Chapter 14.7, Problem 81P

a)

To determine

The rate of heat transfer of air stream.

Answer to Problem 81P

The rate of heat transfer of air stream is $250.9\text{Btu}/\min$.

Explanation of Solution

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

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

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 $\left({\dot{m}}_a\right)$.

${\dot{m}}_1=\frac{{\dot{V}}_1}{v_1}$ (II)

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.

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

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.

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

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

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 $\text{90°F}$.

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

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

Refer Fig A-31E, “Psychrometric chart at 1 atm total pressure”, at inlet temperature $\left(T_1=90\text{°F}\right)$ and relative humidity $\left(\varphi =60\%\right)$, write the value of enthalpy at state 1 $\left(h_1\right)$ as $41.8\text{Btu}/\text{lbm}\text{dry air}$, specific humidity of air entering the cooling section $\left({\omega }_1\right)$ as $0.0183\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$, and specific volume $\left(v_1\right)$ as $14.26{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}$.

Refer Fig A-31E, “Psychrometric chart at 1 atm total pressure”, at exit temperature $\left(T_2=70\text{°F}\right)$ and relative humidity $\left({\varphi }_2=100\%\right)$, write the value of enthalpy at state 2 $\left(h_2\right)$ as $34.1\text{Btu}/\text{lbm}\text{dry air}$, and specific humidity of air leaving the cooling section $\left({\omega }_2\right)$ as $0.0158\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ and specific volume $\left(v_2\right)$ as $13.68{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}$.

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 (I).

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

Substitute $471{\text{ft}}^{\text{3}}/\min$ for ${\dot{V}}_1$, and $14.26{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}$ for $v_1$ in Equation (II).

$\begin{array}{c}{\dot{m}}_{a1}=\frac{471{\text{ft}}^{\text{3}}/\min }{14.26{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}}\\ =33.0\text{lbm}/\min \end{array}$

Substitute $33.0\text{lbm}/\min$ for ${\dot{m}}_a$, $0.0158\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_2$, and $0.0183\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_1$ in Equation (III).

$\begin{array}{c}{\dot{m}}_w=33.0\text{lbm}/\min \left(\begin{array}{l}0.0158\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}-\\ 0.0183\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\end{array}\right)\\ =0.083\text{lbm}/\min \end{array}$

Substitute $38.08\text{Btu}/\text{lbm}$ for $h_w$, $0.083\text{lbm}/\min$ for ${\dot{m}}_w$, $33.0\text{lbm}/\min$ for ${\dot{m}}_{a1}$, $41.8\text{Btu}/\text{lbm}\text{dry air}$ for $h_1$, and $34.1\text{Btu}/\text{lbm}\text{dry air}$ for $h_2$ in Equation (IV).

$\begin{array}{c}{\dot{Q}}_{\text{out}}=33.0\text{lbm}/\min \left(\begin{array}{l}41.8\text{Btu}/\text{lbm}\text{dry air}-\\ 34.1\text{Btu}/\text{lbm}\text{dry air}\end{array}\right)-\left(0.083\text{lbm}/\min \right)\left(38.08\text{Btu}/\text{lbm}\right)\\ =250.9\text{Btu}/\min \end{array}$

Thus, rate of heat transfer of air stream is $250.9\text{Btu}/\min$.

b)

To determine

The mass flow rate of cooling water.

Answer to Problem 81P

The mass flow rate of cooling water is $17.9\text{lbm}/\min$.

Explanation of Solution

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

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

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 $250.9\text{Btu}/\min$ for ${\dot{Q}}_w$ in Equation (V).

$\begin{array}{c}{\dot{m}}_{\text{cooling}\text{water}}=\frac{250.9\text{Btu}/\min }{\left(1.0\text{Btu}/\text{lbm}\cdot \text{°F}\right)\left(14\text{°F}\right)}\\ =17.9\text{lbm}/\min \end{array}$

Thus, the mass flow rate of cooling water is $17.9\text{lbm}/\min$.

c)

To determine

The exit velocity of the airstream.

Answer to Problem 81P

The exit velocity is $576\text{ft}/\min$.

Explanation of Solution

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

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

Conclusion:

Substitute $13.68{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}$ for $v_2$, $14.26{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}$ for $v_1$, and $600\text{ft}/\min$ for $V_1$ in Equation (VI).

$\begin{array}{c}{V}_2=\frac{\left(13.68{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}\right)}{\left(14.26{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}\right)}\left(600\text{ft}/\min \right)\\ =576\text{ft}/\min \end{array}$

Thus, the exit velocity is $576\text{ft}/\min$.