Question

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.

Answer 1

Answer:

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