Chapter 14.7, Problem 130RP

Air enters a cooling section at 97 kPa, 35°C, and 30 percent relative humidity at a rate of 6 m3/min, where it is cooled until the moisture in the air starts condensing. Determine (a) the temperature of the air at the exit and (b) the rate of heat transfer in the cooling section.

To determine

The temperature of the air at the exit.

Answer to Problem 130RP

The temperature of the air at the exit is $\text{14}\text{.8°C}$.

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 partial pressure.

$\begin{array}{c}{P}_{\nu 1}={\varphi }_1{P}_{g1}\\ ={\varphi }_1{P}_{\text{sat@35°C}}\end{array}$ (II)

Here, relative humidity at state 1 is ${\varphi }_1$, initial vapor pressure is $P_{g1}$ and saturation pressure at temperature of $\text{35°C}$ is $P_{\text{sat@35°C}}$.

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.

$h_1=c_p{T}_1+{\omega }_1{h}_{g1@35\text{°C}}$ (IV)

Here, specific heat at constant pressure is $c_p$ and initial specific enthalpy saturated vapor at temperature of $\text{35°C}$ is $h_{g1\text{@35°C}}$.

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 $T_1$.

As the air at the final state is saturated and the vapor pressure during the process will remain constant, thus the dew point temperature is the exit temperature.

$\begin{array}{c}{T}_2=T_{\text{dp}}\\ =T_{\text{sat@}{P}_v}\end{array}$ (VI)

Here, exit temperature is $T_2$, dew point temperature is $T_{\text{dp}}$ and saturation temperature at vapor pressure is $T_{\text{sat@}{P}_v}$.

Conclusion:

Refer Table A-4, “saturated water-temperature table”, and write the saturation pressure and initial specific enthalpy saturated vapor at temperature of $\text{35°C}$.

$\begin{array}{l}{P}_{\text{sat@35°C}}=5.629\text{kPa}\\ h_{g1\text{@35°C}}=2564.6\text{kJ}/\text{kg}\end{array}$

Substitute $0.3$ for ${\varphi }_1$ and $\text{5}\text{.629}\text{kPa}$ for $P_{\text{sat@35°C}}$ in Equation (II).

$\begin{array}{c}{P}_{\nu 1}=\left(0.3\right)\left(\text{5}\text{.629}\text{kPa}\right)\\ =\text{1}\text{.69}\text{kPa}\end{array}$

Substitute $\text{1}\text{.69}\text{kPa}$ for $P_{\nu 1}$ and $\text{97}\text{kPa}$ for $P_1$ in Equation (III).

$\begin{array}{c}{\omega }_1=\frac{0.622\left(\text{1}\text{.69}\text{kPa}\right)}{\text{97}\text{kPa}-\text{1}\text{.69}\text{kPa}}\\ =0.0110\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $0.0110\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ in Equation (I).

${\omega }_2=0.0110\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$

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 $c_p$, $\text{35°C}$ for $T_1$, $0.0110\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ and $2564.6\text{kJ}/\text{kg}$ for $h_{g1\text{@35°C}}$ in Equation (IV).

$\begin{array}{c}{h}_1=\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(\text{35°C}\right)+\left(0.0110\right)\left(2564.6\text{kJ}/\text{kg}\right)\\ =63.44\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $R_a$, $\text{35°C}$ for $T_1$, $\text{1}\text{.69}\text{kPa}$ for $P_{\nu 1}$ and $\text{97}\text{kPa}$ for $P_1$ in Equation (V).

$\begin{array}{c}{v}_1=\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(\text{35°C}\right)}{\text{97}\text{kPa}-\text{1}\text{.69}\text{kPa}}\\ =\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(\text{35}+273\right)\text{K}}{\text{95}\text{.31}\text{kPa}}\\ =\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(308\text{K}\right)}{\text{95}\text{.31}\text{kPa}}\\ =0.927{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $\text{1}\text{.69}\text{kPa}$ for $P_v$ in Equation (VI).

$T_2=T_{\text{sat@1}\text{.69}\text{kPa}}$ (VII)

Refer Table A-5, “saturated water-pressure table”, and write the saturation temperature or exit temperature at pressure of $\text{1}\text{.69}\text{kPa}$ using an interpolation method.

Write the formula of interpolation method of two variables.

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

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

Show the specific enthalpy at state 2 corresponding to specific entropy as in Table (1).

Pressure $P\left(\text{kPa}\right)$	Saturation or exit temperature $T_{\text{sat}}\left(\text{°C}\right)$
1.5 $\left(x_1\right)$	13.02 $\left(y_1\right)$
1.69 $\left(x_2\right)$	$\left(y_2=?\right)$
2 $\left(x_3\right)$	17.50 $\left(y_3\right)$

Substitute $\text{1}\text{.5}\text{kPa,}\text{1}\text{.69}\text{kPa}\text{and}\text{2}\text{kPa}$ for $x_1,x_2\text{and}{x}_3$ respectively, $\text{13}\text{.02°C}$ for $y_1$ and $\text{17}\text{.50°C}$ for $y_3$ in Equation (VIII).

$\begin{array}{c}{y}_2=\frac{\left(\text{1}\text{.69}\text{kPa}-\text{1}\text{.5}\text{kPa}\right)\left(\text{17}\text{.50°C}-\text{13}\text{.02°C}\right)}{\left(\text{2}\text{kPa}-\text{1}\text{.5}\text{kPa}\right)}+\text{13}\text{.02°C}\\ =\text{14}\text{.8°C}\\ =T_{\text{sat@1}\text{.69}\text{kPa}}\end{array}$

Substitute $\text{14}\text{.8°C}$ for $T_{\text{sat@1}\text{.69}\text{kPa}}$ in Equation (VII).

$T_2=\text{14}\text{.8°C}$

Hence, the temperature of the air at the exit is $\text{14}\text{.8°C}$.

To determine

The rate of heat transfer in the cooling section.

Answer to Problem 130RP

The rate of heat transfer in the cooling section is $\text{134}\text{kJ}/\min$.

Explanation of Solution

Express the enthalpy of air at exit.

$h_2=c_p{T}_2+{\omega }_2{h}_{g2@14.8\text{°C}}$ (IX)

Here, temperature at exit is $\text{14}\text{.8°C}$ and initial specific enthalpy saturated vapor at temperature of $\text{14}\text{.8°C}$ is $h_{g2\text{@14}\text{.8°C}}$.

Express mass flow rate of air.

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

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

Express the rate of heat transfer in the cooling section.

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

Conclusion:

Refer Table A-4, “saturated water-temperature table”, and write the final specific enthalpy saturated vapor at temperature of $\text{14}\text{.8°C}$ using an interpolation method.

$h_{g2\text{@14}\text{.8°C}}$ (XII)

Show the final specific enthalpy saturated vapor corresponding to exit temperature as in Table (2).

Exit temperature $T_2\left(°\text{C}\right)$	Final specific enthalpy saturated vapor $h_{g2}\left(\text{kJ}/\text{kg}\right)$
10 $\left(x_1\right)$	2519.2 $\left(y_1\right)$
14.8 $\left(x_2\right)$	$\left(y_2=?\right)$
15 $\left(x_3\right)$	2528.3 $\left(y_3\right)$

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

$h_{g2\text{@14}\text{.8°C}}=2528.1\text{kJ}/\text{kg}$

Substitute $2528.1\text{kJ}/\text{kg}$ for $h_{g2\text{@14}\text{.8°C}}$ in Equation (XII).

$h_{g2\text{@14}\text{.8°C}}=2528.1\text{kJ}/\text{kg}$

Substitute $1.005\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $\text{14}\text{.8°C}$ for $T_1$, $0.0110\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ and $2528.1\text{kJ}/\text{kg}$ for $h_{g2\text{@14}\text{.8°C}}$ in Equation (IX).

$\begin{array}{c}{h}_2=\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(\text{14}\text{.8°C}\right)+\left(0.0110\right)\left(2528.1\text{kJ}/\text{kg}\right)\\ =42.78\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $\text{6}{\text{m}}^{\text{3}}/\text{min}$ for ${\dot{\nu }}_1$ and $0.927{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}$ for $v_1$ in Equation (X).

$\begin{array}{c}{\dot{m}}_a=\frac{\text{6}{\text{m}}^{\text{3}}/\text{min}}{0.927{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}}\\ =\text{6}\text{.47}\text{kg}/\text{min}\end{array}$

Substitute $\text{6}\text{.47}\text{kg}/\text{min}$ for ${\dot{m}}_a$, $63.44\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_1$ and $42.78\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_2$ in Equation (XI).

$\begin{array}{c}{\dot{Q}}_{\text{out}}=\left(\text{6}\text{.47}\text{kg}/\text{min}\right)\left(63.44\text{kJ}/\text{kg}-42.78\text{kJ}/\text{kg}\right)\\ =\text{134}\text{kJ}/\min \end{array}$

Hence, the rate of heat transfer in the cooling section is $\text{134}\text{kJ}/\min$.