# Pumping water on reservoirs

One method used to store energy during times of low demand is by pumping water uphill into a reservoir. When energy demand exceeds the output produced by power plants, the water is released, moves downhill, and is converted into electricity by a turbine. During a 142.0 minute long period of low usage, 62.0 MW is used to pump water up an elevation of 56.7 m in a series of 1.00 m diameter pipes. The friction loss in the pipe going to the upper reservoir is $30.2 m^2/s^2$.

What is the mass flow rate , of water during the low usage period? in [kg/s]

What is the total mass of water raised 56.7 m during this 142.0 minute period? in [kg]

How long can the turbines produce 335.0 MW of power from the release of the mass of stored water calculated above? Assume the water sustains a friction loss of 30.2 $m^2/s^2$ on the way down to the turbine. in [min]

What is the efficiency of this process? (Assume that the pump and turbines operate at 100% efficiency, and the only loses are due to frictional loses within the pipes.) in [%]

Answers

$power = Energy /time$

$Energy =62*10^6*142*60 =5.2824*10^{11}$ Joule

$Energy = M*g*H +M*V^2$

where H is the height and $V^2 =30.2 m^2/s^2$ are the friction losses

Total mass of water is

$M = E/(g*H+V^2) = 5.2824*10^11/(9.81*56.7 +32.2) =8.977*10^8 kg$

Mass flow rate is

$M/t = 8.977*10^8/142/60 =1.0536*10^5 kg/sec$

Now the water is released

$Energy = M*g*H – M*V^2 = 8.977*10^8 *9.81*56.7 -8.977*10^8*32.2 =4.7*10^{11}$ Joules

or (there are losses up and down the pipes) ($losses = M*V^2$)

energy released = initial energy -2*losses $= 5.2824*10^8 -2*8.977*10^5*32.2 =4.7*10^{11}$ Joules

$P = energy/time$

$time = Energy/power =4.7*10^{11}/335*10^6 =1403 sec =23.38$ min

$efficiency = E_{out}/E_{in} =4.7/5.2824 =0.8897 =88.97%$