Chapter 14.7, Problem 88P

Atmospheric air at 1 atm, 32°C, and 95 percent relative humidity is cooled to 24°C and 60 percent relative humidity. A simple ideal vapor-compression refrigeration system using refrigerant-134a as the working fluid is used to provide the cooling required. It operates its evaporator at 4°C and its condenser at a saturation temperature of 39.4°C. The condenser rejects its heat to the atmospheric air. Calculate the exergy destruction, in kJ, in the total system per 1000 m³ of dry air processed.

To determine

The exergy destruction in the total system per $1000{\text{m}}^{\text{3}}$ of dry air.

Answer to Problem 88P

The exergy destruction in the total system per $1000{\text{m}}^{\text{3}}$ of dry air is $\underset{\_}{6758\text{kJ}}$.

Explanation of Solution

Show the T-s diagram for the simple ideal vapour-compression refrigeration system.

Express the mass of air.

$m_a=\frac{V_1}{v_1}$ (I)

Here, volume at state 1 is $V_1$ and specific volume at state 1 is $v_1$.

Express the water mass balance and energy balance equations to the combined cooling and dehumidification section.

For water mass balance:

$\begin{array}{l}{\displaystyle \sum {\dot{m}}_{w,\text{in}}}={\displaystyle \sum {\dot{m}}_{w,\text{out}}}\\ {\dot{m}}_{a1}{\omega }_1={\dot{m}}_{a2}{\omega }_2+{\dot{m}}_w\\ m_w=m_a\left({\omega }_1-{\omega }_2\right)\end{array}$ (II)

For energy balance:

Write the energy balance equation using steady-flow equation.

${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}$ (III)

Here, the rate of total energy entering the system is $E_{\text{in}}$, the rate of total energy leaving the system is $E_{\text{out}}$, and the rate of change in the total energy of the system is $\Delta E_{\text{system}}$.

Substitute $0$ for $\Delta E_{\text{system}}$ in Equation (III)

$\begin{array}{l}{\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_{\text{in}}{h}_{\text{in}}}={\dot{Q}}_{\text{out}}+{\displaystyle \sum {\dot{m}}_{\text{out}}{h}_{\text{out}}}\\ {\dot{Q}}_{\text{out}}={\dot{m}}_{a1}{h}_1+\left({\dot{m}}_{a2}{h}_2+{\dot{m}}_w{h}_w\right)\\ Q_{\text{out}}=m_a\left(h_1-h_2\right)-m_w{h}_w\end{array}$ (IV)

Here, the specific enthalpy at the state 1 and 2 are $h_1\text{and}{h}_2$.

Write the formula for mass flow rate of refrigerant-134a.

$m_R=\frac{Q_L}{h_1-h_4}$ (V)

Write the formula for amount of heat rejected from the condenser.

$Q_H=m_R\left(h_2-h_3\right)$ (VI)

Calculate the exergy destruction in the components of the refrigeration cycle.

For the process 1-2,

$\begin{array}{c}{X}_{\text{dest,12}}=m_R{T}_0\left(s_2-s_1\right)\\ =0\end{array}$

Here, the process 1-2 is an isentropic.

For the process 2-3,

$X_{\text{dest,23}}=T_0\left(m_R\left(s_3-s_2\right)+\frac{Q_H}{T_H}\right)$ (VII)

For the process 3-4,

$X_{\text{dest},34}=m_R{T}_0\left(s_4-s_3\right)$ (VIII)

Write the entropy change of water vapour in the air stream.

$\Delta S_{\text{vapor}}=m_a\left({\omega }_2{s}_{g2}-{\omega }_1{s}_{g1}\right)$ (IX)

Write the entropy of water leaving the cooling section.

$S_w=m_w{s}_{f@28\text{°C}}$ (X)

Determine the partial pressure of water vapour at state 1 for air steam.

$\begin{array}{c}{P}_{vap,1}={\varphi }_1\times P_{g1}\\ ={\varphi }_1\times P_{\text{sat@100°F}}\end{array}$ (XI)

Determine the partial pressure of dry air at state 1 for air steam.

$P_{a1}=P_1-P_{vap,1}$ (XII)

Here, the pressure at the state 1 is $P_1$.

Determine the partial pressure of water vapour at state 2 for air steam.

$\begin{array}{c}{P}_{vap,2}={\varphi }_2\times P_{g2}\\ ={\varphi }_2\times P_{\text{sat@50°F}}\end{array}$ (XIII)

Determine the partial pressure of dry air at state 2 for air steam.

$P_{a2}=P_2-P_{vap,2}$ (XIV)

Here, the pressure at the state 2 is $P_2$.

Write the formula for entropy change of dry air.

$\begin{array}{c}\Delta S_a=m_a\left(s_2-s_1\right)\\ =m_a\left(c_p\ln \frac{T_2}{T_1}-R\ln \frac{P_{a2}}{P_{a1}}\right)\end{array}$ (XV)

Write the formula for entropy change R-134a in the evaporator.

$\Delta S_{R,41}=m_R\left(s_1-s_4\right)$ (XVI)

Write the formula for an entropy balance on the evaporator.

$S_{\text{gen,evaporator}}=\Delta S_{R,41}+\Delta S_{\text{vapor}}+\Delta S_a+S_w$ (XVII)

Write the formula for an exergy destruction in the evaporator.

$X_{\text{dest}}=T_0{S}_{\text{gen,evaporator}}$ (XVIII)

Write the formula for the total exergy destruction.

$X_{\text{dest,total}}=X_{\text{dest,compressor}}+X_{\text{dest,condenser}}+X_{\text{dest,throttle}}+X_{\text{dest,evaporator}}$ (XIX)

Conclusion:

Refer Figure A-31, “psychometric chart at $\text{1 atm}$ total pressure”, and write the properties corresponding to total pressure of 1 atm.

$\begin{array}{l}{h}_1=106.8\text{kJ}/\text{kg}\text{dry}\text{air}\\ {\omega }_1=0.0292\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\\ v_1=0.905{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\end{array}$

$\begin{array}{l}{h}_2=52.7\text{kJ}/\text{kg}\text{dry}\text{air}\\ {\omega }_2=0.00112\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Refer Table A-4, “saturated water-temperature table”, and write the specific enthalpy of condensate water at temperature of $\text{28°C}$ using an interpolation method.

$h_w=h_{f@28\text{°C}}$ (XX)

Here, entropy of saturation liquid at temperature of $\text{28°C}$ is $h_{f@28\text{°C}}$.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (XXI)

Here, the variables denote by x and y is temperature and specific enthalpy of condensate water at state 2 respectively.

Show the specific enthalpy of condensate water at corresponding to temperature as in Table (1).

Temperature $T\left(\text{°C}\right)$	Specific enthalpy of condensate water $h_{w@h_f}\left(\text{kJ}/\text{kg}\right)$
25 $\left(x_1\right)$	104.83 $\left(y_1\right)$
28 $\left(x_2\right)$	$\left(y_2=?\right)$
30 $\left(x_3\right)$	125.74 $\left(y_3\right)$

Substitute $\text{25°C,}2\text{8°C}\text{and}3\text{0°C}$ for $x_1,x_2\text{and}{x}_3$ respectively, $104.83\text{kJ}/\text{kg}$ for $y_1$ and $125.74\text{kJ}/\text{kg}$ for $y_3$ in Equation (XXI).

$\begin{array}{c}{y}_2=\frac{\left(\text{28°C}-2\text{5°C}\right)\left(125.74\text{kJ}/\text{kg}-104.83\text{kJ}/\text{kg}\right)}{\left(\text{30°C}-25\text{°C}\right)}+104.83\text{kJ}/\text{kg}\\ =117.37\text{kJ}/\text{kg}\\ \cong 117.4\text{kJ}/\text{kg}\end{array}$

Substitute $117.4\text{kJ}/\text{kg}$ for $h_{w@h_f}\left(\text{kJ}/\text{kg}\right)$ in Equation (XX).

$h_w=117.4\text{kJ}/\text{kg}$

Substitute $1000{\text{m}}^{\text{3}}$ for $V_1$ and $0.905{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{}\text{air}$ for $v_1$ in Equation (I).

$\begin{array}{c}{m}_a=\frac{1000{\text{m}}^{\text{3}}}{0.905{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{}\text{air}}\\ =1105\text{kg}\end{array}$

Substitute 1105 kg for $m_a$, $0.0292\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$, and $0.00112\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (II).

$\begin{array}{c}{m}_w=\left[\begin{array}{l}\left(1105\text{kg}\right)\\ \times \left(\begin{array}{l}0.0292\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\\ -0.00112\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}\right)\end{array}\right]\\ =\left(1105\text{kg}\right)\left(0.018\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\right)\\ =19.89\text{kg}\end{array}$

Substitute 1105 kg for $m_a$, $106.8\text{kJ}/\text{kg}$ for $h_1$, $52.7\text{kJ}/\text{kg}$ for $h_2$, $19.89\text{kg}$ for $m_w$, and $117.4\text{kJ}/\text{kg}$ for $h_w$ in Equation (IV).

$\begin{array}{c}{Q}_{\text{out}}=\left[\begin{array}{l}\left(1105\text{kg}\right)\times \left(106.8-52.7\right)\text{kJ}/\text{kg}\\ -\left(19.89\text{kg}\right)\times \left(117.4\text{kJ}/\text{kg}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(1105\text{kg}\right)\times 54.1\text{kJ}/\text{kg}\\ -\left(19.89\text{kg}\right)\times \left(117.4\text{kJ}/\text{kg}\right)\end{array}\right]\\ =\left[\begin{array}{l}59780.5\text{kJ}\\ -\left(2335.086\text{kJ}\right)\end{array}\right]\\ =57,445\text{kJ}\end{array}$

From the Table A-11 “Saturated Refrigerant-134a-Pressure Table”, obtain the value of the specific enthalpy and entropy at state 1 of $4°\text{C}$ of temperature as

$\begin{array}{l}{h}_1=h_{g@4°\text{C}}=252.77\text{kJ}/\text{kg}\\ s_1=s_{g@4°\text{C}}=0.92927\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refer Table A-13 “Saturated Refrigerant-134a-Pressure Table”, and write the specific enthalpy at state 2 in 1 MPa of pressure and entropy of $0.92927\text{kJ}/\text{kg}\cdot \text{K}$ using an interpolation method Equation (XXI).

$h_2=275.29\text{kJ}/\text{kg}$

From the Table A-12 “Saturated Refrigerant-134a-Pressure Table”, obtain the value of the specific enthalpy and entropy at state 3 of $39.4°\text{C}$ of temperature as

$\begin{array}{l}{h}_3=h_{f@1\text{MPa}}=107.32\text{kJ}/\text{kg}\\ s_3=s_{f@1\text{MPa}}=0.39189\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, the specific enthalpy at the state 4 and 3 are equal in the throttling.

Calculate the value of $x_4$

$\begin{array}{l}{h}_4=h_f+x_4{h}_{fg}\\ x_4=\frac{h_4-h_f}{h_{fg}}\end{array}$ (XXII)

From the Table A-11 “Saturated Refrigerant-134a-Pressure Table”, obtain the value of the specific enthalpy and entropy of saturated liquid and change upon vaporization at state 4 of $4°\text{C}$ of temperature as

$\begin{array}{l}{h}_f=57.23\text{kJ}/\text{kg}\\ h_{fg}=195.58\text{kJ}/\text{kg}\\ s_f=0.22381\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=0.70565\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $107.32\text{kJ}/\text{kg}$ for $h_4$, $57.23\text{kJ}/\text{kg}$ for $h_f$, and $195.58\text{kJ}/\text{kg}$ for $h_{fg}$ in Equation (XXII).

$\begin{array}{c}{x}_4=\frac{107.32\text{kJ}/\text{kg}-57.23\text{kJ}/\text{kg}}{195.58\text{kJ}/\text{kg}}\\ =\frac{50.09\text{kJ}/\text{kg}}{195.58\text{kJ}/\text{kg}}\\ =0.2561\end{array}$

Calculate the value of specific entropy of the state 4.

$s_4=s_f+x_4{s}_{fg}$ (XXIII)

Substitute $0.22381\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, $0.70565\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$, and 0.2561 for $x_4$ in Equation (XXIII).

$\begin{array}{c}{s}_4=\left(0.22381\text{kJ}/\text{kg}\cdot \text{K}\right)+\left(0.2561\right)\times \left(0.70565\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =\left(0.22381\text{kJ}/\text{kg}\cdot \text{K}\right)+\left(0.180717\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =0.4045\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $57,445\text{kJ}$ for $Q_L$, $252.77\text{kJ}/\text{kg}$ for $h_1$, and $107.23\text{kJ}/\text{kg}$ for $h_3$ in Equation (V).

$\begin{array}{c}{m}_R=\frac{57,445\text{kJ}}{\left(252.77-107.32\right)\text{kJ}/\text{kg}}\\ =\frac{57,445\text{kJ}}{145.45\text{kJ}/\text{kg}}\\ =394.9\text{kg}\\ \cong 395\text{kg}\end{array}$

Substitute $395\text{kg}$ for $m_R$, $275.29\text{kJ}/\text{kg}$ for $h_2$, and $107.32\text{kJ}/\text{kg}$ for $h_3$ in Equation (VI).

$\begin{array}{c}{Q}_H=\left(395\text{kg}\right)\left(275.29-107.32\right)\text{kJ}/\text{kg}\\ =\left(395\text{kg}\right)\times \left(167.97\text{kJ}/\text{kg}\right)\\ =66348\text{kJ}\end{array}$

Substitute $32°\text{C}$ for $T_0$, $395\text{kg}$ for $m_R$, $0.39189\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3$, $0.92927\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$, and $66348\text{kJ}$ for $Q_H$ in Equation (VII).

$\begin{array}{c}{X}_{\text{dest,23}}=\left(32°\text{C}\right)\left(\begin{array}{l}\left(395\text{kg}\right)\left(0.39189-0.92927\right)\text{kJ}/\text{kg}\cdot \text{K}\\ +\frac{\left(66,348\text{kJ}\right)}{\left(32°\text{C}\right)}\end{array}\right)\\ =\left(32°\text{C+273}\right)\left(\begin{array}{l}\left(-212.265\right)\text{kJ}/\text{kg}\cdot \text{K}\\ +\frac{\left(66,348\text{kJ}\right)}{\left(32°\text{C+273}\right)}\end{array}\right)\\ =1607.145\text{kJ}\end{array}$

Substitute $32°\text{C}$ for $T_0$, $395\text{kg}$ for $m_R$, $0.39189\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3$, and $0.4045\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4$ in Equation (VIII)

$\begin{array}{c}{X}_{\text{dest,34}}=\left(395\text{kg}\right)\left(32°\text{C}\right)\left(0.4045-0.39189\right)\text{kJ}/\text{kg}\cdot \text{K}\\ =\left(395\text{kg}\right)\left(32°\text{C+273}\right)\left(0.01261\right)\text{kJ}/\text{kg}\cdot \text{K}\\ =1519\text{kJ}\end{array}$

Refer Table A-4 “Saturated water-temperature Table”, and write the specific entropy at state 1 at $32°\text{C}$ of temperature using an interpolation method Equation (XXI).

$s_{g1}=s_{g@32°\text{C}}=8.4114\text{kJ}/\text{kg}\cdot \text{K}$

Refer Table A-4 “Saturated water-temperature Table”, and write the specific entropy at state 1 at $24°\text{C}$ of temperature using an interpolation method Equation (XXI).

$s_{g2}=s_{g@24°\text{C}}=8.5782\text{kJ}/\text{kg}\cdot \text{K}$

Substitute $1105\text{kg}$ for $m_a$, $8.4114\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{g1}$, $8.5782\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{g2}$, $0.0292\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$, and $0.00112\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (IX).

$\begin{array}{c}\Delta S_{\text{vapor}}=\left(1105\text{kg}\right)\left(\begin{array}{l}0.00112\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\times 8.5782\text{kJ}/\text{kg}\cdot \text{K}\\ -0.0292\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\times 8.4114\text{kJ}/\text{kg}\cdot \text{K}\end{array}\right)\\ =\left[\left(1105\text{kg}\right)\left(0.096076\text{kJ}/\text{kg}\cdot \text{K}-0.245613\text{kJ}/\text{kg}\cdot \text{K}\right)\right]\\ =\left(1105\text{kg}\right)\left(-0.14954\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =-165.238\text{kJ}/\text{K}\end{array}$

Substitute $19.89\text{kg}$ for $m_w$ and $0.4091\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{f@28°\text{C}}$ in Equation (X).

$\begin{array}{c}{S}_w=\left(19.89\text{kg}\right)\times \left(0.4091\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =8.136\text{kJ}/\text{K}\\ \cong 8.14\text{kJ}/\text{K}\end{array}$

Substitute $0.95$ for ${\varphi }_1$ and $4.760\text{kPa}$ for $P_{\text{sat}@32°\text{C}}$ in Equation (XI).

$\begin{array}{c}{P}_{vap,1}=\left(0.95\right)\times \left(4.760\text{kPa}\right)\\ =4.522\text{kPa}\end{array}$

Substitute $101.325\text{kPa}$ for $P_1$ and $4.522\text{kPa}$ for $P_{vap,1}$ in Equation (XII).

$\begin{array}{c}{P}_{a1}=\left(101.325-4.522\right)\text{kPa}\\ =96.80\text{kPa}\end{array}$

Substitute $0.60$ for ${\varphi }_2$ and $2.986\text{kPa}$ for $P_{sat@50°\text{F}}$ in Equation (XIII).

$\begin{array}{c}{P}_{vap,2}=\left(0.60\right)\times \left(2.986\text{kPa}\right)\\ =1.792\text{kPa}\end{array}$

Substitute $101.325\text{kPa}$ for $P_2$ and $1.792\text{kPa}$ for $P_{vap,2}$ in Equation (XIV).

$\begin{array}{c}{P}_{a2}=\left(101.325-1.792\right)\text{kPa}\\ =99.53\text{kPa}\end{array}$

Substitute $1105\text{kg}$ for $m_a$, $1.005$ for $c_p$, $4°\text{C}$ for $T_2$, $32°\text{C}$ for $T_1$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, $99.53\text{kPa}$ for $P_{a,2}$, and $96.80\text{kPa}$ for $P_{a1}$ in Equation (XV).

$\begin{array}{c}\Delta S_a=\left(1105\text{kg}\right)\times \left[\left(1.005\right)\ln \frac{4°\text{C}}{32°\text{C}}-\left(0.287\right)\ln \frac{99.53\text{kPa}}{96.80\text{kPa}}\right]\\ =\left(1105\text{kg}\right)\times \left[\left(1.005\right)\ln \frac{4°\text{C+273}}{32°\text{C+273}}-\left(0.287\right)\ln \frac{99.53\text{kPa}}{96.80\text{kPa}}\right]\\ =\left(1105\text{kg}\right)\times \left[\left(1.005\right)\ln \left(0.908197\right)-\left(0.287\right)\ln \left(1.028202\text{kPa}\right)\right]\\ =-38.34\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $395\text{kg}$ for $m_R$, $0.92927\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$, and $0.4045\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4$ in Equation (XVI).

$\begin{array}{c}\Delta S_{R,41}=\left(395\text{kg}\right)\left(0.92927-0.4045\right)\text{kJ}/\text{kg}\cdot \text{K}\\ =\left(395\text{kg}\right)\left(0.52477\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =207.28\text{kJ}/\text{K}\\ \cong 207.3\text{kJ}/\text{K}\end{array}$

Substitute $207.3\text{kJ}/\text{K}$ for $\Delta S_{R,41}$, $-165.2\text{kJ}/\text{K}$ for $\Delta S_{\text{vapor}}$, $-38.34\text{kJ}/\text{K}$ for $\Delta S_a$, and $8.14\text{kJ}/\text{K}$ for $S_w$ in Equation (XVII).

$\begin{array}{c}{S}_{\text{gen,evaporate}}=\left(207.3+\left(-165.2\right)+\left(-38.34\right)+8.14\right)\text{kJ}/\text{K}\\ =11.90\text{kJ}/\text{K}\end{array}$

Substitute $32°\text{C}$ for $T_0$ and $11.90\text{kJ}/\text{K}$ for $S_{\text{gen,evaporate}}$ in Equation (XVIII).

$\begin{array}{c}{X}_{\text{dest}}=\left(32°\text{C}\right)\left(11.90\text{kJ}/\text{K}\right)\\ =\left(32°\text{C+273}\right)\left(11.90\text{kJ}/\text{K}\right)\\ =3630\text{kJ}\end{array}$

Substitute 0 for $X_{\text{dest,compressor}}$, $1609\text{kJ}$ for $X_{\text{dest,condenser}}$, $1519\text{}\text{kJ}$ for $X_{\text{dest,throttle}}$, and $3630\text{KJ}$ for $X_{\text{dest,evaporator}}$ in Equation (XIX).

$\begin{array}{c}{X}_{\text{dest,total}}=\left(0+1609+1519+3630\right)\text{kJ}\\ =\text{6758}\text{kJ}\end{array}$

Thus, the exergy destruction in the total system per $1000{\text{m}}^{\text{3}}$ of dry air is $\underset{\_}{6758\text{kJ}}$.