How is Work Potential Destroyed in Heat Transfer Between Reservoirs?

[mullzer]

Hi I am studying this topic at the moment and came across this question but cannot solve it. Any help would be much appreciated.

Heat is transferred from a resevoir at 1200C to a second resevoir at 350C at a rate of 500 KJ/s.
Calculate the rate at which work potential is destroyed during this process.

I know the formula for irresversibility is: i = [w][/rev] - <sub>w[/actual] = [T][/0][/gen] but i cannot fit these numbers into it?

thanks in advance</sub>

 

[mullzer]

sorry that's : irreversibilty = reverible work - actual work

 

[lalbatros]

I don't see how you could answer this question.
The work that can be done depends on the cold source which is available.
Being in Groenland or in Sahara could make a difference.
And any way, the available cold source could depend on more than the geographic location.

Did the exercice not mention something about the cold source?

 

[mullzer]

yes you're right. I believe that you presume the cold temperature to be room temperature, or 18C. does that help?

 

[Andrew Mason]

I don't see how you could answer this question.
The work that can be done depends on the cold source which is available.
Being in Groenland or in Sahara could make a difference.
And any way, the available cold source could depend on more than the geographic location.

Did the exercice not mention something about the cold source?

To determine loss of potential work use 350C as the cold temperature. The loss of work potential is the maximum amount of work one could get out of a heat engine operating between those two temperatures.

AM

 

[lalbatros]

mullzer,

This is easy to figure out, based on the Carnot efficiency:

(see http://en.wikipedia.org/wiki/Exergy" )

Let's assume, for your example, these values:

Tc = 273 + 18 = 291 K
Th2 = 273 + 1200 = 1473 K
Th1 = 273 + 350 = 623 K
Qtransfered = 500 kW

For the highest and lowest temperatures,
the maximum work that could be done is respectively:

Wmax2 = 500 * (1-Tc/Th2) = 401 kW
Wmax1 = 500 * (1-Tc/Th1) = 266 kW

Therefore the work destroyed by heat transfer from Th2 to Th1 is:

Wmax2 - Wmax1 = 401-266 = 135 kW

You can also see the relation to entropy production by observing that:

Wmax2 - Wmax1 = Tc (Qtransfered /Th2 - Qtransfered /Th1) = Tc (DS2 - DS1)

 

[mullzer]

Thanks. I see that the heat transfer to the environment is important

So Wmax2 is the heat transferred from the hottest resevoir to the environment and Wmax1 is the heat transferred from the 350C resevoir to the enviroment. The difference between these is the maximum amount of heat that could be transferred from Th2 to Th1, am I right?

But surely this maximum (135kW) should be larger than the actual (500kW)?

 

[mullzer]

oh no I think I understand now.

It is like comparing 2 engines operating at room temp, one with a higher Th (Th2) than the other (Th1). If you use the lower Th than the max amount of work potential lost is the difference bewteen the max amount of work for the engine at Th2 and Th1.

 

[Andrew Mason]

mullzer,

This is easy to figure out, based on the Carnot efficiency:

(see http://en.wikipedia.org/wiki/Exergy" )

Let's assume, for your example, these values:

Tc = 273 + 18 = 291 K
Th2 = 273 + 1200 = 1473 K
Th1 = 273 + 350 = 623 K
Qtransfered = 500 kW

For the highest and lowest temperatures,
the maximum work that could be done is respectively:

Wmax2 = 500 * (1-Tc/Th2) = 401 kW
Wmax1 = 500 * (1-Tc/Th1) = 266 kW

Therefore the work destroyed by heat transfer from Th2 to Th1 is:

Wmax2 - Wmax1 = 401-266 = 135 kW

You can also see the relation to entropy production by observing that:

Wmax2 - Wmax1 = Tc (Qtransfered /Th2 - Qtransfered /Th1) = Tc (DS2 - DS1)

You can replace a single Carnot engine operating between 1473K and 291K with two Carnot engines: one operating between 1473K and 623K and the other operating between 623K and 291K using the output of the first as the input of the second. The two engines are thermodynamically equivalent to the first (https://www.physicsforums.com/showthread.php?t=173879"). Let's assume that the first engine draws 500Kw of heat flow from the hot reservoir. The efficiency is 1 - 623/1473 = .577 so it does work at the rate of 500*.577 = 289Kw and delivers heat at the rate of 211Kw to the intermediate 623K reservoir. The second Carnot engine uses 211 Kw of heat flow as input to produce work at the rate of 211*(1-291/623) = 112 Kw. The total work is 401Kw.

So by "wasting" 500Kw from 1473K to 623K, you lose the output of the first engine. You are then left only with the second. So instead of 401Kw you get only 112Kw, a loss of 289Kw of power.

AM

 

[lalbatros]

Andrew,

The question reads:

"Heat is transferred from a resevoir at 1200C to a second resevoir at 350C at a rate of 500 KJ/s. "

This implies that all the heat is transferred, not only that part that would be released by some intermediate engine.

Michel

 

[Andrew Mason]

Andrew,

The question reads:

"Heat is transferred from a reservoir at 1200C to a second reservoir at 350C at a rate of 500 KJ/s. "

But none of that heat does work. Each second you transfer 500 KiloJoules of energy and you get no work out of it. If you take that same 500 KiloJoules and run it through a Carnot engine you get 289 KiloJoules of work out of it. So it seems to me that the loss of work potential is 289 KJ/sec.

AM

 

[lalbatros]

Andrew,

These 289 kW are not completely lost.
This heat is still available at 623K to produce work down to 291K.
This amount of work is 289*(1-291/623) = 154 kW,
and the heat actually lost is 289 - 154 = 135 kW.

The second point of view leads to the same result.

Michel

 

[Andrew Mason]

Andrew,

These 289 kW are not completely lost.
This heat is still available at 623K to produce work down to 291K.
This amount of work is 289*(1-291/623) = 154 kW,
and the heat actually lost is 289 - 154 = 135 kW.

The second point of view leads to the same result.

Perhaps we should have the complete question as it is given. I agree that you can theoretically reduce the heat waste to nil and get all of the heat flow out as work if you make the cold reservoir cold enough (down to absolute 0). But the question as posed mentioned 2 reservoirs, one at 1200C and the other at 350C. There are no other reservoirs mentioned.

AM

 

[lalbatros]

In real life 350°C is more than enough to produce power.
Some applications are using the Organic Rankine Cycle to produce power from waste heat or from geothermal sources. The hot source can then be at 150°C.

That's why I asked mullzer about the cold source.
He indicated it was 18°C.

 

[RonL]

In real life 350°C is more than enough to produce power.
Some applications are using the Organic Rankine Cycle to produce power from waste heat or from geothermal sources. The hot source can then be at 150°C.

That's why I asked mullzer about the cold source.
He indicated it was 18°C.

I have looked at a few of your post's, threads and questions, I'll not try to describe the mechanics of operation, rather let you build in your mind what I see in mine.

When looking at the Organic Rankine Cycle, a diagram is shown that illustrates T-high, a work device, and T-cold.
The challenge is to build a vision of the work device inside T-high, then insulate T-high and install it inside T-cold. If T-cold is in an environment of warmer temperature a spontaneous flow of thermal energy will move in as long as T-cold is lower.
This can be sustained by moving work out of the system.(best is likely electric output)

A gas flow cycle inside will keep temperatures at a design point of best performance.

Jet ejectors, vortex tubes, compressors and expander's, flywheels, and generators can be assembled in many configurations to keep this gas flow continual or a pulse cycle system.

I see at least two supplies of thermal energy, (1) liquid to liquid transfer through the shell of T-cold (2) If air is taken in and compressed inside of the T-cold, then exhausted at a much colder temperature, the sum of thermal energy ingested by the system has to be balanced by the work sent out.

Waste heat from any source is continually recycled, not LOST.

I think you seem to have the mechanical aptitude to be able to plumb this together, my efforts are slow, as I continually change the components and their locations and sizes.

It's all about heat producing an ability to keep a cycle of gas flowing.

RonL

 

[Andrew Mason]

In real life 350°C is more than enough to produce power.
Some applications are using the Organic Rankine Cycle to produce power from waste heat or from geothermal sources. The hot source can then be at 150°C.

That's why I asked mullzer about the cold source.
He indicated it was 18°C.

I agree. But the original question did not mention a third reservoir. It just asked for the rate at which potential work is lost in flowing the heat to the 350C reservoir. So that is why I asked for the complete question.

AM

 

[lalbatros]

Andrew,

The original question was indeed puzzling, why my first question.
Strictly speaking, if there were only two reservoir, the work lost would have been 500 kW.
All the heat transferred would then have been missed for power production.
This is 100% loss (of exergy).

In your first answer, why did you come with 289 kW?
I think this indicate you had already assumed the existence of a third reservoir.

Michel

 

[Andrew Mason]

Andrew,

The original question was indeed puzzling, why my first question.
Strictly speaking, if there were only two reservoir, the work lost would have been 500 kW.
All the heat transferred would then have been missed for power production.
This is 100% loss (of exergy).

But that is the point of the question. Not all of the heat can be used to generate work. The most efficient heat engine will still waste 211KJ of the heat flow that occurs each second. If more work could be generated and less wasted heat flow, the second law would be violated.

In your first answer, why did you come with 289 kW?
I think this indicate you had already assumed the existence of a third reservoir.

No. That is the maximum amount of work you can get out of a heat engine connected between two reservoirs at those temperatures. By letting the heat flow freely, that work potential with respect to those two reservoirs is lost

AM

 

[lalbatros]

You are right Andrew.
The rules of the game were not very clear.
Answering this minimal information question with practical applications in mind reminds me of an old french joke:

military examinator: "how long does the gun barrel to cool down"
soldier: "for a big barrel, I think, it must be one minute by cold weather"
military examinator: "no, this not right, try again"
soldier: "..."
examinator: "..."

finally, the examinator gives the right answer to the soldier, reading the manual:
examinator reading: "the gun barrel takes a certain time to cool down"​

Personally, my preferred answer is that no work was lost,
since all the heat can still be converted to work by using a very cold reservoir.
(as you suggested)