Calculation natural vaporization (offtake) LP Gas cylinder 3

[hoangnp]

thread135-111334
Hello friends,
I am trying to calculate the natural vaporization Q for a LP Gas cylinder (99 liters or 45 kg LPG). If I do the Q=(m*k*As*Psat)/Rg*Tl equation it turns out 11,9 kg/hr.
If I do Q=K*Aw*(Ta-Ts)/Cv is 1.38 kg/h
So what am I missing?

Thanks
Including:
m (molecular weight): 44
k (mass-transfer coefficient): 1.127
As (area of liquid surface): 0.1256 m2
Psat (saturation vapor pressure of the pure liquid at TL): 9 0C
Rg (ideal gas constant): 0.1885 J/kg 0C
Tl (absolute temperature of the liquid): 25 0C
K (Coefficient of heat exchange with exterior): 500 W*m2*K
Aw (Wet Area): 0.7 m2
Ta (Ambient Temperature): 28 0C
Ts (Temperature correspondent to the vaporization pressure inside the vessel): 25 0C
Cv (Latent heat for the liquid to the temperature inside the vessel): 1.5 kJ/kg*0K
LP Gas comercial: 50/50, density 0.54 kg/liter.
LP Gas cylinder (45kg): OD 360 * H 1050 (mm)

 

[LittleInch]

As far as I can follow this, your first equation doesn't include heat tranfer from ambient, but assumes a fixed liquid temp of 25C

Your second one as far as I can see does include heat, but can't see how you're translating that into vaporisation rate. The Cv used is simply one for heat capacity, not latent heat of vaporisation

Generally you find that as the liquid vapourises, the temperature falls thereby increasing the heat transfer rate until you achieve equilibrium, Also as the tank empties the wetted area falls and hence heat flow in falls. So your initial 25C will fall to something more like 15 or 10C if you start drawing off at a faster rate until heat in = heat out from vapourisation. This is one reason it's not possible to get standard rates. Also as the liquid level falls, the area of liquid to get heat input reduces so vaporisation rates fall again.

If you want higher or fixed flows, then you need to send the liquid through a vaporiser like they do on hot air balloons and other industrial units. Like this

https://www.andrianos.gr/en/products/lpg-natural-gas/lpg-vaporizers

Or the temp falls so much that the pressure falls as well and can't supply the load.

What are you trying to find here?

We've done this before

https://www.eng-tips.com/viewthread.cfm?qid=492250

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[georgeverghese]

On a 45kg cylinder, google tells me propane vaporisation rate should be approx 5kg/hr at 70degF ambient.

Based on heat input through the walls of the cylinder by natural convection (your 2nd expression), I cannot get this rate of 5kg/hr either. Heat input by radiation would be rather low. Some of your values are not correct by the way.

 

[hoangnp]

Dear LittleInch, Georgeverghese;
I also think people using 2rd method. But heat tranfer formulas (conduction vs radiation) is deference.
How to calculate 2,37 kg/h? 5 kg/h?
What my values wrong?

 

[LittleInch]

That sheet you attached gives you the formula and the data.

Just adapt it for your situation.

But you seem to have sued the Heat capacity not the latent heat of vapourisation.

Also your fluid temp is very high. At a delta of only 3C you won't be getting much vapourisation....

They use a delta of more than 20C.
This is normal - when you draw off a lot of LPG, the cylinder gets very cold and you get a lot of condensation or even ice forming.
But if the draw off rate goes too high or the ambient temp falls to 0C or lower, you get very little LPG out without adding extra heat.

Why are you doing this when data is already freely available, but also =the rate varies as the tank empties or as a result of flow?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[pierreick]

Hi,
Q=kA(Ta-Ts)/Cv. this will give you the vaporization capacity of a LPG Tank.

Q [m3)h]
k [kcal/h ºC m2]
Ta ambient Temperature.
Ts- Temperature correspondent to the vaporization pressure inside the vessel
Cv- Latent heat for the liquid to the temperature inside the vessel
A- Wet Area

What is important to notice is the wet area (area of liquid concerned by heat transfer), changing over the time, when the cylinder is full or quasi empty.
Wet area is not a constant.
Pierre

 

[hoangnp]

Dear LittleInch;
I used temperature gauge to measure liquid LP Gas in cylinder. It is 25 0C at tropical country.
I don't understand how to calculate it.

https://www.ito-europe.com/wp-content/uploads/2011/12/UK-LPG-Cylinder-Offtakes.pdf

I think law's LP Gas are similar across countries.
Can you show me how to calculate it?
Thanks.

 

[hoangnp]

Dear Pierreick;
Is it true with the LP Gas cylinder?
Why not calculate the way Conduction heat transfer?
What different when the cylinder is vertical and the bottle is horizontal offtake?

 

[Compositepro]

Perhaps you will get more useful answers if you explain why you are interested in calculating anything. That will determine what equations to use and what conditions or other data you need to use in any equations.
If you fully open the valve on the tank to atmosphere you will get a very high flow rate that is limited by the valve size. The temperature of the tank will drop to -40C (assuming 100 percent propane). The actual flow capacity will vary from some maximum at ambient temperature with tank full to zero at -40C and tank empty.

 

[georgeverghese]

Values I get which differ a lot from yours are :

Cv = latent heat of vaporisation = 350kJ/kg at 20-25degC for C3
K = external heat transfer coeff due purely to natural convection = 4w/m2/degK based on external surface temp of 20-25degC
R = 189J/kg/degK for propane

Can you explain where you got the 1st expression from ?

 

[hoangnp]

@ Georgeverghese:
My teacher gave me a long time "Evaporation rate between gas - liquid phase". I don't remember... )
@ Compositepro, @ all:
Natural vaporization (off-take) of LP gas cylinder is maximum.
it means if Customer uses more than 20 kg/h then vaporizer must be installed (Max 700 kg LP Gas indoor).
Of course, Tropical Countries have an advantage.

 

[Compositepro]

As I said, the max. flow rate will vary depending on tank level and required pressure. If you need 20 kg/hr. continuously until the tank is empty, you must draw-off liquid from the tank and vaporize it in a separate vaporizer that supplies heat from the air, water or electricity. There are no other options, except to heat the bottom of the tank. You can probably use half of your tank before needing additional heat. Again depending on the supply pressure required.

Well, one other option is to add more lpg tanks. When the first tank cannot supply enough flow open the next tank, or use pressure regulators that are set differently. They can supply an intermediate pressure, which is then regulated to the final supply temperature so the end user will not see any pressure fluctuation.

 

[hoangnp]

@ Compositepro;
I think you're misunderstanding this topic.
Do you have a formula to calculate how many kg/h a LP gas cylinder (45kg/47kg) of natural vaporization maximum can hold?
With 20 kg/h, u can installed 16 cy;inder (8 use and 8 stand_by), still ensuring legal conditions. If used vaporizer then this topic doesn't exist.
with low pressure equipment, They can still use the last drops of vapor LP Gas.

 

[crshears]

with low pressure equipment, They can still use the last drops of vapor LP Gas

If done, beware the possibility of higher percentages of ethyl mercaptan in the final stages; may or may not matter

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[hoangnp]

@ Crshears;
Yeahhhh.
In fact, if the pressure in the cylinder is less than 0.2 bar, it is considered empty. (You can't use more than). residue remaining at the bottom of the cylinder considered too small to ignore.
It has a negligible effect on forklift's cylinder.

 

[georgeverghese]

Think the first expression is not relevant at all to propane boiling inside this cylinder. This expression you've got (if it is correctly written) is meant for plain diffusion controlled evaporation from the liquid - vapor interface at the tank top in a liquid that is NOT boiling (ie a subcooled liquid). Instead, in this cylinder, its probably nucleate mode boiling of propane from the warmer vertical sides of the cylinder. Propane bubbles rising up from the warm vertical sides of the cylinder.

There are 2 resistances (heat transfer coeffs) involved in this cylinder, both acting in series. One is external free convection on the outside vertical walls, and the other is the nucleate boiling mode (I suspect) heat transfer coeff on the inside warm vertical surfaces. The unknowns here are the surface temp on the outside, and the surface temp on the inside. This is an iterative and laborious calculation to see if the Ts assumed on the outside leads to a Ts on the inside which also matches the Q traversing these resistances in series. You should run these calcs to see what results you get.

Perry lists a few equations for natural convection, of which I used the simpler one to compute the htc I got. There is a more complicated equation I did not use which may or may not yield the same result as the simpler expression. This is equation 5-33a (Nishikawa - Hirata) in Perry Chem Engg Handbook 7th edn. Try if you can.

Other past threads on E-Tips I've come across so far on this topic state explicitly what vaporisation rates to expect, but there is no discussion on the numerical heat transfer verification of these thumb rule vaporisation rates.

 

[hoangnp]

@ Georgeverghese.
Thank you for your kindness.
I think there is an error somewhere, formulas, units...
You can refer "HEAT AND MASS TRANSFER STUDIES IN LIQUEFIED PETROLEUM GAS STORAGE OPERATIONS, Zainal Zakaria, 2006". A long-term experiment about vaporization.
Truly sincerely thanks again.

 

[terrybivins]

thanks, everyone for the explanations! such a useful thread it is.
Btw, I would like to share a tool that I have found recently:

https://oneconvert.com/unit-converters/viscosity-kinematic-converter

it has plenty of metric units