Query-9

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course Phy 232

8/2/136:18

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Question: univ phy problem 20.45 11th edition 20.44 (18.40 10th edition) ocean thermal energy conversion 6 C to 27 C

At 210 kW, what is the rate of extraction of thermal energy from the warm water and the rate of absorption by the cold water?

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Your Solution:

(300-279)/300=.07

Q in=work/.07

210 kW/.07=3,000 kW

The cold water absorbs 3000-210 =2,790 kW

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Given Solution:

** work done / thermal energy required = .07 so thermal energy required = work done / .07.

Translating directly to power, thermal energy must be extracted at rate 210 kW / .07 = 3,000 kW. The cold water absorbs what's left after the 210 kW go into work, or 2,790 kW.

Each liter supplies 4186 J for every degree, or about 17 kJ for the 4 degree net temp change of the water entering and exiting the system. Needing 3,000 kJ/sec this requires about 180 liters / sec, or about 600 000 liters / hour (also expressible as about 600 cubic meters per hour).

Comment from student: To be honest, I was surprised the efficiency was so low.

Efficiency is low but the energy is cheap and environmental impact in the deep ocean can be negligible so the process can be economical. **

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