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Author: Subject: Efficiency of burning fuel for heat vs internal combustion or turbine motors
RogueRose
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[*] posted on 11-12-2016 at 06:05
Efficiency of burning fuel for heat vs internal combustion or turbine motors


I'm trying to figure out the difference in fuel efficiency (turning fuel into movement) for something like a standard gas/diesel engine or even turbine engine as compared to burning fuel for heat. In the IC/turbine engines I've heard efficiency is anywhere from 20-50% as the rest is lost as heat. IDK if those numbers are accurate or if they may be old and efficiencies have increased.

I guess the best comparison would be to look at a generator in KW produced using a car engine, turbine engine or I guess a steam (boiler/heat powered) engine to turn the generator.

As in the steam engine, heat is the goal of the fuel where IC's, heat is more of a by-product.

It would seem that the steam turbine is the most efficient as that is what is used for electricity generation with oil/gas/coal/nuclear - but how much better is it than the best of other methods?
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[*] posted on 18-12-2016 at 18:28


Fires can create nearly 100% heat from the fuel, if burned efficiently, just like electricity can be converted to heat at near 100% efficiency. But that is not that special because heat is the by-product formed from any form of energy transfer or usage. So a incandescent light bulb makes ~`10% light and 90% heat (some like IR is kinda both in a sense).

In a car, the fuel is converted to about 30% kinetic energy and about 70% heat that is removed via the exhaust and radiator. In the winter that helps heat the car, in the summer, it makes the A/C work harder. So the most efficient use of most fuel is to burn it in a process that both makes usable kinetic energy as well as usable heat. Thus modern gas power plants use a gas turbine engine to drive a generator, plus they use the heat to boil water and drive a steam turbine also. Combined cycle systems can derive almost 50% of the fuel into useful power. And if you can heat a building or water with the remaining heat, that is even better. For nuclear, the heat can only be used for boiling water, so there is less efficiency, but the amount of heat available is much more per pound of fuel.

Thermodynamics unfortunately says that getting more that you put in is tough, but there are ways to cheat a little, mostly if you want heat, since heat pumps can generate more heat than the electricity used by them. But if you want to make electricity, it is hard to make more than about 50% efficiency.
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[*] posted on 18-12-2016 at 19:10


You need to define the terms of the question a little better, but-

Generally, efficiency = energy in ÷ useful work out, and you never quite reach unity.

In a heat engine, this derives from the difference between the input and final output temperatures. Hotter gas in and/or cooler gas out, higher efficiency.

How is your math and physics background?




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[*] posted on 18-12-2016 at 19:31


Last I checked, the best diesel engines were slightly better than the best turbine engines, but it's important to remember that best diesel engines are usually turbocharged (with an exhaust turbine powering an air compressor at the intake). Diesel engines usually use a high compression ratio; a high compression ratio translates to higher maximum theoretical efficiency. Diesel engines have a lot more moving parts and require higher maintenance than simple turbines, so turbine engines are usually used in large power plants despite slightly lower efficiency.

If you add heat generation to energy efficiency calculations, you can greatly increase efficiency.

[Edited on 19-12-2016 by JJay]




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Metacelsus
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[*] posted on 18-12-2016 at 20:27


If you want to heat something by burning fuel, you could get better than "100%" efficiency by using the fuel to run a heat pump, and making somewhere else colder.



As below, so above.
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[*] posted on 18-12-2016 at 21:11


Quote: Originally posted by Metacelsus  
If you want to heat something by burning fuel, you could get better than "100%" efficiency by using the fuel to run a heat pump, and making somewhere else colder.


Last week, I was using a rope- But when I got to the end, it was too short!

No problem, just went back to the other end, cut off what I needed and tied it onto the short end...

https://youtu.be/VnbiVw_1FNs

[Edited on 19-12-2016 by Bert]




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[*] posted on 18-12-2016 at 21:29


Heh heh.

But I think that metacelsus is correct. If you have a good heat pump that for example moves 3 joules of heat energy for every joule of energy used by the device, and you run that heat pump on electricity that was made from fuel at 50% energy efficiency, you come out at 150% compared to burning the fuel and heating directly.




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[*] posted on 18-12-2016 at 22:06


You could just use a passive heat pump and achieve infinite percent efficiency.



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[*] posted on 19-12-2016 at 01:08


You are bringing energy into this from elsewhere, we were talking about efficiencies of particular heat engines?

Broadly stated, the 3 laws of thermodynamics are:

You can't win.

You can't break even.

You can't quit the game.

https://en.m.wikipedia.org/wiki/Ginsberg's_theorem







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[*] posted on 19-12-2016 at 03:00


ROFLMAO! That's the best explanation I've ever heard.



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[*] posted on 19-12-2016 at 07:22


A heat pump is just a specially configured refrigerator (or if you refer, an air conditioner working in reverse).

Think about it.

What a refrigerator does is move heat from the inside of the refrigerator, requiring electricity to do it. That heat, plus all the work done by the refrigeration system is radiated from the cooling coil.

When you make a heat pump the heat coming out of the coil is the product, and you are refrigerating the Great Outdoors. A commercial heat pump can move 3 to 4 J of heat for each J of electricity consumed.

Like all heat engines, the theoretical performance of all refrigerators and heat pumps is simply calculated from the Carnot cycle relationships.

Cp = Th/(Th - Tc)

In the special (but common) case where the initial temperature is the same on both sides:
Cp = (T + Δh)/((T + Δh) - (T + Δc))
=(T + Δh)/(Δh - Δc)

Δc is a negative number so (Δh - Δc) can be replaced by the absolute value of the combined temperature change of the two sides.

But in the limiting case of slight cooling to a very mass of the outside, then it becomes approximately:
(T + Δh)/Δh

[Edited on 19-12-2016 by careysub]




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Metacelsus
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[*] posted on 19-12-2016 at 11:55


Good explanation; I was going to give one myself, but I don't really have anything more to add.

Back on the topic of engines:
For things like steam engines, the efficiency generally increases the hotter the "hot" side of the engine becomes. Practically, this is limited by the ability of the engine to withstand the heat.
The best efficiency of steam turbines is ~50% (but these are the really big supercritical turbines). High compression diesel engines have ~45% efficiency, and gasoline engines have ~30% (because the compression ratio is smaller).




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