Question

4. Water is to be moved from one large reservoir to another a higher elevation as indicated in the Figure. The flow rate, Q=2 ft/s and the head loss hz = 60 where Vis the average velocity. the pump power required 15000 ft.lb/s. Determine the pump head loss hp. Section (2) p=1.94 slugs/ft3 P2 = P1 = V1 = V2 = 0 8-in. inside diameter pipe 50 Ft P2 + v +24thp = Section (1) Pump P2 vž ==+ + Z₂th 2g 29

Answer 1

as given,

$\small \\\\\frac{P_1}{\rho}+\frac{v_1^2}{2g}+Z_1+h_P=\frac{P_2}{\rho}+\frac{v_2^2}{2g}+Z_2+h_l \\\\V_1=V_2=P_1=P_2=0 \\\\Z_1+h_P=Z_2+h_L$

$\small \\\\0+h_P=50+60\frac{V^2}{2g}$

Now as we know,

$\small \\V=\frac{Q}{A}=\frac{2}{\frac{\pi}{4}\left (\frac{8}{12} \right )^2} \\\\V=5.7324 \ ft/s$

so,

$\small \\\\h_p=50+60\frac{5.7324^2}{2*9.81} \\\\h_p=150.490 \ ft$

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