2 replies to this topic

[Eric Nguyen]

Dear forum

Could any experts here guide me How to calculate vapor volume from 1 m3 LGO liquid at 290degC to be contacted with hot catalyst at 750degC ?

Thank you

[PingPong]

Using liquid density of LGO you can convert 1 m³ to number of kg mass.

 

Using mol weight, T and P of LGO vapor you can calculate LGO vapor density.

 

From kg mass of LGO and its vapor density you can calculate LGO vapor volume (assuming no part of LGO cracks).

[P.K.Rao]

Vapor volume at 290 deg C of LGO

Let LGO density be d

Weight of 1 m3 LGO = 1000 x d kg

Let mean mol weight of LGO = M (You can estimate molecular weight by UOP 375 or ASTM D 2502 or by ASTM D 2503)

So, 1000 x d kg of LGO = (1000 x d)/M kg mols

1 kg mol of LGO = 22.4 m3 vapor at NTP theoretically

So, (1000 x d)/M kg mols of LGO = (1000 x d) x 22.4/M  m3 at NTP =  v1 say

Now apply PVT relation, p1v1/T1 = p2v2/T2

Assuming p1 = p2 (because pressure is not stated),

v1/T1 =  v2/T2

v2 = v1 x T2/T1 = v1 x (273+290)/273  = v1 x 1.696 m3. This would be the volume of LGO vapor at 290 deg C and one ata pressure before it comes in contact with the catalyst

Edited by P.K.Rao, 09 November 2014 - 11:01 PM.