v8landy
Chemical
- Jan 3, 2008
- 81
Hi
I need guidance in the calculation required to calculate flow/pressure or Nitrogen fill a tank (5m^3) with Nitrogen.
The tank/Nitrogen is dual purpose
1 it is the blow of tank should the RD go
2 it (the Nitrogen) is used to replace the vacumm of another vessel.
This opperation is preformed every 5 -8 hours, so quite enought time for a steady steam of nitrogen to fill back tank again.
PS tank is open vented to atms due to it as stated ahaing dual purpose of RD dump tank
Is it just as simple as to say tank is 5m^3 = 177 ft^3
therfore if I wanted the tank purged and ready in 2 hours (4 hours between opperations)
(177 ft^3 / 2hrs)60 = 1.48 CFM N2 purge rate (cubic feet per minute) ????????
have used an equation that works out number of purges, and from this it equated to 3.5 to ensure full purge.
From this I would then assume by tank volume to be 3.5 time greater than actual i.e. 615.96 ft^3
i.e (619.5/2hr)/60 mins = 5.12 CFM
???????????
I need guidance in the calculation required to calculate flow/pressure or Nitrogen fill a tank (5m^3) with Nitrogen.
The tank/Nitrogen is dual purpose
1 it is the blow of tank should the RD go
2 it (the Nitrogen) is used to replace the vacumm of another vessel.
This opperation is preformed every 5 -8 hours, so quite enought time for a steady steam of nitrogen to fill back tank again.
PS tank is open vented to atms due to it as stated ahaing dual purpose of RD dump tank
Is it just as simple as to say tank is 5m^3 = 177 ft^3
therfore if I wanted the tank purged and ready in 2 hours (4 hours between opperations)
(177 ft^3 / 2hrs)60 = 1.48 CFM N2 purge rate (cubic feet per minute) ????????
have used an equation that works out number of purges, and from this it equated to 3.5 to ensure full purge.
From this I would then assume by tank volume to be 3.5 time greater than actual i.e. 615.96 ft^3
i.e (619.5/2hr)/60 mins = 5.12 CFM
???????????
