A town is planning on using the water flowing through a river at a rate of 4.9 106 kg/s to carry away the heat from a new power plant. Environmental studies indicate that the temperature of the river should only increase by 0.50° C. The maximum design efficiency for this plant is 32.0%. What is the maximum possible power this plant can produce?

Answer 1

Thermal power is rate of thermal energy.

P = (thermal energy) / (time) = Q/t =
= m•c•ΔT / t
= (m/t) • (c) • (ΔT)
Here m/t = 4.9•10^6 kg/s
c = specific heat of water = 4186 J/kg•C
ΔT = change in the temperature = 0.50 °C
Substituting all these values in the equation we get
P = 4.9•10^6 •4186 •0.50 =
= 1.02557•10^10 W =10.2557 GW.

Ans. 0.32 •10.2557 GW = 3.28GW

Answer 2

Thermal power is rate of thermal energy.

P = (thermal energy) / (time) = Q/t =
= m•c•ΔT / t
= (m/t) • (c) • (ΔT)
Here m/t = 4.9•10^6 kg/s
c = specific heat of water = 4186 J/kg•C
ΔT = change in the temperature = 0.50 °C
Substituting all these values in the equation we get
P = 4.9•10^6 •4186 •0.50 =
= 1.02557•10^10 W =10.2557 GW.

Ans. 0.32 •10.2557 GW = 3.28GW