
William A. answered 10/31/24
A friendly nerd specializing in math, science, and technical writing.
Hi John!
We can compute energy from pressure drop and volume. The equation reads:
E = (delta P) (V)
where
delta P = rho g h
We plug in variables and convert from liters to cubic meters:
E = (998 kg / m^3) (10 N / kg) (803 m) (1628 x 10^9 L [1 m^3 / 1000 L]) = 1.305 x 10^16 J
But the problem asks for GWh (not Joules) so we must convert:
1.305 x 10^16 J [1 GWh / 3.6 x 10^12 J] = 3,624 GWh