Isothermal Combustion

4. Isothermal Combustion

A fuel cell and biological reactions, such as photosynthesis, are examples of situations involving isobaric and isothermal chemical reactions. Here,

) d(E cv -T 0 S cv )/dt = Σ Q ˙

Rj , ( 1 − 0 ) +( Σ N ˙ k ψ k ) i - ( Σ N ˙ k ψ k ) e - W ˙ ˙I (22)

cv -

Rj ,

Assuming steady state conditions in the absence of work, and heat exchange only with the am- bient, i.e., ˙ Q R,j = ˙ Q 0 and T R,j =T o ,

) ˙I )

(23) In this case the irreversibility is caused by entropy generated during the combustion process

σ ˙ c , and during the irreversible heat transfer due to the temperature gradient within the system σ ˙ HE (i.e., the combustion chamber temperature T and the ambient temperature T o ). The total entropy generated

(24) We can evaluate these entropies by using the entropy balance equation, i.e.,

σ ˙ cv = σ ˙ c + σ ˙ HE .

+( Σ N ˙ k s k ) i -( Σ N ˙ ) k s k ) e - ˙ σ cv .

dS cv /dt = Σ

(25) T bj ,

If the system boundary lies just within the reactor walls, leaving out the thin thermal boundary layer, then T b,j =T

˙ dS ) cv /dt = Σ +( Σ N k s k ) i -( Σ N k s k ) e - ˙ σ

c (26) Employing the First Law, i.e.,

(27) And eliminating ˙ Q 0 between Eqs. (26) and (27),

dE cv /dt = ˙ Q 0 – W ˙ cv + Σ k,i N ˙ k e T,k – Σ k,e N ˙ k e T,k ,

) ) σ ˙ c =-(dE cv /dt) (1/T) + W ˙ cv /T+ Σ k,i N ˙ k e T,k /T– Σ k,e N ˙ k e T,k /T+dS cv /dt+( ∑ N ˙ k s k ) e –( ∑ N ˙ k s k ) I . (28)

At steady state conditions, for pure combustion process with negligible ke and pe,

(29) The irreversibility for an isothermal and isobaric process due to chemical reaction alone can be

T σ ˙ c =( Σ N ˆg ˙ k ) e –( Σ N ˆg ˙ k ) I , reaction at temperature T

expressed as T 0 σ ˙ c = (T o /T)(( ΣN k ˆg k ) i –( ΣN k ˆg k ) e ), reaction at temperature T

For ambient temperature reactions (e.g., in a fuel cell or in plant leaves), T 0 σ ˙ c =( Σ k N k ˆg k ) e –( ΣN k ˆg k ) I , reaction at ambient temperature T 0 (30) The irreversibility due to heat transfer alone is

σ ˙ HE =– ˙ Q 0 (1/T o – 1/T).

f. Example 6 The stoichiometric combustion of molecular hydrogen in air proceeds in a premixed state. The reactants enter a combustor at 298 K and 1 bar and the products leave at the

same temperature and pressure. Determine the values of σ and ∆h c if the water formed exists in a liquid or in a gaseous state.

Max Work

Max Work, MJ/kmole 100

T exh, K

Figure 3: Work output with respect to T HE. for stoichiometric methane–air combustion considering variable specific heat c po (T).

Solution The overall chemical reaction is

H 2 + 1/2 O 2 + 1.88 N 2 →H 2 O + 1.88 N 2 .

Recall that σ=W opt /T o , where

w opt = Ψ 1 – Ψ 2 .

With the values

X H 2 =1 ÷(1+0.5+1.88) = 0.296, X O 2 = 0.148, X N 2 = 0.556, ˆg = (0–298 ×(130.57 – 8.314 × ln (1×0.296÷1)) = –41926 kJ kmole –1 H 2 , ˆg O 2 = 0–298 ×(205.03 – 8.314 × ln (1×0.148÷1)) = – 65832 kJ kmole –1 , ˆg = –298

N 2 ×(191.5 – 8.314 × ln (1×0.556÷1)) = –58521 kJ kmole , Ψ 1 =G 1 = (–41,926) + 1/2 ×(–65,832) + 1.88×(–58,521) =

–184861 kJ per kmole of H 2 . Treating H 2 O as a gas,

X HO 2 =1 ÷(1+1.88) = 0.347, X N 2 = 0.653,

ˆg –1 HO 2 = –241820–298 ×(188.71–8.314×ln(1×0.347/1)) = –300678 kJ kmole , ˆg –1 N 2 = –298 ×(191.5–8.314×ln(1×0.653/1)) = –58123 kJ kmole ,

Ψ 2 =G 2 = (–300768) + 1.88 ×(0 –58 123) = –4100392 kJ per kmole of H 2 , w opt = –184861 –(–4100392) = 22178 kJ per kmole of H 2 .

All of this work is lost. Therefore, σ = 225090 ÷298 = 755.34 kJ kmole –1 K, and

∆h c = 0 –(–241820) = 241820 kJ per kmole of H 2 .

Treating H 2 O as a liquid, ˆg HO 2 = g HO 2 = –285830 – 298 × 69.95 = –306675 kJ per kmole of H 2 ,

1.88 (0 – 298 × (191.5 – 8.314 × ln (1×1÷1))) = (–413961) kJ per kmole of H 2 , and

w opt = 229098 kJ per kmole of H 2 .

Consequently, σ = 229098

÷298 = 768.8 kJ kmole –1 , and

∆h c = 285830 kJ per kmole of H 2 .