BUTANE BOTTLE
Statement
A type of commercial butane bottle has a capacity of 26.1 litres and holds 12.5 kg of butane (in
practice it is a hydrocarbon mixture as obtained from crude-oil distillation, not a pure component).
Evaluate:
a)
If it is possible to have this amount of butane inside this volume.
b)
The pressure inside for the three cases: cold operation (10 ºC), nominal operation (20 ºC) and
maximum allowed (50 ºC).
c)
The volume of liquid and vapour.
Una botella de butano comercial de las antiguas tiene una capacidad de 26,1 litros y contiene 12,5 kg
de butano (supóngase puro, aunque en realidad es una mezcla obtenida en el destilado de petróleos). Se
pide:
a) Determinar si es posible que haya esa cantidad de butano en ese volumen.
b) Presión interior en los tres casos siguientes: operación en frío a 10 ºC, operación normal a 20 ºC y
máximo permitido a 50 ºC.
c) Volumen de líquido y de vapor.
Solution.
a)
If it is possible to have this amount of butane inside this volume.
The density of liquid n-butane is found from Thermal Data tables to be 584 kg/m3, so that up to
584∙0.0261=15.2 kg might be put inside this volume; however, no bottle should be fully filled
because of a possible burst by thermal expansion of the liquid.
b)
The pressure inside for the three cases: cold operation (10 ºC), nominal operation (20 ºC) and
maximum allowed (50 ºC).
Using Antoine equation:
p
B
ln
 A
T
p0
C
T0
F
IJ
G
HK
with A=13.98, B=2292 and C=27.86 for T0=1 K and p0= 1 kPa, one gets pv(10 ºC)=150 kPa,
pv(20 ºC)=210 kPa and pv(50 ºC)=500 kPa. Notice that butane would not go out of the bottle
for temperatures below its normal boiling point (0 ºC for n-butane).
c)
The volume of liquid and vapour.
The liquid level varies a little with temperature. Let us find its value at 50 ºC. One may
establish a set of equations m=mL+mV=12.5 kg, V=VL+VV=0.0261 m3, mL/VL=L=584 kg/m3
(with the perfect liquid model), and mV/VV=V=p/(RT)=150103/((8.3/0.058)283)=3.7 kg/m3
with the perfect gas model. These are 4 equations with 4 unknowns (mL,mV,VL,VV), but many
times it is simpler and accurate enough to think that all the mass is liquid and thus it occupies
VL=m/L=12.5/584=0.022 m3 and the rest (26-22=4 litres) is vapour with a mass
mV=VVV=pVV/(RT)=0.046 kg.at 50 ºC (the vapour mass would be 0.015 kg at 10 ºC, and 0.021
kg at 20 ºC, with this ideal gas model).
The perfect substance models just used may be good enough in many cases, but a better model
might be necessary, at least to judge the degree of approximation. The corresponding state
model gives for the vapour at 50 ºC mV=VVV=pVV/(ZRT)=0.052 kg, where Z(pR,TR)=0.87 has
been found using Z-pR diagram (in Thermal Data) with pR=p50/pcr=0.5/3.8=0.13 and
TR=T50/Tcr=323/425=0.76. For the liquid density, the dilatable liquid model yields
=0[1(TT0)]=584∙[11.8∙10-3(5015)]=547 kg/m3. The best available data (NIST) gives,
for T=323 K, pv=494.8 kPa, L=542.3 kg/m3 andV=12.22 kg/m3, i.e. 23.0 litres of liquid and
3,1 litres of vapour.
Comments.
Typical commercial butane in Spain has 56% n-butane, 25% propane, 17% iso-butane, 2% pentane,
0,1 g/kg H2O and 1 mg/kg mercaptans, with Tb=0.5 ºC, L=580 kg/m3 at 20 ºC and pv(50 ºC)=520
kPa.
The traditional butane bottle (Fig. 1) is made from 3.2±0.2 mm steel plate (two halves welded in the
middle, with a welded collar and screwed valve on the top, and a welded standing ring at the bottom).
All bottles are check for resistance with water at 3 MPa, and for air-tightness at 0.8 MPa, and a sample
from the lot suffers a hydraulic end test to check that it breaks at >8.5 MPa.
Fig. 1. Domestic butane bottle UD-125, with pressure regulator K30 (30 mbar).
Fig. 2. Details of the collar, valve and regulator (K30) in a domestic butane bottle.
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