Electrical Generating Plant problem

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An electrical generating plant currently cools its turbines with cooling towers at 96 F, costing $55 million annually in fuel. A proposal suggests using lake water to cool the system to 50 F, potentially increasing thermal efficiency. Initial calculations indicated a savings of approximately $2.979 million, but this was deemed incorrect due to errors in the math. After converting temperatures to Kelvin and applying the Carnot efficiency formula, the estimated savings were recalculated to be around $2.4 million. Accurate calculations and conversions are crucial for determining the potential cost benefits of changing the cooling method.
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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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