For both the fuel and oxidant streams, utilization is an important design factor. The utilization of fuel plays a key role in the fuel efficiency, and the utilization of the oxidant, typically air, affects the mechanical efficiency of the system and polarization losses in the cell. We’ll first consider the oxidant by examining the cathode of a PEMFC to illustrate the concepts. If the oxidant is air, and oxygen is being reduced in an acid medium:

The utilization of oxygen is defined as

(10.18)

where A is the total combined active area of the cells and the current density is the same for all of the cells in series.  is the rate of mass flow of air to the cathodes. n is four for the reduction of oxygen.

As the oxidant is typically air and air is “free,” one might conclude that a high flow rate of air, and hence low oxygen utilization, is preferred. However, power is required to supply air to the stack, directly affecting the efficiency of the fuel-cell system. In contrast, a high utilization of oxygen results in a lower flow rate of air and a reduced ancillary power requirement. What then are the disadvantages of high utilization? We can answer this question in part by considering what happens along the direction of air flow (normal to current flow in the stack) as illustrated in Figure 10.7. Imagine that our planform is divided into three segments that are no longer electrically connected to each other. Simultaneously, we apply the same current density to each segment and measure the corresponding voltage. By repeating this procedure, the polarization curve for each segment can be measured while oxygen flows sequentially through the segments. As oxygen is consumed in the first segment, the concentration of oxygen reaching the second segment will be lower, and still lower in the third segment. As we studied in the previous chapter, the lower oxygen partial pressure is expected to increase both kinetic and mass-transfer polarization. Three hypothetical polarization curves are shown in Figure 10.7, one near the air entrance, one in the middle, and one near the air exit. The greater the utilization of oxygen, the lower the partial pressure of oxygen at the exit, and the lower the limiting current.

Of course, rather than isolated segments, the cell is connected to a highly conductive bipolar plate. This connection allows the current density to redistribute to mitigate the effect of oxygen depletion along the flow direction. Thus, the current density would be higher than the average value near the air entrance and lower at the exit. Controlling the total current to the cell and measuring the potential results in single polarization curve. This curve can be thought of as an amalgamation of the polarization curves for the segmented cells. Under special conditions, we can calculate the effect of utilization easily as illustrated in Problem 10.10. However, most of the time, the effect of oxygen utilization is best measured experimentally. The two typical representations of the experimental data are as a set of polarization curves measured at different utilization values (Figure 10.8), or as a utilization sweep, where the current density is held constant and the flow rate varied (Figure 10.9).

Increased polarization at higher oxygen utilization levels is evident from both figures. At low to moderate current densities, there is a kinetic effect due to the lower average partial pressure of oxygen. At higher current densities, mass-transfer effects dominate as evidenced by the reduced limiting current. A well-designed system will avoid operation near the limiting current. In other words, operation at a point with high sensitivity to oxygen utilization should be avoided.

The oxidant flow rate impacts fuel-cell operation in other important ways; these effects will change depending on the type of fuel cell. For a proton exchange membrane fuel cell, water is removed with the vitiated air (exiting air stream). Starting from a steady state, if the flow rate of air increases, the rate of water removal initially increases. However, any new steady state must also be in water balance. This limitation has an important impact on system performance as discussed in Section 10.7. In contrast, for solid oxide fuel cells, water is produced at the anode, and the flow rate of oxidant doesn’t influence the water balance. However, air flow plays a key role in cooling the solid oxide cell stack. This situation illustrates a second constraint: the cell, the cell stack, and the system must be in energy balance for steady operation.

Let’s now consider the fuel. The utilization of fuel in the stack, u_f, is defined similarly to that for oxygen. If the anodic reaction is the oxidation of hydrogen, then

(10.19)

Utilization and stoichiometry are two methods of expressing the flow rate of reactants relative to the current in the cells. Both terms are used commonly, and are simply the inverse of each other.

High flow rates of reactants mean low utilization or high stoichiometry.

A utilization of 100% means that all of the fuel is used in the cell or, conversely, that just the stoichiometric amount of fuel needed for the required current is provided to the stack. Clearly, the utilization of hydrogen in the stack impacts the fuel efficiency of the system. The consequence of supplying too much fuel is that it is wasted. On the other hand, providing too little fuel can cause severe damage as will be examined more closely in the next section.

ILLUSTRATION 10.4

Calculate the flow rate of hydrogen and air into 5 kW stack operating at a current of 100 A and 47 V. The stack consists of 70 cells (N_c) connected in series with an individual cell area, A_c, of 600 cm². The utilizations of oxygen and fuel are 0.7 and 0.95, respectively.

For the oxidation of hydrogen, n_a = 2 and we assume pure hydrogen: y_H2 = 1.

Solving for the mass flow rate, 

For the air, n_c = 4 and y_O2 = 0.21.

Hence, 

In both cases, we have neglected the impact of water vapor in the feed on the inlet mole fraction, which is significant for some types of fuel cells. This is equivalent to doing the calculation on a dry basis.

The utilization we have defined refers to the fraction of reactant consumed in the cell stack. Other definitions of utilization are possible and useful. One can express a more local utilization, for instance, such as the utilization of a single cell in the cell stack. As noted previously, the reactants travel through the stack in parallel. Although the stack is designed to have uniform flow through each cell, there are inevitably variations in flow to the cells. Likewise, within each cell there are often many flow channels in parallel, and the utilization in each channel can vary. Each of these differences in utilization changes the current distribution through the cells.

Looking beyond the cell stack, we can have instances where we are interested in the utilization of fuel in the overall system. As was suggested in Figure 10.2, not all of the fuel provided to the system contributes to electrical current. Some fuel may be combusted, some fuel may be exhausted from the system, and some may permeate across the separator. The overall utilization will depend on the particular design, and there are a great variety of designs and configurations used. Here we consider just two that illustrate the concepts.

To begin, let’s imagine we have a bipolar PEMFC stack operating on pure hydrogen. At first thought, you might anticipate that we could easily use all of the hydrogen in the cell stack so that u_f = 1. Unfortunately, this proves to be difficult primarily because water vapor and nitrogen dilute the hydrogen. Recall that the membrane requires water for good conductivity, which necessitates that the gas stream be humidified. There is also a small amount of nitrogen from the air side that permeates across the membrane; therefore, there must be a means to remove this nitrogen from the fuel side. Clearly, we don’t want to throw away valuable fuel. To compensate for only partial utilization of the fuel entering the stack, we can recycle a portion of the fuel stream that exits the stack as shown in Figure 10.10. With this configuration, we can conceive of two utilizations:

(10.20)

and

(10.21)

Whereas the consumption terms are identical in Equations 10.20 and 10.21, the supply terms are different (denominators). In this example, we could have pure hydrogen as the feed, but the flow into the cell stack would be diluted with water vapor and a small amount of nitrogen. Additionally, the flow rate through the stack is higher than the feed rate. The recycle ratio, R_r, is the molar flow rate in the recycle loop divided by that of the feed:

(10.22)

Because of recycle, each molecule of hydrogen, on average, may make multiple passes through the fuel-cell stack before either reacting or exiting the system. Use of recycle lowers the concentration of hydrogen in the stack, but enables the utilization of fuel in the system to be increased. Thus, recycle allows us to increase the overall utilization of fuel in the system, u_f,s, while maintaining a lower single pass utilization of fuel, u_f, in the stack. Note that earlier we introduced the fuel efficiency,  (Equation 10.5), which we see is identical to  for this example of a hydrogen fuel cell.

The second case we want to consider is for a fuel-cell system using a hydrocarbon fuel and that includes a reformer or fuel processor that we will treat as a black box. Using reformed fuel implies that the reactants entering the fuel-cell stack will be diluted with carbon dioxide, water, and perhaps nitrogen; therefore, the overall utilization of fuel must be less than 1. A common practice is to take unreacted fuel exhausted from the stack and feed it back to the reformer as shown in Figure 10.11. The fuel stream supplied to the cell stack may contain H₂, N₂, CO, CO₂, H₂O, and some unreformed fuel. Before defining the utilization for this case, it is necessary to introduce a couple of new ideas. First, depending on the type of fuel cell, CO can be a fuel just like hydrogen. While a specific utilization could be defined for each reacting species in the fuel, it is difficult to distinguish the contribution of an individual reaction of a multiple reaction set based on the composition of the fuel-cell effluent. The difficulty is complicated further at higher temperatures because reforming reactions may continue to take place inside the fuel cell. We can nonetheless define the utilization for the fuel-cell stack with just a small modification.

(10.23)

Think of the equivalents as the amount of electrons that would be supplied by the oxidation of the fuel. For hydrogen, two equivalents per mole of hydrogen are supplied. Earlier, Equation 10.2 was used to describe hydrocarbon oxidation with use of the generic formula C_xH_yO_z, where the number of equivalents per mole of fuel is 4x + y − 2z. This concept can be extended to fuel mixtures. The rate at which fuel is consumed is directly related to the current in the fuel-cell stack. The utilization in Equation 10.23 can be related back to the fuel efficiency used in Equation 10.5 by defining a fuel processing efficiency,

(10.24)

Finally, if we multiply these two together,

(10.25)

we see that this product is the same as the fuel efficiency used earlier. There is an alternative definition for fuel-processing efficiency that is sometimes used:

as explored further in Problem 10.12.

ILLUSTRATION 10.5

The fuel-cell system described in Illustration 10.1 uses a steam reformer and a high-temperature SOFC. In a solid oxide fuel cell, both CO and H₂ can be oxidized at the anode. From the 11 g s⁻¹ of methane fed to the system, the output of the fuel processor is a total of 3.678 mol s⁻¹ and has the following composition.

Species	Mole fraction	(4x + y − 2z) CxHyOz	Equivalent [mol·s−1]
CO	0.1421	2	1.045
CO2	0.2131	0	0
H2O	0.06729	0	0
CH4	0.009346	8	0.275
H2	0.5682	2	4.180

Assume the methane does not react in the cell stack; that is, the methane does not undergo further reformation and is not oxidized electrochemically. As defined in Equation 10.24, what is the fuel-processor efficiency? If the utilization, u_f, in the stack is 0.80, what is the fuel efficiency?

SOLUTION:

The fuel processor efficiency is (Equation 10.24)

First, we determine the rate in equivalents of oxidizable fuel supplied to the stack (numerator). The methane cannot be oxidized, but hydrogen and carbon monoxide react at the anode according to

The numerator can be found using the equivalents (mol·s⁻¹) from the table for H₂ and CO. For the denominator, we need the molar flow rate of methane to the fuel processor, which is 11 g s⁻¹/16 g mol⁻¹ = 0.6875 mol s⁻¹. Methane has 8 equivalents per mole of fuel as shown in the table, where x = 1, y = 4, and z = 0. Therefore, the fuel processor efficiency is

Note that in the numerator we did not include methane based on the problem statement. Alternatively, we can do the same calculation by subtracting off the methane equivalents exiting the reformer (see table) from the methane equivalents in the original fuel

Finally, using Equation 10.25,