As you may have noted, some of the industrial processes discussed in this chapter operate at high temperatures. The most extreme example considered is the electrowinning of aluminum, which takes place in molten salt at temperatures of almost 1000 °C. How much heat is required to maintain the required temperature and how is this heat supplied? In this section, we consider heating and cooling of electrochemical systems since temperature control is a critical part of any industrial process.
Let’s begin with an overall energy balance that applies to a system in which multiple reactions take place. For open systems, the following energy balance applies:
(14.22)
where
= mass of the system, assumed constant [kg]
= average heat capacity of the system [J·kg−1·K−1]
= enthalpy of outlet stream p [J·mol−1]
= enthalpy of inlet stream m [J·mol−1]
= molar flowrate of inlet stream m, [mol·s−1]
= molar flowrate of outlet stream p, [mol·s−1]
= heat transferred to the system from the environment [W]
= Rate of work done by the system on the environment [W]
= Rate of reaction of species i [mol·s−1]
= Heat of reaction j per mole of species i [J·mol−1]
For the electrolytic systems considered in this chapter, is positive and equal to the power added to the cell in order to carry out the reaction, IVcell. The heats of reaction apply to full reactions rather than to half-cell reactions, and are determined as described in Chapter 2. When operating at steady state, the term on the left side of Equation 14.22 is zero. Consequently, one important use of the energy balance is to determine the rate of heat () that must be added to or removed from the system in order to maintain a steady temperature. Let’s illustrate the use of the balance by applying it to aluminum electrowinning (see Illustration 14.12).
As the illustration demonstrates, heat must be removed from cells used to produce aluminum. This situation is typical for an industrial electrolytic process. Therefore, the focus is on rejecting heat in order to maintain the desired temperature. The high operating temperature for aluminum production facilitates heat transfer from the cell to the environment, which directly provides the necessary cooling. The formation of a solidified molten salt insulating layer on top of the melt permits the system to flexibly maintain the needed temperature. However, these characteristics are not typical of industrial electrolytic cells. For operation closer to room temperature, heat is removed by placing heat exchangers in the anolyte and catholyte loops. In other words, the cooling takes place outside of the reactor with use of heat exchangers placed in the flow loops as illustrated in Figure 14.12. Evaporative cooling towers are frequently used to provide a heat sink for large systems. If operation is assumed to be steady, Equation 14.22 can be used to calculate the heat that would need to be removed to maintain a constant temperature. Similarly, Equation 14.22 can be used to approximate the increase in electrolyte temperature in the electrochemical reactor by assuming adiabatic operation or a known finite heat loss and then calculating the outlet temperature.
Figure 14.12 Process diagram for Zn electrowinning that emphasizes cooling of the electrolyte.
ILLUSTRATION 14.12
Energy Balance for Hall–Héroult Cell. An aluminum cell is operating at 4.2 V and 200 kA. The faradaic efficiency is 95%. Assume that the system is at steady state and that the reaction occurs as follows:
Determine the amount of heat that would need to be added to the reactor in order to maintain the operating temperature of 970 °C. The Al2O3 and C added to the reactor are at room temperature. The carbon dioxide and molten aluminum leave at the operating temperature of the reactor. The following physical data are known:
Heat capacities [J·mol−1·K−1]:
For aluminum: Tmelt = 933.47 K, and ΔHfus = 10,700 J·mol−1.
First, we determine the molar flow rates. Noting that six electrons are transferred for every mole of Al2O3 that reacts,
We now calculate the heat of reaction at 25 °C with use of heat of formation data from standard tables.
For this problem we choose a pathway where the reactants enter and react at 25 °C, and then the products are heated to the outlet temperature of 1243 °C. All enthalpies are referenced to 25 °C. Therefore, the inlet enthalpies are zero since they are at the reference temperature. The exit enthalpies are calculated as follows:
where the expression for the heat capacity was inserted into the integral and evaluated.
For aluminum, we must account for the change in phase and the heat of fusion.
We can now apply the full energy balance, noting that the reaction rate of Al2O3 is equal to . Solving for
The work term in the equation is negative because it is done on the system rather than by the system. Importantly, we note that is negative. Therefore, heat must be removed from the system rather than added to the system, even for this high-temperature reaction. This is the typical case for electrochemical systems.
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