Question

3. Consider a water fall with a flow rate of 106 kg/s and a height of 60 m. If all of the kinetic energy of the water is used in turning a turbine generator to produce 400 MW of electrical energy, what is the change in temperature of the water between top of the falls and the outlet of the turbine at the bottom?

Answer 1

given
mass flow rate, dm/dt = 10^6 kg/s
h = 60 m

The rate at which potential energy lost, dU/dt = (dm/dt)*g*h

= 10^6*9.8*60

= 588*10^6 W

= 588 MW

The rate at which thermal energy dissipated = the rate of potential energy lost - the rate of kinetic energy gained

= the rate of potential energy lost - the rate of electrical energy generated.

= 588 - 400

= 188 MW

Thermal energy generated per second while water falling down,

Q = 188*10^6 J

let delta_T is the increase in temperature of water.

use, Q = m*c*delta_T

==> delta_T = Q/(m*c)

= 188*10^6/(10^6*4186)

= 0.0449 degrees Celsius <<<<<<<--------------Answer