Vaporization rate of liquid propane 1

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.

 

[whylie]

Dear VS.1

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

 

[David]

vs.1

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