Assignment 9

course Phy 232

009. `query 8

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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:

In the first part of the problem, we were asked to find the efficiency of the power plant. I got an efficiency of 7%.

Using the equation: Q(H) = W/e = 210 KJ/0.07 = 3000 kJ

So we need to find the amount of heat per second to extract this amount of power.

Work = Q = mc*delta T = rho*volume*c*delta T

Solving for volume gives us: V = Q/(rho*c*delta T) = (3000 kJ)/(4190) J/kg*K *21K *1.03*10^3 kg /m^3) = 0.033 m^3/s

We need this t be a flow rate, so it would give us 0.033 m^3/s.

Confidence Rating: 2

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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. **

Your Self-Critique:

My answer is very close to this one. I got 33.1 L/s. It seems like my process is right though>

Your Self-Critique Rating: 2

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&#Your work looks good. Let me know if you have any questions. &#