Lecture 12

1. Recap & Objective

Following the derivation of the Bottom Segment Enthalpy Balance in Lecture 11, the objective of this lecture is to:

1. Define the Heat Demand ($D_E$) and Heat Supply ($S_E$) terms explicitly.

2. Combine the Enthalpy Balance and Oxygen Balance equations to create a unified framework.

3. Map these equations onto the Rist Diagram (Operating Line) to allow for graphical solution of the Blast Furnace variables.

---

2. The Heat Demand Term ($D_E$)

The Heat Demand represents the energy required by the process per mole of Iron ($Fe$) produced. It is a specific property of the bottom segment (Wustite Reduction Zone to Hearth).

Formula for $D_E$

$D_E=D_{WRZ}=\sum (\text{Endothermic Requirements})-\sum (\text{Sensible Heat Inputs})$

Components:

1. Reduction of Wustite: Energy to break $Fe-O$ bonds ($F{e}_{0.947}O\to Fe+O$).

2. Sensible Heat of Iron: Heating Iron from $1200\text{ K}$ (Solid) to $1800\text{ K}$ (Liquid).

3. Sensible Heat of Dissolved Carbon: Heating Carbon from $1200\text{ K}$ to $1800\text{ K}$ and dissolving it in Iron.

4. Impurities (Non-Ideal Case): Reduction of $Si{O}_2,MnO,P_2{O}_5$, and calcination of limestone.

5. Slag Formation: Sensible heat to raise slag components to $1800\text{ K}$.

Numerical Value (for Ideal Hematite Operation):

For the standard case discussed (Pure Hematite, no Slag):

$D_E\approx 116,000\text{ kJ/kg-mole Fe}$

(Note: The instructor approximates this value in derivations. In real operations with slag, $D_E$ is significantly higher.)

---

3. The Heat Supply Term ($S_E$)

The Heat Supply is the energy provided by the Blast and the Combustion of Coke per mole of Active Carbon ($N_{AC}$).

Formula for $S_E$

$S_E=\text{Blast Enthalpy Term}+\text{Combustion Enthalpy Term}$

Derivation steps from Board:

1. Combustion Heat:

   • Formation of $CO$: Exothermic ($C+\frac{1}{2}{O}_2\to CO$).

   • Formation of $C{O}_2$: Exothermic ($C+O_2\to C{O}_2$).

2. Weighted Average by Gas Composition:

   The gas leaving the bottom segment is at equilibrium with Wustite at $1200\text{ K}$.

   • $O/C_{gas}\approx 1.3$

   • This implies $\approx 70\%CO$ and $30\%C{O}_2$.

   • Heat Evolution:

     $0.7\times \Delta H_f(CO)+0.3\times \Delta H_f(C{O}_2)\approx 198,000\text{ kJ/kg-mole C}$

3. Blast Contribution ($E_B$):

   Sensible heat brought in by the hot blast (above $1200\text{ K}$).

Combined Supply Equation:

$\text{Total Supply}=N_{AC}\times [283,000+E_B]$

(Note: The coefficients depend on the precise heats of formation used. The transcript mentions 198,000 as the combustion part.)

---

4. The Combined Characteristic Equation

We now have two equations describing the bottom segment:

1. Oxygen Balance: $N_{OB}+1.06=1.3\cdot N_{AC}$

2. Enthalpy Balance: $D_E=N_{OB}\cdot E_B+N_{AC}\cdot 198,000$

By eliminating $N_{OB}$ (Blast Rate) between them, we solve for Coke Rate ($N_{AC}$) directly.

Final Coke Rate Equation

$N_{AC}=\frac{D_E+1.06\cdot E_B}{198,000+1.3\cdot E_B}$

Physical Interpretation:

• Numerator ($D_E+1.06E_B$): Total thermal load adjusted for blast enthalpy carried by the oxygen associated with Wustite reduction.

• Denominator ($198,000+1.3E_B$): Effective heat generation per mole of carbon, accounting for blast heat.

---

5. Graphical Representation: The Rist Diagram

This is the core of the lecture—mapping the math onto the visual operating line.

Axes:

• Y-Axis: $O/Fe$ (Moles of Oxygen per mole of Iron).

• X-Axis: $O/C$ (Atomic Ratio in Gas).

Key Points on the Diagram:

1. Point W (Wustite Point):

   • Coordinates: $(X=1.3,Y=1.06)$

   • Represents the state at the Thermal Reserve Zone (TRZ).

   • This is the Pivot Point for the bottom segment operating line.

2. Point P (The Pivot/Pole):

   To satisfy the Enthalpy Balance graphically, the operating line must pass through a specific point P on the X-axis.

   • Abscissa of P ($X_P$):

     $X_P=\frac{H_{supply}^{total}}{H_{demand}^{total}}$

   • The instructor derives that the enthalpy balance forces the operating line to rotate around W such that it intercepts the X-axis at a specific value determined by $E_B$ and $D_E$.

Construction of the Operating Line:

1. Plot Point W at $(1.3,1.06)$.

2. Calculate Slope: The slope of the line is the Active Coke Rate ($N_{AC}$).

3. Draw the Line: Pass a line through W with slope $N_{AC}$.

   • Intersect with Top ($Y=1.5$): Gives Top Gas Composition.

   • Intersect with Bottom ($X=0$): Gives Blast Rate ($-N_{OB}$).

Visualizing the Impact of Parameters

• Increasing Blast Temp ($T_B\uparrow$):

  • $E_B$ increases (Supply increases).

  • Coke Rate ($N_{AC}$) decreases (Slope becomes flatter).

  • Operating line rotates clockwise around W.

• Increasing Demand ($D_E\uparrow$):

  • E.g., Higher slag volume or adding limestone.

  • Coke Rate ($N_{AC}$) increases (Slope becomes steeper).

  • Operating line rotates counter-clockwise around W.

---

6. Solution Algorithm (Summary)

To solve any Steady State Blast Furnace problem:

1. Calculate Demand ($D_E$): Sum all endothermic heats (Reduction, Melting, Slag) minus inputs (Sensible heat of Ore/Coke).

2. Calculate Blast Enthalpy ($E_B$): Based on Blast Temperature ($T_{blast}$) and Moisture.

3. Solve for Coke Rate ($N_{AC}$): Use the derived algebraic equation:

   $N_{AC}=\frac{D_E+1.06\cdot E_B}{198,000+1.3\cdot E_B}$

4. Solve for Blast Rate ($N_{OB}$): Use the Bottom Segment Oxygen Balance:

   $N_{OB}=1.3\cdot N_{AC}-1.06$

5. Solve for Top Gas ($O/C_{gas}$): Use the Overall Oxygen Balance:

   $(O/C{)}_{gas}=\frac{N_{OB}+1.5}{N_{AC}}$

Instructor’s Final Remark:

“We have now completely solved the Blast Furnace. Give me the inputs (Ore chemistry, Blast Temp), and I will give you the outputs (Coke Rate, Blast Volume, Top Gas Analysis) using these three equations.”