Coal fired plant Thermodynamics help

In summary, the conversation discusses a coal fired plant that generates 600MW of electric power using 4.8x10^6kg of coal per day. The heat of combustion is 3.3x10^7 J/kg and the plant's efficiency for generating electric power is being questioned. One person suggests a formula to calculate efficiency and another person provides a different formula and calculates the efficiency to be 32.7%. The conversation also includes some humorous remarks about the plant's energy usage and potential for switching to windmills.
  • #1
Dx
Hello!

A coal fired plant generates 600MW of electric power. The plant uses 4.8x10^6kg of coal a day. The heat of combustion is 3.3x10^7 J/kg. The steam that drives the turbines is at temp of 300 degrees C and the exhaust water is at 37 degrees C. Whar is the overall effiecy of the plant for generating electric power?

I have a formula of and have substituted like so e = W/Q_H = (Q_H - Q_L)/ Q_H = 1- Q_L / Q_H.

I came up with 37% is that correct?

Thanks!
Dx :wink:
 
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  • #2
That's amazing! That plant generates enough power to light 6 bulbs, and uses ONLY 4.8 million kilograms of coal per day.

Maybe they should consider switching to windmills.

(I'm guessing it's supposed to be 600 megaWatts. Try again; show YOUR work, not just the formula you copied out of the book...)
 
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  • #3
I don't have a calculator on me, but your method is right.
 
  • #4
It's a good formula, but how are you using it?

Why not just:
(600*10^3 kW * 24 h * 3.6 * 10^6 J/kWh)/(4.8*10^6 kg * 3.3*10^7 J/kg) = 32.7%

(In this problem, aren't the steam and exhaust temperatures red herrings?)
 
  • #5
Originally posted by gnome

That's amazing! That plant generates enough power to light 6 bulbs, and uses ONLY 4.8 million kilograms of coal per day.

Maybe they should consider switching to windmills.

LMFAO!

Good Guess, Gnome!

Thanks Tom!
its right.
 
  • #6
So please enlighten me. Exactly how did you get 37%?
 
  • #7
Originally posted by gnome
It's a good formula, but how are you using it?

Why not just:
(600*10^3 kW * 24 h * 3.6 * 10^6 J/kWh)/(4.8*10^6 kg * 3.3*10^7 J/kg) = 32.7%

(In this problem, aren't the steam and exhaust temperatures red herrings?)

I did it just the way you showed me here, gnome.
 

What is a coal fired plant?

A coal fired plant is a type of power plant that uses coal as its main fuel source to generate electricity. It works by burning coal in a boiler to produce steam, which then drives a turbine to generate electricity.

How does a coal fired plant work?

A coal fired plant works by burning coal to produce heat, which is then used to boil water and create steam. The steam then turns a turbine, which is connected to a generator, and produces electricity. The remaining steam is then cooled and converted back into water, ready to be used again.

What is the efficiency of a coal fired plant?

The efficiency of a coal fired plant depends on various factors such as the type of technology used, the age of the plant, and the quality of the coal. On average, coal fired plants have an efficiency of around 33%, meaning that for every 100 units of energy produced by burning coal, only 33 units are converted into electricity.

What are the environmental impacts of coal fired plants?

Coal fired plants emit large amounts of carbon dioxide, a greenhouse gas that contributes to climate change. They also release other pollutants such as sulfur dioxide, nitrogen oxides, and particulate matter, which can have negative impacts on human health and the environment.

What is the future of coal fired plants?

The future of coal fired plants is uncertain, as there is a growing global movement towards renewable energy sources. However, coal fired plants still play a significant role in electricity generation, especially in developing countries. Efforts are being made to improve the efficiency and reduce the environmental impacts of coal fired plants through technological advancements and stricter regulations.

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