Vaporization rate of liquid propane
Vaporization rate of liquid propane
(OP)
Referring to the 11/29/00 post.
The tank pressure will probably be a maximum of 150 psi.
The smallest diameter pipe in the run will be 21/2".
The farthest length of the run will be 800ft.
The tank pressure will probably be a maximum of 150 psi.
The smallest diameter pipe in the run will be 21/2".
The farthest length of the run will be 800ft.





RE: Vaporization rate of liquid propane
I made a quick calculation for your case using the Panhandle formula:
(P1-P2)^0.5394 D^2.6182
Q = 18583 x --------------- x ---------- where
(Tf L )^0.5394 G^0.4606
Q = Gas flow in cu.ft/day [ @ 14.7 psi ,60 F ]
P1= Initial gas pressur,psi ( 150)
P2= Final gas pressure , psi ( 15 )
Tf= Average gas flowing temperature ,deg R(60 F= 520 R)
L = Pipeline length in miles ( 800/5280 = 0.1515)
D = Pipe diameter,in ( 2.5 in)
G = Gas gravity relative to air ( C3=1.51 )
Substituting above values,we get
Q = 226972 cu.ft/day = 9457 cu.ft/hr
Heating value of Propane is 2558 Btu/cu.ft
Therefore you can provide
9457x 2558 = 24,191,006 Btu/hr (required 12,500,000)
and in the worst case you can get (20,000,000 Btu/hr)
At the stated rate the 25,000 gals tank has to be topped every :
25000 gal x 92290 Btu/gal( C3 heating value )
---------------------------------------------= 184.58 hrs
12,500,000 Btu/hr
or ( 8 days)
I hope this meets your requirement
Regards Whylie
12,500,000 Btu/hr
RE: Vaporization rate of liquid propane
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