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Chapter 6
6-1. The flash point of the mixture occurs when the partial pressure of methanol is equal to the
vapor pressure of pure methanol at the flash point temperature.
From Appendix B, the flash point temperature for methanol is 11oC = 284.5 K
34.27
T
At the flash point temperature of 284.5 K
The partial pressure is given by
sat
i i i
p x P
3636.4
34.3 13.80
297 K 24.1 C 75.4 F
o o
T
T
6-2. First compute the mole fraction on a combustible basis only, ignoring the non-combustible
air. We can obtain the flammable limits from Appendix B. So we have:
Then we use Equations 6-2 and 6-3 to compute the LFL and UFL of the mixture.
3
1
3
1
1 1
11.4%
0.60 0.40
12.5 10.0
1 1
28.8%
0.60 0.40
74.0 15.0
mix
i
i
mix
i
i
LFL y
LFL
UFL y
UFL
6-3. For carbon monoxide, the stoichiometric combustion equation is:
1
From Equations 6-10 and 6-11:
55 55 55
16.3%
4.76 1.19 2.38 1 4.76 1 1.19 0 2.38 1 3.38
350 350 103%
4.76 1.19 2.38 1 3.38
LFL m x y
UFL m x y
Clearly the UFL is in error – it cannot be greater than 100%. This compares to the LFL of
12.5% and UFL of 75.0% from Appendix B. The prediction is not very good.
For heptane the balanced combustion equation is:
From Equations 6-10 and 6-11:
55 55 55
1.03%
4.76 1.19 2.38 1 4.76 7 1.19 7 2.38 0 53.36
350 350 6.56%
4.76 1.19 2.38 1 53.36
LFL m x y
UFL m x y
This compares to the values in Appendix B of 1.0% to 7.0%. Much better than for CO.
6-4. From Appendix B,
Applying LeChatelier’s equation, Equations 6-2 and 6-3:
1 1
21.2%
0.5 0.5
15 36
i
i
UFL y
UFL
6-5. Ethylene: C2H4
Substituting into Equations 6-10 and 6-11
Actual values for ethylene from Appendix B are 2.7% and 36.0%. The estimation
equations do not work all that well in this case.
6-6. Refer to Example 6-6 in the text. Ethylene: C2H4
2
Mole % O
LOC LFL
Mole % fuel
Also, from Equation 6-9:
CmHxOy + (m + x/4 – y/2) O2 mCO2 + (x/2) H2O
Ethylene, C2H4
m = 2, x = 4, y = 0
LFL = 2.7% from Appendix B
(m + x/4 – y/2) = 2 + 4/4 - 0 = 3
LOC = (2.7%)(3) = 8.1% compared to experimental LOC of 10.0% in Table 6-3.
6-7. For carbon monoxide, the stoichiometric combustion equation is:
1
4
6-8. Need data for methyl alcohol.
The stoichiometric line is plotted from the 60% oxygen on the oxygen axis to the 100%
inert vertex.
6
6-9. Propane: C3H8
20 lbm propane = 9.07 kg of propane
From Appendix B, the energy of explosion for propane is given by the lower heating value,
or 2043.1 kJ/mole.
The molecular weight for propane is 3x12 + 8x1 = 44 gm/mole = 0.044 kg/mol
The number of moles of propane is 9.07 kg/0.044 kg/mol = 206 mol
From Equation 6-30:
C
206 mol 2043.1 kJ/mol
n H
6-10. Methane (CH4) VCE. House destroyed 100 feet from blast. 100 ft = 30.5 m
Molecular weight of methane is 16 gm/mole
e
From Equation 6-27,
1/3 1/3
30.5 m
r
c
Substituting,
119 kg 4684 kJ/kg 11.1
8
1.00 since fully confined
0.577 kg
m
TNT
6-12. Worst case stability conditions: F-stability, 2 m/s wind speed, rural conditions
Apply Equation 5-17:
y z
Q
u
Solve for
m
Q
:
y z
Need to determine concentration in kg/m3:
1
4
Convert to PPM:
6
10 ppm
100 vol. %
Then use Equation 2-7 to convert to kg/m3:
3 3 3
PPM 1 atm 72 gm/gm-mole
3250 PPM
C PM
m
6-13. a. T = 25oC = 298 K
2911.32
ln 15.8737
298 56.51
sat
P
b. Use Equation 6-30:
C
m H
Use the ideal gas law to determine the total number of moles in the storage vessel:
1 atm 100,000 L
4089 gm-moles
0.082057 L atm/gm-mole K 298 K
g
PV
nR T
The total moles of heptane in then vessel is:
0.0599 4089 moles 244.9 moles heptane
Hept
n
Substituting into Equation 6-30:
TNT
TNT
4686 kJ/kg
c. Use Equation 6-27:
1/3
50 m
r
6-14. First, determine mass of TNT that will result in a 3 psi overpressure. Use Equation 6-28
to determine the scaled overpressure:
o
p
10
9.5 m/kg
1,166 kg TNT
TNT
m
Now use Equation 6-30 to determine the mass of butane. Since this is a vapor cloud
0.02 2657.5 kJ/mol 1 mol/0.058 kg
5,962 kg butane
c
m
6-15. Equation 6-22 applied: 1/3
G
MAX
dP
V K
dt
1/3
(19)(20) 51.6 bar-m/s
G
K
See drawings below:
12
6-16.
Procedure:
1. Use Antoine equation (Appendix C) to determine the saturation vapor pressure.
a. Acetone
2940.46
sat
b. Benzene
94.5 mm Hg
V%=12.4%
UFL=7.1%
NOT flammable
sat
P
c. Cyclohexane
96.9 mm Hg
V%=12.7%
UFL=8.0%
NOT flammable
sat
P
d. Ethyl alcohol
58.6 mm Hg
V%=7.71%
LFL=4.3%
UFL=19%
Flammable!
sat
P
13
e. Heptane
45.51 mm Hg
V%=6.0%
LFL = 1.0%
UFL=7.0%
Flammable!
sat
P
f. Hexane
150.3 mm Hg
V%=19.8%
UFL=7.5%
NOT flammable
sat
P
g. Pentane
UFL=7.8%
NOT flammable
sat
h. Toluene
28.22 mm Hg
V% = 3.7%
LFL = 1.2%
UFL = 7.1%
Flammable!
sat
P
6-17. The reaction is: 2KNO3 + 3C + S K2S + 3CO2 + N2
The required information is shown in the table below:
Species
f
H
(kJ/mol) Molecular Weight
r PROD REAC
14
This is for the reaction as written.
Thus, per kg of gunpowder:
