Answer:
In summer the power available is 357.55 kW and in winter the power available is 59.59 kW
Explanation:
Given data:
height = 1500 ft = 457.2 m
30 gallon = 0.114 m³
5 gallon = 0.019 m³
In summer the power available is:
[tex]P=\mu \rho ghQ[/tex]
Where
μ = efficiency = 0.7
ρ = density of water = 1000 kg/m³
g = gravity = 9.8 m/s²
Q = 0.114 m³
Replacing:
[tex]P=0.7*1000*9.8*457.2*0.114=3.575x10^{5} W=357.55kW[/tex]
In winter the power available is
[tex]P=0.7*1000*9.8*457.2*0.019=59591.45W=59.59kW[/tex]
