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May 16th, 2016, 08:34 AM   #1
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Math Focus: Certainty
Temperature Gradient

You are part of a team of engineers designing an electric power plant that makes use of the temperature gradient in the ocean. The system is to operate between 20.0°C (surface-water temperature) and 5.00°C (water temperature at a depth of about 1 km).

I'm asked to find energy/hour in TJ/hour

$\displaystyle P=\frac{W}{t} --> W=Pt$

$\displaystyle e=\frac{W}{Q_H} --> Q_H=\frac{W}{e}$

$\displaystyle Q_H=\frac{Pt}{e}=\frac{70\times 10^6\cdot 3600}{0.051}=4.9\times 10^{12} J$

$\displaystyle e = 1 - \frac{(273+5)}{(273+20} = 0.051$

$\displaystyle Energy(E)=Q_H=4.9 TJ$ correct?

Then again TJ/h therefore $\displaystyle Energy(E)=Q_H=4.9/3600=1.36\times10^{-3} TJ/h$

Thank you in advance

Last edited by hyperbola; May 16th, 2016 at 08:39 AM.
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May 17th, 2016, 04:59 AM   #2
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Where did you get the value for P?
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