Chapter 14.7, Problem 80P

Repeat Prob. 14–79 for a total pressure of 88 kPa for air.

To determine

The rate of heat transfer.

Answer to Problem 80P

The rate of heat transfer is $\text{451}\text{.7}\text{kJ}/\text{min}$.

Explanation of Solution

Express initial partial pressure.

$\begin{array}{c}{P}_{\nu 1}={\varphi }_1{P}_{g1}\\ ={\varphi }_1{P}_{\text{sat@32°C}}\end{array}$ (I)

Here, relative humidity at state 1 is ${\varphi }_1$, initial vapor pressure is $P_{g1}$ and saturation pressure at temperature of $\text{32°C}$ is $P_{\text{sat@32°C}}$.

Express the dew point temperature of the incoming air.

$T_{\text{dp}}=T_{\text{sat@}{P}_{\nu 1}}$ (II)

Here, initial specific humidity is ${\varphi }_1$.

Express initial partial pressure.

$P_{a1}=P_1-P_{\nu 1}$ (III)

Here, initial pressure is $P_1$.

Express initial specific volume.

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

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

Express initial specific humidity.

${\omega }_1=\frac{0.622P_{\nu 1}}{P_1-P_{\nu 1}}$ (V)

Express initial enthalpy.

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

Here, initial specific enthalpy at saturated vapor is $h_{g1}$, specific heat at constant pressure of air is $c_p$ and temperature at state 1 is $T_1$.

Express final partial pressure.

$\begin{array}{c}{P}_{\nu 2}={\varphi }_2{P}_{g2}\\ ={\varphi }_2{P}_{\text{sat@20°C}}\end{array}$ (VII)

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

Express final partial pressure.

$P_{a2}=P_2-P_{\nu 2}$ (VIII)

Here, final pressure is $P_2$.

Express final specific volume.

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

Here, final temperature is $T_2$.

Express final specific humidity.

${\omega }_2=\frac{0.622P_{\nu 2}}{P_2-P_{\nu 2}}$ (X)

Express final enthalpy.

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

Here, final specific enthalpy at saturated vapor is $h_{g2}$, specific heat at constant pressure of air is $c_p$ and temperature at state 2 is $T_2$.

Express initial volume rate of air.

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

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.

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

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.

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

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.

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

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

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.

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

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 (XVI).

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

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).

$\begin{array}{c}{P}_{\nu 1}=\left(0.7\right)\left(\text{4}\text{.76}\text{kPa}\right)\\ =3.332\text{kPa}\end{array}$

Substitute $3.332\text{kPa}$ for $P_{\nu 1}$ in Equation (II).

$T_{\text{dp}}=T_{\text{sat@}3.332\text{kPa}}$ (XVII)

Here, saturation temperature at pressure of $\text{3}\text{.332}\text{kPa}$ is $T_{\text{sat@}3.332\text{kPa}}$.

Refer Table A-5, “saturated water-pressure table”, and write saturation temperature at pressure of $\text{3}\text{.332}\text{kPa}$ 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 $\left(y_1\right)$
3.332 $\left(x_2\right)$	$\left(y_2=?\right)$
4 $\left(x_3\right)$	28.96 $\left(y_3\right)$

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

$T_{\text{sat@}3.332\text{kPa}}=25.8\text{°C}$

Substitute $25.8\text{°C}$ for $T_{\text{sat@}3.332\text{kPa}}$ in Equation (XVII).

$T_{\text{dp}}=25.8\text{°C}$

Substitute $\text{88}\text{kPa}$ for $P_1$ and $3.332\text{kPa}$ for $P_{\nu 1}$ in Equation (III).

$\begin{array}{c}{P}_{a1}=\text{88}\text{kPa}-3.332\text{kPa}\\ =84.67\text{kPa}\end{array}$

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

$\begin{array}{l}{R}_a=0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\\ c_p=1.005\text{kJ}/\text{kg}\cdot \text{°C}\end{array}$

Substitute $0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}$ for $R_a$, $\text{32°C}$ for $T_1$ and $84.67\text{kPa}$ for $P_{a1}$ in Equation (IV).

$\begin{array}{c}{v}_1=\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\right)\left(\text{32°C}\right)}{84.67\text{kPa}}\\ =\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\right)\left[\left(32+273\right)\text{K}\right]}{84.67\text{kPa}}\\ =\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\right)\left(305\text{K}\right)}{84.67\text{kPa}}\\ =1.034{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $\text{88}\text{kPa}$ for $P_1$ and $3.332\text{kPa}$ for $P_{\nu 1}$ in Equation (V).

$\begin{array}{c}{\omega }_1=\frac{0.622\left(3.332\text{kPa}\right)}{85\text{kPa}-3.332\text{kPa}}\\ =0.02448\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

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

Show the specific enthalpy saturated vapor corresponding to temperature as in Table (3).

Temperature $T\left(°\text{C}\right)$	specific enthalpy saturated vapor $h_{g1}\left(\text{kJ}/\text{kg}\right)$
30 $\left(x_1\right)$	2555.6 $\left(y_1\right)$
32 $\left(x_2\right)$	$\left(y_2=?\right)$
35 $\left(x_3\right)$	2564.6 $\left(y_3\right)$

Use excels and tabulates the values from Table (3) in Equation (XVI) to get,

$\begin{array}{c}{h}_{g1@32\text{°C}}=h_{g1}\\ =2559.2\text{kJ}/\text{kg}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{°C}$ for $c_p$, $\text{32°C}$ for $T_1$, $0.02448$ for ${\omega }_1$ and $2559.2\text{kJ}/\text{kg}$ for $h_{g1}$ in Equation (VI).

$\begin{array}{c}{h}_1=\left(1.005\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(\text{32°C}\right)+\left(0.02448\right)\left(2559.2\text{kJ}/\text{kg}\right)\\ =94.80\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

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

$P_{\text{sat@20°C}}=2.339\text{kPa}$

Substitute $1$ for ${\varphi }_2$ and $2.339\text{kPa}$ for $P_{\text{sat@20°C}}$ in Equation (VII).

$\begin{array}{c}{P}_{\nu 2}=\left(1\right)\left(2.339\text{kPa}\right)\\ =2.339\text{kPa}\end{array}$

Substitute $\text{88}\text{kPa}$ for $P_2$ and $2.339\text{kPa}$ for $P_{\nu 2}$ in Equation (VIII).

$\begin{array}{c}{P}_{a2}=\text{88}\text{kPa}-2.339\text{kPa}\\ =85.661\text{kPa}\end{array}$

Substitute $0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}$ for $R_a$, $\text{20°C}$ for $T_2$ and $85.661\text{kPa}$ for $P_{a2}$ in Equation (IX).

$\begin{array}{c}{v}_2=\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\right)\left(\text{20°C}\right)}{85.661\text{kPa}}\\ =\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\right)\left[\left(20+273\right)\text{K}\right]}{85.661\text{kPa}}\\ =\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\right)\left(293\text{K}\right)}{85.661\text{kPa}}\\ =0.9817{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $\text{88}\text{kPa}$ for $P_2$ and $2.339\text{kPa}$ for $P_{\nu 1}$ in Equation (X).

$\begin{array}{c}{\omega }_2=\frac{0.622\left(2.339\text{kPa}\right)}{85\text{kPa}-2.339\text{kPa}}\\ =0.01699\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

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

$\begin{array}{c}{h}_{g2@20\text{°C}}=h_{g2}\\ =2537.4\text{kJ}/\text{kg}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{°C}$ for $c_p$, $\text{20°C}$ for $T_2$, $0.01699$ for ${\omega }_2$ and $2537.4\text{kJ}/\text{kg}$ for $h_{g2}$ in Equation (XI).

$\begin{array}{c}{h}_2=\left(1.005\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(\text{20°C}\right)+\left(0.01699\right)\left(2537.4\text{kJ}/\text{kg}\right)\\ =63.20\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

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 (XII).

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

Substitute $15.08{\text{m}}^{\text{3}}/\text{min}$ for ${\dot{\nu }}_1$ and $1.034{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}$ for ${\nu }_1$ in Equation (XIII).

$\begin{array}{c}{\dot{m}}_{a1}=\frac{15.08{\text{m}}^{\text{3}}/\text{min}}{1.034{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}}\\ =14.59\text{kg}/\text{min}\end{array}$

Substitute $14.59\text{kg}/\text{min}$ for ${\dot{m}}_{a1}$, $0.02448\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ and $0.01699\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (XIV).

$\begin{array}{c}{\dot{m}}_w=\left(14.59\text{kg}/\text{min}\right)\left(0.02448-0.01699\right)\\ =0.1093\text{kg}/\text{min}\end{array}$

Substitute $14.59\text{kg}/\text{min}$ for ${\dot{m}}_{a1}$, $94.80\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_1$, $63.20\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_2$, $0.1093\text{kg}/\text{min}$ for ${\dot{m}}_w$ and $83.915\text{kJ}/\text{kg}$ for $h_w$ in Equation (XV).

$\begin{array}{c}{\dot{Q}}_{\text{out}}=\left[\begin{array}{c}\left(16.87\text{kg}/\text{min}\right)\left[\left(94.80-63.20\right)\text{kJ}/\text{kg}\text{dry}\text{air}\right]-\\ \left(0.1093\text{kg}/\text{min}\right)\left(83.915\text{kJ}/\text{kg}\right)\end{array}\right]\\ =451.7\text{kJ}/\text{min}\end{array}$

Hence, the rate of heat transfer is $451.7\text{kJ}/\text{min}$.

To determine

The mass flow rate of the water.

Answer to Problem 80P

The mass flow rate of the water is $\text{18}\text{.01}\text{kg}/\text{min}$.

Explanation of Solution

Express the mass flow rate of the water.

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

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 $\text{451}\text{.7}\text{kJ}/\text{min}$ 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 (XVIII).

$\begin{array}{c}{\dot{m}}_{\text{cooling}\text{water}}=\frac{\text{451}\text{.7}\text{kJ}/\text{min}}{\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(\text{6°C}\right)}\\ =18.01\text{kg}/\text{min}\end{array}$

Hence, the mass flow rate of the water is $18.01\text{kg}/\text{min}$.

To determine

The exit velocity of the airstream.

Answer to Problem 80P

The exit velocity of the airstream is $\text{113}\text{.9}\text{m}/\text{min}$.

Explanation of Solution

Express the exit velocity of the airstream.

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

Conclusion:

Substitute $1.034{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}$ for ${\nu }_1$, $0.9817{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}$ for ${\nu }_2$ and $\text{120}\text{m}/\text{min}$ for $V_1$ in Equation (XIX).

$\begin{array}{c}{V}_2=\frac{0.9817{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}}{1.034{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}}\left(\text{120}\text{m}/\text{min}\right)\\ =\text{113}\text{.9}\text{m}/\text{min}\end{array}$

Hence, the exit velocity of the airstream is $\text{113}\text{.9}\text{m}/\text{min}$.