and 2 is $H_1$ and $H_2$ respectively.

The entropy balance equation around the reversible compressor is,

$\frac{d}{dt}\left\{N\left(\underset{\_}{U}+\frac{v^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\displaystyle \sum _{j=1}^{j=J}{\dot{m}}_{j,in}\left({\underset{\_}{H}}_j+\frac{v_j^2}{2}+g{h}_j\right)}-{\displaystyle \sum _{k=1}^{k=K}{\dot{m}}_{k,out}\left({\underset{\_}{H}}_k+\frac{v_k^2}{2}+g{h}_k\right)}+\\ {\dot{W}}_S+{\dot{W}}_{EC}+\dot{Q}\end{array}\right]$

Here, time is t, number of moles in the system is N, molar internal energy of the system is $\underset{\_}{U}$, velocity of the system is v, height of the system is h, acceleration due to gravity is g, molar flow rate for inlet and outlet streams is ${\dot{n}}_{j,in}$ and ${\dot{n}}_{k,out}$, molar enthalpies of streams inlet and outlet is ${\underset{\_}{H}}_j$ and ${\underset{\_}{H}}_k$, heights at which streams enters and leave the system is $h_j$ and $h_k$, rate at which work is added to the system through expansion or contraction of the system is ${\dot{W}}_{EC}$, rate at which shaft work is added to the system is ${\dot{W}}_S$, and the rate at which heat is added to the system is $\dot{Q}$.

The expression of compressor efficiency is,

${\eta }_{compressor}=\frac{W_{S,reversible}}{W_{S,actual}}$

Here, work done on the shaft by reversible process is $W_{S,reversible}$ and actual work done on the shaft is $W_{S,actual}$.

Explanation of Solution

Given information:

Specific heat at constant volume is $C_V=14R$.

Write an entropy balance around the reversible compressor.

$\frac{d\left(N\widehat{S}\right)}{dt}={\dot{n}}_{in}{\underset{\_}{S}}_{in}-{\dot{n}}_{out}{\underset{\_}{S}}_{out}+\frac{\dot{Q}}{T}+{\dot{S}}_{gen}$        (1)

Rewrite Equation (1) for an entropy balance equation for an adiabatic, steady state turbine with only one stream entering and leaving the system.

$0={\dot{n}}_{in}{\underset{\_}{S}}_{in}-{\dot{n}}_{out}{\underset{\_}{S}}_{out}+\frac{\dot{Q}}{T}+{\dot{S}}_{gen}$        (2)

At steady state $\dot{n}={\dot{n}}_{in}={\dot{n}}_{out}$ and turbine is reversible, ${\dot{S}}_{gen}=0$.

Rewrite Equation (2).

$\begin{array}{l}0={\widehat{S}}_{in}-{\widehat{S}}_{out}+0\\ {\widehat{S}}_{in}={\widehat{S}}_{out}\end{array}$        (3)

Write the molar entropy difference across the compressor modeling the gas as ideal.

$d\underset{\_}{S}=\left(C_V^{*}+R\right)\ln \left(\frac{T_2}{T_1}\right)+R\ln \left(\frac{P_1}{P_2}\right)$        (4)

Substitute 14R for $C_V^{*}$, $8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}$ for R, 278 K for $T_1$, 0.02 MPa for $P_1$, 0 for $d\underset{\_}{S}$, and 0.06 MPa for $P_2$ in Equation (4).

$\begin{array}{l}d\underset{\_}{S}=\left(14R+8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\ln \left(\frac{T_2}{278\text{K}}\right)+8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\ln \left(\frac{0.02\text{MPa}}{0.06\text{MPa}}\right)\\ 0=\left(14\left(8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)+8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\ln \left(\frac{T_2}{278\text{K}}\right)+8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\ln \left(\frac{0.02\text{MPa}}{0.06\text{MPa}}\right)\end{array}$

Simply for $T_2$,

$\begin{array}{l}{T}_2=\frac{278\text{K}}{e^{\frac{4157\ln \left(\frac{0.02}{0.06}\right)}{62000}}}\\ T_2=\text{299}\text{.25}\text{K}\end{array}$

Write the expression of the enthalpy of effluent of the reversible compressor.

${\underset{\_}{H}}_{2,rev}-{\underset{\_}{H}}_1=\left(C_V^{*}+R\right)\left(T_2-T_1\right)$        (5)

Substitute 299.1 K for $T_2$, 278 K for $T_1$, 14 R for $C_V^{*}$, $8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}$ for R, and $50\frac{\text{kJ}}{\text{mol}}$ for ${\underset{\_}{H}}_1$ in Equation (5).

$\begin{array}{l}{\underset{\_}{H}}_{2,rev}-50\frac{\text{kJ}}{\text{mol}}=\left(14\left[8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\right]+8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\left(299.1\text{K}-278\text{K}\right)\\ {\underset{\_}{H}}_{2,rev}-50\frac{\text{kJ}}{\text{mol}}=\left(14\left[8.314\frac{\text{J}\times {\text{10}}^{-3}\frac{\text{kJ}}{\text{J}}}{\text{mol}\cdot \text{K}}\right]+8.314\frac{\text{J}\times {\text{10}}^{-3}\frac{\text{kJ}}{\text{J}}}{\text{mol}\cdot \text{K}}\right)\left(299.1\text{K}-278\text{K}\right)\\ {\underset{\_}{H}}_{2,rev}=52.63\frac{\text{kJ}}{\text{mol}}\end{array}$

Write the entropy balance around the reversible compressor.

$\frac{d}{dt}\left\{N\left(\underset{\_}{U}+\frac{v^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\displaystyle \sum _{j=1}^{j=J}{\dot{m}}_{j,in}\left({\underset{\_}{H}}_j+\frac{v_j^2}{2}+g{h}_j\right)}-{\displaystyle \sum _{k=1}^{k=K}{\dot{m}}_{k,out}\left({\underset{\_}{H}}_k+\frac{v_k^2}{2}+g{h}_k\right)}+\\ {\dot{W}}_S+{\dot{W}}_{EC}+\dot{Q}\end{array}\right]$

At steady state, the accumulation term is 0, and the time independent energy balance simplifies.

$0={\dot{n}}_{in,1}\left({\underset{\_}{H}}_{in,1}+\frac{v_{in}^2}{2}+g{h}_{in}\right)-{\dot{n}}_{out}\left({\underset{\_}{H}}_{out}+\frac{v_{out}^2}{2}+g{h}_{out}\right)+{\dot{W}}_S+{\dot{W}}_{EC}+\dot{Q}$        (6)

Rewrite Equation (6).

$0={\dot{n}}_{in}\left({\underset{\_}{H}}_{in}\right)-{\dot{n}}_{out}\left({\underset{\_}{H}}_{out}\right)+{\dot{W}}_S$        (7)

Here, there is only one material stream entering and leaving the system.

Rewrite Equation (7).

$\frac{{\dot{W}}_S}{\dot{n}}={\underset{\_}{H}}_{out}-{\underset{\_}{H}}_{in}$        (8)

Substitute $52.63\frac{\text{kJ}}{\text{mol}}$ for ${\underset{\_}{H}}_{out}$ and $50\frac{\text{kJ}}{\text{mol}}$ for ${\underset{\_}{H}}_{in}$ in Equation (8).

$\begin{array}{c}\frac{{\dot{W}}_S}{\dot{n}}=52.63\frac{\text{kJ}}{\text{mol}}-50\frac{\text{kJ}}{\text{mol}}\\ =2.63\frac{\text{kJ}}{\text{mol}}\end{array}$

Substitute $50°\text{C}$ for $T_2$, $5°\text{C}$ for $T_1$, 14R for $C_V^{*}$, $8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}$ for R, and $50\frac{\text{kJ}}{\text{mol}}$ for ${\underset{\_}{H}}_1$ in Equation (5).

$\begin{array}{l}{\underset{\_}{H}}_2-50\frac{\text{kJ}}{\text{mol}}=\left(14\left[8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\right]+8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\left(50°\text{C}-5°\text{C}\right)\\ {\underset{\_}{H}}_2-50\frac{\text{kJ}}{\text{mol}}=\left(14\left[8.314\frac{\text{J}\times {\text{10}}^{-3}\frac{\text{kJ}}{\text{J}}}{\text{mol}\cdot \text{K}}\right]+8.314\frac{\text{J}\times {\text{10}}^{-3}\frac{\text{kJ}}{\text{J}}}{\text{mol}\cdot \text{K}}\right)\left(\left(50+273\right)\text{K}-\left(5+273\right)\text{K}\right)\\ {\underset{\_}{H}}_2=55.61\frac{\text{kJ}}{\text{mol}}\end{array}$

Write the energy balance around the actual compressor.

$\frac{d}{dt}\left\{N\left(\underset{\_}{U}+\frac{v^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\displaystyle \sum _{j=1}^{j=J}{\dot{m}}_{j,in}\left({\underset{\_}{H}}_j+\frac{v_j^2}{2}+g{h}_j\right)}-{\displaystyle \sum _{k=1}^{k=K}{\dot{m}}_{k,out}\left({\underset{\_}{H}}_k+\frac{v_k^2}{2}+g{h}_k\right)}+\\ {\dot{W}}_S+{\dot{W}}_{EC}+\dot{Q}\end{array}\right]$

The accumulation term is 0at steady state and hence,

$0={\dot{n}}_{in,1}\left({\underset{\_}{H}}_{in,1}+\frac{v_{in}^2}{2}+g{h}_{in}\right)-{\dot{n}}_{out}\left({\underset{\_}{H}}_{out}+\frac{v_{out}^2}{2}+g{h}_{out}\right)+{\dot{W}}_S+{\dot{W}}_{EC}+\dot{Q}$        (9)

Neglect the terms and rewrite Equation (9).

$0={\dot{n}}_{in}\left({\underset{\_}{H}}_{in}\right)-{\dot{n}}_{out}\left({\underset{\_}{H}}_{out}\right)+{\dot{W}}_S$        (10)

Here, there is only one material stream entering and leaving the system.

Rewrite Equation (10).

$\frac{{\dot{W}}_S}{\dot{n}}={\underset{\_}{H}}_{out}-{\underset{\_}{H}}_{in}$        (11)

Substitute $55.61\frac{\text{kJ}}{\text{mol}}$ for ${\underset{\_}{H}}_{out}$ and $50\frac{\text{kJ}}{\text{mol}}$ for ${\underset{\_}{H}}_{in}$ in Equation (11).

$\begin{array}{c}\frac{{\dot{W}}_S}{\dot{n}}=55.61\frac{\text{kJ}}{\text{mol}}-50\frac{\text{kJ}}{\text{mol}}\\ =5.61\frac{\text{kJ}}{\text{mol}}\end{array}$

Write the expression of compressor efficiency.

${\eta }_{compressor}=\frac{W_{S,reversible}}{W_{S,actual}}$        (12)

Substitute $5.61\frac{\text{kJ}}{\text{mol}}$ for $W_{S,actual}$ and $2.63\frac{\text{kJ}}{\text{mol}}$ for $W_{S,reversible}$ in Equation (12).

$\begin{array}{c}{\eta }_{compressor}=\frac{2.63\frac{\text{kJ}}{\text{mol}}}{5.61\frac{\text{kJ}}{\text{mol}}}\\ =0.469\end{array}$

Thus, the efficiency of compressor is $0.469$.

(B)

Interpretation Introduction

Interpretation:

The required flow rate of the refrigerant.

Concept Introduction:

The expression to obtain the required flow rate of the refrigerant is,

$\frac{{\dot{Q}}_R}{\dot{n}}=\frac{\frac{{\dot{Q}}_C}{\dot{n}}}{\frac{\dot{Q}}{\dot{n}}}$

Here, cooling rate is ${\dot{Q}}_C$.

Explanation of Solution

Write the energy balance around the boiler.

$\frac{d}{dt}\left\{N\left(\underset{\_}{U}+\frac{v^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\displaystyle \sum _{j=1}^{j=J}{\dot{m}}_{j,in}\left({\underset{\_}{H}}_j+\frac{v_j^2}{2}+g{h}_j\right)}-{\displaystyle \sum _{k=1}^{k=K}{\dot{m}}_{k,out}\left({\underset{\_}{H}}_k+\frac{v_k^2}{2}+g{h}_k\right)}+\\ {\dot{W}}_S+{\dot{W}}_{EC}+\dot{Q}\end{array}\right]$

The accumulation term is 0at steady state and hence,

$0={\dot{n}}_{in,1}\left({\underset{\_}{H}}_{in,1}+\frac{v_{in}^2}{2}+g{h}_{in}\right)-{\dot{n}}_{out}\left({\underset{\_}{H}}_{out}+\frac{v_{out}^2}{2}+g{h}_{out}\right)+{\dot{W}}_S+{\dot{W}}_{EC}+\dot{Q}$        (13)

Neglect the terms and rewrite Equation (13).

$0={\dot{n}}_{in}\left({\underset{\_}{H}}_{in}\right)-{\dot{n}}_{out}\left({\underset{\_}{H}}_{out}\right)+\dot{Q}$        (14)

Here, there is only one material stream entering and leaving the system.

Rewrite Equation (14).

$\frac{\dot{Q}}{\dot{n}}={\underset{\_}{H}}_{out}-{\underset{\_}{H}}_{in}$        (15)

Substitute $50\frac{\text{kJ}}{\text{mol}}$ for ${\underset{\_}{H}}_{out}$ and $20\frac{\text{kJ}}{\text{mol}}$ for ${\underset{\_}{H}}_{in}$ in Equation (15).

$\begin{array}{c}\frac{\dot{Q}}{\dot{n}}=50\frac{\text{kJ}}{\text{mol}}-20\frac{\text{kJ}}{\text{mol}}\\ =30\frac{\text{kJ}}{\text{mol}}\end{array}$

Write the expression to obtain the required flow rate of the refrigerant.

$\frac{{\dot{Q}}_R}{\dot{n}}=\frac{\frac{{\dot{Q}}_C}{\dot{n}}}{\frac{\dot{Q}}{\dot{n}}}$        (16)

Substitute $3000\frac{\text{kJ}}{\min }$ for $\frac{{\dot{Q}}_C}{\dot{n}}$, $30\frac{\text{kJ}}{\text{mol}}$ for $\frac{\dot{Q}}{\dot{n}}$, and $400\frac{\text{g}}{\text{mol}}$ for $\dot{n}$ in Equation (16).

$\begin{array}{l}\frac{{\dot{Q}}_R}{400\frac{\text{g}}{\text{mol}}}=\frac{3000\frac{\text{kJ}}{\min }}{30\frac{\text{kJ}}{\text{mol}}}\\ {\dot{Q}}_R=\left(\frac{3000\frac{\text{kJ}}{\min }}{30\frac{\text{kJ}}{\text{mol}}}\right)\left(400\frac{\text{g}}{\text{mol}}\right)\left(\frac{1\text{kg}}{1000\text{g}}\right)\\ {\dot{Q}}_R=40\frac{\text{kg}}{\min }\end{array}$

Thus, the required flow rate of the refrigerant is $40\frac{\text{kg}}{\min }$.

(C)

Interpretation Introduction

Interpretation:

The coefficient of performance of the refrigeration cycle.

Concept Introduction:

The expression to obtain the coefficient of performance of the refrigeration cycle is,

$\text{C}\text{.O}.P=\frac{{\dot{Q}}_C}{{\dot{W}}_S}$

Explanation of Solution

Write the expression to obtain the coefficient of performance of the refrigeration cycle.

$\text{C}\text{.O}.P=\frac{{\dot{Q}}_C}{{\dot{W}}_S}$        (17)

Substitute $3000\text{kJ}/\text{min}$ for ${\dot{Q}}_C$ and $5.61\text{kJ}/\text{mol}$ for ${\dot{W}}_S$ in Equation (17).

$\begin{array}{c}\text{C}\text{.O}.P=\frac{3000\text{kJ}/\text{min}}{5.61\text{kJ}/\text{mol}}\\ =\frac{3000\text{kJ}/\text{min}}{5.61\text{kJ}/\text{mol}\left(40\text{kg}/\min \right)\left(\frac{1000\text{g}}{1\text{kg}}\right)\left(\frac{\text{mol}}{400\text{g}}\right)}\\ =5.35\end{array}$

Thus, the coefficient of performance of the refrigeration cycle is $5.35$.

(D)

Interpretation Introduction

Interpretation:

Design a refrigeration cycle using R-134a as the refrigerant.

Concept Introduction:

The energy balance around the reversible compressor is,

$\frac{{\dot{W}}_{S,rev}}{\dot{m}}={\widehat{H}}_{out,rev}-{\widehat{H}}_{in}$

Here, mass flow rate is $\dot{m}$.

The expression for compressor efficiency is,

${\eta }_{compressor}=\frac{W_{S,reversible}}{W_{S,actual}}$

Here, work done on the shaft by reversible process is $W_{S,reversible}$ and actual work done on the shaft is $W_{S,actual}$.

The energy balance on the boiler is,

$\frac{d}{dt}\left\{M\left(\widehat{U}+\frac{v^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\displaystyle \sum _{j=1}^{j=J}{\dot{m}}_{j,in}\left({\widehat{H}}_j+\frac{v_j^2}{2}+g{h}_j\right)}-{\displaystyle \sum _{k=1}^{k=K}{\dot{m}}_{k,out}\left({\widehat{H}}_k+\frac{v_k^2}{2}+g{h}_k\right)}+\\ {\dot{W}}_S+{\dot{W}}_{EC}+\dot{Q}\end{array}\right]$

Here, time is t, total mass of the system is M, specific internal energy of the system is $\widehat{U}$, velocity of the system is v, height of the system is h, acceleration due to gravity is g, mass flow rate for inlet and outlet streams is ${\dot{m}}_{j,in}$ and ${\dot{m}}_{k,out}$, specific enthalpies of streams inlet and outlet is ${\widehat{H}}_j$ and ${\widehat{H}}_k$, heights at which streams enters and leave the system is $h_j$ and $h_k$, rate at which work is added to the system through expansion or contraction of the system is ${\dot{W}}_{EC}$, rate at which shaft work is added to the system is ${\dot{W}}_S$, and the rate at which heat is added to the system is $\dot{Q}$.

The expression to obtain the coefficient of performance of the refrigeration cycle is,

$\text{C}\text{.O}.P=\frac{{\dot{Q}}_C}{{\dot{W}}_{S,actual}}$

Explanation of Solution

Assume the condenser and boiler temperatures are the same.

Write the entropy balance around the reversible compressor.

${\widehat{S}}_{in}={\widehat{S}}_{out}$        (18)

Refer Appendix F, “Thermodynamic diagrams of pressure-enthalpy”, obtain the specific entropy for inlet state $\left({\widehat{S}}_{in}\right)$ as saturated vapor R-422A at $5°\text{C}$ as $0.225\frac{\text{BTU}}{{\text{lb}}_{\text{m}}\cdot °\text{F}}$.

Condenser operates at 50°C = 122°F so the compressor outlet pressure must equal vapor pressure at this temperature, which according to the figure is slightly below 200 psia.

Write the energy balance around the reversible compressor.

$\frac{{\dot{W}}_{S,rev}}{\dot{m}}={\widehat{H}}_{out,rev}-{\widehat{H}}_{in}$        (19)

Refer Appendix F, “Thermodynamic diagrams of pressure-enthalpy”, obtain the specific enthalpy for inlet state at saturated vapor R-422A at $41°\text{F}$ as $108\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}$.

Obtain the specific enthalpy of the outlet by following the line of constant entropy from the inlet condition to pressure of 195 psiaas $120\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}$.

Substitute $120\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}$ for ${\widehat{H}}_{out,rev}$ and $108\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}$ for ${\widehat{H}}_{in}$ in Equation (19).

$\begin{array}{c}\frac{{\dot{W}}_{S,rev}}{\dot{m}}=120\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}-108\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}\\ =12\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}\end{array}$

Write the compressor efficiency.

${\eta }_{compressor}=\frac{W_{S,reversible}}{W_{S,actual}}$        (20)

Substitute 0.468 for $\eta$ and $12\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}$ for $W_{S,reversible}$ in Equation (20).

$\begin{array}{l}0.468=\frac{12\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}}{W_{S,actual}}\\ W_{S,actual}=25.6\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}\end{array}$

Write the energy balance on the boiler.

$\frac{d}{dt}\left\{M\left(\widehat{U}+\frac{v^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\displaystyle \sum _{j=1}^{j=J}{\dot{m}}_{j,in}\left({\widehat{H}}_j+\frac{v_j^2}{2}+g{h}_j\right)}-{\displaystyle \sum _{k=1}^{k=K}{\dot{m}}_{k,out}\left({\widehat{H}}_k+\frac{v_k^2}{2}+g{h}_k\right)}+\\ {\dot{W}}_S+{\dot{W}}_{EC}+\dot{Q}\end{array}\right]$

The accumulation term is 0at steady state and hence,

$0={\dot{m}}_{in,1}\left({\widehat{H}}_{in,1}+\frac{v_{in}^2}{2}+g{h}_{in}\right)-{\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{v_{out}^2}{2}+g{h}_{out}\right)+{\dot{W}}_S+{\dot{W}}_{EC}+\dot{Q}$        (21)

Neglect the terms and rewrite Equation (21).

$-{\dot{Q}}_C={\dot{m}}_{in}\left({\widehat{H}}_{in}\right)-{\dot{m}}_{out}\left({\widehat{H}}_{out}\right)$        (22)

Here, there is only one material stream entering and leaving the system, and $\dot{Q}$ is not equal to zero.

Rewrite Equation (22).

$\frac{{\dot{Q}}_C}{\dot{m}}={\widehat{H}}_{out}-{\widehat{H}}_{in}$        (23)

Refer Appendix F, “Thermodynamic diagrams of pressure-enthalpy”, obtain the specific enthalpy for outlet state at saturated vapor R-422A at $5°\text{C}$ as $108\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}$.

Refer Appendix F, “Thermodynamic diagrams of pressure-enthalpy”, obtain the specific enthalpy for inlet state at saturated liquid of $50°\text{C}$ as $52\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}$.

Substitute $108\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}$ for ${\widehat{H}}_{out}$ and $52\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}$ for ${\widehat{H}}_{in}$ in Equation (23).

$\begin{array}{c}\frac{{\dot{Q}}_C}{\dot{m}}=108\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}-52\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}\\ =56\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}\end{array}$

Write the expression to obtain the coefficient of performance of the refrigeration cycle.

$\text{C}\text{.O}.P=\frac{{\dot{Q}}_C}{{\dot{W}}_{S,actual}}$        (24)

Substitute $56\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}$ for ${\dot{Q}}_C$ and $25.6\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}$ for ${\dot{W}}_{S,actual}$ in Equation (24).

$\begin{array}{c}\text{C}\text{.O}.P=\frac{56\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}}{25.6\frac{\text{BTU}}{{\text{lb}}_{\text{m}}}}\\ =2.18\end{array}$

Thus, the coefficient of performance of the refrigeration cycle is $2.18$.

(E)

Interpretation Introduction

Interpretation:

Comment on the inventor’s claim that the proprietary refrigerant is better than R-134a.

Explanation of Solution

The inventor’s refrigerant is good when compared to the COP of each cycle. But neither COP is extremely large as the inventor’s system has compressor efficiency of less than 50%.