A farmer is using a pumped storage system to provide electrical power to his cow

Question

A farmer is using a pumped storage system to provide electrical power to his cow shed. Power is required between the hours of 6am and midnight only, during which the demands of the shed are 40 LED lights of 10W each, a water pump requiring 3kW of power that is operating for 10% of the time and electrical machines of 5kW that are used to provide feed for the cows, clean them and scratch their backs and are operating for 5% of the time. The difference in height between the high reservoir, which has a length of 20m, a width of 15m and a depth of 3m, and the low reservoir, which is a large lake, is 18m. The high reservoir is filled to capacity overnight by pumping water from the bottom reservoir to the top reservoir using a pump that is 85% efficient. The pump capacity is 150m3 per hour and operates from 1am in the morning on low price electricity. The cow shed is powered by a turbine of 85% efficiency located at the bottom reservoir and driven by water flowing from the top reservoir. Friction in the flow of water from top to bottom reservoirs in a pipe of 10cm diameter lead to an energy loss of 2.5%. Water has a density of 1000kg.m-3.

Calculate: a. The total daily energy demand.

b. The mass of water that must flow daily from the top reservoir to the bottom reservoir.

c. The velocity of the water flow in the pipe.

d. The reduction in the water level of the top reservoir at the end of the period of operation.

e. The electrical energy required to return the top reservoir to full capacity.

f. The overall efficiency of this pumped storage system.

(Please answer all parts, Thanks)

Explanation / Answer

Power required = for 6hrs,

demand

LEds = 400W*6hrs

3000W pump for 0.6 hrs

5000W for 0.3hrs

total energy demand= 2400Whrs+1800Whrs+1500Whrs= 5.7kwhrs daily energy demand

differencein height =18m =head

volume of reservoir= 20*15*3= 900m3

Pump rateof filling = 0.85* 150 m3/hour= 127.5 in 5 hrs now,1am to 6 am volume fille d is 127.5 *6 = 765m3

turbine efficiency= 85%, loss of energy is 2.5%, ne efficeincy = 0.85*(1-0.025)=0.82875

water density = 1000kg/m3

Pth = q g h

5700=1000*q*9.81*18

Q= velocity of water flow =0.0322m3/s

in 6 hrs we consume= 0.0322*60*60*6= 695.52m3

volume left= 765-695.52= 69.48 m3

leve now= volume left /length*width= left = 69.48/(300)=0.23metres out of 3 meteres

we need to fill 695m3 water again

power required = volume * head/ 9.8 = 1.276 kW

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