Fuel cell voltage, efficiency and hydrogen use

(a) Reversible voltage and efficiency

Two electrons are transferred per H₂ molecule ($n = 2$).

$$E_{rev} = \frac{-\Delta G}{nF} = \frac{237\,200}{2(96\,485)} = 1.229\ \text{V}$$ $$\eta_{th,max} = \frac{\Delta G}{\Delta H} = \frac{237.2}{285.8} = 0.830 \;(83\ \%)$$

(b) Stack power and voltage efficiency

$$V_{stack} = 50 \times 0.65 = 32.5\ \text{V}, \qquad P = V I = 32.5 \times 40 = 1300\ \text{W}$$ $$\eta_V = \frac{0.65}{1.229} = 0.529 \;(52.9\ \%)$$

Efficiency on the basis of the HHV: $\eta = \dfrac{0.65}{285\,800/(2 \times 96\,485)} = \dfrac{0.65}{1.481} = 43.9\ \%$.

(c) Hydrogen consumption (Faraday's law)

$$\dot n_{H_2} = \frac{N I}{nF} = \frac{50 \times 40}{2 \times 96\,485} = 0.01036\ \text{mol/s}$$ $$\dot m = 0.01036 \times 2.016 = 0.02090\ \text{g/s} = 75.2\ \text{g/h} = 0.0752\ \text{kg/h}$$

(d) Heat generated

The heat generated is the difference between the energy released at the thermoneutral (HHV) voltage, 1.481 V per cell, and the electrical output at 0.65 V:

$$Q = N I (1.481 - 0.65) = 50 \times 40 \times 0.831 = 1662\ \text{W}$$

Answer: $E_{rev} = 1.229$ V, $\eta_{max} = 83\ \%$; $P = 1.30$ kW, $\eta_V = 52.9\ \%$ (43.9 % on HHV); H₂ use $= 0.0752$ kg/h; heat $\approx 1.66$ kW.