Chapter 14.7, Problem 83P

a)

To determine

The rate of heat transfer of air stream.

Answer to Problem 83P

The rate of heat transfer of air stream is $255.6\text{Btu}/\min$.

Explanation of Solution

Write the expression to obtain the vapor pressure of air $\left(P_{v1}\right)$.

$\begin{array}{c}{P}_{v1}={\varphi }_1{P}_{g1}\\ ={\varphi }_1{P}_{\text{sat@90}°\text{F}}\end{array}$ (I)

Here, saturated pressure is $P_{\text{sat}}$, relative humidity is ${\varphi }_1$, and saturation pressure of water is $P_{g1}$.

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, vapor pressure at state 1 is $P_{v1}$ and dry air partial pressure at state 1 is $P_{a1}$.

Write the expression to obtain the specific volume of circular 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 incoming air.

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

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 exit vapor pressure of air $\left(P_{v2}\right)$.

$\begin{array}{c}{P}_{v2}={\varphi }_2{P}_{g2}\\ ={\varphi }_2{P}_{\text{sat@70}°\text{F}}\end{array}$ (VI)

Here, saturated pressure is $P_{\text{sat}}$, relative humidity is ${\varphi }_1$, and saturation pressure of water is $P_{g1}$.

Write the expression to obtain the atmospheric pressure of an ideal gas mixture $\left(P_2\right)$ at outlet conditions.

$P_2=P_{a2}+P_{v2}$ (VII)

Here, vapor pressure at state 2 is $P_{v2}$ and dry air partial pressure at state 2 is $P_{a2}$.

Write the expression to obtain the specific volume of circular duct $\left(v_2\right)$ at outlet conditions.

$v_2=\frac{R_a{T}_2}{P_{a2}}$ (VIII)

Here, final temperature is $T_2$ and gas constant of air is $R_a$.

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

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

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

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

Here, specific heat of air is $c_p$, final temperature is $T_2$, and final condition of enthalpy at saturation vapor is $h_{g2}$.

Write the expression to obtain the volume flow rate $\left({\dot{V}}_1\right)$.

$\begin{array}{c}{\dot{V}}_1=V_1{A}_1\\ =V_1\left(\frac{\pi D^2}{4}\right)\end{array}$ (XI)

Here, diameter is D, and velocity of air is $V_1$.

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

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

Here, volume flow rate of air entering the cooling section is $V_1$.

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

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

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

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

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

$\begin{array}{l}{\displaystyle \sum {\dot{m}}_i{h}_i}={\dot{Q}}_{\text{out}}+{\displaystyle \sum {\dot{m}}_e{h}_e}\\ {\dot{Q}}_{\text{out}}={\dot{m}}_{a1}{h}_1-\left({\dot{m}}_{a2}{h}_2+{\dot{m}}_w{h}_w\right)\\ {\dot{Q}}_{\text{out}}={\dot{m}}_{a1}\left(h_1-h_2\right)-{\dot{m}}_w{h}_w\end{array}$ (XIV)

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-4E, “Saturated water – Temperature table”, write the value of saturation pressure of liquid $\left(P_v\right)$ as 0.69904 psia at a temperature of $\text{90°F}$.

Substitute 0.69904 psia for $P_{\text{sat@90}°\text{F}}$ and 0.6 for ${\varphi }_1$ in Equation (I).

$\begin{array}{c}{P}_{v1}=\left(0.6\right)\left(0.69904\text{psia}\right)\\ =0.42\text{psia}\end{array}$

Refer Table A-5E, “Saturated water – Temperature table”, obtain the properties of water at a temperature of $\text{90°F}$.

$\begin{array}{c}{T}_{\text{dp}}=T_{\text{sat@}{P}_v}\\ =T_{\text{sat@0}\text{.69904 psia}}\\ =74.2°\text{F}\end{array}$

Rewrite Equation (II) and substitute 14.4 psia for $P_1$ and $0.42\text{psia}$ for $P_{v1}$ in Equation (II).

$\begin{array}{c}{P}_{a1}=P_1-P_{v1}\\ =14.4\text{psia}-0.42\text{psia}\\ =13.98\text{psia}\end{array}$

Convert the unit of $T_1$ from $\text{°F}$ to $\text{R}$.

$\begin{array}{c}{T}_1=90\text{°F}\\ =\left(90+460\right)\text{R}\\ =550\text{R}\end{array}$

Refer the Table “Ideal gas specific heats of various common gases”, write the value of $R_a$ for air as $0.3707\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$.

Substitute $0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R_a$, 550 R for $T_1$, and 13.98 psia for $P_{a1}$ in Equation (III).

$\begin{array}{c}{v}_1=\frac{\left(0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(550\text{R}\right)}{\left(13.98\text{psia}\right)}\\ =14.57{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}\end{array}$

The mass flow rate of dry air remains constant, but the amount of moisture in the air decreases due to the dehumidification $\left({\omega }_2<{\omega }_1\right)$.

Substitute 0.42 psia for $P_{v1}$ and 14.4 psia for $P_1$ in Equation (IV).

$\begin{array}{c}{\omega }_1=\frac{0.622\left(0.42\text{psia}\right)}{14.4\text{psia}-0.42\text{psia}}\\ =0.0187\text{lbm}{\text{H}}_2\text{O}/\text{lbm}\text{dry}\text{air}\end{array}$

Refer Table A-4E, “Saturated water – Temperature table”, write the value of $h_{g1}$ as $1,100.4\text{Btu}/\text{lbm}$ at a temperature of $\text{90°F}$.

Refer the Table “Ideal gas specific heats of various common gases”, write the value of $c_p$ for air as $0.24\text{Btu}/\text{lbm}\cdot °\text{F}$.

Substitute $0.24\text{Btu}/\text{lbm}\cdot °\text{F}$ for $c_p$, $90\text{°F}$ for $T_1$, $0.0187\text{lbm}{\text{H}}_2\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_1$, and $1,100.4\text{Btu}/\text{lbm}$ for $h_{g1}$ in Equation (V).

$\begin{array}{c}{h}_1=\left(0.24\text{Btu}/\text{lbm}\cdot °\text{F}\right)\left(\text{90°F}\right)+\left[\begin{array}{l}\left(0.0187\text{lbm}{\text{H}}_2\text{O}/\text{lbm}\text{dry}\text{air}\right)\\ \left(1,100.4\text{Btu}/\text{lbm}\right)\end{array}\right]\\ =42.18\text{Btu}/\text{lbm}\text{dry}\text{air}\end{array}$

Refer Table A-4E, “Saturated water – Temperature table”, write the value of saturated pressure $\left(P_{\text{sat}}\right)$ as 0.36 psia at a temperature of $\text{70°F}$.

Substitute 0.36 psia for $P_{\text{sat@70}°\text{F}}$ and 1.00 for $\varphi$ in Equation (VI).

$\begin{array}{c}{P}_{v2}=\left(1.00\right)\left(0.36\text{psia}\right)\\ =0.36\text{psia}\end{array}$

Rewrite Equation (VII) and substitute 14.4 psia for $P_2$ and 0.36 psia for $P_{v2}$.

$\begin{array}{c}{P}_{a2}=P_2-P_{v2}\\ =14.4\text{psia}-0.36\text{psia}\\ =14.04\text{psia}\end{array}$

Convert the unit of $T_2$ from $\text{°F}$ to $\text{R}$.

$\begin{array}{c}{T}_2=70\text{°F}\\ =\left(70+460\right)\text{R}\\ =530\text{R}\end{array}$

Substitute $0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R_a$, 530 R for $T_1$, and 14.04 psia for $P_{a2}$ in Equation (VIII).

$\begin{array}{c}{v}_2=\frac{\left(0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(530\text{R}\right)}{\left(14.04\text{psia}\right)}\\ =14.00{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}\end{array}$

Substitute 0.36 psia for $P_{v2}$ and 14.4 psia for $P_2$ in Equation (IX).

$\begin{array}{c}{\omega }_2=\frac{0.622\left(0.36\text{psia}\right)}{14.4\text{psia}-0.36\text{psia}}\\ =0.0159\text{lbm}{\text{H}}_2\text{O}/\text{lbm}\text{dry}\text{air}\end{array}$

Refer Table A-4E, “Saturated water – Temperature table”, write the value of $h_{g2}$ as $1,091.8\text{Btu}/\text{lbm}$ at a temperature of $\text{70°F}$.

Substitute $0.24\text{Btu}/\text{lbm}\cdot °\text{F}$ for $c_p$, $70\text{°F}$ for $T_2$, $0.0159\text{lbm}{\text{H}}_2\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_2$, and $1,091.8\text{Btu}/\text{lbm}$ for $h_{g2}$ in Equation (X).

$\begin{array}{c}{h}_2=\left(0.24\text{Btu}/\text{lbm}\cdot °\text{F}\right)\left(\text{70°F}\right)+\left[\begin{array}{l}\left(0.0159\text{lbm}{\text{H}}_2\text{O}/\text{lbm}\text{dry}\text{air}\right)\\ \left(1,091.8\text{Btu}/\text{lbm}\right)\end{array}\right]\\ =34.16\text{Btu}/\text{lbm}\text{dry}\text{air}\end{array}$

Refer Table A-4E, “Saturated water – Temperature table”, write the value of enthalpy at saturation liquid $\left(h_{f@70\text{°F}}\approx h_w\right)$ as $38.08\text{Btu}/\text{lbm}$ at a temperature of $\text{70°F}$.

Substitute 1 ft for D and $600\text{ft}/\min$ for $V_1$ in Equation (XI).

$\begin{array}{c}{\dot{V}}_1=600\text{ft}/\min \left(\frac{\pi {\left(1\text{ft}\right)}^2}{4}\right)\\ =471{\text{ft}}^{\text{3}}/\min \end{array}$

Substitute $471{\text{ft}}^{\text{3}}/\min$ for ${\dot{V}}_1$, and $14.57{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}$ for $v_1$ in Equation (XII).

$\begin{array}{c}{\dot{m}}_{a1}=\frac{471{\text{ft}}^{\text{3}}/\min }{14.57{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}}\\ =32.3\text{lbm}/\min \end{array}$

Substitute $32.3\text{lbm}/\min$ for ${\dot{m}}_{a1}$, $0.0159\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_2$, and $0.0187\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_1$ in Equation (XIII).

$\begin{array}{c}{\dot{m}}_w=32.3\text{lbm}/\min \left(\begin{array}{l}0.0159\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\\ -0.0187\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\end{array}\right)\\ =0.0904\text{lbm}/\min \end{array}$

Substitute $38.08\text{Btu}/\text{lbm}$ for $h_w$, $0.0904\text{lbm}/\min$ for ${\dot{m}}_w$, $32.3\text{lbm}/\min$ for ${\dot{m}}_{a1}$, $42.18\text{Btu}/\text{lbm}\text{dry air}$ for $h_1$, and $34.16\text{Btu}/\text{lbm}\text{dry air}$ for $h_2$ in Equation (XIV).

$\begin{array}{c}{\dot{Q}}_{\text{out}}=\left(\begin{array}{l}32.3\text{lbm}/\min \left(\begin{array}{l}42.18\text{Btu}/\text{lbm}\text{dry air}-\\ 34.16\text{Btu}/\text{lbm}\text{dry air}\end{array}\right)\\ -\left(0.0904\text{lbm}/\min \right)\left(38.08\text{Btu}/\text{lbm}\right)\end{array}\right)\\ =255.6\text{Btu}/\min \end{array}$

Thus, rate of heat transfer of air stream is $255.6\text{Btu}/\min$.

b)

To determine

The mass flow rate of cooling water.

Answer to Problem 83P

The mass flow rate of cooling water is $18.3\text{lbm}/\min$.

Explanation of Solution

Write the expression to obtain the mass flow rate of cooling water $\left({\dot{m}}_{\text{cooling}\text{water}}\right)$.

${\dot{m}}_{\text{cooling}\text{water}}=\frac{{\dot{Q}}_w}{c_p\Delta T}$ (XV)

Here, specific heat of water is $c_p$, rate of heat transfer of water is ${\dot{Q}}_w$, and change in temperature of air is $\Delta T$.

Conclusion:

Substitute $1.0\text{Btu}/\text{lbm}\cdot \text{°F}$ for $c_p$, $14\text{°F}$ for $\Delta T$, and $255.6\text{Btu}/\min$ for ${\dot{Q}}_w$ in Equation (XV).

$\begin{array}{c}{\dot{m}}_{\text{cooling}\text{water}}=\frac{255.6\text{Btu}/\min }{\left(1.0\text{Btu}/\text{lbm}\cdot \text{°F}\right)\left(14\text{°F}\right)}\\ =18.3\text{lbm}/\min \end{array}$

Thus, the mass flow rate of cooling water is $18.3\text{lbm}/\min$.

c)

To determine

The exit velocity of the airstream.

Answer to Problem 83P

The exit velocity is $577\text{ft}/\min$.

Explanation of Solution

Apply the conservation of mass of dry air to calculate exit velocity $\left(V_2\right)$.

$\begin{array}{l}{\dot{m}}_{a1}={\dot{m}}_{a2}\\ \frac{V_{1}A}{v_1}=\frac{V_{2}A}{v_2}\\ V_2=\frac{v_2}{v_1}{V}_1\end{array}$ (XVI)

Conclusion:

Substitute $14{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}$ for $v_2$, $14.57{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}$ for $v_1$, and $600\text{ft}/\min$ for $V_1$ in Equation (XVI).

$\begin{array}{c}{V}_2=\frac{\left(14{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}\right)}{\left(14.57{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}\right)}\left(600\text{ft}/\min \right)\\ =577\text{ft}/\min \end{array}$

Thus, the exit velocity is $577\text{ft}/\min$.