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approximate required liquid mass to tank cool down

approximate required liquid mass to tank cool down

approximate required liquid mass to tank cool down

(OP)
Hi.
I have a question.
We have a 36000 m3 propane storage tank. Initial tank conditions is as follows:

T=20 C
P=0 barg.

We are going to cool down the tank to -44 C using liquid propane at -44 C. How much propane required?

thanks in advance.

RE: approximate required liquid mass to tank cool down

Can only assume your tank is insulated?

Can you ignore heat input?

What is the tank made of?

How much mass of solid material is there to cool down?

Do you have spray nozzles?

What is your vapour handling rate? That will limit your maximum flow rate and coiling time.

This is a transient thing so difficult to do as a simple calculation.

What do you think it is? This is supposed to be a forum to check things and provide advice, not do free engineering. The people who built this and designed it should be giving you the answer.

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

RE: approximate required liquid mass to tank cool down

Quote:

to -44 C using liquid propane at -44 C.

You do realize that this means infinite time, right? Although, you can probably get a degree or more by pumping on the vapors, basically wind chilling

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RE: approximate required liquid mass to tank cool down

Especially as propane is normally -42C at atmospheric pressure... that is though -44F...

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RE: approximate required liquid mass to tank cool down

(OP)
I meant -42 c, sorry!.
Dear irstuff, no!, that would be finite time using large of liquid propane

RE: approximate required liquid mass to tank cool down

(OP)
Dear littleinch, i am agree with you, it's difficult to be calculated, but our head engineer insist on doing that!

RE: approximate required liquid mass to tank cool down

Well then you're going to need to make some assumptions or estimations. We can't do that as we have no data.

But this is essentially an energy equation in terms of heat to be lost from the steel in the tank by evaporation of the propane. Then times by 2 once you found that volume.

But make sure your vapour handling system can cope with all this propane coming off it as it flashes into the warm tank. And I think you really need to use spray nozzles or drop it in from the top to avoid cold spots in the tank.

36,000 m3 is about 20,000 tonnes propane.

My guess - and it is a guess - is that you're looking maybe at 10% of that - 2,000 tonnes to get the temperature down to -42C. But I could be out by -50 to + 100%.

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

RE: approximate required liquid mass to tank cool down

Dear Jack Nicholson,

I think the most efficient path forward would be for you to do and attach your calculations and sketch/drawing, since you have access to all the data, and we'll review the assumptions and calculations and help with observations and replies.

Good Luck,
Latexman

RE: approximate required liquid mass to tank cool down

It might be off the subject. But you need to make sure the vessel material has adequate toughness for the brittle fracture for the temperature you are after. You can harm the plant and people around if the material does not have the toughness. Be careful.

RE: approximate required liquid mass to tank cool down

(OP)
Dear littleinch
3600 m3 is equal to 20000 ton propane? Which density have you used? (During cooldown, liquid didn't form)

RE: approximate required liquid mass to tank cool down

The original amount was 36,000 m^3 not 3,600 m^3; supposing it's for gaseous propane.

For gas the specific volume is roughly 0.552 m^3/kg; so 1.8 kg/m^3 so 6500 kg for 3600 m^3 or 65,000 kg for 36,000 m^3

What's missing is the allowable temperature rise in the liquid propane, or is the idea that there is a delta T of 2C when the liquid propane is sprayed or otherwise introduced into the tank to drop the final temp to -42C.

It seems like the goal is to get the gas to the saturation temperature by allowing the liquid to expand/boil to reach the saturation temperature. I'd say that the thermal mass of the tank will also have to be considered in this as well as thermal gains through the tank walls.

You are looking at a delta T of 62 C for a material that has a specific heat, c sub p, of 1630 J/kg/K; so

3600 kg * 62 * 1630 -> 3.6E8 Joules.

Heat of vaporization is 428000 Joules/kg, so 3.6E8/428000 = 850 kg of liquid propane. (or 8,500 for 36,000 m^3 as originally stated.)

I'm ignoring the heat required to warm the liquid from -44 to -42; that's all the calcs I'm interested in.
I'm also ignoring the potential that propane is not a perfect gas and that the specific heat may vary - I don't have access to tables showing what the value is across the range of interest. This calculation might be far off if that's the case.

Values of constants from https://www.engineeringtoolbox.com/propane-d_1423.... so check to see these match what you expect.

RE: approximate required liquid mass to tank cool down

Jack.

Please define your tank size.

Your OP did say 36,000 m3. I took that to be a tank capable of storing 36 thousand cubic metres of liquid propane.

Now it's suddenly 3,600 m3, but is this liquid storage or equivalent gas volume?

If liquid then it's 2000 tonnes of propane, so maybe 200 tonnes to cool it all down over a period of say 6 hours.

I'm working on the basis that you have a pepper cryogenic storage tank which you want to commission. Please let me know if you're doing something different

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