Propane/Butane vaporization calculation 9

[poseilus]

Hi,

I am trying to find out a way to calculate the rate of vaporization of butane/propane inside a storage tank at various temperatures.

Can you please help me?

Happy new year

 

[MortenA]

1) Estimate your heat input. This is a question of definition. You could e.g. go to your local metological institute. They may have a reference design solar radiation rate. Remember to include correction for the surface treatment of the storage tank. But other cases e.g. fire case may also be your design case - that is for you to decide.

2) Latent heat for propane and butane should be easily avaiable in various reference books (CRC data book or perhaps Perry) or search a little on the web.

Use the lowest latent heat value and you have a design rate. A more refined estimate would be acheived if you know the exact composition and use e.g. a process simulation tool to estimate the latent heat.

Best regards

Morten

 

[poseilus]

Thank you for your help,

I have a LPG manual that has a way to determine propane vaporization capacity.

It says if D outside diameter L overall length in inches of the storage tank and K a constant for percent volume of liquid in container then e.g.

when 60 percent of liquid is in the storage tank it gives me an equation DxLx100 where I get propane vaporization capacity in BTU/hr

if 30 percent of liquid the equation becomes DxLx70.

if 50 percent then DxLx90

I cannot understand how they get these equations. Also these equations are for propane. How about butane mixture with propane.

 

[Montemayor]

poseilus:

Your description of what you are seeking is not well written, poorly communicated, or lacks basic data describing what you want to do. As a result, it is difficult to make out what you want to do.

First and foremost: you say you have "butane/propane inside a storage tank"; I have to presume you mean a nominal 50/50 mixture of each (commonly called "LPG") in the LIQUID state. I also have to presume the LPG liquid is SATURATED (to non-chemical engineers this means that the liquid is in equilibrium with its vapor above it) and the pressure in the vapor space of the storage tank is the mixture's corresponding VAPOR PRESSURE. You also say you want to calculate the rate of LPG vaporization at various temperatures. If you lack a complete understanding or key learnings over the use of a Mollier Diagram (or T-S Chart), you are not going to understand what is going on in the vaporization process nor will you be able to design an LPG vaporizer. I'm going to assume you know what I'm writing about.

In order to vaporize the LPG, you have to apply its required Latent Heat of Vaporization (which is easily read on the Mollier Diagram). The rate of the heat applied determines the rate of the vapor LPG produced. It's that simple. However, you can't vaporize any LPG inside the storage tank without venting the resultant product - assuming you're applying the heat to the tank directly and not using an LPG vaporizer like most engineers do. The reason you can't vaporize LPG inside a closed tank is that upon applying heat to the tank, the vapor pressure of the LPG increases (the pressure inside the tank goes up) but the LPG remains basically in the liquefied state (in equilibrium with its vapor - but at a higher pressure & temperature. As you continue to allow the tank to achieve a higher and higher temperature, the vapor pressure inside the tank continues to climb - until it reaches and exceeds the MAWP (Maximum Allowable Working Pressure) of the tank (which is an ASME pressure vessel). I caution you on carrying out or allowing this type of operation - in case this is what is up your sleeve. As I stated before, you haven't really told us what you are trying to do and I caution you with the basic thermodynamic facts I have described above.

Depending on your operation, needs, and scope of work you normally do not apply heat to an LPG tank in order to vaporize its liquid contents. What is normally done is that an external, steam-heated vaporizer is used to control the rate of vaporized liquid and the product is taken directly to where it is needed. Sometimes, if needed, the vaporized product is piped up to the tank vapor space - but the vapor produced is always drawn off and not allowed to accumulate in the tank. Otherwise, the storage tank could have its pressure increased to an un-safe and hazardous level. You don't want to go there.

The nature of your query is so basic and simple to most of us that it concerns me that you have not told us really what it is that you intend (or plan) to do. Additionally, if you lack the experience and key learnings of vaporizing LPG, I would advise you to obtain the direct help and consultation of an experienced engineer.

I hope I have been of some help.

Art Montemayor
Spring, TX

 

[25362]

Pay good attention to previous posts, in particular to the one by Mr Montemayor.

Are you, however, trying to estimate the bubble points of a given (known mol fractions) hydrocarbon mixture at different temperatures ?

One of the common exercises in distillation units is to determine the top reflux drum pressure conditions assuming air-cooled or water-cooled condensers.

This is generally done by predetermining the temperatures and calculating by trial-and-error (ie, an iterative procedure using tabulated K[sub]i[/sub] values) the equilibrium pressure that would result in the sum of all vapor mol fractions y[sub]i[/sub] = K[sub]i[/sub]x[sub]i[/sub] = 1.

Once the pressure is determined, the distillation tower top temperature is estimated by a similar method using "dew points".

Sometimes liquid mixtures are "sub-cooled" (low bubble point pressures). The required pressure levels are maintained, for example, by fuel gas streams (as in reflux drums of crude ditillation towers in refineries) or by the use of nitrogen (in some LPG storage tanks).

 

[poseilus]

You are actually right montemayor.

I talk about propane/butane mixture (this mixture varies, it usually is 80%propane 20%butane and is called propane, but when it is 20%propane and 80%butane it is called butane). These are the two typical products that LPG storage facilities handle.

For each product I want to calculate the vaporization rate inside an ASME cylindrical pressure vessel with ellipsoidal heads or inside cylinders.

Now what I mean with the vaporization rate? That is actually what I found in various literature on books and internet. For a better understanding of my problem I guess I can give you a graph of what I am looking for if you give me your email. That will help mostly I think.

What I specifically want is the vaporization rate of propane inside various types of storage media, including asme pressure vessels and LPG(butane or propane mixtures) cylinders in terms of the temperature of the
liquid and the wet surface area of the container.

I do not apply any heat on the storage media. Neither I want to.

To understand more what I am trying to find I will explain why I want this rate. The LPG mixtures are consumed at the final stage of consumption eg. burners for central heating, various kitchen cooking devices, space heaters not in a liquid state but in a gas state (those devices have consumption rates eg. 2Kg/h of LPG at gas state). The liquid in the storage media is evaporated at a gas state and this gas is transfer through piping at the device for consumption. If I know what is the evaporation rate of LPG in the storage media and the friction losses of the piping I can then calculate and find out if the evaporation rate of LPG inside the tank will be enough to feed the appliances.

If it is not enough I should consider a bigger tank with greater wet surface so that I can achieve higher values of vaporization rate.

I think I cannot understand the tables and graphs I have on my literature. I mean that I dont know under what conditions they calculate the vaporization rate. They could assume that there is no top of the tank and the vaporization is at its highest value. This is less likely to happen.

But they could also assume that under a constant pressure of eg. 1bar at certain temp. and volume of liquid what is the vaporization rate. Obviously in order to achieve that there should be a piping open that carries the gas to the appliances for consumption.

I guess the second assumption is more logical.

I want to thank you montemayor and 25362 for your help and especailly monte for his great advice and the time spend for my problem.

I wish you happy new year.

 

[Mik1]

Hi,Poseilus

Did you solve your rate vaporization problem?
I can´t find out bibliography about this theme. So, you can solve this problem on this way:

Vaporization rate is function of wet surface and latent heat of your mixture. So, at first you must to determine the wet surface of a cilinder tank with ellypsoidal or espheric heads (at different levels). This point determine your coefficient affecting the formula.(100/70/50 at differents levels) (You can see ASME code)

At the same time, the heat over the tank is taken of atmosphere, so the difference of temperature between air sorrounding tank and equlibrium temperature of the mixture at pressure of your sistem (we take 1.5 kg/cm2)(I am not considering Radiation)

But is important to take note abut the different rate of vaporization of butane and propane. (different latent heat, and different phase gas composition over the time).So, you could calculate by steps, considering the different compositions of your gas phase in equilibrium with liquid phase (at equilibrium pressure)(for example, at first 50/ 50. Probably you will consume at first high quantities of propane, and your liquid phase will be rich in butane soon. Properties could be taken of NIST.
It will change your vaporization, because of If you change composition (richer in butane) you will have less vaporization at the same temperature.

One point that I cant solve quitely, is the possible stratification by density. This subject is important if you take liquid phase (vaporizator) and gas phase.

 

[Stones1973]

Dear Poseilus, Maybe you have already solved your problem though I leave the solution: 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 insedi the vessel
Cv- Latent heat for the liquid to the temperature insede the vessel
A- Wet Area

Stones1973