'

[15.!> J
Co m bu s tio n Pr
oc es se s
l
Ca rb o an d 14 er p ), an d
th e c
n P ce nt hy dr o (~: Ca lc u late t he value o f du o ( va
lu es of b.hge
o rresp on di ng va o iq ) an d b. ho (v
ap ) .
.
Firs t, 1 ko of fuel produc es 0. 86 ( 44 / 12 -
o
an dQ J - 3. 15 kg of CO02
, _, .1 4( 18 / _2 ) = 1.2 6 kg
of H •
A t .... 5 C, llr g = 23 04 kJ / k
g and h - 2 44 2 kJ / kg . So that
2

rg
.
"'
uu 0
(
va p) _ "' (I . + m
- uu 0 1q) H 2O Ufg

= - 46 890 + ( I X 2304 ) = -4 39 8 7 kJ / kg
.26
.
Th
m bu st1 0n eq ua tio n can be written as
e co
0 0
1 kg fu el + 3.41 k g 2 ~ 1. 26k gH 2O + 3 .1 5k gC. 2
. .
n 11.h (Ii from equa ti on
Th e re la n be twee g th o q) a nd 11.uo(liq) can be obtained
tio
be
(15.11). Remem berin s and products shou ld
th at t~~ only . g~ se ~us. re ac ta nt
included, an d given th e eq uat ion becomes
fuel is Ill liqmd form,

.1ho (l iq ) = .1uo(ii q) + Rro( L :• - I~•)

p r R m; gas

= -4 6 890 + { 83145 x 298. IS C ~S _ );:!)}

6 977 kJ / kg
= -4 6 890 - 87 = -4
) + m H2ohrg
.1ho(vap ) = Llho(liq
x 2442) = -4 39 00
kJ / kg
= - 46 97 7 + (1 .2 6
(v ap ) using eq ua tio
n ( 15 .11 ):
h (v ap ) fro m t1u
find .1 0 0
As a ch ec k, we ca n

.1h 0 (v ap ) = Llu 0 (v ap
) + RT0 (I ~i - L~i)
p m; R 111 ; ga~

26
298. 15 (1 . + J. l S _ 32
3.41)}
= - 43987 + { 8.3145
X
18 44

= - 43 900 kJ / kg r ce nt ,
t1u (li q) an d .1 u 0 (vap) is 6.2 pe
rence be tween 0
0.2 per cent. Jn an
y
N o te th at th e diffe r on ly by ab ou t
d Llh 0 (Iiq) diffe rt an t to decide wheth
er
\\- here as Llu 0 (li q) an ev id en tly m or e im po
stion da ta , it is or va po ur than wheth
er
ap pl ic at io ns of co m bu re ga rd ed as liq u id
uc ts sh ou ld be
the H 2 0 in th e pr od
e ap pr op riat e.
Llu 0 or .1 h 0 da ta ar

co m bu st io n pr oc es ses, engineers have in

energy tr an sfe rs in rific valll e of the fuel
Whe n de al in g w ith sfer s to so m e ca lo
related these tr an es refe r di re ctly to
th e pa st co m m on ly t1h C alor ific va lu
pr op er ties Llu 0 and
.
rn t completely in a
0
ra th er th an to the m as s of fue l is bu
er at ed when unit e defined in
qu an tit ie s of hea t lib nd iti on s. B ec au se calori fic va lues ar
ec ified co end states
ca lo ri m et er un de r sp de fin iti o n m u st sp ecify not merel y the
he a t, th e
terms of qu an titie s of
345
 '!
App licat ions to Part icula r Fluids (15.6 l

, . , f th ~ process connecting these end states. Provided

but also the d etd l 1s O c: d. · dh
. . . d and the specific con itions are a erect lo th
standard ised eq uipment is use . f h· h . ' e
, b . . J will be accu rately reproducible. The ac~ t at t e qua ntities are
1
res u ts o tame d fi d opert1es ~u and 111 ·
not identical with the more rigoro usly e ne pr ffi . fo lo is not
important fo r such purposes as compa ri ng the e c1ency o power plant, or
r th hea ting value of a gas supply
meeting' statutory requirements
·
ior e ·
·fi
The four ca 1on c va1ues tha t have been in common use are as follows, together
wi t t e va1ues or L.1AUo and OA/1o to which they correspond most closely:
· h h

Gross ( or hi gher) calorific va lue at constant volume: Q11r .v == -i1u2s (liq)

Net (or lower) calorific value at constant volume : Qn ct.t• == - i1u25 (vap)
Gross (or higher ) calori fic va lue at constant pressure: Q gr,p == -i1h2s( liq)
Net (or lower) calorifi c value at constant pressu re: Qne1 .p == -tih 25( vap )

Two points should be noted. First, while values of ~uo and_~h o for an exothermic
reaction in accord with our sign conven tion are negative numbers, calorific
values are always quoted as positive numbers. Secondly, gross ( or higher) and
net (or lo wer) refer to the phase of H 2 0 in the products, the former to liquid
and the latter to vapour.

15.6 Efficiency of power plant and of combustion processes

The cycle efficiency of a heat engine is calculated without reference to the source
of heat. When considering the efficiency of the power plant as a who le, however,
we are also concerned with the efficiency of the means of producing the heat,
i.e. whether the fuel is in fact being completely burnt under the most
advantageous conditions. The overall thermal efficiency of a power plant can
be sui tably defined as the ratio of the work produced to the latent energy in
the fu el supplied. The question arises as to how this latent energy should be
assessed.
Any power plant, considered as a whole, is an open system undergoing a
steady-flow process- with air and fuel entering the system, and the products
of combustion leaving it, at a stead y rate. This is illust rated in Fig. 15.8a. The
fu el and air enter at at mospheric temperature, and the largest quantity of energy
that can be transferred to the surroundings will be produced if (a) the fuel is
burnt completely, (b) the products leave at atmospheric temperature, and (c)
any water in the products is in the liquid phase. The energy transferred under

Fig . 15.8
Schematic combu stion
pow er plan t w Fu eI ---...-,r-

Fuel __,.
1------- Air

I
I
Air--cL...'.._

(a)

346
-
ses
[1 5. 6] Combustion Proces

.
these conditions ' per unit mass ff uel burnt, is tlh :'m (liq) , where 'a ' and 'atm'
fi ference ~
:e er lo the am bien t re
z a1 m(l' ) _ m pe r~ tu re and press ure respectiv
ely. To all
effective
intents and purposes t11 0 (1iq). lf tlho(li q) is regarded as the
;h
ffiiq .-
tlh
latent energy in the fu el' e e c1ency of th e power plant is suitably defined by
w ( 15.13)
1J=--
L1'1o(liq)
t
where W is thteernwork ransfer per unit mass of fue l.
N ow the peraturef, of h. the p d uc ts _1.s never reduced to atmospheric
ro a t exchange
tem pe ra tu re in practic ts wo uld en ta il an in finitel y lar ge he
e~h or \ steam plant).
surface somewhere in the economiser of a
•t· ~ pa nt ( e.g . in
Moreove r th is cond 1 10 n ts no t even approach ed , 1'f 1ror no other reason than
' .
th e un de sirability of 11 r va po ur o
t_ c?nd en ~e . Combust io n products
owmg_ wate in the
usually include a sm~l hu r d1 0x 1de which would di sso lve
:~I ~u an ti ty of su lp -called
wa ter an d fo
su lp hu ric ac id (b ut note the reference to so any
cor~oS iv e ve led m
, con dens. 01 er s' m secti on 11.6 ). Consid
erations of this type ha
. mg 1 ffi . en cy in terms of tlh 0 (vap ),
i.e.
th
e ngmeers to prefer to defi ne erm a e c1
w (15.14)
1J= -- --
t1ho(vap)
er va lue s fo r th e th ermal efficiency because
slightly hi gh
This defin _ition yields q).
r in magnitude than tlh 0 (li all
t1ho (v ap ) 1s sm al le
eff ici en ci es ar e pr oc ess efficien cies and lik e
plan t some
It is clear th at power on s be tw ee n an achieved re sult a~d
are co m pa ris ard,
such efficiencies, they po rta nt alw ay s to sta te clearly which stand
is mos t im plant
ar bi tra ry standard. It ed . A m or e ra tio nal definition of power
), is be in g us at , of the two
t1h 0 (li q) or L\h 0 ( vap .5, fro m wh ic h it is apparent th
in se ct io n 17
efficiency is di sc usse d re, de fi ni tio n (15.14) is to be prefer
red.
pres en te d he lly to steam
ar bi tra ry defin iti ons e fo re go in g discussion applies equa
d th at th
It must be emphasise en gi ne s. For th e form er, Fi g.
15.8a can
rn al-c om bu sti on
power plan t and inte A of th e syste m is re
garded as the
in Fi g. 15 .8b ; pa rt
be an a lysed fur th er as through a
B as th e w or ki ng fluid which is taken
rt t in the conden se r
boiler furn ace and pa he a t re jec te d Q 2 pa sses ou
Most of the
thermodynamic cycle. on pr oc es s its elf ca n be iso lated from th e rest
the combu sti is
cooling water. Whe n r or ga s tu rb in e co mbustion chamber, it
steam boile of
of the plant, as in a of its eff ici en cy so that the performance
e me asur e ost
ne cessary to have som n be co m pa re d. Co m bu stion efficiency is m
systems ca tively
different combustion n of the la ten t en er gy input wh ich is effec
e fracti o on th e
generally de fined as th Th e de ta ile d de finiti on wi ll depend up
man ne r. tput .
utilised in the de sired en er gy in pu t an d the effecti ve ly used ou
the la te nt to produc e
method of designati ng fo r ex am pl e, th e combu stion is used
bo ile r, ca n the refore
In the case of a steam am . Th e combustion efficiency
w ate r in to ste
heat and so transform
d as
be al ternativ ely defi ne ( 15.1 5)
Q Q
- or
= - - - q) - Llh o(vap)
11 - Llho(li
34 7
 . ular Fluids
App li cation s t o Par t ic

. , C"rrcd 10 the water, i.e. the increase in ent halp\,

\\.here Q 1s the hea t trans e . J or the:
, , ·c nnss of fuel sup plied.
wa ta pe r um ~ h t of the adi abatic combust' ·
T o t.1kc ,1 second examp le, t e o b. _~ec .
. ion in a gel
.
tu rhme comb ust10n c am h ber is to increa se the sensib le enthalpy of th fl s
. d . e U1d
flowin g thro ug h it- so that this can be used to pro uce work In the tu b.
• f h ,
. . . r ine.
The efficie ncy may b e defi ned, the refo re' as the rat10
. o t e actual
. incre ·
sensib . le
ent ha IPY per k'logr
1 am of fuel to the mcrea se obtained when<-1se, 0 1
. . each
kil. ogra m offue 1 1.s burn t. comp letely. The energy equat ion 1s
(Hr2 _ HRi) = (HP2 - H p0 ) + '1.h 0 + (HR o - HRi) = 0
With complete combustion , the increase in sensible entha lpy is give n
by
(Hr2 - Hro ) - (HR 1 - HRo) =- '1. ho
With incomplete combu stio n, the increase in sensible entha lpy will be

The sum mation is of unburn t or partially burnt constituents; mi is the

mass of
constituent i per unit mass of fuel, and '1.hoi is its entha lpy of combustion.
The
choice of reference temperature T0 has little effect on the efficiency,
but the
phase of the H 2 0 in the products must be specified. The expres
sion for
comb ustion efficiency becomes

{(Hr2 - Hro) - (HRl - HRo )}actual

11 = -'1.ho ( 15.16)

The enthalpy of combustion can be taken as either .1h (liq ) or

0 6h 0 (vap)
accord ing to conve ntion.
The numera tor in equat ion (15.1 6) is best determined from the res ults
of an
experimental analysis of the products. A second definition of efficiency freque
ntly
employed is the fraction of the actua l fuel used which would have been
sufficient
to achieve the actual produ cts temperature with complete combustion
. Further
discussion of these and oth er ways of expressing gas turbine combu
stion
efficie ncies can be found in Ref. 26.

Exam ple 15.9 Fin d the efficien cy of the comb ustion process of Example 15.5
on the basis
of '1.ho ( vap ) and the percentage of comp lete comb ustion
atta ined. The
follow ing data can be assumed:

'1.ho ( vap ) of the coal =- 24 490 kJ / kg

'1.h 0 of solid carbo n =- 32 790 kJ / kg
'1.h 0 of ca rbon mono xide = - 1 0 11 o kJ ; kg

The efficie ncy may be expressed by

f/=
- 6h0 (vap) - '\' m-( - "h )
L. I l.l 0/
-'1.h 0 (vap)

3
 [ 15 .7 J n P ro ce ss es
C om bu st io

t· f tituents
F ro m th e so lu io n o E x a m PIe 15-5, the u n b u rn t an d PJrtialJy b u rn t con::.
s k u rn t , u" re
in th e p ro d u ct ' p er g o f eo aJ b
in ti d u ct s
0.17 k g o f C O ie g<1seous p ro
.
0 .0 86 8 k g o f C in th e refuse
0.4 X 0.217 =
T h er ef o re
= 4565 k l kg
= (0 1 7 x 10 110 32 790)
L, 117 ; ( - !l lio);
. ) + (0.0 868 x

24 4 9 0 - 45 65
17 = = 0.814
24490

n
15. 7 D iss o c ia ti o
re re ac h ed in
se ct io n 15 th e m ax im u m tem p er at u
d o u t in
lt w as p o in teco m b u st io n ro -~ t~ a~ ss o ci at io n .
by th e ph e n o m en o n o f di
an a d ia b at ic am h p ces~ti oisn lim
ited
d e r fo r ex 1 e re ac
C o n si ' P e, t

c o + t o 2- c o 2 it is ac co m p an ie d b
y

co:
· . ar ro w
ed s in th e d ir · ti·o n indi ca ted by the d ir ec ti o n
W h en it p ro ce
ec
e to p ro ce ed in th e re v er se
ca n be m ad e o f th e
1 se ~ f e n e rg t T h e re ac ti o n o le cu le s of C O 2 . Som
a re ea su p p li ed to m co ll is io n
1 su ff ic ie n t en er ~ y is
ce iv e su ff ic ie nt en er g y in
h o w ev er, ~
d u ct s d o re re ac ti o n
s m th e ~o m b u st10n p ro le s u n d er g o th e
m o le ~ u le of th e m o le cu
u r, 1. e. so m e
fo r this to o cc
C O 2 -+ C O +
f0 2
u at io n as
b o li se d by wri tin g th e eq
is sy m
This p o ss ib il it y
C O + ½0 2 ~
CO 2 y , an d it is
an ab so rp ti o n o f en er g
p an ie d by m p er at u re
se d re ac t io n is ac co m th at a t an y p ar ti cu la r te
T h e re v er fo u n d ti l th e
d o th er m ic re ac ti o n . It is 0 ad j u st th em se lves u n
te rm ed a n en o rt io n s o f C O 2
, C O an d 2
b e r of CO 2 m o
le cu le s
th e p ro p ti l th e n u m
a n d p re ss u re a t th e sa m e ra
te, i.e . u n
e o f st ab le ch
em ic a l
o n s p ro c ee d a ti ng . A st at
tw o re ac ti m b e r d is so ci st a tic o n e,
rm e d is e q u al to th e n u te o f eq u il ib ri u m is n o t a
is
b ei n g fo
en sa id to ex ist. T h e st a sl y a n d si m u lt an eo u sl y. It
is th n co n ti n u o u
eq u il ib r iu m o n s a re g o in g o si gn ifica nt p ro p o
rt i o n
tw o re ac ti , th at a
b ec au se th e u t 1500 K m mi x tu re.
te m p e ra tu re s, ab o v e a b o e a n eq u il ib ri u
ly a t h ig h a te to p ro vi d m b u q io n
on s m u s t di ss o ci le cu les in co
m o le c u le H 0 m o
o f th e C O 2
m a rk s a p p ly
eq u al ly to 2
co m b u st io n
ce d in g re te w h y th e ~d ia b a tic
Th e p re ci a 15. 7.
n o w p o ss ible to ap p re le ca lc ul at io n o f E x am p le
p ro d u ct s. It
is by th e sim p d 0 2•
Je ss th a n th a t p re d ic te d m ix lu re o f CO 2 • CO a n
te m p e ra tu re
is eq u il ib ri u m an d H 2
th e p ro d u c t"> co n tai n an .J T h e p re se n ce o f C O
h e q u es ti o n
2.
E v id en tl y O , H 2 a nc 0

u il ib ri u m m ix ru rc o f H 2 in th e fu el is rele a sed. T a ll
an d a n eq th e ch em ic a l en erg
y
be p re d i ct ed . W e sh
t n o t a ll a ca n
in d ic at e s th a ic al eq u il ib ri d ed u ce d fr o m
th e
h o w th e st at es o f ch em ri u m ca n be
a ri se s as to for eq uil ib th er m o -
sh o w th a t a co n d itio n st ep is to vi s u a lise h o w a
p ro ce ed to cs. T h e first
L aw o f T h e rm o d y n am i
S ec o n d 3 49
 App licat io ns to Pa rti c ula r Fluids [15.7]

d) n..1m1c.i lly re, ers1ble i..: 11l;m1c:. 1 1 . ion cad□ hbe

. • . . react achie ved, and the argurn ,
. ·re cnt
sugge stcJ by \'an ' t Ho IT 111 1887 will be w,e e .

75 . 7. 7 The van 't Hoff equi libriu m box

Cons
. . .~ ·
ider a ~to1ch10metrn.: reaction between CO andCO 0 2 , each initiall y at h
. h t e
~tanda rd press ure pt) and a tem pera ture T.' to .
fo rm
. 2 a t t e same pres sure
and tempera ture. This reaction can be earned o_ut m man
y ways . In the gas
. r
ca Ion mete r, 1or examp , le all the energ y relea sed 1s tra nsfer red as heat and the
. • • 'bl Whe .
rea ct ion 1s 1rreve rs1 e. n the react ion is made reversible, ho wever, we shall
fi nd that some of the energy is transferre d a~ work .
. Fig._ 15-9 illust rates
one concei vable stead y-flo w open system in whic h th e
react wn can proceed
reve rsibly. The react ion itself occu rs in the reactio n _cham
ber, or ' eq~ilib ri um
box', whic h contain s CO , CO and 0 in such prop o rt ions
2 2 that the reactions
CO + ½0 2 -+ CO 2 and CO 2 -+ CO + ½0 2
are proce eding simulta neo usly at equal rates. Th e box is main
tai ned at a constant
tempera tu re T by surro unding it with a reser voir at T. An
y heat exchanges will
therefo re take place reversibly. The total press ure in the
box may have an y
arbit rary value p, whic h may be greater or smaller than
Pe, dete rmin ed by the
tota l mass of gas prese nt. The CO, 0 and CO enter or
2 2 leave the box throu gh
semi permeable membranes, the mass fl ows at entry and
exit bein g equal. If
these mass transfers are to take place reversibly (see secti
on 14.3 ), the pressure
of each cons ti tuent outsi de the box must be equal to its
part ial press ure inside
the box (i.e. Pco , p0 2 or Pco) and the temperatu re of each
cons titue nt outsi de
the box must be equa l to T Since the pressure and temp
erature of the CO, 0
and CO 2 are pe and Tat the boundary, reversible isoth 2
ermal com pressors and
expa nders must be included in the system to main tain the
respecti ve pressures
at Pco , Po 1 and Pco2 outside the semiperm eable mem bra
nes.
Imagi ne that CO and 0 2 are transported slow ly into the
reaction box at the
stead y mola r fl ow rates of nand n/ 2 respecti vely, and tha
t the reacti on proceeds
I 15 9
i van t Hoff equtl1bnu m Surro und ing rese rvo ir at T
Wco Qco
Q
Pco
I r----+----
Expa nder
Mixture of
Compressors CO, 0 and
2 co 2 , .........---4
I at p, T

Equilibrium box
Wo . .
l Qol wCO 2 Qco1
[Drawn for the case where the net reaction
proceeds accor ding to co + J.. O, _ CO
and any partial pressure > pk] - i

350
p

[ 15. 7] Combusti on Processes

!n t~e normal direction. The equilibri um mixture will remain unchanged if CO 2

~s d ischa rged at the steady molar flow rate of n.
Only an infi nitesimal change
m the ex terna l conditions is needed to reverse the expanders a nd co mpressors
and to cause the reaction to proceed in the oppos ite di rection. All the quantities
of hea t and wo rk tra nsferred across the bounda ry t hen have the same magni tude
but are reversed in sign. T h e whole iso thermal steady-flow process can therefore
be regarded a s reversible.
We shall proceed to show that the power output from thi s hypothetical
reversible plant is the maximum a ttainable from this chemical reaction in
surroundi ngs at the temperatu re T In the course of so doing we shall find th a t
the relative pro portions of constituen ts in the equilibrium box, expressed in
terms of mole fractions , depend upon a p arameter K which is a functi on of T
but not of p. Fo r simplicity we shall assume here that the reaction involves
on ly perfect gases, for which a mole fraction is propo rtional to the partial
press ure, equation ( 14.17).

f 5. 7. 2 Maximum work of a chemical reaction and equilibrium constants

The rate of work tran sfer when a perfect gas undergoes a reversible isot herm al
steady-flo w process is given by equation ( 10.6). When the gas flows at a molar
rate of ri, it ca n be written as
· - ln -P2
= riRT
W
P1
The net p o wer output from the sys tem shown in F ig. 15.9 is therefore
0
e p ) 1
WI= nRT ( 2 ln -
ln Peo2 + In£_ +-
I p0 Pea Po2
0
(Pco)P ) (15.17a)
= riRTln (Pco / P0)( Po l / po)l /2
in the reactio n box is changed to p', but all
N ose that the pressure P . , ,
ow SUPP . h n ged The new partial pressures wil 1 be Pea, Po2,
ot he r cond itions remain unc a .
a nd Pea 2 such that I

Pea + Po 2 + co = p
P 2

a nd the net power output becomes

_ (p'c 0 ) p0 ) (15.17b)
I W'I = riRTln (P'co / pe)(p'o)Pe )1 12
. .,, . I W'I . If the latter, the system produ cing the l~sser
Ei ther I WI = I vJt I or I WI ~ h tw o sys tems co upled together. The combined
power could be reve rsed ~n t e I d produ ce a net amou n t of wo rk while
a te m a eye e a n Th .
system will then oper . I rvoi r of uniform temperatu re T is
·th a singe rese . , h
excha n ging h eat w1 d I WI must be eq ua l to I W. 1. T e same.
nd Law an so
contradict s t h e Seco ' .bl sys tem th at ca n be conceived operating
arg ument would a ppl y to a ny :~vers1 f e e d T. and the power I WI is there fo re
ond1t1ons o P an ,
with the sa m e boun d a ry c
351
 · far Fluids
Applic ations to Particu [1s,7)

. d r,-vm this chemi£:al reaction. When the

· Jar can be obcazne J ' • Proce
the maximum c 1 11 be less·' in the cal orimeter process 1·t 1s
. ~"
· h
is irreversible. t e pow er output WI ·
tero.
Smee I ~v I = Irv I, it foll ows that
I

(Pea, Ip") =
(P co/ Pe)
e , / e )1 ;2
(Pco / pd)(Po ) pe) 112 (p'co/ p HP02 p
Thus the com bination of press ure ratios (Pi ! p1=' ) ab~ve ~s independent of the
total pressu re p m . th e equili brium box. This combmatwn, or pararnete .
.. . r, is
called t he stan da,·d or tliernwdynamic eqwllbrium . constant
. . . for the react·ion
cons1.d ered. Th e symbol used is K e' the superscript 0md1catmg
Th that all part··
1a1
press ures are normalised by the standard pressure p . us

e_ (Pea/ Pe)
K - e I 0 )1 12
(15.1 8a)
(Pea l P HP02 P
It is sometimes usefu l to gather all Pe terms toge th er a nd rewrite equati on
( 15.18a) as
Ke = (Pco) (p e)1/ 2
12 (15.18b)
(PcoHPo/
(as is done in the table headings of Ref. 17 to save space). Although K-:, is
dimensionless, its numerical value will depend on the choice of pe. It is
independen t of pe only for reactions in which np = nR, when the net exponent
of the pe term becomes zero. * In common with other chemical-thermodynamic
da ta which have to be referred to a standard pressure (e.g. ~ut ~hn
the
standard press ure agreed upon internationally is 1 bar. There is nothing in the
fo regoing to suggest that K e is independent of the temperature T of the mix ture,
and ex peri ment sho ws th at it varies strongly with temperature, as we shall see
later.
As anticipated earl ier, we have not on ly show n that f,.;_ t:' -:f. f(p) , but also
derived the important res ult that equati on l I 5. l 7a) exp resses the maxim um
work I Wlmax attaina ble from th is particul ar reactio n. The result can be
generalised to apply to any other chem ica l react ion involving perfect gases for
which the eq uilibri um constant K 0 is kn own and WI can be related to Ke by
J
' max

I W/max = n.R Tln K e (15.19)

This result is discussed furth er m section 17.7 and, for the more advanced
st udent, in section 7.9.3.
In generalising the foregoi ng arguments to other chemical equations, let us

• It useJ to be com~on practice lo define the equilibrium constant in terms of partial pressures.
e.g. in the case com1dered here as K = (p ) (p )( )1 2 Th b • added to imp!~
the above mod O f d fi . . P coi . co Po, . e su script p was _
T hi d e e n1t ion. KP would then in ge neral not be dimensionless unless np - "R ·
c/10:~r~t:m:°o~~ernt ~alues of log KP (o r _In Kp), an inadmissible practice because ar~ume~l;
an; ra nsce ndental fun ction, should always be dimensionless quanti ties. W -
a~e ~::s~~:~
d'd ti · b . 'h

K - K'
is"~~~:~dt to ~rrors? It is easy to show that if the unit in\.\ hic h the partial pressu~;;
d h b'
0 1
e sta nd a rd pressure adopted (it used to be I atm ), then oumi'ru u
P - an I e reach ts th us swept under th e carpet.

352
 ss es
(1 5 .7] Co mb us tio n Pro ce

etr ic eq ua r·ion s uch as

co ns ide r an y sto ich iom
I f B ;;::::::- \, km o l of C + km o l of D
\1 km o l of A +\ · :: 1-..rn 0 o
Y4

effici en b
a re call ed t h e sl ot Lh wm et ric col.'jj 1ue n1 ,·' the co
wh e re V i' vz ' 1'3 an d v4 · k tho se of the rea cta n ts
o f t h e pr.od uc t s a re by
co en t 1on
nv ta en as pos1t1v. c ' an d
th en a rri ve at th e ex prcs<;1on
as ne ga t ive. W e wo uld
)' •
K o = (pc / p'' )'' (Po , p''
6 3
(pA lp J' ' (Ps / P' )' '
ly as
itt en ev en mo re ge ne ral
An d th is co uld be wr
( 15.20)
(nPd P
0
Y' or In K "' = L In( P, p"" )'-
Ke = i
ue nts i, an d L,
IS the su m o f th
e
wh ere ni me an s the pro
te rm s.
d uc t for all co ns tit
i t hm ic for ex a mp le.
on H 2 + ½0 2 ~ H 2 0,
log ar
( 15 .20 ) to the rea cti
Ap p lying eq ua tio n
ri um co ns tant
we ge t fo r th e eq u il ib 1 2
0
_ (PH 2 0HP '
) ( 15.2 1)
K e = ( PH 2o f P )
2
0 0 112
(PH J( Po Y '
(PH ) P HPo ) P ) -
va lue s of
en tab ula tin g ex pe ri me nta lly de ter m ine d
t wh co ns tan t K ~ wo u ld
It sh o uld be cle a r th a rea ction. Th e eq uil ibr ium
ry to sp ec ify the ~2 H 2 0 or
K , it is ne ce ssa en as, sa y, 2H 2 + 0 2
0

rea cti on we re wr itt

if the ua re of th at gi ve n
ha ve dif fer en t va l ue s ca se K wo uld be the sq
0
fo rm er
H 2 0 ~ H 2 + ½0 2 In the
.
rec ipr oc al of it.
a nd in the lat ter it wo uld be the
by eq ua tio n ( 15.2 1 ), mo le fra ct io ns
the pa rti al pr ess ure s are pr o po rti on al to the
Re me mb e r ing th a t th e am ou nt
t K is a me as u re of
0

( 14. 17 ), it is ev ide nt tha

as g ive n by eq u ati on la rge pr op or tio n
va lue im pl ies th at the mi xtu re co nt a ins a
o f di sso cia t ion . A h igh th er mi c
d so the re is li ttle dis so cia tio n. Fo r e xo
uc ts an ex pe ct K e to
of un d iss oc iat ed pr od en co ns ide r ing , we wo ul d
os e we ha ve be
re ac tio ns, su c h as th be ca use mo re m ole cu les
of pr od uc ts wo uld
as e of tem pe ra tu re
de crea se wit h in cre
I I -
15 10
so I'~
(Values of K with pe
0 = l ba r )
: ;i
40 ~
: -::n of ec u br um

"' ~
30
·s•a · • 1• 'h ter rpe rat ure
20
ln K '~ ~ ........
"' -" ~, 1-,
.___ -..,;;'. -;.. I
10

"' ... - ~o

,~ ----
1 .!:!.20
8
() C..... ".,..~ ......

➔
ro ,>

2
~
2500 30 00
20()( l
I 1000 1500
Te mp era ture [ K J
0 500

35 3
 . Particular Fl ui ds
App licati ons to
.. . ·. Th " b (1 5)
, . ,LJllision!:> to dissociate. at K co anct ·;
, , ( >f1Cfg} 111 C • . -· 2 /(_ •
reL~I\ C•rnfi ,H.: n 1; _ h , p... rJturcs is indicated rn F 1g. 15.10. It . 1-iio dee
11
, r } h1g tt:m .., . 1s us rt
strongly Jt ,e f K'' rather th an values of K ' . Thi !:> is b Uai ta '\
1
Jnd ta buJ,1r~ _\'alu~!:, lol I~ope more easily wit h the wide range efcause g/lal
d I lJ11thn11ca Y l: o v · <tpt
plotte o~ . t, pola tion from tables of In K e provides valu ariation ,~
f... ·ind lrncar rn t.:r b • d f es Wh· or
·' , 1 es than wo uld be o tame rom linear . 1ch .
clo,er to the true va u . Inter c1rt
, . f ;.:_ e Moreover, thermodynamic theory norrn 11 Polc1ti 0
bet,, een \ alucs o . . fI e a y re n
. f l K ' (see eq uation (15.19)). Values. o n K for several reacti 9Uire,~
, J lue:, o n •
Om b ustion are tabulated against temperature in R f 0 ns of
importance ll1 c . e . 17
r ,· g example ill ustrates the method of calculatmg . ·. .
the equ1Jib
Th e 1011o~ m . . .
the percentage dissociatwn has been obtained by a riuni
cons t an t When .. . Proct
. , , .• and also how this eqw1Ibnum constant can then be used t o Prect. Ucts
ana1}-,1s, . . . .
the percen tage dissociat10n m other cJrcumstances. Jct

:xample 15.10 The products fro m the combustio n of a stoichiometric mixture of CO

· T and
o2 are at a press ure of 1 ba r and a certain tem perature. he products a 1 .
. d' . nays1s
shows th at 35 per ce nt of each kmo I o f Co 2 1s 1ssoc1 at ed. Determin
.. . . d h t· e the
equr l1bn um constan t for thrs tem perature, an t ence ind th e percen ta
dissociati on w hen th e products are at the same temperature but compress;e
to 10bar. d

The combustion equation for the reaction to products in equilibrium is

CO+ ½0 2 ➔ (1 - a)CO 2 + aCO + ½aO 2
where a is the fraction of CO 2 dissociated. At 1 bar the products consist
of
0.65 kmol CO 2 + 0.35 kmol CO+ 0.175 kmol 0 2
The total amoun t of substance of products is l.1 75 kmol, and the partial
pressures of the constituen ts in bar are therefore

2
nco 0.65
Pea 2 = - p = - - x 1[bar] = 0 553 ba1·
n 1.175 ·
ncp 0.35
Pea = -;;P = l.1 x l [bar] = 0.298 bar
75
no 2 0. 175
P02 = - n P = 1. l 7 x l[b ar J = 0.149 bar
5
Based on the equilibri um equation CO + 10
2
_,, CO 2 ,
z...--
Ke = (Pco)( pe)112 0.553
(Pco)(Po) 112 = 0.298( 0.149)112 = ·
4 81
At any pressure p th •
' e partJal pressures will be
1- a
Pco === - --
1 + a/ 2 P, Pco =- a
- a
/2
1 + a/2 p, = -1+-a/-2 p
2
Po 2

354
[ 15. 7]
Comb ustion Processes

And the equilibrium constant will be

l - a )
(
K ti = 1 + a/ 2 p(p'-' )112
( 15. 22)

C/a;2)C : :;J x 1 12
Px p'"
Since the tem perature is unchanged, Ke wi ll still equal 4.81. At 10 bar,
therefo re,

4. 81 = (l - a )( 2 + a )1 12 I
a3 / 2 X 101 '2

Squ aring and simplifying, we have

230.4a 3 + 3a = 2
a = 0.185. Thus the
a?d t~e ~olu~ion, by trial and error or graphical mean s, is
d1ssoc1 at1on 1s 18.5 per cent of the co 2 .

that although the

The preceding exam ple illustra tes the important point
reacting mixture, the
equilibrium constant is indep endent of the pressure of the
pressure. Inspection
actual fraction of the disso ciated product usually varies with
containing p only
of equa tion ( 15.22) in the example shows that the term
g comp ound and of
va nishes when the amount-of-s ubstance of dissociatin
dissociated constituents is the same.
ants enables the
We may now see how a know ledge of the equili brium const
ustion temperature
combustio n equation to be esta blishe d, and the adia batic comb
react ants are liquid
to be pred icted, for the reaction of Example 15.7. The
that some of the CO 2
hepta ne and air in stoichiometric proportion s. Assum ing
tion will be
and H O mo lecules have dissociated, the co mbustion equa
2

79 79
C 7 H 16 + I 10 2 +- 11 N 2 -, aCO 2 + hH 2 0 + cCO + dO 2 + eH 2 + -21 11 N 2
21
can be written do wn
There are five unknowns: a, b, c, d and e. Three eq uati ons
gen and oxygen on
directly by eq ua ting the numb er of atom s of carbon, hydro
each side of the eq uation :

a+c = 7
2b + 2e = 16
2a + b + c + 2d = 22
aid of th e equilibrium
The remaining equa tions can be formul ated with the
2 , CO and 0 2 , and
cons tants whic h establish the proportions in which CO
amount-of-substance
H o, H ·and 0 , can exist in equilibriu m. If n is the total
2
2 2
of prod uc ts. given by
79
11
n=a +b+ c+d +e+ 21
355
 Application s to Particula r Flu ids [15 .7]

then the partial pressures are

a C
Pco 2 = - P, Pco = - P, etc.
n n
The two equations are therefo re
a (p '=' )1 12
K ~o2 = c(d/ n)1 12 p
b (Po)112
K ~20 = e(d/n)1 12 p
Since p appears in these equations, the pressure at which the reaction occurs
must be known. (In Example 15. 7 this was irrelevan t because the enthalpies
can be assumed to be independent of pressure.) Assuming the pressure is
specified, there is still the difficulty that the dissociation constants are functions
of the unknown temperature. The method of solution is to assume a fin al
temperatu re, obtain the appropria te values of K e from tables, e.g. Ref. 17, and
then solve the five simultaneous equations for a, b, c, d and e. Once the
composition of the products is known, the final temperature can be computed
from the energy equation. If this does not agree with the assumed temperature,
a better approximation to the values of K e can be obtained and the calculation
repeated until the desired agreement is reached.
The energy equation for the reaction in Example 15.7 must now be written
more generally as
(Hpr - H p29s ) + fiH 29s + (HR 29 s - HR 2ss ) = 0
tiH 2 98 replaces t,Ji298 because, although we are still relating the solut ion to
1 kmo1 of heptane, ~h298 must be modified to make allowance for the unburnt
CO and H 2 in the producls; that is, l~h298 I must be decreased by the magnitudes
of the enthalpies of combustio n of c kmol of CO and e kmol of H . The
2
numerica l solution of such a dissocia ti on pro blem is too lengthy to be given
here, but the reader can find some simple problems in Appendi x B and a variety
of worked examples in Ref. 7.
When the fuel / air mixture is appreciab ly richer than stoichiometric, it will
be found that the amount-of•substanc e d of oxygen in the products is negligible.
A somewha t simpler trea tment involving only one equilibrium constant can
then be used. The argument res ts on the fact that the oxygen produced by the
dissociatio n of CO 2 can be regarded as being wholly used in the combustion
of H 2 . The l wo reaction equati ons

CO+ ½0 2 ~ CO 2 and H2 + ½0 2 ~ H 2 O
can then be combi ned to yield the so-called water-ga s reaction
CO 2 + H 2 ~CO + H 2 0

Wi th the 0 2 elimi~~ te~ fro m the products, one fewer equation is required. and
the water- gas eq u11I bnum constan t is all that is needed for the solution of the

356
 (15 .7 ]
C omb u st ion P roc esse s

sim ult a neou s equ a tio ns. Va lues of

K r' = ( pH 2<?.l_( Pco)

( PH2)( Pco )
a re a lso give n in Ref. 17.
adia ba tic stead y-flow p roce ss a t
The fore goin g meth od of so lutio n is for an
st
a k ~ow n con ant press ure, such as occu
rs in a ram -jct com bu stio n cham ber.
ine co mb us tio n ch am bers, bu t the
(T hts type of process a lso occu rs in gas turb
to be signi fica nt. ) In th e cyl inde rs
tem pera tu res . are too lo w for d issoc ia ti o n
en gines the reac tion occu rs a t
o f m a n ~ recip roca ting inter na l-com bus tion
pressure is un know n . Th e followi ng
subs ta n tially cons ta n t volu me and the final
mon oxid e as the fuel for simp licity .
exam ple d eals wi th t his case, usin g carb on

fro m th e ad iaba tic com bu stio n of a

Exa mp le 1 5.11 The an a ly si s o f the p ro du cts obta ined
c on stan t vo lume sho wed that th ere
stoi chi o met ric m ixt ure of CO and air at
CO and 0 .0 9 6 km o l of 0 2, per km o l
we re 0 .808 k mol of CO 2 , 0 .192 km ol of
re and temp eratu re of the reac ta nts
of CO in the reac tant s. The initial pressu
eratu re wa s foun d to be 2 742 K.
w ere 1 atm and 333 K, and the f inal temp
atom bala nce an d by verif y ing that
Ch ec k t he accu racy of the resu lts by an
the eq ui li briu m con ditio n is satis fied.

per kmo l of CO, a re

The reac tant s for a stoic hiom etric mix ture,

co + t o , + G)~N,
is there fore
The com bust ion eq uatio n fo r the reac tion
+ 0.09 6 0 + 1.881 N 2
CO + 0.5 0 + l. 881 N 2 -0.808 CO 2 + 0.19 2 CO
2
2

C heck ing the a tom bala nce we ha ve

(C) 0. 808 + 0. 192 = I
( o) 1.616 + 0. 192 + 0.192 = 2

The re le vant p a rtial p ressu res a re

0.1 92 0.096
808 = -n- p, Po i= - n- p
0.
Pco = - - p, Pco
i n
perfe ct gase s. and
reac ta nt s a nd pro d uc t s a re assu med to be
Sinc e the
t he rea cti o n occ urs a t con sta nt volu me.
nR T
n RT1 = -
l -
P1 P
u bsta nce of
re '"ers to the ini ti a l cond i ti o ns. Th e am o unt -of-s
\.\ here subscrip t 1 1 '

rea cta nts is

n = 1 + 0 _5 + t. 881 = 3.38 1 kmo l
1
35 7
 App licat ions to Part1c ular Fluid s

-1 5 bar and T1 == 333 K. Hence

and P1 = I ··1tn1
== J.l) 13 -
(' T 1.0l3 25 [ba r ] x 2742[ K ] = .4
1 2 68 bar
p = - _- == [1-.mol] x 333 [K ]
/l fl1f1 3. 381 kmo l

The e4u1 libri um co(;),s (~p)

K" = ( p;~ )(:~• r = ( n~o p)(11~, ;0 r
0.808 [kmol] =8
= - - - --(;- [ba r] 0.0 96[ kmo l]) 112 ·
65
0. 192 [ kmol ] 2.468 [ kmol]
1[ba r]
This is in good agreement with the linearly
interpol ated valu e obta ined
from Ref. 17, which is 8.64.
The check can be completed by showing
that the energy equation is
satisfied . For this adiabatic constant volume proc
ess, the energy equation will be
( Up2 4 72 - UP2 9s ) + L~..U 29a + (UR 29B
- UR3 33) = 0
tlU 2 5 is the inte rnal energy of combustion
of 0.808 kmol of CO , per kmol of
CO supplied in the reactant s.

The treatme nt of the combustion of hydrocarbon

fuel s in air given here requires
some qualifica tion. First, at high temperature
s nitrogen is not the inert gas we
have hith er supposed it to be; some of it com
bines with oxygen to form nitric
oxide acco rding to the equation ½0 + ½
2 N 2 ~ NO . The combination is
end othermi c and causes a reduction in produc
ts tempera ture. Seco ndly , some
of the molecules of water vapour dissociate no
t only into hydrogen and oxygen,
but also into hydrogen and hydroxyl (½H +
2 OH ~ H 2 0 ). And finall y, some
of the mol ecules of oxy gen, nitroge n and hydroge
n dissocia te into their resp ective
atoms. The dissociation con stants for all thes
e reaction s are known, and their
effect on the products temperature can be take
n into account in more accurate
calculat ions.
It mus t be emp hasi sed that dissociation does
not affect the internal energies
and enthalpies of combustion , which are dete
rmined at about atmo spheric
te?1perature. In general, if the products of com
bus tion of a hydrocarbon fuel-air
r:11xture a:e coo led below 1500 K before leaving
the system, the chemical ene:g y
~rberated 15 not ?1uc~ affected by dissocia tion
. As the products of combu~tio~
are cooled, the d1 ssoc1 ated prod ucts recombine
and the combustion is effect1vel_)
complete below l 500 K .* The efficiency of
th ercfo re not the com bustion process itsel~is
aITected by any diss ocia tion that may occ
ur due to a high
tempera ture reac hed at so me inte rmedi ate stag
e. The efficienc y of a power plant
* When the product s are cooled
wate r-coo led sarn Jin .d
very rnp1 ly, however, as when a sample is through a
. . p g tube , th e react ion withdrawn ,er , llttl~
recom binat,on takes place. rates may be reduced to such an exten t th at' )

358
 es
[15 .8] Combustion Process

process oc
in wh ich the combust ion rkin fl ~ur~ may be affect ed, howeve r, becau se the
heat transferred to the wo now transferre~ at a lower
average
ma xim um ~e uid is the
tempe ra ture - the mp era tur e reached bemg reduced by
. · t· I th word s th · ted is the same as if
d1ssoc1a 10n. n. o er . . e quantity of energ y libera
1

there were no d1ssoc1at1o n but it s effiect1·ve ne ss fo r producin g wo rk is dimin ished .

. ' · the
For steam .or gas turbin.e
plant s th ·is point· 1s of no im portance owing to· ·
ma.ximum tempera, tur es im · posed by the metallurgic al limi t.
use of .relati ve ly low
. · n
· t10
· socia
.
4 a d 18 4 2 how ever ind· ica· te that d1s
ss10ns m sec tio ns 17 · 'n . . , · n
The d1 . scu lim iti ng the ffi · · ern al-combust10
· oc' ati·ng mt
f rec ipr
is an. important fac tor e c1e ncy o
roc ke t mo tor s.
en gmes and

ta
od yn am ic re ac t ion da
15.8 Ta bu la tio n of th er m
with
sin g tha t alt ho ug h thi s chap ter has bee n concerned
It is wo~th emphasi on known as combustion, mo
st of the
ss of ch em ica l rea cti
the pa rticular cla ica l reaction s in gene ral. Th
e enthalpy
ap pli es eq ua lly to ch em
theory presented ed to as the en thalpy of reacti
on. In
ex am ple , is the n ref err
of comb ustion, for tab ulate the enthalpies of all
reaction s
xt , it is im pra cti ca ble to
this wide r conte s. Instead it is
ica l co mp ou nd ma y un de rgo with other substance
which a chem shown, it is
of for ma tio n D.h rn wh ich is tabulated and, as will be
the enthalpy rticular case
tte r to cal cu lat e the en tha lpy of reaction M10 in any pa
a simple ma on .
s of D.h rn of the sub sta nc es taking part in the reacti
from the va lue
tio n is de fin ed as the inc rease in enth alpy when a
The enthalpy of forma ele me nts in their natural forms and
m its co nst itu en t
co mpound is forme d fro pre sse d in energy units per un
it of
It is us ua lly ex
in a standard state. d is referred to the ca se where
each of
ce of co mp ou nd , an
amount -of-substan an d T0 = 25 °C = 298.15 K
and the
nts is at p 0
= 1 ba r
the reacting eleme tem pera tu re. The qua lification
'natural
sa me pr ess ure an d
product is at the us hydrogen
tha t it is the en tha lpy of fo rm ation of, say, gaseo
fo rms' impli es na tornic gas H, which is pu
t equ al to
th at of the dis soc iut ed mo
H 2 , and not ms ' and on ly
ur se som e ele me nts ma y ex ist in severa l ·nat ura l for
zero. Of co
sta te; fo r ex am ple , for ca rbon it is graphite and
one ca n be us ed as a datum
for this purpose .
not diamond which is used co mp ou nd s at T0 = 25 °Care as follows:
tio n of thr ee
The enthalpies of for ma
hf25)c H. = -7 48 70 kJ / km
ol of CH 4
ite ) + 2H --+ CH ; (!l
C( graph 2 4

(tl hr~ 5 )co2 = -3 93 520 kJ

/ kmol of CO 2
C( grap hit e) + 0 2
--+ CO 2 ;

o = -2 4l 830kJ / kmoJ
of H 2 0
O( va p); (tlh ~ s)H
H 2 + ½0 2 --+ 2 H 2

lp y is ind ep en de nt of the process by which the end

Since the ch ange of entha on in two stage s: a breaking up
of the
can vie w a rea cti
state is reached, we a rec ombin ation to form the pro
ducts.
me nts , fol low ed by
reacta nts into ele of methane and oxygen we
can write
the en tha lpy of rea cti on
Thus to find
~H + 20 2 ....... CO 2 + 2H1
O(vap)
CH 4 --+ 20 2 ....... C + 2

359
 .c ular Fl u ids
. Part 1
Applications to

and hence ( Ah- ) _ (~hf25)c H,. - 2 (~ hr2 5 )0

- ) . + 2 Ll f25 H "O 2

D.h2s = (~hm co 2
_ ( - 74 870) - 0
= (-393 520) + 2( - 2418 30 )

=- 802 310 kl / kmol of CH4

. porn
t
. ts must be made here. First, the. reader is reminded that
Two 1mportan b t died on a second readmg of the book. The same
· · tended to e s u
7 15 .
Chapter m .d { this chapter which requires an understandin 9, in.
applies to the remam er q
79
particular, of sectiohn · d.3. f section !5.4.1 it was suggested that the superscr
Secondly near t e en o e 1b ipt
. h ; rs to properties at the standard press ure p = ar, can safel y be
a wh1c re,e · th t · h
' . d. Th'is was because we were gomg to assume a 'dm t e absence of
om1tte
changes of phase, the Properties u and H of . reactants an products . .can be
· depen dent of pressu re· In the follo wmg we shall
ta ken as m . be deahng With two
properties, namely the entropy S and the Gibbs funct10n G ( or their molar
equivalents g and g), and even for perfect gases these are st~ongly dependent
on pressure. In view of this we shall revert ~o the more ngorous notation
incorporating the superscript e for propert1e_s at standard pressure. For
consistency the superscript will be attached not Just to S _a n~ G, b~t also to U
and H , although in the latter case we shall neglect vanat10ns with pressure
where appropriate.
We have seen in section 7.9.3 that data for enthalpy of reaction ~nt change
in Gibbs function t1gi and equilibri um constant Ki are in terchangeable. Thus
from equations (7.72), (7. 73) and (7.75),
- ~gi = - t1hi + To(s;o - s:o ) (15.23)
A ~e
lnK 0
e
= - -u go ( 15.24)
RT0
By virtue of the Third Law (sec section 7.9. 1}, which enables a bsolute entro pies
to be determined for the rcactan ls and products rela ti ve to a common datum
at absolu te zero of temperature, the entro py term in equation (15.23) can be
calculated. The Gibbs function of fo rma tio n /j gf0 can then be defi ned in an
analogous manner to the value of /j }zf0 ; a nd the values of /j g for any reaction
6
can be fou n~ by adding the fo rmation values as for /j)zi . Conseq uently it follows
fro~ . e~uatwn ( 15.24) that it is also possible to define logarithms of th e
equ~l~br~um constant of formation In K?0 , from which the logarithm of th e
m con stant Ki for any reaction can be calcula ted from a similar
sequJhbr~u
ummation procedure.* A short compilation of various chemical- thermodynamic
data fro m Refs 36 and 37 15· · • . ~e A-➔
given in Fig. 15. 11 ; it contains values of ~h ro , uBro•
* Because values of In K ~ can be C
foll ows that values of Ke can b C nd
. .
ound fro m add1t1ons of the appropria te values of In K'iear
-
e. ic

·l~b ~u by multiplication of the K~ values. Th is sho uld also be c '-le

from the defini tion of th~ equ 11
·1 ·b · c . constant in
n um . •
· 15.7.2 since by muluphcat10n
· section of su1tJvnt
equ i 1 n um constants and
b ance11 at1on of par t·1a 1pressure terms, any req
' .
uired equ11lbnum
consta
can e foun d.

360
[15 . 8 ] ses
Co mb ustio n Pro ces

at P - 1 ba r and T0
-3
298. IS K
-e - t-•
n1 t:.ii ;~ t:.g~ CpO So

[k:goi] [k:l~ll] [k~oi] ln K ~ [km~I K] [kmk~I K]

12.011 0 8.53 5.69
C (grap hi te) 0 0
C (diamond) 12.011 J 900 2 870 - 1.1 57
-27 0.098
6.06
20.84
2.44
158.10
C (ga s) 12.0 11 714 990 669 570
20.498 35.64 186.26
CH 4 (gas) 16.043 - 74 870 -5 0 810 219.33
28.054 52470 68 350 - 27.5 73 42.89
C 2 H 4 (gas ) 197.65
28. 01 05 - J 105 30 - 137 160 55.33 1 29. 14
CO (gas ) 213.80
44.01 0 -39 3 520 -3 94 390 159.09 3 37. 13
CO 2 (ga s) 114 .71
1.008 217 990 203 290 - 82.003 20.79
H (gas) 28.84 130 68
H 2 (ga s) 2.0 16 0 0 0
29.99 183 61
OH (gas) 17.005 39 710 35 010 -1 4.122 70.00
- 23 7 150 95.660 75.32
H 2 0 (liq) 18.01 55 -28 5 820 188 .8 3
-22 8 590 92.207 33. 58
H 2 0 (va p) 18.0155 -2 4 1 830 20.79 153 .30
14. 0065 472650 455 500 - 183 .740
N (gas) 29.21 191.61
28.01 3 0 0 0 21 0.76
N 2 (gas ) - 34.933 29.84
NO (ga s) 30.006 90 290 86 600 161.06
- 93.48 1 21. 91
15.999 5 249 170 23 [ 750 205. 14
0 (gas) 0 29.3 7
31. 999 0 0
0 2 (gas )

he at ca pa cit y ce an d the ab solute molar entro py

lar
In K fo and also of the mo
O

of pe = l bar a;d T0 = 29
8.15 K.
sta nd ard sta te are
~ . all in the
an d the eq uil ibr ium co nstant for perfe ct gases
Both the enthalp y the da ta for L\hf0 and ln Kfo
can be
ure , an d in pr ac tic e
independen t of press an d va pours over wide ranges of pre
ssure.
tan t for rea l ga ses
assumed to be cons ar ly ind ep endent of pressure for liquid
s an d
lue s are ve ry ne
Similarly , th ese va r, inv olves an entropy term wh
ich varies
s fu nc tio n, ho we ve s and
so lid s. The Gibb s su bs tan ces (although not for liquid
ess ur e fo r ga se ou
apprecia bly with pr int o ac count when any gaseou s rea
ctant s
tio n mu st be tak en
solids) , and thi s varia r.
e stand ard pressure of I ba
or products arc not at th reg ard to substances which cannot
exist
ap pe ar s wi th
A parti cular di fficulty ca nn ot ex ist at 1 bar at 25 °C, and
us 2 H 0 va po ur
at the standard state. Th on ly as a co m pressed liquid . The table
s
te ca n ex ist
H 2 O in the sta ndard sta nd ar d sta te, obtained by assumin g
vapour at the sta
mak e us e of a fictitiou s s be tw een the saturated vapour
state
ve s as a pe rfe ct ga
that the substance beha H O) an d the standard state. This
is
in the ca se of
at 25 °C (0.03166 bar
2

es ted on ly in us ing tab ulated data to calcul ate

inter
pe rm iss ibl e beca use we are us co nsider the differences betw
een the
am ple , let
changes in Ii or g. As an ex n,
va lue s for wa ter . Fo r the enthalp y of form atio
ur
tabu lated liq uid and vapo
for brevity, we have
omitt ing th e su bscrip t 25
1 bar
(dl1f tap at l ba r - (dhf} 1iq at
hrtap at 0.03166bar }
-- ).r( uAfi8f )vap at 1 ba r - (d
}
(d h) at O 03 l 66 ba r - (dhr )liq at 0.031 66 bar
+ l.r f va p ·
- t?
l
b 1.J
ar
r - (d hr ) iq at
+ {(dhr)liq at 0.031 66 ba
1

of H 2 0
= O+ hrg + 0 = 43 990 kJ / kmol
361
 p · far Fluid s (15_
Appl icatio ns to artic u
81
enthalpy of vapo ur anct of li
. d term s are zero becau!>e the 'ddl
d h
The fir~t an t ir d nt of pressure· the m1 e term repre sen ts t he la9U1d
. . d ' •d •
1s tak en as in cpen e
. ation per kmo l of H 2 0. Cons1 enng the Gibb s fu nct1.onten t
en thalpy o f va po ns of
formati on, aga in at 25 °C, we have
(.tlgn ap at I bar - (.tlJ f )liq at l bar
= {(.tlgf)vap at I bar _ ('1firtap at 0.03 166 bar }
- (L1iJr)1iq a t 0.011 66 bar}
+ 'l (i1g~ f )vap at 0.03 166 bar
) . at O· 031 66 bar - (.tlg?)iiq at 1 bar }
+ l' (i1c'i::1 filq
1
= 8.3 145 x 298.1 5[ _!!___ ] 1n + 0 + 0 = 8559kJ/ kmolofH 2 0
kmo l 0.03166
s vapour state using ( 15_23 !
The fi rs t term represen ts extrapolation to the fict itiou
ct gas over that rang e. The
and ass uming tha t the vapo ur beha ves as a perfe
ins constant in an isothermal
seco nd term is zero because th e Gibb s function rema
the third term is zero becau se
chan ge of phase (see secti ons 7. 7 and 7.9. l ), and
ent of press ure.
the Gibb s function of a liqui d is taken as independ
writ e
Fina Jly, considering the entropy at 25 °C we can
6
(s'0l ,ap at 1 bar - (s )1iq at 1 bar
= {(s )vap at 1 bar - (s)va p at 0.03166 bar }
6

+ {(s)vap at 0.031 66 bar - (s)liq at 0.031 66 bar}

+ {(.s) 1iq at 0.03 166 bar - (s0 )liq at I bar}

= _ 8_31 45 [ kJ Jin 0.03I166 + 43 990[ kJ / kmolJ + 0

298. IS[K J
kmo J K

= (-28 .7 1 + 147.54 )[kJ / kmo l KJ = 11 8.8.l kJ knwl K

t:1 111 at 1:i, the chan ge in
Or, sin ce g = ii - TI and th e lcmp cra turc is cons
entropy can also be found from

i1ii
0
i1g" (43 990 -8559 )/ kJ / k mol J
- - -= - - - =- 11 8.83 kJ; kmolK
To T0 298. I 5 [ K J
thermodynamic reaction data
We conclude this brief explanation of the way
use to calcu late the value s of
J re presented with an example illustrating their
re and press ure which
.1Fi, J~7 Jnd K for a reaction at a specified temperatu
differ., from th e standard sta te.

Exa mpl e 15.1 2 Calc ulate Jh, Jg and In K • at 298 .15 K for the reac tion

CO 2+ H2 ---+ CO + H2O(v ap)

and prod ucts is 1 ba r each .
wh ere th e tota l pressure of the reactants
for cons isten cy. .
Use the data on Fig. 15.11 and chec k them tion
Also calc ulate the value of In K.=. at T
= 400 K, us ing first equa

362
[1 5.8 ]
Combustion Processes

( 15.2 4 ) an d se co nd ly eq ua tio ( 7· 77) . ass · lar

· · ~
ns ta
n
nt , um e in each case tha t the mo
ac1 t1e s cP are co n
hea t cap ov er the ter perature ran ge considered.

equilibrium
Enthalpy, Gibb s f un ction ande h , con5 rant_ at 298 .15 K
Th e en tha lpy of reaction !).h 2 98 w en each co nstitu ent is at I bar is
_
830) - ( -3 93 520 ) - 0 }[k l ]
0

d h2 9a = {( - 110 530) + ( - 241

= +41 160 kl
also be equa l to dh2 98 ~or th e specified pressu re of each rea ctanen
· t and
This wi ll (.
d h.
e 0.5 ba r each), bec a use the enthalpy can be tak as
pro uct m t is cas
mdependen t of pr essu re.
dg ! 98 is giv en by
The ch ange of Gibbs fun cti on
8590 )- ( -3 94 39 0) - O}[kJ J
d i}!9s = {( - 137 160) + (- 22
= + 28 640 kl
eac h at
ch an ge of Gi bb s fun cti on fo r the reactants and produ cts
To find the l ha ve to be
9
modified by nR T0 ln (p I p),
of eac h ter m wil
0.5 ba r, the Gibbs fun ction ob vio us tha t these mod ificati ons (with
d p = 0.5 bar . It is
wh ere p = 1 ba r an
0
l be a
ns) wil l can cel in thi s cas e and dg 298 = t1g~98 ; there wil
ap pr op ria te sig h tha t the am ount-of-substanc
e of the
ly wh en the rea ction is suc
ne t effect on
tha t of the gaseou s reactants.
gaseous pr od ucts differs from
Th e equilibrium constant is
.09 3 - 0 = - 11.555
ln K 9 = 55.33 1 + 92.207 - 159
23 );
To check for co nsiste ncy,
let us first apply equation ( 15.

- ~l"i29 8 = - d92 98 - To (I
p
11 iSi29 B - L l1;Si 2 9B )
R

(l x 197 .65)
= - 28640[ kJ ] - 298.15[KJ{
130.68) }[ ~]
+ (1 X 188 .83 ) - () x 2 13.80) - (J X

= {- 28 640 - (298.15 X 42.00 )}[ kJ ] = -4 11 60 kJ

4
Also, from eq ua tio n ( 15.2 ),
28 640 = - 11. 553
~U29s
x2 98 .15
ln K = - R To = -8 .3145
0

at 400 K e at T = 0 K usi ng equa ti on ( 15.24 ), ~e will

Equilibrium constant 40 cP from
the val ue of In K r S ) at T.' taking the va lues of
To calc ula te . Ah d (S PT - R T
ua tio ns for l.l T an
fi rst write eq
15. 10),
Fig . 15. 11. Fr om eq uatio n ( 41
~
33.)B )} K
[k.,J](T -
193_ 15 [K J) +
-
l60 [K J

t1hT = {( l x 29. 14) + ( I x

36 3
 . Pa rtic ul ar Fl ui ds
Applications to

,., 84 )}[kJ ](298.15 [KJ - T)

I 1)
, 37 .. + (I x - 8· K
Tl ( I X

. ,, ( _!_ _ 298.15)} [ kJ ] = (42130 - 3.2s_!_)kJ

= {-+ I 160 - 3. .. 5 \ [ K] [ K]
. h ressure te rms cancel in thi s case,
Jnd remembering that t e p

(sPT -
_ , .c .ln-2::,
SRT ) - ~p 11 rr
1
'O
+pl
(I:
p
1l;S;29s - L n; 5;29s) + Ln/ .In To
R
R
pi - -
T
T )klK
= (42.00 - 3.25 ln 298.15[K]

At T = 400 K,
l!:.gT = l!:.hT - T(Sn - SRT)

= {42 130 _ ( 3.25 x 400)} - 400 ( 42.00 - 3.25 In 2 :i~s )rkJJ

= +244 10 kJ
and from ( 15.24 ),
24410
In K; = - - - -- = - 7.340
8.3145 X 400
Alternatively, using equation (7. 77) and treating ~h as a variable between T0
and T,

I d(ln
T
To
K e )= :
R
J !lh d(~)
T
T
To

and hence, using the intermediate result for !lhT above,

1n K 940 0 - ln Ke29 a = - --I

8.3 145 ,,~
JT(
--) d --
42 130 - -3.25
[K] / T T
([K])
_
- -
{ l 421 30
8.3145T/ [KJ - 3.lS ln
([K])}T
T To

=_ 1 (42130 42130 298.15)- +4213

8.3 145 - 4-00- - -29-8.-15 - 3·25 In -4-00- - .

Thu s, using In K e = - 11.555 at 298.15 K from the first part of the example,
In K1oo = 4.2 13 - 11.555 = _ 7_ 342
which agrees closely with the result obt . d a bove.
ame
15.9 Li mitat ions of the th erm 0 d .
vnam,c analysis
In the previous chap ter it was e . . I •sis of
the processes occ urring . mphasised that the th ermodynamic a~a _) d to
. . . . plant 15
1n co ndens ers and a1r-cond1t10mng · 1,rn1te
364
[ 15 ,9]
Co mbus tion Proc esses
state s and energy t "
th e pred ictio n of. final . rans 1ers from k nowledge of the ini tial
tates and certa m oper atmg cond ition s. A
.a
led analysis of the way in
:hic h heat is trans ferre d mus t be und ert;~ re detai
designed. n before an actual plant can be
The ther mod ynam ic anal ysis of com bust ion ·
• • processe s, given · · h.
a simi lar way : it does not e bl mt 1s chap ter,
. 5 limit ed in ·· s
speci·fY th e con d 1t10n
b na. e us to .
1
der whic h the proc ess can actu all
. Y e earn ed out m practice For exam ple'
un air /fuel ratio and . ·t· ·
it cann ot be used to pred ict the m1 1a 1 temp eratu re necessary
. . . . .
for self-1gmt10n to occu r. Nor. has the simple therm o d ynam1c treatment anyt hmg
·d " •
to say abou t. the rate at whic h the proc ess can be c·arne out ; 1or examp 1e, 1t
r which a stable flame is
cann ot pred ict flam e spe~ ds, or the cond itions unde
omb ustion engines.
possible, or the dela y ~e~iods expe rienced in inter nal-c
the years to explain
A grea t deal of emp inca l data has been amassed over
ustion systems such as
the deta iled phen ome na enco unte red in practical comb
ing engines, gas turbi nes,
coal and oil fired furn aces , gas burn ers, reciprocat
for a synthesis of this mass
ram-jets and rock et mot ors. The concepts necessary
the fundamenta l sciences
of data are now bein g form ulate d with the growth of
y. Quantitatively there is
of com bust ion kine tics and com bust ion wave theor
be applied directly so
still a wide gap to be brid ged before these scien ces can
drawing boar d, but even
that new com bust ion syste ms can be designed on the
to reduce the amo unt of
now they can be used to aid consisten t think ing and
design can be mad e to
ad hoc deve lopm ent whic h is still required before a new
comb ustion kinetics and
work efficiently. Ref. 30 prov ides an intro ducti on to
com bust ion wav e theo ry.

365