December 2, 2020

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

At the flash point temperature of 284.5 K

The partial pressure is given by

sat

i i i

p x P

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.

6-3. For carbon monoxide, the stoichiometric combustion equation is:

From Equations 6-10 and 6-11:

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:

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:

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

6-7. For carbon monoxide, the stoichiometric combustion equation is:

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:

6-10. Methane (CH4) VCE. House destroyed 100 feet from blast. 100 ft = 30.5 m

Molecular weight of methane is 16 gm/mole

From Equation 6-27,

Substituting,

7

0.02

6-11. Tank Truck Compartment: 11,860 Liters

T = 25oC, P = 1 atm.

a. Stoichiometric concentration of octane: C8H18.

2. Basic stoichiometry:

Assume 100 moles of air

b. Use Equation 6-30 in the text to determine the equivalent mass of TNT:

8

6-12. Worst case stability conditions: F-stability, 2 m/s wind speed, rural conditions

Apply Equation 5-17:

Solve for

m

Q

:

Need to determine concentration in kg/m3:

Convert to PPM:

Then use Equation 2-7 to convert to kg/m3:

6-13. a. T = 25oC = 298 K

b. Use Equation 6-30:

Use the ideal gas law to determine the total number of moles in the storage vessel:

Substituting into Equation 6-30:

TNT

c. Use Equation 6-27:

6-14. First, determine mass of TNT that will result in a 3 psi overpressure. Use Equation 6-28

to determine the scaled overpressure:

10

TNT

Now use Equation 6-30 to determine the mass of butane. Since this is a vapor cloud

6-15. Equation 6-22 applied: 1/3

G

MAX

dP

V K

dt

See drawings below:

12

6-16.

Procedure:

1. Use Antoine equation (Appendix C) to determine the saturation vapor pressure.

a. Acetone

b. Benzene

c. Cyclohexane

d. Ethyl alcohol

13

e. Heptane

f. Hexane

g. Pentane

sat

h. Toluene

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

14

This is for the reaction as written.

Thus, per kg of gunpowder: