Question

A dam is situated at a height of 550m above sea level and supplies water to a power house which is at a height of 50m above sea level. 2000kg of water passes through the turbine per second. What would be the maximum electrical power output of the power house if the whole system were 80% efficient?

Answer 1

Power = rate of transfer of energy = rate of loss of PE in this case ..

PE = mgh
PE rate of loss = (mg∆h) / t = (m/t)g∆h .. .. (m/t) = mass of water per sec (kg/s)

= 2000kg/s x 9.80 x (550-50)m = 9.80^6 J/s (W) .. at 100% efficiency

At 80% efficiency, power output = 0.80 x 9.80^6W .. .. ►P = 7.84^6 W

Answer 2

Answer:

Step-by-step explanation:

2000kg/s x 9.80 x (550-50)m = 9.80^6 J/s (W) .. at 100% efficiency  

At 80% efficiency, power output = 8MW