C H A P T E R
29 Bottom Segment Calculations With Natural Gas Injection O U T L I N E 29.1 Replacing Tuyere Injection of CH4(g) With Natural Gas Injection 29.2 Comparison of CH4(g) and Real Natural Gas 29.3 Natural Gas Injection Equations 29.3.1 Injected Natural Gas Quantity Equation 29.3.2 Amended Hydrogen Balance Equation 29.3.3 Amended Carbon Balance Equation
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29.1 REPLACING TUYERE INJECTION OF CH4(g) WITH NATURAL GAS INJECTION Chapter 11, Bottom Segment with CH4(g) Injection, described tuyere injection of CH4(g) with the CH4(g) standing in for industrial natural gas. This chapter repeats calculations of
Blast Furnace Ironmaking DOI: https://doi.org/10.1016/B978-0-12-814227-1.00029-4
29.3.4 Amended Oxygen Balance Equation 29.3.5 Amended Nitrogen Balance Equation 29.3.6 Amended Enthalpy Balance Equation
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29.4 Results
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29.5 C-in-Coke Replacement by Natural Gas of Appendix Q
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29.6 Summary
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Exercises
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Chapter 11, Bottom Segment with Injection, but using the composition natural gas (Fig. 29.1). The objective of fuel injection is to expensive C-in-coke with inexpensive gas reductant/fuel. In this chapter, we determine;
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CH4(g) of real replace natural
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29. BOTTOM SEGMENT CALCULATIONS WITH NATURAL GAS INJECTION
TABLE 29.1 Composition of and 25 C Enthalpy of CH4(g) and Real Natural Gas
FIGURE 29.1 Conceptual blast furnace bottom segment with tuyere injection of 25 C real (industrial) natural gas. The composition and enthalpy of the natural gas are given in Table 29.1.
Element (mass%)
CH4(g)
Real Natural Gas(g)
C
74.9
73.4
H
25.1
24.0
N
0
1.7
0
1.0
2 4.66
2 4.52
O
25 C Enthalpy (MJ per kg)
Composition from Appendix Q. Enthalpy from Appendix R.
• the amount of C-in-coke (kg) that is saved by each kg of injected real natural gas.
29.2 COMPARISON OF CH4(g) AND REAL NATURAL GAS Table 29.1 compares the composition and 25 C enthalpy of CH4(g) and real natural gas. They are quite similar.
29.3.2 Amended Hydrogen Balance Equation With 24.0 mass% H in our natural gas, hydrogen balance equation (11.3) of Chapter 11, Bottom Segment with CH4 Injection, becomes; mass tuyere injected 052 0:240 natural gas mass H2 O out mass H2 out 11 0:112 1 in ascending gas in ascending gas
(29.2)
29.3 NATURAL GAS INJECTION EQUATIONS
where the first right-hand term is new and where 0.240 is 24.0 mass % H in natural gas/100%.
29.3.1 Injected Natural Gas Quantity Equation
29.3.3 Amended Carbon Balance Equation
As with CH4(g), a straightforward natural gas quantity equation is;
With 73.4 mass% C in our natural gas, carbon balance equation (11.4) of Chapter 11, Bottom Segment with CH4 Injection, becomes;
mass tuyere injected natural gas
5 60 kg=1000 kg of Fe in product molten iron
or, in matrix form 60 5
mass tuyere injected 1 natural gas
(29.1)
mass tuyere injected 052 0:734 natural gas mass C in mass CO out 2 11 0:429 descending coke in ascending gas mass CO2 out mass C out 0:273 1 1 1 in ascending gas in molten iron (29.3)
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29.4 RESULTS
where the first right-hand term is new and 0.734 5 73.4 mass % C in natural gas/100%.
Chapter 11, Bottom Segment with CH4(g) Injection. The term;
29.3.4 Amended Oxygen Balance Equation
becomes;
With 1.0 mass% O in our natural gas, oxygen balance equation (11.5) of Chapter 11, Bottom Segment with CH4(g) Injection, becomes;
and the steady-state enthalpy balance equation becomes;
mass tuyere injected 05 2 0:01 natural gas mass Fe0:947 O into mass O2 1 0:232 2 2 bottom segment in blast air mass CO out 1 0:571 in ascending gas mass CO2 out 1 0:727 in ascending gas mass H2 O out 1 0:888 in ascending gas (29.4)
where the first term is new and 0.01 5 1.0 mass % O in natural gas/100%.
With 1.7 mass% N in our natural gas, nitrogen balance equation (7.5) of Chapter 7, Conceptual Division of the Blast Furnace - Bottom Segment Calculations, becomes; 1
mass tuyere injected natural gas mass N2 out in ascending gas
0:017 2
mass N2 in blast air
mass tuyere-injected natural gas ð4:52Þ
2320 5 ½mass tuyere-injected natural gas ð 24:52Þ ½mass Fe0:947 O into bottom segment ð 23:152Þ ½mass C in descending coke 1:359 ½mass O2 in blast 1:239 ½mass N2 in blast 1:339 1½mass Fe out in molten iron 1:269 1½mass C out in molten iron 5 1½mass CO gas out in ascending gas ð 22:926Þ 1½mass CO2 gas out in ascending gas ð 27:926Þ 1½mass N2 out in ascending gas 1:008 1½mass H2 gas out in ascending gas 13:35 1½mass H2 O gas out in ascending gas ð 211:49Þ (29.6)
The six amended equations are now introduced into matrix Table 29.1 and the;
29.3.5 Amended Nitrogen Balance Equation
052
mass tuyere-injected CH4 g ð4:667Þ
• C-in-coke, and • O2-in-blast requirements for steady production of 1500 C molten iron are calculated.
1
29.4 RESULTS
(29.5)
Bottom segment calculated values of Table 29.1 show that steady-state production of 1500 C molten iron with 60 kg of natural gas injection requires;
1
where the first right-hand term is new and 0.017 5 1.7 mass % N in natural gas/100%.
29.3.6 Amended Enthalpy Balance Equation Natural gas injection changes only one term in the enthalpy balance equation (11.7) of
• 337 kg of C-in-coke, and • 322 kg of O2-in-blast air as compared to: • 335 kg of C-in-coke, and • 323 kg of O2-in-blast air with 60 kg of CH4(g), Chapter 11.
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29. BOTTOM SEGMENT CALCULATIONS WITH NATURAL GAS INJECTION
This representation does require calculation of the gas’s elemental composition and enthalpy as described in Appendices Q and R. These calculations can be automated in Excel if, for example, the steel company’s gas supply varies in composition or if the company has some choice between natural gases of different composition. Representation of industrial pulverized coal injectant is little more difficult. This is because it contains alumina-silicate ash. This task is tackled in Chapter 37, Bottom Segment Calculations With Pulverized Coal Injection. FIGURE 29.2 C-in-coke requirement for steady production of 1500 C molten iron as affected by injection of 25 C natural gas. As expected, the coke requirement decreases with increasing natural gas injection. The line is straight.
29.5 C-IN-COKE REPLACEMENT BY NATURAL GAS OF APPENDIX Q Fig. 29.2 shows the effect of natural gas injection on steady-state C-in-coke requirement. Each kg of injected natural gas saves 0.92 kg of C-in-coke. This is slightly less than the 0.95 kg of C-incoke saved by injected CH4(g) of Chapter 11, Bottom Segment with CH4(g) Injection (Fig. 11.2). This difference is due to all the equations in matrix Tables 11.1 and 29.1 but we may postulate that is mainly due to; 1. the natural gas’s smaller concentrations of C and H, Table 29.1, kg per kg of injectant, and 2. the natural gas’s oxygen which is mainly in the form of CO2(g) (Appendix Q), which is not a reductant/fuel.
29.6 SUMMARY Industrial natural gas injection is readily represented in our matrix calculations. We do this for the remainder of the book.
EXERCISES These exercises all refer to natural gas of Table 29.1, which contains 0.734 kg C, 0.24 kg H, 0.017 kg N, and 0.01 kg O, and an enthalpy content of 24.52 MJ (all per kg of gas). All the masses in this set of exercises are in kg per 1000 kg in product molten iron. 29.1. Table 29.2 blast furnace team wishes to increase their natural gas injection quantity to 140 kg/1000 kg of Fe in product molten iron. They would like to know how much; 1. C-in-coke, 2. O2-in-blast, 3. N2-in-blast, and 4. air will be required for steady production of 1500 C molten iron while injecting this 140 kg of natural gas. Please calculate these for the team. 29.2. Tuyere-injected natural gas can sometimes be cheaper than C-in-topcharged coke and C-in-pulverized coal. For this reason, Exercise 29.1 team wishes to maximize natural gas injection quantity.
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TABLE 29.2 Bottom-Segment Matrix With 60 kg of Injected 25 C Natural Gas
All masses are per 1000 kg of Fe in product molten iron.
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29. BOTTOM SEGMENT CALCULATIONS WITH NATURAL GAS INJECTION
The team knows, however, that proper gas flow in the blast furnace requires at least 250 kg of C-in-top-charged coke (per 1000 kg of Fe in product molten iron).
Please calculate the maximum amount of natural gas that they can inject into the furnace without lowering the furnace’s steady-state Cin-coke input below this 250 kg minimum. Use two calculation methods.
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