Solve this equation for M, the water mass that must be pumped to a height of H = 500 m.
M g H = 160*10^6 W * 5400 s = 8.64*10^11 Joules
Use g = 9.8 m/s^2
Divide the mass by the density for the required volume.
To store the energy produced in 1.5 hour by a 160 MW (160 10^6 W) electric-power plant, how many cubic meters of water will have to be pumped from the lower to the upper reservoir? Assume the upper reservoir is 500 m above the lower and we can neglect the small change in depths within each. Water has a mass of 1000 kg for every 1.0 m^3. Express your answer using two significant figures.
5 answers
where would I use the 9.8m/s^2? And the density is which number?
EMily, your question baffles me.
MgH=some number
Mass=some number/gH
Mass/volume=density
Volume=Mass/density
MgH=some number
Mass=some number/gH
Mass/volume=density
Volume=Mass/density
I understand the mass should be divided by the density for the required volume. I don't think that I am using the correct numbers. I came up with a volume of 172800 or 1.7 x 10^5 but the answer is incorrect. Can you please help?!! I did (1000V)(10)(500) = 1.7x 10^5
Use g in the first formula I gave you, to get M. They gave you the density nof water, 1000 kg/m^3.
I get 1.763*10^8 m^3 for the mass and
1.763*10^5 m^3 for the volume
I get 1.763*10^8 m^3 for the mass and
1.763*10^5 m^3 for the volume