Filling a high pressure bottle

[DesmoFan]

I need to fill a bottle submerged in LH2 with 800lbm of -290F GHe in about 30 minutes. My bottle temp limit is somewhere around 130F. I have tried to model the problem with adiabatic compression and isothermal. The adiabatic heats the gas in the bottle too quickly and the isothermal of course works but I know this is not representative of the problem since I have filled high pressure bottle before. I have tried a new approach using energy balance mdotin * hin = dU/dt but not sure if this is the right approach. Anyone have any suggestions on how to approach this problem?

Thanks in advance

 

[sailoday28]

Your formulation for adiabatic flow is correct. However calclation of final conditions are not time dependent.
Using perfect gas, const specific heats and multiplying both sides of you differential equation by dt should yield

dU=ho dm where ho is stag enthalpy

with constant stagnation of source
d(mT)=gamma*To*dm
mT-miTi=gammaTo(m-mi) where m is mass in bottle and subscript i relates to initial conditions. Subscript o is for stagnation conditions of source

As a rough approx neglection initial mass compared to added mass, the above equation reduces to
T=gamm*To Temperatures are absolute

If To= -290F, 170R and gamma approx 1.7
T final=289R which is much less than 130F bottle temp limit
Also, does your integration include realistic initial conditions in the bottle? Is your source enthalpy constant?

Regards

 

[iainuts]

Sounds like a rocket application. What's the temperature of the helium going in? What final pressure?

Just to add to what sailoday mentioned, I'd suggest creating either a spreadsheet or computer program and doing this iteratively. As mentioned, for this application, and assuming no heat transfer, the first law reduces to dU = ho dm. I'd also suggest adding a heat transfer to that though, dU = ho dm + Q.

If you set up a spreadsheet, you'd start with initial pressure, temperature and mass in the bottle. That gives you the state of the helium (ie: internal energy) and thus the total internal energy inside the bottle, U. Recalculate conditions with each small increment of helium going into the bottle. If it takes 30 minutes, I might suggest doing something like 10 second increments. Add 10 seconds worth of helium, calculate the heat into or out of the bottle, and then recalculating internal energy per the equation above.

The helium coming in has some enthalpy based on its pressure and temperature. That energy (ho dm) is added to the fluid in the bottle, and you arrive at a new internal energy, U. Note that the specific internal energy will rise slightly, and from that you can recalculate the temperature. So far you're doing what sailoday said, dU = ho dm.

You can also add/subtract heat transfered into/out of the helium. That's another energy term, the Q in dU = ho dm + Q. This requires calculating the convective heat transfer coefficient, but if it's not overly important, you can calculate the upper and lower limits and use those values in your program just to be conservative.

If your computer has a properties database, you can use these directly. If not, I'd suggest you get one rather than trying to calculate enthalpy and internal energy based on pressure and temp. I've been using one for 10 years now which is linked to Excel, and I've never tried to use anything else. It works great.

If you're just looking for a very rough WAG, you can do what sailoday suggests:

As a rough approx neglection initial mass compared to added mass, the above equation reduces to
T=gamm*To Temperatures are absolute

Sometimes the WAG gives enough information, and is also a good way of seeing if you've done something wrong if you do a more detailed analysis.

 

[DesmoFan]

Thank you both for the input, iainuts you are correct this is for a rocket application, the temp of the GHe is coming in -290F and the final pressure will be around 3200 psi. I do have a spreadsheet that I have been working on which is uses some gas property routines (GASP) and I am in the progress of adding in the heat transfer which is somewhat of an unknown since I dont know material or thickness yet but I took a guess. Sailoday, to answer your questions, I do have initial conditions of the bottle in the spreadsheet and have factored in mixing before calculating a new pressure. I am also using a constant ethalpy for the source gas.

Thank you both

 

[iainuts]

Hi DesmoFan. Note that the enthalpy of your gas source isn't constant even if the temperature is. As the pressure increases, enthalpy increases. The amount can't be neglected.

Also, if you have access to the Atlas RSB charging analysis done, you'll find all the data on heat transfer in those files. I left that company about 10 years ago, but had done considerable work using a Fortran program called Charge and another called Outgas.

If this is inside the LH2 tank, I'd assume you're going to take advantage of the reduced temperature. Thus, the final temperature of the LHe will continue to drop well below -290 F. That has to be considered also.

 

[sailoday28]

DesmoFan (Mechanical)You state the source is gas.
As I sense from the reply of "iainuts (Mechanical)"the source seems to be a finite volume.
If the source is reasonably insulated and the process fast enough, the source process can be approximated as isentropic.
For the source dE=hdm where E is the internal energy of the source, h the specific enthalpy and m, the mass in the source. mde+edm=hdm m(Tds-pdv)=pvdm
V=const dm/d=dv/v Tds-pdv=-pdv ds=0
For a perfect gas, const specif heats pv^gamma=constant

Note h of source tank is input stagnation enthalpy of my previous post.

Regards

 

[DesmoFan]

The source gas configuration is unkown at this point. I assumed that the enthalpy of the gas coming into the bottle would be constant to simplify the problem. Maybe this is a bad assumption. Sorry if I seem a little confused,my background is hydraulics so I am clearing the cobwebs on this compressible stuff. Iainuts I dont have access to any Atlas data, Is the fortran program charge a Lockheed proprietary program or a commercially available program? Thanks again to both of you for your input.

 

[iainuts]

Hi DesmoFan. The Fortran program (Charge) was written at General Dynamics Space Systems Division back when Atlas belonged to that company. That whole product line was bought out by Martin Marietta about 14 years ago. Shortly thereafter, Martin merged with Lockheed. Now we talk about the Atlas being a Lockheed product, but it started out at General Dynamics where all the work was done on liquid nitrogen refrigerated helium bottles. The Atlas used/uses helium bottles, charged to high pressure, that are jacketed using liquid nitrogen to increase the helium density. The analysis was done using those Fortran programs way back when. I don't know what they've been using in the past 10 or 11 years though since I left the company.

The way the analysis is done is to do your best at creating a single heat transfer resistance coefficient for the helium/bottle/LH2 interface. It can of course vary with temperature and pressure, but you need to come up with a single equation that can be used to model the heat transfer. Then you model the bottle using the first law. You can always come back to the heat transfer equation and modify it after you test and refine the values.

If you haven't decided on how this helium will be injected into the bottle yet, I think you have 3 options.
1. Ambient temperature, high pressure storage: Storage of helium at 6000 psi. Provide restrictions in line to control flow. Put liquid nitrogen heat exchanger in line to cool gas to -290 F just before gas goes into bottle. Heat exchanger could be located on the ground of course, and a VJ pipe brought over to the disconnect.
2. Cryogenic helium storage: Pressurize liquid helium using ambient temperature helium as the pressurant in a high pressure cryogenic tank. Flow the cryogenic helium through transfer lines into the bottle.*
3. Pumped liquid helium: Use a pump to take low pressure liquid helium and compress it to 3200 psi. Control pump flow rate using VFD motor.

The first option means you will have 800 lbm of helium at -290 that needs to be cooled to roughly -420 F. Aprox. 100 btu/lbm of energy will be required to cool it. Parahydrogen will boil at a rate of roughly 150 btu/lbm, meaning you'll be boiling 530 lbm of liquid hydrogen to cool the helium down to -420 F, so the liquid hydrogen vent system must take that into account. The other two options don't require any boil off of LH2. In fact, they will result in the hydrogen becoming denser due to the lower temperature the helium will be at (about -440 F). These are all rough numbers.

Sailoday is correct in saying:

If the source is reasonably insulated and the process fast enough, the source process can be approximated as isentropic.

But that really wasn't what I was thinking. In the case of #1 above, the helium is assumed to always come out of the liquid nitrogen heat exchanger at -290 F. The pressure it comes out at is dependant on the back pressure from the bottle you're charging, but if we assume the pressure drop is small, we can assume the pressure is roughly the same as bottle pressure. In this case, enthalpy is not constant. At a temperature of -290 F, the enthalpy varies according to:
h = 3.77 E-7 P^2 + 5.91 E-3 P - 4.55 E2 (btu/lbm)
Similarly, internal energy changes, but only very slightly according to:
u = 3.18 E-7 P^2 + 2.26 E-3 P - 5.40 E2 (btu/lbm)
(Pressure in psia, temperature -290 F)
I included internal energy because the values of internal energy and enthalpy are relative to each other, so giving one without the other is nonsense. If you chart these two, you'll find internal energy is fairly flat (slowly decreasing with increasing pressure) with enthalpy significantly increasing with increasing pressure. The point is, that if doing an analysis, you can't assume a constant enthalpy coming in. You need to do it iteratively, and account for the change in the enthalpy of the helium coming into your bottle.

If you create the spread sheet using some estimate for heat transfer, you can always check that value by test. Wet dress rehearsal (WDR) is one way, but knowing how the industry likes to test things long before that stage, doing an off-line test on the actual hardware is certainly a possibility. At least by providing some analysis to start things off, you can provide some bounds on how long it will take to charge and the temperature/pressure curve expected. Thus, you will have sufficient information to size pipes, valves and other components to design the system. The test data can be refined further during WDR to ensure the bottle charging system will meet the timeline expected during launch.

*Note: liquid helium, compressed this way, may result in near perfect isentropic compression. If this happens, helium will solidify to some degree. Chunky helium may not flow well through valves and filters. Note also that the helium pump concept isn't close enough to perfectly isentropic to create an issue with solidifying helium. That concept has already been proven.

PS: Have you checked patents and patent applications on this? Seems to me this might already be a patented design.

 

[sailoday28]

For perfect gas isentropic flow, const gamma in SOURCE tank
ho/hoi=(m/mi)^(gamma-1) where ho is stag enthalpy
m is mass in source, m decreases
i refers to initial conditions.
Knowing flow rate, stagnation enthalpy is readily caculated.

For insulated receiver
Change in Internal Energy in receiving tank (mass*spec internal energy= Internal Energy) = integral of ho*dm where m is mass in receiver. m increases


Regards

 

[iainuts]

Hi Sailoday,
Not sure what you mean by "perfect gas isentropic flow", but I think this is a good discussion point to talk about how this process must be modeled. I mean no disrespect in disagreeing on some points of yours here. Anyway, are you suggesting the flow through the pipe is isentropic? Perhaps you mean adiabatic (ie: no heat transfer). Certainly the flow is not reversible, so it's not isentropic. And why "perfect gas"? If the OP is using GASP, they are doing the right thing by not assuming an ideal gas. I don't think you'd want to use ideal or perfect gas laws for any of this when real properties are readily available for use in a computer program.

Also, you consistantly refer to stagnation enthalpy, which is simply the enthalpy of the gas when brought to rest. Technically you're correct, but the difference between stagnation enthalpy of a gas flowing through a pipe and the enthalpy of the moving gas only differs by the energy locked up in the velocity. That energy is negligable and generally neglected. Realistically, heat transfer from the line will have a much greater affect even on a very well insulated line. For example, helium at -290 and 3000 psi, flowing through a pipe with a velocity of 50 ft/sec (which is fairly high velocity, but reasonable) will only result in a 0.05 Btu/lbm difference between stagnation enthalpy and the enthalpy of the flowing gas. This results in a 0.04 degree F rise in temperature when the helium is brought to rest. Such a small rise in enthalpy is hardly worth the effort. Note also the difference between 100 psi helium and 3000 psi helium at -290 F is roughly 20.6 Btu/lbm, so that affect is very large and must be considered. If the gas going into the bottle is always coming out of a LN2 heat exchanger at -290 F as pressure rises in the bottle, the enthalpy going into the bottle is not constant, and can't be treated as such.

Depending on how the source gas is set up, the source could come from a large pressure vessel, 6000 psi, which will experience a gradual decay in pressure over time. The final pressure will of course have to be higher than or equal to the final pressure in the receiver, 3200 psi. As pressure decays, we can say it is isentropically expanding if there is no heat transfer, but that is far from realistic. In reality, the gas comes out slowly enough, and there is enough heat transfer from the storage cylinder, such that the temperature is roughly isothermal. Thus, the gas going into the line has an enthalpy which can be determined from the state of the fluid at the pressure and temperature of the source. Again, I'd generally neglect kinetic energy increases in the gas as they are exceedingly small.

For insulated receiver
Change in Internal Energy in receiving tank (mass*spec internal energy= Internal Energy) = integral of ho*dm where m is mass in receiver. m increases

Yes, I'd agree with this. As you say, we need to integrate ho*dm (or just h*dm) where dm is the mass entering the reciever. If this is done numerically using a spreadsheet or a Fortran program for example, h changes with pressure, especially if temperature remains constant. In fact, temperature would need to constantly decay with a rise in pressure to compensate for the increase in PdV energy if enthalpy were to remain constant.

Note also, this reciever is not intended to be insulated, but is submerged in liquid hydrogen, presumably because there is an advantage in reducing rocket weight by increasing helium density and thus having a smaller reciever. Also, rocket manufacturers generally take advantage of, and take credit for, the increased strength of the material used for the bottle at cryogenic temperature, meaning they can make the walls thinner to reduce weight further.

Best regards.

 

[sailoday28]

Purpose of my source model was to make it simple. This may not be realistic. That is, if only vapor from source, the hg is source enthalpy which will vary with the vapor flow from source vessel.
With a homogenous source, and negligible Q, then process is isentropic. I uses perfect gas, const gamma just for simplication which yields p/(m^gamma)=const.

If a subcooled source is used, I would then use properties of fluid based on isentropic process. For example, plot specific enthalpy vs speicif volume to obtain a relation of sp. enthaply vs mass in tank

My model has assumed connection pipe to have no effect on process. If that pipe has effect on Q, then it should be considered.

Dependent on how approximate models give results, then one might consider slow processes in source and receiver and a transient model (ie method of characteristics) for piping.

You have shown no disrespect since your response/criticism was offered in a positive or constructive way.

Regards

 

[DesmoFan]

I drew out the system again today and understand what you are saying about the source gas enthalpy not being constant due to the resistance of the bottle changing as it fills. I will add this into the spreadsheet. I originally started using ideal gas but gamma for helium is around 1.7 at -290, if I remember correctly, which is too far from 1 to neglect. I will also have to go back and look at the sizing calcs for the vent system, I think we sized that according to chilling down the tank assuming all the liquid in the transfer line was changing to gas. You are correct the reason for the helium bottle submerged in LH2 is for weight savings purposes.

Thanks again to you both for the valuable input.