Chapter 14.7, Problem 86P

Saturated humid air at 70 psia and 200°F is cooled to 100°F as it flows through a 3-in-diameter pipe with a velocity of 50 ft/s and at constant pressure. Calculate the rate at which liquid water is formed inside this pipe and the rate at which the air is cooled.

To determine

The rate at which liquid water is formed inside this pipe and the rate at which the air is cooled.

Answer to Problem 86P

The rate at which liquid water is formed inside this pipe is $\text{0}\text{.0670}\text{lbm}/\text{s}$ and the rate at which the air is cooled is $\text{83}\text{.2}\text{Btu}/\text{s}$.

Explanation of Solution

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

${\omega }_1={\omega }_2$

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

Express initial partial pressure.

$\begin{array}{c}{P}_{\nu 1}={\varphi }_1{P}_{g1}\\ ={\varphi }_1{P}_{\text{sat@200°F}}\end{array}$ (I)

Here, relative humidity at state 1 is ${\varphi }_1$, initial vapor pressure is $P_{g1}$ and saturation pressure at temperature of $\text{200°F}$ is $P_{\text{sat@200°F}}$.

Express initial humidity ratio.

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

Here, pressure at state 1 is $P_1$.

Express initial enthalpy.

$h_1=c_p{T}_1+{\omega }_1{h}_{g1@200\text{°F}}$ (III)

Here, specific heat at constant pressure is $c_p$ and initial specific enthalpy saturated vapor at temperature of $\text{200°F}$ is $h_{g1\text{@200°F}}$.

Express specific volume at state 1.

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

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 final partial pressure.

$\begin{array}{c}{P}_{\nu 2}={\varphi }_2{P}_{g2}\\ ={\varphi }_2{P}_{\text{sat@100°F}}\end{array}$ (V)

Here, relative humidity at state 2 is ${\varphi }_2$, final vapor pressure is $P_{g2}$ and saturation pressure at temperature of $\text{100°F}$ is $P_{\text{sat@100°F}}$.

Express final humidity ratio.

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

Here, pressure at state 2 is $P_2$.

Express final enthalpy.

$h_2=c_p{T}_2+{\omega }_2{h}_{g2@100\text{°F}}$ (VII)

Here, final specific enthalpy saturated vapor at temperature of $\text{100°F}$ is $h_{g2\text{@100°F}}$.

Express volume flow rate of the dry air at inlet.

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

Here, velocity at state 1 is $V_1$, cross-sectional are at inlet is $A_1$ and diameter is $D$.

Express mass flow rate of air.

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

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 ${\dot{m}}_a$.

Express the rate at which liquid water is formed inside this pipe by using an water mass balance.

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

Here, mass flow rate of water at inlet and exit is ${\dot{m}}_{w,i}\text{and}{\dot{m}}_{w,e}$ respectively, and the rate at which liquid water is formed inside this pipe is ${\dot{m}}_w$.

Express the rate at which the air is cooled by using an energy balance.

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

$\begin{array}{c}{\dot{Q}}_{\text{out}}={\dot{m}}_{a1}{h}_1-\left({\dot{m}}_{a2}{h}_2+{\dot{m}}_w{h}_w\right)\\ ={\dot{m}}_a\left(h_1-h_2\right)-{\dot{m}}_w{h}_w\\ ={\dot{m}}_a\left(h_1-h_2\right)-{\dot{m}}_w{h}_{f@150°\text{F}}\end{array}$ (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, the rate at which the air is cooled  is ${\dot{Q}}_{\text{out}}$, enthalpy at inlet and exit is $h_i\text{and}{h}_e$ respectively and enthalpy at state 1, 2, water is $h_1,h_2\text{and}{h}_w$ respectively and enthalpy of saturation liquid at temperature of $\text{150°F}$ is $h_{f@150°\text{F}}$.

Conclusion:

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

$\begin{array}{l}{c}_p=0.240\text{Btu}/\text{lbm}\cdot °\text{F}\\ R_a=0.3704\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\end{array}$

Refer Table A-4E, “saturated water-temperature table”, and write the saturation pressure and initial specific enthalpy saturated vapor at temperature of $\text{200°F}$.

$\begin{array}{l}{P}_{\text{sat@200°F}}=11.538\text{psia}\\ h_{g1\text{@200°F}}=1145.7\text{Btu}/\text{lbm}\end{array}$

Substitute $1$ for ${\varphi }_1$ and $11.538\text{psia}$ for $P_{\text{sat@200°F}}$ in Equation (I).

$\begin{array}{c}{P}_{\nu 1}=\left(1\right)\left(11.538\text{psia}\right)\\ =11.538\text{psia}\end{array}$

Substitute $11.538\text{psia}$ for $P_{\nu 1}$ and $70\text{psia}$ for $P_1$ in Equation (II).

$\begin{array}{c}{\omega }_1=\frac{0.622\left(11.538\text{psia}\right)}{70\text{psia}-11.538\text{psia}}\\ =0.1228\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\end{array}$

Substitute $0.240\text{Btu}/\text{lbm}\cdot °\text{F}$ for $c_p$, $\text{200°F}$ for $T_1$, $0.1228\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_1$, and $1145.7\text{Btu}/\text{lbm}$ for $h_{g1\text{@200°F}}$ in Equation (III).

$\begin{array}{c}{h}_1=\left(0.240\text{Btu}/\text{lbm}\cdot °\text{F}\right)\left(\text{200°F}\right)+\left(0.1228\right)\left(1145.7\text{Btu}/\text{lbm}\right)\\ =188.7\text{Btu}/\text{lbm}\text{dry}\text{air}\end{array}$

Substitute $0.3704\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$ for $R_a$, $\text{200°F}$ for $T_1$, $11.538\text{psia}$ for $P_{\nu 1}$ and $70\text{psia}$ for $P_1$ in Equation (IV).

$\begin{array}{c}{v}_1=\frac{\left(0.3704\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)\left(\text{200°F}\right)}{70\text{psia}-11.538\text{psia}}\\ =\frac{\left(0.3704\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)\left[\left(\text{200}+460\right)\text{R}\right]}{58.462\text{psia}}\\ =\frac{\left(0.3704\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)\left(660\text{R}\right)}{58.462\text{psia}}\\ =4.182{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}\end{array}$

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

$\begin{array}{l}{P}_{\text{sat@100°F}}=0.95052\text{psia}\\ h_{g2\text{@100°F}}=1104.7\text{Btu}/\text{lbm}\end{array}$

Substitute $1$ for ${\varphi }_1$ and $0.95052\text{psia}$ for $P_{\text{sat@200°F}}$ in Equation (V).

$\begin{array}{c}{P}_{\nu 2}=\left(1\right)\left(0.95052\text{psia}\right)\\ =0.95052\text{psia}\end{array}$

Substitute $0.95052\text{psia}$ for $P_{\nu 2}$ and $70\text{psia}$ for $P_2$ in Equation (VI).

$\begin{array}{c}{\omega }_2=\frac{0.622\left(0.95052\text{psia}\right)}{70\text{psia}-0.95052\text{psia}}\\ =0.00856\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}\end{array}$

Substitute $0.240\text{Btu}/\text{lbm}\cdot °\text{F}$ for $c_p$, $\text{100°F}$ for $T_2$, $0.00856\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_2$, and $1104.7\text{Btu}/\text{lbm}$ for $h_{g2\text{@100°F}}$ in Equation (VII).

$\begin{array}{c}{h}_2=\left(0.240\text{Btu}/\text{lbm}\cdot °\text{F}\right)\left(\text{100°F}\right)+\left(0.00856\right)\left(1104.7\text{Btu}/\text{lbm}\right)\\ =33.46\text{Btu}/\text{lbm}\text{dry}\text{air}\end{array}$

Substitute $\text{50}\text{ft}/\text{s}$ for $V_1$ and $\text{3}\text{in}$ for $D$ in Equation (VIII).

$\begin{array}{c}{\dot{\nu }}_1=\left(\text{50}\text{ft}/\text{s}\right)\left[\frac{\pi {\left(\text{3}\text{in}\right)}^2}{4}\right]\\ =\left(\text{50}\text{ft}/\text{s}\right)\left[\frac{\pi {\left\{\left(\text{3}\text{in}\right)\frac{\text{ft}}{12\text{in}}\right\}}^2}{4}\right]\\ =\left(\text{50}\text{ft}/\text{s}\right)\left[\frac{\pi {\left(\text{0}\text{.25}\text{ft}\right)}^2}{4}\right]\\ =2.454{\text{ft}}^{\text{3}}/\text{s}\end{array}$

Substitute $2.454{\text{ft}}^{\text{3}}/\text{s}$ for ${\dot{\nu }}_1$ and $4.182{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}$ for $v_1$ in Equation (IX).

$\begin{array}{c}{\dot{m}}_a=\frac{2.454{\text{ft}}^{\text{3}}/\text{s}}{4.182{\text{ft}}^{\text{3}}/\text{lbm}\text{dry}\text{air}}\\ =0.5868\text{lbm}/\text{s}\end{array}$

Substitute $0.5868\text{lbm}/\text{s}$ for ${\dot{m}}_a$, $0.1228\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_1$, and $0.00856\text{lbm}{\text{H}}_{\text{2}}\text{O}/\text{lbm}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (X).

$\begin{array}{c}{\dot{m}}_w=0.5868\text{lbm}/\text{s}\left(0.1228-0.00856\right)\\ =\text{0}\text{.0670}\text{lbm}/\text{s}\end{array}$

Hence, the rate at which liquid water is formed inside this pipe is $\text{0}\text{.0670}\text{lbm}/\text{s}$

Refer Table A-4E, “saturated water-temperature table”, and write the enthalpy of saturation liquid at temperature of $\text{150°F}$.

$h_{f@150°\text{F}}=117.99\text{Btu}/\text{lbm}$

Here, enthalpy of saturation liquid at temperature of $\text{150°F}$ is $h_{f@150°\text{F}}$.

Substitute $0.5868\text{lbm}/\text{s}$ for ${\dot{m}}_a$, $188.7\text{Btu}/\text{lbm}\text{dry}\text{air}$ for $h_1$, $33.46\text{Btu}/\text{lbm}\text{dry}\text{air}$ for $h_2$, $\text{0}\text{.0670}\text{lbm}/\text{s}$ for ${\dot{m}}_w$ and $117.99\text{Btu}/\text{lbm}$ for $h_{f@150°\text{F}}$ in Equation (XI).

$\begin{array}{c}{\dot{Q}}_{\text{out}}=\left(0.5868\text{lbm}/\text{s}\right)\left[\left(188.7-33.46\right)\text{Btu}/\text{lbm}\right]-\left(\text{0}\text{.0670}\text{lbm}/\text{s}\right)\left(117.99\text{Btu}/\text{lbm}\right)\\ =\text{83}\text{.2}\text{Btu}/\text{s}\end{array}$

Hence, the rate at which the air is cooled is $\text{83}\text{.2}\text{Btu}/\text{s}$.