my 2 cents....
going back to #8 by breizh, the calculation procedure shown in https://www.eng-tips....cfm?qid=163305 , would say:
- LEL: 2,7 %v in air (for C2H4 in GPSA physical properties table)
- RL = (100-LEL)/LEL = 36.04
- RST = 3/.21 = 14.29 (stoichiometric, .21 is the fraction of O2 in air))
- RXS = RL-RST = 21.75
- NE = 1 (for nitrogen)
- RI = RXS/NE = 21.75
- FOS = 1.5
- RSN = RI*FOS = 32.63
C2H4 load in kmol/h is 11000/28.05 = 392.16
N2 min snuffing flow should be 392.16*32.63 = 12796 kmol/h or 358.3 t/h (MW N2 = 28)
it seems a lot....
Edited by gegio1960, 26 September 2019 - 11:49 PM.