Temperature after isenthalpic expansion

[chemter]

Hello. I have a question concerning the isenthalpic expansion of steam across a control valve (please assume that the pressure drop across the valve would result in a temperature drop and result in dry superheated steam). Say for example the upstream steam temperature was 400°F and the downsteam temp. was 300°F superheated. Where would this temperature actually occur? Would it be immediately downstream of the CV or would it take sometime to decrease from the upstream temperature of 400°? I understand that for this example the temperature would drop, but I don't think I understand where the removed heat goes and how it is not transferred to the steam passing through the CV. If the heat is lost to friction, it seems the body of the CV would increase and keep adding to the downstream temperature and that the effects of the 400°F upstream temperature would be felt on the outlet of the CV. Any explanation on this would be greatly appreciated. Thanks.

 

[25362]

You say "removed heat" and at the same time "isenthalpic expansion". Isn't that an oxymoron ? A visit to thread378-106900 may be of help.

 

[sailoday28]

25362 (Chemical) 13 Jan 07 2:25

You say "removed heat" and at the same time "isenthalpic expansion". Isn't that an oxymoron ?

Consider the steady state process and neglect change in elevation.
h1 + u1^2 +Qadded = h2 +u2^2
Generally for the adiabatic process and neglecting KE changes the change in enthalpy =0
However Q added for the original question would be determined from----
isenthalpic u1^2 +Qadded = +u2^2

I think an appology is due.

Regards

 

[25362]

Steam throttling (isenthalpic) processes are generally assumed to take place without any appreciable change of kinetic energies so that u1^2 [≈] u2^2, and produce no shaft work. Or may be I didn't entirely grasp chemter's meaning ?

 

[chemter]

What I'm trying to ask is after an isenthalpic expansion across a control valve, the steam temperature will be 300°F (superheated steam) and about 100°F less than the upstream temperature (please assume this is correct for the sake of my question). If you could measure the temperature immediately downstream of the control valve (say within 2 feet or less), would the temperature be 300°F? Or does it take a certain amount of time/distance in order to actually lower to 300°F from the upstream 400°F? Thank you.

 

[rmw]

The time you are referring to is only the time it takes to get from stability at the upstream condition to stability at the downstream condition. In the valve or orifice, there are some velocity considerations.

rmw

 

[dcasto]

Heat is not temperature in thermodynamics. Heat is energy. The total energy of the molecules. One form of energy we observe is locked up in the molecules vibration and we can correlate this energy as temperature. Another form is from the motion of the molecules and we correlate this energy through pressure/velocity. The last is random energy, entropy, which we measure from differences is the other measured values.

SO,When a gas has its pressure reduced across a valve, there is no loss in energy. Energy in temperature is transfer to entropy and velocity energy, not lost, transfered. The transfermation is nearly instataneous.

This maybe over simplified and the PhD's may get excited. Look up Joules-Thompson or JT effect in thermo books

 

[25362]

There are two steps taking place, including the vena contracta effect, during steam throttling that make it somewhat different from a true J-T effect which I mentioned in thread378-106900.

When speaking of orifice-type restrictions, the full effect of the isenthalpic temperature drop would be felt some two or three diameters downstream.