Question

The UAE government is building a pumped storage dam 155 m above the Hatta dam. Water is pumped up from the Hatta dam during off-peak hours and released back into the dam (passing hydroelectric turbines) during peak-hours of electricity use. The total volume of the elevated storage dam is 800 million gallons (1 gallon is = 3.78 liters and the density of freshwater is 1000kg/m3).

a) Calculate the total stored energy (in Joules) in the elevated storage dam when it's full

b) Calculate the power output (in Watts) given that the pumped storage dam is emptied down into the Hatta dam in 5 hrs - given a 90% efficiency

Answer 1

(1) formula ;; density(d)= (m)mass/(v)(volume).

;; d=m/v ; m=dxv....(1)

Parameters from above.

density (d) of water= 1000kg/m^3

v=800 million gallons convert to cubic metre gives us; 3028329.427m^3.

From formula above.

m(mass) =1000x3028329.427

m(mass)=3028329427kg.

Formula for stored energy(potential energy)= mgh

Where m= mass ; g=gravity, h= hiegth

Parameters ; m=3028329427kg

g=10m/s^2(approx). , h=155m

Stored energy = (below)

3028329427 x 10 x 155

Es=4693910611850 J....(ans)

Es= 4.6tJ (tethra joules)

Couldn't do the second one due to insufficient time anymore problem pls let me know, thank you.