Good morning experts,
thank you for the message you sent for my preoccupation cited above
Concerning firefighting calculation, I opted the following codes:
- NFPA 11, 15 et 30
- API650 édition (Annex C)
The worst case scenario is:
The worst case is to calculate the firefighting flow (cooling and foam) of a tank independently of the other tanks.
Firefighting flow calculation
Calculation conditions
- The Crude oil is classified as Class II Liquid as per NFPA 30 Chapter 1-7.3.2.
- Fuel flash point is between 100 º F (37.8 º C) and 140 º F (60 º C).
- Product is crude oil,
- Diameter of the tank is 67020 m,
- Height of the tank is 17.18m,
- Nominal capacity of the tank is 50000m3,
As per NFPA 30 table 2-7:
The radius of the imaginary circle is the radius taken from the shell of the tank supposed to fire more then 15m
The formula is = 67.02 / 2 + 15 = 48.5 m
The (02) adjacent tanks are outside the imaginary circle that requires no cooling.
Tank on fire
1)Cooling water :
As per NFPA 15 Item 4.3.1.3
Required water flow (6.1 l / min / square meter) of the surface of the shell IN TOUCH OF FIRE OR THIS IS ANOTHER AREA (OTHER THAN THE TOTAL AREA OF THE SHELL)
Q1 = 6.1 x 67.02 x 3.14 x 17.18= 22065 l/min Or 1324 m3/h !!!!!!!!!!!!!
This flow seems very right to cooling the shell of the tank supposed on fire
2)Foam:
a.Calculate the total area of the roof of the tank
S1 =3528 m²
b.Calculation of the internal surface ofter foam dam
S2 =3403 m²
c.Calculation of the internal surface of foam dam
Premixture flow rate (water + emulsifier) necessary for the production of the foam
As per NFPA 11 TABLE B-1
12.2 liter / min / m ² of the annular surface between the shell and the foam dress.
Q2pm = 12.2
S = 12.2 x 126 = 1537 L/min 93 m3/h (water + emulsifier)
Water flow is 94% with 6% emulsifier
Q2 = Q2pm x 0.94 = 93 x 0.94 = 88 m3/h
Please your comments
Thanks and have a nice day
R.B (Mechanical Engineer)