Query 9

course Phy 232

009. `query 8*********************************************

Question: univ phy problem 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: Work done/0.07 which is the thermal energy required. Then to find the rate of extraction we would take 210/0.07 = 3,000 kW. The cold water would be 3000-270 = 2790 kW. Each liter supplies 80 kJ per 19 degree temperature change so the 3000 kJ/sec would require about 40 liters/sec.

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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 80 kJ for the 19 deg net temp change. Needing 3,000 kJ/sec this requires about 40 liters / sec, or well over a hundred thousand liters / hour (a hundred 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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