The Grand Coulee Dam in north central Washington state is the largest hydroelect
ID: 1461068 • Letter: T
Question
The Grand Coulee Dam in north central Washington state is the largest hydroelectric dam in the USA, capable of producing 7,079 MW of electric power (several times what a modest coal-fired power plant is capable of). The energy is supplied by gravitational potential energy. Water flows through a hydrostatic “head” of 380 feet (the height drop) through generators that convert gravitational potential energy to electricity with 85% efficiency.
(a) What is the pressure at the level where the water enters the generators, 380 feet below the surface?
b) What must be the flow rate of water, in cubic feet per second, through the generators in order to supply the maximum power of 7,079 MW? Treat water as an incompressible fluid of density 1000 kg/m3.
Explanation / Answer
SOLUTION=
Power = Change in Energy/time
The change in energy in this case is PE, m*g*h
Power = m*g*h/t
Mass is defined by the density... mass = density*volume
power = densty*volume*h/t
Volume/t = volume flow rate = v_dot = the answer we want
power = density*v_dot*h
v_dot = power / (density*h)
From efficiency
Efficiency = Power out / power in
power in = power out/efficency
Subbed in:
v_dot = (power out / efficiency)/(density*h)
Convert 380 ft to m
380 ft * (1 m / 3.28 ft) = 116 m
plug in numbrs
v_dot = (7079*10^6 W / 0.85) / (1000 kg/m^3 * 116 m) = 71800 m^3/s
Convert to ft^3/s
v_dot =71800 m^3/s * (3.28 ft / 1m)^3 = 2.53*10^6 ft^3/s <---- ANS!
Now, did you mean MW or just W? Because the first power number is 7079 MW. The 2nd is just 7079 W. I ran this a 7079 MW. You can rerun it as just W if that is the correct.
*****
I forgot to cary a 'g' term
I had:power = densty*volume*h/t
Should be: power = density*volume*g*h/t
So v_dot should be: v_dot = (power out/effciency) / (density*h*g) = 2.59 *10^5 ft^3/s
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