Chapter 14.7, Problem 78P

Air enters a 40-cm-diameter cooling section at 1 atm, 32°C, and 70 percent relative humidity at 120 m/min. The air is cooled by passing it over a cooling coil through which cold water flows. The water experiences a temperature rise of 6°C. The air leaves the cooling section saturated at 20°C. Determine (a) the rate of heat transfer, (b) the mass flow rate of the water, and (c) the exit velocity of the airstream.

FIGURE P14–79

To determine

The rate of heat transfer.

Answer to Problem 78P

The rate of heat transfer is $\text{478}\text{.7}\text{kJ}/\text{min}$.

Explanation of Solution

Express the dew point temperature of the incoming air at a temperature of $\text{32°C}$.

$T_{\text{dp}}=T_{\text{sat@}\left({\varphi }_1\text{×}{P}_{\text{sat@32°C}}\right)}$ (I)

Here, the saturation pressure at temperature of $\text{32°C}$ is $P_{\text{sat@32°C}}$ and initial specific humidity is ${\varphi }_1$.

Express initial volume rate of air.

$\begin{array}{c}{\dot{\nu }}_1=V_1{A}_1\\ =V_1\left[\frac{\pi D^2}{4}\right]\end{array}$ (II)

Here, initial volume and area is $V_1\text{and}{A}_1$ respectively, and diameter is $D$.

Express the mass flow rate of air at inlet.

${\dot{m}}_{a1}=\frac{{\dot{\nu }}_1}{v_1}$ (III)

Here, initial specific volume is $v_1$.

As the process is a steady flow and thus the mass flow rate of dry air remains constant during the entire process.

$\begin{array}{c}{\dot{m}}_{a1}={\dot{m}}_{a2}\\ ={\dot{m}}_a\end{array}$

Here, mass flow rate of dry air at exit is ${\dot{m}}_{a2}$ and mass flow rate of dry air is ${\dot{m}}_a$.

Express water mass balance to the combined cooling to obtain the mass flow rate of water.

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

Here, mass flow rate of water at inlet and exit is ${\dot{m}}_{w,i}\text{and}{\dot{m}}_{w,e}$ respectively, specific humidity at state 1 and 2 is ${\omega }_1\text{and}{\omega }_2$ respectively and mass flow rate of water is ${\dot{m}}_w$.

Express the cooling rate when the condensate leaves the system by applying an energy balance on the humidifying section.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}\\ {\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_i{h}_i=}{\dot{Q}}_{\text{out}}+{\displaystyle \sum {\dot{m}}_e{h}_e}\end{array}$

$\begin{array}{c}{\dot{Q}}_{\text{out}}={\dot{m}}_{a1}{h}_1-\left({\dot{m}}_{a2}{h}_2+{\dot{m}}_w{h}_w\right)\\ ={\dot{m}}_{a1}\left(h_1-h_2\right)-{\dot{m}}_w{h}_w\end{array}$ (V)

Here, rate of heat rejected or cooling rate when the condensate leaves the system is ${\dot{Q}}_{\text{out}}$, the rate of total energy entering the system is ${\dot{E}}_{\text{in}}$, the rate of total energy leaving the system is ${\dot{E}}_{\text{out}}$, the rate of change in the total energy of the system is $\Delta {\dot{E}}_{\text{system}}$, initial and exit mass flow rate is ${\dot{m}}_i\text{and}{\dot{m}}_e$ respectively, enthalpy at inlet and exit is $h_i\text{and}{h}_e$ respectively, enthalpy at state 1 and 2 is $h_1\text{and}{h}_2$ respectively, and enthalpy of water is $h_w$.

Conclusion:

Refer Table A-4, “saturated water-temperature table”, and write saturation pressure at temperature of $\text{32°C}$ using an interpolation method.

Write the formula of interpolation method of two variables.

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

Here, the variables denote by x and y is temperature and saturation pressure respectively.

Show the saturation pressure corresponding to temperature as in Table (1).

Temperature $T\left(°\text{C}\right)$	Saturation pressure $P_{\text{sat}}\left(\text{kPa}\right)$
30 $\left(x_1\right)$	4.2469 $\left(y_1\right)$
32 $\left(x_2\right)$	$\left(y_2=?\right)$
35 $\left(x_3\right)$	5.6291 $\left(y_3\right)$

Substitute $\text{30°C,}\text{32°C}\text{and}\text{35°C}$ for $x_1,x_2,\text{and}{x}_3$ respectively, $\text{4}\text{.2469}\text{kPa}$ for $y_1$ and $\text{5}\text{.6291}\text{kPa}$ for $y_3$ in Equation (VI).

$\begin{array}{c}{y}_2=\frac{\left(\text{32°C}-\text{30°C}\right)\left(\text{5}\text{.6291}\text{kPa}-\text{4}\text{.2469}\text{kPa}\right)}{\left(\text{35°C}-\text{30°C}\right)}+\text{4}\text{.2469}\text{kPa}\\ =\text{4}\text{.76}\text{kPa}\\ =P_{\text{sat@32°C}}\end{array}$

Thus, the saturation pressure at temperature of $\text{32°C}$ is,

$P_{\text{sat@32°C}}=\text{4}\text{.76}\text{kPa}$

Substitute $0.7$ for ${\varphi }_1$ and $\text{4}\text{.76}\text{kPa}$ for $P_{\text{sat@32°C}}$ in Equation (I).

$\begin{array}{c}{T}_{\text{dp}}=T_{\text{sat@}\left(0.7\text{×4}\text{.76}\text{kPa}\right)}\\ =T_{\text{sat@}3.33\text{kPa}}\end{array}$ (VII)

Here, saturation temperature at pressure of $\text{3}\text{.33}\text{kPa}$ is $T_{\text{sat@}3.33\text{kPa}}$.

Refer Table A-5 , “saturated water-pressure table”, and write saturation temperature at pressure of $\text{3}\text{.33}\text{kPa}$ using an interpolation method.

Show the saturation temperature corresponding to pressure as in Table (2).

Pressure $P\left(\text{kPa}\right)$	Saturation temperature $T_{\text{sat}}\left(°\text{C}\right)$
3 $\left(x_1\right)$	24.08 $\left(y_1\right)$
3.33 $\left(x_2\right)$	$\left(y_2=?\right)$
4 $\left(x_3\right)$	28.96 $\left(y_3\right)$

Use excels and tabulates the values from Table (2) in Equation (VI) to get,

$T_{\text{sat@}3.33\text{kPa}}=25.8\text{°C}$

Substitute $25.8\text{°C}$ for $T_{\text{sat@}3.33\text{kPa}}$ in Equation (VII).

$T_{\text{dp}}=25.8\text{°C}$

Refer Figure A-31, “psychometric chart at $\text{1 atm}$ total pressure”, and write the properties corresponding to dry bulb temperature of $\text{32°C}$, dew point temperature of $\text{25}\text{.8°C}$ and relative humidity of $70\%$.

$\begin{array}{l}{h}_1=86.35\text{kJ}/\text{kg}\text{dry}\text{air}\\ {\omega }_1=0.02114\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\\ {\nu }_1=0.8939{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\end{array}$

Refer Figure A-31, “psychometric chart at $\text{1 atm}$ total pressure”, and write the properties corresponding to dry bulb temperature of $\text{20°C}$ and dew point temperature of $\text{25}\text{.8°C}$.

$\begin{array}{l}{h}_2=57.43\text{kJ}/\text{kg}\text{dry}\text{air}\\ {\omega }_2=0.0147\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\\ {\nu }_2=0.8501{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\end{array}$

Here, final specific volume is ${\nu }_2$.

Refer Table A-4, “saturated water-temperature table”, and write the enthalpy of water at temperature of $\text{20°C}$.

$\begin{array}{c}{h}_w=h_{f\text{@20°C}}\\ =83.915\text{kJ}/\text{kg}\end{array}$

Here, specific enthalpy saturation liquid at temperature of $\text{20°C}$ is $h_{f\text{@20°C}}$.

Perform unit conversion of diameter from $\text{cm}\text{to}\text{m}$.

$\begin{array}{c}D=40\text{cm}\\ =40\text{cm}\left[\frac{\text{m}}{\text{100}\text{cm}}\right]\\ =\text{0}\text{.4}\text{m}\end{array}$

Substitute $\text{120}\text{m}/\text{min}$ for $V_1$ and $\text{0}\text{.4}\text{m}$ for $D$ in Equation (II).

$\begin{array}{c}{\dot{\nu }}_1=\left(\text{120}\text{m}/\text{min}\right)\left[\frac{\pi {\left(\text{0}\text{.4}\text{m}\right)}^2}{4}\right]\\ =15.08{\text{m}}^{\text{3}}/\text{min}\end{array}$

Substitute $15.08{\text{m}}^{\text{3}}/\text{min}$ for ${\dot{\nu }}_1$ and $0.8939{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}$ for ${\nu }_1$ in Equation (III).

$\begin{array}{c}{\dot{m}}_{a1}=\frac{15.08{\text{m}}^{\text{3}}/\text{min}}{0.8939{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}}\\ =16.87\text{kg}/\text{min}\end{array}$

Substitute $16.87\text{kg}/\text{min}$ for ${\dot{m}}_{a1}$, $0.02114\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ and $0.0147\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (IV).

$\begin{array}{c}{\dot{m}}_w=\left(16.87\text{kg}/\text{min}\right)\left(0.02114-0.0147\right)\\ =0.1086\text{kg}/\text{min}\end{array}$

Substitute $16.87\text{kg}/\text{min}$ for ${\dot{m}}_{a1}$, $86.35\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_1$, $57.43\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_2$, $0.1086\text{kg}/\text{min}$ for ${\dot{m}}_w$ and $83.915\text{kJ}/\text{kg}$ for $h_w$ in Equation (V).

$\begin{array}{c}{\dot{Q}}_{\text{out}}=\left[\begin{array}{c}\left(16.87\text{kg}/\text{min}\right)\left[\left(86.35-57.43\right)\text{kJ}/\text{kg}\text{dry}\text{air}\right]-\\ \left(0.1086\text{kg}/\text{min}\right)\left(83.915\text{kJ}/\text{kg}\right)\end{array}\right]\\ =478.7\text{kJ}/\text{min}\end{array}$

Hence, the rate of heat transfer is $\text{478}\text{.7}\text{kJ}/\text{min}$.

To determine

The mass flow rate of the water.

Answer to Problem 78P

The mass flow rate of the water is $\text{19}\text{.09}\text{kg}/\text{min}$.

Explanation of Solution

Express the mass flow rate of the water.

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

Here, mass flow rate of the water is ${\dot{Q}}_w$, specific heat at constant pressure of water is $c_p$ and rise in temperature is $\Delta T$.

Conclusion:

Refer Table A-2, “ideal-gas specific heats of various common gases”, and write specific heat at constant pressure of water.

$c_p=4.18\text{kJ}/\text{kg}\cdot \text{°C}$

Substitute $\text{478}\text{.7}\text{kJ}/\text{min}$ for ${\dot{Q}}_w$, $4.18\text{kJ}/\text{kg}\cdot \text{°C}$ for $c_p$ and $\text{6°C}$ for $\Delta T$ in Equation (VIII).

$\begin{array}{c}{\dot{m}}_{\text{cooling}\text{water}}=\frac{\text{478}\text{.7}\text{kJ}/\text{min}}{\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(\text{6°C}\right)}\\ =19.09\text{kg}/\text{min}\end{array}$

Hence, the mass flow rate of the water is $\text{19}\text{.09}\text{kg}/\text{min}$.

To determine

The exit velocity of the airstream.

Answer to Problem 78P

The exit velocity of the airstream is $\text{114}\text{m}/\text{min}$.

Explanation of Solution

Express the exit velocity of the airstream.

$V_2=\frac{{\nu }_2}{{\nu }_1}{V}_1$ (IX)

Conclusion:

Substitute $0.8939{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}$ for ${\nu }_1$, $0.8501{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}$ for ${\nu }_2$ and $\text{120}\text{m}/\text{min}$ for $V_1$ in Equation (IX).

$\begin{array}{c}{V}_2=\frac{0.8501{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}}{0.8939{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}}\left(\text{120}\text{m}/\text{min}\right)\\ =\text{114}\text{m}/\text{min}\end{array}$

Hence, the exit velocity of the airstream is $\text{114}\text{m}/\text{min}$.