# Smelting furnace refractory and freeze lining heat transfer model

In this post I would like to illustrate how to model heat transfer through the refractory of a smelting furnace, and how to estimate the thickness of a freeze lining on the hot face. This is targeted at readers familiar with smelting furnaces, and would like to utilize measured data to evaluate the heat transfer performance, or alternatively to investigate refractory options. Although complex phenomena exists in the furnace influencing heat transfer and the formation of freeze lining which are in some cases  largely unknown, a simple approach is demonstrated to model the most important effects observed.

Refractory domain
The image below is a typical view through the smelting furnace sidewall when considering heat transfer, with the lower part of the sidewall refractory in line with the slag bath of particular interest.  In the lower part of the furnace heat is transferred from the electrodes to the slag bath, and conducted through the slag/metal baths and then refractory to where it is lost by whatever convective cooling means.

Build-up and wear of the freeze lining and refractory is driven by thermal (heat transfer), chemical (reactions), and mechanical (wear) phenomena. Should the sidewall be highly conductive and sufficient cooling provided on the shell, slag might solidify on the inside of the sidewall refractory. Increased heat conduction cools the slag close to the sidewall, causing its viscosity to increase and solidify, forming a freeze lining layer. The increased resistance will have heat transfer decrease, as well as the total heat loss from the shell. Also, chemical reactions occurring between the slag bath and freeze lining, and movement in the slag bath influences the freeze lining formation or wear.

Model overview
Heat transfer through the refractory in line with the slag bath, and freeze lining formation is of interest. As a simplified solution, one-dimensional (1D) heat transfer through this section is modelled. Heat transfer through the freeze lining and refractory wall can be modelled by applying conductive heat transfer principles, while cooling on the shell can be modelled by convective heat transfer principles. A larger simplification is however required on the freeze lining interface. In this case a constant temperature boundary condition will be assumed.

An important assumption/simplification made for this formulation is that of the hot-face temperature. In theory this is the temperature where the slag has cooled enough to ‘stick’ to the sidewall refractory and to withstand mechanical wear. This however unknown to some extent, and becomes a tuning factory in which the model is ‘tuned’ to predict more accurate results.

With the above assumptions it should be possible to model heat transfer, with the limitation that to determine the thickness of the freeze lining the heat loss through the sidewall should need to be known. This could easily be calculated however either from thermocouple data, or temperature differential and cooling water flow rate data.

1D heat transfer formulation
In this simplification heat transfer through the refractory in line with the slag bath is formulated. Despite the typical furnace being cylindrical, the equations used here will not be based on cylindrical coordinate system, but rather Cartesian, with x depicting the radial direction through the refractory.

The following illustrate the domain for which heat transfer will be formulated.

Considering conduction, the transient heat equation is given by the following, with k the heat transfer coefficient (W/mK), ρ the refractory density (kg/m3), and Cp the refractory heat capacity (J/kgK) . $-k dfrac {d^2 T}{d x^2} = rho C_p$

We will only consider steady state conditions, for the heat transfer (q) in W/m2 is calculated by the following equation in differential form. In this, and equations to follow, q refers the heat transfer in the radial direction in units of W/m2, meaning the heat conducted (W) per meter squared of outer surface. The total heat loss is equal to q multiplied with the outside surface area. $q = -k dfrac{dT}{dx}$

The equation for heat conduction could be discretized, into the following, here shown with the equation for calculating convective heat transfer in terms of the cooling temperature, the surface temperature on the shell, and h the heat transfer coefficient (W/m2K). $q_{conduction)i}=-k_i dfrac {T_2-T_1}{r_i} ;; and ;; q_{convection}=-h (T_{infty}-T_{surface})$

To further simplify the model, and reduce the need to calculate the outer shell temperature, the thermal resistance analogy is used. This method is similar to electrical resistance, and makes it possible to calculate thermal resistances for the different sections of the domain (sidewall) and then adding them together to yield the total thermal resistance. The total resistance in then given by the following equation, for which it has been assumed that any resistances except that of conduction and convection are negligible (e.g. contact resistance, etc.). $R_{total}=R_{conduction ; layer ; 1}+R_{conduction ; layer ; 2}+{dots}+R_{convection ; cooling}$

The thermal resistances for conduction through each layer and convection from the shell can be calculated by the following, derived from previous equations: $R_{conduction}=dfrac{r_i}{k_i} ; ; and ; ; R_{convection}=dfrac{1}{h}$

Following the above, the heat transfer through the conducting refractory layers, and convection from the shell are: $q_{conduction)i}=-dfrac {T_2-T_1}{R_{conduction)i}} ;; and ;; q_{convection)i}=-dfrac {T_{infty}-T_{surface}}{R_{convection}}$

Example
For example, consider a furnace with the following particulars regarding the slag, refractory layers, and sidewall cooling.
• Slag hot face temperature of 1500°C.
• Freeze lining thickness to be determined, thermal conductivity 1 W/mK.
• Inner refractory thickness 125 mm, thermal conductivity of 1.5 W/mK.
• Outer graphite layer thickness 500 mm, thermal conductivity of 17.5 W/mK.
• Steel shell thickness 50 mm, thermal conductivity of 50 W/mK.
• Shell water cooling with a heat transfer coefficient of 250 W/m2K, and water temperature of 25°C.
• Thermocouple at 290 mm from steel inside measuring 185°C, and one at 90 mm from steel measuring 100.5°C.

This is illustrated by the following.

We would like to estimate the freeze lining thickness from data provided, by firstly estimating the heat  flow through the entire domain in the radial direction (from left to right). If the cooling water flow rate and temperature differential was provided this could also have been used, but in this case we will have to use the temperature difference and distance between the thermocouples in the graphite layer as follows. $R_{conduction}=dfrac {r_i}{k_i} ;; and ;; q_{conduction} = - dfrac {T_2-T_1}{R_{conduction}}$ $R_{conduction}= dfrac {0.2}{17.5}=0.0114 ; m^2 K / W$ $q_{conduction} = - dfrac {100.5-185.5}{0.0114}= 7438 ; W/m^2$

In the above the heat transfer rate has been estimated as 7438 W/m2, this figure could be validated using the outer surface of the furnace sidewall, and the cooling water data discussed earlier. The next step however is to calculate the thickness of the freeze lining, by calculating the thermal resistances of all the layers and convective cooling, and then using the temperature difference between the hot face and cooling water, and heat loss rate calculated above to estimate the thermal resistance of the freeze lining. With this known, the thickness of the freeze lining can be estimated. The following illustrates the calculation of the total thermal resistance, excluding that of the freeze lining. $R_{total} = R_{refractory}+R_{graphite}+R_{steel}+R_{cooling}$ $R_{total} = dfrac {r_{refractory}}{k_{refractory}} + dfrac {r_{graphite}}{k_{graphite}} + dfrac {r_{steel}}{k_{steel}}+ dfrac {1}{h}$ $R_{total}= dfrac {0.125}{1.5} + dfrac {0.5}{17.5} + dfrac {0.5}{50} + dfrac {1}{250}=01169 ; m^2 K / W$
In the following the thermal resistance of the freeze lining is calculated. $q = - dfrac {T_{infty} - T_{hot ; face}}{R_{total}+ R_{freeze ; lining}}$ $R_{freeze ; lining}= - dfrac {T_{infty} - T_{hot ; face}}{q} - R_{total}$ $R_{freeze ; lining}= - dfrac {25-1500}{7438}-0.1169=0.0814 ; m^2 K/W$

If the above yielded a negative value, it means that the freeze lining will have a negative thickness calculated, implying wear of the refractory and not build up. To determine the amount of wear, the heat transfer coefficient of the refractory need to be used in the following, and not that of the freeze lining. The thickness of the freeze lining is estimated as follows. $R_{freeze ; lining} = dfrac{r_{freeze ; lining}}{k_{freeze ; lining}}$ $r_{freeze ; lining} = k_{freeze ; lining} times R_{freeze ; lining}= 1 times 0.0814= 0.081 ; m$

In the above the freeze lining thickness is estimated as being 81 mm thick. All of the calculations were made in an Excel spreadsheet, with the grey background cells marking the values being calculated.
In the above, the table towards the lower right corner depicts the distances from the front of the freeze lining and temperatures calculated using the same model used above. This is used to plot temperature versus distance from the hot face, showing the temperature distribution throughout the sidewall. In the following it can be seen how the temperature slope is steeper in areas with high thermal conductivities.

Conclusion
In this post simplified one-dimensional heat transfer was used to illustrate how heat transfer could be modelled in a smelting furnace refractory and how the freeze lining thickness could be estimated. This approach could be used to trend the freeze lining thickness over time, using thermocouple data and/or shell cooling water data.

This approach was based on some assumptions and simplifications to cater for uncertain phenomena, and resulted in a model of which the parameters could be ‘tuned’ to predict results closer to observed values. The model could also be improved with thermal conductivities measured of the freeze lining by research facilities, as well as calculating more a accurate convective cooling heat transfer coefficients from cooling data.

To investigate this system in more detail, the formulation could be extended to two or three dimensions, and some phenomena included such as movement in the slag bath and slag solidification/melting. This will model the system more realistically catering for the actual phenomena observed. This formulation can be solved following a finite element approach, which is likely to be covered in a post in future.

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