Lecture 10

1. Recap & Model Framework

The goal is to develop a fully predictive steady-state model for the Blast Furnace (BF) to calculate performance indicators based on input parameters.

Basis of Calculation: 1 kg-mole of Iron ( $F e$ ) product.
Known Parameters (4):
1. Iron Input: $n_{F e}^{in} = 1$ (since output is 1 mole).
2. Iron Output: $n_{F e}^{o u t} = 1$ .
3. Ore Composition: $(O / F e)_{or e}$ (e.g., 1.5 for Hematite).
4. Metal Composition: $(C / F e)_{m e t a l}$ (Dissolved carbon, $\approx 0.21$ for saturated hot metal).
Unknown Variables (The “Big Three”):
1. Coke Rate ( $n_{C}^{a c t i v e}$ or $N_{A C}$ ): Moles of carbon participating in gas-phase reactions per mole $F e$ .
2. Blast Rate ( $n_{O}^{b l a s t}$ or $N_{O B}$ ): Moles of oxygen supplied by the blast per mole $F e$ .
3. Top Gas Composition ( $(O / C)_{g a s}$ ): The atomic oxygen-to-carbon ratio in the top gas.
Objective: Derive 3 Characteristic Equations to solve for these 3 unknowns.
- Lecture 10 focuses on the first two equations derived from Oxygen Balances.

2. Equation 1: Overall Oxygen Balance

This equation conserves oxygen across the entire furnace.

2.1. Derivation

Consider the control volume of the entire Blast Furnace:

Inputs of Oxygen:
1. From Blast: $N_{O B}$ (Atomic oxygen moles).
2. From Ore: $(O / F e)_{or e}$ (Stoichiometric oxygen associated with iron).
  - For Hematite ( $F e_{2} O_{3}$ ): $(O / F e) = 1.5$ .
Outputs of Oxygen:
1. In Top Gas: All oxygen leaves as $C O$ or $C O_{2}$ .
2. The amount is defined by the total active carbon ( $N_{A C}$ ) and the gas composition ratio $(O / C)_{g a s}$ .
- $Output Oxygen = N_{A C} \times (O / C)_{g a s}$

Equation 1 (Standard Form):

$N_{O B} + (O / F e)_{or e} = N_{A C} \cdot (O / C)_{g a s}$

For Ideal Hematite Operation:

$N_{O B} + 1.5 = N_{A C} \cdot (O / C)_{g a s}$

2.2. Graphical Representation (The Operating Line)

This equation can be rearranged into a linear form $y_{2} - y_{1} = m (x_{2} - x_{1})$ to be plotted on an Rist Diagram (Oxygen vs. O/C Ratio).

Rearrangement:

$1.5 - (- N_{O B}) = N_{A C} \cdot [(O / C)_{g a s} - 0]$
- Slope ( $m$ ): $N_{A C}$ (Active Coke Rate).
- Point 1 (Bottom/Blast): Coordinates $(0, - N_{O B})$ .
- Point 2 (Top/Gas): Coordinates $((O / C)_{g a s}, 1.5)$ .
Interpretation:
- The slope of the line connecting the Blast point and the Top Gas point represents the Coke Rate.
- Efficiency: A flatter slope indicates a lower Coke Rate (Higher efficiency). A steeper slope indicates high fuel consumption.

2.3. Handling Impurities (Non-Ideal Case)

In real operations, other elements (Si, Mn, P) are reduced, consuming carbon and releasing oxygen.

Example: Silicon Reduction ( $S i O_{2} \to S i + 2 O$ )
- If Hot Metal contains 1 wt% Si:
- This requires extra carbon (Endothermic load) and releases extra oxygen into the gas.
- Modification: The term $(O / F e)_{or e}$ increases to account for oxygen from silica.
- Revised Eq: $N_{O B} + [1.5 + Δ O_{S i}] = N_{A C}^{'} \cdot (O / C)_{g a s}^{'}$ .

3. The Conceptual Line of Division (CLD)

To derive the second equation, the furnace is conceptually divided into two zones based on temperature and reaction types.

3.1. The Line of Division

Location: Located at the Thermal Reserve Zone, approximately 1200 K (approx. 900-1000°C).
Physical Significance:
- At this boundary, the descending solids and ascending gases are in Thermal Equilibrium ( $T_{g a s} \approx T_{so l i d} \approx 1200 K$ ).
- Below this line: High temperature (Endothermic reactions active).
- Above this line: Lower temperature (Exothermic/Mildly Endothermic reactions).

3.2. Upper Segment (Preparation Zone)

Reactions: Indirect reduction of higher oxides ( $F e_{2} O_{3} \to F e_{3} O_{4} \to F e_{x} O$ ).
Key Assumption: Carbon is INERT.
- Temperature is too low for the Boudouard Reaction (Solution Loss: $C + C O_{2} \to 2 C O$ ).
- Therefore, the moles of active carbon entering the top ( $N_{A C}$ ) equal the moles crossing the CLD into the bottom.

3.3. Bottom Segment (Wustite Reduction Zone - WRZ)

Reactions:
- Final reduction of Wustite to Iron: $F e_{x} O + C O \to F e + C O_{2}$ .
- Coke gasification (Solution Loss) and combustion occur here.
Iron Input Form: Iron enters this zone not as $F e_{2} O_{3}$ , but as Wustite (Non-stoichiometric FeO).

4. Equation 2: Bottom Segment Oxygen Balance

We perform an oxygen balance specifically for the Bottom Segment (below the CLD).

4.1. Wustite Stoichiometry

Wustite is iron-deficient: $F e_{0.947} O$ .
To produce 1 mole of Fe product, we need more than 1 mole of $F e_{0.947} O$ .
Moles of Oxygen in Wustite per mole Fe:

$(O / F e)_{w u s t i t e} = \frac{1}{0.947} \approx 1.056 \to 1.06$
Input Oxygen from Ore term becomes 1.06.

4.2. Chemical Equilibrium Constraint

In the Thermal Reserve Zone (at the CLD), the gas composition is fixed by the Chemical Equilibrium of wustite reduction:

$F e O + C O ⇌ F e + C O_{2}$

At ~1200 K (1000°C), thermodynamics dictate the equilibrium ratio of $C O$ to $C O_{2}$ .
Equilibrium Gas Composition:
- $% C O \approx 70%$
- $% C O_{2} \approx 30%$
Equilibrium O/C Ratio ( $(O / C)_{W R Z}$ ):

$(O / C)_{W R Z} = \frac{1 \cdot n _{C O} + 2 \cdot n _{C O_{2}}}{n _{C O} + n _{C O_{2}}} \approx 1 + 0.3 = 1.3$

4.3. Derivation

Balance Oxygen for the Bottom Segment:

Input:
1. Blast Oxygen: $N_{O B}$ (Same as overall).
2. Wustite Oxygen: $1.06$ (Oxygen entering with solid).
Output:
1. Gas leaving the bottom segment (at equilibrium).
2. Oxygen = $N_{A C} \times (O / C)_{W R Z}$ .

Equation 2 (Bottom Segment Balance):

$N_{O B} + 1.06 = N_{A C} \cdot 1.3$

5. Summary of Derived Equations

We now have two linear equations linking the unknowns:

Overall Oxygen Balance:

$N_{O B} + 1.5 = N_{A C} \cdot (O / C)_{g a s}$

(Note: Contains 3 unknowns: $N_{O B}, N_{A C}, (O / C)_{g a s}$ )
Bottom Segment Oxygen Balance (Wustite Reduction Zone):

$N_{O B} + 1.06 = 1.3 \cdot N_{A C}$

(Note: Contains 2 unknowns: $N_{O B}, N_{A C}$ . This fixes the linear relationship between Coke Rate and Blast Rate.)

Instructor’s Note:

These equations are “Conceptually Fabulous.” They demonstrate that for large blast furnaces, the existence of a Chemical and Thermal Reserve Zone allows us to fix the gas composition at the zonal boundary ( $O / C \approx 1.3$ ), significantly simplifying the complex reactor into solvable algebraic expressions.

Next Step: The third equation (Enthalpy Balance) will be derived in the next lecture to fully solve the system.

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