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# hw26a - aldawsari(ma28683 HW26 Betancourt(16856 This...

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aldawsari (ma28683) – HW26 – Betancourt – (16856) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 20.3-20.7 001 10.0 points The efficiency of a 905 MW nuclear power plant is 21 . 5%. If a river of flow rate 3 . 19 × 10 6 kg / s were used to transport the excess thermal energy away, what would be the average temperature increase of the river? Correct answer: 0 . 247451 C. Explanation: Given: P output = 905 MW = 9 . 05 × 10 8 W and e = 21 . 5% = 0 . 215 . 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 9 . 05 × 10 8 W 0 . 215 parenrightbigg (1 - 0 . 215) = 3 . 3043 × 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 . 3043 × 10 9 W (3 . 19 × 10 6 kg / s) (4186 J / kg ·

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hw26a - aldawsari(ma28683 HW26 Betancourt(16856 This...

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