Chapter 6, Problem 6.114P

To determine

The volume that flows in each duct and the cross-section size of the longer duct.

Answer to Problem 6.114P

The flow rate in small duct is equal to $6.924f{t}^3/s$

The flow rate in long duct is equal to $20.772f{t}^3/s$

The cross-section size of long duct is equal to $10.4in$

Explanation of Solution

Given information:

The length of the small duct is $100ft$ and the cross-section of $6in$ by $6in$

The length of the longer duct is $200ft$

The plenum pressure is equal to $5lbf/f{t}^2$

The volume flow rate in the longer duct is three times that of small duct.

The energy equation for this problem can be written as,

$\Delta P=\frac{1}{2}\rho V^2\left(1+f\frac{L}{D_h}+K_{sharpedgedentrance}\right)$

In the above equation,

$\Delta P$ -pressure difference

$\rho$ - Density of the fluid

$V$ - Velocity of the flow

$L$ - Length of duct

$D_h$ - Hydraulic diameter

$K$ - Loss co efficient

The Reynolds’s number is defined as,

${\mathrm{Re}}_{D_h}=\frac{\left(\rho V{D}_h\right)}{\mu }$

$\mu$ - Dynamic viscosity

Calculation:

Assume,

$\rho =0.00237slug/f{t}^3$

$\mu =3.759\times {10}^{-7}slug/fts$

$K_{sharpedgedentrance}=0.5$

Calculate the hydraulic diameter of the duct,

For a square duct,

Hydraulic diameter = Side length of the duct

Therefore,

$D_h=\frac{6}{12}ft$

Assume, $f=0.02$

According to the energy equation,

$\begin{array}{l}\Delta P=\frac{1}{2}\rho V^2\left(1+f\frac{L}{D_h}+K_{sharpedgedentrance}\right)\\ 5lbf/f{t}^2=\frac{1}{2}\left(0.00237slug/f{t}^3\right)V^2\left(1+\left(0.02\right)\frac{100ft}{\frac{6}{12}ft}+0.5\right)\\ V=27.697ft/s\end{array}$

Calculate Reynolds’s number

${\mathrm{Re}}_{D_h}=\frac{\left(\rho V{D}_h\right)}{\mu }=\frac{\left(0.00237slug/f{t}^3\right)\left(27.697ft/s\right)\left(\frac{6}{12}ft\right)}{3.759\times {10}^{-7}slug/fts}=87312.96$

Calculate the roughness ratio

$\frac{\epsilon }{D_h}=\frac{0.00016ft}{\left(\frac{6}{12}ft\right)}=3.2\times {10}^{-4}$

Calculate the exact value of friction factor

$\begin{array}{l}\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\\ \frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{87312.96}+{\left(\frac{\text{}3.2\times {10}^{-4}}{3.7}\right)}^{1.11}\right)\\ f=0.0196\end{array}$

Not a considerable amount of change in friction factor, therefore take

$f=0.02$

Therefore,

$V=27.697ft/s$

Calculate the flow rate in small duct,

$Q=VA=\left(27.697ft/s\right){\left(\frac{6}{12}ft\right)}^2=6.924f{t}^3/s$

To find the flow rate in longer duct,

We know that,

$\begin{array}{l}{Q}_{long}=3Q_{small}=3\left(6.924f{t}^3/s\right)\\ Q_{long}=20.772f{t}^3/s\end{array}$

To find the cross-section size of longer duct,

We need to find the hydraulic diameter of the longer duct.

Therefore, rewrite the energy equation for longer duct,

$\begin{array}{l}\Delta P=\frac{1}{2}\rho V^2\left(1+f\frac{L}{D_h}+K_{sharpedgedentrance}\right)\\ 5lbf/f{t}^2=\frac{1}{2}\left(0.00237slug/f{t}^3\right){\left(\frac{20.772f{t}^3/s}{D_h{}^2}\right)}^2\left(1+f\frac{200ft}{D_h}+0.5\right)\end{array}$

Assume $f=0.02$,

$\begin{array}{l}5lbf/f{t}^2=\frac{1}{2}\left(0.00237slug/f{t}^3\right){\left(\frac{20.772f{t}^3/s}{D_h{}^2}\right)}^2\left(1+\left(0.02\right)\frac{200ft}{D_h}+0.5\right)\\ D_h=0.885ft\\ =10.63in\end{array}$

Calculate the velocity of the flow

$V=\frac{Q}{A}=\frac{20.772f{t}^3/s}{{\left(0.885ft\right)}^2}=26.52ft/s$

Calculate the Reynolds’s number

${\mathrm{Re}}_{D_h}=\frac{\left(\rho V{D}_h\right)}{\mu }=\frac{\left(0.00237slug/f{t}^3\right)\left(26.52ft/s\right)\left(0.885ft\right)}{3.759\times {10}^{-7}slug/fts}=147976.52$

Calculate the roughness ratio

$\frac{\epsilon }{D_h}=\frac{0.00016ft}{\left(0.885ft\right)}=1.8\times {10}^{-4}$

Calculate the exact value of friction factor

$\begin{array}{l}\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\\ \frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{147976.52}+{\left(\frac{1.8\times {10}^{-4}}{3.7}\right)}^{1.11}\right)\\ f=0.0175\end{array}$

Therefore, according to the new friction factor, find the exact value of the cross-section size

$\begin{array}{l}\Delta P=\frac{1}{2}\rho V^2\left(1+f\frac{L}{D_h}+K_{sharpedgedentrance}\right)\\ 5lbf/f{t}^2=\frac{1}{2}\left(0.00237slug/f{t}^3\right){\left(\frac{20.772f{t}^3/s}{D_h{}^2}\right)}^2\left(1+\left(0.0175\right)\frac{200ft}{D_h}+0.5\right)\\ D_h=0.867ft\\ =10.4in\end{array}$

Conclusion:

The flow rate in small duct is equal to $6.924f{t}^3/s$

The flow rate in long duct is equal to $20.772f{t}^3/s$

The cross-section size of long duct is equal to $10.4in$.