FUEL AVAILABILITY

D. FUEL AVAILABILITY

At steady state, the maximum work under thermo-mechanical equilibrium conditions is expressed by the relation

W ˙ opt =( Σ N ˙ k ψ ˆ k (T i ,p k,i )) i –( Σ N ˙ k ψ ˆ k (T e ,p k,0 )) e .

where exit partial pressures are not necessarily same as the ambient. Recall that in case of combustion this equation can be divided by the number of moles of fuel N ˙ F flowing into the combustor to yield the availability under thermomechanical equilibrium conditions.

( w opt = Ψ R – Ψ P ) TM , where (37) Ψ R or P, TM =( Σ ν k ψ ˆ k ) i or e, TM =( Σ ν k ψ ˆ k (T o, p k,0 )).

(38) Here, the exit is considered to be in thermomechanical equilibrium at a total pressure of P 0 ,p k0

denotes the partial pressure of a species k in the thermomechanical equilibrium state with the ambient (or standard atmosphere), but each component is not necessarily in chemical equilib- rium with the corresponding components in the ambient.

Consider a special case in which fuel and air, although not premixed, both enter a combustor separately at the state (T o ,P o ). The reaction products are discharged at partial pres-

sures p k,∞ so that products are in thermo-mechanical-chemical equilibrium with the ambient. The optimum work under such a condition where each species is discharged at the partial pres-

sure corresponding to the ambient is called the fuel availability. In this case,

Ψ R =( Σ ν k ψ ˆ k ) i, = ( ψ F (T o ,P o )+ ν O 2 ψ O 2 (T o ,p O 2 ,o )) =

g F (T ( o ,P o )+ ν O 2 g O 2 (T o ,p O 2 ,o )), and

(39a)

(39b) Then,

Ψ P,TMC =( Σ ν k ψ ˆ k ) e,TM =( Σ ν k g k (T o ,p k, ∞ )) e .

w opt,TMC = Avail F = Ψ R - Ψ P,TMC

= ( g F (T o ,P o )+ ν O 2 g O 2 (T o ,p O 2 ,o )) – ( Σ ν k g k (T o ,p k, ∞ )) e . Nitrogen need not be considered, since the partial pressures and moles of N 2 at the inlet and

exit are equal. Table A-27B tabulates the fuel availability assuming X O 2 ,∞ = 0.2035, X CO 2 ,∞ =0.0003, X HO 2 ,∞ = 0.0303, X N2, ∞ = 0.7659

h. Example 8 Determine the fuel availability for methane. Assume that T o = 298 K, p O 2 ,o = 0.2055,

p CO 2 ,o =0.003, and p HO 2 ,o = 0.0188. If the lower heating value LHV of methane is 802330 kJ kmole –1 , determine the ratio of the fuel availability to LHV. Solution

Consider the overall reaction

CH 4 + 2O 2 → CO 2 + 2H 2 O.

The CO 2 and H 2 O so produced joins the atmosphere in gaseous form at partial pres- sures corresponding to the environment. The fuel availability

Avail F = w opt = Ψ R – Ψ P , TMC . Ψ R =( ψ CH 4 (T o ,P o )+ 2 ψ O 2 (T o ,p O 2 ,o )) , where ψ CH 4 (T o ,P o )= h CH 4 –T o s CH 4 (T o ,P o ) = –74850–298 × 186.16 = (–1.303 ×10 5 ) kJ kmole –1 , ψ O 2 (T o ,p O 2 ,o )) = h O 2 –T o s O 2 (T o ,p O 2 ,o )= × (205.04 – 8.314× ln (0.2055÷1)) = 0 – 298

(–65022) kJ kmole –1 , i.e.,

Ψ R =1 ×(–130300) +2×(–65022) = –260300 kJ per kmole of CH 4 . Likewise,

ψ CO 2 = h CO 2 –T o s CO 2 (T o ,p CO 2 ,∞ )=

(–393520 – 298 × (213.7 – 8.314 × ln (0.003÷1)) =

(–472000) kJ per kmole of CO 2 , ψ HO 2 = h HO 2 –T o s HO 2 (T o ,p HO 2 ,∞ )=

(–241820) – 298 × (188.7 – 8.314 × ln (0.0188÷1)) =

–307900 kJ per kmole of H 2 O, i.e.,

Ψ P,TMC =1 ×(–472000) + 2×(–307000) = – 1088000 kJ per kmole of fuel. Therefore,

Avail F = –260300 – (–1088000) = 827500 kJ per kmole of fuel. The ratio Avail F /LHV = 827500 ÷802330 = 1.031.

Remarks This procedure can be repeated for butane, for which Avail F =2767296 kJ per kmole of fuel and the ratio Avail F /LHV = 2767296 ÷2708330= 1.0218.For most hydrocarbon

fuels, the ratio of fuel availability to the lower heating value is in the range 1.02–1.07. An empirical relation for (Moran)

Avail F /LHV = 1.033 + 0.0169 (H /C) – 0.0698/C, gaseous hydrocarbon, C: carbon atom, H: hydrogen atom and Avail F /LHV = 1.0422 + 0.0119 (H /C) – 0.042/C, liquid HC

The fuel availability Avail F = w opt,TMC , where

w opt,TMC =( g F (T o ,P o )+ ν O 2 g O 2 (T o ,p O 2 ,o )) – ( Σ ν k g k (T o ,p k, ∞ )) e . For ideal gases

g (T,p k )= h k (T o )–T o s k (T,p k )= h k (T o )–T o ( s k (T,1) – R ln (p k /1)) =

h k (T =( o )–T o s k (T,1)) + R T o ln (p k /1) =

g o k (T o )+ R T o ln (p k (bar)/1), i.e.,

w opt,TMC =( g o F (T o )+ ν O 2 g O o 2 (T o )+ ν O 2 R T o ln (p O 2 /1)) –

Σ ν k o g k ( (T o ) e –( Σ ν k R T o ln (p k, ∞ /1)) e , i.e.,

o /( R T ∆G ν o ) + ln ((p , /1) O2 /( Π(p k, ∞ /1) k O 2 ∞ ) e ). (40) Typically, ν ∆G o /( R T

w ν opt,TMC /( R T o )=–

o ) » ln ((p O 2 , ∞ /1) O2 /( Π(p k, ∞ /1) ν k ) e ), so that

opt,TMC ≈ – ∆G .

Further, since o ∆G = ∆H –T

oo

o ∆S and ∆H »T o ∆S w opt,TMC ≈ –∆H o = –(LHV).

Therefore, the ratio of fuel availability to LHV is roughly unity.

The chemical availability of a fuel Avail F is a measure of the maximum possible work. The actual work output W is generally lower than Avail F , and the exergetic efficiency η avail = W/Avail F .