Mean specific heat and combustion-air
requirement:
If the temperature of the combustion products is known, then a mean
molal specific heat can be calculated as follows:
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|
The mean molal specific heat data is available for a number of
industrial gases over a wide temperature range.
Example 3.3: It is necessary that complete combustion of
propane yield a temperature of 1600 °C. The propane supply temperature is 25 °C. Determine the air-to-fuel ratio on a molar
basis that must be supplied to the burner to attain the desired
temperature. Heat of combustion of
propane is 530605.6 cal/mole. Latent
heat of vaporization of water at 25 °C is 10500
cal/mole. Mean molal heat capacity of
the compounds between 25 °C and 1600 °C
is given as below:
C3H8
= 6.601 cal/(mol·°C), O2 = 8.269 cal/(mol·°C),
CO2 = 12.75 cal/(mol·°C),
N2 = 7.844
cal/(mol·°C),
H2O = 9.95 cal/(mol·°C), air =
7.929 cal/(mol·°C)
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Solution: Basis of calculations: 1 mol of gas,
Feed temperature, TFeed = 298.15 °K,
Product temperature, TProducts = 1873.15 °K.
$
Enthalpy of feed, SHFeed
= 0
C
Heat added to the system, q = 0
$
Standard heat of reaction, ΔH298
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ΔH298 = -530605.6 + 4(10500) =
-488605.6 cal
C
Enthalpy of products, SHP = SHF
- ΔH298
- q = SnpCpm(TP - TF) =
488605.6 cal
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|
nP |
Cpm |
npCpm |
|
CO2 |
3 |
12.75 |
38.25 |
|
H2O |
4 |
9.95 |
39.8 |
|
N2 |
5(3.764)
=18.82 |
7.844 |
147.62 |
|
Air |
X |
7.929 |
7.929X |
|
S |
|
|
225.67
+ 7.929X |
C
Excess air, X
(225.67 + 7.929X)(1873.15 - 298.15) = 488605.6
X = 10.664 mol
C
Stoichiometric air = 5(4.764) = 23.82
mol
C
Total air requirement = 10.66 + 23.82 =
34.48 mol