Advantages of melting concentrate in air blast
Melting in a suspended state has a number of advantages:
1) The smelting process is autogenous;
2) obtaining a constant stream of concentrated exhaust gases (16 - 80%), suitable for obtaining liquid sulfur dioxide, sulfuric acid or elemental sulfur;
3) the possibility of full automation of the process;
4) a high degree of desulfurization (70 - 80%), which makes it possible, respectively, to obtain a matte rich in copper content and even blister copper;
5) high degree of sulfur recovery (about 90%) for the plant;
6) increase in labor productivity.
2.Metallrugical calculations
2.1. Flash smelting of sulfide copper concentrates on matte with heated air blast
2.1.1 Calculation of the rational composition of the concentrate
The composition of the copper concentrate, %: Сu -22.0; Fe - 26.0; S-34.0, CaO-2.0; MgO-2.0; SiO2-11.0; Al2O3-2.0; others - 1.0.
We accept that in the concentrate Fe, Cu and S are represented by minerals CuFeS2, CuS, FeS2
We calculate for 100 kg of copper concentrate.
Let us denote the number of kg of minerals contained in 100 kg of copper concentrate through the letters: x-CuFeS2, y-CuS and z-FeS2.
Then the mass of copper contained in CuFeS2 and CuS will be; kg:
63.54x + 63.54y = 20
The mass of iron contained in CuFeS2 and FeS2 will be, kg:
55.85x + 55.85z = 26.1
The mass of sulfur contained in CuFeS2, CuS and FeS2 will be, kg:
64.12x + 32.06y + 64.12z = 33.4
Thus, we have a system of three equations:
63.54x + 63.54y = 20 (3.1)
55.85x + 55.85z = 26.1 (3.2)
64.12x + 32.06y + 64.12z = 33.4 (3.3)
We solve equations (2) and (3) with respect to z
55.85x + 55.85z = 26.1 × 64.12
64.12x + 32.06y + 64.12z = 33.4 x 55.85
64.12∙55.85x + 64.12∙55.85z = 64.12∙26.1
55.85∙64.12x + 55.85∙32.06y + 55.85∙64.12z = 55.85∙33.4
Subtracting equation (2) from equation (3), we obtain
1790.55 y = 1865.39 - 1673.53 = 191.86
Where do we get
y=0.107
Substituting the value of y into equation (1) we get
x = 0.236
Substituting the value of x into equation (2), we get
z = 0.231
Thus, 100 kg of copper concentrate contains minerals:
CuFeS2 - 0.236 kMol, CuS - 0.107 kMol and FeS2 - 0.231 kMol.
Then each mineral contains components:
CuFeS2: Cu - 14.99 kg, Fe - 13.18 kg, S - 15.13 kg.
СuS: Cu - 6.80 kg, S - 3.43 kg
FeS2: Fe - 12.90 kg, S - 14.81 kg
The data obtained are summarized in table 1.
Table 1 - Rational composition of sulfide copper concentrate
Mineral
Component
|
CuFeS2
|
CuS
|
FeS2
|
CaO
|
MgO
|
SiO2
|
Al2O3
|
Others
|
Total
|
Cu
|
14,99
|
6,80
|
|
|
|
|
|
|
21,79
|
Fe
|
13,18
|
|
12,90
|
|
|
|
|
|
26,08
|
S
|
15,13
|
3,43
|
14,81
|
|
|
|
|
|
33,37
|
CaO
|
|
|
|
2,1
|
|
|
|
|
2,1
|
MgO
|
|
|
|
|
1,5
|
|
|
|
1,5
|
SiO2
|
|
|
|
|
|
13,8
|
|
|
13,8
|
Al2O3
|
|
|
|
|
|
|
1,0
|
|
1,0
|
Others
|
|
|
|
|
|
|
|
0,3
|
0,3
|
Total
|
43,3
|
10,23
|
27,71
|
2,1
|
1,5
|
13,8
|
1,0
|
0,3
|
100
|
2.1.2 Dust calculation
We accept the removal of dust from the furnace 10%. We calculate for 100 kg of concentrate
The amount of CuFeS2 carried away by the exhaust gases will be:
39,81∙0,1 = 3,98 kg.
It contains:
Cu – 13,8∙0,1 = 1,38 kg,
Fe – 12,1∙0,1 = 1,21 kg кг,
S – 13,91∙0,1 = 1,39 kg
The amount of CuS carried away by the exhaust gases will be
12,33∙0,1 =1,23 kg.
It contains
Cu – 8,2∙0,1 = 0,82 kg,
S – 4,13∙0,1 = 0,41 kg.
The amount of FeS2 carried away by the exhaust gases will be
29,86∙0.1 = 2,99 kg.
It contains
Fe – 13,9∙0,1 = 1,39 kg,
S – 15,96∙0,1 = 1,60 kg.
Other components of the concentrate will pass into dust in an amount
СаО - 2∙0,1 = 0,2 kg, MgО - 2∙0,1 = 0,2 kg, SiO2 – 11,0∙0,1 = 1,1 kg, Al2O3 - 2∙0,1 = 0,2 kg.
others – 1,0∙0,1 = 0,1 kg.
Table 2 - Quantity and composition of dust from concentrate
Минерал
Компонент
|
CuFeS2
|
CuS
|
FeS2
|
CaO
|
MgO
|
SiO2
|
Al2O3
|
Others
|
Total, kg
|
Cu
|
1,38
|
0,82
|
|
|
|
|
|
|
2,2
|
Fe
|
1,21
|
|
1,39
|
|
|
|
|
|
2,6
|
S
|
1,39
|
0,41
|
1,6
|
|
|
|
|
|
3,4
|
CaO
|
|
|
|
0,2
|
|
|
|
|
0,2
|
MgO
|
|
|
|
|
0,2
|
|
|
|
0,2
|
SiO2
|
|
|
|
|
|
1,1
|
|
|
1,1
|
Al2O3
|
|
|
|
|
|
|
0,2
|
|
0,2
|
Others
|
|
|
|
|
|
|
|
0,1
|
0,1
|
Total, kg
|
3,98
|
1,23
|
2,99
|
0,2
|
0,2
|
1,1
|
0,2
|
0,1
|
10
|
The rational composition of the concentrate minus dust is presented in table 3.
Table 3 - Rational composition of sulfide copper concentrate, taking into account dust
Mineral
Component
|
CuFeS2
|
CuS
|
FeS2
|
CaO
|
MgO
|
SiO2
|
Al2O3
|
Others
|
Total, kg
|
Cu
|
12,42
|
7,38
|
|
|
|
|
|
|
19,8
|
Fe
|
10,89
|
|
12,51
|
|
|
|
|
|
23,4
|
S
|
12,52
|
3,72
|
14,36
|
|
|
|
|
|
30,6
|
CaO
|
|
|
|
1,8
|
|
|
|
|
1,8
|
MgO
|
|
|
|
|
1,8
|
|
|
|
1,8
|
SiO2
|
|
|
|
|
|
9,9
|
|
|
9,9
|
Al2O3
|
|
|
|
|
|
|
1,8
|
|
1,8
|
Others
|
|
|
|
|
|
|
|
0,9
|
0,9
|
Total, kg
|
35,83
|
11,1
|
26,87
|
1,8
|
1,8
|
9,9
|
1,8
|
0,9
|
90
|
2.1.3 Calculation of the rational composition of the matte
During the smelting process, higher sulfides decompose according to the equations
2CuFeS2 = Cu2S + 2FeS + 0,5S2
This creates
Сu2S - = 15,54 kg.
FeS - = 17,16 kg
S - = 3,13 kg
2CuS = Cu2S + 0,5S2
This creates
Cu2S - = 9,24 kg
S - = 1,86 kg.
FeS2 = FeS + 0,5S2
This creates
FeS - = 19,68 kg
S - = 7,18 kg.
Total formed as a result of the dissociation of higher sulfides
Cu2S – 24,78 kg, FeS -36,84 kg, S – 12,17 kg
We accept that the copper content in the matte is 60.0%, and the extraction of copper in the matte is 94.5%.
Then copper will pass into the matte
19,8 ∙0,954 = 18,89 kg
The mass of the matte will be
= 31,48 kg
The copper content in the matte according to the Mostovich rule is 25%. Then the mass of sulfur in the matte will be
31,48∙0,25 = 7,87 kg
The copper in the matte is in the form of Cu2S. The amount of sulfur in the matte associated with copper will be
= 4,77 kg.
The amount of Cu2S in the matte will be
18,89 + 4,77 = 23,66 kg
Oxygen is always dissolved in copper matte, which is bound with iron in the form of Fe3O4. the amount of oxygen in the matte containing 60% copper is 1.24%. The mass of oxygen in the matte is
0,0124∙31,48 = 0,39 kg.
The amount of iron associated with oxygen in the matte will be
= 1,02 kg
The mass of magnetite in the matte will be
1,02 + 0,39 = 1,41 kg
We accept the content of others in the matte 1%, which is 0.31 kg.
The rest of the matte mass will be represented by FeS, the amount of which will be
31,48 – 23,66 – 1,41 – 0,31= 6,1 kg
The amount of iron associated with FeS will be
= 3,88 kg.
The amount of sulfur in the matte associated with FeS will be
= 2,22 kg
The rational composition of the matte is given in table 4.
Table 4 - Rational composition of the matte
Component
|
Cu2S
|
FeS
|
Fe3O4
|
Others
|
Total
|
kg
|
%
|
Cu
|
18,89
|
|
|
|
18,89
|
60,0
|
Fe
|
|
3,88
|
1,02
|
|
4,9
|
15,57
|
S
|
4,77
|
2,22
|
|
|
6,99
|
22,20
|
O2
|
|
|
0,39
|
|
0,39
|
1,24
|
Others
|
|
|
|
0,31
|
0,31
|
1,0
|
Total, kg.
|
23,66
|
6,1
|
1,41
|
0,31
|
31,48
|
100,01
|
2.1.4 Calculation of the composition and yield of slag without the addition of fluxes
The amount of copper that passes into the slag will be
19,8 -18,89 = 0,91 kg
Copper is present in the slag in the form of Cu2S. The amount of sulfur in the slag associated with Cu2S will be
= 0,23
The amount of Cu2S in the slag will be
0,91 + 0,23 = 1,14 kg
The amount of sulfur passing into the gas phase will be
Sgas.= Scon.- Sdust -Ssl..-Spcs.= 34.0 - 3.4 - 0.23 -6.99 = 23.38 kg.
The amount of FeS oxidized to oxides
36,84 -6,1 = 30,74 kg
It contains iron
= 19,53 kg
and sulfur
= 11,21 kg
Sulfur passes into the gas phase
12,17 + 11,21 =23,38
The oxidation of FeS in the flame proceeds according to the equation
FeS + 1,5O2 = FeО + 3SO2
As a result, FeO is formed.
= 25,12 kg
Part of FeO is oxidized to magnetite according to the equation
6FeО + О2 = 2Fe3О4
We accept that 30% FeO is oxidized to magnetite, which will be
25,12∙0,3 = 7,54 kg FeО
The amount of magnetite formed will be
= 8,1 kg
Of this amount of magnetite, 1.41 kg goes into matte. The rest of the magnetite
8,1 -1,41 = 6,69 kg
turns into sludge.
It contains oxygen
= 1,85 kg.
and iron
= 4,84 kg
FeO will also pass into the slag in the amount
25,12 – 7,54 = 17,58 kg
It contains oxygen
= 3,91 kg
and iron
= 13,67 kg
The composition and amount of smelting slag without flux addition are given in table 5.
Table 5 - Composition and amount of slag without the addition of fluxes
Component
|
Cu
|
Fe
|
S
|
O
|
SiO2
|
CaO
|
MgO
|
Al2O3
|
Others
|
T
|
kg
|
%
|
Cu2S
|
0,91
|
|
0,23
|
|
|
|
|
|
|
1,14
|
|
FeO
|
|
13,67
|
|
3,91
|
|
|
|
|
|
17,58
|
|
Fe3O4
|
|
4,84
|
|
1,85
|
|
|
|
|
|
6,69
|
|
SiO2
|
|
|
|
|
9,9
|
|
|
|
|
9,9
|
|
CaO
|
|
|
|
|
|
1,8
|
|
|
|
1,8
|
|
MgO
|
|
|
|
|
|
|
1,8
|
|
|
1,8
|
|
Al2O3
|
|
|
|
|
|
|
|
1,8
|
|
1,8
|
|
Others
|
|
|
|
|
|
|
|
|
0,59
|
0,59
|
|
Total, kg
|
0,91
|
18,51
|
0,23
|
5,76
|
9,9
|
1,8
|
1,8
|
1,8
|
0,59
|
41,3
|
|
%
|
|
|
|
|
|
|
|
|
|
|
|
2.1.5 Calculation of the amount and composition of slag with the addition of fluxes
The recommended content of SiO2 in suspended melting slag is about 30%.
Quartz ore is used as a flux with composition, %: SiO2- 78.4; FeO- 4.9; MgO 6.0, Al2O3 - 6.0 others - 4.7
3,8∙16/55,85 =1,09
Denote by X the mass of added quartz ore.
We draw up a balance for silica, assuming that all silicon ore goes into slag
0,784X + 9,9 = 0,3(41,3 + X)
0,784X -0,3X = 12,39 – 9,9
0,484X = 2,49
X = 5,14 kg
The mass of slag will be
41,3 + 5,14 = 46,44 kg
With flint ore, SiO2 enters the slag
5,14∙0,784 = 4,03 kg
With flint ore, FeO enters the slag
5,14∙0,049 = 0,25 kg
It contains iron
= 0,19 kg
and oxygen
= 0,06 kg.
With silicon ore, the following also enters the slag:
MgO - 5.14∙0.06 = 0.31 kg, Al2O3 - 5.14∙0.06 = 0.31 kg, others - 5.14∙0.047 = 0.24 kg
Thus, the slag will contain oxides:
SiO2 - 9,9 + 4,03 = 13,93 kg, Fe3O4 – 6,69 kg, FeO – 17,58 + 0,25 = 17,83 kg,
(Fe -13,86 kg, О – 3,97 kg)
CaO – 1,8 kg, MgO – 1,8 + 0,31 = 2,11 kg, Al2O3 – 1,8 + 0,31 = 2,11 kg,
others – 0,59 + 0,24 = 0,83 kg.
The composition and amount of slag, taking into account the addition of flux, is presented in Table 6.
Table 6 - Composition and amount of slag, taking into account the addition of fluxes
Component
|
Cu
|
Fe
|
S
|
O
|
SiO2
|
CaO
|
MgO
|
Al2O3
|
Others
|
Total
|
kg
|
%
|
Cu2S
|
0,91
|
|
0,23
|
|
|
|
|
|
|
1,14
|
|
FeO
|
|
13.86
|
|
3,97
|
|
|
|
|
|
17,83
|
|
Fe3O4
|
|
4,84
|
|
1,85
|
|
|
|
|
|
6,69
|
|
SiO2
|
|
|
|
|
13,93
|
|
|
|
|
13,93
|
|
CaO
|
|
|
|
|
|
1,8
|
|
|
|
1,8
|
|
MgO
|
|
|
|
|
|
|
2,11
|
|
|
2,11
|
|
Al2O3
|
|
|
|
|
|
|
|
2,11
|
|
2,11
|
|
Others
|
|
|
|
|
|
|
|
|
0,83
|
0,83
|
|
Total, kg
|
0,91
|
18,7
|
0,23
|
5,82
|
13,93
|
1,8
|
2,11
|
2,11
|
0,83
|
46,44
|
|
%
|
|
|
|
|
|
|
|
|
|
|
|
2.1.6 Calculation of the amount and composition of dust
We believe that 10% of quartz ore passes into dust. Then it is necessary to introduce quartz flux into the charge
= 5,71 kg
Turns into dust
5,71 – 5,14 = 0,57
quartz ore. It contains
SiO2 -0,57∙0,784 = 0,45 кг, FeO - 0,57∙0,049 = 0,03 кг (Fe – 0,02 kg, О – 0,01 kg).
MgO – 0,57∙0,06 =0,03 kg, Al2O3 – 0,57∙0,06 = 0,03 kg, others –0,57∙0,047= 0,03 kg.
Then the dust contains
SiO2 – 1,1 + 0,45 = 1,55 kg; FeO 0,03 kg (Fe – 0,02 kg, О – 0,01 kg);
MgO – 0,2 + 0,03 = 0,23 kg; Al2O3 – 0,2 + 0,043 = 0,23 kg; others – 0,1 + 0,03 = 0,13 kg
The final dust composition is given in Table 7.
Table 7 - Quantity and composition of dust
Mineral
Component
|
CuFeS2
|
CuS
|
FeS2
|
CaO
|
MgO
|
SiO2
|
Al2O3
|
FeO
|
Others
|
Total, kg
|
Cu
|
1,38
|
0,82
|
|
|
|
|
|
|
|
2,2
|
Fe
|
1,21
|
|
1,39
|
|
|
|
|
0,02
|
|
2,62
|
S
|
1,39
|
0,41
|
1,6
|
|
|
|
|
|
|
3,4
|
О
|
|
|
|
|
|
|
|
0,01
|
|
0,01
|
CaO
|
|
|
|
0,2
|
|
|
|
|
|
0,2
|
MgO
|
|
|
|
|
0,23
|
|
|
|
|
0,23
|
SiO2
|
|
|
|
|
|
1,55
|
|
|
|
1,55
|
Al2O3
|
|
|
|
|
|
|
0,23
|
|
|
0,23
|
Others
|
|
|
|
|
|
|
|
|
0,13
|
0,13
|
Total, kg
|
3,98
|
1,23
|
2,99
|
0,2
|
0,23
|
1,55
|
0,23
|
0,03
|
0,13
|
10,57
|
2.1.7 Calculation of the amount and composition of the gas phase
We assume that sulfur in the gas phase is in the form of SO2. The amount of sulfur transferred to the gas phase will be
34,0 - 3,4 – 0,23 - 6,99 = 23, 38 kg
Then the degree of desulfurization of the melt will be
= 76,4 %.
The amount of oxygen spent on the formation of SO2 will be
= 23,34 kg
The amount of SO2 in the gas phase will be
23,38 + 23,34 = 46,72 kg.
The amount of oxygen spent on the formation of FeO due to the oxidation of FeS will be
= 5,59 kg
The amount of oxygen spent on the oxidation of FeO to Fe3O4 will be
= 0,28 kg.
Total oxygen consumption
23,34 + 5,59 + 0,28 = 29,21 kg
The amount of air nitrogen supplied to the furnace will be
= 97,79 kg.
The amount of air spent for melting will be
29,21 + 97,79 = 127 kg.
For melting in a suspended state, the charge is dried to a moisture content of about 0.1%. Then, with the charge, moisture enters the furnace
= 0,11 kg.
Considering the exhaust gases as an ideal gas, we calculate the volume of gas using the Mendeleev-Clapeyron equation
V = ,
where g – mass of gas, kg;
М – molecular weight of gas, kg
22,4 – volume occupied by 1 kmol of gas, м3;
V – gas volume, м3.
The volume of exhaust gases will be:
SO2 - = 16,3 м3; N2 - = 78,23 м3; Н2О - = 0,14 м3.
The amount and composition of exhaust gases are presented in table 8.
Table 8 - Quantity and composition of exhaust gases
Gases
|
kg.
|
м3
|
%( voluminous)
|
SO2
|
46,72
|
16,3
|
17,22
|
N2
|
97,79
|
78,23
|
82,63
|
H2O
|
0,11
|
0,14
|
0,15
|
Total
|
144,62
|
94,67
|
100
|
2.1.8 Material balance of flash melt
On the basis of the performed calculations, we draw up a material balance for the smelting of copper concentrates in suspension. The material balance is presented in table 9.
Table 9 - Material balance of heat
|
Total kg
|
Cu
|
Fe
|
S
|
O
|
SiO2
|
CaO
|
MgO
|
Al2O3
|
H2O
|
N2
|
Other
|
Loaded
|
Kt
|
100,11
|
22,0
|
26,0
|
34,0
|
|
11,0
|
2,0
|
2,0
|
2,0
|
0,11
|
|
1,0
|
Ore
|
5,71
|
|
0,21
|
|
0,07
|
4,48
|
|
0,34
|
0,34
|
|
|
0,27
|
Air
|
127
|
|
|
|
29,21
|
|
|
|
|
|
97,79
|
|
Total
|
232,82
|
22,0
|
26,21
|
|
29,28
|
15,48
|
2,0
|
2,34
|
2,34
|
0,11
|
97,79
|
1,27
|
Received
|
Matte
|
31,48
|
18,89
|
4,9
|
6,99
|
0,39
|
|
|
|
|
|
|
0,31
|
Slag
|
46,44
|
0,91
|
18.7
|
0,23
|
5,82
|
13,93
|
1,8
|
2,11
|
2,11
|
|
|
0,83
|
Dust
|
10,57
|
2,2
|
2,62
|
3,4
|
0,01
|
1,55
|
0,2
|
0,23
|
0,23
|
|
|
0,13
|
Gases
|
144,62
|
|
|
23,38
|
23,34
|
|
|
|
|
0,11
|
97,79
|
|
Total
|
233,11
|
22,0
|
26,22
|
34
|
29,56
|
15,48
|
2,0
|
2,35
|
2,35
|
0,11
|
97,79
|
1,27
|
The discrepancy is
= 0,12%.
2.1.9 Calculation of heat balance of melting
2.1.9.1.1 Initial data for calculation
We accept the values of heat capacities for
Сsht. = 0,941 kJ/(kg∙deg.)
Сshl.. = 1,42 kJ/(kg∙deg.)
Сshiht. = 0,84 kJ/(kg∙deg.)
Сdust. = 0,84 kJ/(kg∙deg.)
СSO2 = 2,298 kJ/(nм3∙ deg.)
СN2 = 1,423 kJ/(nм3∙ deg.)
СH2O(gas) = 1,803 kJ/(nм3∙ deg.)
СH2O(zh) = 4,186 kJ/(kg∙deg.)
Сair = 1,443 kJ/(nм3∙ deg.)
The heat of evaporation of water is 2300 kJ/kg.
The temperature of the materials entering the furnace is 25 °C.
The temperature of the produced matte is 1180 °C.
The temperature of the produced slag is 1250 °C.
Exhaust gas temperature 1300 °C.
Blast heating - 500°C.
2.1.9.2 Articles of the arrival of heat
2.1.9.2.1 Heat of exothermic reactions
FeS + 1,5O2 = FeO + SO2 ∆H = - 461,33 kJ
Due to the oxidation of 30.74 kg of FeS, heat will be released
= 161316 kJ
3FeO + O2 = Fe3O4 ∆H = - 322,48 kJ
Due to the oxidation of 7.54 kg of FeO, heat will be released
= 11280 kJ
S2 + 2O2 = 2SO2 ∆H = - 593,8 kJ
Due to the oxidation of 12.17 kg of S2, heat will be released
= 117797 kJ
2FeO + SiO2 = 2FeO∙SiO2 ∆H = - 34,3 kJ
Due to the slagging reaction, 17.83 kg of FeO will release heat
= 4256 kJ.
СаО + SiO2 = СаО∙ SiO2 ∆H = - 37,54 kJ
Due to the CaO slagging reaction, heat will be released
= 1339 kJ.
Exothermic reactions generate heat
161316 + 11280 + 117797 + 4256 + 1339 = 295 988 kJ.
2.1.9.2.2 Heat supplied to the furnace with loaded materials
The heat supplied to the furnace with the loaded materials is calculated by the equation
Q = C i∙gi∙ti (26)
where Ci - heat capacity of the 9th component, kJ/(kg∙deg.);
gi - component weight, kg;
ti - temperature, оС
The heat supplied to the furnace with the charge will be
105,82∙25∙0,84 = 2222 kJ
The heat supplied to the furnace with air will be
127 ∙ 500∙1,343 = 85281 kJ.
The heat supplied to the furnace with loaded materials will be
2222 + 85291 = 87503 kJ
2.1.9.3 Heat consumption
2.1.9.3.1 The heat spent on the decomposition of higher sulfides due to
2CuFeS2=Cu2S+2FeS+0,5S2 ∆H = 100,4 kJ
Due to the dissociation reaction, 35.83 kg CuFeS2 absorbs heat
= 9801 kJ
2CuS = Cu2S + 0,5S2 ∆H = 53,56 kJ
Due to the dissociation of 11.1 kg of CuS, heat is absorbed
= 3,109 kJ
FeS2 = FeS + 0,5S2 ∆H = 71,18 kJ
Due to the dissociation of 26.8 kg FeS2
= 16115 kJ.
The heat expended on heating and evaporation of moisture will be
4,186∙75∙0,1 + 0,11∙2300 = 288 kJ
Heat is absorbed through endothermic reactions.
9801 + 3109 + 16115 + 288 = 29313 kJ
2.1.9.3.2 Heat lost by melt products discharged from the furnace
Heat lost with matte matte
31,48∙0,941∙1180 = 34955 kJ
The slag heat lost with the slag will be
46,44∙1,42∙1250 = 82431 kJ.
The heat lost with dust is
10,57∙0,84∙1300 = 11542 kJ
The heat lost with the exhaust gases will be
(16,3∙2,298 + 78,23∙1,423 + 0,14∙1,803)∙1300 = 193739 kJ.
Heat loss with melting products will be
34955 + 82431 + 11542 + 193739 = 322667 kJ
The total heat loss will be
29313 + 322667 = 351980 kJ.
We accept that heat loss to the environment will be 5% of the total loss
= 18525 kJ
The heat balance of melting is presented in table 10
Table 10 - Heat balance of the melting process
|
The arrival of heat
|
|
Heat consumption
|
parish article
|
kJ
|
%
|
Expenditure
|
kJ
|
%
|
1
|
Heat supplied with feed materials
|
87503
|
|
1
|
Heat carried away by the matte
|
34955
|
|
2
|
Heat of exothermic reactions
|
295988
|
|
2
|
Heat carried away by the slag
|
82431
|
|
3
|
|
|
|
3
|
Heat carried away by flue gases
|
193739
|
|
4
|
|
|
|
4
|
Heat of endothermic reactions
|
29313
|
|
5
|
|
|
|
5
|
Heat carried away by dust
|
11542
|
|
6
|
|
|
|
|
Losses to the environment
|
18525
|
|
|
|
|
|
|
Unaccounted losses
|
3985
|
|
|
Total
|
383490
|
|
|
Total
|
379505
|
Total
|
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