The analysis of 15000 litre of gas mixture at standard conditions is as follows:
CO2 = 9.5% ; SO2 = 0.5% ; O2 = 12.0% ; N2 = 78.0%.
How much heat must be added to this gas to change its temperature from 25oC to 700oC?
Data: Specific heat values in kcal/(kmol.oK)
Gas CO2 SO2 O2 N2 Cp at 25oC 8.884 9.54 7.017 6.961 Cp at 700oC 11.303 11.66 7.706 7.298
Calculations:
Basis: 15000 litre of gas
22.4 litre is occupied by 1 gmol of gas at at 0oC.
therefore, number of gmol of gas in the volume of 15000 litre at 25oC, is estimated as:
PV = nRT (Ideal gas equation)
From the above equation,
n2P1V1/T1 = n1P2V2/T2
here P1 and P2 are the same.
Therefore,
n2 x 22.4 / 273 = 1 x 15000 / (273 + 25)
n2 = 613.5 gmol = 0.6135 kmol
|
Component |
Mole fraction |
No of moles, kmol |
|
CO2 |
0.095 |
0.6135 x 0.095 = 0.0583 |
|
SO2 |
0.005 |
0.6135 x 0.005 = 0.0031 |
|
O2 |
0.12 |
0.6135 x 0.12 = 0.0736 |
|
N2 |
0.78 |
0.6135 x 0.78 = 0.4785 |
Heat required (H) to raise the temperature from 25oC to 700oC is given by,
H = Σ niCPmi(T2 - T1) = Σ niCPmi (700 - 25)
Where ni is the moles of component 'i', and CPmi is the mean molal specific heat of component 'i'.
The calculations are shown in the following table.
|
Component |
kmol |
CP at 25oC kcal/(kmol.oK) |
CP at 700oC kcal/(kmol.oK) |
CPm kcal/(kmol.oK) |
Heat to be added, kcal |
|
CO2 |
0.0583 |
8.884 |
11.303 |
(8.884 + 11.303)/2 = 10.0935 |
0.0583 x 10.0935 x (700 - 25) = 397.2 |
|
SO2 |
0.0031 |
9.54 |
11.66 |
10.6 |
22.18 |
|
O2 |
0.0736 |
7.017 |
7.706 |
7.3615 |
365.72 |
|
N2 |
0.4785 |
6.961 |
7.298 |
7.1295 |
2302.74 |
|
Total heat to be added |
3087.84 kcal |
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Last Modified on: 04-Feb-2022
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