82

INTERNAL COMBUSTION ENCINES

INTRODUCTION TO INTERNAL COMBUSTTON ENGINBS 83

(ui ) Measurement of air consumption by air bor method :

l' Any type of eugine or machine whiü <Ierives heat
_
mergr from tüe mmbution of fuel or any other sou¡ce Ih
_ and converts this energy into mechanical work is
terurñ as a freat engine. Volume of air paesing through the orifice, { = 8a0AC, J f
2 The function of a carbuettor is to atomise and
meter the liquid fuel anct mix it with air 6 it enters ttre
injection svstem ofthe engine maintaining unaer and mass ofair passing through the orifice'
aii c-aitior" orop"""tion fuel airproportion approxi-
mate to those conditions, m. = 0.066 C¿ x d .fi,lp1 tg*i"
& 'Ihe two basic ignition systems in onent use an :
(i) Battery orcoil ignition sysrem where, A = Area of orifice, mz
(ii) Magxeto ignition system. d = Diameter oforifice, cm
h = Head ofwater in 'cnf causing the flow
4 Following are the methods ofgoverning I.C. engines
: o = Density ofair in kgi/m! under atmospheric conditions.
(j) Hit and miss method (ii) qua¡t"tot"*tt
(iii) Quantitygoveming.
6' Pre'igrrition is the premature combmtio¡ which starts
before the appücation ofsprk. overheat€d spark
OBJECTTVE TYPE QI,]ESTIONS
plugs and exhaust valv* which are.tlt.
-"i" o*;;f;;;-ignition should be carefully avoided in engines.
ñ A very sudden rise to pressure dwing mmbustior Choose the comect answer :
detonation' The region in whicl¡ detnation oeurs ly metallic hamme¡ ,'ke eound is called are
"."o-pái"a
i"J"lt".t from the sparking plug, and is In a four stroke cycle engine, the four o¡rerations namely suction, compression, expmsion and exhaust
named the'detonation zone'and even with.*"t" ""-ored completed in the number ofrevolutions ofcank shaft equal to
a"t"n"iioo this zone is rarely more than that one
quarter the clearance volume. (o) four (ó) tlree

octane in the nüture (c) two (d oue.

" Tl",::3i:_...Tf;*..T.1Tr"":ntaseof
heptane (low rating), byvolu¡el which
urte" nu."-"
[of isomtane (high raring) and normal
Tli:i In a two stroke cycle engine, the operations namely suctim, compression, expansion and exhaust are
knocks
Delay period or ignition lag is the time imrneüately -iiiffiff
¡on"wi"g iq¡""tio" of fuel during which the ignition completed in the number ofrevolutions ofcrank shalt equal to
prmess is beinginitiated and theprersure (b) ü¡ee
does notris'e buyouá-ñre"atoe itwouldhave due to compression (o) four
(c) üwo (d) o¡e.
Higher the cet"ne rating ofthe fuel leseer is the pmpensity
for diesel knock In general a high octane value In a four stroke cyde S.I. engine the cam shaft ru¡g
implies a low cetane value. (ü) at halftüe sPeed of crank shaft
(o) at the same speed as r¿nk shaft
lo. The, purpose of supercharging-is to rai.ae
the volumetric efEcimry above that value that which can (c) at twice the speed ofcrank shaft (d) at any speed inespective ofcrank shaft speed.
obtained bv nomal aspiration. superc'arging be
of p"t-t ü"."." orit, po- Á"i"."""riv,
very popular and r used only when a to"ge
r*uot ""ei";; 4 Ttre following is an S.I. engine
comperuate altitude los.
orpo*i" J í"ua.a
o" wh"nio." po*""lil¿"¿" t" "" (o) Diesel engine (ó) Pehol engine
ll, Dissociation refers to disintegratio-nof bumt gms (c) Gas engine (d) none ofthe above'
at high temperatures. It is a revemible process a¡d
rnceases with temperature. Dissociation, in geieral,
ar"r'"a . fo"'u of po*u" *a um"ie."y.
- The following is C.L engine
Performance ofLC. engines. Some important (a) üesel engine (ó) Petrol engine
relations :
(c) Gas engine (d) none of the above.
(i) Indicated power (I.p.) np^rr'ANk xro
=
6
kw & In a four stroke cycle petrol engine, during suction strcke
(c) only airis sucked in (ó) only pet¡ol is suc}ed in
(ii) Br¡¡s p.*.' q:6o d)lv mixture ofPet¡ol and air is sucked in (d) none ofthe above.
{n. p.) =
r{E#' o* ". [=ffi**) 7,
(c)
In a four stroke cycle diesel engine, during suction stroke
\ (ó) o¡ly fuel is sucked in
(o) only air is sucked in
(iii) _J,,,,d=
Mechanical efnciency,r¡-*
I.p. fj (c) mixtue of fuel ud air is sucked in (d) none on the above.
(iu) Ihermal eficiency (indicated),r¡r,n I.P. & The two stroke cycle engine has
=
rh, xC (¿) one suction valve ud one exhaust valve operated by one cam
(ó) one suction valve and one exhaust valve operated by two cámr
and thermal elficiency (brake),r¡*.", B.P. (c) only ports covered md mcovered by pision to effect üarging and exhausting
=

where ñf = mass of fuel used in kg/sec. ^¡a 9. lbr

(d) none of the above'
same output, same speed md same compression ¡atio the thermal efficiency of a two stroke cycle
petrol
lthemal engine as compared to that for four stroke cycle petrol engine is
(u) n (ó)less
= tlair-steda¡d (¿) more
(c) same aslong as coirpressionratiois same (d) same as long ro outPutis same'
10. The ratio ofbrake power to indicated power ofar I.C. engine is called
(c) mechanical effi ciency (ó) themal efficiencY
(c) volumetric eflrciencY (d) relative efficiencY

ii
1l
84
r
INTERNAL COMBUST¡ON ENGINES

| ^{mrrnr I
r. (c) z (d) & (¿) 4 (ó) ó (a) 6. (c) I
& (c) e. (ó) lO (c). 7. (a)

_
:nrronnucAr er.r'sTr;G-l
l. Name the two gen"r" r".r""
| Air Standard CYcles
z Discuss thererative adva¡taro "ffieylasicauy
d aris"¿r*t"g&-"ii"üi"JJ.b.ouoo
sp-
difrain principle ?
& baeic ty{s u_t"-¡ *Ju*["]?ñ#'z w,.t mginss. 3.1. Definition of a cycle.3.2. Air standard efrcienry. S'3' The Carnot rycle'3'4",Comtant
ill;H"Jlii:? "od"rt"-"Ioorrhuti-on
*," tuadamental cycle'
"". difiereacss volume or Otto cycle. 3.5. Constant Preasure or Diesel cycle' 3'6' Dual combustio¡
¿L What is the function á.i. bo*pu.o" Jf ot¡o, Diesel antt Dual combustion rycles_Efiiciency versus courpression
porte¡governor.ofa Eoven
- nor?Enumeratethetypeeofgovemorsanddiscusswiüaneatsketcht¡e
ratio--Fór the same compression ratio md the same heat input-For coDstant maximm
5. Dilfermtiate between a flywheel and a govemor. pressure and heat supplieil. 3.8. Atkinson cycle' 3.9. Ericsson cycle' 3'l0 Brayton-cycle'
6, (al the fuction of a carburettor i.fr-Sti"tlrg 3.12. Miller cycle. 3.13' lcnoir cvcle-Highlights-Objective lYpe Quegtiorp
_Jr¿rt€ in a petml engine. "y4". Examples.
(á) Describe a simple Theoretical QuestioeUnsolved
carburettor
with u o""t
Explain with neat sketches the construction "t"t¿r-LJio state its limitaüons.
and worling ofthe following :
(d) Fuel p'mp
& Exptain tie roitowing rerms as apptied 8.1. DEFINITION OF A CYCLE
r" I.c.
Bore, stroke, T.D,C., B.D.C., clearane ,h,*t"t #':: A cycle is defined as a repeated, series of operations occuring in a certain ord.er, lt may be
volume, mept volurne, cor perfect
9. Dxplain withsuit"¡i".r"áo tr" *""H"c;;;;:;;"*k" repeated 6y repeating the procJsses in the samó oriler. The cycle may be of imaginary
ffiffñ:":-*tioandpietonspeed_ In ideal
to. Discuss the difference between i uüo" o. átout errgii". lbe former is called ideal cycle and the latter actual cycle'
11. i.:Tff t"::i$fr h:[*:*,,m;Tlroii:"*:,1;m;* .yJtu att teat losses are prevented and the working substance is assurned to behave like
,,ffiTi ::fi#n:: cyc,e spark a perfect"."i¿"ntal
working substance.
tz Disruss t,'e difference between trreoreticar
l& a¡d actual valve timing diagrams
of a diesel engine ?
the developmmt of two
ü; 3.2. AIR STANDAND EFFICIENCY
ffiJ!"-""" "rote "rg;;;;'; are rhe two main t¡rpes of two sroke
it is ofparamount importancethat the effect ofthe
t4 To compare the effects ofdifferent cycles,
calorific value of the fuel is altogether eiiminated uttd thit can be achieved by considering air
15. iñ"T?:#1T:',r',i*:[:t]ffi::ffi:Tcvcresparkignition(sr)ensine.Howüsindicarorüagran
compare the relative advutages and disadvantages (which is assumed to behave as a f,erfect gas) as the working substance.in the engine cylinder. ?he
offour stroke and two stroke cycle
engines. efficiency of engine usíng aír as ti" workíng medium is kniwn os on "Air standard
efiiciency"'
This efliciency is oftenly called ideal efficiency.
of that cvcle
The actual efñciency of a cycle is always less than the ai¡-standard efficiency
under ideal conditions. This is trken into account by introducing a ne$' term "Relative
effi'
ciency" which is defined as :
Actual thermal efficiencY ...(3.1)
lrehtivc =
Air standard effrciency
The analysis ofall air standard cycles is based upon the following assutnptions:
Assumptions :
1. The gas in the engine cylinder is a perfect gas i.e., it obeys the gas laws and
has con-
stant specific heats.
2. The physical constants ofthe gas in the cylinder are the same as those ofair at
moder'
ate temperatures i.e., the molecular weight of cylinder gas is 29'
cp = 1.005 kJ/kg-K, cp = 0'718 kJ&g-K'
place without
3. The compression and Lxpansion pro."".", are adiabatic and they take
internal friction, i.e., these p¡ocesses are isentropic.
4. No chemical reaction takes place in the cylinder. Heat is supplied or rejected-by
bring-
or a body in contact with cylinder at appropriate points during the
ing a hot body cold
Drocess.

85
86 INTERNAL COMBUSTION ENC¡NES 87.
AIR STANDARD CYCLES

5. The cyclc is considert:d r:lr¡sed with the same'air' always remaining in the cylinder
to Stage (1). Line 1-2 fFig, 3.1 (c)l reprerente the ieotbermal expansion whic,b take¡ place at
repeat the cycle.
temperatuie ?, when source ofheat lf is applied to the end ofcylinder. Heat eupplied in this caee
It.S. THE CARNOT CYCLE is given by R?, log" r and where r ie tbe ratio of oxpansion.
Stage (2). Lino 2-3 represents the epplication of non-conducting cover to the end of the
This cyclc has the lrrgñcsf ¡ressible fficienq and consists of four simple operations namely, cylinder. This is followed by the adiabatic €apansion and the temperature falls from Trta T2,
(a) Isothe¡mal expansion Stage (3). Line 3-4 represents the isot¡ermal compression which takes place when sump
(ó) Adiabatic expansion ,S' is applied to the end ofcylinder. Heat is rejected during this operation whose vslue is given by
(c) Isothermal compression RTrlog" r where r is the ratio ofcompression.
(d) Adiabatic compression. Stage (4). Line 4-1 represents repeated appücation of non-conducting cover and adiabatic
The condition of the ca¡not cycre may be imagined to occur in the foilowing way : compression due to which temperature incceas€s fron ?, to ?t.
One kg of a air is enclosed in the cyliuder which (except at the end) is made of perfect non- It may be noted that ratio of €xpaDEion during isotherrnal 1-2 and r*tio of conpression
conducting material' A source of heat 'II' is supposed to provide unlimited quantity of during isothermal 3-4 must be equal to get a closed cycle.
heaü, non- Fig. 3.1 (b) represents the Carnot cücle on ?'s coordinates.
conducting cover 'C, and a sump 19 which is of infrnite capacity so that its ierrpeiature
remains Now according to law of conservatim of energy,
unchanged irrespective ofthe fact how much heat is supplied to it. The temperature
of source I/ is
?t and the same is of the working substance. Ttre working substance while rejecting heaü to sump Heat supplied = Work done + Heat rejected
',S'has the temperature Tri,e.,tLLe same as that of sump S. Work done = Ileat supplied - Heat rejected
Following are the fozr sú4ges ofthe Ca¡not cycle. Refer Fig. 8.1 (o). =R?r.log"r-RTrlog"r
Work done Rlog.r (T1-72)
Efñciency of cycle = ll*t""pp¡j*d=- n1-t*t
Tt-T,
--TL ...(B.z)

From this equation, it is quite obvious that iftenperature ?, decreasos, efficiency increases
and it if ?, becomes absolute zem which, ofcourae is impossible to attain. Further
becomes 100%
more iú is not possible ío prod.uce an enginc that should work on Cornot's qrcle o.8 it would.
necessitate the píston to tráuel uery sloui$r furing first portion of the forword. stmlrc (isothermal
erpansian) ond. to travel more quíchly duing the renainder of the strohe @d'iabatít ezpansiPn)
which however is not practízable.
Example 8,1. A carnot engine worhing between 400'c dnd, 40"c produces 130 hJ of work.
Determíne :
(í) The engine thermol efficiency.
(íi\ The heat ad.d.ed',
(iíi) The entropy changes during lual rejectinn proc¿ss'
Solution, Temperature, T1= Tr= ¿100 + 273 = 673 K
Temperature, Tr= Tr= 40 + 273 = 313 K
Work produced, lV = 130 hI.
(i) Engine thermel efñciency' q," :

n,^. =
gHg = 0.6:|6 or 68.ó7o.
(Ans)
(ii) Heaü added :
Workdone
rl¿. =
He¡t added
(c) Four stages of camot cycle (ó)T-s üagram
0.5s5 =
Fig.3.1 Healk;¡
.'. Heat added = # = 243 kr. r,lns)
 INTERNAL COMBUSTTON ENG¡NES
AIR STANDARD CYCLES
(iii) Entropy change durlng the heeü rejection proccss,
(S, - S.) : (ii) The volume aú the end of isothermel expansion V,
Heat rejected = Heat added - Work done
:
Heat transferred during isothermal expansion
= 243 - 130 = 118 kJ

313 K

T-S diagram
p-V d¡agram

FiC.3.2 Fig. 3.3. Camot cycle.

Heat rejected
= ?r (Sr - S.) = 113
= p1v, ln(r) - mlTrt' |.9] = 4o x ld ..,..(Given)
(ss - s{) = =;|} = o.B6r rar/K. (Ans.)
\ vr,/
Example 3,2. 0.s
# or 0.5 x 282 x 58b.4 ln fg) l¡a
r¿s of air (ideat gas) e,,zcutes a carnot powera thermarcycre having = 40 x
of 50 per cent. Trú ruot transfer'ili \0.12l
"[ni:n"v the ísoth¿rmar exponsion is 40 hJ. At
the beginning of the isothermar expansíon the"t"-á"¿ra
pressure is 7 bar and, the uorume is 0,12 4ox1o3
Determine mB, or nl]\ = os'zst;s8r4 = o 47e
\0.12./
(i) The mazímum and minimum temperd,tures or
for the cycle ín K ; Vz = O.lZ x (e)o.476
= 0.193 m3. (Ans.)
(ií) The uolume at the end. of isothermal (ril) The heat
e:;pansíon in ms ; transfer for each of the four processes :
(iii) 7¡" heat transfer for each of the
four processes in hJ. Process Classífication Heat transfer
For air cu = 0,721 kJ / hC K, and co = I.0OB kJ I kg K. (U:p.S.C. 1998) 7-2 Isothermal expansion 40 kJ
Solution. Refer Fig. 8.S.,Giuen : ¡n = 0.5-kg i
mal expansion = 40 kJ ; p1= 7 \6 = geo¡o; Heat transfered
during isother_ 2-3 Adiabatic reversible expansion zero
Vt = O.t'o,r ;-", ='b.ZZf kJtS K-, ;r-= i.'Oói iin
_bar, sf at Isotherrnal cornpression
(ú) The maximum and minin,n tem¡reratures, \, T, : ' - 40 kJ
4-r Adiabatic reversible compression zeto. (Ans.)
P1V, =
'¡17,
Z x 10ó x 0.12 O.S x
¡$Example 3,3, In a Carnot cycle, the matitnum pressure and, temperature are limited. to
= 287 x T,
18 bar and 410"C The ratio of isentropíc compression ís 6 and isothermal expansion is 1.5.
... Maximum remperature, r, = t;:T;#, = 585.4 K. Ans.
Assuming the volurne of the air at the beginning of isothermal expansion as 0.18'm3, d,etermine :
(i) The ternperature and pressures at main points in the cycle.
(íi) Change in entropy during isothermal expansíon.
no^"=L# + o.s= q#a (iii) Mean thermal efficiency of the cycle.
.'. Minimum temperature, Tz= EgS.4 _ 0.5 x 885.4 = 22,2 K. (Ans.)
(iu) Mean effectiue pressure of the cycle.
(u) The theoretical poruer if there are 210 worhing cycles per minute.
 9l
AIR STANDARD CYCLES
90 INTERNAL COMBUSTION ENCINBS
/¡¡ \1= t, I'v,q=ql
Solution. Refer Fig. 8.4. Ps=Pzx
tü,J " lb),
\ro)
fL¡ %l
Maximun pressure, pr = 18 bar
Maximurn temperature, ?t= (Tz) = 410 + 2ZB 693 ¡¡ /r \L4
= t' " [;j = o'e7 bar' (AnsJ
Ratio ofisentropic (or adiabatic) conpression, 3V1 =6
Hence pr = 18 bar \=?¡--685K
I
Ratio of isothermal ex¡ransion,
pz=t2bat I 0\ns.)
? = t.U.
Pe = 0.9i1 bar Ts = Tn = 333'2 K
I
Volume of the air at the begiming of isothermal expansion,
% = 0.1g ms. p¿ = 1.46 bar )
(d) Temperatures and pressuree at the ¡n¡in pointe in the cycle : (ii) Change in entroPY :
For the isenúropit process 4-l Change in entropy during isothermal expansíon,
1.. pV --mRT

= (G)r'1-r = (G)04 - 2.05 s2- sr = mJ,tos"(+\=+'"u

\v1,/ ¡r
l9l
tVt/ lor ,nR= PV
I

e =
[+)" L T
r,=#=#=ae3.2K=?s -
18 x 10o x 0'18
loe- (1.5) = 0.192 ¡¡¡6 $ns.)
10o x 683

(iii) Mean thermal efficiency of the cycle :

lv-\
Heat supplied, e" = n1Y1t.*
lu, ,J

= 11 (s2 - sl)
= 683 x 0'192 = 131.1 kI
/¡¡ \
Heat rejected, e,=pava." l.ü]
= Ta (Ss - Sa) becauee increase
in entropy during heat adilition
il eqiral ti d"".uu"" in entropy during heat rejection'
Q" = 333.2 x 0.192 = 63.97 kJ

.'. EfficiencY,
a"-a, .
,,=-A- Q,

63 97 ot 51.27o. (AnsJ
FiC.3.1 - 1- - 131.1
= 0.512 i

Mean effective pnessr¡re of the cycle' p- :

(iu) :i
Also,
..
Pt
pL = (fl =(6)t¿=12.2e The mean effective pressure ofthe cycle is given by
Work done per rycle
P-= --3t of"ffie-
P¿= =!!==
t229 # = 1.46 bar
v"
Fot the isothennal process L-2
Vl =6x1'5=9
P1V1= P2Y2 8 x 0'18 = 1'44 mg
% = V, - Vt= 9Vt- Vt= 8Vt=
o'=$= 18
l5 = 12 bar .- (Q"-Q,)*.¡ _ =
(Ar_-QJf] (... J=1)
v2
For isentropic process 2-3, we have
P^=
-- v" -i"
68.97) x 103
p2Y21 = psYrr =(131.1-t¿,+;¡A-
_ - 0.466 bar. (AnsJ
=

rI
I

I
 92
¡NTERNAL COMBUST¡ON ENGINES
r AtR sTANDARD cYcLEs 93
(u) Power of the engine, p :
Power of the engine working on tbis Example 3.G. An id'eal engine operotes on the Carnot cacle usíng a pet'ects gos as the
cycle is given by
workíng flui.d. The ratio of the greatest to the leost uolume is fired atd ís r : 7, the lower tempera-
Examp, e s.4. A,_"::!,,;ü;;;;W:#i:i,,-*T# ,1!; ,#::,0
ture of the cycle is also fi.xed, but the uolurne compression ratio'r'of tln reuersible d.iabatic
compression is uariable. Thc ratio of the specific heats is ^¡.
X:I:;:T:;",:¡,'!t:";:o
is reduced' av zo"c,;;;-"tu';,; is doubrcd. Find, the teiperature
when the
'
of the Show thot if the worh dnne in the cycle ís a manimum tlen,
Solution. Let ?, = Temperature ofthe source (K),
,,
and (Y- 1) Ioe L *J--t=0.
rl-t
= Temperaüure of the sink (Il).
Firct case , Solution. Refer Fig. 3.1.
vs v1
\-Tz _-6! =xi vL=r
Tr vr
t.€., 6Tr- During isotherms, since compression ratio = e4¡ansion ratio
GTr= T,
or 5?, = 67" ot Tr = 1.272
vs
Second case : ...G) -Yz
Vr-vt
Tt-ÍTr_(70+27a)l | vt vl lx
TL _
-3 Also -vs
va-vt x;; =&x fr
Work done per kg ofthe gas
T1_T, +343 I
Tr -3 = Heat supplied Heat rejected = F?rlo;, t
3Tt- gTz +
- : -RTrlog,
LO2g = Tr

,, ,,#],=xT; _ilJlf (.''

= R(rt - r,) tog"
| n, (ft - )
= rt" i
Tt = 7'2 Tz)
or
2'47-, = Sfi - rc2g r, '= (r)'-l
,,
O.6T, = 1929

T"= = l?16 K or
But

.'. Work done per kg of the gas,

T2
=
[fl
and # 1442"C. (Ane.)
I
. . Exaup,e o^,,,"^lo:--",!;l,i-:ü;;::";"r;ffi"i;,,if}e
".u.min' rhe temperature o¡ hroi-ilu,"" nw ¡o" o n"o,
* = RTz 1¡r-r -1¡ ¡og.
í!!'Jiu:k'li!";frhr,t is reel K and thai of sinh is 8s0 K. rs
Differentiating l{'w.r.t. 'r' and equating to zero
Solution. Temperature of heat
Temperature of sink
source, ?r = lgg0 f
# = rr,f -' - t,{;,.,- ",-')} * r"s" 1¡11 - r>r -'? t] = o
"
Heaisupplied, ' ?z = 850 K
fo*,"" auu"top"d engine, (/-r-1)(-1) +(r-1) x rr-2 ros"L =o
by the
The most effícient engine is one
: r:?iry^*
that " Carrnt qcle
*ork" oo
1
n T, -T, r99o - 850 - ¡-z*!*¡-2 (y-r)log" =o
-
4 = --1ee0 = o'573 or 67'3vo
Arso, thermar efficiency
"rrJl;;;", ,r-r[-t* ..,'1r+(y-1)
' "l = o
toge -J

0.4
[
.'',h. =- _\ü91k done _ 0.4 x 60 I
H;rG;;,lied = CttTrr) -t*;7- +(y-1)roa | =o
whi¿h is not feasibte as no e1i
ca." ;'::::;:::7,han that workins on carnot cycte. (t ÍL
Hence claims of the inventor
is not true. (Ans.) - 1) lo& -l=0. Proved.
 94
INTERNAL COMBUSTION ENCTNES
AIR STANDARD CYCLES
9.4. CONSTA¡¡'T VOLUME OR OTNO CYCL,E
ul
This cycle is so namcd. as_it was conceived by 'otto'. on this
cycre, petrol, gas and maoy 'Let compression ratio, r r-.r-
types of oil engines work. rú b th¿ sbndord of comiarison U2
for internál engines.
"áiuurt¡oi
Fig' 3'5 (o) and (ü) shows the theoreticalp-V diagram and ?-s diagrans
of t¡is cycle respec- g
tively. expansion ratio, r" (= r) =
t)g
The point 1 represents that cylinder is full of ai¡ with volume (These úpo rotios ore sorne in this cycle)
temperature ?r.
Vr, pressure p, and absolute
ñ | \'f-f
Line 1-2 represents the adiabatic compression of air due to whichpr,
V, and T, respectively, v, and ?, change to pr, As '2 - l,r I
T, \,2)
Line 2-3 shon's the supply ofheat to the air at constant volume
p3 and ?, (V, beingthe san as Vr).
so thatp, and T, change to Then, T"=Tr' ftf -'
Line 3-4 represents the adiabatic expansion of the air. During expansion pr,
change_to a finalwalue of p,vror y, and f. respectivety. V, and ?, Ts 1,,)t-t
;-=l-l
t4 \us,/
I¡ne 4-1 shows the rejecüion ofheat by air at constant vorume t'l original
state (point 1) -r
reaches. T'= Tn ' Q)r
Consider 7 hg of air (rryorkiug subsüance) : Inserting the values of T" and' ?r in equation (i), we get
Heat supplied at con¡tant vol.-e c,(Tg_ T2), T, _7, .L- Ta-Tt
= flo
Heat rejected at constqñt volu-" - =i__ =
tin _ í¡. rr-\\-T,)
But'workdone ",
:;"ü;:?ji"j;i;:'-?,";"' = 1_ :-:;-
I

(r)' '
Efficiency = c, (Tt -Tl) - cu (T¿ -Tt)
=!e4!99- -
flear supptrecl This erpression is known as th'e o;ir stor.dard. efficiency of the Otto cycle'
cu (T3 _ T2)
It is clear from the above expression that effrciency increases with the increase in the value
Tt-Tt of ¡, which means we can have maximum effrciency by increasing r to a considerable extent, but
-'- T"-T"
-i
due to practicoJ d.fficulties its ualue is limited, to about 9.
The n¿ü worh dnne per.kg in the otto cycle can also be expressed in terms ofp, u. Ifp is
expressed in bar i.¿. 105 N/m2, then work done

y¡=(o"ur-poro_prur-lülx102kJ ...(s.4)
\ 1-I T-I )

Arso *=r= X
.. ¿q=U=r_
P2PLP
where r, stands for pressure ratio.

and Ur= frZ = U¿ rUg [.. o=.='l

= Luzu3l
- J
Swept volurne

Total volume--J
*=*[,,,(ffi-')-",.(f ')]

*1,,.[#-')-,,,,t* ')]
(rr-1-1)-pr i.r-l- r)]
(b) #[o,
Fis.3.5

T-
I I
I
 96
r
¡NTERNAL COMBUSTION ENGINES AIR STANDARD CYCLES

= Solution. Bore of the engine, D=250mm=0.25m

tt[,"'-'-t)(p¿-pt)f Stroke of the engine, L = 375 mm = 0.3?5 m
Clearance volume,
-rxro -r)] % = 0.002ffi ms
1"¡ Initial pressure, Pr=1ba¡
,*, ,ttlir
'...13.a
Mean effecrive pre.sure
Initial temperature, ?r=50+273=323K
=
". [¡*i -f. - *'i- í,^)-r,, -,¿rl ¡* ...(3.5)

?{L1"1-r _t)(r, _1)

AIso
P-=
@t -rz)
ffi("t-'-r)(" -tr
,r-?
-&!¡1"r-t-r)(t -r)l
--------
,,lrl)
'\ r /
i.e., .- Prrfir7-t-f)(1 -f)
o_-
17_1lr_t) ...(3.6)
t.7. The efficiency of an Otto qrcle is
60Vo and T = 7.5. What is the
,oüo ?E*r^ple compressinn
Solution. Eñiciency of Otto
Ratio of specific heats,
cycle, \ = 60Vo Maúmum pressure, ,, =::;:t
Compression ratio,
1= 1.5 Swept volume, V,= t¡14 D2L = trl4 x 0.252 x 0.3?5 = 0.O184 mg
r =?
, = L#=
Efficiency of Otto cycle is given by compression ratio, o'otil¡tf;T'ut
= r,
llouo=l-:+-
(r)1- I
(i) Air standard efñciency :
The air standard efficiency of Oüto cycle is given by

0.6 = 1_ (r)+:r rotto=1 ,h =t- r+- =r- *-.

or1 = 1 - 0.435 = 0.665 or 56,67o. (AnsJ
GIt =0.4 or (¡)oo= 2.5 or r-6.25 (ii) Mean effectivo presaure, pñ :
# =
Hence, compressíon ratin = 6.25. (Ans.)
For adiabatic (or isentropic) process 1-2
Example B.g. An engíne g[- zl0 nn tor" oriTZS. mm strohe works on P1V¡I = PrVrr
clearance uorum¿ is 0.0026s-mr.
The initiar p-r"es"ll"4 Otn cycle..The
maximum pressure is timited to zs tunperd,ture are I bar and 50"c. If
bal, ¡iá'rl"|"h"ií¡r, ,
(í) The air stand,ard efficiency
the
,r= o'(h)' = 1 x (r)r'4 = 1 x (g)r'r = 18.88 bar
of the cycle'.
(ii) T¡" *"on effective pressure
Assume the id,eat conditíons.
for the cycle. .'. Pressure ratio.

The mean effective pressure is given by "=f:=ff6='au

prr[(rr-1 -lXrp -1)] _ 1x 8t{(8)14-1 -1}(1.g6-1)
,_ = (r-lXr-1) ...tEqn. (3.6)l
(1.4-l)(8-1)
98 INTERNAL COMBUSTION ENGINES AIR STANDARD CYCLES

8 (2.297 - lX0.36) or 0.72 (\- 689.1) = 1500

= -------:u.4x / = 1.334 bar 1500
- or '3 - +689.1 = 2772'4Yv (An¿'l
Hence mean effective pressure = 1.334 bar. (Ans.) "^ o.7z
Example 3.9. The ¡ninimunt pressure and. ternperature in an Otto qrcle dre 700 kPo ond p2 ps - - pzTs - 18379x2772'4 bar. (Ans.)
,r= ,t - Pt= 7, =
= 73.94
27'C. The amount of heat a.d.ded, to the air per cycle is 150O kJ I hg. Also, 689'r
(i) Deter¡níne the pressures and ternperatures at all points of tle air stonddrd. Otto cycle. 34 :
Ad.iabatic ExPansion Process
(ii) Also calculate the specific work and thermal fficienq of thc qcle for a compression z rY-l
ratioofS:1. 4-=[s{] =(r)r-r = 2.297
Take for air : cu = 0.72 kJ I hg K, and 1 = 1.4. (GATE,1998) T4 l.rsl
Solution. Refer Fig. 3.7. Giuen: pr = 100 kPa = 105 N/m2 or 1 bar ;
T Ts -2772'4 1206.9 K. (Ane.)
- 4=
Tt = 27 + 273 = 300 K ; Heat addecl = 1500 kJ/kg ; 2.279 2.297 =
r = 8:1i cu=0.72kJkS;t=t.+. /.. \'Y ¡/t\La
I a pt="'|.?,J =4o23 ber' (Ans')
Consider kg of air. Also, p,¿u3r = P4v41 =za.e4' l;J
(ü) Speciñc work and thermal efñciengy :

specnc work =
"".,.u11]-,{:";,",":lü _ r,, = c,r(rs_ ry _ g::-r!)
(1206.9 - a00)l = 847 kJ/kg. (Ans.)
= tílz\<zttlz.t_689.1) -
Thermal efficiency, n* = 1-
#:T
=1- = 0.5647 or 56.479o' (Ans)
-+-;
(8)"'-'
ExampleS.lo.Anairstand.ardottocyclehasauolumetriccornpressíonratíoof6,the
Iowest iiia ond. operotes betuteen temperature límits of 27"c and 1569"C'
qrcle presswe of 0.1
0) Cotculote the temperoture and pressure after
the isentropíc expansion (ratio of specific
¡¿s¡s = L4).
cycle operating condí-
(ü) Since it is obserued that vohns in (i) are well aboue the lowest
process w@s ollowed' to continue d'own to a pressure of 0'1 MPa' Which
tions, the expansion
-irá"J"í
Fig.3.7 i".equíred' ti
compt'ete the clcle ? Nome the cycle so obtaíned"
(íií) Determitu by whai percenage the cycle eff't'cicncy has been
improued" (GATE' 1994)
(i) Preseures and temperatures at all points :
Adiabatic Compression process 7-2 : MPa= l bar;?' =27 +273
Solution-ReferFig' u2 vg =r=6;pl =0'1
3'8'Given'.}=3
'i /' \'f-I
; I =l?l
\%/
=1"¡'-' =(8)La-L =2.2e7 K
= 300 K ; ?g = 1569 + 273 = !842 ; T=
1'4.
(i) Temperature and prcsrure after the lsentropic elpansion' T4r P4 :
.l \ Tz = 3OO x 2.297 = 689.1 K. Gns.) air:
Consider 7 hg of
AIso pp{ = pzuzl Ft the comPression Process 7-2:

* =,,.(tl
a
lz=lrtl'=(8)La pru1l = p¡u] + = 1x(6f =123 bar
=18.8?e
Pt \uz)
:. pz = 1 x 18.379 = 18.379 bar. (Ans.)
Constant uolume process 2-3 : t, =lyr')t-t =(o)r.r-r = z.o¿a
Heat added durin¡ the process,
Also Tt lrr)
cu(l¿- ?r) = 1500 Tr= 300 x 2.048 = 614.4 K
 r
INTERNAL COMBUSTION ENGINES AIR STANDARD CYCLES
For the constant uolume process 2-J :

+=+
4 ¡B
+ pr=hri =12.sr1842
ri2 614¿ = 36.9 ba¡

Fig. 3.9. Alkinsm cycle.

j,
- -t-1 *
L4-t

Fig.8.8
Now,
A=tfr)" or ?u= 184,
" (#) =65?K
For the expansíon process 3-4:
li
r. rr. 1='1(9=57;3P']
amson -r - (1842-614.4) = 0.5929 or 89.299o.

.'. Improvement ln efficiency = 59.29 - 51.16 = 8.137o. (Ans.)

+:(f)"=(6)La-l= 2.048
Example 3.11. A certain quantity of air at a ptr;ssur:e of 1 bar and temperature of 70'C is
compressed, od,iobaticalQ until the pressure is 7 bar ín Otn qcle engine. 465 kJ of heat per kg of
1842
^7"
' zo48
¿¡=<-
900 IL (Ans.) air is now odded. at cot sta.nt uolumc. Determine :
2"048 = (i) Cornpression ratio of the engine,
(ii) Temperature at the end. of compression.
Atso psqt = plvor * o. = r, , (iii) Temperature at the end. of h¿at ddd,iüon.
[f.) Take for air c, = 1.0 kJtkg K, c, = Q./ffi kJthg K.
Show ea¿h operation on p-V and T-s diagrans,
p¿ =86.e, f+l'
\o./
=8 bar. (AneJ
Solution. Refer Fig 3.10.
(ii) Process required to complete
the cycle : Initial pressure, Pr=1bar
Process required to complete the cycle
is lie constant pressure scavenging. Initial temperature, Tt= 70 + 273 = 343K
The cycle js called Atklnson cycle (Refer
Fig. 3.9). Pressure after adiabatic compreseion, pz= 7 bar
(ili) Percentage improvemenUincrease
irrlffi"i"rr"y, Heat addition at constant volume, Q = a6S kJ/kg ofair
n -r I - 1 Specific heat at constant pressure, cp = 1.0 kl&g K
'rouo=r- (rt:f =t_ GF:T =0.5116 or 5l.t6qo. (Ans.) Specific heat at constant volume, c,, = 0.706 kJkg K

Heat supllied - Heat rejected c^ 1.0

lArki,"o, = -.YI\4#= -
ueat supplied ---.- r=;=o-6=1.41
Heat supplied
(i) Compression retio of engine, r :
- ",(rz-\).-"{4-T) =r_"r!.,r.: -Ir!=- t_
^ ?(4 -2i) According to ad.iabatíc cotnpression L-2
q(Ts -T) cu(T3 -Tz) <E _f"¡
P1V1r = PrVt'l
 INTERNAL COMBUSTION ENGINES AIR STANDARD CYCLES

(i) Compression ratio.

(íi)Thermal efficíency of the cycle.
(iii) Y{orh done.
Tohe'1 for air = 1.4.
Soluüion. Refer Fig. 3.11.

p (ba4 T (K)

p-V diagram V(m.) T-S diagram s(kJ/kgK)

Fig.3.r0

Ivrl P2
t;it = -Pl
\vz J
¡-s - !.2 (v,)
Dt
l' h-') s(kj/kgK)

or --l¿.)i
r?'\* =#-: t tt
=
Hene compression ratio lr=r^l;:1,;l?rr.
Fig.3.ll
(ii) Temperature at the end
of comprcssion, T, : Initial temperature, ?r = 38 + 273 = SLLK
In case of ad,iabatíc compression I-2,
Maximum temperature, ?s = 1950 + 273 = 2223If..
T2 lv, )t-t (i) Conpression ratio, r :
= l.üJ = (3'e7)r'4r -' = 1'76 For adiabati¡ compression 7-2,
"'Tz 1.76 Tt PlVl = PrVrr
:: = = 1.76 x B4S = 609.? K or ggO.?.C
nence temperdture at the end of cornpressioz ggg.7g. (Ans.). (y'\' o"
\v") -
(lll) Ts¡ps¡¿lure at the end =
of heat addition, T, : nr
According to constant uolume heating
operation 2-S Po
oi = tu
Q=c,(?3-T)=4ob But "'(Given)
-\,
0.706 (?3 _ 603.?) 465
=
or \-608.7= # "'
(r)r = 15
(r)l'a = 15
[ "=#]
or ^ = 465 + 603.7 = 1262.3 K or 989.3.C 7
o,'e r= (15)il = (15)0.?11 - 6.9
Hence ternperature at the" end, of heat a.dditían 9E9j.C. (Ans.)
Example 3'rz' In a constant uorume 'otto cycre',.=
Hence compression ratio = 6.9. (Ans.)
the pressure at the end of compression ís (ii) The¡mal efficiency :
at the start, t.he tempe-rature o¡ oir"iil'n,
!_l:y:: th.at
tnaxttnum tenaperature attained. ín the cycre
beginning of conpression is S8"C and 11
i 1s50"ó. Dete¡mine : Thermal effrciency, 4r¿ = 1- (rf:T = 1- (e9F4-1 = 0.538 or 53.87o. (Ans.)
 I
I

r,
INTERNAL COMBUSTTON ENG¡NES
AIR STANDARD CYCLES
(ill) IVq¡k {q¡s .

Again, for ad.iabatí¡ compression I_2, p (bar)

?[ _(vrl"-,
=
14., =
r.r_l-
(rr, =(6.e)1.1_1=(6.e)0.1=2.16

= 311 x 2'16 = 671'? K or 398'7"c

For ad.í¿htic epansian
or!:r:1i," ''16
T^ /rr \7-l
d lüJ= = (r)r-t =(d.e)oa=2.rd

-tn= T" = 2223 = 1029 K or 756'c

TR 2t6
Heat supplied per hg ofair
= co(Tt- Tr) = O.717(2228 _ 67I.7)

= f112.3 kJlkg ofair

I n 0.287
lc.=-=-.
1
v (*t)
l" T-l t.4-'
I

Heat rejected, per hg of aír

| =o.zrznrgr j

= c,(Tr-- Tr) = 0.7t7(LO29 _ g1t)

= 514.8:kl&g of air
.'. Worh done
= Heat supplied - Heat rejected
x =(+)'= (r)' or
'= [*) = (+)* = (11)0'714 - 5.5
= 1112.3 - 514.8
= 59?.6 l¡.I or 592600 N-m, (Anc.) '= (,)1-r
:"::::*jj":!::y:f
*.,rkins on otto qrcte has a votume of 0.45 nr, pressure I bar
Also
? =(+l = 1.977= 1.98

i#;:x::::'::;i{S T,liri:;r::i":i::ii:l:;;':;;;.x';ff"T;7"#,if,-Y,i,!í:,
n¡lr
the pressure is 11 bor. zn Tz= Tt x 1.98 = 303 x 1.98 = 6fl) K. (Ans,)
niá,fx"',;;;í';";;;;;iT;,xkf,rÍ,i",il!,i?Wressi Applying gas laws to points 1 and 2, we have
(i) Pressures, temperdtures and. uolumes
at salíent points in the qcle, ptVt
(ii) Percentage cleara,nce. _ pzYz
(iiü Efficicncy. TL-Tz

+,X, u, = flffi€
t, (iu) Net work per c1rcle.
(u) Mean effectiue pressure. ,r= = o.oel ¡¡s. (Ans.)
1r
(ui) Ideal power dcveloped' by The heat supplied during the process 2-3 is given by
engine if the number or-working qrcles per minute
is 270.
:
.the
Assume the.qrcle is reuersible.
i
I
Q" = tn cu (Ts T2)-
I Solution. Refer Fig. B.12 I
ptVt 1x106x0.45
l Volume,{=0.45m9 1 where
^=ffi= ,8?-goa =o'517ke
Initial pressure, p1 = 1 6u"
210 = 0.517 x 0.71 (?, - 600)
Initial temperatr"",?, = B0 + 2?B = 303 K
Pressure at the end ofcompression stroke, p, 210
= 11 bar = o¡-iilT.z--f + 600 = u72 IL (Ans.)
Heat added at constant volüme 210 kJ " "¡
= For the cozsú¿nt volume process 2-3,
Number of working cycles/min.
=21O.
(i) Pressures, temperaúures
and volu¡¡es at salient points le=22
For adiabatic compression 1-2,
: /B t2
PlVl = prVrt Ta p, :áÉ
IL72
or= x 11 = 21.48 bar. (Ane.)
; n ' =
Vs = Vz = o.o81 m3. (Ans.)
i

-- =J+.*
 TÍT_

AIR STANDARD CYCLES t07

INTBRNAL COMBUSTTON ENGINES

For the adi.abatb (or isentropb) prress 34, (b) Determine the aír-standard. effíciency of the cycle when the cycle dzvelops n'Laxtmurrl
worh with the tetnperature limits of 310 K and 1220 K and, working fluid is oir. What wíll be the
PsVsr = PrYrr percentage change in efficiency if helium is used. as working fluid. instead of oir ? The cycle
operates betueen the same temperature lirnits for maxí¡nutn work dcuelopment.
P.=Psx (f)' ="' (i)' Consid.er that all conditions are id.eal.
Solution. Refer Fig. 3.13.
/r\L{
= Zr.a8 ' li\o.D,,f| = 1.97 bar. (Ans.)

Arso r _ lyr)'-t _=1.;i I r \"-' =o'505

f1)'-' _=l*,1
4=l%J
t' Tt = 0'505 fs = 0.505 x 1172 = 591.8 K. (Ane.)
Vr= Vr= O.45 m3. (Ans.)
(ii) Percentage clearance :
Percentage clearance
i'09!-
= t';= ' 16e =
r-*E -- ' 1oo

= 21.95%. (AnsJ
(lii) Effieiency :
The heat rejected per cycle is given by
Qr= mco(Tt- Tt)
= 0.517 x 0.71 (591.8 - 303) = 106 kJ
Fig.3.1.3
The ai¡-standard efficiency of the cycle is given by
(o) The woik done per kg of fluid in the cycle is given by
210 - 106
=W
,l*, -Q.-zt1
- = 0^495 or 49.6%. (Ans.) W= Q- Q,= cu(Tr-7")-cu(Tl-T)
Altematively : I '=
Iotto = 1-
11
= 1- ('sf----.¡:r =
0.495 or 49.64o. {lnsJ
But
?= [fi (r)1-l

Gt-r I Tr= Tr. (r)1 -7

(iu) Mean effective p¡es{¡ur¡e, pr 3 Similarly, Tr=To. Q)Y-r
The mean effecüive pressure is given by

P^=
Tf (work done)
- Q"-Q,
w=",1r,-!.(r)1 '-#.n]
V"(swept volume) (Vr -Vz) This expression is a function of r when ?, and ?, are ñxed. The value of Wwill be maximum

..
\ =,!',T-,t99lltot=
(0.,15 - 0.081) x 10" = 2.818bar. (Ans.) dw
=0.
d'r
(u) Power developed, P l
Power developed, P = Worh done per second +dr =-7,.(^t-1) (r)Y-2 -?s(1 -T) (r)-1 =0
= Wort done per cycle x Nurnber of cycles per second ?, {r¡-r =Tr(7)t-2
OI
= (210 - 106) x (210/60) = 364 kW. (Ans.)
' Example 3.14. (o) Shout that tlv compression ratío for the marimum work to be d.one TB ¡,-\2(y-1)
per hg of aír in an Otto qtcle between upper and lower límits of absolute ternperatures T, and. T, ry = \')
is giuen by / m \U2(^r -r')
I l. I

r= Pr,ved'
--(bl'"'-"
'- l.4l l4l
 I
, l0g
INTERNAL COMBUST¡ON ENGINES
lr' AIR STANDARD CYCLES
(ó) Change ln efñciency I
For aír .l L.4
= Tz=Tt= Jryr. Proved.
I /^\v2(L4-1, /--- \t/o.E (ó) Power developed, P :
.=l1.:l
'=lrr) _ [L22o\'
=|.sloj =tM 4=310K I
l The air-standa¡d efñciency is given by 4 = ra50Kl ....(Given)
m =o¿e kgJ
nouo = 1 #- = t - **r= = o.498 or 40.67o. (Ans.) Work done W = c, (?s - Tz) - (Tt- Tr)l
I
lf helium is used, then the values of
Tr= Tn = lEn = 1610 l45o = 6?0.1 K
cp = S22WikCiK and c, = g.1g ktkC K "
li W = o.vL K1450 - 670.4) - (670.4 - 310I
... c" 5.22
Y= =t'sz = 0.?1 (?79.6 - 360.4) =297.6kJtkr,
ii
ii
*=g,' Work done per second = 297,6 x (0.38/60) = 1.88 kJ/s
The compression ratio for ¡1o!-u6 work for the temperature limits ?, and ?, is given by Hence power developed, P = 1.88 kW. 6ns.)
- -', E¡qmple 3.16. For the same compression mtin, show that tlw fficiency of Otto cycle ís
,= (+)"' " ry1u',u,
1.4/ = ¡
[sroJ = 2.77 greater than that of Diesel qcle.
Soluüion. Refer Fig. 3.14.
The air-standard efficiency is given by

Ioto=l- =0.49d or
,h=t- ¿r9.59o.

Hence change in elficiency is nil. (Ans.) "#-

Exanple 8.16. (a) A¿ engire.workir* on otto cycle, in whi¿h the
sotieni points are 1, 2, S
and. 4, has upper ond, Iower temperature tiiits To and.i
be done, show that the intermed.iate temperatui is giuLn r. If the mo*ímum-*""ií"i'ns of air ís to
by
Tz=\= JTtTs.
(b) If an erqire
works on otto cycle between temperature rí¡nits 1450 K
ad gI0 K, find. the
maximum porter deueloped by the engine assuming tie círcur.atinn
of air per ;¡n;¡" * 0.Jg hg.
Solution. (¿) Refer Fig. B.tB.(Example 3.14).
Using the equation (iii) of example 8.14.

w=
"1"
",1r"-n.,"n-, (rr'- --h.r,l |

and differenbiating Ww.r.t. r and equating to zero Diesel cycle

/a
I Ir
\u2(t-r) I
F8.3.r4
r= lrl
l?r / We know that

= r/(r)"r -L rlotto = 1_r+_

substituting rhe varue *, ,lio=l'ot I:*i"3
-' -
r¿"*r=r- 1 1fol-rl
,, = r,V+)'n-
"]' = n(?,)"' = lM G)Fxrt-j
As the compression ratio is same,

Similarly, ,-=
vr vr'
v"=W='
¡oft_,ry"=#=Jrn v"' rr, then cut offratio, p= v"' r
ltrl J 14 J O
t= ü=;
 INTERNAL COMBUSTION ENGINES
11t
Putting the value of p in 4u,*,, we get

l¿""¡ = I -

From above equation, we observe

lrt
11

Let r, - r- 6, where 6 is á small quantiüy.

66263
Then
i=*=ó=('-i)-'='. -+-+-
,1213

Fig.3.15
volume increases from V, to V, and temperature Tztn ?3, conesponding to point 3. This point (3)
is called tbe poínt of cut off. The air then expands adiabatically to the conditions pn, Vn and Tn
fr.6 . r(r+1) 62 respectively corresponding to point 4. Finally, the air rejects the heat to the cold body at constant
... rrü""c,=l--'il#
.

volume till the point 1 where it returns to its original state.

Consid¿r 7 hg of oin
|
Lrr' -+-;+...... Heat supplied at constant pressure = cp(73- Tz)
Ileat rejected at constant volume = cu(Tn- T)
f9*r+r 62.
='-#='l#l
L;+P+"""
I

J
Work done

n..
= Heat supplied - Heat rejected
= cp(Ts- Tr) - c"(To- Tt)
.= Work done
'drese¡ Heat supplied
The ratio inside the bracket is greater than 1 eince the co-efficients ofterms ü/r2 is greater
than 1 in the numerator. It means that.something more is subtracteil in case ofdiesel cycle than _ ci(Ta-Tz)-c,(Tt-T)
in Otto cycle. co(Ts -?12)
Hence, for sctme cornpressíon ratio 4o* > ü¡n¿.
-, (T4 -Tr) l.t :e=t¡
= '- rt4lD "'(;) lr'
3.5. CONSTANT PRESSI'RE OR DMSEL CYCLE L-ul
' This cycle was introduced by Dr. R. Diesel in 189?. It üffers frtm Otto cycle, in, that heat Volume at cut - off
Let compression ratio, r = 5 and cut offratio. o = !3- i.".
is supplied at constant pressure instead ofat constant volume. Fig. 3.15 (o and ó) shows the p-V u2 u2 Clea¡ance volume
and ?-s diaglarns of this cycle respecüvely. Now, during a.d.iabatic compression L-2,
This cycle comprises of the following opeltions:
Q) 7-2..... Ad.iobatic compression. ?7\= lo)t-'=(r)r-r or rr=rr.(")r-'
(¿i) 2-3.....Addition of heat at const@nt pressure. 1,, )
(iii) 3- 4..... Ad.iabatíc erpansion. During consúont pressure process 2-3,
' ^:
=p.rr=p. f,.
(iu) 4-l.....Rejection of heat at consto,nt volu¡ne. lc U.
or ?, (r)1-r
Point 1 represents that the rylinder is full ofair. Lct pr,V, and ?, be the corresponding t = t=p
pressure, volume and absolute temperature. The piston then compresses the air adiabatically (i.e. Dunng adinbatb expansion 3-4,
pVY = constant) till the values become pr, V, and, T, respectively (at the end of the stroke) at point ñ / \1-1
2. Heat is then added from a hot body at a constanü pressure. During this addition of heat let
ts lu¡ |

T4 l.rrj
 INTERNAL. GQMBUSTJON f NSINES 113
/ \ltj1
lr I. (... n=o=st,.4='rl
\P,/, \ us ug u2 ug p)

...(3.9)

= pquo(p - a PgPuz-.p¿t'ttz P*z- Ptntz

T_l , -
1¡
T_1

Fis'
''rG
-
Compression *tio,
l \
v-
' = rs
l= ¡fJ
lfor air = 1.4
=(:)'=r" '1 Air standard efficiency ofdiesel cycle is given hy

¡l¿¡ ' I fot-r-l

ser=r-th¡r:Tlr-J ..(,)

|L *=lr.')'
P, \uz) ", Pz=Pt'rrandS=r or uz-utr_,I
-
where p =
",rt-of.atio = f
] j-
prurrl -1 h(p - l) - rt -r (pr - 1) f^-v^=
ózloo V" (V" = stroke volume,
------- I
(1 - 1) ,(3.8)
I
¡
I
f
?'
,ri( _
 AIR STANDARD CYCLES ll5
114 INTERNAL COMBUSTION ENGINES

rai*a=1
= 0.06 (Yr - Vr) = 0.06 (15
[,
Vz- V2)
*h[H]='
...
= 0.84 Vz or V, = 1.${

o=Y,=s=LYvz =r.r^
'Vzv2 When the fuel is cut-off at 87o.
= 1 - 0.248 x 1.563 = O.6L2
we have
"*n-'lq#]
or 61.2%

Putting the value in eqn. (i), we get P-l 8 P-l R

or ffi=*o
1
= t* =o'oe
-
\¿¡"*t=L-
l{r.an)tn-rl' ,'' = P=1+1'04=2.04
l-4GIr=fl 1s4L l lpt-rl . l
I
ñ _i
¡ldi"""r=r itz.o¿to-ll
=I- 0.2417
' r.eó¡ = 0.612 or
81.2%. (Ane.)
J=t-;¡"t=-L m4L I
Example 3.La. The strohe and cylinder d.iameter of a. cornpression ignition engine are 250
mm and 150 mm respectively. If the clearance volume is 0.0O04 m3 and fuel injectian takes place = 1 - 0.248 x 1.647 = O.69f oi 69.17o. (Ans.)
at constant pressure for 5 per cent of th.e strohe determine the effuienca of thc engine. Assume the Hence percentage loss in effrciency due to delay in fuel cut-ofr
engíne working on the d.iesel cycle.
= 61.2 - 59.1 = 2.1%. (AneJ
Soluüion. Refer Fig. 3.16. Eranple 5.2O. The meon effective pressure of a Dicsel cycle is 7.5 bar and compression
Length of stroke, L = 25O mm = 0.25 m ratio is 12.5. Fütd the percentage cut-offof the eyle if its initial pressure ís I bar.
Diameter of cylinder, D = 150 ¡nm = 0.15 m Soluüion. Mea¡ effective pressr¡re, p- = 7.5 ba¡
Clearance volume, Vz = 0.0üX m3 Compression ratio, r = 12.5
Swept volume, V" = ttJ4 ÚL = Í/4 x 0.152 x 0.25 = 0.004418 mg Initial pressure, Pr=lbar
Total cylinder volume Refer Fig. 3.15.
= Swept volume + Clea¡ance volume
The mean effective pressure is given by
= 0.004418 + 0.0004 = 0.004818 mg
P^=
prrl[T(p-1)-rr-r(pr -1) ,,.tEqn. (8.9)l
Volume at point of cut-off, Vt = Vz + V, (7_r)(r_1)
ft
- L1 (pt1
1 x (125)11 [1.4 (p - 1) - (125] - 1)
* ,0.004418=0.fi)0621 me
=0.0004* 7.5 =
(1.4-1xr2.5-1)

.'. Cut-off ratio,

% 0.000621
p=il.=-=r.Do 7.5 =
3433[1.4 p - 14 - 0.364p1¿ + 0.364]
4.6
7.5 = 7.46 (1.4 p - - 0.364 pl1)
1.036
Vt 0.0041418 + 0.0004
--
'-
Compression ratio,
v2=V"+Vz
v2
=
0.0004 = L2.O4 1'005 = 1.4 P - 1'036 - 0.364 Ptr
2.O4= 1.4 p - 0.364 pr'a 0.346 pr'{ or - 1.4 p + 2.04 = 0
r [o1-rl - r -rl
ltr¡slln SoMng by trial and error method, we get
Hence, üuut =I - y(¡.)ñrle-t l='- *;*¡¡F =l rsr-r I p = 2.24

= t- x 1.54 = 0.593 or 69.87o. (Ans.)

O.264 Vo cut-off= r t* = # x irXl = 1O.784o. (Ans.)
¡¡n\nple 3.19. Calculate the percentage loss ín the idzat fficiency of o di.esel engíne with "-LJ
compression ratio 14 if the fuel cut-off is d¿layed. from íVo to 8Vo.
exlvcnple 8.21. An engíne with 200 mm eflinder d,iameter and. 300 mm stroke works on
tlleoreti¿,al Di.esel eycle. The inítial pressure and temperature of air used, are 1 bar and 27"C. The
Solution, Let the clearance volume (Vr) be unity. cut-off is 8/o of tle stroke. Determíne : t
Then, compression ratio, r = t4 (0 Pressur* ond. temperotures at all salient points.
(ii) Theoretical air standard. fficiene1.
Now. when the fuel is cut-offat S4o.wehave
(iíi) Mean effective pressurc.
p-l 5 (iv) Pouer of the engine if the worhing gtcles per minute are 380.
r-1 = 1oo o¡
fr| =0.05 or p-l=13x0.05=0.65 Assume that compressíon ratio is 15 and. working fluid. is air.
P = 1.65
. Consid.er all conclitions to be iá¿al.

I
 iló INTERNAL'CC'[¡ÉÜSiIdÑi IDNOTNE$ 1t7

ii i For the ad.í.abatic (or ísentropic) process I-2, : r.{i ,.,.ñrri,ir.r..i r:

,..t t.j::. , ' ,j,.. /t¡ \r

.i,j : i!:1,: j
. ,, 1{,t1,n{zj e' ,f ={f,,J =t"t'
l]g l-- ri

i:ü ii = ¡i¡,1 .,. ¡

ll_ 0.00e42
Hf(rAns.)
,'¡ ;I.; ¡ft; tnJ:,
",!t$i:g;!Jffi
_
= b3ó66tlB
1
;i;' t¡'i'}
,¡i.1.,. ) t ;;t;,..
^
-p¿-=!b-j
p;-=p¡-.=-44;81 bg..v i*r., :'

E cútfif iaiitJ
:t'¡: i' ,:,¡.r:i'1 r 'ft-.,{';:,¡ :rii:¡' .'
:',r,:r,
?t.,i 'iS: j.i. .. i\l-,-i ]i;l'¡a:!, {r|\n¡ii. .: í,ii1 } .:
8 :'r)r :''- :':
': Pl+-!
¿I- ir " : l
' 100;',15=1' i, ,

t:e. ,'t
.l]..''!i.'f,\'i;i.,:p:-5ry.6gi\¿¡r1=i.t¿
fltlilitit!l\i!
-'r.a
\',ij ,:tl i,,:ii1,.. r%: ó
'.i¡itiril.\':
ü :'z:rz* b.Oooetzs:'o.0oi4% ms.
11r:.,ir: i,!. q1i¡.11,1-1,r\: ..¡.ri.1,. .
(Ans.)

{.4) il;i,a ;r: .'t;,11-.:. :,'i., fvs

can alsó rÉ'calcuta¡$ SfffrhVe:-. . .: l
,- ",,'
Fis.S.ú
:
/, : r.: ii :, ly¡i ='ó.¡e¡frr ¡..4, = 0J8 x Oi0094? + 0i00O6?38 = 0.00 1ae6 m3 ]
For the '", :, 1'.| ,
:
,
Cylidder diameter, D=ZOOmmor0.2n ,'i ' t¡ r\"iiiíantt¡li,&iiEpt'áóbls'z.}, r:
Stroke length, ' Yi;i"t ':':' :i " "
Z=300mmor0.Bm
Initial pressure, P¡ = l'0 bar
Ts Tz
Initial temperature, Tl= 27 + 273 = 30O K Ts=rr, v. 0.001426
= 1878'8rL (Ane.)
ñ=886'2x
Cutolf 8 For the isentropic process 3-4 "
= iñ' Y' = o'08 Y' psVsr = paVaI
(i) Pressures and temperatures d salient polnts :
Now, stroké volume, V,,='r44 D2L = rl4 x o,22 xb.3 = 9.66942 - -..(v,'l' 1 f.
v4
=v4 xv2 =Lrb
v3v2vsv2vsl I
-t e{
lüJ
x
=,Ds x
(?¡tT = PB

Vr=V,+V"=V,+ jj\/f v.=l--l

t" " r-L) =ffi=r*'bar. (An*) t
I
=i,' u=v=fi=tot)
=v ft*-L);-¡ rn _ (vr\'-'
":rJ=;'%
16 .,,, 15 .:.
Also,
4 [%] =l?,?J
= =ll_)'0,-' =o'457
L.€.
Y, _
,,
i
.

'
iEJ*% =fr x 0.00942 = 0.0101 m8. (Ans.) Tr = ?¡ t 0.457 = 1878.3 x O.457 = 858,38 IL (Ans.)
Mass ofthe air in the cylinder ca{r be calct¡lated
by usrng the gas qquation,
V¿=%=0.0101 m3. (Ans.)
(ii)'Ttreoretical air standaid efficiency :
PTV.=
^37,
-= ¿¡Vt=lxlou"o.010t T"¡ne, =' *hi#]=' o#-
= 1 - 0.2418-x-1.663 = 0.598 or 89,87o. (Ans.)
l#l
¡
I

I
I
 lt8 INTERNAL COMBUSTION ENCINES AIR STANDARD CYCLES
(iii) Mean effective pressure, pñ : Me¡¡ effective p¡eaaure' pn :
Mean effective pressure ofDiesel cycle is given by Work done
_ _
P- =
by the cycle.
--3*6fioiñd-
Pl(r)l[fo-1)-rr-r(Pr -1)
^ - ----
'- (i_lx_lj-- Work done = Heat added - Heat rejected
x (15)u[r.4(2.12 (1s)1- L4 (2.1214 Heat added = mcp(Tt- ?¡), and
_ 1 - t) - - 1)]
(1.4-1X15-1) Heat rejected = mcu (Tt- T)
Now assume air as a perfect gas and mass of oil in the air-fuel mixture is negligible and is
_ aa.3rt1.56t:_0.9,,,,,,,,,,,,,,,.38 x t.8631 not taken into account,
= 7.424bar. (Ans.)
Prress 1-2 is an ad.iabati.c compression process, thus
(iu) Power of the engine, P :
- ¡,, r1-1
workdonepercycle
Work done per second
=p^v"===*ft*t =6.99kJ/cycle rt \Y2 J|
+=l+ o" rz= rt,
?¡ = 300 x (15.3P'a = 893.3 K
Iar (since 1 = 1,4)

= Work done per cycle x No. of cycles per second

= 6.99 x 380/60 = 44.27 hlls = 44.27 kW
Hence power of the engine = M27 klf,. (Ans.)
Also, p1v{ = p2v2r ¿ pz= pt"
- f+lt = 1 x (16.s)1'a = 46.5G bar
Example 3.22. The uolume ratios of compression and, apansíon for o d.iesel engine as
\vz)
tneasured. from an indicator diagram are 15.3 and 7.5 respectively, The pressure and, ternpera- Preess 2-3 is a constant pressure process, hence
ture at the beginning of the compressíon are 1 bar and 27.C,
Assurning an ideal engine, determine the mean efftzctive preasure, the ratio of marimum '12 'ti + r^=u-{'
2=3 " Yz
=r.onx8es.3= LB2z.sK
pressure to meo,n effectiue pressure and, cycle fficizncy, Assume that the volume at point 2 (Vr) is 1 m3. Thus tbe mass of air involved in the precess,
Also find the fuel comsurnption per hWh if the indiroted, th¿rmol fficiency is 0.5 of ideal
effiiiency, mechanical efficiency is 0.8 and. the calorific value of oil 42000 hJ lkg. l'. Y¿
l =vt =Lrb
Assutne for air : cp = 7.005 kJlhg K; cu = 0.718 hJthg K "f 1.4.
= (u.P.s.c. 1996) -=*=tf;;i#'
Rrz =,777w l"_ L 'n"'É_l;,:.^.1
=rr=z.otl
Solution. ReferFig. 3.18. Giuen, !v2= fS.e ' 3vs =r.U 1". ü=ü.ü
Pless 34 is an ad.iabaüc epansíon process, thus
Pr=lbar;Tr=27+273=300K;r¡*,r,=0.5xr¡*-"*,u¡¡i4necr,.=0.8;C=42000kJ&g.
The cycle is shown in the Fig. 3.18, the subscripts denote the respective points in the cycle. t' l]h')t-t =ll)"-' = 0.4466
Ts= [Vrl U5/
T¿ = L822.3 x 0'4466 = 813.8 K
"' work crone ==T?rfirolu'r;#i:t1l;rfi-
0.?18 (Bls.B - s'o)r = 100a6 kJ
Workdone 10035 10035 10035
P- = ffiptiJiñá = (vr-vrr= erst/r4rr= 148
= 701.7 kN/m2 = 7.017 bar. (A¡s.)
('.' Vz = 1m" aseumed)
Ratio of maximum pr€esrure to mean effective p¡easure

=6.4e. (Ans.)
=
X=m
Cycle efficiencyr lla"a¡¿ 3

Work done
l"y¿" =
Heat supplied

Fig. 3.18. Diesel cycle.

l¿ü
¡NTERNAL uot{BosrloNrENctNEs
AIR:-STI(ND*ID;,G1EIH¡ i r/ :j:]'j ri.ir
1009_ _ 10095 : 6e cvileslls
rrl.
= ;;64 =
@-;r;leb¡b
tearr€a€?e "l'J-ó.48%.
r¡BsM.
(Ans.) Consid¿r I kg of air. .i.:g *vr .riit ;;0i"+:;¡;ls lri ,ll t-, .,uilr sr-Í:l glliliut
FueI consunption per tWf,
i,¡glo,, lqruE- Total heat supplied = Ileat suppüed duringlhe operation 2-B
rtr,oi = 0.5 r¿r4" = 0.5 ,Iuto,&yrtry.Hoz4sfth"
bg:[¿y, i;!G¡ ;i.!rj;/ - during the operation B-4
It<ar = 0.3bg{ lsnf"r$i?4ir = bsbbstsel{
'r*,Eqfrsuppüed V
- nT)
ai bas4l8gitgan , s'..amrtrn'
ei grurxic-I I'r . .- -
ñ -AÁl "T - -^ 3oo0 t'sr"sici rnsH
H;F&=ffif;Silñ#üiq::ff
Heatrejectedduringoperarion;?!:;íl:r.?"$'-4'
work done noi"Jiñu¿á i¡r;.fl¡**a
*,nl;,,
riioónib¡ tnj 3i s-!' ¿?\)olq ",
=

or r-Irl,,,o.rnr'-1tl@foiazsrqmor
(S.I = y o:rrriz) | L". l ,. . - 14¡ xa2000'l :t\
or ia¿l^,.='"1[* l.ij-,-i "T'

ir18ffi se2-trIq¡Eóco0re0*3*514 kskwh. (Ars.) j'.,r ,;r,r+Jffi

iQtzii!1,I*"" (r4-r3)
;i tetu =--+++-
cr(76 -T1) I c-\
...G) lr Y=¿. 1.
,j',,.',(Tt+721,+'¡(Ta+\¡)
,,.;
"= "

, -,i ,,.Jnq.1:
.,, .)..!. , - 12., ,i -,. ,; ..,

i ri _a. p,
(u) í-L-Rejection of heat at constqnt, rolu^". , . ' "
,, = ;, = where pis knowl aa pnessune or erplocion raüo,
.,: ,o-!-!t'¡r,_
'2- a, ...(iii)
Dunng adiabatic epagsion p.roc*s (. ,
T1 lru'\t-t
;=l;l
1l \u4 /
, .tll
_ trl ...(iu)
'\p/
,f'' x lz lt lc l,
' \. 3'$t- +
Vt= +
tk u4= u2x u4= p''p being ratiol
--- -- ---'-/
-- cut-off
-----' the
During consúant prcEsure.heating proceis S-2,
u"='n ::
Ta Tt
.I
i,=7.\=oT^.
Fig.3.19 Putting the value of fn in tl. uq*tiolttirl, *; ,*
z rY- I
f=6)'-',' or t,= r r. [+.,l
 INTERNAL COMBUST¡ON ENCINES
AIR STANDARD CYCLES 123
Putting the value of ?, in equation (ji), we get
Ts
¡¡anple 3.23. The swept volume of a dizset engitu worhing on dual eycle is 0.0053 m3 and.
clearance uolume is 0,00035 n3. Th¿ ma.timum pressure ís 65 bar. Fuel injection ends at 5 per
F -¡r¡r-r cent of th¿ stroke. The t¿mperature and pressure ot tlu start ofthe comptession are 80'C and' 0'9
Tr bar. Determin¿ the air stondard effteicncy of the qcle. Tahz l for air = 1.4,

= 4t
Solution. Refer Fig. 3.20.
p-'t¿r-t
Now inserting the values of "Z:1, T2, ?. and ?, in equation (i), we get
| (" 1)
-
rl¿u¡ = I - _,
-¡--I
Gt:rle'
tl
e.,

T(e - r)
l\1- p.,l+
L€., tlo*r=1-",! ...(3.10)
Work done is given by,

1¡r=pr(ua-ur)* ry
=psu3(p -r, *
@ova - pínz):!p2aa - pLrus)

_N
tl
p',rte -rxr -u. p,,, t^I
lr_ V. = 0.0053 m
e.,,, - -
_ [e f")- (1
#,') Vs = V. = V. - 0.00035 m-
7-l Fis.3.20
Also zE=[,.')'=fef and p, =("\\' =., Swept volume, V = 0.0053 m3
P4 \u¡,/ \r/ h lrz)
also, Clearance volume, V"=V"= Vz = 0.0O035 ms
Ps= P$ ut = us, u5 = ul Maximum Pressure, Ps = P1 = 65 ba¡
¿b[p5 (p. 1) (y - 1) + ps (p - prrr-1) - p2 (1 - rr-r)
w= Initial temperature, ?t = 80 + 273 = 353 K
(y-D Initial pressure, Pt= 0.9 bar
pPe[9(p -1)(T -1) F(p - pl¡l-]) - (t -r1-r)
+ fl¿o¡t - ?
- (Y_1) The efñciency of a dual combustion cycle is given by
_ a(rIqtfPr(p -r)+ (F - r) -rt-l,(Bpv- r) ó _-1- F,'Pt=-,t ,,.|
¡ _ .t 'f,.
.,rduar ...au
Y-1 (r[_1 ltF_rl* Ffp_l)J
(
- Aq/-rtFT(p-1)+(p-1)-rr-\pp1-1) ...(3.11) rario, ,=
o'o0x1if;0J03u
= re.rn
Mean efiective preasu¡e (p-) is g1ven by,
"t -1 compression
+"="#=
t,' Vz =V" = Clearance volumel
w w arur[rr-1gr(p-r)+(p-r)-rr-r (Fpr -r)
P^= lÁ=--1"4=-
",1;/ (t-t)*[+) Cut-offratio, p=
vs vs vc
V2=Vs =Vc)

pr(r)Y[9(p-r)+(9-r)-rr-r (ppr
o
¡E- (r_tXr_l)
-t)] ...(3.12)
0.05x0.0053+0.00035
= 1.757 say L.76

t
 r"24
INTERNAL icoritBftlslfó¡¡4rt*Gtr{¡Es
AIR] STANPARD OTO¡.D8 J,/r If f !¡TI4 i
leonrprusp_Jffiqpeacr{gna bA:h.i,,r, :ririiro,:
l:.::"r:r.Y.i9**sit\g
r6r'c:!) ¿5ri' ''oI::si-ni l':u;v_rred¡[.Vrai r.ú...!,;:, si\
.tX,* síqrcexit
Initial pressure, ii a.¡iqFJ{b.! x É0ú ', 80.::.il x ,T = ui
I.i.ri', tr'r'.)'(lf\ 'r"o .r.,izz."(sri.--i 91ri.¿3,ri rn.,,,n,rr,,,, r(-i, r;1. ¿.,..rt(!í-).\)..1
:,¡n,::,¡,,\:i..rj¡r:.
or -'t = sl\!{,:Í¡lrie lil} ln:.r¡rr,,:rli i,.,;,,r)¡i,,,r.r..,lilt:,i,T ..i,.:t, y:\, r.., ,.
*=t
,t}tr¡b:R¡1'linr:,',t";;i1;"i=
u. 1"*i"*1 '," .,,
Initial temperature, f, = 30 + 273 = 303 K -?-1i ?.?r:')"q :rnsut¡r;
Maximumpressure,
Cut-ofr .
pg=p.=5¿¡"t'
voluDe .s-l=é
tsr¡t¿¡'..1r) er.Í") '!i)';l

or = 4% of shoke 3,{ sq
Number of working cycledsec'g 3.
'i :r ü.tSi' = r.,. T. -_ i¡
B= !c= 95= = f.¿Z
Pressureorexplosionratio, 1{ úS0S = =;;
(bar) I r-r.l: '.t
puningthevalue | .l-q
ofr, p anit p h:il"f":'dr, ;;r
r,
*-ó.+-'i.l
i
¡¡
¡c.fj util l- 1

- t" g
Ít.I = q ru $U,U = ';-
IE
;
.¿'itt¡1q lliJi;¿lYIq :i 11 ¡) ! a i!r'-r

ratio 14 and, the explosion ratio

soluúion' Refer Fip.,,sn]-p,l*"*{l

----- ; [¿;\'." rJ i,rl$ lí..
ratio, 1' vÚ)
Comr,r'ession |
it-' - -. I
- . :- . I ü.a ¿\¡ ,oiler roierr*qxs ctir1-
r?.:\r!i'i(: )!i$(\:)iLD aO t
{¡s.l-r. ¡l
P = r.78

j..,.¡ii¡,'i¡.,; ;¡ .¡l-1

. r.',':::i')1 ;,::: t',

,,,-p*"-p,e3.25rhe,"*;;Hm;,y;;::;::-:#",du,a,cyc,e
is e. The maxi¡num pressure,in the ': )i0.ó', rt
cyti"tu úi;il";"t1b!.to;, rh";;;r;;;;;";ftumper4ture
the air at the besiniing of tle ct!1a11 ¡ t"i i#rt.b. of
od.d."d. during coÁnnt pressure
e =r
-r %,-.,
f:r, ,!:
t¡l:q,-t¡i¿"":¿;o;,;; ;;á'fiíi
';ff::;3:;,':;' :":;';:l:!;;"t##i!¿ü ,uo
*
o%t¿f.;
.',., :,..(i\.{!re : air pn4d,ar@ effigiency of the qcle. ^*,^ ^ = g oplgs,i",, .,,¡, ,,, ¡*r..', ..,.,r,
r;
(ii) The pou:er aer"i.n"1.-iytny_numbei Vt= V"+ Y"= 0.0L47 + 0.0018' j 9.6¡66:¡ie ;'
of worki4g cycles are B per second,. For the ad.inbatic Q1 iyn!rop!g),.8¡99esg 1,Q,
, '.,- air cu = Q./7 hJlkg:K a@ : cp r.0 O.liW*.., ;

Tg!".f!,
Soh¡tion. P=
' ..-.:' l
Refer Fig. 8.21.
Cylinder diameter, D = 250 mm = Q.!g ¡¡
--:,'i'1;'.
Compression ratio; r -..¡ ;;g I , I r'r'' :' :.
Stroke length, . ¿ = 300 mm = 0.3 i,r¡ '

I
t

I
 121
AIR STANDARD CYCLES .
126 INTBRNAL COMBUSTION ENGINES
cExample3.26.InanengíneworkingonDuoleycle,thetemperatureand,pressureatthe
Tz= Ttr
2.408 = 303 x 2.408 = 729.6K compression r'atí9-í::',The maximum
beeinning of the clcle are 90'C ood I bo' respectiuely'.Theper ig of air is 1750 kJ' Determine :
For the cozsúont uolume ptzr,es 2.9, i"Í""rií7 i i¿Á¡id. to ea a*-""¿1otit n"Lt
"uppu"a
Ts (t) Pressure and temperotures at oll salient points
=T" (ii) Air standard efficiency
hh
60 (iii) Mean effective Pressure'
To=T". ¿l
pz =?29.6, ñF =20zox Solution. Refer Fig' 3'22'
p-l 1
Also,
¡J=lgO or 0'04
p-1
9-l =0.(X c e=1.32
For the consf¿nt pressure prws 34,

+=+ * t=* ='

Tt= fs x P =2fl?oxL'32=2666'4K
'' v- V, t
Also exPansion ratio' =V5 xVt
vr =Vr,.
v2 vl = p ['.' V¡ =Vr andV2 =V3]
r v2
For ad.inbatic process 4-5,

a=fyrI-'=ls)'-'
Tt lvsl \"/
r¡=rtx(:l'=2666.4i (T)'' ' =rz37r
Also PlVl = P"Vtr
FE.3.Z2
/v. Y I'Y lm2\'n
ps=pt.lüJ =*'l.oj =60'[.9J =4.08bar itial pressure, pr=1bar
Heat supplied, In\al lemperature' ?, = 90 + 273 = 363 K
Q = co(?¡ - Tr) + cr(To- Tt)
Compression ratio, r=9
= O.It (Wn - 729.6) + 1.0 (2666.4 - 2O2O) = 1562.58 kJ/t<s
Maximum Pressure' Ps = Pa = 68 bar
Heat rejected, Q,= co(T¡- T) = 1750 kJ/kg
Total heat suPPlied
= O.7r (1287 - 303) = 663.14 kJ&g
(i) Pressures and üemperaüures at salienü points :

For ttre isenúroPic Process L-2,

\ l¡¡-¡t¡¡tla¡<t = 0.6766 o¡ 67.56%. (Ans.) prVl=Prv]
(¿) Power developed by the engine, P : . .,y
¿rt
(Ans.)
1x (9)14 = 21'67 bar'
Mass of ai¡ in ühe cycle is given by " [ir) =lx(ry =
p¡=pr
rx195 x9..0-roo
^ = +- = 0.0189 kg '
Arso, T2=f]i)'-t =(r)1-r =(s)La-1 =2'408
.'. Work done per cycle = m(Q,- Q)
i'fU K' (Ans')
.. T2 = Trx 2'408 = 363 x 2'4Og = 814'l
= 0.0189 (1562.58 - 663.14) = 16.999 kJ bar' (Ans')
Ps = P¿ = 68
Power deueloped = Work done per cycle x No, ofcycles per second
Fo¡ the consúont uolurne process 2-3,
= 16.999 x 3 = 60.99 say 61 kW. (Ans.)
 128
INTERNAL coMBUsTIoN
ENGINES
A¡R STANDARD CYCLES
PL 129
=b-
724
b :' p- = h @v" -v¡ !!Úsr-&J lz - *'"=,
Heat added at constant voluoe
T¡=?¡x =874'7.
# =2742pr( (Anc.) -il
r=9,p=1.16,y=1.4
i
+
-?*.f
- T2) = o'71 (2742'9 - 874'1) = 1326.8 kJrkg Pr = I bar, pz = 21.67 bar, p, = pn = 6g bar, p6 = B.g1 bar
.'. Heat added at consranr,xs Substituting the above values in the above equation, we get
= Total heat added _ Heat added aü
constant volume 68"1i*_sr.B1*e
,.
"r("4 "B)
I,O(T|_
- :
l75o - t326'8 = 423'2 kJ/*c r_ = o!loacrrs-tr* tr?l
.\
= f tro.z + 109.?? - 91.6?) = 11,04 tm¡
2742.9) = 423.2

;^- T¡ = 8166 K. (Ans.) Henen, mean effective pressure

ror consta,nt pressure process B-4, = 11.04 bar. (An*)
Example 8.27. An I.C. engíne operating on the dual cyck
Aimited pressurc cycle) the
temperdture of tlw working fluí'd (air) at the beginning of comjressian
l
,=+=+=ffi=rra maritnum and' minimum pressures of the qrcte * zO aid compressian
is zz;c. The ratin of the
ratio is 15. The amounts of
For adiabatic (or isentropic) process --'-'- heat added. at constant volume and, a.t constant pressure are
LS, equar, compute the air standard,
t!.e1mal efficiency of the cycle. State three maiti rea,sons why
the actucll therrnal efficiency is

AIso
+=t-+=#,"+=; d,ifferent from the th¿oretical ualue.
Tahc y for aír = 1.4.
PqvaT = povur
(u.P.s.c. 1997)

" P5='.,[F,J =",(:)'=ee;ffi'n=s.s¡ = =

P¡üUI
Solution. Refer Fig. 3.28. Giuen : T, 27 + 2ZB 3(X) K; '.-
u2 u3 = fs
?l =70, -=-'

Again, !L l&y-'-lp)t-l
11= (vu) -t;/ =l./r.15\11-r
,j =o'4ge
t T, = T. x 0.4Ítg = 8166 x O.nr,
(ii) Air standard efñciency
:
= ,rrr., K. (Ans.)
Heat rejected during constant
volume process 5-1,
= ,, = 0.71(l38e.8 _ s6B)
.'. "f:ru,_ -Tr) = 72s kJtkg
nair.¡ra¡dod ==gl$ga = a" -e"
a,
T#,"-.?X"',
(rii) Mean effective
= -l?5t- = 0'68i14 or 68'$47o. (Ans.)
o..o.
M".' .;;",i";;;;"1l,ffi;"ü'
p- = %IF d-on" P"t cy.l"
st¡oke volume

psvs Fig. 3.23. Dual cycle.

o^ = P¿vL -
ffo, <v, - vr¡ + 1ir standerd efiñciency, Iah-¡t¡¡dard i
Consider I kg ofair.
í1:[;;;;'u=va=u",v¿ =pv",
f' "=\:=r-;]"l
Adiabatic compression process 7-2 :
l

1." %=(r-r)%
=(t¡)ta-1 = 2.gs4
130 INTERNAL COMBUST¡ON ENCINES
AIR STANDARD CYCLES I31
. T2 = 3fi) x 2.954 = 886.2 K
3. Effect ofvariable specific heat, heat loss through clinder walls, inlet and exhaust vejoci-
et ties of airlgas etc. have noú been taken into account.
pr -[qf -,rr¡n =+ p2= 44.3 pr
\u2 ,, o<rExample 3.28. A Diesel engine worhing on a dual combustion cycle has a stroke uolume
Constant pressure process 2-3 : of 0.0O85 m3 and. a cotnpression ratio 15 : 7. The fitel has a calorific ualue of 43890 hJlhg. At the
end of suction, the air is at 1 bar ond, 100"C. Th¿ ma*í¡num pressure in the qcle is 65 bar
lz=b and ai.ir fúet ratio i.s 21 : 1. Find for ideal cycle th¿ thermal efficiency. Assume cp = 1.0 hJ / hg K and
T2 Ts
c" = 0.71 kJ lkg K.
Solution. Refer Fig. 3.24.
Ts= Tzx '4a = aeo z
pz ' ' +!
44.3p1 = 14oo K
p (bar)
Also, Heat added at :onstant volume = Heat added at constant pressure ...(Given)
or cu(Tt- T) = cp(T+- Ts)
or Ts- T2= 1(T. - ?r)
ro=r^* Y
"T = 14oo +
tlrtfrg = u6? rc
Constant uolume process 34 :

:+
fr=t t=+=ffi =,,u
Also, = 1'26 or u¿ = o'084 ur
f = dt
Also, us = ul
Ad,íobatic etponsion process 4-5 :
^ / \r-l¡1.4-l I -f- 1(1oo"c)
Ts lur / =l-+-l
+=lYLl ( 0.08au1 =2.6e
./
I

To
" = 2.69=1767
T. 656.9 K V" = 0.0085 m3
L69 =
Work done Heat supplied - Heat rejected Fig.3.24
'rair'shndard - Heat supplied Heat supplied ?1= 100 +273=373}{
=
. Heat rejected
Pr=1bar
'- Heat s"ppliud

1- cu(Ts -Tt) Stroke volume, V" = 0'0085 mg

=
%(\ -T) + co(Ta -Ts) Air-fuel ratio =21l.l
(Ts-\) Compression ratio, r=l'c:I
,
(\ -T) + 1(Ta -Ts) Calorific value of fuel' C = 43890 kJ/tg
cp = 1.0 kJ/ke K cu = 0.7I kJilrg K
(656.9 - 3oo)
(1400
= 0.653 or 65.37o. (Ans.) Thermal efficiency :
- 886.2) + 1.4(1767 - 1400) % = Vt - Vz = 0'0085 m3
Reasons for actual thermol effi.ciency being dffirent from the theoretical ualue : v,
1. In theoretical cycle working substance is taken air whereas in actual cycle air with fuel and as r= 7f
V2 = 15, then V, = 15V,
acts as uorhing substance.
75V2- V2 = 0.0085
2. The fuel combustion phenomenon and associated problems like dissociation of gases,
dilution of charge during suction stroke, etc. haye not been taken into account. or 14V, = 0.0085
0'0085
or v^=
z v^=V
ó c- aÁ = o.ooo6 m3
 INTERNAL COMBUSTION ENCINES
A¡R STANDARD CYCLES
For adiabatb compression process l-2,
P1V{ = prv{ or ,r=
h=# =L447.5Kor 1124.5.C

tfi =rxos)L' lt=;=#=t.4

pz=pt Heat rejected during constant volume process 5-1,
=mcu(Tu-Tr)
= 45.5 bar = (0.00854 + 0.0004) x O.7l (1447 ,5 - 373) = 6.713 kJ
Work dooe = Heat supplied - Heat rejected
= (2.898 + 14.66) - 6.?13 = 10.845 ltt
Also, (ry-l=(101.41-r :.
?=(ül'= = B.o4 Thermal effzcíency,
,. Tz= ?r x 9.04 = BZ3 x 3.04 = 1184 K or 861"C ¡.. =- Work done 10 R14
=O,6126o161.167o. (AnsJ
Fot constant volume process 2-3, Heat supplied (2.898 + r+.oo.r
¡eExample 3.29. The cotnpression ratio ond erpansian rutío of an oil engine worhing on
Pz _Ps
tfu dual qtcle are I ancl 5 respectiuely. The inítial pressure ond, Emperature of the air are 1 bar
Tz Tt and 30'C' TIE hfut líberated dt ccinstant press¿re is twíce the h¿at liberqted at constant volume.
The expansion dnd. compression folLow the law pVL25 - consta,nt, Determine :
Ts= rz' fr =1184x fr =rezox (i)Pr¿ss¿r¿s and temperotures at all salient points.
According to characteristic equaüion ofgas, (ii) Mean effective pressure of the cycle.
PlVr= (iii) Efficíene1 of the cycle.
^¡¿7, (iu) Pouer of the engine if working cycles per second. a¡e g.
m _ .pü _ 1xld xo.q)g
RTt 297xg7g = 0.0084 ks (atu) Assume : Cylind.er bore = 250 ¡nm and strohie length
= 400 mm,
Heat add¿d. during constant volum¿ process 2-3, Solution. Refer Fig. 3.25.
-mtcu(\-Tz)
= 0.0084 x 0.?l (1G20 - 1194)
= 2.898 kI
Amount of fuel added during the constant volume process 2-3,
2.898
= 4g890 = 0.000066 kg
Also as air-fuel ratio is 21 : 1,

.'. Total amount of tuel added = = 0.0004 kc

W
Quantity offuel added during the process B-4,
= 0.0004 - 0.000066 = 0.O0OB34 kc
:.Heot added d.uring the constant pressure operation B-4
= 0.000334 x 43890 = 14.66 kJ
But (0,0084 + 0.0004) c, (?, - ?.) = 14.66
0.0088 x 1.O (74 - 1620) = 14.66

t, = + 1620 = 3286 Kor

Again for operation 3-4,
# 3018"C

b=v, or Vo= Ys4 _ 0.0006x9286 = 0.001217 m3

Ts 1620
For adiabatic expansion op"r"tio? ¿-# Fig.3.25
-r Initial temperature, T, = 30 + 273 = 303 K
A =f%./
= | o.oog
l&)'-' -=lororn?J '1tar
Initial pressure,
4 =2-27 Pl=1bar
Compression and expansion law,
pllr.zs = constant
 I35
134 INTERNAL COMBUSTION ENGINES STANDARD CYCLES

Compression ratio, r.=9 ly.)"_- t =g:-=4.?zbar.

- aü (Ans.)
Expansion ratio, ¡¡=5 P5= Pax = P1' (5Fr5 -
Number of cycles/sec. = 8
l%J
Cylinder diarneter, D = 2ffi mm = 0.25 m
Arso +=f}I '=+-="*- =o',,,
Stroke length, tr=400mm=0.4m T1 [Vsl Q,Y-' (5r- ^

x 0'668 = 1445 K or 1172'c' 6ns')

Heat liberated at constant pressure
i¿¿l uu"o effective '0'66s:2163'4
= 2 x heat liberated at constant volume
(i) Pressure and tem¡reretures aü all salient pointe : """"lii"]';:
Mean effective Pressure is given bY
For compression process 7-2,
r-= {frrtv. -vi*ú#
vrVi = nzY{
p¡=p1 x
lv,'\"
=1x(e)1.26=15.59ber. (Ane.)
=riof*(p-l)+W
-rrL(rc "#l
tüJ Now,t"=p,p=1-8't=t'25'pr=1bar'p'=16'59bar'p3=35'7ba¡'
-'=(e)1.26-, na = 35.7 bar, po 4.77 bar
3=fE) " =
A1so,
Tr lvz)
= r.z'2
,- =
¿¡" tr e - r¡ + ellffifre - "11h-'i']
.'. T, = Tt x 1.732 = 303 x 7.732
= 624.8 K or 251.8'C. (Ans.) 1
tza.se + 85.32 - 26.361 = 10.e4 bar
= i
Also, cD(Tl- Tr) = 2 * cu(Ts- T2) (nven)
For constant pressure process 3-4, Hene mean effective pressure = 10'94 ber' (Ang')
v... u u EfficiencY of the cYcle :
(iiü)
Tt _Vt ^ _ Compression ratio (l;) uo"'= ur"ñ Work done per cycle is given by W = p^V,
4 V3' Expansionratio(¿) u1
=iX-
here

=9= tt v3P = 19!g#!g9q kJ/cycle = 21.44 kltcycre

D V'lr- "
Tt=t.8Tz =-x-=i
Vz P P /eat supplied per cYcle = mQ,
r^ r^
p=É=; - i" th" mass of air per cycle which ie given uy
, v"
v, =V"+v"=
v4 ^= # where
f1
Substituting the values of ?, and Tn in the equation (i), we get
1.0(1.8?s -T¡l=2 x 0.71(?s -524.8) '"=L{='*fr * u"=# I
0.8Tr= 1.42(Tr- 524.8)
.. vt=v,.#=*[t.
O.8Tr=l.42Tr-745.2
0'62Tt=745'2
#)=*'.1
q
"' T¡ = 1201.9 K or 928.9'C. (Ans.) 0.02205 m3
\ ni= '0.0196 =
Arso, +=t ..,,..for process 3-2
;. m= 1# = o.o2bs5 kg/cycle
Ts 1201 9
Ps=P2 x T, = tr.ro, - 85.7 bsr. (Ans.) Heat supplied per cycle
-524.g
"' me"=0.02585tc,r?3- Tr) + co(Tn- Tr)l
=
P¿ =P, = 36.7 bar. (Ans.)
0.0i535[0.71(1201.9 524.8) + 1.0(2163'4- - 1201'9)]
=
T¡ = 1.8?s = 1.8 x 1201.9 = 2183.4 X or 1890.4'C. 6ns.)
For etpansion process 4-5, = 36.56 kJ/cycle
Work done Per cYcle 21.4
PaVl^ = PsVun
Efficíenqr =
E*t *ppit"d p"t tytt" = 3656
= 0.5864 or 58.647o. (AnsJ
 INTERNAL COMBUSTTON
ENG¡NES AIR STANDARD CYCLES
(iu) Power of the engine, p :
Power of the engine, P = Work done per second S.7.2. For tüe Same Compreeelon Ratio and the Same Eeat Input
A comparison of the rycles (Oüto, Diesel and Dual) on thep-u.and ?-s diagrams for the s¿/¿e
= Work done per cycle x no. ofcycles/sec
omprusinn ratia and, luat eupplied, is shown in the Fig. 3.27.
= 27.44 \ 8 = 17t.62 k\y. (tns.)
3.7. COMPARISON oF OIto, DIESEL AND DUAL
COMBUSTION CycLEs Otto = 1,2,3',4'
are the important variable factors 'Diesel
which are used as a basis for = 1,2, 3, 4
.""t", lo"o*tng comparison ofthe Dual = 1',2', t, 4'
.Compression ratio.
o Maximum pressure
. Heat supplied
o Heat rejected
r Net work.
Some of the above mr
¿,"t .o-¡u"tio;;ñ:'; ;fl:iT;XHfo'"'are fixed when the performance of otro, Diesel and
3.7.1. Efficiency Versus Compression
Ratio
Fig. 3.26_shows the comparis
combustion
i¿ron cycles
cycres at orthe
of-the otto,
ouo, Dieset
.""rr':lj:;"t1""i::"::11T9."qciencies
ar various vv¡r¡v¡so'rur¡
""ti"*;;;;;il";il';"4i"ffÍ,"{1"i"".i""
ranos and *lll,4y"" cut Diesel and Duat
Dual combusiion
combustion cycles. tiil" jllgy"" cut.off-ratio
It is evident from rha Fia r-.1.L_r11,1n.a¡"1a,¿"1á
a o¡
off_ratio for the Diesel aJ
gl.!1" rncrease "y.t"".
wEn rne i191ea¡e i" t¡"
í" ,h" the air standard erficiencies
.m"iJi,.r". io"""usi
increase
efficient
eff "";;::::1"1J:Tjy:*; j"fi:jjt:1]
""-p""Jil;:1il';;.'r;";íf,:nu
ícíentuthítetheDirr"t"!;i;;";;';;;f;n;i:"i.fi
uthile the Diesel qtue rs the ted.st cffraiont h t compr*sion ratio otto cyc¿e is the most
:: Fig. 3.27. (a)p-u diag¡am, (b) ?-s diag¡am.

We know that. ñ=l- - Heat reiected

- ,..(3.13)
Heat supplied
Since all the cycles reject their heat at the same specific volume, process line from state 4 to
1, the quantity of heat rejected, from each cXcle is represented by the oppropríate area under the
line 4 to 7 on the T-s diogram. As is evident from the eqn. (3.13) the cycle which has the least heat
rejected will have the highest efliciency. Thus, Otto cycle is the most efFcient and Diesel cycle is
FAA
the least eflicient ofthe three cycles.
*"" i.e., Sott¡ ) l'ldual > Idiqa .

For Constanü Meximun Pressur€ and Heat Supplied

3.7.8.
Fig. 3.28 shows the Otto and Diesel cycles on p-u and ?-s diagranrs for constant maximum
pressure and heat input respectively.

operatino c-.-.I engtne opefating

S.l. engine
,mnracc¡^h rari^
compression -^_-_^
ratio range corpre"iiln iÍü?n"ou
Compression ratio, r-_-____+
Fig. 3.26. Comparison ofelficiency
at various compression ratio.

.,rt. .,il""11;ffi,il:iü,:i:,:iliff'"ilT"*'r"rg petrorcnsine is rimited by detonation. rn their

ll: respective

Fig. 3.28. (o)p-u diag¡am, (b) ?-s diagram

 138 AIR STANDARD CYCLES
TNTERNAL coMBusrroN ENcrNEs
For the maximum pressure the points 3 and (i) l-2-Eeot rejeetion at constdnt pressure
- B'must lie on a constant pressure line.
on ?-s diagram the he¿ü rejected from the Diesel (ü) 2-?-Adi¿b atic c omp re s sio n
- the line 4 to I and this a¡éa is less than the
cycle is represented byibe area under
hence the Dieset ycle is more
otto cycle ú;d", tí"-curve ( to 1 ; (iii) s-4-Addition of heat at constant uolume
"6"i"it-ino, the otto qcb for tre cond.ition of maxi-
mu¡rl pressure and heat eupplied. (iv) A-l-Mizbat ic ex pan s io n.
B¡qmple a'w. (a) with.tlz h'etp of p-u ond r-s d.iagran
otto,.d'iesel and d.uar combustian compne tlp cold air standard
"wt;" f";
;";;;""imu-m presstie or¿ tenperd.-
^ii^u^
(AMrE Sunmer, 1998)
solution. Refer Fig. 3.29. (a, b).
The air-standard otto, Dual and Dieser cycles
for the same maximum erTlu: and ;;;;;;;;".ture, are drawn- on commotrp-u and ?-s diagrams
for tbe purpose of comparison.
Otto 1-2-8-,1-1, Dual 1-218,-8_4-1, Diesel t-2._S_4¡L(Fig
8.29 (c)).
volume lines on ?-s diagram is higher than
t,""r. ,Flls.llconstant thaü of constant pressure

Fis.3.30

Consider 1 kg of air
Compression ratio =2 =o

Expansion ratio =1t=.

U1

Heat supplied at constant volu¡¡g = s,(?. - ?3)

Fig. S.29
Here the otto cycle must be limited to a low
Heat rejected = cu(Tt- Tz)
compression ratio (r) to fulfill tüe
point B (same maxirnurn pressure and
temperl;;;.ñ; b" . .o-.ooo át"ii rl"'"ilti"condition
tt
that
cycles.
Work done = Heat supplied - Heat rejected
The constructio" of^rJ:r"" o'. ?-s diagram proves ""e = cu(Tt- Tt) - c,(Tt- Tt)
rejected is same for all the three cycles (""""-;;d;"';;essthat for tüe given conditione the heat
line 4-1). Since, by definition.
¡' = Workdone -_c,(T+-Ti-cr(Tt-Tz)
const. Heat supplied cu (Ta - T3)
n=t - E3ljggctgg,a, -.| _
(Tr-T2)
the cvcre, with srearer heat addition *rff:ti::il*,.1r. *"*?;" ?.s diagram, _ ,.
=1_r (\_\) ... (¿.,
A",¿¡or¡ = Area under 2,13
Dunlrg ad¿abatic compression 2-3,
Qt¿ua¡ = A¡ea under 2-318
Q<q$o, = Area under 2-3. ^ / r1-l
5=l2l
_ It can be\eem tnat, tdn*i"r,, e<¿u"u, e"(or") T2 l.,t/ =r.')r-1
and thus, tld¡"""r ) Tldo"l ) lo.¡. -t
Tt = T, 1sY ...ui)
3.8. ATKINSON CYCLE During constant pressure operation 1-2
This cycle consists of two ad,iabatics, a constant UUq
uorume and a consta,nt pressure process.
p-v diagram of this cycle is shown in Fig. TL r,
3.30. It consists of the folrowing
/our operations :

I
 140
INTERNAL COMBUSTION ENGINES ATR STANDARD CYCLE,S

L=rt =y p (bar)
Trull ...(iii)
(o=oxuB=%x%=g)
Dwng adiabatic expansian 4-1, \ur U9 Ut lt ur r)
/ \1-l
A=l3ll -\"
Tr lr.J -r-rr-r
r,= *-
(rr' - ...(iu)
Putting the value of ?, in equation (iii),
{i we get
T,= Tt-r' a
I G)f ,
ll
tl
=ú,
Substituting the value of ?, ir, (di), we get
"q,riloo v(mt)

"s=411s¡r-r=lS)'.
t Fig.3.31
Finally putting rhe varues r,, rfl"a * ,, Te=Tz* 1.284=800 x1.284= gB5.2K or!12.2.C
"r "l,llu"" ,r'r, *, ,o For constant uolume process J-4,
Pq _Pa
T4 T3

16x3852
- PtTz -
, t=;= =1540.8Kor1267.8'C
4
Hene.e, air stand,ard, effi.ciency For adiabatic expansion process
=1 - t. l+a+') - 4_1,
)rn , ,y-l L22_L
s.st. A pe,fect ,", ,^**""\l'r"7" r6).F
,"u", Ír*]#!"ri Tt \p'J '
2=ta)
of the fottowíngr-:::: =f = r.ena
"uco¿sisrs
(a) Heat rejection at
\r/
constant pressure.
(b) Ad.iabatíc compressíon
I
_ To = 1540.8
Tt=T.648 934.9 K or 661.9.C.
from bar and 27.C to 4 brlr. 1"648 =
(c) Heat addition at t (i) Work done per kg of gas, W :
to a finat pressure of 16 bar.
(d) Ad.iabatic
:::ume Heat supplied = cu (To- T")
Calculate : (í) ""r";;::::'
Work donelkg of gas.
(ií) Efficíency of the qrcle. = 0.75 (1540.8 - 385.2) = 866,? kJ/tte
Heat rejected c, (Ty-
= ?2) = 0.92(934.9 -
300) = 584.1 kJ¿rC
?!" ' "o Refer kJlke K, c, = o.7s kJl ks K.
Solution.
= 0.92 Worh done / kg of gas, W = Heat supplied - I{eat rejected
Fig. 8.81.
Pressure,
=Pt = [¿¡
-
= 866.? 584.1 = 282.6 kJ/hg = 282600 N-m,4<g. (Ans.)
Temperature,
P2 1 (ll) E¡¡¡.'.n"" of the cycle :
Tz=27+2?3=300X
Pressure after adiabatic compression, pz= 4bar Effícíency, n' = J:*!"1:= =r=3! (Ans.)
Heat srrpplied 866.2 =0.326o1 82.67o.
-
Final pressure after heat addition,
For adíabatic compression 2-3,
ii = fe U".
3.9. ERICSSON CYCLE
It is so named as it was invented byEricsson. Fig. B.32 shows p-vdiagram of this cycle.
 INTERNAL COMBUSTION ENGINES ATR STANDARD CYCLES I43
It comprises ofthe following operations : The variouE operations are as follows : ,

(i) l-%-Rejection of heat at consto,nt pnessure Operation 1-2.Ihe air is conpressed isentropically from the lower pressure p1 to the
(ii) 2-3-I s ot her mal c o mp r e s sion upper pressure p2, the temperature rising from ?t to Tr. No heat flow occurs.
(ii¿) !-4-Additíon of heat at constant pressure O¡reration 2.9. Heat flows into the aysten increasing the volume from Vz¡n Vs and tem-
(iu) 4 - L-Isot her mal exp an sion. per'ature fuom, Trto ?, whilst the pressure renains constant at pr. Heat received = mcr(T"- Tr).
Operaüion 8.4. the air is expanded isentropically from p, to p' the temperature falling
from 4 to ?.. No heat flow occurs.
Operetion 4-1. Heat is rejected from the system a8 the volume decreases from V{to Vl and
the temperature from To to ?, whiht the pressure rernai¡sconstantatpr.Heatrejectcd=mco
(T1-T).
do,o"
r¡¡.¡t¿rd¡r¿ = ,
=.*ltkrecelveo
rleeE
_ Heat ¡eceiveücycle - Heat ¡ejecteiVcycle
Heat receiveücycle
mco (Ts -Tl) - nrc, (Tt -T) _, _ Tt -Tt
_ =,_84

Fis.3.32
Consider 1 kg ofair.

Volumeratio, ,=h=\
ug u4

Heat supplied to air from a¡t external source tii

= Heat supplied during the isothermal expansion 4"1

= RTrlog.r C=Comp¡essor T=Tuüine
Heat rejected by air to an external source (a)
= RTz . log" r
Work done = Heat süpplied - Heat rejected lll
= fiIr - RT". log"
. lo& r r = I log. r (TL- T2) l,;

_'' Work done Rlog. r (T1 - T2)

Heat supplied RT1.log" r
T'-T' ...(3.15)
= Tr
which is ühe sime as Carnot cycle.

BRAITON CYCLE
t
S.10.
Brayüon cycle is a constant pressure cycle for a perfect gas. It is also called Joule cycle,
The heat transfers are achieved in reversible constant pressure heat exchangers. An ideal gas
Fig. 3.33. Brayto¡ c,ycle : (o) Basic omponeats of a gas turbine power plmt
turbine plant would perform the processes that make up a Brayton cycle. The cycle is shown in the
Fig. 3.33 (a) aod it is represented on p-u and ?-s diagrams as shown in Fig. 3.33 (ó, c). (ó)p-V diag¡a¡n (c) ?-S diagram, t:i:
jir
.ril
!tl
nli
ifl
ii;
 INTERNAL COMBUSTTON
ENGINES
Now, from isentropic expansion AIR STÁNdARD CYCLES 145

TI h p*lofj given turbine the minimum temperature ?, and übe maximum temperature ?,
a,=(a\ f p.rTcribed'
ar.e. being the temperatu¡e of the ahosphere irnd ?, the rraximum temperature
rr la/ -Tt
which the metals ofturbine would with¡tand. Consider the specific héat at constant pressure co to
be conetant' Then'
II ,-l
Tr= T1 b) r , where ro = h.essu¡e ratio. n ñ
Since, t=rrr' ='ñ
Similarly a=l¿)f ot
rl )
rr laj Ts=T¿bbr'l
Using the consta¡t 2'=
+,
.'. 1- T" -T' we have, work outpuUcycle
naü*ta¡da¡d =
---ti------E -' - ---T:T 1

| ( ,\
rat) r -T{rolr G)T ...(3.16)
I
L T *T;)
w = K lrsl t - l-t, (rn" - L)l
J
Differentiating with respect to
1

Is ' o ror a maxir¡um

#=*Fr,#^-rruou-,,1=
F
¡e
' '
A
rPQ+i
=TF(rPl" -r\

," =+
1
rp =(T{\)ub i.e. ro = (Ts¡)zl-r) ...(3.17)
Thus, the pressure ratio for manimum work is o functíon of the limiting tenperature
Fíg. 8.84. Eflect of preszure ratio
on the efficiency of Brayton
cycle. Work Ratio
The eqn. (3.16) shows that
!!1T::Z "t,ih"
,*"t ¡:yt, qrcle inoeoses wíth the pressure worh ratía is defined as tl¡e rcüdo of net worh output to tlt¿ itorh dnw by the turbíne.
i¡^ii¡ü-i,tí"ra,ure o*he
ii,?,,íii;i;:'::i::::,"1;:;:;,r.;';iir
process atone, no
turtherieltipg
X:*::! ? 'i:
loii";";;::r;::n;;:#:;rr"r::X{S;:;":;
.'. Work ratio =
W'.;%
wT
sÁ
'::;;':"':,::í::;';,-#:*::::"t*"t"; b"r;;;;;';;,n a*)on,,ft;,*í; excess work fwhere \f¡ = Work obtained ñom this turbine,l

Pressure ratio for maxinun worA

land Wc = Work suppüed to the compressorJ
: mco(Ts-T)-mcr(Tz-T) Tz_Tt
tu*p"I::r::"shall
prove that the pressu re rotio for maximum worh is a mco(Ts-Ta) =t' _ Ts-Tt

f'.
furction of the limítíng
Work output during the cycle
I
= Heat receiveücycle _ Heat rejecteücycle
= mco(Tr- T") _ mco e1_ T) =r- ;l_b+-rl=,_fto,,? ...(3.18)
= mco (Ts - Tr) _ nco ez _ T)
's¡t--f¡- |

=mc,r,(,_+) -(+ ,) |- ,r,TJ

Eranpfe 8Jl2. Air enters tln crlmpressor of a gas turbinc plant operating on Brayton qtcle
at 101.325 kP_a'
!7"C. The preseure ratio in the cycle is 6. Calcutáte the matimim temperature in
tIc cyde and th¿ qch ffiicncy._Asswrn Wr= 2.6 W" where Wrand W"are the íurbine ond
thz eompressor uorh respectiuely. Tahe "¡ - L4. (p"U.)
 INTERNAL COMBUSTION ENGINES 147
AIR STANDARD CYCLES
Solution. Pressure ofintake ai¡, p¡ 101.825 kpa
Temperature of intake air,
= Eranple 833. A gas turbíne is sttpplid with gas at 5 bar and 1000 K and' etpand's it
= 27 + ?lB = B(X) K ad.iabatinllX to 7 bar The mean specific heat at constont pressure and. constant aolume are 7.O425
", rD = 6
The pressure ratio in the cycle,
hJ I kg K and. A.7662 h,I I kg K respecüaely.
(i) Maximum temperature in the cycle,
T, :' (i) Draut the temperature-entrow di.ryrcm to represent the processes of the simple gos
Refer Fig 3.35. turbine systen,
(ii) Catculete the power d'eueloped. in hW per kg of gds per second' and the erhaust gas
tetnperoture. (GAT8,1996)
Solution, Giuen ; p, = bar ; Pz = 5 bar ; ?, = 1000 K i cp = 1.0425 kJ/kg R
1 ;

cu = 0.7662 kJ&g IL
l'uT
t- cD -
'=b= -
0?662 =r.so
(i) Temperaüure-enüropy diagtan :
Temperature-entropy diagram represenüng the processes of the simple gas turbine system
is sbown in Fig. 3.36.
p2=5bar
Fig.3.35

m / tt-t 1-l r{-r

!:=l?zl '
tt 1fr,l =r".1 ' =(6)T = 1.668

?z = 1.668 = 1.668 x 300 = 500.4 K

.Ll "r
Ll-l
*=<r; 1 =16¡ rr
Also, Pr=1bar
= 1.uut
Ts
T'=
* 1668
But Wr=2.8 Wc (Given)
mco (7, - Ta) = 2.6 mc, (T" - Tr)
-T-lr\
",- = 2.5 (500.4-300) = sor * = sor
refu
", |.t-¡fuj
Ts= = 1251 K or 9?8'C. (Ans.) FIg.3.36
1- I
L668 (u) Power required :
(íj) Cycle efifrcienc5l,.r1."o,. : r-l 136-1
T- 9xr
ra -= l¿-lr
r
Now, ?, 1" =ll)-r"- s 0.658
=
úÉr=ffi =?5oK
lp2) \5/
\ É
Network _ ztc, (4 -4)- mc, (4 -fi) ?. = 1000 x 0;653 = 653 K
" r.y.t"
Heatadded -
- - rr"r7rl) Power'developedper kg ofgas per second
trz¡r-ibqt-rpo¿ - gobl - = co(Ts- Ta)
(1251_500.4) = 0.4. or 4A%. .
(Ans.)
- 653) = 861.7 kW. (Ans.)
= t.0425 (1000
T
E¡ample .3.84 , An isentropit aír turbine is used. to suppty 0.1 hg I s of air at 0.1 MN I tn2
cne*,ncyae=1--7f,¡=1--+=T= o.4 ot 4vo, (Ans.) and at 285 k to a inbín. The pressvre at ínlet to the turbine is 0.4 MNlmz' Determine the
I tenperature at turb.ine inlet and. the power deoeloped, by the turbine. Assume co= 1.0 kJlhg K.
|
r ,")('J (6)-14
v D/ (GATE' 1999)
 r48
INTERNAL COMBUST¡OJ{
ENC¡NES
AIR STANDARD CYCLES
Solution. Gioen: i6 - 0.f hels i pr = 0.1
YN/*
cp : f Ut
1.0 k.lng r
T, = N K ; p"= o.¿ l¡rúbt=i¿ bar ;
(iu) Heat rejected in ühe coole,r, Q":
= (W)
Tenperature
Work output = Ileat eupplied (e) - Heat rejected (e2)
¡t úr¡¡üt¡o tnlef T, : :. ez= Ql- W = 466.27 - 136.88 = 819.39 k l/kg. (Ans.)

g=(a\i-rnl#
Y-l

-(il Tuüno
(u) Temperaüure of air leaving the turbine; Tn :
Fot e*pansíon (isentropb) pruess 3-1, we have
11 (n/ = 1.486
ñ IJ r¡r-1
:. Ts = 2&5 x 1.486
= {SSj X. (AE.) r =(8.5) 11 = 1.¿a \
Power developedn p : á=Or,
p = rhocr(Is_T)
t = *=Ir* = 610.5 K. (Ans.)
= 0.1 x 1.0 G?3.6_2u5)
- - [Check : Heat rejected ai¡ cooler at constant preseure during the process 4-l can also
in. the
= tS.8E l¡W. (Anr) Fis.3.37
_
be calculated as : Heat rejected = mx cp(T1-T)
= 1 x 1.00b x (610.b: ZSS) = 819.1 kJ/kgl
Er¡nPle 836. Conaid¿¡ ur ai¡ éwwtt
*tt n t^*t o"n :- _-., ! , ttu_ür
Á,^r^ ,o.-?!i"l Exanple 3.36. Air enters thc compressor of a gas turbine plant operating on Brayton cycle
1.0-y" 'q zec, nn-i*i*
and, áeC. Ttn preasure ff mr,?:k:-y!bl tlu-air enhn tle ompresnr
enten the ompresnr at at 7 bar,27'C. The pressure ratb itt ttu eXcle is 6. If Wr= 2.5 Wc, rh"r" Wrand.W" are the
ttb amnressor ¡¡ J.6 toi'ü-ütrp",wuo
turbin¿
curbúu ialet h 6UfC.
inJet is b*#JSHp
6oec. b"¿r ;;ü;;r:
"f ot turbine and compressor worh.rcsputively, calculote thá marimuk tumperatire andíhe c.vcle
A Etrzcienel of tlu qcl.e, (ü Hat supplbd, to air, efficienóy. (GATE, 1ó96)
(ü)Worh evailúle at tlu.W,
(iv) Heat rqjected
(v) Tempe¡atutz in the wbr, qd, Soluüion. Giuen:pr=1bar; Tt=27 +2?B=gg9 X; ? =6;W,=21.5yy"
of otr talüg tle turbin¿. PI
lolair t= L4 and, cp= 1.06 hJlhg K Maximum temperature, T, :
Solutlon. Refer Fig. g,gS.
m / .1-t L{-r
Pressure' of
air enteriag the ompreaaor, pr Now, f = l¿.l t =(6)T=(6)ü
o.{

=1.668
= 1.0 ba¡ tp|/
Temperature at the inlet of
rn K t
"r*of ;;ñ;m:';'#J""rtt
hessuto = t. Tz = 300 x 1.668 = 500.4 K
Temperature air at tuñinc idet, i iOo + Zzg = AZg X ^ /
Pzl'
.1-t o.a
(r) Etflcteacy of tüe cycle q*. : ", Arso, $
t4 =l\pr,, = (G)il = 1.668

1- I ( ?3 = 1.668 4
l"v¡" = -E
,
= 1- -+-, = o.3b or g0%.
(An¡J
l" o=ff-#=*J Now, compressor work
("r) (g5)-r¡- W" = mc, (Tz- Tr),
0¡) Eeat cuppüed to ¡ir: and turbine work,
Ft compreesbn prcess I-Nwe bave W, = mco (Ts - T)
ñ f r1-l - L¡l-l Since wf = 2.5 Wc ......(Given)

t,-l'") =(;
frltfJ'=(Tl-='* :. mco(Tr- T) = 2.5 x men(Tr- T1)
Tz= Tt x lry\
a¡r. e1= Co(T, ;;;;;il
29Í x 1.48 = 1.43
= 293 lrs - :*l
Fis.3.38
,.
i* orr¡K= 16s.27 t¡,nrs. (Ans.)
¿¡EáL Eqppue(¡ ¡o 419 = 2.5 (500.a - 300)
\ r-bbó/
, Work
0¿) #:l'"T$1': il;""1i.f,Í.:
¡vail¡ble atihe ch¡ü W :
r.fu- I
" \ 1.668/)=sor
We know that,
n_,.=Hi##
.. Te = 1251 K. (Ans.)
g.39
= -I- or W = 0.3 x 456,27
= 136.88 hr¿rg Cycle efficiency, Tlayq¡¿ :

W, -W" _ mc o(7,¡ - T¿) - mc o(Tz - Tt)

nrcYcle
.=
mcr(Tg-?2) mcr(Tg-?:2)

-A
 -T

INTERNAL COMBUSTION ENGINES AIR STAND{RD:CYCLES

Fv¡nFle 8.38. In a gas turbine plant working on Brayton cXcle, the ab at inlzt is ZZ"C, 0.1
_
f*t-l*'l-('001-soo)
r L668' MP.. TItc pnessure ratíp is 6.25 and. the ma¿imum tem¡rcrature is 9MC, Tl¿e turbínc and, com-
(u5r_500.4) prcssor effu:iznci¿s are each &ub. F¡n¿ compressor work, turbine worh, hcat supptíed, qrcle effi-

=
tHPn o.r0 or 1o%. 6$)
ciengt an'd' turbine exhpust temperature. Mass of oir may be considered, as 7 hg, Draw T-s d.ia-
gratn.
76(16 = (AllfIE Summen 20fi))
Erample 8.37. A cro*d qcre idar gu pbnt operates betwen temperaturc timits Solution Refer Fig. 3.40.
of 9p-'c and 30'c and prd,uces a puEr lm-turbüu
tw.'iru ptantrs dcsigned, euch tr,,t trere is ta
ú
need, for d' regen¿rator. A fwl of cala¡iftc asm
n t t *g is uod. calc¿ai" tn" iÁ
through the plant and tzte of'f"el rxlÁuniá";- " - rto, mte of ,,i.r
tAssume cp = I kJlhg K ard = Ll.
.l
. (GAIE, 19e8)
Solution. Given : Tr= 3O + II3 = fl)3 K; ?. g00 + nA 1073I(; C
_
cp = 1 lcllks K; T = t.¿ ; W;- = = = 45000 kZkg ;
W* = róOÉW.
th¡r rir¡ :
since no regenerator is used we c¡- assr¡me the tur. bine
a way that ühe exhaust gas temp€rature a,o tüe Spands the gases upto ?. in such
tu¡bine is equal L tn" t"-ñ*r-t."J"rair coming
out of the comp¡essor Le., T"= T.
_1 . ,l
Pz Ps p,
pt - pl , p¡=[&_)rt
l\) ""0 =[+j'-' f T-s dlagram

&-=r" =&.
'. Tr T1 T2 Giuen :
Fis.3.4O
T, = 27 + 273 = 900 K ; pr = 0.1 lttPo ; r, = 6.%, Ts= 800 + 275 = LO73 K
("' Tt=7, """assumed) Tlo-p.=lturuirr=0'8'
Tz2 = TtTs ot Tr= lffi For the cornpressíon praesa I-2, we have
i-:!
t{-l
?r= SGffi =6?0.2K =(625) 1' = 1.688
Now, W*r*"=
+=e)+=r',,?
Wru*r"- rz¡ x Cx¡ (f Tz= 3OO x 1'688 = 506'4 K

loo= m¡ x45(xx)'
-- ---vv ^
' ft-3:]il
L 4-"¡l
Arso, r.o.p.=
ffi or 0.8=
Tf,|# Iil
roo
=,itrx4**tt-ffi:il Fis.3.39
?r, = !9%; + Boo = 868 K li:

= rhs x ZlOg5.9 "' compreesor worle ==1:"í;;1il1'l*,

"*' = Z'e.*ekr/kg. (Ans.)
*f. = ,ñi
100
= 1.71 x, lüstrg/a (Anc.) For exponsion process 34, we have fit
¡suin{ wr*br." - w*o* = 1{)0 kw
(rito + rh¡XTs
-T1, - tho xlx(72 _ ?i) = IOO +=l+lf
T1 tp./ =(",)f =1u.ru¡ti = r..* fil
trt
(rit" + 0.0O474X10?g ii!
- SZO.2) - zr" (5?0.2.-.BOB) = 100 ¡¡,1

r. =
or
(rh"+ 0.00479 x 502.8 -267.2 tiz" = l0O #=# = 635.66 K líJ

602.8 t4 +2.383-267.2
or *
= 1O0
235.6 ñd = 97.61?
Also, rlt-bi,"= ffi or 0.8=
#-#; tü{

ir" = 0.414 k;gs. (A¡e.) |:

NI
tl

-l
ür
ti
I

,'t 152
TNTERNAL COMBUSTTON STANqARD CYCLES
ENctNEs
or T¡, = l0?8 _ 0.8 (1OZg _ 6q5.66)
153

,'. T\rrblne wor! W¡,+_ = 1 x co x (TB_ Tí)

= ?23.13 K 4rl:soo
. fuel mass) +3oo= 482.2K
Ner work o, = 1 ¡ 1.006 (10?S _ ?2B,lB)
= 35I.6 kr/kg. 6inB.)
";=
Norr', Heat supplied by the fuel = Heat taken by the burning gases
'r."t - 't*;Á'=ezsr krks
"opffil'. i:= I*FlI:^¡;:361'6
=I x 1.O06x(10?B_658)
O.9 x m¡ x C = (mo + m¡)x co x(Ts -?z)

=6t7.E7hfAg. t¡ns.l
cvcle
rurbine
efñcienc5r, n*, = F=# = o.I?ttit or
t?ai|%. (An¡.) "=W)'4#=[ff.'),.u%1
T:yi:*TT*tFig. S,i0.
lhe ?;s diagram is ehown in = 72s.rs Kor 450.1sc. (Ans.) 42000(+.)* r'oo(rrt=¿se¿z)
= l*.'l = ?se.?8
f,¡¡npre g.gg. r¿nd thc
required oir-fire. ratb-in a gas \',? )
pressor efficíencies are turbinc wru*
a¿ co*, ,""p""a""t1. M*i^;; ;;"ííii"orrr" turbüe and, com-
1 = 65'77 say 56

and'
-85%
working flui.d m" u:..!:.*.:"."i
%: l;í;;4 k,
27'C, The nressurS rutia is 4.
.r
is 875.c. The
= t.¿iii¿:;;;;;, il" compressor at 1 bar
+=#t-
fu f*, l.
*'*{ri?#:"ii:,:
't" - €t i
**'lla,f"íi^0",.
^ituu "otori¡" valuc of d2000Iurths'
,m;,i#ü 8.11. SITRIJNG CYCI,E
A/F ratio = 66 : (An¡J

800 K; =l.i [j;":irh-bi.e l*o'".n =,20% : Tr= 278 + 875 = 1trf8I! Tt 27 + 278
:
kcE¡-1'#(iltriH,';il1'h=ri"'ftra;
_=
"" jff,f *;i."ilLuir¿rló'=alódn*i
= EiC. 3,42 shows a stirling cycle (182?) on p-u and T-s
and two constant volume ptocesses.
diagrams. It coosbts of ilrlo isotherms
For isentroprh rrlmpresrlbn T_2 : P¡¡ocess 1'2 is the i&thcmal ampression with heat rqiection
temperatue fr. Q, to the tiilroundings at
P¡¡ocese 8'1 is the inthennat expansion wirh hcat addition
ture ?* e¡¡ from a Bou¡ce at tempera-

Fig. .4t

, .l-1
?
+=l?l
¿1 \A/ =r,,tii=r.¿se
\É,/ -
?¿ = 300 x 1.486 = 445.8 K
_
rfmposq = n-n (c)p-u diagram
(ó) ?-s diagram
ffi Fig. 3.42. StirliDgcycle.
08=## Process 2.8 ¡nd 1-l a¡e constant aolum¿ heat transfer pr@esses.
For t
.

hg of idea.l gas,

Ul
Q7-2(Heat rejected) = Wt.z= - RTLI¡ (mmpression)
u2
 ¡NTERNAL COMBUSTION ENCINES
AIR STANDARD CYCLES I55

Qzt= cu(TH-TLt; W¡¡ = 0 (since u = const.) = 287 x 310 ln (5.33) = 148828 J/kg = 1a8.88 kJ&C
Procese 2-B : Qz_s= W,-s = 0 (since volume is constant)
Qs.a (Heat supplied) = Ws,t= RTE6 ? = RTx ln | (Expansion)
-¿b"u2 Process 3-4 : Heat supplied, Qs-t= Wt-t = R?a ln(r)
(.' pa
= u¡ and u¿ = r¿) = 28? x 930 ln (5.33) = 146634 J/kg or 446.63 ktkg
Qa-1=-cu(T"-T*) or c,(T, -T"l; W¿-r = 0 (since u = const.) Process 4-1 : Q+r = - c, (T¡- Tr) or c,(7, - Tr)
The efticiency ofthe Stirling cycle is lecs '\¡n that ofthe Carnot cycle due to heat tmnsfers Heat supplied during the process 2-3 = Heat rejected during the process 4-1.
at constant uolume prxesses. However, if a regeneratíve arrangeme¿t is used euch that
Qqt= i.e. area under 1-4 is equal to area under 2-3, then the cycle efficiency becomes
.'. Work done = Neü heat exchange during the isotherms
Qz-s
= 446.63 - 148.88 = 297.?1kJtkr
nqrnS-n4rn5
" v2 t)2 work done 2917F.
rl' = -.- =Tr:7, ...(g.19) ... Thermal ef;ñcien<
J,r,¡¡= HffiÑ = = 0.6676 or 86.7to.
RT¡1ln\ TH ñr*
Since Stirling cycleis completely reversible, its efficiency is also given as,
which means the regeneratiue Stirlíng grcle has sarne effrcieney as the Carnot c7rcle.
Tu -T, 930-310
The following points are worth noting : n=-
2b: = t* = o'667 ór 66'79o' (Ane')
As far as the impractibility of actomplishing isothermal compression and expansion
processes with a gas is concerned, the Stirling cycle suffers frorn the lirnitdtinn of the
S.12. MILLER SYCLE
Carnot cycle. But, it does n¿ú suffer from oüher drawbacks of the Carnoü cycle, uiz, very
low m.e.p., the narrow p-u diagran a¡¡d great susceptibility to tbe internal efficiencies The Miller cycle (named after R. H. Miller) is a modern modification of the Atkinson cycle
of the conpressor and the expander. and has an erpansian ra'tio grcater than the compression rafüo, which is accomplished, however,
The mean effective pressu¡e (m.e.p.) of the Stirling cycle is much greater ühan that of in much a different way, whereas a complicated, mechanical linkage system of some kínd, is
the Carnot cycle. required for an engine designed, to opero,te on the Atkinson clcle, d Miller qrcle engine uses
a A reversed Stirling cycle with regeneration can sinilarly attain Camot C.O.P. uníque ualue tining to obtain the same desired results.
a The Stirling cycle can suitably replace Otto cycle (haüng two constant volume proc-
esses) in reciprocating I.C. engines.
Example 8.40. A¿ aír stondard Stírling clcle is equipped with a 100 percent fficient
regenera,tor syátem. The isothenna.I compresbn drmtwnces ftom I bar and, 310 K and. eubse.
quent keat od.dition ot constant uolu¡ne rai.ss tlre preseure dnd, temperaturc to 16 bar 930 K The
cycle íe ft,nally completed. through an isotlt¿trtrlrl
"-pansian and. constdnt uolume h¿o,t rejection.
Arnlyse e@h of the four processes for worh er.d lnot trdnsfer and d.etermine the engine ¿fficíén;ct.
Solution. Refer Fig.3.42.
Give'n: p, = l bar; T1=T"= 310 I{;p, = 16 bar iTs=Tx = 930 It The closing rycles for :

Consfiler t hg ofair. .Early intahe valv e : 67 -L-7 -2-3-4-5-7 -6

L¿úe intake valve : 6-7 -5-7 -2-3-4-5-7 -G
Pp1= RTt

', =
#=u:*# = o'88e7 m'/rrg

similarly uz= * = = 0.1668 ms/kg

#=H#
r\ ')=
o.*rt u.rrl
|.,. compression ratio. r = 5 llr =
t u2 (usl 0.1668=
I
Process 1-2: Tr=T"= 310K= ?¿

Heat rejected,
T.D.C B.D,C.
Fig. 3.43. Air-standard Müer cycleformth¡ottled a naturallyaspirateelfour-stmke rycle S.I. engine.
 156
NTERNAL COMBUSTION ENGINES AIR STANDARD CYCLBS
In a Miller cvcle air i¡Jake is zn thr-ottled.lhe quantity
of air ingested into each cylinder
bv ctosing the intdhe valve a,t i;-;;op", time, is
'i7 ;:;:;:t" tong'bero" ¡.o.c. (poinr ? in ----t-\

with the movemenü

of the piston towarde B.D.c. during -t--
'
the later part of the intake stroke,
the cylinder pressure is reduced along the p.o""""
Z--i. /',/
r,,"""u#;,:"'l:fJ',filJf;:
B'D'c' and sta¡ts back towards the r.D.c.,

Then, the resulting rycle is: G7_l-7-2-A4i+6.

The work produced during 6-? (inr'Le process)
rhe pressure again
,V
- is caneeiled by ?-6 (exhaust process).
The process Z-1 is cancelled by 1-?.
-
*t
indicared work = A¡ea within the toop 7-2_B_4_5-7; there being no pump
i;:;:
rt
The compression ratio = !_
u2
i
...(3.20)

The larger compressioq ratio =

!!=vt
...(3.21)
A greater net indicated work
' r"1 9raf,, of*ro* as a result of the shorter compression
súroke which absorbs work, combined with
the loog",
""p"orio"
,i-ü *u"¡ produces
Fig.3.44. Airet¿ndard approximation forhistoric I¿noire.g¡ne qycle, 1_2€
i
l+%1. .

o Further' by permitting-air to.flow through the-intake

system unthrottled, a mqior ross
experienced by most S.I. Thermodynanic analysir :
engines ie elimírlo,ted..
r Due to the absence of pump work, the Miller
cycle eagine has a higher thermal
Consider the clearance volume to be esseutially nit, Cancelling the intake process 1-2 and
latter half of the stroke 2-1 thermodynamically on p-u coordinat-eg, the cycle then becomes
.ciency. efti-
2-3-4-6-2.
o Milrer cycle en*ines are^u.rrllv supercharged
or turbocharged with peak intake mani- Process 2'3. Constant uolume heat input (combust¡b¿); All valves closed :
fold pressures of 160-200 kpa.
Pz = P t = Poan', U g = ú z i Wz-s = 0, Qz_g = Q ¡o = c u(Ts - T") = (u, - u2\
r Automobiles with Mille-r cycle engines were
first marketed in the latter half of the Proceeg 3-4. Isentropic power or expansían sürole ; All valves closed : -
1990s. A typical varue ofthe compression
ratio i,
"¡""t , l, *ltr, ." "i0"".", ratio of
g Qs¿ = O ; Tt = Tsfu{v¿y, ; pn = p, (v{u n)r ; .
about 10 : 1.
Ws¿ = @p+ _ psus)l(t_ 1) = A(?. _ T"ye _,¡)
P¡ocess *5, Constant uolume heat rejection (exhaust blow d.own); Exhaust valve open
3.13. LENOIR CYCLE and intake valve clos€d ;
Fig 3'46shows an air standard approximation fo¡ a historic uE= ú1= us.o.c.i W¿-¡= 0, Qr-¡ = Qout = cu(To-Tr)= ut-
2-1.
Lenoir engine cycre, 1-2-3-4-5- uÉ
Proe¿es 5'2- constant pressure uhaust stroke ái p"¿^ Exhaust op"t, ¿nd intake
;
o Thefirsthalfofthestrokeisintake,withair-fuelenteringthecyrinderatatmospheric valve closed. "rvJ
presstrro ,.,',,,,,.) _ proCesS l_2. Ps= Pz= Pt = Pon. i Wu-r= po^(vu- uz) ; Qs-z= Qout = (h5- h2)
¡ At about halfway through the first stroke, the intake
varve is crosedand the air-fuel
= cp(Ta- Tz).
mixture is ígnited.without d.ny cotnpressrioz. combustio" Thus, thermal efficiency oflenoir cycle,
in the clindero rrnost at constant-vorume ""i";, il; ;ierature and
inrherro*--o.,iig (q¿),.oo¡- [*t =Q-;Q*
;::r"* "rr'il+process q4" Qio =1-9tt
- Qi"
r r""onu half of the first stroke then becomes the power _ , _ fc,(\ -Tl) + co(Ts -T)l
l\1 or expansion process
cr(Ts -72)
o Near B.D.C., the exhaust valve opens and
blowdown occurslrroceec 4_5. _, _ Í(T¿ - Ts) + "t(Ts - 4)'l
¡ Then follows the exhaust stroke 6_1, thus
completing tbe two stroke cycle. Ts-Tz
...(3.22)

.-t
I
 INTERNAL COMBUSTION ENGINBS
AIR STANDARD CYCLES
ilfi
EICHLIGHTS Z The air standard efficiency of Otto cycle ie given by

L A cycle is defined as a repeated series of o¡rerations mrring in a certain order. -l

2.lhe elliciency of an engine ueing air u the working medium ie linown a8 an 'Air atandard efficiency'.
(4.,4= I+ (ó)n= l- (,rh
-
& ñ€tanve emosw.i =
Actual themal efficienry
Air ct¿rd¡rd efñeienw
(c)n= 1- --
(¡)r "
1
(d)¡=2-
(,)h
Ii
& The ti¡ermal efficienc¡, of theoretical Otto cycle
mm
Carnot cycle efficieacy, rr"*= -A-. (a) increases with incease in ompression ratio
(á) incremes with increroe in isentropic hdexT iii.
(¿) does not depend upon the pressure ¡atio
Otto cycle efEcienc¡r, neo=r-
rt 4.
(d) follows all tl¡e above.
The work output oftheoretical Otto cycle
A ¡f(rr-r -lXr, -1)l (¿) inaeaaes witl¡ increase in compression ratio
Mean effective pressure,
- P,*,= .
iil
-dtix_ii-- (ó) inseases with inceaee in pressue ratio

Diesel cycle efñciency, 4m,=r-

r for -rl (c) i¡creases with increase in adiabatic indexy
(d) follows all the above. lt
,1,t=TLeL l 5. Forsamecompressio¡ratio rll ür
[y(p - 1) - r1- 1(pr - tI
Mean eflective pressure, P¡r.¡= Arr (T-U,_l) .
(a) themal efficiency of Otto cycle is greater than that
(ó) themal efñciency of Otto cycle is less tha¡ that
of Diesel cycle
of Diesel cycle

7. Dualcycleefficiency, rm -- t I
=r
ro.ot -rl I
(c) thermal elñcieacy of Otto cycle is eame as that
for Diesel rycle
(d) ther¡nal efñciency ofOtto cycle caaaot be predicted.
{lH
(",t:il(p-:ilE6:tj 6 In air stmdard Diesel cycle, at fixed compression ratio mcr fixed value :ilfi
of ailiabatic index (p
(o) thermal eftcien.}r inceases with increase in
Mean efective pressure, ., _ - _
Arrlp(p- r)+(p-l)-.r-1 (ppr -t)l (ó) thermal efficiency decreases with increase in
heat atrdition cut offratio
'(,ru (r_lXr_l) heat addition mt offratio
(c) themal efñciency remains sarne with
increase ia heat aalarition cut ofiratio
(r - cr) (d) none of tlle above.
& Atkinson cycle eñcieocy, tl¡rm= 1-7.
/ _at
where o = compression ratio, r = erpaasion ratio.
fANñERs I
9. Braytoncycle, larwton =r - whei€ rp = p¡essute ratio. (ó) z (b) 3. (d) 4 (d) n(a) c (b).
-''!',
(rr\ r
THEORETICAL QUESTIONS
Ts -Tt
1(). Stirlingcycle, t*w='E-. rt
1L Millercycle engrnes areuuallysuperchargedorturbocüargedwitlrpeak intake manifoltlpressm of 150 l. What is a cycle ? What is the differe¡ce between an ideal and
actual cvcle ?
-20OkPa. ?- What is a¡ air-sta¡da¡d efiicieucy ?
3. What is relative efficiency ?
It Leloircycle; (tr);* Q4 - T¿+JS6 - rü
= L - + Derive expressiom ofefrciency ia the following cases :
fd) Canot cyde (ii) Diou¡.t"ru (iii) Dual combwtion cycle.

oBJECTM TYPE QL'EstflONS " *y"1?:r"*"rjlr,*analysis"whichhasbeenadoptedforl.C.enginerycles.Statetheassumprionsmade

for air stanclard rycles.
e Derive an exprssion for.Atkinson
rycle',
Chooee the cotl€ct a¡raw€r ! Derive an expression for the themal efficiency of Stirling
cycle.
& Exdain the following cycles briefly and deri-ve upressiÁ of
(a) two coustant prgLeúó procssseE a¡d two co¡8tant volume proceaccs (i) Miilercycle "fñ"iuo"y.
Q9 coaelgntfreasur€ and two consta[t e¡tropy pr@sses (ii) Lenoir cycle.
(c) two constant volume proceesea and two constant entropy plooegSGs
(d) no¡e ofthe above.
 160
INTERNAL CoMBUSTION ENCINES
AIR STANDARD CYCLES

UNSOLVED UAMPI,ES *
fffffifrñmt*:""::*1l.*:::X11"^?10 -y, ."g a cylinrter rlia:neter or 165 mm. nre crearance
A carnot eagine working betweea 3z?"c and B?.c producas
120 kJ of work. Detemine : ;Sffi "hHf .'""atr,"n"ire,t"""t"rli'ü1fiAñffi?1ff
l;ü,T",T"ffi$,T.:Xr"ffi;;
(i) Theheataddedi¡hl. (üi) The entropy change durirgheat 16. lte followiag:::,rrh";;si";;;",,;;;ñT"ff
H."li:_"f data belong to a Diesel lAne.6l.
d
,i:1.3"""".H:l:*:i:.::_
jr j\
^L
^ ^_ j_ -
rejection prccess. cycle :
rAna.(i)z2s.5kr;(joo.ssshr/K (¿¡;Daz.snl *a"a
^
Find the thermal efficiency ofa carnot engine whoee hot
and 15'Creapectively.
aad.cold.bsdies have tem¡reratures of r54.c
;ffiffi:il"f,:;ii"'=t;rY;:t
Dete¡:rrine:
= z5¿6 ¿¡¡"; Lowesr pressure in tre
rycre = I bu ; Lowest

n-r..^ an tAns.B2.55%lj
& Derive ^_expres"¡oo ¡q¡ ch"'ge-i-¡r-efficienry for a üange incom.pressiou ratio. Ifthe compression rautn
(i) thernal efrciency oftñe
cycle. (ii) Mean efi'€ctive pressure.
The efficiency of an Otto ¿ycle i|*
ttre¡e.r1n1_as_l rnilease in effi ciency
4, ;.':T:::"^T:i.1:111 is 60% and.lis 1.6. What is the cornpression 16, rhe rT,g:-:,,T31l-Dual
ratio ? ffi;; co_mpreseion rario
G An engine worting on otto cycle,has a volume of-0.5 mr, pressure l¡nited to?0 bu' rhe presu. re and temperature
crcte is t2:ld.rhe. maxi.ffi,:,1"1t"ÍJ',3:Hl"1l
1 bar and tenpuature 2z.C at the áf cycre ai tl" r"gioning orompression proce's are r
comencement of compression stroke. At the e¡d of comireesion
-- prÁirre¡
stroke, the ----'# t0
bar and 800 K Cat".t.t , (i) Th"*"i"ft1"ü i"iii"""l,"r"ctiye prersure.
added duringthe coutantvolme proceds i8 200 k I. Detenoiue ^" bar.
"-' Heat Ass'.ie; ry¡¡¿"'bore
(i ) Percontage clearance
: = 260mm, st-ke t""sor:;00;,;"= iñ;;=;;18 andy= 1.4.
(ii) Air
(iii) M""o p¡assuro ",-Ur"U "ffi
cimcy
u'ltecompYilr"1tT:H*r*":lj*:::ryg:-q*ry:*ffilí"1ffi IAn*(i)67.92% i(jj) 9.84? ba¡l
"O"*oe
;:HÍ;:lj:
u,"rrnall-dlñffiffi:k-. .;;1;üff,ln:trffif ;,#;,J
(du) Idu¿ **"t ueveloped by the engine if the engine runs at 4gonp.m. bo thsüüere $e 200 complete
cyclesprminutes. tA¡re,(i)2g.76%iiii)4i.zq;iülinár (iu)32lkw
;ñ,#;""f
675 fftre*1i*T,rxy_",'}.:;;i#ffi
#.ftFin¿
h/kg of air.
A¡airsta¡dardDualcyclehasacomo""r": . tAD'.og.bzol
IA¡s,59.5Vo1
6 1he conFreesiou ratio in a¡ air..sland.ar-d Olto cyclá ia g.
At .lre beginning ofco,¡p¡essi6¡
18,
.ri"'a;;
.üli';?ffi
*ji"
zo r,". ,¡q1T1f::::li "l16, "-r.
and compreaeion begi¡s ai I ba¡, 60"C.
process, the iszó¡a,.r¡"rrü;;ffiffi
mumrun D¡eaam i.
,l?ffi The
pressueis lbaratrdthe temperatureie 300Ii rt¡e heat'h"*r*t"-tli"
Jip"""y"¡it"-rsirj0k Ifuofair. friffiSJffi
volume. Determine: ffi f ffi ij,',j"T;igl;fi :
(i) Themalefficiency (ii) The mean effective pressure.
(i) The cycle efrcie¡cy. (ij) flr" _"*
*"ctive pressure of the cycle.
Take : c, = 1.005 kJ/kg-I! c, O.?18
= lüykC_K
z An ensine 200 m rg' computl t'e
ora B-ra¡on cycle operaringbe**"
bm mtr 300 mm stroke works on ono cycle. rrre
n*** and temperarure are l bar an¿r 60rc. ftñu ","j*l.t::A-Jf ;:;¿#;iHi pressure of12"i" "t""a"J"maenry
bar. Ta.kel= 1.4. "
f"i,1?"Ti?;:'*:t"t Hi
HÍlt -"*i-,- ñ";;;;;;ed ro 24 bar, tA¡a.50.g%l
(i) lhe ainstqndard efliciency ofthe
rycle (ii) The nea¡ effective pressure for the rycle.
Assume ideal cond¡tions.
& calculate rhe air standard erriciency of a four srroke ouo
o*'
Piston diarneter (bore) = IBZ mm ; Length ofstroke
cycle engin" .jf"T3rfr::i|t;'Lt"lt'
Clearance volume = 0.0002g mr,
= IBO mm ;
Express clearance as a percentage ofswept volume.
0' lAns,56.LVoll4.6Vd
In a¡ ideal Diosel cycle, the temperatw^s at the beginning
of compression, at the eod of compression and
at the end of the heat addition are g?"C, ?89.C má f
$9:C. finá ,he th;;;
"ffi;;";"f
lo' An air-standard Diesel cycle hts a mmpression ratio lAns.69.67ol
of 18, aad the heat transfeneal to the working fluid
per cvcle is 1800 kJlg. Atthe beginning ofthe
compilsio".trot"rll*;r;.irJi"-r'¡r" u.a ¡r.
temperature is 300 K Calculate : (i) Themal efficien.i, (¡¡)
tfr" ,rr"* .ffud;"-pr;;;".
t1. r)<g orair isiaken th:ough
lpiesd cv_cle. rnitiany the air is ar 15.. ,
jf;d3"t#;:tlr::'::rH
15 aad the hear added is 1850 kJ. ód"olate , í¿ r¡"r¿"J"v"1. "".
pTessu¡e. "r#ü,i;;; d;|Hm effecriv¿
12' what will be loss in the ideal erriciency of a Diesel engine
delayedftom$votog%? -- with
"-"¡ comn*r".f;Íl:?;ií.,'rÍ'l,ll"Hl
vv¡¡'F^voe u""'
¡
l& The pressures on the compression cune of a üesel
".r.r*l
engine are at th stroke 1.4 bar aad at
14 bar' Estimate the compression ratio. calculate
f f th stroke
the air standa¡d efñciehcy ofthe engine ifthe ot ofi
occurs at ft of the stroke.
[Ans. 18.54;63.79ú]