Chapter 14, Problem 56EP

For the duct system and fan of Prob. 14-55E, partially closing the damper would decrease the flow rate. AU else being unchanged, estimate the minor loss coefficient of the damper required to decrease the volume flow rate by a factor of 2.

To determine

The minor loss coefficient of the damper.

Answer to Problem 56EP

The minor loss coefficient of the damper is $137$.

Explanation of Solution

Given information:

Inner diameter of the duct is $9.06\text{in}$, the average roughness of the duct is $0.0059\text{in}$, total length of the duct is $34.0\text{ft}$, number of elbows is $3$ each having minor loss coefficient of $0.21$, hood entry loss coefficient is $4.6$, loss coefficient is $1.8$, given equation for available head is $H_{avialable}=H_o-a{\dot{V}}^2$, shutoff head is $2.30\text{in}$ of water column, volume flow rate is $226\text{SCFM}$, and constant $a$ is $8.5\times {10}^{-6}\text{in}$ of water column.

Expression for the required head using the energy balance equation.

  $H_{required}=\frac{P_2-P_1}{\rho g}+\frac{{\alpha }_2{V}_2^2-{\alpha }_1{V}_1^2}{2g}+\left(z_2-z_1\right)+h_{Ltotal}$....... (I)

Here, the initial pressure is $P_1$, the final pressure is $P_2$, the initial velocity is $V_1$, the final velocity is $V_2$, the density of the water is $\rho$, the acceleration due to gravity is $g$, the potential head of water at section 1 is $z_1$, the potential head of the water at section 2 is $z_2$, the total loss is $h_{Ltotal}$.

Expression for the capacity.

  $\dot{V}=\frac{\pi D^2}{4}V$....... (II)

Expression for the available head

  $H_{availabe}=H_o-a_{air}{\dot{V}}^2$....... (III)

Here, the shutoff head is $H_o$, the coefficient is $a$ the capacity is $\dot{V}$ and the velocity of the water is $V$.

Substitute $\frac{\pi D^2}{4}V$ for $\dot{V}$ in Equation (III).

  $\begin{array}{l}{H}_{availabe}=H_o-a{\left(\frac{\pi D^2}{4}V\right)}^2\\ H_{availabe}=H_o-a\left(\frac{{\pi }^2{D}^4}{16}{V}^2\right)\end{array}$....... (IV)

Expression for the Reynolds number

  $\mathrm{Re}=\frac{VD}{\nu }$....... (V)

Here, the velocity of the water is $\rho$ and the kinematic viscosity is $\nu$.

Expression for the roughness factor

  $R=\frac{\epsilon }{D}$..... (VI)

Here, the diameter of the pipe is $D$

Expression for the minor losses

  ${\displaystyle \sum F_L=K_{L,entrance}+K_{L,damp}+3K_{L,elbow}}$..... (VII)

Here, the minor loss coefficient at pipe entrance is $K_{L,entrance}$, the minor loss coefficient at each elbow is $K_{L,elbow}$, the minor loss coefficient at damp is $K_{L,damp}$.

Expression for the total head loss

  $h_{Ltotal}=\left(f\frac{L}{D}+{\displaystyle \sum K_L}\right)\frac{V^2}{2g}$....... (VIII)

Here, the friction factor is $f$, the length of the pipe is $L$.

Expression for the friction factor using the Colebrook equation

  $\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{R}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)$....... (IX)

Expression to convert the shutoff head from inches of water column to inches of air column

  $H_{o,air}=\frac{H_{o.water}\times {\rho }_w}{{\rho }_a}$...... (X)

Here, the density of the water is ${\rho }_w$, the shutoff head is $H_o$ and the density of the air is ${\rho }_a$.

Expression to convert the $a$ from inches of water column to inches of air column

  $a_{air}=\frac{a_{water}\times {\rho }_w}{{\rho }_a}$...... (XI)

Calculation:

Refer to the Table-A-9E, "Properties of air at $1\text{atm}$ pressure" to obtain the density of the air as $0.07392\text{lbm}/{\text{ft}}^{\text{3}}$, the dynamic viscosity of air as $1.242\times {10}^{-3}\text{lbm}/\text{ft}\cdot \text{s}$, the value of kinematic viscosity as $1.681\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}$.

Substitute $226\text{SCFM}$ for $\dot{V}$ and $9.06\text{in}$ for $D$ in Equation (II).

  $\begin{array}{l}226\text{SCFM}=\left(\frac{\pi {\left(9.06\text{in}\right)}^2}{4}\right)V\\ 226\text{SCFM}\left(\frac{1{\text{ft}}^{\text{3}}/\text{s}}{60\text{SCFM}}\right)=\left(\frac{\pi {\left(9.06\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^2}{4}\right)V\\ V=\frac{3.766{\text{ft}}^{\text{3}}/\text{s}}{0.4477{\text{ft}}^{\text{2}}}\\ V=8.413\text{ft}/\text{s}\end{array}$

Substitute $8.413\text{ft}/\text{s}$ for $V$, $1.681\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}$ for $\nu$ and $9.06\text{in}$ for $D$ in Equation (V).

  $\begin{array}{c}\mathrm{Re}=\frac{8.413\text{ft}/\text{s}\left(9.06\text{in}\right)}{1.681\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}}\\ =\frac{8.413\text{ft}/\text{s}\left(9.06\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}{1.681\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}}\\ =\frac{8.413\text{ft}/\text{s}\left(0.755\text{ft}\right)}{1.681\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}}\\ =37785.93\end{array}$

Substitute $\mathrm{0..0059}\text{in}$ for $\epsilon$ and $9.06\text{in}$ for $D$ in Equation (VI).

  $\begin{array}{c}R=\frac{0.0059\text{in}}{9.06\text{in}}\\ =6.51\times {10}^{-4}\end{array}$

Refer to Figure A-12 "Moody chart" to obtain the friction factor as $0.024$ at relative roughness $R=6.51\times {10}^{-4}$ and the Reynolds number $\mathrm{Re}=37785.93$.

Substitute $0$ for $V_1$, $0$ for $\left(z_2-z_1\right)$, $P_{atm}$ for $p_1$ and $P_{atm}$ for $p_2$ in Equation (II).

  $\begin{array}{c}{H}_{required}=\frac{P_{atm}-P_{atm}}{\rho g}+\frac{{\alpha }_2{V}_2^2-{\alpha }_1{\left(0\right)}^2}{2g}+0+h_{Ltotal}\\ =0+\frac{{\alpha }_2{V}_2^2}{2g}+h_{Ltotal}\\ =\frac{{\alpha }_2{V}_2^2}{2g}+h_{Ltotal}\end{array}$....... (XII)

Substitute $4.6$ for $K_{Lentrance}$ and $0.21$ for $K_{Lelbow}$ in Equation (VII).

  $\begin{array}{c}{\displaystyle \sum F_L}=4.6+K_{L,damp}+3\left(0.21\right)\\ =4.6+K_{L,damp}+\left(0.63\right)\\ =5.23+K_{L,damp}\end{array}$

Here, the exit is equal to the total velocity of the water.

Substitute $\left(f\frac{L}{D}+{\displaystyle \sum K_L}\right)\frac{V^2}{2g}$ for $h_{L,Total}$ in Equation (XII).

  $\begin{array}{c}{H}_{required}=\frac{{\alpha }_2{V}_2^2}{2g}+\left(\left(f\frac{L}{D}+{\displaystyle \sum K_L}\right)\frac{V^2}{2g}\right)\\ =\frac{{\alpha }_2{V}^2}{2g}+\left(\left(f\frac{L}{D}+{\displaystyle \sum K_L}\right)\frac{V^2}{2g}\right)\\ =\left({\alpha }_2+\frac{fl}{D}+{\displaystyle \sum K_L}\right)\frac{V^2}{2g}\end{array}$....... (XIII)

Let us consider that the air is flowing through the duct turbulent at ${\alpha }_2=1.05$.

Substitute $1.05$ for ${\alpha }_2$, $0.024$ for $f$, $34\text{ft}$ for $l$, $9.06\text{in}$ for $D$, $5.23+K_{L,damp}$ for $K_L$, $8.413\text{ft}/\text{s}$ for $V$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ in Equation (XIII)

  $\begin{array}{c}{H}_{required}=\left(1.05+\frac{0.024\left(34\text{ft}\right)}{\left(9.06\text{in}\right)}+\left(5.23+K_{L,damp}\right)\right)\frac{{\left(8.413\text{ft}/\text{s}\right)}^2}{2\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)}\\ =\left(\left(6.2593+1.0293K_{L,damp}\right)\right)1.099\text{ft}\\ =6.879+1.1312K_{L,damp}\end{array}$....... (XIV)

Substitute $2.3\text{in}$ for $H_{o,water}$, $62.24\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_w$ and $0.07392\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_a$ in Equation (X).

  $\begin{array}{c}{H}_{o,air}=\frac{2.3\text{in}\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =\frac{\left(2.3\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =161.38\text{ft}\end{array}$

Substitute $8.5\times {10}^{-6}\text{in}$ for $a$, $62.24\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_w$ and $0.07392\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_a$ in Equation (XI).

  $\begin{array}{c}{a}_{air}=\frac{8.5\times {10}^{-6}\text{in}\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =\frac{\left(8.5\times {10}^{-6}\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =5.753\times {10}^{-4}\text{ft}\end{array}$

Substitute $161.38\text{ft}$ for $H_o$, $5.753\times {10}^{-4}\text{ft}$ for $a_{air}$ and $3.767{\text{ft}}^{\text{3}}$ for $\dot{V}$ in Equation (III).

  $\begin{array}{c}{H}_{avialable}=161.38\text{ft}-\left(5.753\times {10}^{-4}\text{ft}\right){\left(3.767{\text{ft}}^{\text{3}}/\text{s}\right)}^2\\ =161.38\text{ft}-\left(5.753\times {10}^{-4}\text{ft}\right)\left(14.19{\text{ft}}^6/{\text{s}}^2\right)\\ =161.38\text{ft}-\left(8.1636\times {10}^{-3}\text{ft}\right)\\ =161.37\text{ft}\end{array}$

Substitute $161.37\text{ft}$ for $H_{available}$ in Equation (XIV).

  $\begin{array}{l}161.37\text{ft}=6.879\text{ft}+\left(1.1312\text{ft}\right)K_{L,damp}\\ K_{L,damp}=\frac{161.37\text{ft}-6.879\text{ft}}{1.1312\text{ft}}\\ K_{L,damp}=\frac{154.49}{1.1312}\\ K_{L,damp}\approx 137\end{array}$

Conclusion:

The minor loss coefficient of the damper is $137$.