Roaster Gas — Heat Content of 1 kmol above 25 °C: 23-Chem-A1 Process Balances and Chemical Thermodynamics, Undated paper, Q5

Question B2: Roaster Gas — Heat Content of 1 kmol above 25 °C (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The gas leaves at 502 °C (775.15 K), and the reference state is 25 °C (298.15 K). The composition is by mass: 3.57% SO₂, 1.08% O₂, 0.18% SO₃ and 95.17% N₂. The heat content is the sensible enthalpy $\sum y_i\int_{298.15}^{775.15}C_{p,i}\,dT$ per kmol of mixture.

Find. the heat content of 1 kmol of roaster gas relative to 25 °C.

Data check: printed SO₃ and N₂ polynomials. The N₂ polynomial has the same sign typo as in Question A3; as printed it gives 23.3 kJ/kmol·K at 775 K against a true value of about 31.2. The printed SO₃ polynomial has a negative $T^3$ term, which makes $C_p$ fall to 27 kJ/kmol·K at 1000 K; the true value is about 76. The sign-corrected forms, $C_{p,SO_3}=\ldots+24.3691\times10^{-9}T^3$ and $C_{p,N_2}=29.5909-5.1141\times10^{-3}T+1.31829\times10^{-5}T^2-4.968\times10^{-9}T^3$, reproduce the reference values: SO₃ 50.8 and 76.2, N₂ 29.1 and 32.7 at 298 and 1000 K. They are used below.

Approach. Convert the mass analysis to mole fractions on a 100 kg basis. Integrate each polynomial from 298.15 to 775.15 K, then mole-weight the results.

Mass to moles (basis 100 kg). $$\mathrm{SO_2}\ \frac{3.57}{64.058}=0.05573,\quad \mathrm{O_2}\ \frac{1.08}{31.998}=0.03375,\quad \mathrm{SO_3}\ \frac{0.18}{80.057}=0.00225,\quad \mathrm{N_2}\ \frac{95.17}{28.014}=3.39723\ \text{kmol}.$$ The total is 3.48896 kmol, so the mean molar mass is 28.66 kg/kmol. Mole fractions: SO₂ 0.01597, O₂ 0.00967, SO₃ 0.00064, N₂ 0.97371.
Integrated heat capacities, 298.15 → 775.15 K. With $\int C_p\,dT=a\Delta T+\tfrac b2\Delta(T^2)+\tfrac c3\Delta(T^3)+\tfrac d4\Delta(T^4)$: $$\Delta h_{SO_2}=22{,}440,\quad \Delta h_{O_2}=15{,}031,\quad \Delta h_{SO_3}=30{,}346,\quad \Delta h_{N_2}=14{,}297\ \text{kJ/kmol}.$$
Mixture heat content. $$H=\sum y_i\,\Delta h_i=0.01597(22{,}440)+0.00967(15{,}031)+0.00064(30{,}346)+0.97371(14{,}297)$$ $$=358.5+145.4+19.6+13{,}921=\boxed{14{,}445\ \text{kJ per kmol of gas above 25 °C}}.$$ That is about 504 kJ/kg of gas.

Sensitivity: the question prints “mass composition”, and that basis is used. If the same numbers were read as mole percent, the heat content would be 14,625 kJ/kmol (+1.2%); the nitrogen-dominated gas makes the result insensitive to the basis. If both printed polynomials were used uncorrected, the result would be 13,233 kJ/kmol, about 8% low, almost entirely from the N₂ term.

Species	Mole fraction	$\int_{298}^{775}C_p\,dT$ (kJ/kmol)	Contribution (kJ/kmol mix)
SO₂	0.01597	22,440	358.5
O₂	0.00967	15,031	145.4
SO₃	0.00064	30,346	19.6
N₂	0.97371	14,297	13,921
Heat content of 1 kmol above 25 °C			14,445 kJ

Nitrogen supplies 96% of the heat content. Getting its polynomial right therefore matters far more than the minor species do, which is why the printed N₂ coefficients had to be checked before use.

Concept & Theory

Topic. Sensible enthalpy of an ideal-gas mixture from polynomial heat capacities.

Key relations. Ideal-gas mixture: $H_{mix}=\sum y_i\,\Delta h_i$ with $\Delta h_i=\int_{T_{ref}}^{T}C_{p,i}\,dT$. Mass to mole fraction: $y_i=(w_i/M_i)/\sum(w_j/M_j)$.

Why this works. In an ideal-gas mixture there is no heat of mixing, so the enthalpy is the mole-weighted sum of the pure-component enthalpies at the same temperature. Each of those is just the heat-capacity integral from the reference temperature.

Common pitfall. Weighting by the mass fractions directly while using molar heat capacities. Another is integrating from 25 to 502 in °C with a kelvin-based polynomial.

Source. Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes §8.3 (sensible heat and heat capacities, Table B.2 form).

Question 5 of 6: Roaster Gas — Heat Content of 1 kmol above 25 °C

Part A — Process Mass and Energy Balances

Question B2: Roaster Gas — Heat Content of 1 kmol above 25 °C (equal value)