Asked by Chelsea
A nuclear power plant has an electrical power output of 1200 MW and operates with an efficiency of 31%. If excess energy is carried away from the plant by a river with a flow rate of 1.0 multiplied by 10^6 kg/s, what is the rise in temperature of the flowing water?
Answers
Answered by
drwls
If the plant efficiency is 31%, 69% of the nuclear reaction heat is carried away in cooling water. That would be (.69/.31)*1200 MW or 2670*10^6 Watts.
Transferring that heat to water flowing at 10^6 kg/s will raise the temperature by an amount deltaT, such that
2670*10^6 W = dM/dt*C*(deltaT)
deltaT = (2670*10^6 J/s)/(10^6 kg/s*4184 J/kg*C)= 0.64 C
Transferring that heat to water flowing at 10^6 kg/s will raise the temperature by an amount deltaT, such that
2670*10^6 W = dM/dt*C*(deltaT)
deltaT = (2670*10^6 J/s)/(10^6 kg/s*4184 J/kg*C)= 0.64 C
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