Modern Engineering Thermodynamics

620 CHAPTER 15: Chemical Thermodynamics
EXAMPLE 15.11
We want to measure the heat of combustion of liquid octane by burning
it in the rigid, sealed, adiabatic bomb calorimeter shown in Figure 15.9.
The internal volume is of the combustion bomb is 50.0 × 10 −3 ft 3 .We
insert 10.0 grams of fuel and fill the bomb with enough pure oxygen to
have 50.0% excess oxygen for the combustion reaction. Determine the
maximum possible explosion pressure inside the bomb when the fuel is
ignited.
Solution
The molecular mass of octane ðC 8 H 18 Þ is 114 kg/kgmole, so at 10.0 g
(0.0100 kg), it contains
T = T A
p = p max
ð0:0100 kgÞ/ ð114 kg/kgmoleÞ = 8:77 × 10 −5 kgmole
The reaction equation for 50.0% excess pure oxygen is
C 8 H 18 + 1:5ð12:5ÞðO 2 Þ ! 8CO ð 2 Þ+ 9H ð 2 OÞ+ 6:25ðO 2 Þ
and 8.77 × 10 −5 kgmole of octane yields
or
8:77 × 10 −5 ð8 + 9 + 6:25Þ = 2:00 × 10 −3 kgmole of product
FIGURE 15.9
Example 15.11.
2:00 × 10 −3 kgmole ð2:2046 lbmole/kgmoleÞ = 4:41 × 10 −3 lbmole of product
We can estimate the adiabatic flame temperature for this closed system reaction from Eq. (15.18) using the temperatureaveraged
molar constant volume specific heats found in Table 15.6:
T A closed
system
= T°+
h f ° fuel −∑ R
ðn i /n fuel
∑
P
ÞRT° −∑
R
ðn i /n fuel Þðc vi
ðn i /n fuel Þðh° f − RT°Þ i
Þ avg
Since the fuel is in liquid form at a low pressure here, we use the approximation ðu f Þ fuel ≈ h f ° fuel : Since
RT° =ð0:0083143Þð298:15Þ = 2:4789 MJ/kgmole is a constant, the numerator of Eq. (15.18) becomes
and the denominator is
Then, Eq. (15.18) gives
and Eq. (15.19) gives
P max = n pRT A
V p
− 249:952 − 1:5ð12:5Þð2:4789Þ− 8ð−393:522 − 2:4789Þ
− 9ð−241:827 − 2:4789Þ− 6:25ð0 − 2:4789Þ = 5090 MJ
8ð0:04987Þ + 9ð0:03419Þ + 6:25ð0:02468Þ = 0:861 MJ/K
ðT A Þ bomb = 25:0 + 5090 = 5930°C = 6210 K = 11,200 R
calorimeter 0:861
ð
= 4:41 × 10−3 lbmoleÞ 1545:35 ft . lbf/ ðlbmole . RÞ ð11:2 × 10 3 RÞ
50:0 × 10 − 3 ft 3 144 in 2 /ft
2 = 10,600 psi
Exercises
31. Determine the maximum explosion pressure in Example 15.11 for a bomb volume of 10.0 × 10 −3 ft 3 instead of 50.0 ×
10 −3 ft 3 . Assume all the remaining variables are unchanged. Answer: p max explosion = 53,900 psi.
32. If the amount of fuel used in the bomb calorimeter in Example 15.11 is 50.0 grams instead of 10.0 grams, determine
the maximum explosion pressure in the bomb. Assume all the remaining variables are unchanged.
Answer: p max explosion = 53,900 psi.
33. If the bomb calorimeter in Example 15.11 is filled with 100.% excess oxygen instead of 50.0% excess oxygen, determine
the maximum explosion pressure in the bomb. Assume all the remaining variables are unchanged.
Answer: p max explosion = 11,700 psi.