1000.0 gpm (gallons-per-minute) of water is released from the reservoir down the hill in order to generate electricity during the day (peak hours) for a 12-hour period each day. The drop of the water is 50.0 m and has negligible friction. Calculate how much energy (in kWh/day) can be generated on a daily basis, if the turbine is 85% efficient. Assume the density of the water is 1000.0 kg/m3.

E = 378.7 kWh / day

Explanation:

Formula to calculate the energy generated (E):

E=P*t Formula (1)

P = Power in kilowatts (kW)

t = period the time to generate (h)

The formula for calculating hydraulic power of the turbine is as follows:

P = Q * p*η Formula (2)

Where:

P = Power in kilowatts (W)

Q: Water flow (m³/s)

p : Water pressure in Pascals (Pa)

η : efficient of the turbine (%/100)

The formula to calculate the Water pressure is:

p = ρ*g*H Formula (3)

p: pressure in Pa (N/m²)

ρ = density of the fluid (kg / m³)

g: acceleration due to gravity (m/s²)

H : The drop of the fluid (m)

Data

Q = 1000 gpm

1 gpm = 7,5768 * 10⁻⁵ m³/s

Q = 1000 * 7,5768 * 10⁻⁵ m³/s

Q = 7,5768 * 10⁻² m³/s

H = 50 m The drop of the water

η = 85% = 85/100 = 0.85

ρ = 1000.0 kg/m³ : density of the water

g = 9.8 m/s²

t = 12-hour

Calculation of the Water pressure

We replace data in the formula (3)

p = (1000 kg/m³) * (9.8 m/s²) (50 m)

p = (1000 kg/m³) * (9.8 m/s²) (50 m)

p = 490000 (kg*m/s²) / m² = 490000 (N/m²) = 490*10³ Pa

Calculation of the Power

We replace data in the the formula (2):

P = Q * p*η

P = (7,5768 * 10⁻² m³/s) * (490*10³N/m²) * (0.85)

P = 31557.37 (N*m) / s = 31557.37 W

1 W = 1 kW / 1000

P = (31557.37 / 1000) kW

P = 31.56 kW

Calculation of the Energy

We replace data in the the formula (1):

E=P*t

E=31.56 kW * 12 h

E = 378.7 kWh / day