Question

2.37 Fuel cells are a promising alternative energy technology. They are based on producing energy by the following reaction:

H 2(g)+1/2O2(g) -->H2O(g)

One type of fuel cell, the solid oxide fuel cell, operates at high temperatures. A solid oxide fuel cell is fed 0.32 mol/s H2 and 0.16 mol/s O2, and the reaction goes to completion. The heat loss rate (in W) is given by:

Q=-7.1(T-T0)-4.2*10^-8(T^4-T0^4)

where T is in K and T0 is the ambient temperature, which can be taken to be 293 K. (Thanks to Prof. Jason Keith for providing the information for this problem.)
(a) Explainphysicallywhatthefirstandsecondtermsontheright-handsideoftheprecedingheat rate equation represent.

(b) Calculate the temperature at which the rate of heat loss is equal to the rate of heat generation.

Answer 1

(a)

The first term on the RHS of heat loss equation represents convective heat loss to surroundings:

Q_conv = - hA(T-T₀);

h = heat transfer coefficient, A = Area

The second term on RHS represents radiation heat loss:

Q_rad = - αAσ(T⁴-T₀⁴);

α = Emissivity/Absorptivity, σ = Stefan-Boltzmann constant

(b) Rate of heat generation = No. of moles * ∆H_rxn

∆H_rxn = -285.84 kJ/mol of H₂ reacted

= -285.84 kJ/mol * 0.32 mol/s = -91.47 kW

Rate of heat loss = -7.1(T-293) -4.2*10^-8 (T⁴-293⁴) W

⇒ 7.1(T-293)+4.2*10^-8 (T⁴-293⁴) = 91.47*1000

By hit and trial method, we get

T = 1194.1 K

Therefore at a temperature of 1194.1 K, the rate of heat loss is equal to the rate of heat generation.