The analysis of 15000 litre of gas mixture at standard conditions is as follows:
CO_{2} = 9.5% ; SO_{2} = 0.5% ; O_{2} = 12.0% ; N_{2} = 78.0%.
How much heat must be added to this gas to change its temperature from 25^{o}C to 700^{o}C?
Data: Specific heat values in kcal/(kmol.^{o}K)
Gas CO_{2} SO_{2} O_{2} N_{2 }C_{p} at 25^{o}C 8.884 9.54 7.017 6.961 C_{p} at 700^{o}C 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 0^{o}C.
therefore, number of gmol of gas in the volume of 15000 litre at 25^{o}C, is estimated as:
PV = nRT (Ideal gas equation)
From the above equation,
n_{2}P_{1}V_{1}/T_{1} = n_{1}P_{2}V_{2}/T_{2}
here P_{1} and P_{2} are the same.
Therefore,
n_{2 }x 22.4 / 273 = 1 x 15000 / (273 + 25)
n_{2} = 613.5 gmol = 0.6135 kmol
Component |
Mole fraction |
No of moles, kmol |
CO_{2} |
0.095 |
0.6135 x 0.095 = 0.0583 |
SO_{2} |
0.005 |
0.6135 x 0.005 = 0.0031 |
O_{2} |
0.12 |
0.6135 x 0.12 = 0.0736 |
N_{2} |
0.78 |
0.6135 x 0.78 = 0.4785 |
Heat required (H) to raise the temperature from 25^{o}C to 700^{o}C is given by,
H = Σ n_{i}C_{Pmi}(T_{2} - T_{1}) = Σ n_{i}C_{Pmi} (700 - 25)
Where n_{i} is the moles of component 'i', and C_{Pmi} is the mean molal specific heat of component 'i'.
The calculations are shown in the following table.
Component |
kmol |
C_{P }at 25^{o}C kcal/(kmol.^{o}K) |
C_{P }at 700^{o}C kcal/(kmol.^{o}K) |
C_{Pm} kcal/(kmol.^{o}K) |
Heat to be added, kcal |
CO_{2} |
0.0583 |
8.884 |
11.303 |
(8.884 + 11.303)/2 = 10.0935 |
0.0583 x 10.0935 x (700 - 25) = 397.2 |
SO_{2} |
0.0031 |
9.54 |
11.66 |
10.6 |
22.18 |
O_{2} |
0.0736 |
7.017 |
7.706 |
7.3615 |
365.72 |
N_{2} |
0.4785 |
6.961 |
7.298 |
7.1295 |
2302.74 |
Total heat to be added |
3087.84 kcal |
Last Modified on: 04-Feb-2022
Chemical Engineering Learning Resources - msubbu
e-mail: learn[AT]msubbu.academy
www.msubbu.in