## Cell Thermal Balance

In an aluminum reduction cell the alumina is reduced to aluminum according to the following reaction:

The energy requirement for this reaction is equal to the change in enthalpy between the products and the reactants. In our case:

Assuming a 100% current efficiency and being included also the heat necessary to increase the temperature of the reactants from the room temperature up to the operating temperature of the cell. More in general, considering a real case where current efficiency is less than 100%, we have:

Where CE is the current efficiency expressed as a fraction between 0 and 1. As an example, for a normal current efficiency of 93% we have ΔH=6.45 kWh/kg.

Let’s now calculate the energy requirements to produce aluminum expressed in terms of voltage. First, the specific energy requirement expressed in Joule is equal to:

While the production rate of aluminum per second, given a line current of I amperes and a current efficiency equal to CE (expressed as fraction of 1 and not in percentage), is given by:

Multiplying and rearranging the two equations we have the energy per second (hence, power) needed to produce aluminum:

This energy is supplied from the external as electrical energy:

or:

We have used the symbol V_{Al} to say that this is the voltage required to produce aluminum.

The following chart depicts the equation above:

From the above calculations it turns out that the voltage required to produce aluminum is around 2V. But we know that a pot works with a voltage higher than this, typically 4.0 ÷ 4.5 volts. The difference between the voltage of a real pot and V_{Al} is equal to the energy that is lost in the ambient as heat. This is also the heat that keeps melted the aluminum and the bath in the pot.

Let’s consider the various voltage components:

where:

- E
_{rev}is the voltage to apply in reversible conditions for the basic reaction to occur - The terms indicated with η are the so called “overvoltages”:
- η
_{cc}: concentration overvoltage at cathode - η
_{aa}: concentration overvoltage at anode - η
_{ac}: reaction overvoltage at anode (0.6 ÷ 0.9 V) - The terms indicated with the letter V denotes ohmic voltage drops. In detail:
- V
_{A}is the voltage drop in the anodes - V
_{B}is the voltage drop in the electrolyte - V
_{C}is the voltage drop in the cathode - V
_{X}is the voltage drop in the buss bars external to the pot but which contributes to the total pot voltage

The voltage components that are located entirely into the ACD space are:

The amount of voltage generated into the interelectrode space and used to produce aluminum is equal to V_{Al}. So, the amount of voltage that generates heat that must be subsequently dissipated into the external environment is equal to:

The terms V_{A}, V_{C} and V_{X} represents too heat that needs to be dissipated outside from the pot, but they are constant, do not change over a short time and are located outside the interelectrode space.

This heat leaves the pot basically following two main paths:

- Through the anodes and the cathodes, following a vertical path
- Through the cryolite ledge, following an horizontal path

Let’s analyze more in detail these two heat fluxes, starting from the one following a vertical path.

The heat exiting the cell upward has to go through the anodes and the crust covering them. Some fraction of this heat is released also through the anode stubs. The heat exiting the cell downward has to pass through the metal (with a thermal resistivity which is negligible compared to the resistivity of all the other materials present in a cell) then the cathodes, the bricks layers and the potshell. What is important to understand is that the thermal resistivity of these parts as well as the areas crossed by this heat do not change in the short period:

with:

- R
_{Cathode}: thermal resistivity of the cathodes - T
_{M}: metal temperature - T
_{A}: ambient temperature - T
_{B}: bath temperature - A
_{Cathode}: cathode area - R
_{Anode}: thermal resistivity of the anodes - A
_{Anode}: anode area

Because the metal temperature is close to the bath temperature, we have:

And we can express the heat escaping from anodes and cathodes as:

Because, under normal operating conditions, the terms A_{Cathode}, R_{Cathode}, A_{Anode}, R_{Anode} do not change and the interval of operating cell temperatures (T_{b}) is not so big, we can say that:

For the heat exiting the pot from the cryolite ledge (horizontal path) we can write:

with:

- α: liminar coefficient of heat exchange between electrolyte and ledge
- T
_{Liq}: electrolyte liquidus temperature - A
_{Ledge}: ledge area - R
_{Side}: heat resistance including cryolite ledge + cell side insulations + potshell + ambient air

Because Q_{V} is approximately constant, all the variations in the heat generated by Joule effect in the interelectrode space must be absorbed by the component Q_{O}.

Changes in the horizontal heat Q_{O} are automatically realized changing the area and thickness of the cryolite ledge.** **