Electrical Generating Plant problem

In summary, the conversation discusses the fuel cost savings of an electrical generating plant if it were to use a nearby lake's water to cool its cold reservoir instead of using cooling towers. By applying the Carnot cycle and calculating the initial and final thermal efficiencies, it is estimated that the plant can save approximately $2.4 million in fuel costs per year.
  • #1
huybinhs
230
0

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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  • #2
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!
 
  • #3
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.
 
  • #4
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.

[tex]\eta[/tex]at 96F = 1 - Tc/Th = 1 - 308.555/788.555 = 0.6087 = 60.87%

[tex]\eta[/tex]at 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?
 
  • #5
Actually I got: $55 million -( $55 million * .61/.64) = $2.4 million savings
 
  • #6
RTW69 said:
Actually I got: $55 million -( $55 million * .61/.64) = $2.4 million savings

Thanks, man ;)
 

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