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# hw30b - aljabr(faa335 Hw30 Ross(89251 This print-out should...

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aljabr (faa335) – Hw30 – Ross – (89251) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The efficiency of a 846 MW nuclear power plant is 18 . 5%. If a river of flow rate 2 . 8 × 10 5 kg / s were used to transport the excess thermal energy away, what would be the average temperature increase of the river? Correct answer: 3 . 17979 C. Explanation: Given: P output = 846 MW = 8 . 46 × 10 8 W and e = 18 . 5% = 0 . 185 . The excess thermal energy transported per second by the river is P excess = P input (1 - e ) = parenleftbigg P output e parenrightbigg (1 - e ) = parenleftbigg 8 . 46 × 10 8 W 0 . 185 parenrightbigg (1 - 0 . 185) = 3 . 72697 × 10 9 W where e is the efficiency and P output the power output of the plant. Thus the temperature of the river is increased (per second) by d m d t c Δ T = d Q d t = P excess where c is the heat capacity of water and d m d t is the flow rate of the water. Thus Δ T = P excess d m d t c = 3 . 72697 × 10 9 W (2 . 8 × 10 5 kg / s) (4186 J / kg · C) = 3 . 17979 C .

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