Chapter 14.7, Problem 132RP

a)

To determine

The rate of dehumidification.

Answer to Problem 132RP

The mass flow rate of condensate water is $0.01861\text{kg}/\min$.

Explanation of Solution

Write the expression to obtain the vapor pressure at inlet conditions $\left(P_{v1}\right)$.

$\begin{array}{c}{P}_{v1}={\varphi }_1{P}_{g1}\\ ={\varphi }_1{P}_{\text{sat}@30\text{°C}}\end{array}$ (I)

Here, saturation pressure of water at $\text{30°C}$ is $P_{g1}$ and relative humidity at state 1 is ${\varphi }_1$.

Write the expression to obtain the atmospheric pressure of an ideal gas mixture $\left(P_1\right)$.

$P_1=P_{a1}+P_{v1}$ (II)

Here, dry air partial pressure at state 1 is $P_{a1}$.

Write the expression to obtain the specific volume of duct $\left(v_1\right)$.

$v_1=\frac{R_a{T}_1}{P_{a1}}$ (III)

Here, inlet temperature is $T_1$ and gas constant of air is $R_a$.

Write the expression to obtain the specific humidity $\left({\omega }_1\right)$ of air incoming.

${\omega }_1=\frac{0.622P_{v1}}{P_1-P_{v1}}$ (IV)

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

Write the expression to obtain the enthalpy at state 1 $\left(h_1\right)$.

$h_1=c_p{T}_1+{\omega }_1{h}_{g1}$ (V)

Here, specific heat of air is $c_p$, initial temperature is $T_1$, and initial condition of enthalpy at saturation vapor is $h_{g1}$.

Write the expression to obtain the vapor pressure at second inlet conditions $\left(P_{v2}\right)$.

$\begin{array}{c}{P}_{v2}={\varphi }_2{P}_{g2}\\ ={\varphi }_2{P}_{\text{sat}@20\text{°C}}\end{array}$ (VI)

Here, saturation pressure of water at $\text{20°C}$ is $P_{g2}$ and relative humidity at state 2 is ${\varphi }_2$.

Write the expression to obtain the specific humidity $\left({\omega }_2\right)$ of air incoming from second duct.

${\omega }_2=\frac{0.622P_{v2}}{P_2-P_{v2}}$ (VII)

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

Write the expression to obtain the enthalpy at state 2 $\left(h_2\right)$.

$h_2=c_p{T}_2+{\omega }_2{h}_{g2}$ (VIII)

Here, initial condition of enthalpy at saturation vapor at state 2 is $h_{g2}$.

Write the expression to obtain the mass flow rate of dry air $\left({\dot{m}}_{a1}\right)$ in each stream.

${\dot{m}}_{a1}=\frac{{\dot{V}}_1}{v_1}$ (IX)

Here, volume flow rate is ${\dot{V}}_1$.

Apply the water mass balance equation to the combined cooling and dehumidification section.

$\begin{array}{l}{\displaystyle \sum {\dot{m}}_{w,i}}={\displaystyle \sum {\dot{m}}_{w,e}}\\ {\dot{m}}_{a1}{\omega }_1={\dot{m}}_{a2}{\omega }_2+{\dot{m}}_w\\ {\dot{m}}_w={\dot{m}}_{a1}\left({\omega }_2-{\omega }_1\right)\end{array}$ (X)

Here, initial and final mass flow rate of dry air is ${\dot{m}}_{a1}$ and ${\dot{m}}_{a2}$, specific humidity of air at entering  and leaving the cooling coils is ${\omega }_1$ and ${\omega }_2$, and mass flow rate of vapor is ${\dot{m}}_w$.

Conclusion:

Refer Table A-4, “Saturated water – Temperature table”, obtain the properties of water at a temperature of $30°\text{C}$.

$\begin{array}{l}{P}_{sat@30°\text{C}}=4.247\text{ kPa}\\ h_{g1}=2,555.6\text{kJ}/\text{kg}\end{array}$

Substitute 4.247 kPa for $P_{\text{sat}@30\text{°C}}$ and 0.7 for ${\varphi }_1$ in Equation (I).

$\begin{array}{c}{P}_{v1}=\left(0.7\right)\left(4.247\text{kPa}\right)\\ =2.973\text{kPa}\end{array}$

Refer Table A-5, “Saturated water – Pressure table”, obtain the properties of water at a pressure of $\text{2}\text{.973}\text{kPa}$.

$T_{\text{dp}}=23.9°\text{C}$

Rewrite Equation (II) and substitute 90 kPa for $P_1$ and 2.973 kPa for $P_{v1}$.

$\begin{array}{c}{P}_{a1}=P_1-P_{v1}\\ =90\text{kPa}-2.973\text{kPa}\\ =87.03\text{kPa}\end{array}$

Convert the unit of $T_1$ from $\text{°C}$ to $\text{K}$.

$\begin{array}{c}{T}_1=30\text{°C}\\ =\left(30+273\right)\text{K}\\ =303\text{K}\end{array}$

Refer Table A-2, “Ideal gas specific heats of various common gases”, obtain the value of $R_a$ for air as $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ and $c_p$ for air as $1.005\text{kJ}/\text{kg}\cdot °\text{C}$.

Substitute $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R_a$, 303 K for $T_1$, and 87.03 kPa for $P_{a1}$ in Equation (III).

$\begin{array}{c}{v}_1=\frac{\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(303\text{K}\right)}{\left(87.03\text{kPa}\right)}\\ =0.9992{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute 2.973 kPa for $P_{v1}$ and 90 kPa for $P_1$ in Equation (IV).

$\begin{array}{c}{\omega }_1=\frac{0.622\left(2.973\text{kPa}\right)}{90\text{kPa}-2.973\text{kPa}}\\ =0.02125\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $\text{30°C}$ for $T_1$, $0.02125\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$, and $2,555.6\text{kJ}/\text{kg}$ for $h_{g1}$ in Equation (V).

$\begin{array}{c}{h}_1=\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(\text{30°C}\right)+\left[\begin{array}{l}\left(0.02125\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}\right)\\ \left(2,555.6\text{kJ}/\text{kg}\right)\end{array}\right]\\ =84.45\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Refer Table A-4, “Saturated water – Temperature table”, obtain the properties of water at a temperature of $20°\text{C}$.

$\begin{array}{l}{P}_{sat@20°\text{C}}=2.3392\text{ kPa}\\ h_{g2}=2,537.4\text{kJ}/\text{kg}\end{array}$

Substitute 2.3392 kPa for $P_{\text{sat}@20\text{°C}}$ and 1.00 for ${\varphi }_2$ in Equation (VI).

$\begin{array}{c}{P}_{v2}=\left(1.00\right)\left(2.3392\text{kPa}\right)\\ =2.3392\text{kPa}\end{array}$

Substitute 2.3392 kPa for $P_{v2}$ and 95 kPa for $P_2$ in Equation (VII).

$\begin{array}{c}{\omega }_2=\frac{0.622\left(2.3392\text{kPa}\right)}{95\text{kPa}-2.3392\text{kPa}}\\ =0.01660\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $\text{20°C}$ for $T_2$, $0.01660\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$, and $2,537.4\text{kJ}/\text{kg}$ for $h_{g2}$ in Equation (VIII).

$\begin{array}{c}{h}_2=\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(\text{20°C}\right)+\left[\begin{array}{l}\left(0.01660\text{kg}{\text{H}}_2\text{O}/\text{kg}\text{dry}\text{air}\right)\\ \left(2,537.4\text{kJ}/\text{kg}\right)\end{array}\right]\\ =62.22\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $4{\text{m}}^{\text{3}}/\min$ for ${\dot{V}}_1$ and $0.9992{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}$ for $v_1$ in Equation (IX).

$\begin{array}{c}{\dot{m}}_{a1}=\frac{4{\text{m}}^{\text{3}}/\min }{0.9992{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}}\\ =4.003\text{kg}/\min \end{array}$

Substitute $4.003\text{kg}/\min$ for ${\dot{m}}_{a1}$, $0.01660\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$, and $0.02125\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ in Equation (X).

$\begin{array}{c}{\dot{m}}_w=4.003\text{kg}/\min \left(\begin{array}{l}0.02125\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\\ -0.01660\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}\right)\\ =0.01861\text{kg}/\min \end{array}$

Thus, the mass flow rate of condensate water is $0.01861\text{kg}/\min$.

b)

To determine

The rate of heat transfer.

Answer to Problem 132RP

The rate of heat transfer of air stream is $87.44\text{kJ}/\min$.

Explanation of Solution

Apply the energy balance equation to the combined cooling and dehumidification section.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}{}^{↵\left(\text{steady}\right)}\\ {\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\end{array}$

$\begin{array}{l}{\displaystyle \sum {\dot{m}}_i{h}_i}={\dot{Q}}_{\text{out}}+{\displaystyle \sum {\dot{m}}_e{h}_e}\\ {\dot{Q}}_{\text{out}}={\dot{m}}_{a1}\left(h_1-h_2\right)-{\dot{m}}_w{h}_w\end{array}$ (XI)

Here, the amount of energy rate required for cooling coils is ${\dot{E}}_{\text{in}}$, amount of rate of energy rejected from cooling coils is ${\dot{E}}_{\text{out}}$, change in the energy rate of a system is $\Delta {\dot{E}}_{\text{system}}$, and rate of heat removal from air is ${\dot{Q}}_{\text{out}}$.

Conclusion:

Refer Table A-4, “Saturated water – Temperature table”, obtain the value of enthalpy at saturation liquid $\left(h_{f@20\text{°C}}\approx h_{w2}\right)$ as $83.915\text{kJ}/\text{kg}\text{dry air}$ at a temperature of $20°\text{C}$.

Substitute $83.915\text{kJ}/\text{kg}\text{dry air}$ for $h_w$, $0.01861\text{kg}/\min$ for ${\dot{m}}_w$, $4.003\text{kg}/\min$ for ${\dot{m}}_{a1}$, $84.45\text{kJ}/\text{kg}\text{dry air}$ for $h_1$, and $62.22\text{kJ}/\text{kg}\text{dry air}$ for $h_2$ in Equation (XI).

$\begin{array}{c}{\dot{Q}}_{\text{out}}=\left(\begin{array}{l}4.003\text{kg}/\min \left(\begin{array}{l}84.45\text{kJ}/\text{kg}\text{dry air}\\ -62.22\text{kJ}/\text{kg}\text{dry air}\end{array}\right)\\ -\left(0.01861\text{kg}/\text{min}\right)\left(83.915\text{kJ}/\text{kg}\text{dry air}\right)\end{array}\right)\\ =87.44\text{kJ}/\min \end{array}$

Thus, rate of heat transfer of air stream is $87.44\text{kJ}/\min$.

c)

To determine

The mass flow rate of the refrigerant.

Answer to Problem 132RP

The mass flow rate of the refrigerant is $0.620\text{kg}/\min$.

Explanation of Solution

Write the expression to obtain the inlet enthalpy of the refrigerant $\left(h_3\right)$.

$h_3=h_f+x_3{h}_{fg}$ (XII)

Here, enthalpy evaporation is $h_{fg}$, enthalpy saturation liquid is $h_f$, and dryness fraction of refrigerant is $x_3$.

Write the expression to obtain the mass flow rate of a refrigerant $\left({\dot{m}}_R\right)$.

${\dot{m}}_R=\frac{{\dot{Q}}_R}{h_4-h_3}$ (XIII)

Here, heat lost by the air is ${\dot{Q}}_R$.

Conclusion:

Refer Table A-12, “Saturated refrigerant 134a – Pressure table”, obtain the properties of refrigerant 134a at a pressure of 700 kPa.

$\begin{array}{l}{h}_f=88.82\text{kJ}/\text{kg}\\ h_{fg}=176.26\text{kJ}/\text{kg}\end{array}$

Substitute $88.82\text{kJ}/\text{kg}$ for $h_f$, 0.2 for $x_3$, and $176.26\text{kJ}/\text{kg}$ for $h_{fg}$ in Equation (XII).

$\begin{array}{c}{h}_3=88.82\text{kJ}/\text{kg}+0.2\left(176.26\text{kJ}/\text{kg}\right)\\ =124.07\text{kJ}/\text{kg}\end{array}$

Refer Table A-12, “Saturated refrigerant 134a – Pressure table”, obtain the value of $\left(h_g=h_4\right)$ as $265.08\text{kJ}/\text{kg}$ at a pressure of 700 kPa.

Substitute $87.44\text{kJ}/\min$ for ${\dot{Q}}_R$, $124.07\text{kJ}/\text{kg}$ for $h_3$, and $265.08\text{kJ}/\text{kg}$ for $h_4$ in Equation (XIII).

$\begin{array}{c}{\dot{m}}_R=\frac{87.44\text{kJ}/\min }{265.08\text{kJ}/\text{kg}-124.07\text{kJ}/\text{kg}}\\ =0.620\text{kg}/\min \end{array}$

Thus, the mass flow rate of the refrigerant is $0.620\text{kg}/\min$.