3) A pumped hydro energy storage facility has a water reservoir of 19.5 billion

Question

3) A pumped hydro energy storage facility has a water reservoir of 19.5 billion gallons of water. The elevation of the reservoir is approximately 90 meters above the generator/pump that exits into a very large lake. The change in elevation of the water in the reservoir is minimal during the reservoir drainage during power generation. There are three generators (which also serve as pumps) operating during full power production with each fed by a penstock with a diameter of 4 meters. a) Determine the total potential energy stored in the filled reservoir. (2 points) b) If the system is 93% efficient in electric output-power generation, then determine the peak MW (electric) power production capacity of the pumped storage power generation system. (Hint Use the Bernoulli equation to determine a velocity and use this to calculate answers for b through d.) (4 points) c) If the reservoir is never drained to no less than 20% of full reservoir capacity then determine how many hours the pumped system can produce power at the peak MWs (electric) you computed in part b above, and also compute the total kW-hours (electric) available from this system. (4 points) d) One particular afternoon in the summer the pumped hydro energy storage system produces rated peak power for 5 hours. It is known that to refill the reservoir to its full water-storage capacity (at its drainage flow rate each generator acting as a pump must operate at 1.5 x 10

Explanation / Answer

Amount of water = 19.5*10^9 gallons = 73.8155 *10^6 m^3
height, h = 90 m
no change in height while power generation

a) Total PE in reservoir = mghmax/2
m = mass of water = 73.8155*10^6 * 1000 kg
PE = 73.8155*10^9*9.81*90/2 = 3.2585*10^13 J
b) pump diamtere = 4 m
radius, r = 2 m
Area, A = pi*r^2
mass flow rate = rho*A*v = 1000*A*v
from bernoulli's theorem
rho*g*H/2 = 0.5*100*v^2
v = 93.96 m/s
KE transfer rate = mass flow rate * v^2 / 2 = 55.446 MW
POWER output = 0.93*KE = 51.564 MW
c) maximum volume to be drained = 0.2*73.8155 *10^6 m^3 = 14.7631*10^6 m^3
time = mass / mass flow rate = 14.7631*10^6 m^3*1000 / 1000*A*93.96 = 3.474 hours
Net kWh = 55.446 MW * 3.474 hours = 192669.35 kWh

d) n tuen around = power drained from dam / power required for refill
Power= 1.5*10^6 hors epower
power drained = 51.564 MW
so, n = 51.564*10^6/1118.55*10^6= 0.0460

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