Chapter 6, Problem 73P

To determine

The volume flow rate of air in the steam heating system.

Answer to Problem 73P

The volume flow rate of air in the steam heating system is $299.964{\text{ft}}^3/\text{s}$.

Explanation of Solution

Given:

The inlet pressure of air $\left(P_1\right)$ is 14.7 psia.

The inlet temperature of air $\left(T_1\right)$ is $80°\text{F}$.

The outlet temperature of air $\left(T_2\right)$ is $130°\text{F}$.

The inlet pressure of steam $\left(P_3\right)$ is 30 psia.

The inlet pressure of steam $\left(T_3\right)$ is $400°\text{F}$.

The outlet pressure of steam $\left(P_4\right)$ is $25\text{ psia}$.

The outlet temperature of steam $\left(T_4\right)$ is $212°\text{F}$.

Calculation:

Draw the schematic diagram of the steam heating system.

Write the expression for the mass balance.

  $m_{in}-m_{out}=\Delta m$

  $m_{in}-m_{out}=0\left(\text{For steady state}\right)$        (I)

Here, mass of the fluid entering into the system is $m_{in}$, the mass of fluid leaving out of the system is $m_{out}$, and the change in mass is $\Delta m$.

Write the energy rate balance equation for a control volume.

  ${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=d{\dot{E}}_{\text{system}}/dt$

  ${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\left(\text{steady}\right)$        (II)

Here, total energy rate at inlet is ${\dot{E}}_{\text{in}}$, total energy rate at outlet is ${\dot{E}}_{\text{out}}$ and change in net energy rate is $d{\dot{E}}_{\text{system}}/dt$.

Write the ideal gas equation for specific volume of air $\left(v\right)$.

  $v=\frac{RT}{P}$        (III)

Here, gas constant for air is $R$, temperature of air is $T$ and pressure of air is $P$.

Calculate the volume flow rate of the air at the inlet $\left({\dot{V}}_1\right)$.

  ${\dot{V}}_1={\dot{m}}_a{v}_1$        (IV)

Here, mass flow rate of air is ${\dot{m}}_a$ and specific volume of air is $v_1$.

Refer Table A-1E, “Gas constant of common gases”, obtain the gas constant of air as $0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$.

Refer Table A-2E, “Ideal – gas specific heats of common gases”, obtain the constant pressure specific heat of air as $0.240\text{Btu}/\text{lbm}\text{.}°\text{F}$.

Refer Table A-6E, “Superheated water”, obtain the inlet specific enthalpy $\left(h_3\right)$ of steam at pressure $\left(P_3\right)$ of $\text{30 psia}$ and temperature $\left(T_3\right)$ of $400°\text{F}$ as $\text{1237}\text{.9}\text{Btu}/\text{lbm}$.

Refer Table A-4E of “Saturated water-Temperature table”, obtain the final specific enthalpy of saturated water at temperature $\left(T\right)$ of $212°\text{F}$.

  $h_4=h_f=\text{180}\text{.21}\text{Btu}/\text{lbm}$

Rewrite Equation (I) for the mass flow rate of air $\left({\dot{m}}_a\right)$.

  ${\dot{m}}_1={\dot{m}}_2={\dot{m}}_a$

Here, mass flow rate of air at inlet is ${\dot{m}}_1$ and mass flow rate of air at outlet is ${\dot{m}}_2$.

Rewrite Equation (I) for the mass flow rate of steam $\left({\dot{m}}_s\right)$.

  ${\dot{m}}_3={\dot{m}}_4={\dot{m}}_s$

Here, mass flow rate of steam at inlet is ${\dot{m}}_3$, and mass flow rate of steam at outlet is ${\dot{m}}_4$.

Rewrite Equation (II) for the energy balance in the steam heating system.

  ${\dot{m}}_1{h}_1+{\dot{m}}_3{h}_3={\dot{m}}_2{h}_2+{\dot{m}}_4{h}_4$        (V)

Substitute ${\dot{m}}_a$ for ${\dot{m}}_1$, ${\dot{m}}_a$ for ${\dot{m}}_2$, ${\dot{m}}_s$ for ${\dot{m}}_3$, and ${\dot{m}}_s$ for ${\dot{m}}_4$ in Equation (V).

  $\begin{array}{l}{\dot{m}}_a{h}_1+{\dot{m}}_s{h}_3={\dot{m}}_a{h}_2+{\dot{m}}_s{h}_4\\ {\dot{m}}_a=\left(\frac{h_3-h_4}{h_2-h_1}\right){\dot{m}}_s\end{array}$

  ${\dot{m}}_a=\frac{h_3-h_4}{c_p\left(T_2-T_1\right)}{\dot{m}}_s$        (VI)

Here, specific enthalpy of air at the inlet is $h_1$, specific enthalpy of air at the outlet is $h_2$, constant pressure specific heat of air is $c_p$, temperature of air at inlet is $T_1$ and temperature of air at the outlet is $T_2$.

Substitute $h_3=\text{1237}\text{.9}\text{Btu}/\text{lbm}$, $h_4=\text{180}\text{.21}\text{Btu}/\text{lbm}$, $c_p=0.240\text{Btu}/\text{lbm}\text{.}°\text{F}$, $T_2= 130°\text{F}$, ${\dot{m}}_s=15\text{lbm}/\text{min}$, and $T_1=80°\text{F}$ in Equation (VI).

  $\begin{array}{c}{\dot{m}}_a=\frac{\text{1237}\text{.9}\text{Btu}/\text{lbm}-\text{180}\text{.21}\text{Btu}/\text{lbm}}{0.240\text{Btu}/\text{lbm}\text{.}°\text{F}\left(130°\text{F}-80°\text{F}\right)}\times 15\text{lbm}/\text{min}\\ =\frac{\text{1237}\text{.9}\text{Btu}/\text{lbm}-\text{180}\text{.21}\text{Btu}/\text{lbm}}{0.240\text{Btu}/\text{lbm}\text{.}°\text{F}\left(130°\text{F}-80°\text{F}\right)}\times 15\text{lbm}/\text{min}\times \frac{1\text{ min}}{60\text{ s}}\\ =22.04\text{lbm}/\text{s}\end{array}$

Substitute $R=0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$, $T_1=80°\text{F}$, and $P_1=14.7\text{psia}$ in Equation (III).

  $\begin{array}{c}{v}_1=\frac{0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\times 80°\text{F}}{14.7\text{ psia}}\\ =\frac{0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\times \left(80+459.67\right)\text{R}}{14.7\text{psia}}\\ =13.61{\text{ft}}^3/\text{lbm}\end{array}$

Substitute ${\dot{m}}_a=22.04\text{lbm}/\text{s}$ and $v_1=13.61{\text{ft}}^3/\text{lbm}$ in Equation (IV).

  $\begin{array}{c}{\dot{V}}_a=22.04\text{lbm}/\text{s}\times 13.61{\text{ft}}^3/\text{lbm}\\ =299.964{\text{ft}}^3/\text{s}\end{array}$

Thus, the volume flow rate of the air in steam heating system is $299.964{\text{ft}}^3/\text{s}$.