Forums
Blogs
Articles
File Library
FAQ
Search
FB
5
2 votes
Boil-Off Gas (Bog) Calculations For Cryogenic Liquefied Natural Gas Tanks
03 September 2012
Boil-off Gas has been discussed on "Cheresources" in several posts in the past.Recently, I came across an article on Boil-off Gas from LNG tanks which I found to be quite interesting. Today's blog entry is related to boil-off gas calculations based on this article for cryogenic LNG storage tanks of some standard capacities such as 140,000, 160,000, 180,000 and 200,000 m3 (mentioned as kL or kilo-liters in the article).
The article provides heat leakage (in other words heat ingress) values for the above mentioned four tank capacities for cryogenic LNG storage tanks and the BOG is calculated based on a simple heat balance from the heat leakage to the tanks. The heat leakage values are provided in Table 1 of the article and the link for the article is:
http://www.enggjourn...10-02-04-30.pdf
Please note that the blog entry does not provide calculations or basis of the heat leakage values as provided in the article and the values for heat leakage have been considered on an "as-is-where-is" basis provided in the article.
Let us go to the calculations for the various tank capacities:
140,000 m3
The heat leakage value (Q) as provided in the article for this capacity is 169,919 W.
1 W = 0.001 kJ / s
or
169,919 W = 169.919 kJ / s
The latent heat of vaporization (λ) of commercial LNG is assumed to be 512 kJ / kg
A phase change heat-balance can be represented by:
Q = m*λ
or
m = Q / λ
where:
m = Boil-off Gas rate, kg/s
Q = Heat Leakage, kJ / s
λ = Latent Heat of vaporization of LNG, kJ / kg
Calculations
m = 169.919 / 512 = 0.3319 kg / s
or m = 0.3319*3600*24 = 28,676 kg / day
160,000 m3
The heat leakage value (Q) as provided in the article for this capacity is 168,243 W.
1 W = 0.001 kJ / s
or
168,243 W = 168.243 kJ / s
The latent heat of vaporization (λ) of commercial LNG is assumed to be 512 kJ / kg
A phase change heat-balance can be represented by:
Q = m*λ
or
m = Q / λ
where:
m = Boil-off Gas rate, kg/s
Q = Heat Leakage, kJ / s
λ = Latent Heat of vaporization of LNG, kJ / kg
Calculations
m = 168.243 / 512 = 0.3286 kg / s
or m = 0.3286*3600*24 = 28,391 kg / day
180,000 m3
The heat leakage value (Q) as provided in the article for this capacity is 166,552 W.
1 W = 0.001 kJ / s
or
166,552 W = 166.552 kJ / s
The latent heat of vaporization (λ) of commercial LNG is assumed to be 512 kJ / kg
A phase change heat-balance can be represented by:
Q = m*λ
or
m = Q / λ
where:
m = Boil-off Gas rate, kg/s
Q = Heat Leakage, kJ / s
λ = Latent Heat of vaporization of LNG, kJ / kg
Calculations
m = 166.552 / 512 = 0.3253 kg / s
or m = 0.3253*3600*24 = 28,105 kg / day
200,000 m3
The heat leakage value (Q) as provided in the article for this capacity is 163,253 W.
1 W = 0.001 kJ / s
or
163,253 W = 163.253 kJ / s
The latent heat of vaporization (λ) of commercial LNG is assumed to be 512 kJ / kg
A phase change heat-balance can be represented by:
Q = m*λ
or
m = Q / λ
where:
m = Boil-off Gas rate, kg/s
Q = Heat Leakage, kJ / s
λ = Latent Heat of vaporization of LNG, kJ / kg
Calculations
m = 163.253 / 512 = 0.3188 kg / s
or m = 0.3188*3600*24 = 27,544 kg / day
It is interesting to note that as the capacity of the LNG tank increases the BOG rate decreases.
Well this is what I have today for the readers of my blog and I hope to get some comments on this blog entry by process engineers who are actively involved in LNG storage and terminal design.
Regards,
Ankur.
I have seen different rules of thumb (or calculation methods) in LNG plants where I used to work, and most often the designers used a fixed figure of 0.05 wt% of maximum tank volume as the BOG generation rate. It makes sense to fix the evaporation percentage irrespectively of actual LNG volume inside the tank since heat gain is essentially always the same, making the equivalent mass of evaporated LNG to balance the heat gain also the same.