Respuesta :
Answer:
The answer is '[tex]2.9584 \ \frac{lbf}{ft^2}[/tex]'
Explanation:
Given:
[tex]\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}[/tex]
energy balance equation:
[tex]\ P_1 - P_2 = \rho gh_{L}[/tex]
[tex]V= \frac{\ Q}{ A}\\\\V= \frac{\frac{5000}{60}}{2 \times \frac{9}{12}}\\\\V= 55.55 \frac{ft}{s}[/tex]
calculating [tex]D_h[/tex]:
[tex]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\\\\[/tex]
calculating [tex]R_e[/tex]:
[tex]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\\\\[/tex]
calculating relative vovghness:
[tex]\frac{k_s}{D_h}= \frac{\frac{0.006}{12}}{1.0909}\\\\[/tex]
= 0.0006875
If [tex]R_e= 3.366 \times 10^5 \ and \ \ \frac{K_s}{D_h} = 0.0006875 \ \ \ so, \ \ f= 0.019[/tex]
Calculating [tex]H_f[/tex]:
[tex]H_f= \frac{f \ l\ v^2}{ 2 \ g \ D_h}\\\\[/tex]
[tex]=\frac{0.019 \times 50 \times 55.55^2}{2\times 32.2\times 1.0909}\\\\= 41.727 \ ft[/tex]
Calculating pressure drop:
[tex]P_1-P_2= \gamma h_{L}\\\\[/tex]
[tex]= 0.0709 \times 41.727\\\\= 2.9584 \ \frac{lbf}{ft^2}[/tex]
