10 replies to this topic

[Julien123456789]

Dear All,

 

Could you please explain me why temperature increases while expanding? (sketch attached).

 

The fluid in this case is water, in an application of boiler.

 

 

Thank you in advance.

Attached Files

•  MB.png   8.68KB   4 downloads

Edited by Julien123456789, 13 April 2016 - 02:14 AM.

[breizh]

Julien ,

Consider this resource to support your work .

http://www.questcons...ermo_properties .

 

Probably also good to check into Water tables .

 

Breizh

[Julien123456789]

Thank you Mr. Breizh,

 

Julien.

[Mahdi1980]

Dear Julien

According to the 1st law of thermodynamics internal energy of fluid is equal to receiving heat minus receiving work. In other words,

(delta) E= Q-W.

 In expansion valves we can assume that the process is adiabatic due to the sudden expansion. So, Q=0.

Also, W= V. (delta)P in which delta P=P2-P1.

As here P2-P1<0 we can get W>0.

So (delta)E>0 .

As (delta)E~ (delta) T or T2-T1 we can obtain that T2-T1>0 or T2>T1

So, Expansion associated with temperature increase.

 

I Hope this is helpful for you.

 

Mahdi

[Julien123456789]

Mahdi,

Thank you for your explanation.

However, It did not convince me. Mostly, I am persuaded that you are forgetting V changes during expanding.

Thermodynamic say normally T decreases while expanding.

However, in some applications of boiler, It does not respect the above thermodynamic rule.

That's why I posted this thread.

 

Julien

Edited by Julien123456789, 13 April 2016 - 10:33 AM.

[Mahdi1980]

Julien

You are right.

However, what you are explaining is about gases that their volume is severely affected by pressure changes.

In your case, this is liquid water that its state will be remained during throttling process.

So, its volume may not have a dramatic change and the above equations that I cited would be true.

Please note that throttling may associated with cooling or heating. Not only here also in liquid state of many substances, Joule-Thomson coefficient is negative which means temp. change is opposite to pressure change.

This coefficient, on the other hand, is negative for gases.

 

I hope you are convinced.

 

Mahdi

Edited by Mahdi1980, 14 April 2016 - 03:36 AM.

[Julien123456789]

Mahdi,

This is a perfect answer.

I understand the problem now.

 

Thank you so much Mahdi !

 

Julien

[shantanuk100]

@ Mahdi,

 

1. I would like to suggest a correction in your answer. The work done is defined as always PdV, it can never be VdP. 

So you cannot consider the work done to be VdP.

 

2. Also, in this case, the work done will be =0, since there is no volume change in incompressible liquids.

 

---------------------------------------------

@ Julien,

 

1. The process described here would be adiabatic, since no heat is added to or removed from the system externally.

Also, the work done by the fluid would ideally be =0, since liquids can be considered to be incompressible.

 

2. If you were to consider it in simple terms, 

dQ = dU + d(PV)

dQ (Heat change) = dU (internal energy change) + PdV (work done by system) + VdP

 

3. Now in your case, since it is an adiabatic process (dQ=0) and a liquid, assume certain incompressibility (dV=0).

 

So we get PdV = 0

Thus, from, dQ = dU + PdV + VdP, we get,

and dU = -VdP

 

4. But, we know that the dP<0 since your pressure at the outlet is lesser. 

So, we thus get 

dU > 0

 

5. Internal energy is nothing but a measure of the inherent energy your fluid possesses by virtue of its temperature, and thus it is only dependent on the temperature of the fluid. 

Since your internal energy is increasing here (as seen from dU>0 ), your temperature thus has to proportionally increase.

That is why your outlet temperature is higher.

 

6. In the Joule Thomson effect, the fluid in consideration expands too, so there is also an additional work term involved (PdV). This is why the internal energy change there has also got a possibility of becoming negative, which is why we see lower outlet temperatures too in JT effect.

 

But in this case, there is no work done due to incompressibility of fluid, so your temperature of the fluid can only increase when outlet pressure is lesser, as per the equation.

 

7. Also, if I were to put it in simple terms, in the JT effect, the heat contained in a fluid is utilized in doing work on that fluid, thus lowering overall heat content of the fluid itself and thus reducing the temperature.

But in this case, there is no work done. So whatever reduction in pressure is observed has to be compensated only by increasing the temperature of the fluid since work cannot be done.

 

Regards,

Shantanu

Edited by shantanuk100, 15 April 2016 - 01:54 AM.

[breizh]

http://processengine...damentals.html#

 

Some info about Thermo !

 

Breizh

[Julien123456789]

@ Mahdi,

 

1. I would like to suggest a correction in your answer. The work done is defined as always PdV, it can never be VdP. 

So you cannot consider the work done to be VdP.

 

2. Also, in this case, the work done will be =0, since there is no volume change in incompressible liquids.

 

 

@ Julien,

 

1. The process described here would be adiabatic, since no heat is added to or removed from the system externally.

Also, the work done by the fluid would ideally be =0, since liquids can be considered to be incompressible.

 

2. If you were to consider it in simple terms, 

dQ = dU + d(PV)

dQ (Heat change) = dU (internal energy change) + PdV (work done by system) + VdP

 

3. Now in your case, since it is an adiabatic process (dQ=0) and a liquid, assume certain incompressibility (dV=0).

 

So we get PdV = 0

Thus, from, dQ = dU + PdV + VdP, we get,

and dU = -VdP

 

4. But, we know that the dP<0 since your pressure at the outlet is lesser. 

So, we thus get 

dU > 0

 

5. Internal energy is nothing but a measure of the inherent energy your fluid possesses by virtue of its temperature, and thus it is only dependent on the temperature of the fluid. 

Since your internal energy is increasing here (as seen from dU>0 ), your temperature thus has to proportionally increase.

That is why your outlet temperature is higher.

 

6. In the Joule Thomson effect, the fluid in consideration also expands, so there is also an additional work term involved (PdV). This is why the internal energy change there has also got a possibility of becoming negative, which is why we see lower outlet temperatures too in JT effect.

 

But in this case, there is no work done due to incompressibility of fluid, so your temperature of the fluid can only increase when outlet pressure is lesser, as per the equation.

 

7. If I were to put it in simple terms, in the JT effect, the heat contained in a fluid is utilized in doing work on that fluid, thus lowering overall heat content of the fluid itself and thus reducing the temperature.

But in this case, there is no work done. So whatever reduction in pressure is observed has to be compensated only by increasing the temperature of the fluid since work cannot be done.

 

Regards,

Shantanu

 

Shantanuk,

Thank you so much for your very clear answer.

I completely agree with you !

 

Julien.

[Julien123456789]

http://processengine...damentals.html#

 

Some info about Thermo !

 

Breizh

 

Mr. Breizh,

Thank you so much for your document.

You always provide me the interesting resources.

 

Julien.