Electrical Generating Plant problem

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Homework Help Overview

The problem involves estimating the yearly savings in fuel costs for an electrical generating plant when changing the cooling method from cooling towers to a nearby lake. The temperatures involved are significant, with steam at 960 F and the cold reservoir initially at 96 F, with a proposed reduction to 50 F.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conversion of temperatures to Kelvin and the implications for calculations. There are attempts to calculate thermal efficiencies based on the Carnot cycle, and questions arise regarding the correctness of the original calculations and formulas used.

Discussion Status

Some participants have provided guidance on the need for temperature conversion and have suggested using the Carnot efficiency to approach the problem. There are multiple interpretations of the calculations being discussed, with differing results being presented without a clear consensus on the correct savings amount.

Contextual Notes

Participants emphasize the importance of showing all working steps and question the assumptions made in the calculations. There is a focus on the accuracy of mathematical reasoning and the need for clarity in the problem setup.

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Homework Statement



An electrical generating plant operates so that the steam driving its turbines is at 960 F, cooling towers keep the cold reservoir at a temperature of 96 F.
If the plant's fuel cost for one year is 55.0 million dollars when using the cooling towers, estimate the yearly saving in fuel cost if the plant used a nearby lake's water to cool the cold reservoir to 50 F while producing the same amount of electricity (in millions of dollars).

2. The attempt at a solution

I tried:

960 F = 516 C ; 96 F = 36C ; 50F = 10C.

[(516 - 36) - (516-10) ] (516-36) = -0.0542 = -5.417 = save 5.417 % = 2.979 millions of dollars save each year => Incorrect

Please help!

Thank you!
 
Last edited:
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Firstly, when working with temperatures in Physics or Chemistry, ALWAYS convert them to degrees Kelvin.

Also, what formulae are you using when you try to solve this? Your maths doesn't add up. [(516 - 36) - (516-10) ]*(516-36) does not equal -0.0542 which definitely does not equal -5.417. You must show all your working!
 
Can you assume this is a Carnot cycle? If so can you calculate the initial and final thermal efficiency? Knowing the annual costs and these two efficiencies you should be able to calculate the dollars saved.
 
joriarty said:
Firstly, when working with temperatures in Physics or Chemistry, ALWAYS convert them to degrees Kelvin.

Also, what formulae are you using when you try to solve this? Your maths doesn't add up. [(516 - 36) - (516-10) ]*(516-36) does not equal -0.0542 which definitely does not equal -5.417. You must show all your working!

Ok. Then 960F = 788.555K; 96F = 308.555K.; 50F = 283K.

\etaat 96F = 1 - Tc/Th = 1 - 308.555/788.555 = 0.6087 = 60.87%

\etaat 50F = 1 - Tc/Th = 1 - 283/788.555 = 0.6411 = 64.11%

64.11 * 55 / 60.87 = 57.93 millions dollars - 50 = 7.93 in saving. Correct?
 
Actually I got: $55 million -( $55 million * .61/.64) = $2.4 million savings
 
RTW69 said:
Actually I got: $55 million -( $55 million * .61/.64) = $2.4 million savings

Thanks, man ;)
 

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