Quick question re engineering

Soldato
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19 Jan 2005
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Hi, just revising for my exams and memorising some engine cycles, in a few of the cycles it says "isothermal heat addition".

Isothermal means constant temperature so to me that sounds slightly oxymoronic, I know its not a mistake because its everywhere, not just in my notes but on other websites and things, I cant find a rational explanation though, am I just bein a ****?
 
Isothermal means at a constant temperature.

Temperature is a system property, and heat is a method of energy transfer to or from the system. So it's quite possible to have isothermal heat addition if the heatsink is large enough.:)

edit: or if the system is losing energy via doing work.
 
It means during expansion work and compression that the temperature in the system stays constant, this occurs at high RPM. It just simplifies the thermodynamics so that you can do the calculations.
 
I still dont get what it means though, how can you have a constant temperature while heat is added is added to it? Surely that means the temperature is going up?

I dont get this at all.
 
It means during expansion work and compression that the temperature in the system stays constant, this occurs at high RPM. It just simplifies the thermodynamics so that you can do the calculations.


No its not in the expansion or compression stage, thats adiabatic compression/expansion, where no heat is transferred.

This is what the steps say:

1-2 Adiabatic Compression
2-3 Isothermal heat addition
3-4 Adiabatic Expansion
4-1 Isothermal heat rejection

And on the T-s diagram the temperature isn't constant between steps 2-3 and 4-1.
 
I still dont get what it means though, how can you have a constant temperature while heat is added is added to it? Surely that means the temperature is going up?

I dont get this at all.

If you have a very large system, ie: the Earth, then adding heat to it won't significantly increase the temperature, so it can be considered isothermal.

Also, consider the 1st law:

Q-W = deltaU

where Q is heat addition, W is work done and deltaU is change in internal energy of the system. If you add heat then the internal energy increases, thus the temperature does also. However, if the system were to do work at the same rate as the heat is being added, then delta U = 0. Thus the temperature stays constant therefore it's an isothermal process.
 
does it mean that the thing adding heat to the fluid is a constant temperature?

Like I put a litre of water into a pan at 60 degrees and the pan stays at 60 degrees no matter what? I.e. constant temperature heat addition?

Is that a good analogy or am I still way off? :(
 
Also, consider the 1st law:

Q-W = deltaU

where Q is heat addition, W is work done and deltaU is change in internal energy of the system. If you add heat then the internal energy increases, thus the temperature does also. However, if the system were to do work at the same rate as the heat is being added, then delta U = 0. Thus the temperature stays constant therefore it's an isothermal process.

I understand that but I dont think thats what applies in this context, if you're trying to cool the system down, why would you balance that out with work so that the temperature remains constant?

This is messin with my head now.
 
does it mean that the thing adding heat to the fluid is a constant temperature?

Like I put a litre of water into a pan at 60 degrees and the pan stays at 60 degrees no matter what? I.e. constant temperature heat addition?

Is that a good analogy or am I still way off? :(

The temperature of the heat source is irrelevant.

The pan analogy is quite good.

The pan and its contents are the system. Heat is added by the hob, but the same amount is lost by conduction/radiation to the atmosphere. So it's isothermal.
 
The temperature of the heat source is irrelevant.

The pan analogy is quite good.

The pan and its contents are the system. Heat is added by the hob, but the same amount is lost by conduction/radiation to the atmosphere. So it's isothermal.


ok, I sort of get that, but if the same amount of energy being added to the system is being lost by conduction etc. why does the temperature increase?

I swear on my mother's life I am not trying to wind you up on purpose.
 
I understand that but I dont think thats what applies in this context, if you're trying to cool the system down, why would you balance that out with work so that the temperature remains constant?

This is messin with my head now.

I don't understand what you mean. You're not trying to cool the system down, the system remains at a constant temperature, with it being an isothermal process.

Are you getting confused between thermodynamic properties and methods of heat transfer?
 
So what do you need to do to it to maintain constant temperature?

Oh, I see what you did there.:)

I'm diving in trying to give the answers away, sorry...:p

edit: Anksta: think about Jokester's expanding gas situation, it's a pretty good example.
 
No its not in the expansion or compression stage, thats adiabatic compression/expansion, where no heat is transferred.

This is what the steps say:

1-2 Adiabatic Compression
2-3 Isothermal heat addition
3-4 Adiabatic Expansion
4-1 Isothermal heat rejection

And on the T-s diagram the temperature isn't constant between steps 2-3 and 4-1.

Oh sorry I thought that was what you were referring to.
 
I don't understand what you mean. You're not trying to cool the system down, the system remains at a constant temperature, with it being an isothermal process.

Are you getting confused between thermodynamic properties and methods of heat transfer?


I dont know, I'm obviously getting something confused here, I dont get why its called heat addition if the temperature remains constant I think thats my main problem.

I thought I had it for a second, heres what it says on wikiP:

WikiP said:
Reversible isothermal expansion of the gas at the "hot" temperature, TH (isothermal heat addition). During this step (A to B on Figure 1) the expanding gas causes the piston to do work on the surroundings. The gas expansion is propelled by absorption of heat from the high temperature reservoir.

Isentropic (Reversible adiabatic) expansion of the gas. For this step (B to C on Figure 1) we assume the piston and cylinder are thermally insulated, so that no heat is gained or lost. The gas continues to expand, doing work on the surroundings. The gas expansion causes it to cool to the "cold" temperature, TC.

Reversible isothermal compression of the gas at the "cold" temperature, TC. (isothermal heat rejection) (C to D on Figure 1) Now the surroundings do work on the gas, causing heat to flow out of the gas to the low temperature reservoir.

Isentropic compression of the gas. (D to A on Figure 1) Once again we assume the piston and cylinder are thermally insulated. During this step, the surroundings do work on the gas, compressing it and causing the temperature to rise to TH. At this point the gas is in the same state as at the start of step 1.

The diagram they are talking about is basically a square on a T-s diagram with A being top left and labelled clockwise.

See I thought I had it, but as jokester says, as a gas expands, it cools, so why would you add heat to it to get it to expand?
 
I dont get why its called heat addition if the temperature remains constant I think thats my main problem.

You need to seperate properties from energy transfers.

Just because you add heat doesn't necessarily mean the temperature increases. If you simply had a block of metal and you added heat by sticking it in a fire then yes, its temperaure would increase. But there are many situations where something will act to counter this temperature increase by removing energy from the system. Either the system doing work (another method of energy transfer) or by heat loss. Or even if you have an infinitely large sink for which the mass is so great that heat addition only has a negligible effect on temperature.
 
You need to seperate properties from energy transfers.

Just because you add heat doesn't necessarily mean the temperature increases. If you simply had a block of metal and you added heat by sticking it in a fire then yes, its temperaure would increase. But there are many situations where something will act to counter this temperature increase by removing energy from the system. Either the system doing work (another method of energy transfer) or by heat loss. Or even if you have an infinitely large sink for which the mass is so great that heat addition only has a negligible effect on temperature.

alright mate I think I kind of get you now, dont reckon I'm gonna have the same level of udnerstanding on this as you if we discuss for another 3 hours so I reckon its probably best to leave it there before I drive both of us mental.

Cheers for your help mate, its appreciated.

Can close this thread now if you want Donners.
 
No worries. I'm also revising this kind of stuff amongst other things at the moment. It helps to talk through it sometimes.
 
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