Question 1 of 6: Iron Blast-Furnace Charge and Slag Balance

Question 1: Iron Blast-Furnace Charge and Slag Balance (Part A — 25 marks)

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. Basis 1 ton = 1000 kg of pig iron (93.8% Fe, 4% C, 1.2% Si, 1% Mn). Charge: 1750 kg ore and 500 kg limestone of the stated analyses, plus coke (90% C, 10% SiO₂ ash, as printed in the analysis table). The limestone’s 1% is moisture (H₂O), which leaves with the flue gas and adds nothing to the slag. Products: 4200 m³(STP) of flue gas at 58% N₂ / 26% CO / 12% CO₂ / 4% H₂O by mole, pig iron, and slag. The gas volume is evaluated at STP, $V_m = 22.414\ \text{m}^3/\text{kmol}$.

Find. (a) mass of coke per ton pig iron; (b) volume/mass of air blown per ton pig iron; (c) mass composition of the slag.

Approach. The flue-gas moles are fixed by the 4200 m³ datum; a carbon atom balance over all carbon sources and sinks gives the coke, the inert N₂ acts as a tie substance to back out the air, and the slag is the sum of the oxides that entered but were not reduced into the pig iron.

1. Flue-gas moles. Convert the 4200 m³(STP) to kmol and split by the rational analysis: $$n_{gas}=\frac{4200}{22.414}=187.4\ \text{kmol}\;\Rightarrow\;n_{CO}=48.7,\; n_{CO_2}=22.5,\; n_{N_2}=108.7,\; n_{H_2O}=7.50\ \text{kmol.}$$
2. Carbon atom balance → coke. Carbon leaves as CO + CO₂ in the gas and as dissolved C in the pig iron; it enters from the coke and from CaCO₂ decomposition. With $M_C=12.011$: $$n_C^{pig}=\frac{40}{12.011}=3.33,\quad n_C^{lime}=\frac{475}{100.09}=4.75\ \text{kmol.}$$ The coke carbon is therefore $$n_C^{coke}=(n_{CO}+n_{CO_2})-n_C^{lime}+n_C^{pig}=(48.7+22.5)-4.75+3.33=69.8\ \text{kmol.}$$ Since coke is 90% C by mass, $$m_{coke}=\frac{69.8\times12.011}{0.90}=\boxed{931\ \text{kg coke / ton pig iron.}}$$
3. N₂ tie substance → air. All flue-gas nitrogen comes from the blast air (coke/ore N is negligible), so with air at 79 mol% N₂ / 21 mol% O₂: $$n_{O_2}^{air}=n_{N_2}\frac{21}{79}=108.7\times\frac{21}{79}=28.9,\qquad n_{air}=108.7+28.9=137.6\ \text{kmol.}$$ Expressed as a volume and a mass ($M_{N_2}=28.01,\ M_{O_2}=32.00$): $$V_{air}=137.6\times22.414=\boxed{3084\ \text{m}^3\ (\approx3969\ \text{kg}) \text{ air / ton.}}$$
4. Slag — unreduced oxides. The slag collects everything that entered as oxide but was not reduced into the metal. The CaCO₃ calcines fully to CaO, the coke ash reports as SiO₂, and the Fe, Si and Mn that were not taken up by the pig iron remain oxidised (unreduced iron as FeO). Computing each oxide: $$\text{CaO}=n_C^{lime}\,M_{CaO}=4.75(56.08)=266\ \text{kg},\qquad \text{Al}_2\text{O}_3=52.5\ \text{kg (ore only).}$$ SiO₂: input (ore 210 + limestone 20 + ash $0.10\times931=93$) minus the silica reduced to the 12 kg Si in pig iron $\left(\tfrac{12}{28.09}\times60.08=25.7\right)$ gives $323-25.7=297\ \text{kg}$. Likewise MnO $=\left(\tfrac{17.5}{70.94}-\tfrac{10}{54.94}\right)70.94=4.6\ \text{kg}$, and the unreduced iron reports as FeO: ore Fe $=1400\times\tfrac{2(55.85)}{159.69}=979$ kg, so $\left(\tfrac{979-938}{55.85}\right)71.84=53\ \text{kg FeO}$. Summing: $$m_{slag}=266+52.5+297+4.6+53=\boxed{674\ \text{kg slag / ton pig iron.}}$$

Normalising the slag oxides to a percentage basis gives a lime-silica slag typical of iron-making practice, with only small amounts of iron and manganese oxide lost to the slag (good reduction efficiency).