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

The answer is '
`2.9584 \ (lbf)/(ft^2)`'

Step-by-step explanation:

Given:

`\mu=3.96 * 10^(-7) \ (lbfs)/( ft^2)\\\\\gamma=0.0709 \ (lfb)/(ft^3)\\\\v= 1.8 * 10^(-4) \ (ft^2)/(s)`

energy balance equation:

`\ P_1 - P_2 = \rho gh_(L)`

`V= (\ Q)/( A)\\\\V= ((5000)/(60))/(2 * (9)/(12))\\\\V= 55.55 (ft)/(s)`

calculating
`D_h`:

`D_h= (4 * \ scetor \ area )/( wetted \ permetor)\\\\D_h=(4 * 2 * (9)/(12))/(2(2+(9)/(12)))\\\\D_h= 1.0909 ft\\\\`

calculating
`R_e`:

`R_e=(VD_h)/(V)\\\\R_e=(55.55 * 1.0909)/(1.8 * 10^(-4))\\\\R_e=3.366 * 10^5\\\\`

calculating relative vovghness:

`(k_s)/(D_h)= ((0.006)/(12))/(1.0909)\\\\`

= 0.0006875

If
`R_e= 3.366 * 10^5 \ and \ \ (K_s)/(D_h) = 0.0006875 \ \ \ so, \ \ f= 0.019`

Calculating
`H_f`:

`H_f= (f \ l\ v^2)/( 2 \ g \ D_h)\\\\`

`=(0.019 * 50 * 55.55^2)/(2* 32.2* 1.0909)\\\\= 41.727 \ ft`

Calculating pressure drop:

`P_1-P_2= \gamma h_(L)\\\\`

`= 0.0709 * 41.727\\\\= 2.9584 \ (lbf)/(ft^2)`