Question

Estimate:

b) LFL and the UFL at 25 ºC of 1 vol % hexane, 1 vol % ethylene, 3 vol % methane, 95% vol air. c) LFL and the UFL for hexane individually at a temperature of 300 ºC. Use the lower heat of combustion of 921.4 kcal/mol. ( Use flammable mole fraction)

Answer 1

Solution :

Anawer (b) :

Gas	vol %	mole fraction of combustible gases	LFL(Vol %)	UFL(Vol %)
hexane	1	0.2	1.2	7.5
ethylene	1	0.2	3.1	32
methane	3	0.6	5.3	15
total combustibla gas	5
air	95

Here LFL is the lower flammability limit and UFL is the upper flammability limit.

total combustibla gas = hexane + ethylene + methane = 1+1+3 = 5

mole fraction of combustible gases for hexane y1 = 1/5 = 0.2,

for ethylene y2 = 1/5 = 0.2,

for methane y3 = 1/5 = 0.6

• Now LFL of mixer at 25 cel is :

LFLmix = 1/sum(yi/LFLi)

LFLmix = 1 / (0.2/1.2 + 0.2/3.1 + 0.6/5.3)

LFLmix = 2.904 vol %

Now UFL of mixer at 25 cel is :

UFLmix = 1/sum(yi/UFLi)

UFLmix = 1 / (0.2/7.5 + 0.2/32 + 0.6/15)

• UFLmix = 13.71 vol %

Answer : (c)

LFLT = LFL25 - 0.75*(T - 25)/deltaHc  

UFLT = UFL25 + 0.75*(T - 25)/deltaHc  

deltaHc = Heat of comb.

For hexane at T = 300 cel temperature,

LFLT = LFL25 - 0.75*(T - 25)/deltaHc  

LFLT = 1.2 - 0.75 * (300-25) / (-921.4), Here 1 J = 0.239 Cal

LFLT = 1.4238 vol %

UFLT = UFL25 + 0.75*(T - 25)/deltaHc  

UFLT = 7.5 + 0.75 * (300-25) / (-921.4), Here 1 J = 0.239 Cal

UFLT = 7.4259 vol %