User:Masoud Sheykhi
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Copper Smelting& Converting
Sar Cheshmeh 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
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
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;
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
= 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.
'Sar Cheshmeh 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 Sar Cheshmeh 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
length 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 times[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
Therefore, during copper blowing we have;
e = 0.4248*0.84 =0.3568
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 .
Sar Cheshmeh copper converters
" Calculations on Cold material addition during Slag
blowing "
Masoud Sheykhi
Process inspection
Smelter Process
Converter inspection
Technical_inspection@nicico.com
Calculations
q[Watts/cm2]|(slag blowing)[1] = 5.76*0.4832*[(T2/1000)4-
(T1/1000)4]=
2.78*[(T2/1000)4-(T1/1000)4]
Qc[Mcal/cm2.sec]|(slag blowing) = 0.012*0.50*(t2-t1)^1.233
if T1 = 1477[0K] , T2 = 298[0K] ,S = 90000 cm2 , then ;
q =17.185Mcal/min.converter.
if t1 = 25[0c] , t2 = 200[0c] ,S = 1380000cm2 ,then; Qc =
3.498 Mcal/min.converter.
q+Qc = 17.185 + 3.498 = 20.683 Mcal/min.converter = 60.6
Mcal/min.3 converter.
Reverts used during slag blowing have the following analyses
[2];
CU : 25
Fe :22 S :8
Sio2:27
Cao:3
Al2o3:3
Others:12
Terminology for heat characteristics of components of the
Say ;[3];
Tm[0c] ; melting point
Cm[Kcal/Kg]; specific heat
QLMP[Kcal/Kg]; Total liquid heat content at melting point
Tb[0c] ; boiling point
Lg[Kcal/Kg] ; evaporation laten heat
CU :Tm=1084 ;Lf=57.6 ,Cm=0.092 +0.0000125t ,QLMP = 172
,Cm(liquid) = 0.112
Fe:
0-600[0c] ; Tm = 0.104+0.000064t ,QLMP = 85*
0-760[0c] ; Tm = 0.180+0.0003(t-6) ,QLMP = 121*
760-910[0c]; Tm = 0.320-0.000053(t-760) , QLMP = 161*
910-1400[0c] ; Tm = 0.160+0.00001(t-910) ,QLMP = 244*
1400-1534[0c] ;Cm = 0.185 ,QLMP = 334*
S :
Tm = 119 ,Lf = 9.2 ,Tb =445 ,Lg = 79
Sio2 :
Quartz ; Tm = 1470 , Lf = 57
Cristobalite; Tm = 1700 , Lf = 35
Cao :
Tm = 1700 , Lf = 35
Al2O3 :
Tm = 2045 , Lf = 250
heat content of reverts(Cold materials) =
0.27(0.092+0.0000125*1084)+0.27*57.6+0.27*0.112(1100-1084)
+0.24(85+121+161+(0.160+0.00001(1100-910)*1100]+0.10
[30.60+193+0.153(1100-445)]
+0.29*0.19*1100+0.05*0.24*1100+.05*0.28*1100] =
299.75396 Kcal/Kg Cold materials , from which; 59.5 Kcal/
Kg Cold materials belongs to copper in the reverts(Cold
materials). Hence; 59.5/299.75396 =
0.198 =mass of copper / mass of cold materials.
2Fes + 3O2 = 2Feo + 2So2 ,Reaction heat at 298 degree kelvin
= -230000
Kcal/2Kgmol Fes.
2Feo + Sio2 = 2Feo.Sio2 , Reaction heat at 298 degree
kelvin = -10000
Kcal/Kgmol sio2.
2Fes + 3O2 +Sio2 = 2Feo.Sio2 +2So2 , Reaction heat at 298
= -240000
Kcal/Kgmol sio2.
in 176 Kg Fes = 176(0.1355+0.000078*1200)1200= 48.4Mcal/176
Kg Fes
= 10Mcal/176Kg Fes
in 60Kg Sio2 = 60*0.264*1200 = 19Mcal/176Kg Fes
in 3*22.4M3 O2 = 3*22.4(0.302+0.000022*1200)1200 =26.4
Mcal/176Kg Fes .
in 2*22.4 M3 So2 = 2*22.4(0.406+0.00009*1200)1200=27.6
Mcal/176 Kg Fes .
in 264 Kg(2Feo.Sio2) = 264*0.227*1200 =
71.9 Mcal/176Kg Fes .
= 99.5
Mcal/176Kg Fes .
Reaction heat at 1200[0c] = -240-104+99.5 = -244.4 Mcal/176
Kg Fes.
Heat in nitrogen = (79/21)*3*22.4(0.302+0.000022*1200)1200 =
99.6 Mcal/176Kg Fes .
Net heat released = 244.4-10-99.6 = 134.7 Mcal/176 Kg Fes .
Slag blowing times in each sarcheshmeh copper converter[2] =
271minutes .
Daily we have 2.66 cycles for each converter[2].
Hence; daily slag blowing times for 3 converter are 36 hr.
134.7[Mcal/176 Kg Fes]*1110000[Kg matte/day]*0.50[Kg
Fes/matte]/(36[hr]*60[min]) = 196.66 Mcal/min .
heat in nitrogen[Mcal/min] = 99.6*196.66/134.7 = 145.4 .
Accumulated heat in converters during slag blowing = 196.66-
145.4-60.6 =
9.34 Mcal/min .
on the other hand;
Accumulated heat in converters during slag blowing[Mcal/min]
=
m(cold materials)[Kg/min] * 0.29975396 Mcal/Kg
thus;
m(cold materials) = 31.13 Kg/min .
where m(cold materials) be the mass of cold materials needed
Cold materials needed for one day = 31.13*36*60/1000 =
67.248 Tons/day .
Scrap % [4] = 16
Blister production[2] = 420 Tons /day
Scrap needed = 420*0.16 = 67.2 Tons/day
67.2 * mass of cold materials/mass of copper = 67.2 *0.198 =
13.3056 Tons /day cold materials .
thus;
Total reverts (Scrap + cold materials ) needed for Slag and
13.3056 +67.248 =80.6 Tons /Day.
By [2] we need 80 Tons ;reverts /day for Slag and copper
blowing .therefor calculations agreed with designed criteria.
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" ,Parson Jurdan Int .corp,1977.
[3] Liddell ,Donald M ."The metallurgists , and chemists
Hand Book" ,3d ed.,McGraw-Hill Book ,Inc., New York ,1930 .
[4] M.Sheykhi " Sarcheshmeh copper converters- Calculations
on Scrap addition during copper blowing ", technical
inspection division , sarcheshmeh copper
complex ,kerman,iran ,2004 .
Common cause of Regrinding cyclone pumps shaft failure
Sar Cheshmeh Copper Complex
Masoud Sheykhi
Common cause of pump shaft failure is due to breakage caused by the pump ingesting hard particles or foreign objects(Balls,etc).
If the pump shaft does not break from the sudden impact, it could still fail in the near future due to stress cracks. In some instances the hard particles(Balls,etc) will distort the impeller and cause the pump shaft to have excessive run out.
This will result in the impeller rubbing and literally gouging or imbedding itself into the pumps wear ring liner. This excessive force can easily break the pump shaft.
While it is not possible to eliminate this problem, regular inspection and maintenance will extend life and allow for replacement before it becomes a problem and ruins & outing.
