Please help with the below questions The pump in the figure discharges 20degree
ID: 1860950 • Letter: P
Question
Please help with the below questions
The pump in the figure discharges 20degree C water at 65.8 m3 / h through a horizontal pipe of ID 3 cm. The intake to the pump is from a horizontal pipe of ID 9 cm. Neglecting losses, what power in kW is delivered to the water by the pump? (Adapted from F.M. White, Fluid Mechanics) Water at 20degree C is pumped at 2,100 gal/min from the lower to the upper reservoir, as shown in the Figure. Pipe friction losses are approximated by hf = 29 V2 / (2g), where V is the average velocity in the pipe. If the pump is 81% efficient, what horsepower is needed to drive it? Use the head form of the Engineering Bernoulli equation, and perform all the calculations in British units. (Adapted from F.M. White, Fluid Mechanics).Explanation / Answer
Q = 65.8 m^3/h = 65.8/(60*60) m^3/s = 0.01828 m^3/s
V1 = Q/A1 = 0.01828 / (3.14/4*0.09^2) = 2.87 m/s
V2 = Q/A2 = 0.01828/(3.14/4*0.03^2) = 25.87 m/s
P1 + 1/2*rho*V1^2 + rho*g*Hp = P2 + 1/2*rho*V2^2
120*10^3 + 1/2*1000*2.87^2 + 1000*9.81*Hp = 400*10^3 + 1/2*1000*25.87^2
Hp = 62.236 m
Power delivered = rho*g*Q*Hp
= 1000*9.81*0.01828*62.236
= 11160.5 W
= 11.16 kW
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2100 gal = 280.729 ft^3
Q = 2100 gal/min = 280.729 / 60 ft^3/s = 4.6788 ft^3/s
V = Q/A = 4.6788 / (3.14/4*(6/12)^2) = 23.84 ft/s
Since pump inlet and outlet dia are same, we get that V1 = V2. Also, P1 = P2 = P_atm
P1/(rho*g) + V1^2 /2g + z1 + Hp - h_f = P2 / (rho*g) + V2^2 / (2g) + z2
z1 + Hp - h_f = z2
Hp = 150-50 + 29*V^2 / (2g)
Hp = 100 + 29*23.84^2 / (2*32.2)
Hp = 355.95 ft
Power = rho*Q*Hp
= 62.4*4.6788*355.95 ft-lb/s
= 62.4*4.6788*355.95*60 ft-lb/min
= 62.4*4.6788*355.95*60 * (3.03*10^-5) hp
= 188.93 hp
Actual power reqd = 188.93 / 0.81 = 233.25 hp