A river flows into a hydroelectric power plant with a velocity v 1 = 1.15 m/s an
ID: 1261719 • Letter: A
Question
A river flows into a hydroelectric power plant with a velocity v1= 1.15 m/s and emerges from it with a final velocity v2=0.11 m/s. If water flows through the plant at a rate V? =3.22 m3/s and has a head of h=10.35 m.
- Calculate the water mass flow rate through the plant (in Kg/s).
- Calculate the rate of change in the gravitational potential energy through the plant (in kW).
- Calculate the rate of change of the kinetic energy through the plant (in kW).
- Calculate the theoretical rate of production of energy from this hydroelectric plant (in kW).
Explanation / Answer
according to continuty equation
Qinflow=AVI
A=3.22/1.15
=2.8m2
AVI=AVF
A=2.8*1.15/0.11
=29.27m2
in order determine mass
m=rho*V
=1000*10.35*2.8
=28980kg
but time is not given in question
rate of change in the gravitational potential energy through the plant (in kW).
=1000*9.8*10.35
=101.43KW
rate of change of the kinetic energy through the plant (in kW).
=0.5*1000(1.152-0.112)
=0.655KW
Due to energy loss the practically available power will be less than the theoretically power. Practically available power can be expressed as
P = 1000 ? q g h
where
P = power available (W)
? = efficiency (in general ranging 0.75 to 0.95)
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