Homework Answers
Given data for pump:
Fluid flow rate Q = 400 (gal/min)*0.00379 (m3/1 gal) * (1min/60sec) = 0.0252 m3/sec
pump power W= 20hp = 20 hp* 746watt/1hp = 14920 watt.
Given pump efficiency for delivering fluid = 0.75
Then power delivers to fluid Wf = 0.75*14920 = 11190 watt.
We know,
Power Wf = flow rate * pressure drop
1 watt = 1N-m/sec
Pressure drop = Wf/Q = 11190(N-m/sec)/0.0252(m3/sec) = 444047.6 N/m2
Pressure rise = 444047.6 (N/m2)* (1 atm/101325 (N/m2)) = 4.39 atm
We know ( 1atm = 14.7 psia)
So pressure rise = 4.39* 14.7 = 64.533 psia
head required :
gasoline density , rho = 800 kg/m3
so head required = ΔP/rho*g = 444047.6(N/m2)/800(kg/m3)*9.8(m/s2) = 56.68 m
( 1N = 1kg*m/s2)
So head required = 56.68 m
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