temperature drop in one hour in a LNG tank

[LNGuy]

Hi,

I am working on calculating the gas rate formation in a tank into which LNG is being filled. I want to calculate the heat going into the LNG at -163 from the natural gas vapor existing at a high temperature initially, say +20 degrees, in one hour. For this purpose, I am focusing on calculating the temperature of the cooled down vapor from +20 degrees to some temperature in 1 hour and will then calculate the heat it has given out to the LNG for evaporation. How do I arrive at what will this final temperature of the vapor be at the end of one hour, assuming the LNG is at saturated state and whatever heat it absorbs is used in evaporating into vapor. We can neglect the effect of the fresh evaporated vapor for the time-being.

Any insights are welcome. Thank you.

 

[LNGuy]

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')

 

[rishirga1981]

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.

 

[LNGuy]

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).

 

[rishirga1981]

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?