Thermodynamic of Depressurising Water 3

[Nosey]

we're looking at temperature build-up in a pump minimum flow recycle loop and I've been reliably informed that the temperature of water will increase when water is depressurised i.e. when letting water pressure down from 100 bar to 30 bar the temperature could increase by 1 degC (I've used random figures here but you get the idea).

I've never come across this before, so I have the following questions for the forum:
- Has anybody come across this before?
- What're the thermodynamics behind it?
- And, most importantly, where can I do some background reading on it to fully understand it?

Thanks for your help.

 

[MortenA]

if you let water (or any fluid) through a valve/orifice then there will be a temperature increase across the valve. Thats the work performed by the fluid thats "stored" in the liquid. But for the upstream part im notr sure.

Best regards

Morten

 

[sailoday28]

Morten
I don't have a complete table of J-T coef for water, however, in general, it is possible for the J-T coef can be +,0, or negative.
Regards

 

[Bribyk]

What you have is a "constant-enthalpy throttling process", any thermodynamics textbook should cover this. And, yes, dropping water at 40 degC's pressure from 100 bar to 25 bar (didn't feel like using double-interpolation) will result in a temperature increase of about 1.6 degC.

Brian Bobyk - Hoerbiger Canada

 

[ginsoakedboy]

Nosey,

The wikipedia article is very informative:

http://en.wikipedia.org/wiki/Joule–Thomson_effect

(sorry, but this was the best source when I was trying to understand this effect.)

J-T coefficients of water and many other substances can be obtained from the NIST webbook at:

http://webbook.nist.gov/chemistry/fluid/

 

[MortenA]

well i wouldnt consider JT for a liquid - but of course its similar. Theres just very little cooling from the expansion compared to the heating from the work performed over the valve.

Best regards

Morten

 

[FredRosse]

When the liquid flows through the throttling valve, the pressure energy is converted into velocity energy, and this velocity energy (kinetic energy) is then dissipated by the internal friction of turbulence, this is a shearing process that turns directly into heat. Even without significant velocity, there are shear stresses set up with the flow through the restriction (shear between the pipe wall and the liquid, for example), and this shearing process also directly shows up as heated liquid.

The process is not JT, as you are not expanding the liquid much, as does occur with a gas.

 

[dcasto]

look up the enthalpy at P1, T1. The go to p2 and find at what temperature has the same exact enthalpy. There is no work going across a valve. look at the entropy for the two points, it went up huh?

remember our laws of thermo....

 

[25362]

As I am given to understand, practically all fluids (liquids included) undergoing a J-T throttling process, will heat up when their reduced temperatures are below 0.5, regardless of pressure.

 

[ginsoakedboy]

FredRosse,
Where is the energy to shear coming from? From within the fluid itself. Your shearing explanation leads precisely to Joule Thomson effect.

It all depends on your initial state. Whether, J-T effect leads to cooling or heating simply depends on the sign of the J-T coefficient.

 

[MortenA]

I dont see how "shear lead to JT). Its a part of it but JT is a fancy word for an isenthalpic expansion. For a gas there's a big difference between an isenthalpic and an isentropic expansion (e.g. in an expander) but for a liquid is not that much.

And JT coefficient is just the thermo dyn. reduced to a constant. The cooling comes from expansion (PV=nRTZ) and the offset is due to the process as describe by FredRosse.

You e.g. get isentropic expasion (partially) in an oil/gas well - since some of the pressure difference from bottom to top comes from the staic collumn in the well. E.g. a 5000 metres deep well with an average density of 300 kg/m³ woudl experience 150 bar dP just because of the static column (grossly simplified since the density changes from bottom to top but still). Another (partly) isenthalpic process is expansion across an expander (as in a turbo-expander). Here the gas get colder because some of the work is "taken from the system" instead of heating the gas as it will in an expansion across a valve (as several others has explained)

Best regards

Morten

 

[ginsoakedboy]

MortenA, are there any other manifestations of this shear effect? I have not come across this before.

 

[MortenA]

Well it wasnt me who introduced it? But i think it may be splitting hairs:

The valve is a resistance - and work is required to let down the pressure. The work dosnt "leave the system (fluid)" so the enthalphy is constant. The opposite is an ideal expander where all the work is "taken out of the system" (via the shaft) - in this system the enthropy is constant.

The two roads for expansion has a significant temperature difference for a gas - but not for a liquid (som but not a lot).

Best regards

Morten