Mass balance anode side

The results show that, at temperatures below 60 °C and an air feed stoichiometry below three, the cathode exhaust is fully saturated (nearly fully saturated at 60 °C) with water vapor and the exhaust remains saturated after passing through a condenser at a lower temperature. In order to maintain water balance, all of the liquid water and part of the water vapor in the cathode exhaust have to be recovered and returned to the anode side before the cathode exhaust is released to the atmosphere. Because of the low efficiency of a condenser operated with a small temperature gradient between the stack and the environment, a DMFC stack for portable power applications is preferably operated at a low air feed stoichiometry in order to maximize the efficiency of the balance of plant and thus the energy conversion efficiency for the complete DMFC power system. Thermal balance under given operating conditions was calculated here based on the demonstrated stack performance, mass balance and the amount of waste heat to be rejected. 

Two different descriptions are possible. The simplest approach is the balance border around the complete module including all stacks and the joint burner from the inlet I of the fuel F and the air A to the outlet aB of the flue gas G after the burner. The more detailed approach is a balance border which surrounds all stacks from the inlet I to the outlets O of the anode side AnO and of the cathode side CaO. The calculation of this power generating burner is similar to the calculation of a combustor of a gas turbine or of a furnace of a boiler. The calculation of the mass flows of the module does not differ from any calculation of a conventional oxidation. The energy balance of this simpler approach (from / to aB) gives... 

To capture the dynamics of PEM fuel cell startup it is essential to include the dynamic mass balance for the water content in the membrane, TV . If water accumulation in the hydrophobic gas diffusion layer is neglected, then the water accumulation in the membrane is the difference between the water produced and the net water removed by convection from the anode and cathode, as given by Eq. (3.4). We measure all the quantities on the right-hand side of Eq. (3.4), so we can determine the accumulation rate of water. 

Fuel cells must carry the costs of conditioning the two reactant gases as well as their own capital charges. Hydrogen requires transport to the anode side of the fuel cells. This is usually by rotary blower, but it also should be possible to operate membrane cells at some positive pressure and then to deliver the hydrogen without mechanical aid. The temperature and water content of the hydrogen must be considered in the overall heat and mass balance. Air and oxygen are candidates for use at the cathodes. The classical balance between cost and efficiency determines the choice. Wth alkaline fuel cells, the carbon dioxide in the air is of concern. It can consume the hydroxide value and contaminate the end product. It is possible to scrub the air to remove the CO2 before... 

Back diffusion of water depends on water concentration on both sides of the membrane, water diffusivity through the membrane, and membrane thickness. Because water concentration is not uniform, it is not easy to explicitly calculate back diffusion for the entire cell or a stack of cells. For the sake of mass balance, back diffusion may be expressed as a fraction, p, of electroosmotic drag. When P = 1, back diffusion is equal to electroosmotic drag, that is, there is no net water transport across the membrane. The coefficient p may be determined experimentally by carefully condensing and measuring water content in both anode and cathode exhaust streams. 

For a fuel cell from Problem 3, calculate the heat generated at 0.65 V and 1 Acm by using the Equation 6-41 and by doing a detailed mass and heat balance analysis (assume hydrogen is supplied in dead-end mode saturated at 60°C and that net water transport through the membrane is such that there is no water accumulation on the anode side). Explain the difference in results. 

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