Availability Balance Equation

1. Availability Balance Equation

The general availability balance equation is

d(E cv -T 0 S

cv )/dt = Σ Q ˙ Rj , ( 1 −

) +( Σ N ˙ k ψ k ) i -( Σ N ˙ k ψ k ) e - m ˙ - ˙I , (5)

T Rj ,

where ˙I =T o σ ˙ cv ≥ 0 since combustion is typically an irreversible process and the absolute stream availability ψ ˆ k = ( ˆe T –T o ˆs ) k = ( ˆh + ke + pe–T o ˆs ) k . Recall that species stream avail- ability (or species stream exergy or relative availability) is defined as

) ψ k ’= ψ k - ψ k,0 ,

(6a) ) where ψ k,0 (p k,e ,T 0 ) is the thermo-mechanical absolute stream availability of each species and

p k,e is the partial pressure of species k, p k,e =X k,e P 0 . One may write Eq. (5) in terms of species stream exergy also. For ideal gases, s k =s o k – R ln (p k /P ref ), where the reference state is generally assumed to be at P ref = 1 bar. Therefore, neglecting “ke” and “pe”,

ψ k = ψ o k +RT o ln(p k /1) where p k is expressed in units of bar. (6b) Dividing Eq. (5) by the molal flow rate of the fuel, (1/ N ˙ F ) d(E cv –T o S cv )/dt =

( Σ ν k ψ ˆ k ) i + Σ q R,j (1–T o /T R,j )–( Σ ν k ψ ˆ k ) e – w cv – i ,

(7a)

where w cv =( W ˙ cv / N ˙ F ) and i =( ˙I / N ˙ F )=T o σ . In terms of “i”,

i = –(1/ N ˙ F )d(E cv –T o S cv )/dt + ( Σ ν k ψ ˆ k ) i +

Σ q R,j (1–T o /T R,j )–( Σ ν k ψ ˆ k ) e – w .

(7b)

If a gas turbine combustor is started from cold conditions, the energy within it (dE cv /dt) starts accumulating and the entropy increases (dS cv /dt) due to a temperature rise within the combus- tor. Consequently, the stream availability at the exit will be zero initially and starts raising over time so that the work output from turbine increases until a steady flow steady state is achieved, when

(8) The optimum work is obtained when i = 0, i.e., w opt =( Σ ν k ψ ˆ k ) i + Σ q R,j (1–T o /T R,j )–( Σ ν k ψ ˆ k ) e. (9) In the absence of work and thermal reservoirs (that are typical of an adiabatic combustor),

i =( Σ ν k ψ ˆ k ) i + Σ q R,j (1–T o /T R,j )–( Σ ν k ψ ˆ k ) e – w .

(10) In general, we are interested in the change in the availability of streams entering or exiting the

i =( Σ ν k ψ ˆ k ) i –( Σ ν k ψ ˆ k ) e .

reactor. Let

Ψ R = Ψ 1 =( Σ ν k ψ ˆ k ) I , and Ψ P = Ψ 2 =( Σ ν k ψ ˆ k ) e , then

i = Ψ R – Ψ P = Ψ 1 – Ψ 2 = change in absolute availability, where (12) ψ 1,2 (T,P,T o )=e T,1,2 (T,P)–T o s 1,2 (T,P) = (h 1,2 (T,P)+ke 1,2 +pe 1,2 )–T o s 1,2 (T,P).

(13) For a steady flow process from an entry (state 1) to an exit (state 2), under negligible kinetic

and potential energy changes, ψ 1 – ψ 2 =((H 1 –H o )–T o (S 1 –S 0 ))–((H 2 –H o )–T o (S 2 –S 0 ))=H 1 –H 2 –T o (S 1 –S 2 ),

(14a) where on a unit mass basis

(14b) If T o =T 1 =T 2 = T, Eq. (14) assumes the form

ψ 1 – ψ 2 =h 1 –h 2 –T o (s 1 –s 2 ).

(15) For ideal gases, s = s o – R ln (P/P ref ), where the reference state is generally assumed to be at P ref

ψ 1 – ψ 2 =G 1 – G 2 .

= 1 bar. Therefore, = ψ o 1,2 ψ +RT o ln(P 1,2 /P ref ).

b. Example 2

1 kmole of butane enters a steady state steady flow reactor at 298 K and 250 kPa with 50% excess air. Combustion is assumed to be complete, and the products leave the re- actor at 1000 K and 250 kPa. Determine the heat transfer, reactant and product en- tropies, the absolute availability of the reactants and products, the entropy change between the exit and inlet, the entropy generation, the optimum work and the irre- versibility.

Solution The overall reaction can be expressed as

C 4 H 10 + 9.75O 2 + 36.66N 2 → 4CO 2 + 5H 2 O + 3.25O 2 + 36.66N 2 (A) The reaction coefficients are

ν CO 2 = (N CO 2 /N F ) e = 4, ν HO 2 = 5, ν O 2 = 3.25 and ν N 2 = 36.66, for the exit

stream and ν N 2 = 36.66 and ν O 2 = 9.75 for the inlet stream.

The energy equation is dE cv /dt = ˙ Q cv – W ˙ cv + Σ k,i N ˙ k ¯e T,k – Σ k,e N ˙ k ¯e T,k

(B) neglecting the kinetic and potential energies, we obtain the expression

(1/ N ˙ F ) dU cv /dt = q – w +( Σ ν k ˆh k ) i –( Σ ν k ˆh k ) e . At steady state and ideal gas conditions, ˆh k = h k , and since w = 0, we have the sim-

plified relation

(C) q =H p –H R , where

q +( Σ ν k ˆh k ) i –( Σ ν k ˆh k ) e = 0, i.e.

(D)

H p =( Σ ν k ˆh k ) e and H R =( Σ ν k ˆh k ) i . (E)

H R = –126148 kJ kmole –1 of C 4 H 10 , and

H P =4 ×(–393,520 + 33,425) + 5×(–241,827 + 25,978) + 3.25×(22,707) +

36.66 ×(21,460) = -1,659,104 i.e., q =H

P –H R = –1533044 kJ kmole of C 4 H 10 .

Likewise, we may show that S R =( Σ ν k

ˆs –1 k ) i = 9212.3 kJ kmole of fuel K –1 , Ψ

R =( Σ ν k ψ k ) i = –2869892 kJ kmole of fuel,

S P =( Σ ν k ) = 11363.7 kJ kmole –1 ˆs k e of fuel K –1 ,

Ψ =( Σ ν k ) = –5045283 kJ kmole –1 P ψ k i of fuel K –1 , and S P –S R = 11363.7 – 9212.3 = 2154.5 kJ kmole –1 of fuel K –1 .

Note that ˆs k

– R ln (p k /1), p k =X k P Table 1. Units: p are in bar, h in kJ kmole –1 , and the entropies in kJ kmole –1 K k –1 .

s o k ( h –T o ˆs ) k Reactants

p k =X k P

ˆs

C 4 H 10 1 –126148 0.021 0.0527 310.2 334.7 –225887 O 2 9.75 0 0.2057 0.5141 205.1 210.7

1.933 191.6 186.1 –55428 Products CO 2 4 –360470 0.0818

N 2 36.66 0 0.7733

H 2 O(g)

1.874 228.2 222.99 –44973 Applying the entropy balance equation for a steady state steady flow process,

(1/ N ˙ F ) dS cv /dt = ( Σ ν k ˆs k ) i + Σ( q j /T b,j )– ( Σ ν k ˆs k ) e + σ ,

σ =S P –S R – Σ( q j /T b,j ). (F) Using the values T b = 298 K and q = –1533044 kJ,

σ = 11363.7 – 9212.3 –(–1533044) ÷298 = 7294 kJ kmole –1 of fuel K –1 . The optimum work relation is (1/ N ˙ F )d(E cv –T o S cv )/dt = ( Σ ν k ψ k ) i + Σ q R,j (1– T o /T R,j )–( Σ ν k ψ k ) e – w – i .

Under steady state conditions if there is no thermal reservoir, then w opt = Ψ R − Ψ P , where Ψ R =( Σ ν k ψ k ) I , and Ψ P =( Σ ν k ψ k ) e , i.e.,

(G)

opt = 2175291 kJ kmole w –1 of C 4 H 10 .

Therefore,

i = w opt – w cv = 2175291 kJ kmole –1 of C 4 H 10 , since w cv = 0.

c. Example 3 Consider that fuel and air enter an automobile engine at 298 K and 1 bar (state 1), and the car exhaust exits at the same pressure, but at 400 K (state 2). Assume complete combustion with 20% excess air. Show that

w –1 = Ψ

R – Ψ P – i –P v F , kJ kmole of burnt fuel

(A) Determine the optimum work per kmole of burned liquid octane (C 8 H 18 ).

Solution The mass of the automobile is variable, since the liquid fuel weight in the fuel tank decreases over time due to fuel consumption. Hence, this is an unsteady problem. As-

sume that the species in the exhaust are CO 2 ,H 2 O, O 2 , and N 2 . From the availability balance relation

d(E cv -T

0 S cv )/dt = Σ Q ˙ ( 1 −

) +( Σ N Rj ˙ , k ψ k ) i -( Σ N ˙ k ψ k ) e - W ˙

cv - ˙I (B)

Rj ,

Consider the system control surface to exist around the automobile where T R = T o . Neglecting the kinetic and potential energies and considering an adiabatic car, Eq. (B) simplifies to the form

˙ d(U ) cv –T o S cv )/dt = ( Σ N k ψ k ) i -( Σ N k ψ k ) e - W ˙ cv - ˙I

(C) Since U cv =N F u F and S cv =N F s F and u F and s F are constants, then

d(N F ( u F –T o s F ))/dt = +( Σ N ˙ k ψ k ) i -( Σ N ˙ k ψ k ) e - W ˙ cv - ˙I

– N ˙ F,b ( u F –T o s F )=( Σ N ˙ ) k

˙ ψ k ) i -( Σ N k ψ k ) e - W cv - ˙I

(D)

where the fuel burn rate (dN F /dt) = – N ˙ F,b . Now

(E) Dividing Eq. (D) by the fuel burn rate, ψ F +( Σ( N ˙ k / N ˙ F,b ) ψ ˆ k ) i –( Σ( N ˙ k / N ˙ F,b ) ψ ˆ k ) e –( W ˙ / N ˙ F,b )–( ˙I / N ˙ F,b ) = 0, or

( u F –T o s F ) ≈ h F –P v F –T o s F = ψ F –P v

ψ F +( Σ ν k ψ ˆ k ) i, –( Σ ν k ψ ˆ k ) e – w – i –P v F = 0, i.e.,

(F) w = Ψ R – Ψ P – i –P v F , where Ψ R = ψ F +( Σ ν k, ψ ˆ k, ) i , Ψ P =( Σ ν k ψ ˆ k ) e .

Note that ) Ψ R =( Σ N ˙ k ψ k ) i denotes the availability of air crossing the boundary of car. The optimum work Note that ) Ψ R =( Σ N ˙ k ψ k ) i denotes the availability of air crossing the boundary of car. The optimum work

The reaction equation that represents the complete combustion of a kmole of C 8 H 18 with 20% excess air is

C 8 H 18(liq) + 15 O 2 + 15 ×3.76 N 2 = 8CO 2 + 9H 2 O + 2.5O 2 + 15 ×3.76 N 2 . At the inlet

Ψ R =( ψ F + ν O 2 ψ ˆ O 2 + ν N 2 ψ ˆ N 2 ) i , where

ψ F = h F –T o s F = –249910 – 298 ×360.79 = – 357425 kJ per kmole of fuel. (G1) In the case of air

X O 2 = 15 ÷(15 + 15×3.76) = 0.21, and ψ ˆ O 2 = h O 2 –T o s O 2 (T o ,p O 2 ) = 0–298 ×(205.04–8.314×ln (0.21×1÷1)), i.e.,

(G2)

ψ ˆ O 2 = – 64969 kJ per kmole of O 2 .

Likewise,

X N 2 = 0.79, and ψ ˆ N 2 = – 57651 kJ per kmole of N 2 .

Using Eqs. (E)–(G2), Ψ R = ψ F +( Σ ν k ψ ˆ k ) i = (–357,425) + 15 ×(–64969) + 56.43×(–57,651) = –4585118 per kmole of fuel burned.

Ψ P =( ν CO 2 ψ ˆ CO 2 + ν HO 2 ψ ˆ HO 2 + ν O 2 ψ ˆ O 2 + ν N 2 ψ ˆ N N 2 ) e , where (H)

X CO 2 =8 ÷(8+9+2.5+56.43) = 0.1053 and p CO 2 = 0.1053 ×1 = 0.1053 bar, Similarly, X HO 2 = 0.1185, X O 2 = 0.0329, X N 2 = 0.7432.

We can now determine the values of ψ ˆ k,e , e.g., ψ ˆ

CO 2 = (–393520+13372–9364)–298 ×225.225 = –456,629 kJ (kmole CO 2 ) . (I) Hence, Ψ P =( Σ ν k ψ ˆ k ) e = – 9835835 kJ per kmole of fuel burnt, and

(J) w opt = (– 4585118 – (– 9835835)) – 100 ×114÷703 = 5250700 kJ per kmole of burned

fuel, or 122569 kJ gallon –3 assuming the liquid density to be 703 kg m . Remarks

The work that can be developed by a 20 gallon (1 gallon = 3.7854 l) tank of octane is 2451374 kJ. Since 1 kW hr = 3600 kJ, in the case of this example W opt = 681 kW hr, and an automobile with a power output of 100 kW (a 6 cylinder engine) can be ideally driven for 6.81 hrs and at 50 kW can be driven for 13.62 hrs. At a speed of 60 mph for a 6 cylinder car this allows the vehicle to cover a distance of 409 miles at a maxi- mum of 20.5 miles per gallon for 100 kW engine and 41 MPH for 50 kW engine. The typical work output of an automobile is roughly 42500 kJ gallon –1 due to nonideal conditions.

d. Example 4 Five kmole of CO, three kmole of O 2 , and two kmole of CO 2 are fully mixed when they enter a combustor at 3000 K and 1 bar. The products leave the combustor at the same temperature and pressure at equilibrium. Determine the optimum work.

Solution The equilibrium composition for this problem is readily determined as 2.888 CO, 1.944 O 2 and 4.112 CO 2 using a similar procedure given in Example 7 of Chapter 12. The overall reaction can thereafter be represented by the equation

5CO + 3O 2 + 2CO 2 → 2.888CO +1.944O 2 + 4.112CO 2 .

(A)

/HHV= I/HHV

Equiv Ratio

Figure 1: The irreversibility produced by burning methane and air, with both reactant streams at 298.15 K.

W opt = availability in – availability out = Ψ R – Ψ P , where (B)

Ψ R = (5 ψ ˆ CO +3 ψ ˆ O 2 +2 ψ ˆ CO 2 ) i , and

(C)

(D) Now,

Ψ P = (2.888 ψ ˆ CO + 1.944 ψ ˆ O 2 + 4.112 ψ ˆ CO 2 ) e .

ψ ˆ CO, i = h CO (T,p CO )– T o s CO = g CO + (T – T o ) s CO =

(–110530) + 96395.7 – 298 ×(274.6– 8.314×ln ((5÷(5+3+2)) ×1)) = (–0.9772

×10 5 ) kJ per kmole of CO,

ψ ˆ O 2 ,i = 0 + 98152.9 – 298 ×(284.4–8.314×ln ((3÷10) ×1))=

×10 5 0.1042 kJ per kmole of O

2 , and

ψ ˆ CO 2 ,i = –393520 + 152347.9 – 298 × (332.9– 8.314×ln 0.2) =

(=–0.3444 ×10 6 ) kJ per kmole of CO 2 .

Therefore, Ψ 6 R = –1.146 ×10 kJ

(E) At the exit (in units of kJ),

5 ψ 6 ˆ CO, e = –0.9866 ×10 , ψ ˆ O 2 ,e = 9699.5, and ψ ˆ CO 2 ,e = –0.3424 ×10 kJ, i.e.,

(F) Hence,

Ψ P = –1.6215 ×10 6 kJ

Figure 2: Entropy generation per unit amount of heat released (based on the higher heating value of the fuel) during adiabatic combustion of various hydrocarbon fuels under stoichiometric conditions.

W opt = (– 1.146 + 1.6215) ×10 6 = 475500 kJ per 5 kmole of CO consumed, or (G) 95100 kJ per kmole of CO entering the reactor.

This is an example of isothermal availability since T reac =T prod .