Chapter 14.7, Problem 137RP

(a)

To determine

The temperature and relative humidity of the air when it leaves the heating section.

Answer to Problem 137RP

The temperature is $\text{33}\text{.1°C}$ and relative humidity of the air when it leaves the heating section is $18.5\%$.

Explanation of Solution

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, the mass flow rate of air at inlet is ${\dot{m}}_{a1}$, mass flow rate of dry air at exit is ${\dot{m}}_{a2}$ and mass flow rate of dry air is ${\dot{m}}_a$.

The amount of moisture in the air remains constant as it flows through the heating section as process involves no dehumidification or humidification.

${\omega }_1={\omega }_2$ (I)

Here, specific humidity at state 1 and 2 is ${\omega }_1\text{and}{\omega }_2$ respectively.

Express initial pressure of vapor.

$\begin{array}{c}{P}_{\nu 1}={\varphi }_1{P}_{g1}\\ ={\varphi }_1{P}_{\text{sat@15°C}}\end{array}$ (II)

Here, relative humidity at state 1 is ${\varphi }_1$, initial pressure of water vapor is $P_{g1}$ and saturation pressure at temperature of $\text{15°C}$ is $P_{\text{sat@15°C}}$.

Express initial humidity ratio.

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

Here, pressure at state 1 is $P_1$.

Express initial enthalpy.

$h_1=c_p{T}_1+{\omega }_1{h}_{g1@15\text{°C}}$ (IV)

Here, specific heat at constant pressure is $c_p$ temperature at state 1 is $T_1$ and initial specific enthalpy saturated vapor at temperature of $\text{15°C}$ is $h_{g1\text{@15°C}}$.

Express specific volume at state 1.

$\begin{array}{c}{v}_1=\frac{R_a{T}_1}{P_{a1}}\\ =\frac{R_a{T}_1}{P_1-P_{\nu 1}}\end{array}$ (V)

Here, gas constant of air is $R_a$, partial pressure of air at state 1 is $P_{a1}$ and temperature at state 1 is $T_1$.

Express pressure vapor at state 3.

$\begin{array}{c}{P}_{\nu 3}={\varphi }_3{P}_{g3}\\ ={\varphi }_3{P}_{\text{sat@25°C}}\end{array}$ (VI)

Here, relative humidity at state 3 is ${\varphi }_3$, vapor pressure at state 3 is $P_{g3}$ and saturation pressure at temperature of $\text{25°C}$ is $P_{\text{sat@25°C}}$.

Express humidity ratio at state 3.

${\omega }_3=\frac{0.622P_{\nu 3}}{P_3-P_{\nu 3}}$ (VII)

Here, pressure at state 3 is $P_3$.

Express enthalpy at state 3.

$h_3=c_p{T}_3+{\omega }_3{h}_{g3@25\text{°C}}$ (VIII)

Here, specific heat at constant pressure is $c_p$, temperature at state 3 is $T_3$ and specific enthalpy saturated vapor at state 3 and temperature of $\text{25°C}$ is $h_{g3\text{@25°C}}$.

Express enthalpy at state 2.

$h_2\cong h_3$ (IX)

Express enthalpy at state 2 to obtain the temperature at heating section.

$\begin{array}{c}{h}_2=c_p{T}_2+{\omega }_3{h}_{g2}\\ =c_p{T}_2+{\omega }_3\left(2500.9+1.82T_2\right)\end{array}$ (X)

Here, temperature at heating section is $T_2$, specific heat at constant pressure is $c_p$, temperature at state 3 is $T_3$ and specific enthalpy saturated vapor at state 3 and temperature of $\text{25°C}$ is $h_{g3\text{@25°C}}$.

Express pressure of water vapor at state 2.

$P_{g2}=P_{\text{sat@}{T}_2}$ (XI)

Here, saturation pressure at temperature at leaving section is $P_{\text{sat@}{T}_2}$.

Express relative humidity at heating section.

${\varphi }_2=\frac{{\omega }_2{P}_2}{\left(0.622+{\omega }_2\right)P_{g2}}$ (XII)

Here, pressure at state 2 is $P_2$.

Conclusion:

Refer Table A-4, “saturated water-temperature table”, and write the saturation pressure and initial specific enthalpy saturated vapor at temperature of $\text{15°C}$.

$\begin{array}{l}{P}_{\text{sat@15°C}}=1.7057\text{kPa}\\ h_{g1\text{@15°C}}=2528.3\text{kJ}/\text{kg}\end{array}$

Substitute $0.55$ for ${\varphi }_1$ and $\text{1}\text{.7057}\text{kPa}$ for $P_{\text{sat@15°C}}$ in Equation (II).

$\begin{array}{c}{P}_{\nu 1}=\left(0.55\right)\left(\text{1}\text{.7057}\text{kPa}\right)\\ =\text{0}\text{.938}\text{kPa}\end{array}$

Substitute $0.938\text{kPa}$ for $P_{\nu 1}$ and $\text{96}\text{kPa}$ for $P_1$ in Equation (III).

$\begin{array}{c}{\omega }_1=\frac{0.622\left(\text{0}\text{.938}\text{kPa}\right)}{\text{96}\text{kPa}-\text{0}\text{.938}\text{kPa}}\\ =0.006138\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $0.006138\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ in Equation (I).

${\omega }_2=0.006138\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$

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

$\begin{array}{l}{c}_p=1.005\text{kJ}/\text{kg}\cdot °\text{C}\\ R_a=0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $\text{15°C}$ for $T_1$, $0.006138\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ and $2528.3\text{kJ}/\text{kg}$ for $h_{g1\text{@15°C}}$ in Equation (IV).

$\begin{array}{c}{h}_1=\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(\text{15°C}\right)+\left(0.00638\right)\left(2528.3\text{kJ}/\text{kg}\right)\\ =30.59\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $R_a$, $\text{15°C}$ for $T_1$, $\text{0}\text{.938}\text{kPa}$ for $P_{\nu 1}$ and $\text{96}\text{kPa}$ for $P_1$ in Equation (V).

$\begin{array}{c}{v}_1=\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(\text{15°C}\right)}{96\text{kPa}-0.938\text{kPa}}\\ =\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(\text{15}+273\right)\text{K}}{\text{95}\text{.06}\text{kPa}}\\ =\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(288\text{K}\right)}{\text{95}\text{.06}\text{kPa}}\\ =0.8695{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\end{array}$

Refer Table A-4, “saturated water-temperature table”, and write the saturation pressure and initial specific enthalpy saturated vapor at temperature of $\text{25°C}$.

$\begin{array}{l}{P}_{\text{sat@25°C}}=3.17\text{kPa}\\ h_{g3\text{@25°C}}=2546.5\text{kJ}/\text{kg}\end{array}$

Substitute $0.45$ for ${\varphi }_3$ and $\text{3}\text{.17}\text{kPa}$ for $P_{\text{sat@25°C}}$ in Equation (VI).

$\begin{array}{c}{P}_{\nu 3}=\left(0.45\right)\left(\text{3}\text{.17}\text{kPa}\right)\\ =\text{1}\text{.426}\text{kPa}\end{array}$

Substitute $1.426\text{kPa}$ for $P_{\nu 3}$ and $\text{96}\text{kPa}$ for $P_3$ in Equation (VII).

$\begin{array}{c}{\omega }_3=\frac{0.622\left(\text{1}\text{.426}\text{kPa}\right)}{\text{96}\text{kPa}-\text{1}\text{.426}\text{kPa}}\\ =0.009381\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $\text{25°C}$ for $T_3$, $0.009381\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_3$ and $2546.5\text{kJ}/\text{kg}$ for $h_{g3\text{@25°C}}$ in Equation (VIII).

$\begin{array}{c}{h}_3=\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(\text{25°C}\right)+\left(0.00938\right)\left(2546.5\text{kJ}/\text{kg}\right)\\ =49.01\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $49.01\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_3$ in Equation (IX).

$h_2=49.01\text{kJ}/\text{kg}\text{dry}\text{air}$

Substitute $49.01\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_2$, $0.006138\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ and $1.005\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$ in Equation (X).

$\begin{array}{l}49.01\text{kJ}/\text{kg}\text{dry}\text{air}=\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)T_2+0.006138\left(2500.9+1.82T_2\right)\\ T_2=33.1°\text{C}\end{array}$

Hence, the temperature of the air when it leaves the heating section is $\text{33}\text{.1°C}$.

Substitute $\text{33}\text{.1°C}$ for $T_2$ in Equation (XI).

$P_{g2}=P_{\text{sat@33}\text{.1°C}}$ (XIII)

Refer Table A-4, “saturated water-temperature table”, and write the saturation temperature or exit temperature at temperature of $\text{33}\text{.1°C}$ using an interpolation method.

Write the formula of interpolation method of two variables.

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

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

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

Exit temperature $T_{\text{sat}}\left(\text{°C}\right)$	Saturation pressure $P_{\text{sat}}\left(\text{kPa}\right)$
30 $\left(x_1\right)$	4.2469 $\left(y_1\right)$
33.1 $\left(x_2\right)$	$\left(y_2=?\right)$
35 $\left(x_3\right)$	5.6291 $\left(y_3\right)$

Substitute $\text{30°C,}\text{33}\text{.1°C}\text{and}35\text{°C}$ 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 (XIV).

$\begin{array}{c}{y}_2=\frac{\left(\text{33}\text{.1°C}-\text{30°C}\right)\left(\text{5}\text{.6291}\text{kPa}-\text{4}\text{.2469}\text{kPa}\right)}{\left(35\text{°C}-\text{30°C}\right)}+\text{4}\text{.2469}\text{kPa}\\ =\text{5}\text{.072}\text{kPa}\\ =P_{\text{sat@33}\text{.1°C}}\end{array}$

Substitute $\text{5}\text{.072}\text{kPa}$ for $P_{\text{sat@33}\text{.1°C}}$ in Equation (XIII).

$P_{g2}=\text{5}\text{.072}\text{kPa}$

Substitute $0.006138\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$, $\text{96}\text{kPa}$ for $P_2$ and $\text{5}\text{.072}\text{kPa}$ for $P_{g2}$ in Equation (XII).

$\begin{array}{c}{\varphi }_2=\frac{\left(0.006138\right)\left(\text{96}\text{kPa}\right)}{\left(0.622+0.006138\right)\text{5}\text{.072}\text{kPa}}\\ =0.185\\ =18.5\%\end{array}$

Hence, the relative humidity of the air when it leaves the heating section is $18.5\%$.

(b)

To determine

The rate of heat transfer in the heating section.

Answer to Problem 137RP

The rate of heat transfer in the heating section is $\text{636}\text{kJ}/\text{min}$.

Explanation of Solution

Express the rate of heat transfer in the heating section.

${\dot{Q}}_{\text{in}}={\dot{m}}_a\left(h_2-h_1\right)$ (XV)

Here, mass flow rate of air is ${\dot{m}}_a$.

Express mass flow rate of air.

${\dot{m}}_a=\frac{{\dot{\nu }}_1}{v_1}$ (XVI)

Here, volume flow rate at inlet is ${\dot{\nu }}_1$.

Conclusion:

Substitute $\text{30}{\text{m}}^{\text{3}}/\text{min}$ for ${\dot{\nu }}_1$ and $0.8695{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ in Equation (XVI).

$\begin{array}{c}{\dot{m}}_a=\frac{\text{30}{\text{m}}^{\text{3}}/\text{min}}{0.8695{\text{m}}^{\text{3}}/\text{kg}}\\ =34.5\text{kg}/\text{min}\end{array}$

Substitute $34.5\text{kg}/\text{min}$ for ${\dot{m}}_a$, $49.01\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_2$ and $30.59\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_1$ in Equation (XV).

$\begin{array}{c}{\dot{Q}}_{\text{in}}=\left(34.5\text{kg}/\text{min}\right)\left[\left(49.01-30.59\right)\text{kJ}/\text{kg}\text{dry}\text{air}\right]\\ =\text{636}\text{kJ}/\text{min}\end{array}$

Hence, the rate of heat transfer in the heating section is $\text{636}\text{kJ}/\text{min}$.

(c)

To determine

The rate of water added to air in the evaporative cooler.

Answer to Problem 137RP

The rate of water added to air in the evaporative cooler is $\text{0}\text{.112}\text{kg}/\text{min}$.

Explanation of Solution

Express the rate of water added to air in the evaporative cooler.

${\dot{m}}_{w,\text{added}}={\dot{m}}_a\left({\omega }_3-{\omega }_2\right)$ (XVII)

Conclusion:

Substitute $0.009381\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_3$, $0.006138\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ and $34.5\text{kg}/\text{min}$ for ${\dot{m}}_a$ in Equation (XVII).

$\begin{array}{c}{\dot{m}}_{w,\text{added}}=\left(34.5\text{kg}/\text{min}\right)\left(0.009381-0.006138\right)\\ =\text{0}\text{.112}\text{kg}/\text{min}\end{array}$

Hence, the rate of water added to air in the evaporative cooler is $\text{0}\text{.112}\text{kg}/\text{min}$.