I've just replied in the other forum: http://www.eng-tips.com/viewthread.cfm?qid=391438

Dejan IVANOVIC
Process Engineer, MSChE

Thank you EmmanuelTop for your post. I am reading those links. Posted here in response to the other suggestion to post it on this forum.
The papers you suggested look good. Will update here.
Thanks.

Also have a look at this article - it provides a detailed breakdown of heat (cold) loss elements.

Dejan IVANOVIC
Process Engineer, MSChE

• http://files.engineering.com/getfile.aspx?folder=4998d9a5-1253-437e-849c-83

This sounds like the first fill of LNG into a tank??

If so you cannot just pour LNG into a tank without spending hours or days slowly cooling it down. This question is therefore irrelevant. Just search first fill lng storage tank.

Evan if by some miracle your tank can stand going from 20c to –163 in a short period of time, the evolved vapour will not be trivial and should not be neglected.

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

LNG ship cooldown normally lasts 3-4 hours based on my experience. Cooling off the tank shouldn't take much more time.

Dejan IVANOVIC
Process Engineer, MSChE

But it surely doesn't involve dumping liquid lng direct into a tank at ambient temp?? That's what the OP seems to be saying.

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

Pretty much this is the way it is done - at a controlled rate, of course (far below "dumping of liquid").

Also, prior to commissioning the facilities are at ambient temperature, so this is how cooldown is performed for all lines and equipment - circulating (and evaporating) LNG till normal operating temperatures are achieved.

Dejan IVANOVIC
Process Engineer, MSChE

LNG tank cooldown rate would be controlled by injection of LNG through cooldown spray nozzles located on tank roof, which I've seen on LNG tank drawings. Its only after the tank has reached -160degC or so, that liquid is introduced at gradually increasing higher rates through the normal fill line, else tank boil off gas rates would exceed the BOG recycle gas compressor.

Thank you people for the response.

As of now, yes LNG tanks are spray cooled before gradually increasing the LNG inflow. I did some rough iteration in excel and find that the rate of temperature drop for a 12000 m3 tnk is almost 10 degrees per hour. So, it makes the calculation a bit easier. The total heat input to the LNG is from external environment, the heat to cool the tank and the heat from the LNG vapor already in the tank at a higher temperature. All these will then lead to vaporization and temperature rise and expansion of the total vapor present in the tank. From this subtracting the tank empty volume remaining after cargo inflow will give the gas which has to be discharged ashore through the vapor return line.

Additionally, I was doing calculation on the very first loading of a newly built LNG vessel where there is no LNG to pre-cool it before the actual LNG starts being pumped in.

Also remember to include heat gains from the LNG loading pump, piping, and a significant amount of heat coming from heating of the tank bottoms.

Dejan IVANOVIC
Process Engineer, MSChE

LNGuy;

Could you suggest me how to calculate temp. drop per hour to calculate total cooling down rate for LNG tank? I am also in same boat and I need to calculate required amount of LNG for cooling down and how much time it may require to cool down the tank up to -130.C. I was told that it might take 10-12 hours for cooling down operation alone.

Thanks

Rishi

Hello All

I did some kind of calculations for LNG tank cooling down before to start loading process for LNG carrier.

Normally, cooling down time is recommended 12-15 hr based on site conditions. However my calculation sheet is showing bit high time and I am bit confuse whether I am on right track or not.

Attached is calculation sheet for your reference. Your comments are highly appreciable.

Regards

Rish

Sorry;

Here is the attached file.

Regards

Rish

• http://files.engineering.com/getfile.aspx?folder=2b3cfb6e-6dac-4aac-8425-e7

Hi,
There will be a lot of factors coming into play. First of all, as per regulations there should not be a drop of 10 C per hour, so your first calculation should be based on a dT= 1 hour. Then you should find how much liquid you should pump in to extract the above heat. Another interesting part you have to consider and which is why your calculations show such a long time is that you are injecting liquid fluid(LNG) which will first absorb the heat and evaporate to gas, without any change in temperature (latent heat of vaporization) and then this evaporated vapor will further absorb heat and reach certain temperature. Assuming that all the process is so slow that at end of each hour your system is in equilibrium and the temperature drop is no more than one hour. In short, the heat to be extracted from the tank to achieve a temperature drop of 10 degrees per hour is being taken by the liquid (latent heat of vaporization) and then by the evaporated liquid till it reaches a 10 degrees increase, so that the vapor inside and the tank are at the same temperature.
Also, in your sheet you are not considering the fact that the temperature gradient is changing as the tank is cooling. The temperature gradient is not always between 32 and -130. It starts from 32 when the sea and tank are at same temperature. At the end of one hour (since we need only 10 degrees drop), the gradient will be between 32(sea) and 22(the new tank temperature after 10 degrees drop) and so on.

Hope this helps.

So, start with your limiting criterion that you do not need a drop of more than 10 degrees per hour and then do calculations accordingly.

Hi LNGuy

Thank you for your help.

I am still wondering how to restrict the calculation based on dT= 1 hr. Would you mind to share any file or reference calculation on rishi.rga at gmail.com? It will be highly appreciable.

Thanks

Rish

At this stage I can only explain that you find the heat to be extracted from the tank using M_tank*Cp*dT, where dT is your desired temperature drop per hour needed, say 10 degrees (from 25 C to 15 C). Then you equate this to (m*L + m*Cg*dT'), where m is the mass of the liquid(LNG) required to be pumped, L is the latent heat of vaporization of LNG and Cg is specific heat of LNG, dT' is the temperature rise of the LNG after the vaporized LNG has reached the tank temperature i.e. equilibrium e.g. it evaporated at say -163 and reached the new tank temperature of 15 C at equilibrium (dT'=15-(-163)=178). Eauate: M_tank*Cp*dT = (m*L + m*Cg*dT') and find the mass of LNG you should pump in for a temperature drop of 10 degrees. Your dT and dT' will go on changing since the temperature will drop by ten degrees each hour. Then you can modify your sheet gradually to include heat ingress from the outside to the tank as it cools down then heat from the pump then heat from the pipelines and so on once you have your basic sheet ready.

Hope this helps.

Hi LNGuy

Thank you once again. I have few clarifications, would appreciate if you could assist me;
I will calculate Q= M_tank*cp*dT where M-Tank is Mass volume of tank, cp - specific heat of LNG and dT is 10.C ( 32-10= 22.c)

when I want to calculate , ( m*L+ m*cg*dT'). You confirmed m is mass of LNG which need to pump and I want to get this quantity. Here are 2 “m” and I am bit confuse. I applied the formula as bellows
Q= m*L, where I have already calculated Q and I can find the value of m, since latent heat of vaporization is given 511KJ/kg.
For example; Q = 3700000 KJ (based on 10 .C delta T), Mass of tank 1000 kg, cp= 2.02 KJ/kg
M_tank*Cp*dT = (m*L + m*Cg*dT')
3700000 = 3700000/511(m =Q/L) + m*2.02*178
I am using cp & cg value same. Please suggest if my understanding is correct or not.

From the (Q=m*L + m*Cg*dT'), take out common m to get m=Q/(L+Cg*dT'). Not just from m*L.
Your equation should essentially be:

M_tank*Cp*dT = (m*L + m*Cg*dT')= m*(L + Cg*dT')

LNGuy;

Am I supposed to use Cp and Cg values same? Sine you said Cp is specific hear of LNG so Cg suppose to same or different?
Thanks again for your guidance.

No.
In---M_tank*Cp*dT ---Cp is the specific heat of the material you are extracting heat from, here the tank, say steel.
In---(m*L + m*Cg*dT'), L is latent heat of LNG and Cg is the specific heat of the vaporized LNG (gas form).

LNGuy;

Noted. Many Thanks for your guidance.
I guess, I can find LNG consumption from (+45.c to -20.C) and each time consider Delta T 10.c but how I will establish the time for total cooling down.
Also heat ingress due to pump and heat ingress due to loading line length can be calculated but these values will co-relate with cooling down process?