Chapter 14.7, Problem 75P

(a)

To determine

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

Answer to Problem 75P

The temperature and relative humidity of air when it leaves the heating section is $\text{19}\text{.5°C}\text{and}\text{37}\text{.7\%}$ respectively.

Explanation of Solution

Express initial partial pressure.

$\begin{array}{c}{P}_{\nu 1}=P_{\nu 2}\\ ={\varphi }_1{P}_{g1}\\ ={\varphi }_1{P}_{\text{sat@10°C}}\end{array}$ (I)

Here, partial pressure at state 2 is $P_{\nu 2}$, relative humidity at state 1 is ${\varphi }_1$, initial vapor pressure is $P_{g1}$ and saturation pressure at temperature of $\text{10°C}$ is $P_{\text{sat@10°C}}$.

Express initial partial pressure.

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

Here, initial pressure is $P_1$.

Express initial specific volume.

${\nu }_1=\frac{R_a{T}_1}{P_{a1}}$ (III)

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

Express initial specific humidity.

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

The amount or quantity of air moisture will remain constant while flowing through heating section,

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

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

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 partial pressure at state 3.

$\begin{array}{c}{P}_{\nu 3}={\varphi }_3{P}_{g3}\\ ={\varphi }_3{P}_{\text{sat@20°C}}\end{array}$ (VII)

Here, relative humidity at state 3 is ${\varphi }_3$, vapor pressure at state 3 is $P_{g3}$ and saturation pressure at temperature of $\text{20°C}$ is $P_{\text{sat@20°C}}$.

Express specific humidity at state 3.

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

Here, partial pressure at stat3 is $P_{\nu 3}$ and pressure at state 3 is $P_3$.

Express enthalpy at state 3.

$h_3=c_p{T}_3+{\omega }_3{h}_{g3}$ (IX)

Here, specific enthalpy saturated vapor at state 3 is $h_{g3}$, specific heat at constant pressure of air is $c_p$ and temperature at state 3 is $T_3$.

Express the mass flow rate of dry air.

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

Here, initial volume rate is ${\dot{V}}_1$ and initial specific volume is ${\nu }_1$.

Express final specific enthalpy by applying an energy balance on the humidifying section.

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

$\begin{array}{l}{\dot{m}}_w{h}_w+{\dot{m}}_{a2}{h}_2={\dot{m}}_a{h}_3\\ \left({\omega }_3-{\omega }_2\right)h_w+h_2=h_3\\ \left({\omega }_3-{\omega }_2\right)h_{g\text{@100°C}}+h_2=h_3\\ h_2=h_3-\left({\omega }_3-{\omega }_2\right)h_{g\text{@100°C}}\end{array}$ (XI)

Here, 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, initial and exit specific enthalpy is $h_i\text{and}{h}_e$ respectively, mass flow rate of water and air is ${\dot{m}}_w\text{and}{\dot{m}}_a$ respectively, specific enthalpy of water is $h_w$, mass flow arte at exit of air is ${\dot{m}}_{a2}$, specific enthalpy at state 2 and 3 is $h_2\text{and}{h}_3$ respectively, specific humidity at state 2 and 3 is ${\omega }_2\text{and}{\omega }_3$ respectively and specific enthalpy saturation vapor at temperature of $\text{100°C}$ is $h_{g\text{@100°C}}$.

Express temperature leaving the heating section.

$\begin{array}{c}{h}_2=c_p{T}_2+{\omega }_2{h}_{g2}\\ =c_p{T}_2+{\omega }_2\left(h_{g@0.01\text{°C}}+1.82T_2\right)\end{array}$ (XII)

Here, temperature at state 2 is $T_2$.

Express relative humidity leaving the heating section.

$\begin{array}{c}{\varphi }_2=\frac{P_{\nu 2}}{P_{g2}}\\ =\frac{P_{\nu 2}}{P_{\text{sat@19}\text{.5°C}}}\end{array}$ (XIII)

Here, partial pressure at state 2 is $P_{\nu 2}$ and saturation pressure at temperature of $\text{19}\text{.5°C}$ is $P_{\text{sat@19}\text{.5°C}}$.

Conclusion:

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.

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

Refer Table A-4, “saturated water-temperature table”, and write the saturation pressure at temperature of $\text{10°C}\text{and}\text{20}°\text{C}$.

$\begin{array}{l}{P}_{\text{sat@10°C}}=1.2281\text{kPa}\\ P_{\text{sat@20°C}}=2.3392\text{kPa}\end{array}$

Substitute $0.70$ for ${\varphi }_1$ and $1.2281\text{kPa}$ for $P_{\text{sat@10°C}}$ in Equation (I).

$\begin{array}{c}{P}_{\nu 1}=P_{\nu 2}\\ =\left(0.70\right)\left(1.2281\text{kPa}\right)\\ =0.860\text{kPa}\end{array}$

Substitute $\text{95}\text{kPa}$ for $P_1$ and $0.860\text{kPa}$ for $P_{\nu 1}$ in Equation (II).

$\begin{array}{c}{P}_{a1}=\text{95}\text{kPa}-0.860\text{kPa}\\ =94.14\text{kPa}\end{array}$

Substitute $0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}$ for $R_a$, $\text{10°C}$ for $T_1$ and $94.14\text{kPa}$ for $P_{a1}$ in Equation (III).

$\begin{array}{c}{\nu }_1=\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\right)\left(\text{10°C}\right)}{94.14\text{kPa}}\\ =\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\right)\left[\left(\text{10}+273\right)\text{K}\right]}{94.14\text{kPa}}\\ =\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\right)\left(283\text{K}\right)}{94.14\text{kPa}}\\ =0.863{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $\text{95}\text{kPa}$ for $P_1$ and $0.863\text{kPa}$ for $P_{\nu 1}$ in Equation (IV).

$\begin{array}{c}{\omega }_1=\frac{0.622\left(0.863\text{kPa}\right)}{\text{95}\text{kPa}-0.863\text{kPa}}\\ =0.00568\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $0.00568\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ in Equation (V).

${\omega }_2=0.00568\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$

Refer Table A-4, “saturated water-temperature table”, and write the initial and final specific enthalpy saturated vapor corresponding to temperature of $\text{10°C}\text{and}\text{20°C}$.

$\begin{array}{c}{h}_{g1}=h_{g\text{@10°C}}\\ =2519.2\text{kJ}/\text{kg}\\ h_{g3}=h_{g\text{@20°C}}\\ =2537.4\text{kJ}/\text{kg}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{°C}$ for $c_p$, $\text{10°C}$ for $T_1$, $0.00568$ for ${\omega }_1$ and $2519.2\text{kJ}/\text{kg}$ for $h_{g1}$ in Equation (VI).

$\begin{array}{c}{h}_1=\left(1.005\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(\text{10°C}\right)+\left(0.00568\right)\left(2519.2\text{kJ}/\text{kg}\right)\\ =24.36\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $0.60$ for ${\varphi }_3$ and $2.3392\text{kPa}$ for $P_{\text{sat@20°C}}$ in Equation (VII).

$\begin{array}{c}{P}_{\nu 3}=\left(0.60\right)\left(2.3392\text{kPa}\right)\\ =1.40\text{kPa}\end{array}$

Substitute $\text{95}\text{kPa}$ for $P_3$ and $1.40\text{kPa}$ for $P_{\nu 3}$ in Equation (VIII).

$\begin{array}{c}{\omega }_3=\frac{0.622\left(1.40\text{kPa}\right)}{\text{95}\text{kPa}-1.40\text{kPa}}\\ =0.00930\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{°C}$ for $c_p$, $\text{20°C}$ for $T_3$, $0.00930$ for ${\omega }_1$ and $2537.4\text{kJ}/\text{kg}$ for $h_{g3}$ in Equation (IX).

$\begin{array}{c}{h}_3=\left(1.005\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(\text{20°C}\right)+\left(0.0093\right)\left(2537.4\text{kJ}/\text{kg}\right)\\ =43.70\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $\text{35}{\text{m}}^{\text{3}}/\text{min}$ for ${\dot{V}}_1$ and $0.863{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}$ for ${\nu }_1$ in Equation (X).

$\begin{array}{c}{\dot{m}}_a=\frac{\text{35}{\text{m}}^{\text{3}}/\text{min}}{0.863{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}}\\ =40.6\text{kg}/\text{min}\end{array}$

Refer Table A-4, “saturated water-temperature table”, and write the specific enthalpy saturation vapor at temperature of $\text{100°C}$.

$h_{g\text{@100°C}}=2675.6\text{kJ}/\text{kg}$

Substitute $43.70\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_3$, $0.00930\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_3$, $0.00568\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ and $2675.6\text{kJ}/\text{kg}$ for $h_{g\text{@100°C}}$ in Equation (XI).

$\begin{array}{c}{h}_2=43.70\text{kJ}/\text{kg}\text{dry}\text{air}-\left[\left(0.00930-0.0053\right)\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\right]\left(2675.6\text{kJ}/\text{kg}\right)\\ =34\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Refer Table A-4, “saturated water-temperature table”, and write the specific enthalpy saturation vapor at temperature of $\text{0}\text{.01°C}$.

$h_{g@0.01\text{°C}}=2500.9\text{kJ}/\text{kg}$

Substitute $34\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_2$, $1.005\text{kJ}/\text{kg}\cdot \text{°C}$ for $c_p$, $0.00568\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ and $2500.9\text{kJ}/\text{kg}$ for $h_{g@0.01\text{°C}}$ in Equation (XII).

$\begin{array}{l}34\text{kJ}/\text{kg}\text{dry}\text{air}=\left(1.005\text{kJ}/\text{kg}\cdot \text{°C}\right)T_2+0.00568\left(2500.9+1.82T_2\right)\\ 34\text{kJ}/\text{kg}\text{dry}\text{air}=\left(1.005\text{kJ}/\text{kg}\cdot \text{°C}\right)T_2+14.205+0.01033T_2\\ T_2=19.5\text{°C}\end{array}$

Hence, the temperature of air when it leaves the heating section is $\text{19}\text{.5°C}$.

Refer Table A-4, “saturated water-temperature table”, and write the saturated vapor at temperature of $\text{19}\text{.5°C}$ 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 saturated vapor respectively.

Show the saturated vapor corresponding to temperature as in Table (1).

Temperature $T\left(\text{°C}\right)$	Saturated vapor $P_{\text{sat}}\left(\text{kPa}\right)$
15 $\left(x_1\right)$	1.7057 $\left(y_1\right)$
19.5 $\left(x_2\right)$	$\left(y_2=?\right)$
20 $\left(x_3\right)$	2.3392 $\left(y_3\right)$

Substitute $\text{15°C,19}\text{.5°C}\text{and}20\text{°C}$ for $x_1,x_2\text{and}{x}_3$ respectively, $\text{1}\text{.7057}\text{kPa}$ for $y_1$ and $\text{2}\text{.3392}\text{kPa}$ for $y_3$ in Equation (XIV).

$\begin{array}{c}{y}_2=\frac{\left(\text{19}\text{.5°C}-\text{15°C}\right)\left(\text{2}\text{.3392}\text{kPa}-\text{1}\text{.7057}\text{kPa}\right)}{\left(20\text{°C}-\text{15°C}\right)}+\text{1}\text{.7057}\text{kPa}\\ =2.2759\text{kPa}\\ =P_{\text{sat@19}\text{.5°C}}\end{array}$

Substitute $\text{0}\text{.860}\text{kPa}$ for $P_{\nu 2}$ and $2.2759\text{kPa}$ for $P_{\text{sat@19}\text{.5°C}}$ in Equation (XIII).

$\begin{array}{c}{\varphi }_2=\frac{\text{0}\text{.860}\text{kPa}}{2.2759\text{kPa}}\\ =0.377\\ =37.7\%\end{array}$

Hence, the relative humidity of air when it leaves the heating section is $\text{37}\text{.7\%}$ respectively.

(b)

To determine

The rate of heat transfer in the heating section.

Answer to Problem 75P

The rate of heat transfer in the heating section is $391\text{kJ}/\text{min}$.

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)

Conclusion:

Substitute $40.6\text{kg}/\text{min}$ for ${\dot{m}}_a$, $34\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_2$ and $24.36\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_1$ in Equation (XV).

$\begin{array}{c}{\dot{Q}}_{\text{in}}=\left(40.6\text{kg}/\text{min}\right)\left[\left(34-24.36\right)\text{kJ}/\text{kg}\text{dry}\text{air}\right]\\ =391\text{kJ}/\text{min}\end{array}$

Hence, the rate of heat transfer in the heating section is $391\text{kJ}/\text{min}$.

(c)

To determine

The rate at which water is added to the air in the humidifying section.

Answer to Problem 75P

The rate at which water is added to the air in the humidifying section is $0.147\text{kg}/\text{min}$.

Explanation of Solution

Express the rate at which water is added to the air in the humidifying section.

${\dot{m}}_w={\dot{m}}_a\left({\omega }_3-{\omega }_2\right)$ (V)

Conclusion:

Substitute $40.6\text{kg}/\text{min}$ for ${\dot{m}}_a$, $0.0093\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_3$ and $0.00568\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (V).

$\begin{array}{c}{\dot{m}}_w=\left(40.6\text{kg}/\text{min}\right)\left(0.0093-0.00568\right)\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\\ =0.147\text{kg}/\text{min}\end{array}$

Hence, the rate at which water is added to the air in the humidifying section is $0.147\text{kg}/\text{min}$.