Slag from Pasminco's lead-zinc smelters, which is rich in zinc, is transferred in thirteen tonne capacity ladles from the blast furnace to the slag fuming furnace. The cycle time is about three hours. The Study Group was asked to examine whether heat could be conserved by some insulation strategy, without at any stage overheating the steel of the ladle

Jenkins, D.

Stokes, N.

English

Mathematics in Industry Information Service Eprints Archive

Slag from Pasminco's lead-zinc smelters, which is rich in zinc, is
transferred in thirteen tonne capacity ladles from the blast furnace
to the slag fuming furnace. The cycle time is about three hours. The
Study Group was asked to examine whether heat could be conserved
by some insulation strategy, without at any stage overheating the
steel of the ladle.
Pasminco metals-BRAS produce about 200,000 tonnes of lead and 40,000
tonnes of zinc annually at a lead-zinc blast furnace in Port Pirie. Slag produced
from the lead smelting phase of the processing of lead-zinc ores contains a con-
siderable amount of zinc. During the last twenty five years, it has been economic
to process both current and stockpiled slag in a slag fuming furnace to recover
the zinc.
In the smelting process, the molten slag from the lead smelter is passed to the
zinc recovery process as soon as possible. This involves transporting the molten
slag in thirteen tonne capacity ladles; the cycle time for those ladles is about
three hours. The slag may be in transit for up to one and a half hours. Normally
ladles are transported three at a time on a trolley from the blast furnace to the
zinc fuming furnace.
An important part of the economics of the process is the conservation of
heat. The slag freezes to some extent to form a 'skull', and has to be re-melted;
the solid slag can interfere with the handling procedures. Drastic measures like
"knocking" are used to break the crust to allow slag to pour out. Therefore it
would be desirable to conserve heat by insulation as much as possible, and that
was one of the issues to be studied at the MISG.
An extra difficulty is that the temperatures involved are quite high. The
slag freezes at about 1l00°C, while the steel material of the ladle loses strength
at about 600°C. So an insulation strategy which raised the wall temperatures
significantly could not be used. In fact, the layer of frozen slag which forms on
the walls is essential to preserve the steel.
An illustration of the slag ladle is given in figure 1. The ladle is made of 50
mm thick steel, and weighs six tonnes. Its heat capacity is a significant part of
the total heat budget.
Slag
point 1150 C 0.05 m Frozen slag
"skull"
3. Insulation on the return journey only, to conserve the heat retained in the
steel.
The best strategy of all would be to insulate the interior of the ladle, but we
understood that that would be impracticable.
1. Radiation. Because of the high temperatures involved, this is often the
dominant mode of heat transfer. It is governed by the Stefan-Boltzmann
law (Rohsenow et al., 1985)
where e is the surface emissivity, u = 5.67 X 10-8 watt m-2 K-4 is the
Stefan Boltzmann constant, T is the surface temperature, and Ta is the
ambient temperature.
2. Convection. Convection through air boundary layers at vertical walls is
the most important convective transfer process. It is described by a Nus-
selt number Nu and a Rayleigh number Ra. The Nusselt number is a
dimensionless representation of the convective heat flux, defined by:
qL
Nu = AkaT
where q is the flux, A the area (so q/A is the flux density), L is a charac-
teristic length, usually the height of the wall, k is the thermal conductivity
of the boundary layer material (here air), and aT is the temperature dif-
ference across the boundary layer.
The Rayleigh number expresses the role of thermal buoyancy, and is here
given by
where g is the acceleration due to gravity, f3 is the volume coefficient of ex-
pansion of the material, a is the thermal diffusivity and v is the kinematic
viscosity.
The boundary layer heat transfer can be calculated from a semi-empirical
formula relating these two dimensionless numbers. It is:
NUL = C1Ra1/4, C1 ~ 0.5,
NUt = C2Ra1/3, C':! ~ 0.091
N U/ and NUt are the Nusselt numbers for laminar and turbulent con-
vection, respectively. Equation (4) combines these two numbers to man-
age a continuous change from the laminar to turbulent regimes at about
Ra ~ 109•
In the purely turbulent regime, combining equations (2-4) and rearranging
gives
Convection from the top horizontal surface was ignored, because the area
is smaller, the air speed is less than that adjacent to a vertical surface,
and because the temperature was hotter, so that in any case radiation
dominated.
3. Phase change. The slag is a glassy material, without a definite melting
point. We were unable to find any data on latent heat of freezing, and it
is possible that there isn't any, with the lack of apparent crystallisation.
Consequently, we ignored latent heat.
aT
pCat = V.q = V.(kVT)
where p is the density, C is the specific heat, k is the thermal conductivity.
A useful concept from this equation is the time constant, T, for conduction,
given by
As is often the case with hot molten materials, it is difficult to measure many
things, and physical properties can be hard to obtain. In this case, the molten
slag properties, and their variation with temperature, are not well established.
We used the following data, mostly obtained from Pasminco or Rohsenow et al.
(1983):
Slag specific heat
Slag freezing point
Slag Thermal Diffusivity
Emissivity
External wall temperature (bottom)
External wall temperature (middle)
External wall temperature (top)
Temperature of top slag surface
Charging temperature of slag
Average discharge temperature of liquid
Skull thickness (average)
Steel
Density
Specific Heat
Thermal conductivity
Thermal diffusivity
Emissivity of dirty surface
Air 400K
Dynamic viscosity
Kinematic Viscosity
Density
Thermal diffusivity
Thermal conductivity
950 Jjkg
1l00°C
6 X 1O-7m2s-1
0.7
350°C
500°C
350°C
700°C
1200°C
1150°C
0.05 m
7850 kg m-3
670 Jjkg
30 WjmjK
6 X 10-6m2s-1
0.7
2.43 X 10-5 Pa s
2.64 X 10-5 m2 s-1
0.871kg m-3
3.8 X 10-5 m2 S-1
3.36 X 10-5 W jmjK
1. Heat lost by slag during transfer
Here we worked on an observation that on average, of the thirteen tonnes of
slag, three froze, and reached an estimated average temperature of 700°C,
while the remaining ten tonnes dropped by 50°C from its original 1200°C.
With the specific heat of slag (liquid and solid) given at 950 JjKg, the
total heat lost is:
2. Heat lost to environment during transfer
Time of journey - 1 hour.
Side walls:
Radiation flux density = 5.67 X 10-8 * 0.7* (7004 - 3004)
~ 9000 Wjm2
Total radiation loss = 9000 W/m2 X 12m2 * 3600s = 3.9 X 108J
Convection (working at average air temp 400K:
R 10ms-2 X (700 - 300)/500 X 8m
3 10
a = ------------ = 6.4 X 10
2.64 X 10-5 X 3.8 X 10-5
NUt = 0.091 X Ra1/3 = 364, NUL = 0.5 X Ra1/4 = 251
Nu = ((364)6 + (251)6)1/6 = 370
q/A = Nu X k X tJ..T/L = 370 X 3.36 X 10-2 X 400/2 = 2500 Wm-2
Convective loss = 2500 X 12 X 3600 ~ 1.3 X 108 J
Radiation flux density = 5.67 X 10-8 * 0.7 * (10004 - 3004)
~ 36000W/m2
Total radiation loss = 36000W/m2 X 3m2 * 3600s = 3.9 X 108J
3. The return trip
The heat stored in the steel on discharge of the slag is substantial, and it
is shown here that most is lost on the return trip. It could be retained
by insulation, although then the ladle would run significantly hotter at all
stages.
On its return, the surface temperature of the steel is governed by the
differential equation:
dT 4 4MC- = -AelT(T - T )dt a
where M is the ladle mass, Ta the ambient temperature, and A the ladle
surface area (inner surfaces included).
This can be solved: rT dT
iTo (T4 - T:)
AelTt
---
MC
T- To ---(T1To)4 - T~ - Me
This equation determines the temperature decay of the steel shell. The
time constant for conduction in the steel is
indicating that conduction in the steel will not significantly delay heat loss.
We now calculate the radiative losses in 1 hour, using (10):
Inside side walls 240 MJ
Inside bottom 54 MJ
Outside side walls 288 MJ
Outside bottom 43 MJ
It is possible the last figure should not be included, because the bottom
is generally not exposed. However, the heat loss there is quite small. The
total loss in one hour is 625 MJ, which compares with the 1000 MJ gained
by the steel shell while the ladle is full. The values are not expected to
balance, because there are also convective heat losses.
It is interesting to cost this loss, at an energy value of 10 cents per Kw hr,
or about 3 cents per MJ which is probably close to the cost of energy used
to reheat the slag. Each ladle trip would lose about $20 in heat radiated
on the return journey. This is about $1000 per day, or $350 K per year.
So it is certainly worth taking measures to save the energy if possible.
The sidewall skull is the only thing that prevents the steel being heated
to unsustainable temperatures, so it is of interest to see what determines its
thickness and thermal behaviour. There are various considerations which do
ensure that the thickness is fairly stable and predictable.
Firstly it should be noted that the solid slag is quite a poor conductor.
Nevertheless, a large part of the heat lost has to pass through it. That means
that there is a substantial temperature drop across it. It also means that a
significant part of the heat lost does not pass right through the skull, but comes
from the further cooling of its outer layers.
We believed that as a reasonable approximation the liquid slag could be
taken as generally well stirred, with a convective boundary layer close to the
freezing zone. The reason for this belief is that the temperature, being close to
the freezing point, must be held fairly constant, despite the large heat fluxes.
The inner boundary condition, therefore, is approximately that of a constant
flux at the freezing temperature. This is not so different from the condition
that would be obtained in a semi-infinite solid slag conduction solution, and
this is the most reasonable known solution for comparison. The character-
istic distance associated with this problem is ..jOi = V2 X 10-7 X 3.6 X 103
= 0.027 m. The layer thickness is approximately this distance multiplied by
log ((1150 - 400)/(1150 - 1100)) ~ 2.5 or about 0.07m, which is close to the
estimated value.
The skull dynamics, and the requirement not to overheat the steel, ensure
that there is an inevitable loss of heat into the steel. This is the amount of heat
loss required to freeze a skull thick enough to keep the steel temperature down.
So while it is desirable to conserve the heat carried in the ladle, it essential to
ensure that it retains sufficient cooling capacity to create this protective skull.
Since the skull dynamic behaviour is important, and not obvious, it was
modelled using the program FastfloTM being developed at CSIRO Division of
Mathematics and Statistics. Simplifying assumptions made were
• The temperature at the inside of the domain (10 cm from the wall) was
held constant at 1150°C.
• The initial temperature of the steel was 100°C in figure 2(a) and 300°C in
figure 2(b).
• Otherwise, all thermal and fluid properties were independent of tempera-
ture.
Figure 2 shows the course of events during the first hour, in the steel wall of
the ladle and the inner 10cm of slag. It is a temperature contour plot in a space-
time plane; the bottom of the plot is the exterior wall. The top 1100°C contour
represents the edge of the skull. It is clear that this grows rapidly initially and
then stabilises. The extent to which it stabilises the temperature in the steel is
clear from the near horizontal contours at the interface. An interesting effect of
the skull growth is the plateau reached by the outer wall temperature in figure
2(a), and the maximum reached in figure 2(b). This occurs because a large part
of the heat in the steel enters in the first few minutes, and supply at that rate
is not continued because of the relatively insulating skull.
Figure 2(a): Skull formation at side wall. Steel initially at 100 C.
Temperature contours {lOOC, 150C ... 1100C} in space-time plane.
~--- -- ------
';jj~_~_~~-~--~~. ~~·c~=~
Figure 2(b): Skull formation at side wall. Steel initially at 300 C.
Temperature contours {300C, 350C ...1100C} in space-time plane.
• At the higher temperature, the greater effect of radiation ensures that the
temperature rise is smaller, and passes through a peak.
However it should be emphasised that this is a preliminary analysis, not
actually done at the MISG, and that more research would be needed before the
conclusions could be fully relied on. They are however promising, suggesting
that the initial steel temperature does not have a great effect on the maximum
temperature reached, provided free radiation while charged is permitted.
These are some details of the computer model. The solution algorithm was fi-
nite element in both space and time, with an unstructured mesh of quadratic tri-
angle elements. There were 3840 elements, with a total of 7839 nodes. The mesh
was concentrated about the steel/slag interface at t=O. A Newton-Raphson al-
gorithm converged in four iterations.
The roughness of the contours near t=O is caused by the initial temperature
discontinuity, and the roughness of the HOOoe isotherm is caused by the dis-
continuity of the thermal conductivity at the phase interface. A virtue of the
fully finite element procedure used is that it remains stable despite the numerical
difficulties associated with the discontinuities.
Surprisingly, according to our calculations, the amount of heat stored in the
ladle at discharge is almost as much as the heat lost during the transfer of the
full ladle. Most of this heat is then lost to the air in the return cycle.
The considerations in the previous section suggest that while there is a limit
to the extent to which this return heat can be conserved, because of the steel
running hotter, nevertheless there may be more scope for insulation on the return
trip than was originally anticipated at the MISG.
A useful observation was that the majority of the heat lost from the slag was
not from the cooling of the part that remained liquid, nor the cooling represented
by the initial solidification, but rather from further cooling of the slag that had
already frozen. Although this is only about three tonnes, its temperature drop
is on average about 500°K.
1. Insulated lid.
An insulated lid is a good idea. During slag transport it will block the top
radiative flux, which is about 40% of the total heat loss. The coupling to
the side wall temperature is through the liquid, and since this can vary
only slightly, the coupling is weak. So there should be little increase in the
temperature of the steel.
It is likely that with an effective lid, the top skull will almost vanish, with
the temperature at about 1l00°C. The bottom surface of the lid would
have to come into radiative equilibrium with this - i.e. at about the same
temperature. So here also the thermal endurance of whatever material
makes up this surface is of concern.
Both for the lid to be effective, and for the temperature to be tolerated,
an appropriate insulating material would be needed. The aim might be
to reduce the flux density from 100 kW m-2 to about 5 kW m-2, so that
h;::;j5Wm-2K-1• According to the published ratings, about 30 mm thick
Durablanket on the underside should achieve this. Exterior insulation
would also prevent heat loss, but would not protect the steel.
2. Insulate the ladle on the return trip only (perhaps)
A blanket, or insulated storage cavity, could be added to save up to 600
MJ per ladle per trip, but this saving would be more closely coupled to the
steel temperature. It would be hotter on subsequent loading, and would
run at a higher temperature at all stages. The preliminary calculations
of the previous section suggest that this effect may be of only moderate
importance - the rise in peak temperature may be only about half the
rise in initial temperature. However, these calculations would need to be
done more fully, and even then we would strongly advise close monitoring
of the steel temperature during the subsequent loaded phase, even if the
insulating effect had been removed during that phase.
3. Do not insulate, externally, while loaded
Doing so would directly raise the steel temperature, and very little saving
could be achieved before the limiting temperature was reached.
4. Monitor the wall temperature for overheating
Any effort to save heat will have some tendency to raise the steel temper-
ature. The monitoring should allow for the inside wall being at least 60°C
greater than the outside.
W.M. Rohsenow, J.P. Hartnett and E.N. Ganic, (Eds), Handbook of heat transfer
fundamentals (McGraw-Hill, NY, 1985).
The moderators, Nick Stokes and David Jenkins, were helped at all stages by
Ralph Johnson, the Pasminco representative. The problem was well attended,
and contributions were too many to list in full, but particular efforts by Noel
Barton, Paul Cleary, Jim Hill, David Marlow, Peter McGowan, Irene Pestov and
Sam Yang must be acknowledged.


Slag ladle insulation

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