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HW22-1 - piping and turbines or is consumed by the various...

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PROBLEMS 1. In a BWR or PWR, steam is generated with a temperature of about 290 °c. If river water used to receive waste heat has a temperature of 20 °c, what is the maximum possible (ideal) conversion efficiency of the reactor's thermal energy into electrical energy? Nuclear power plants typically have conversion efficiencies of 34%. Why is this efficiency less than the ideal efficiency? Solution: The maximum possible thermodynamic efficiency is the Carnot efficiency given by Eq. (11.1), which, for the given temperatures, yields Tin -Tout Tin = f~~~ \ = 48%. 1]= (290 + 273) -(20 + 273) In practice the actual conversionefficiency is quite a bit less, typically around 34%. One may be tempted to think a power plant using a steam/water cycle must have huge energy losses. However, relatively little energy is lost in the
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Unformatted text preview: piping and turbines or is consumed by the various pumps. Moreover, the generator converts rotational energy to electrical energy with high efficiency. The principal reason a the efficiency of a BWR or PWR power plant is less than the ideal Carnot efficiency is because of the irreversible nature of how energy is transferred from the hot fuel to the water coolant. To obtain the ideal Carnot efficiency in a closed cycle, all energy transfer must be between elements at infinitesimally different temperatures. In a reactor, the average fuel temperature is much higher than the water/steam coolant so that the energy transfer from the fuel to the coolant is done in a very irreversible manner, thereby reducing the overall thermal efficiency of the water/steam cycle. July 24, 2002...
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