Water flows from a reservoir through a 0.8-m-diameter pipeline to a turbine gene
ID: 1829222 • Letter: W
Question
Water flows from a reservoir through a 0.8-m-diameter pipeline to a turbine generator unit and exits to a river that is 30 m below the reservoir surface. If the flow rate is 3 m3/s and the turbine-generator efficiency is 80%, calculate the power output. Assume the loss in the pipeline (including the exit) to be 2(V2/2g). Water is added to the tank shown below through a vertical pipe to maintain a constant water level. The tank is placed on a horizontal plane which has a frictionless surface. Determine the horizontal force, F, required for the tank to remain stationary. (Answer: F = 0)Explanation / Answer
V = Q / A
= 3 / (3.14*0.8^2 /4)
= 5.97 m/s
Head loss = 2 V^2 / (2g)
= 2*5.97^2 / (2*9.81)
= 3.63 m
Turbine head = 30 - 3.63 = 26.37 m
Theoretical Power output = rho*g*Q*H
= 1000*9.81*3*26.37
= 775929.4 W
= 776 kW
Actual power output = 80 % of 776 = 0.8*776 = 620.7 kW
2.
Velocity of efflux = sqrt (2gh)
from 625 mm^2 jet, V = sqrt (2*9.81*(1+1)) = 6.26 m/s
From 1250 mm^2 jet, V = sqrt (2*9.81*1) = 4.43 m/s
Flowrate Q = A*V
Q1 = 625*10^-6 *6.26 = 0.003912 m^3/s
Q2 = 1250*10^-6 *4.43 = 0.0055368 m^3/s
Mass flow rate m = rho*Q
m1 = 1000*0.003912 = 3.912 kg/s
m2 = 1000*0.0055368 = 5.5368 kg/s
Momentum conservation:
F = m1*V1 - m2*V2
F = 3.912*6.26 - 5.5368*4.43
F = 0
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