Physics Physics Forum

 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: 2,143 Thanks: 726 Math Focus: Physics, mathematical modelling, numerical and computational solutions Where did you get the value for P?

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post shunya Elementary Math 8 March 9th, 2016 07:17 PM girlbadatmath Chemistry 2 November 6th, 2014 01:25 PM ChristinaScience Algebra 2 January 25th, 2014 12:23 PM dave daverson Algebra 1 October 28th, 2012 06:35 PM dave daverson Calculus 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top