Copper Smelting& Converting
Sarcheshmeh copper complex
Copper converting
"A calculation on copper converters"
Masoud Sheykhi
Process inspection
Smelter Process
Converter inspection
Technical_inspection@nicico.com
Introduction
Copper ores usually smelted to matte , which for purposes of
calculation is usually taken as a simple molten solution of
CU2S and FeS . The percentage of copper in the matte
commonly termed the "grade" of the matte , thus determines
the percentage composition in CU ,Fe , and S of this ideal
matte ; the percentage of CU multiplied by 160/128 give the
percentage of CU2S , and the balance is FeS.High- grade
mattes commonly have part of their copper content present as
free CU . The action in a copper smelting furnace and also
in a converter is governed by the fact that sulphur has a
greater affinity for copper than for iron. Though the heat
of formation of FeS is greater than that of CU2S , the
oxidation of FeS has greater free energy change than the
oxidation of CU2S ,and the reaction 2CUO + FeS = CU2S+FeO
will take place from left to right as shown . The operation
of copper converter is divided into two stages . the first ,
or slag- forming ,stage consists in the oxidation of FeS to
FeO with SiO2 added as flux:
2FeS +3O2 = 2FeO +2SO2
x FeO+ySiO2 = xFeO.ySiO2
At the end of the first stage the slag is poured off;the
material remaining consists mostly(theoretically entirely)of
CU2S ,Called"white metal". The second stage is the oxidation
of CU2S to Blister copper:
CU2S + O2 = 2CU +SO2.
Theoritically no slag is made during the seconde stage , and
this may be assumed to be the case in calculations.
Calculations
In the room temperature we have:
CU2S + O2 = 2CU +SO2
In 1100[0c] we have;
Heat contents of reactants ;
in 160kg CU2S = 160*0.131*1100
= 23050 Kcal
Laten heat of smelting = 160* 34.5
= 5520 Kcal
in 22.4 M3 O2 = 22.4*(0.302*0.000022*1100)*1100 = 8030 Kcal
Total heat of reactants = 36600Kcal
Heat content of products;
in 128 Kg CU = 128*(0.0916+0.0000125*1083)*1083 = 14600 Kcal
in 128 Kg CU = 128*41.8 = 5360 Kcal
in (1083[0c]-1100[0c]) = 180 Kcal
in 22.4 M3 SO2 = 22.4*(0.406+0.00009*1100)*1100 = 12440 Kcal
Total heat in products
= 32580Kcal
Reaction heat in 1100[0c] = -51990-36600+32580 = -56010Kcal
Heat in nitrogen = (79/21)*22.4*(0.302+0.000022*1100)*1100
= 30250Kcal
net heat released in converter = 56010-8030-30250
= 17730 Kcal
References
[1] A.B.,"Metallurgical problems"McGRAW-HILL,INC,1943.
'Sarcheshmeh copper converters
"Calculations on Scrap addition during copper blowing "
Masoud Sheykhi'
Process inspection
Smelter Process
'Converter inspection'
Technical_inspection@nicico.com
Calculations
q[Watts/cm2] |(copper blowing)[1] = 5.76*0.0.357*[(T2/1000)4-
(T1/1000)4]=
2.06*[(T2/1000)4-(T1/1000)4]
For copper blowing we have:
Qc[Mcal/cm2.sec] = 0.012*0.5*(t2-t1)^1.233
if T1 = 1373[0K] , T2 = 298[0K] ,S = 90000 cm2 , then ;
q =9.42Mcal/min
if T1 = 1443[0K] , T2 = 298[0K] ,S = 90000 cm2 , then ;
q =11.5 Mcal/min
if t1 = 25[0c] , t2 = 200[0c] ,S = 1380000cm2 ,then;
Qc = 3.498 Mcal/min
q+Qc = 9.42+3.498 = 12.92 Mcal/min
Daily[2]; 1110 Tons of matte with 41.4%(take 50% for
calculations) CU charged in 3 converters , copper blowing
times = 113 minutes ,each converter cycle times = 9 hr .
Hence; each converter has 2.66 cycles per day and copper
blowing times for 3 converter is 15.029hr/day . thus;
released heat during copper blowing[3] = (17.73Mcal/160Kg
CU2S)*[(1110000/2)Kg CU2S/day]*[ day/15.029hr]*[hr/60min] =
68.2 Mcal/min.
heat accumulated in each converter =( 68.2/3)-12.92=
9.813Mcal/min.
heat accumulated in 3 converters = 3*9.81 = 29.44 Mcal/min.
heat carried by blister copper for changing its temperature
from 25[0c] to 1100[0c] = (14.6+5.360+0.18)/128 = 0.157
Mcal/Kg.
Daily production[2] ; are 420Tons blister copper.
(0.151Mcal/Kg)*[420000/(3*15.029*60)] =
24.428Mcal/min.converter
24.428/accumulated heat per converter = 24.428/9.813 = 2.5
Hence ; Max of Scrap% = 9.813*100/(2.5*24.42) = 16
Now ,we check the above calculations ;
We have[3];
32.580-12.44-[3.498*17.73/68.2] = 19.23Mcal/160Kg CU2S =
24.66Mcal/min.converter that near to 24.428 .hence
calculations are true.
Min of scrap% =[9.813/(24.66*2.50)]*100 =15.92
Now,We take the average of 24.66 and 24.428 i.e;24.54
24.54/9.813 = 2.50
Hence ;
Average of Scrap% =[9.813/(24.54*2.50)]*100 = 15.995 .
References
[1]M.Sheykhi "Emissivity calculation&convection heat
transfer coefficient on sarcheshmeh copper complex's
converters", technical inspection division , sarcheshmeh
copper complex ,kerman,iran ,2003 .
[2]Sarcheshmeh "Smelter operating manual ,section 7 ;
Converting" ,Paeson Jurdan Int .corp,1977.
[3] M.Sheykhi " A calculation on copper converters",
technical inspection division , sarcheshmeh copper
complex ,kerman,iran ,2004 .
Emissivity calculation & convection heat transfer
coefficient on Sarcheshmeh copper complex's Converters
Masoud Sheykhi
Process inspection
Smelting Process
Converter inspection
Technical_inspection@nicico.com
Calculations
Converter heat losses equation[1] ;
Q[kcal/m2h]= [4.88*e*(T/100)^4]*S ,Let; e = .80
Let assume that converter has 4 meter diameter and 9 meter
lengeth then;
Q[kcal/min]= [0.583*(T/100)^4]
Let temperature of the melt surface be 1200 0c then;
q[Mcal/min] = [4.88*0.80*{1200+273)/1000}^4]*9/(60*1000)=
27.568
Let temperature of the shell surface be 200 0c then;
q'[Mcal/min] = [4.88*0.80*{200+273)/1000}^4]*138/(60*1000)=
4.498
Q' = q'+ q = 27.568+4.498 = 32.066
(q/Q' )*100 = (27.568/32.066)*100=85.97
(q'/Q')*100 = (4.498/32.066)*100=14.03
Qc = convection heat losses from converter shell = 2.66
Mcal/min[2].Thus;
4.498-2.66 = 1.83 Mcal/min is the radiation heat losses from
converter shell.
we have [2];
Qc[Mcal/cm2.sec] = n*(t2-t1)^1.233
t1[0c] ; temperature of air sourrounding converter shell.
t2[0c] ; temperature of converter shell
n ; is a constant depending on geometry and condition of
surface .
A = 1380000cm2
Hence;
n = 0.012 Mcal.cm-2.min-1.[c0]-1.233
Assume that total radiation heat losses from the converter
mouth , thus;
q[Mcal/min] = 27.568+1.83 = 29.398 .
we have[3] ;
q[Watts/cm2] = 5.76*e'*[(T2/1000)4-(T1/1000)4]
T2=1200+273=1473 [0K]
T1=25+273=298[0K]
29.398[Mcal/min] = 22.78[Watts/cm2]
Hence , we correct;
e=e' = 0.84
Radiation heat losses during slag blowing[Mcal/min] = 5.416
Radiation heat losses during copper blowing[Mcal/min] = 4
Convection heat losses during converter blowing[Mcal/min] =
2.66
Mean radiation heat losses during converter blowing
[Mcal/min] = 4.708>4.498
Sarcheshmeh converter Slag blowing tims[hr][4] = 4.52
Sarcheshmeh converter Copper blowing times[hr][4] = 1.88
Sarcheshmeh converter Total blowing times[hr] =
6.4*[5.416/(5.416+4)]*100 = 57.52
100-57.52 =42.48
Therefor during copper blowing we have;
e = 0.4248*0.84 =0.3568
and during slag blowing ;
0.5752*0.84 = 0.4832
Hence;
q[Watts/cm2]|(slag blowing) = 5.76*0.4832*[(T2/1000)4-
(T1/1000)4]= 2.78*[(T2/1000)4-(T1/1000)4]
q[Watts/cm2] |(copper blowing) = 5.76*0.0.3568*[(T2/1000)4-
(T1/1000)4]=
2.06*[(T2/1000)4-(T1/1000)4]
[2.66/(2.66+2.66)]*100 = 50
100-50 = 50
Hence;
Qc[Mcal/cm2.sec] = 0.12*0.50*(t2-t1)^1.233 during slag
blowing
Qc[Mcal/cm2.sec] = 0.12*0.50*(t2-t1)^1.233 during copper
blowing
References
[1] S.GOTO ; The Application of thermodynamic calculations
to converter practice ,paper presented at AIME annual
meting ,1979 .
[2] Nickel Converter " Material And Heat Balance
report" ,Outokumpu Oy co ,1968.
[3] Ann .chim .phys.,7(1817) .
[4] Sarcheshmeh Smelter operating manual ,section 7; Converting ,Parson Jurdan Int .corp ,1977 .