Question

The occupants of a remote village rely exclusively on a single micro hydro facility for all of their electricity needs. The hydro plant diverts water from a mountain stream 1500 feet above the facility. The flow rate varies from a summer maximum of 30 gallons per second to a winter minimum of 5 gallons per second. Determine the power available (in KW) during the summer and winter, assuming the facility is 70% efficient.

Answer 1

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:

$P=\mu \rho ghQ$

Where

μ = efficiency = 0.7

ρ = density of water = 1000 kg/m³

g = gravity = 9.8 m/s²

Q = 0.114 m³

Replacing:

$P=0.7*1000*9.8*457.2*0.114=3.575x10^{5} W=357.55kW$

In winter the power available is

$P=0.7*1000*9.8*457.2*0.019=59591.45W=59.59kW$