Question

A duct for an air conditioning system has a rectangular cross section of 2ft by 9in. The duct is fabricated from galvanized iron (e 0.006in). Calculate the pressure drop in psi for a horizontal 50-ft section of pipe with a flow rate of air of 5000cfm at 100F and atmospheric pressure

Answer 1

Answer:

The answer is '$2.9584 \ \frac{lbf}{ft^2}$'

Explanation:

Given:

$\mu=3.96 \times 10^{-7} \ \frac{lbfs}{ ft^2}\\\\\gamma=0.0709 \ \frac{lfb}{ft^3}\\\\v= 1.8 \times 10^{-4} \ \frac{ft^2}{s}$

energy balance equation:

$\ P_1 - P_2 = \rho gh_{L}$

$V= \frac{\ Q}{ A}\\\\V= \frac{\frac{5000}{60}}{2 \times \frac{9}{12}}\\\\V= 55.55 \frac{ft}{s}$

calculating  $D_h$:

$D_h= \frac{4 \times \ scetor \ area }{ wetted \ permetor}\\\\D_h=\frac{4 \times 2 \times \frac{9}{12}}{2(2+\frac{9}{12})}\\\\D_h= 1.0909 ft$

calculating $R_e$:

$R_e=\frac{VD_h}{V}\\\\R_e=\frac{55.55 \times 1.0909}{1.8 \times 10^{-4}}\\\\R_e=3.366 \times 10^5$

calculating relative vovghness:

$\frac{k_s}{D_h}= \frac{\frac{0.006}{12}}{1.0909}$

    = 0.0006875

If $R_e= 3.366 \times 10^5 \ and \ \ \frac{K_s}{D_h} = 0.0006875 \ \ \ so, \ \ f= 0.019$

Calculating  $H_f$:

$H_f= \frac{f \ l\ v^2}{ 2 \ g \ D_h}$

     $=\frac{0.019 \times 50 \times 55.55^2}{2\times 32.2\times 1.0909}\\\\= 41.727 \ ft$

Calculating pressure drop:

$P_1-P_2= \gamma h_{L}$  

            $= 0.0709 \times 41.727\\\\= 2.9584 \ \frac{lbf}{ft^2}$