Modeling orifice discharge temperature with hydrogen at 6000 PSIG 3

[bgordon8439]

Hi,

Sorry I'm new to these forums but they have been a great reference tool for me in past so thank you all for the useful information you put on the website.

I'm trying to model a flat plate orifice that is throttling 6000 PSIG gaseous hydrogen to approximately 0.5 kg/minute. I'm have no problem modeling the choke vs. non-choke flow across the orifice, so that is not the issue.

The question I have is how do I determine the discharge velocity and gas temperature assuming isenthalpic process (H1 = H2....something has to be constant). I've tried to factor in Joule Thompson but this give me astronomically high temperatures (1000K) which is not correct since I know from physical experience that the gas cools when the dP is great enough and then starts to heat as you reach the non-ideal gas regions or dP across the orifice decreases. I don't want to assume an istentropic process since this process is highly irreversible

Does anyone have some words of wisdom?

 

[PaoloPemi]

Joule Thompson (the derivative of temperature vs. pressure) is not constant so you should calculate the entire cooling curve, as alternative if you assume adiabatic (i.e. hout=hin) then calculate hin (at specified tin,pin) and calculate tout at specified hout=hin @ pout, guess a initial tout and then correct untill you'll find hout=hin , the final error will depend from the accuracy in calculated enthalpies, in moxt simulators or process libraries there are specific methods for solving adiabatic flash operations.

 

[Latexman]

If the 6000 psi H2 is above H2's inversion temperature (~ 200 K), the H2 will heat up.

Good luck,
Latexman

 

[zdas04]

And when you calculate JT cooling, be really careful with your signs. Restricting flow through an orifice results in a pressure drop which will cool the gas not heat it.

This is an Adiabatic process (i.e., it is non-reversible and isentropic [dS=0]), but it isn't isenthalpic (dh is not zero). The iteration that PaoloPemi is talking about is around entropy, not enthalpy.

David

 

[zdas04]

Latexman,
I never seem to think about the whole temp range and keep forgetting the inversion in JT cooling. Good point.

David

 

[Latexman]

Yep. Helium and hydrogen always have stuck in my mind as being "weird". With inversion temperatures below room temperature, most of the time they get expanded, they heat up. Just one of thousands of exceptions to the norm.

Good luck,
Latexman

 

[iainuts]

You are correct in saying this is an isenthalpic process. If you put a control volume around the orifice you have no work done on or by the orifice and because the process is relatively fast, we can neglect heat transfer so the first law of thermo reduces to dH = 0. See for example:

http://web.mit.edu/16.unified/

http://www/FALL/thermodynamics/thermo_4.htm

(Half way down the page, III.C "Adiabatic throttling of a gas")

Note that an isentropic process is one in which work is removed (or added) such as with an expander (or compressor), so for that kind of machine, assuming no heat transfer, dH = W.

We can take the velocity of the gas into account if necessary, but generally the velocity change is slight and we can neglect it. But if we include it, some energy can be converted to kinetic energy and the first law can be written to add this energy. Just add the term for kinetic energy due to velocity (ie: 1/2 m v^2) and consider the change in velocity, then make sure the change in enthalpy is also equal to this change in kinetic energy. Anyway, that generally isn't necessary until the velocity downstream (ie: not at the orifice) approaches mach speed.

If you don't have a database that has enthalpy for hydrogen as a function of temperature and pressure, you can use the NIST web site and create your own, interpolating between values that it spits out. Another option is to use NIST REFPROP, which comes in very handy if you do a lot of thermo work. You can find the NIST web site with an abbreviated set of properties here:

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

 

[zdas04]

Velocity change is slight???????????????

Let's say that we're taking a 14 MPa pressure drop across this plate (pressure drop required for JT effects to heat the gas to 1000 K). Mass flow rate is constant by the continuity equation.

Upstream density at 15.9C and 6000 psig (don't blame me for the mixed up units) is 33.4 kg/m^3, so if the pipe ID is 50 mm, upstream velocity (mass flow rate given as 0.5 kg/min) is 0.123 m/s.

Since temperature went up 1000 K, JT heating would say that the downstream pressure is around 50 kPa, so the new density is 0.011 kg/m^3 and the downstream velocity is 365 m/sec.

A 3,000 fold increase in velocity seems significant.

David

 

[iainuts]

Hi David. The OP didn't say what the inlet temperature was so we don't have the upstream fluid state. Also, the downstream pressure wasn't mentioned so we can't even perform an analysis on this. All I can assume from the statement "I've tried to factor in Joule Thompson but this give me astronomically high temperatures (1000K) which is not correct ..." is that bgordon8439 attempted to calculate the temperature and found it wasn't even close, so they're looking for a way to determine temperature.

Regarding the first law, if we expand it to include kinetic energy, and assuming this is an adiabatic expansion, the first law reduces to:
Hi + KEi = Ho + KEo
Where H is the enthalpy and KE is the kinetic energy.

For the sake of argument, let's say the velocity upstream of the orifice is negligible and the temperature is ambient (70 F). I'll use imperial units just because that's what I'm accustomed to. I'll also be using a computerized database for properties so if you need to follow along, I'd suggest the NIST site. Further, let's say the downstream pressure is 3000 psi, so the orifice is choked and we have a shock wave at the vena contracta. Normally, piping systems are sized so the pressure drop is not too horrific, so let's say the pipe downstream limits the velocity to 300 ft/s. That's a pretty decent velocity change of course, but that isn't the point. The point is to find the change in kinetic energy so it can be used in the calculation to see how it changes the overall energy balance and its affect on the temperature.

To calculate kinetic energy per unit mass, we simply use .5*V^2 and put in the appropriate conversion factor. To convert kinetic energy to Btu/lbm we take the velocity in ft/s, divide by Gc and use 1 Btu/s = 777.477 lb ft/s. The increase in velocity to 300 ft/s equates to an energy of 1.798 Btu/lbm. Mach number is 0.062 for this case (ie: The final temperature will be 81.5 F and sonic velocity is 4827 ft/s). The first law reduces to:
Hi = Ho + KEo
(Note that KEi is zero since we assume velocity is small and can be neglected.)

If this were a purely isenthalpic process and we neglected kinetic energy, the temperature would have risen to 82.0 degrees F (assuming 6000 psi and 70 F inlet and 3000 psi outlet). That's the temperature rise neglecting kinetic energy. If we include the kinetic energy of a flow stream at 300 ft/s (1.798 Btu/lbm) then the temperature drops slightly to 81.5 F.

The point isn't that the velocity change is small, the point is that the kinetic energy change is small. Sorry if that didn't come across properly, but generally we assume the process in a typical piping system is isenthalpic (as stated for example by PaoloPemi) and that assumption is generally correct. It is correct for 'typical' piping systems in my experience because the change in kinetic energy is small and can be neglected. Yes, it sounds like 300 ft/s or even considerably higher velocity should make a difference in the temperature, but we find the difference is slight if we actually do the calculation.

Hope that clarifies things.

 

[PaoloPemi]

The original post (bgordon8439) noted "I don't want to assume an istentropic process" so in my answer I suggested to solve for enthalpy...
Apart from that consideration I think the assumption (isenthalpic) could be quite reasonable depending from the area of application,
in a throttling process we can consider two phases, the first occours during the short period of time through the orifice, the fluid accelerates converting "energy" to kinetic energy, this is the "isentropic" phase, after the orifice the fluid expands to outlet pressure, the expansion requires a reduction of kinetic energy which is then reconverted (apart from friction losses),
the overall process in many cases can be considered isenthalpic (as said not including friction losses).

 

[bgordon8439]

All,
All thanks for the replys! I think that I have what I need from this thread discussion but I will let everyone know the results. To what I've been reading to find out exactly what I want to find out I need to factor kinetic energy into the equation and continue assuming an isenthalpic process. The only reason why I say this is because below a certain orifice dP reagion (which I don't know yet) the orifice acts as an expander and the JT effects are negated. This is due to the velocity, pressure and volume changes from upstream to downstream conditions.

Ultimately what I would like do to is create this model so I can determine how to extend this region where the orifice acts as an expander valve to cool the gas for dispensing applications.

I know eventually to get results that I need I'll need to model the system according to the second law.


I know I wasn't clear about all of the conditions so here they are. I'm assuming ambient conditions (20C) for inlet and the differential pressure across the orifice is dynamic which is the aspect that I'm modeling. It can range from 5750 PSIG to 500 PSIG dP for any given time. I realize that the JT coefficient will definitely play into the gas temperature but from doing research from NIST and national laboratories this is the temperature directly at the exit of the orifice plate or at the venta contracta.

 

[Latexman]

What does NIST say the inversion temperature of hydrogen is? Starting at 20 C, will there be cooling or heating? The data I looked at (

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

) says it will heat up on expanding @ 20 C, so the statement "to cool the gas", isn't going to happen.

Good luck,
Latexman

 

[25362]

It is a general ROT that gases above reduced pressures of about 12 heat up on expansion. As pointed out by Latexman it is apparent that at the given pressure, hydrogen is well above and out of the generalized J-T inversion curve.

 

[PaoloPemi]

hydrogen has a inversion temperature of about 200 K , at room temperature you are well above, for production plants there are alternatives as regenerative cooling which use already cooled gas as a coolant of incoming gas below inversion temperature, other methods may exist which I do not know.

 

[25362]

There is an interesting Google site named:

Joule-thomson inversion curves and related coefficients for several ...ntrs.nasa.gov/archive/.../19720020315_1972020315.pdf

which includes hydrogen among other fluids.

 

[25362]

Inversion temperatures are related to pressures. For normal hydrogen, Perry's Chemical Engineering Handbook 6th Ed., Table 3-157, gives the following approximate data enabling the construction of a curve, outside of which, the gas would heat up on a J-T expansion:


P, bar 0 25 50 75 100 125 150 164
T_L, K (28)* 32 38 44 52 61 73 92
T_U, K 202 193 183 171 157 141 119 92

• Hypothetical limit

Evidently, at the given pressure hydrogen would heat up on a J-T expansion unconstrained by temperatures.

 

[25362]

Then, there is the heating effect that a jet of high velocity fluid can have on compressing the gas downstream.