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 May 16th, 2016, 08:34 AM #1 Senior Member     Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 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.
 May 17th, 2016, 04:59 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,155 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions Where did you get the value for P?

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