No. 60: Steam Flash Tank Sizing
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BEST PRACTICE NO. 60
FLASH TANK SIZING STEAM
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1.1. Steps to Designing a Flash Tank
1. Calculate the amount of condensate entering the flash tank.
The amount of condensate entering the tank will be the sum of the steam-consuming capacity of all equipment discharging into the condensate return line that is going to the flash tank. This could be only one component or multiple components.
Example: 25,000 lbs./hr. (11.3 tons/hr.)
2. Determine the process pressure and flash tank pressure.
Example: Process pressure: 150 psig (10.3 bar)
Flash tank pressure: 10 psig (0.7 bar)
3. Calculate the condensate that flashes to steam.
% Flash = High 150 psig (10.3 bar) sensible heat (–) Low 10 psig (0.7 bar) sensible heat
Low 10 psig (0.7 bar) latent heat
% Flash = 338.4 Btu @ 150 psig (10.3 bar) – 207.9 Btu @ 10 psig (0.7 bar)
952.6 Btu @ 10 psig (0.7 bar)
Example: 13.7% = 338.4 – 207.9 Btu
25,000 lbs. per hr. (11.3 tons/hr.) X 13.7% = 3,425 lbs./hr. (1.7 tons/hr.) flash steam
4. Size the steam space.
If the condensate line is adequately sized, a high percentage of the steam flashing will occur in the condensate line. The steam section of the tank need only be sized to take care of the instantaneous flash. Unfortunately, a high percentage of the condensate lines in industrial plants are undersized for a number of reasons.
Therefore, the following example should be followed in sizing the steam section of the flash tank.
3,425 lbs./hr. x 16.5 (sp. vol. of steam @ 10 psig) cu. ft./lb. = 15.7 cu. ft./sec.
3,600 sec./hr. flashing
To accommodate steam system malfunctions that introduce additional steam to the system, SEA recommends adding a sizing factor of 1.5.
1.5 x 15.7 cu. ft./sec. = 23.55 cu. ft.
A 42 in. tank is capable of 9.65 cu. ft./12 in. Therefore, 23.55 ÷ 9.65 * 12 in. = 29.3 in.
Answer: 29.3 in. length on a 42 in. diameter tank for the steam section
5. Size the condensate section.
25,000 lbs. per hour (total volume) – 3,425 lbs./hr. (flash steam volume) = 21,575 lbs./hr. liquid or condensate
21,575 lbs./hr. ÷ 8.33 lb./gal. = 2,590 gal./hr.
2,590 gal./hr. ÷ 60 min. = 43.2 gpm (gallons per minute)
To provide stability of flow from the flash tank, at least five minutes of water should be provided.
43.2 gpm x 5 min. = 216 gal.
42 in. tank gallon capacity per 12 in. = 72 gal.
216 gal. ÷ 72 gal. * 12 in. = 36 in.
Answer: 36 in. length on a 42 in. diameter tank for the condensate section.
6. Size the tank.
29.3 in. length on a 42 in. diameter tank for the steam section
36 in. length on a 42 in. diameter tank for the condensate section
65.3 in. length on 42 in. diameter tank
5.44 ft. length on 42 in. diameter tank
7. Size the flash vent line off the tank.
Tank outlet flash steam velocities should not exceed 3,000 feet per minute.
V (fpm) = 2.4 x flow (lb./hr.) x v (cu. ft./lb.)
A (sq. in.)
V (fpm) = 2.4 x 3,425 (flash steam) x v 16.5 = 2,712.6 fpm
50 (8 in. vent line)