Nuclear Power Plant Efficiency and Cooling Calculation

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Homework Help Overview

The discussion revolves around the efficiency calculations of a nuclear power plant, specifically focusing on thermal efficiency, actual efficiency, and the cooling process of water in the condenser. The subject area includes thermodynamics and energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the concept of Carnot efficiency and its application to the problem. There are attempts to calculate the actual efficiency of the plant using given power outputs. Questions arise regarding the heat exchange process, particularly concerning the mass of steam and the cooling water's exit temperature.

Discussion Status

Some participants have provided calculations and reasoning regarding the efficiency and cooling temperature. There is an ongoing exploration of the energy conservation principle, with participants questioning the fate of the remaining energy not converted to electrical power. While some calculations have been presented, there is no explicit consensus on the correctness of the final temperature calculation.

Contextual Notes

Participants note the confusion regarding the mass of steam and the assumptions necessary for the calculations. The discussion also highlights the need for clarity on the heat transfer equations and the parameters involved in the cooling process.

bcjochim07
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Homework Statement


A nuclear power plant generates 2000 MW of heat energy from nuclear reactions in the reactor's core. This energy is usd to boil water and produce high pressure steam at 300 degrees C. The steam spins a turbine, which produces 700 MW of electric power, then the steam is condensed and the water is cooled to 30 degrees before starting the cycle again

a)What is the maximum possible thermal efficiency?
b)What is the plant's actual efficiency?
c) Cooling water from a river flows through the condenser ( the low temperature heat exchanger) at the rate of 1.2 * 10^8 L/hr. If the water enters the condenser at 18 degrees C, what is its exit temperature?


Homework Equations





The Attempt at a Solution


a) Carnot efficiency = 1 - Tc/Th
1- (303K/573K) = .471

b) rate of Wout = 700 MW = 7 * 10^8 J/s
Rate of Qh = 2000MW = 2 * 10^9 J/s

Efficiency = (7*10^8J/s)/(2*10^9J/s)=.35

c) This is the part that I am confused about. I was going to use the formula for heat and set up an equation for heat lost = heat gained, but I don't know the mass of the steam. Is there a formula I'm not thinking of?
 
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Ok, I just had an idea

rate of work out = rateQh-rateQc
(7.0*10^8)=(2*10^9J/s)-Qc
Qc= 1.3 * 10^9 J/s

and so 1.3 *10^9J/s * 3600s/1.2*10^8kg = 39000J/kg
Q=mcT
39000 = 1kg*4160T
T= 9.3 so the final temp is 27.3 degrees C Is this correct?
 
2000 MW of heat is available.
700 MW of that becomes electrical power.
Energy must be conserved, what happens to the remaining energy?
 
bcjochim07 said:
final temp is 27.3 degrees C Is this correct?

Looks good.

(I didn't check with a calculator, but doing rough estimations in my head your numbers all look good.)
 

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