Q. A dam is situated at a height of 550 m above sea level and supplies water to a power house which is at a height of 50 m above sea level. 2000 kg of water passes through the turbines per second. What would be the maximum electrical power output of the power house if the whole system were 80% efficient?
A) 8 MW
B) 10 MW
C) 12.5 MW
D) 16 MW
Explanation
(a) $h = 500\ m,\frac{dm}{dt} = 2000\ kgs^{- 1}$
power output $= \frac{80}{100} \times \frac{dm}{dt}gh$
$= \frac{4}{5} \times 2000 \times 10 \times 500W$
$= 8 \times 10^{6}W = 8MW$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.