Chapter 8, Problem 83EP

To determine

The rate of flow of air from dryer.

Answer to Problem 83EP

The rate of flow of air from dryer is $0.784{\text{ft}}^{\text{3}}/\text{s}$.

Explanation of Solution

Given information:

The air discharge rate is $1.2{\text{ft}}^{\text{3}}/\text{s}$, diameter of the vent is $5\text{in}$, length of the duct is $15\text{ft}$, the diameter of the duct is $5\text{in}$ and the friction factor due to the duct is $0.019$.

Write the expression to calculate the energy equation for point $1$ and point $2$.

  $\left(\begin{array}{l}\dot{m}\left(\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1^2}{2}+g{z}_1\right)\\ +{\dot{W}}_{fan}\end{array}\right)=\left(\begin{array}{l}\dot{m}\left(\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2^2}{2}+g{z}_2\right)\\ +{\dot{W}}_{turbine}+\dot{m}g{h}_L\end{array}\right)$ (I)

Here, the inlet pressure is $P_1$, the exit pressure is $P_2$, the inlet velocity is $V_1$, the exit velocity is $V_2$, datum at the inlet is $z_1$, the exit datum is $z_2$, the gravitational acceleration is $g$, the kinetic energy correction factor at inlet is ${\alpha }_1$, the kinetic energy correction factor at exit is ${\alpha }_2$, the density of water is $\rho$, the head loss is $h_L$, the mass flow rate of air is $\dot{m}$, the turbine power is ${\dot{W}}_{turbine}$ and the fan power is ${\dot{\cancel{W}}}_{fan}$.

The pressure at the inlet and exit is atmospheric, the inlet velocity is zero, and the height at both points is same. There is no existing turbine and the head losses are null.

Write the expression to calculate mass flow rate at Point $2$.

  $\dot{m}=\rho {\dot{V}}_2$ (II)

Write the expression to calculate the mass flow rate at point $3$.

  $\dot{m}=\rho {\dot{V}}_3$

Here, the volumetric flow rate at exit is ${\dot{V}}_2$.

Write the expression to calculate the average velocity.

  $V_2=\frac{{\dot{V}}_2}{\frac{\pi }{4}{D}^2}$ (III)

Here, the diameter of the pipe is $D$.

Assume a point $3$ at the exit of the duct.

Write the expression to calculate the energy equation between point $1$ and point $3$.

  $\left(\begin{array}{l}\dot{m}\left(\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1^2}{2}+g{z}_1\right)\\ +{\dot{W}}_{fan}\end{array}\right)=\left(\begin{array}{l}\dot{m}\left(\frac{P_3}{\rho g}+{\alpha }_3\frac{V_3^2}{2}+g{z}_3\right)\\ +{\dot{W}}_{turbine}+\dot{m}g{h}_L\end{array}\right)$ (IV)

Here, the pressure at point $3$ is $P_3$, the kinetic energy correction factor at point $3$ is ${\alpha }_3$, the velocity at point $3$ is $V_3$ and the datum at point $3$ is $z_3$.

The pressure at point $1$ and point $3$ is atmospheric; the height at both points is same.

  $z_1=z_3$

Write the expression to calculate the average velocity at point $3$.

  $V_3=\frac{{\dot{V}}_3}{\frac{\pi }{4}{D}^2}$ (V)

Here, the volumetric flow rate at point $3$ is ${\dot{V}}_3$.

Write the expression to calculate the sum of loss coefficients.

  ${\displaystyle \sum K_L=}+3K_{L,bend}+K_{L,exit}$ (VI)

Here, the loss coefficient for bend pipe is $K_{L,bend}$ and the loss coefficient at sharp edge exit is $K_{L,exit}$.

Write the expression for the head loss.

  $h_L=\left(f\frac{L}{D}+{\displaystyle \sum K_L}\right)\frac{V_3^2}{2g}$ (VII)

Here, the friction factor of the duct is $f$.

Calculation:

Substitute $P_{atm}$ for $P_1$, $P_{atm}$ for $P_2$, $1$ for ${\alpha }_1$, $1$ for ${\alpha }_2$, $z_1$ for $z_2$, $0$ for ${\dot{W}}_{turbine}$ and $0$ for $h_L$ in Equation (I).

  $\begin{array}{l}\left(\begin{array}{l}\dot{m}\left(\frac{P_{atm}}{\rho g}+1\frac{0}{2}+g{z}_1\right)\\ +{\dot{W}}_{fan}\end{array}\right)=\left(\begin{array}{l}\dot{m}\left(\frac{P_{atm}}{\rho g}+1\frac{V_2^2}{2}+g{z}_1\right)\\ +0+\dot{m}g\left(0\right)\end{array}\right)\\ \left(\begin{array}{l}\dot{m}\left(\frac{P_{atm}}{\rho g}+0+g{z}_1\right)\\ +{\dot{W}}_{fan}\end{array}\right)=\left(\begin{array}{l}\dot{m}\left(\frac{P_{atm}}{\rho g}+1\frac{V_2^2}{2}+g{z}_1\right)\\ +0+\dot{m}g\left(0\right)\end{array}\right)\\ {\dot{W}}_{fan}=\dot{m}\frac{V_2^2}{2}\end{array}$ (VIII)

Substitute $\rho {\dot{V}}_2$ for $\dot{m}$ in Equation (VIII).

  ${\dot{W}}_{fan}=\rho {\dot{V}}_2\frac{V_2^2}{2}$ (IX)

Substitute $1.2{\text{ft}}^{\text{3}}/\text{s}$ for ${\dot{V}}_2$ and $5\text{in}$ for $D$ in Equation (III).

  $\begin{array}{c}{V}_2=\frac{1.2{\text{ft}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(5\text{in}\right)}^2}\\ =\frac{1.2{\text{ft}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(5\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^2}\\ =8.80\text{ft}/\text{s}\end{array}$

Substitute $P_{atm}$ for $P_3$, $1$ for ${\alpha }_3$, $z_1$ for $z_3$, $P_{atm}$ for $P_1$, $1$ for ${\alpha }_1$ and $0$ for ${\dot{W}}_{turbine}$ in Equation (IV).

  $\begin{array}{l}\left(\begin{array}{l}\dot{m}\left(\frac{P_{atm}}{\rho g}+1\frac{0}{2}+g{z}_1\right)\\ +{\dot{W}}_{fan}\end{array}\right)=\left(\begin{array}{l}\dot{m}\left(\frac{P_{atm}}{\rho g}+1\frac{V_3^2}{2}+g{z}_1\right)\\ +0+\dot{m}g{h}_L\end{array}\right)\\ \left(\begin{array}{l}\dot{m}\left(0\right)\\ +{\dot{W}}_{fan}\end{array}\right)=\left(\begin{array}{l}\dot{m}\left(0+\frac{V_3^2}{2}+0\right)\\ +0+\dot{m}g{h}_L\end{array}\right)\\ {\dot{W}}_{fan}=\dot{m}\frac{V_3^2}{2}+\dot{m}g{h}_L\end{array}$ (X)

Substitute $\rho {\dot{V}}_3$ for $\dot{m}$ in Equation (X).

  ${\dot{W}}_{fan}=\left(\rho {\dot{V}}_3\right)\frac{V_3^2}{2}+\left(\rho {\dot{V}}_3\right)g{h}_L$ (XI)

Substitute $\rho {\dot{V}}_2\frac{V_2^2}{2}$ for ${\dot{W}}_{fan}$ in Equation (XI).

  $\begin{array}{l}\rho {\dot{V}}_2\frac{V_2^2}{2}=\left(\rho {\dot{V}}_3\right)\frac{V_3^2}{2}+\left(\rho {\dot{V}}_3\right)g{h}_L\\ {\dot{V}}_2\frac{V_2^2}{2}=\left({\dot{V}}_3\right)\frac{V_3^2}{2}+\left({\dot{V}}_3\right)g{h}_L\end{array}$ (XII)

Substitute $5\text{in}$ for $D$ in Equation (V).

  $\begin{array}{c}{V}_3=\frac{{\dot{V}}_3}{\frac{\pi }{4}{\left(5\text{in}\right)}^2}\\ =\frac{{\dot{V}}_3}{\frac{\pi }{4}{\left(5\text{in}\left(\frac{1\text{ft}}{1\text{in}}\right)\right)}^2}\\ =7.33{\dot{V}}_3\text{ft}/\text{s}\end{array}$

Refer to Table 8-4, "Loss coefficients of various pipe components for turbulent flow" to obtain the value of $K_{L,bend}$ as $0.3$ and $K_{L,exit}$ as $1$ for $90°$ bend pipe and sharp edged exit.

Substitute $1$ for $K_{L,exit}$ and $0.3$ for $K_{L,bend}$ in Equation (VI).

  $\begin{array}{c}{\displaystyle \sum K_L}=3\left(0.3\right)+1\\ =1.9\end{array}$

Substitute $0.019$ for $f$, $15\text{ft}$ for $L$, $5\text{in}$ for $D$ and $1.9$ for ${\displaystyle \sum K_L}$ in Equation (VII).

  $\begin{array}{c}{h}_L=\left(0.019\frac{15\text{ft}}{5\text{in}}+1.9\right)\frac{V_3^2}{2g}\\ =\left(0.019\frac{15\text{ft}}{5\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)}+1.9\right)\frac{V_3^2}{2g}\\ =2.584\frac{V_3^2}{2g}\end{array}$

Substitute $2.584\frac{V_3^2}{2g}$ for $h_L$ and $7.33{\dot{V}}_3\text{ft}/\text{s}$ for $V_3$ in Equation (XII).

  $\begin{array}{l}{\dot{V}}_2\frac{V_2^2}{2}=\left({\dot{V}}_3\right)\frac{{\left(7.33{\dot{V}}_3\text{ft}/\text{s}\right)}^2}{2}+\left({\dot{V}}_3\right)g\left(2.584\frac{{\left(7.33{\dot{V}}_3\text{ft}/\text{s}\right)}^2}{2g}\right)\\ {\dot{V}}_2\frac{V_2^2}{2}=96.282{\dot{V}}_3^3\\ {\dot{V}}_3={\left({\dot{V}}_2\frac{V_2^2}{2\left(96.283\right)}\right)}^{1/3}\end{array}$ (XIII)

Substitute $1.2{\text{ft}}^{\text{3}}/\text{s}$ for ${\dot{V}}_2$ and $8.80\text{ft}/\text{s}$ for $V_2$ in Equation (XIII).

  $\begin{array}{c}{\dot{V}}_3={\left(\left(1.2{\text{ft}}^{\text{3}}/\text{s}\right)\frac{{\left(8.80\text{ft}/\text{s}\right)}^2}{2\left(96.283\right)}\right)}^{1/3}\\ ={\left(0.4825\right)}^{1/3}{\text{ft}}^{\text{3}}/\text{s}\\ =0.784{\text{ft}}^{\text{3}}/\text{s}\end{array}$

Conclusion:

The volumetric flow rate at point $3$ is $0.784{\text{ft}}^{\text{3}}/\text{s}$.