Topic: Physics Problem please?
July 20, 2019 / By Colette Question:
A power plant has a power output of 1055 MW and operates with an efficiency of 33.0 percent. Excess energy is carried away as heat from the plant to a nearby river that has a flow rate of 1.1 x 10^6 kg/s. How much energy is transferred as heat to the river each second?
Berenice | 2 days ago
Total Work output of power plant:
W_dot_net = 1055 MW
If that's the total output, use efficiency definition to find heat input:
eta = what we want/what we pay for
eta = W_dot_net/Q_dot_in
Solve for Q_dot_in:
Q_dot_in = W_dot_net/eta
By conservation of energy:
Q_dot_in = W_dot_net + Q_dot_out
Solve for Q_dot_out:
Q_dot_out = Q_dot_in - W_dot_net
Q_dot_out = W_dot_net*(1/eta - 1)
Plug in data:
W_dot_net :=1055 MW; eta:=0.33;
Result for Heat Rejection:
Q_dot_out = 2142 Megawatts
Which means 2142 MegaJoules of heat are rejected to the river each second.
Output Power = 1055e6 watt
Input Power = 1055e6/0.33 = 3197e6 watt
Power lost in river as heat = (3197 - 1055)e6 = 2142e6 watt = 2142e6 J/s
Amount of water flowing = 1.1e6 kg/s
Amount of heat removed by river per sec = 2142/1.1 = 1947 J/kg ANS
add'l info: as the specific heat of water is: 4190 J/kg-C°
The river water temperature would increase by: 1947/4190 = 0.5°C
2142.66 MJ, or 511705.7 kcal.
1% of total power input is 31.979797........... MW which rounds up to 31.98. 67 times this is 2142.66 MW.