Answer:
61.6 kw
Explanation:
Q= 0.35 m^3/s ( flow rate )
Hl = [tex]\frac{2v^{2} }{2g}[/tex] ( head loss)
D = 30 cm = 0.3 m ( diameter of pipe )
∝ = 1
A = area
first we find the V ( velocity )
V = Q / A = [tex]\frac{0.35}{\frac{\pi }{4}(0.3)^{2} } = 4.95 m/s[/tex]
Hl = [tex]\frac{2*(4.95)^{2} }{2*9.81} = 2.497 m[/tex]
to calculate Hp apply the energy equation between water surface tank and the outlet of pipe neglecting pressure head of velocity head
The pipe will be :
Hp = (P1/y) + (V2^2/2g) + Zs + Hl
= (100*10^3 / 9810 ) + (4.95^2 / 2*9.81 ) + ( 10-6 ) + ( 2.497 )
= 17.939 m
to calculate the power the pump must supply we apply the power equation
P = yQHp
= 9810 * 0.35 * 17.939
= 61.6 kw
y = 9810 (constant)
