Internal Combustion Engine

7 (c) The cetane number of

alcohol uels
In air standard Otto cycle the
terminal pressures
low which prevents good co is ve
7.1
at the end of compression,
heat release
and combustion
expansion P,
are respectively P and Pa. lfthe
(d) None of these
1992:2 Marke
corresponding values are P', Pa' and P4, taking 7.6 Brake thermal efficiency of the three basin
asic type
heat
into account the effect of variable specific of reciprocating engines scommonly uSed in
and dissociation of the working fluid, then
vehicles are given in the increasing order as road
(a)P2 Pa'and Pg> P (a) 2 Stroke Sl engine, 4 Stroke Sl engine 4
(6) Pa< Pa' and P > P Cl engine
Stroke
(c)Pa> P' P> P and P< P (b) 2 Stroke Sl engine, 4 Stroke Cl engine.
4 Stroke
(d) Pa> P' Pa> Pg [1989 2 Marks]
Sl engine
7.2 The power output from a spark ignition engine is
(c) 4 Stroke Sl engine, 2 Stroke Sl engine, 4Se
varied by Clengine
(a) Changing the ignition timing (d) 4 Stroke Cl engine, 4 Stroke Sl engine, 2Stre
(b) Regulating the amount of air inducted Sl engine
(c) Regulating the amount of air fuel mixture
[1992 2 Marks
(d) Regulating the amount of fuel 7.7 The figure below shows a thermodynamic cycle
[1990 2 Marks] undergone by a certain system. Find the mean
effective pressure in N/m2
7.3 An air standard diesel cycle consists of
(a) Two adiabatic and two constant volume P
processes kN/m
5
(6) Two constant volume and two isothermal
processes
() One constant pressure, one constant volumne 2

and two adiabatic processes

(d) One constant pressure, one constant volume Vm)
0.01 0.03
and two isothermal processes
[1993:2 Marks
[1990 2 Marks]
7.4 BHP of diesel 7.8 In order to burn 1 kilogram of CH, completey. u
a engine can. be increased by eeded
(a) increasing the pressure of intake air minimum number of kilograms of oxygen
Is (take atomic weights of H, C and 0 as 1,
(b) increasing the temperature of intake air
(c) increasing the density of intake air and 16 respectively)
(d) decreasing the density of intake air (a) 3 (b) 4
(c) 5 (d)6 [1995:2Mand
[1991:2 Marks]
7.5 Alcohols are unsuitable at diesel engine fuels 7.9 With increasing temperature of inlan
because
(a) The cetane number of alcohol fuels is very engine efficiency
lowwhich prevents their ignition by compression (a) decreases
(b) The cetane number of alcohol fuels is very
(6) increases
(C) remains same
high which prevents their ignition by
(d) depends on other factors
Compression [1998:1Ma
EEASY7
Thermodynamics
T h e
523

and stroke of 2 units each. 7.16 A diesel engine is usually more efficient than a
has
Cengineh
a bore
heat loss can be taken as spark ignition engine because
alculate
calculat

(a) diesel being a heavier hydrocarbon, releases

he
a r e a
lo

(b) 5 more heat per kg than gasoline

(d) 8 T [1998: 2 Marks]
(b) the air standard efficiency of diesel cycle is
air craft isflyng at an altitude higher than the Otto cycle, at a fixed
hreathing half the
half the value at ground compression ratio
density Is
e
w h e r e
theair to the ground level, the air (c) the compression ratio of a diesel engine is
With reference

evel
aititude
will be
be higher than that of an Sl engine
at this (d) selfignition temperature of diesel is higher than
ratio
hel
that of gasoline [2003 1 Mark]
2
(b)2 7.17 An automobile engine operates at a fuel air ratio of
(d) 4 volumetric efficiency of 90% and indicated
0.05,
c) 2 [1998 2 Marks] thermal efficiency of 30%. Given that the calorific

value of the fuel is 45 MJ/kg and the density of air

in.
ernal combustion engine
ofa n indicated mean effective
The
silencer
at intake is 1 kg/ms, the
12(a) reduces noise is
specific fuel consumption pressure for the engine
(b) decrease break (a) 6.075 bar (b) 6.75 bar
(d) 243 bar
(BSFC) (c) 67.5 bar
o) increase BSFC [2003 2 Marks]
its efficiency
air standard Otto cycle,
effect on
has no [1999 2 Marks] on
d) 7.18 Foran engine operating 10% of the swept volume.
the clearance volume
is
on the ideal
engine working ratio of air is 1.4. The airstandard
ignition The specific heat
13 tn a spark ratio is 5.5. The work
cycle, the compression cycle efficiency is
Oto is
the P-vdiagram)
per cycle (i.e., area of (a) 38.3%
(b) 39.8%
output where V, is the
x 103 x V, J,
(d) 61.7%
c) 60.2%
equal to 23.625 The indicated
mean
[2003 2 Marks]
clearance volume in ms.
the
efective pressure is test on a 4 cylinder engine,
a Morse
(b) 5.250 bar 7.19 During were
a) 4.295 bar measurements of brake power
(d) 106.300 bar following
(c) 86.870 bar
[2001:2 Marks] taken at constant speed.
3037 kW
All cylinders firing
2102 kW
Number 1 cylinder
not firing
has a 2102 kW
114 An ideal air standard Otto cycle Number 2 cylinder not firing
2100 kW
ratio of the specific not firing
compression ratio of 8.5. If the Number 3 cylinder
2098 kW
thermal efficiency not firing
eals of air (y) is 1.4, what is the Number 4 cylinder
is
efficiency of the engine
(n percentage) of the Otto cycle? The mechanical
(b) 85.07%
a) 57.5 (b) 45.7 (a) 91.53%
(d) 61.22%
c)52.5 (d) 95 [2002:2 Marks] (c) 81.07%
[2004: 2 Marks]
145 Fora equivalence ratio
spark ignition engine, the chamber on air
standard Otto cycle
has

the combustion working of

9o mixture entering 7.20 An engine
d i a m e t e r of 10
cm and
stroke length
has values a cylinder is 1.4.f
heats for air
of specific
991 for idling and > 1 for peak powe 15 cm. The
ratio the heat
volume is
196.3 cc and
Conditions the
is 1800 kJ/kg,
clearance
the
) 1for both idling and peak power conditions of air per cycle
Supplied per kg of air is
C)> 1 for idlin and ¢ < 1 for peak power cycle per kg
work output per (b) 890.2 kJ
Conditions (a) 879.1 kJ
(d) 973.5 kJ
for both idling and peak power conditions (c) 895.3 kJ [2004 2 Marks]
[2003 1 Mark]
 524 GATE Previous Years Solved Papers: M
are 1.4 and 288.8
MADE ED
7.21 The stroke and bore of a four stroke spark ignition JkgK, resp
engine are 250 mm and 200 mm respectively. The
Clearance volume is 0.001 m3, If the specific heat
compression
temperature in the
ratio is 8 and cively. t
the
oycle is 2660 K maim
ratio y = 1.4, the air-standard cycle eficiency of kJ/kg) supplied to the engine is the hea,
the engine is
(a) 46. 40% (b) 56.10%
7.27 A diesel engine has a
12014:2 Marks, Spl
(c) 58.20% (d) 62.80% and cut-off takes place at
compression ratio o
[2007 2 Marks] 10% of the
Assuming ratio of specitic heats (y
7.22 Which one of the following is NOT a necessary air-standard efficiency (in percent) is as 14 te
assumption for the air-standard Otto cycle?
(a) All processes are internally reversible.
2014:2 Marks, Sea
7.28 In a compression ignitic
(6) Intake and exhaust processes are constant engine, the
volume heat rejection processes.
pressure is 1 bar and the
pressure at th inle
isentropic compressic is 32.42 end
expansion ratio is 8. Assuming ratio ofbarsnaesTre
c)The combustion process is a constantvolume
heat addition process.
heats (y) as 1.4, the air
(a) The working filuid is an ideal gas with constant
percent) is.
standard efficiencya
specific heats.
[2008 1 Mark] (2014:2 Marks, Set
7.23 In an air-standard Otto cycle, the compression ratio 7.29 Air enters a diesel engine with a density of 1
is 10. The condition at the beginning of the
kg/m3. The compression ratio is 21. At steat
Compression process is 100 kPa and 27°C. Heat state, the air intake is 30 x 10 kg/s and
added at constant volume is 1500 kJ/kg, while
thens
work output is 15 kW. The mean
700 kJ/kg of heat is rejected during the other
effectie
pressure (KPa) is .
constant volume process in the cycle. Specific 2015 1 Mark, Set1
gas constant for air = 0.287 kJ/kgK. The mean
effective pressure (in kPa) of the cycle is .30 A rope-brake dynamometer attached to te
(a) 103 (b) 310 crank shaft of an 1.C. engine measures abrale
() 515 (a) 1032 power of 10 kW when the speed of rotationdt
the shaft is 400 rad/s. The shaft torque (in Nn
[2009 2 Marks]
sensed by the dynamometer is.
7.24 A turbocharged four-stroke direct injection diesel
[2015 1Mark, Set
engine has a displacement volume of 0.0259 m3
(25.9 liters). The engine has an output of 950 kW 7.31 An air-standard Diesel cycle consists of
at 2200 rpm. The mean effective pressure (in MPa) following processes:
is closest to 1-2: Air is compressed isentropicaly
(a) 2 (6) 1 2-3: Heat is added at constant pressure
(c) 0.2 (d) 0.1 3-4 Air expands isentropically to the ong

[2010: 1 Mark] volume.

4-1: Heat is rejected at constant voluin
7.25 The crank radius of a single-cylinder IC engine is If y and T denote the specific heat rai
ofte
60 mm and the diameter of the cylinder is 80 mm. temperature, respectively, the efficiency
The swept volume of the cylinder is cm3 is
cycle is
(a) 48 (b) 96
(c) 302 (d) 603
(o)1 ()7-
[2011 2 Marks]
T-
In an air-standard Otto cycle, air is supplied at o) 1 - 4 (d) 1- -
7.26
0.1 MPa and 308 K. The ratio of the specific
2015: 1 Mark,
Se
constant (R) of air
heats (y) and the specific gas
 Thermodynamics

AADEEASY
ame alues of peak pressure, peak 525
F o rt h e , 7.37 An air standard
Otto cycle has
and heat rejection, the correct order
mean effectivethermal efficienoy
a ature of 0.5 and the
ciencies
eft
forOtto, Duel and Diesel cycles is pressure of the
cycle is 1000 KPa. For air assume
of
pual Diesel ratio y 1.4 and
= specific heat
(a) lotto
"Dual 1oto R 0.287
specific gas constant
(b)TlDiesel kJ/kg.K. If the pressure
(c)pual
"lDiesel otto
temperature at the beginning of the and
lOtto Dual stroke are 100 kPa and 300 K compression
(0)TDiesel [2015 2 Marks, Set-2] the specific net work respectively, then
output of the
(CH) is burned in an
oxygen
S. kJ/kg (round off to two decimal cycle
Propane places).
1 mosphere with 10% deficit oxygen with respect
[2019:2 Mark, Set-2]
to the ichiometric requirement. Assuming no 7.39 For an air-standard Diesel cycle,
carbons in the products, the volume (a) heat addition is at constant volume
and heat
of CO in the products is rejection is at constant volume
Dercentage
2016: 1 Mark, Set-1] (b) heat addition is at constant pressure and heat
rejection is at constant volume
79% N, and 21% Oz on a molar
contains (c) heat addition is at constant pressure and heat
34 Air is burned with 50% excess
basis. Methane (CH,) rejection is at constant pressure
stoichiometrically. Assuming
air than required (d) heat addition is at constant volume and heat
combustion of methane, the molar rejection is at constant pressure
complete
in the products is_ [2020:1 Mark, Set-2]
percentage of N,
[2017 2 Marks, Set-1] 7.38 The indicated power developed by an engine
with compression ratio of 8, is calculated using
powered by a spark ignition engine
1.35 A vehicle an air-standard Otto cycle (constant properties)
follows air standard
Otto cycle (y 1.4). The =

70 kW while consuming The rate of heat addition is 10 k . The ratio of

engine generates
calorific value of fuel is specific heats at constant pressure and constant
10.3 kghr of fuel. The volume is 1.4. The mechanical efficiency of the
ratio is
44000 kJ/kg. The compression
.

of
to two decimal places). engine is 80 percent. The brake power output
(correct kW (round off to one
the engine is.
[2018 2 Marks, Set-2] decimal place).
7.36 An engine working on air standard
Otto cycle is [2020 2 Marks, Set-1]1
and 35°C. The the
Supplied with air at 0.1 MPa 7.40 Keeping all other parameters
identical,
is 500
compression ratio is 8. The heat supplied an air standard
diesel
Compression Ratio (CR) of
1.005 kJ/kgK, ratio of
kU/kg. Property data for air: c =

cycle is increased
from 15 to 21. Take
0.287 kJ/kgK. The 1.3 and cut-off ratio
of the cycle
0.718 kJ/kgK, R = specific heats
=

the
maximum temperature (in K) of the cycle between the new and
The difference
=2.
S correct to one decimal place). old efficiency values, in percentage,
%. (round
15=.
[2018 2 Marks, Set-1] new CR-21-(Moia lCR-
one decimal
place).
off to Set-2
[2020: 2 Marks,

Answers IC Engine 7.6 (a) 7.7

(1.5) 7.8 6)
11 7.5 (a)
d) 7.2 (c) 7.3 (c) 7.4 (a, c) 7.15 (b) 7.16 ()
18 7.13 (b) 7.14 (a)
7.10 (a) (b) 7.12 (a) 7.23 ()7.24 (a)
7 7.10
7.11 7.22 (b)
l 7.20 (d) 7.21 ()
7.31 () 7.32
7.31 (b)
(b)
7.32 (b)
7.18 (d) 7.19 (c) 7.30 (25)
25 () 7.29 (525)
7.26 O
(1409.6) 7.27
9 . 6 ) 7.27 (59.6) 7.28 (59.6)
7.38 (4.5) 7.39 (6) 7,40(4.8)
7.39 (b) 7.40(4.8)

3(14.28) 383) 7.35 (7.61) 7.36 (1403.97)

7.37 (708.60)
 GATE Previous Years Solved Papers: m
526

Explanations IC Engine
MADEk
74 (a, c)
7.1 (d)
Ihe terminal pressures at the end of compression, By p,TSupercharged t
heat release and expansion i.e. Po Pg and P4 I.P. B.P.T
respectively, will be greater than the correponding 75 (a)
values i.e. P, P and P respectively, when we
Alcohols are unsuitable as diesel
take into account the effect of variable specific following reaons fuels in
heat and dissociationof the working fluid. (a) The cetane number of
alcohol fuels i
72 (of the order of zero to
eight), whichhprene
ne
s'sVen
their ignition by
The power output from a spark ignition engine is
(b) Alcohol fuels have low
compression
causing trouble in injectionlubricatino m
varied by regulating the amount of air-fuel mixture
qualng
i.e. by supplying richer or leaner mixture
(c) There are material problems
pumps anr
harsh reaction of methanol
causeri byr-
POINTS TO REMEMBER towards e
plastics and metals.
I n a spark ignition engine, ignition timing refers 7.6 (a
to the timing, relative to the current piston Brake thermal efficiency of diesel
position and crankshaft angle, of the release engine is hom
than the
petrol engine. At the sametime, foursth
of a spark in the combustion chamber near the
end of the compression stroke. engines have more efficiency than the 2str
The need for advarncing or retarding the timing engines.
of the sspark is because fuel doesn't
completely burn the instant the spark fires. POINTS TO REMEMBER
Volumetric efficiency of eisel engine is also hig
73 ( than the petrol engine.
Air standard diesel cycle:
P
7.7 Sol.
Work done = Area under the cycle
1
=x3x0.02 =0.03kN

Work done 0.03

mep =
Volume 0.02
V2 V3 VV
1.5 kPa
PConstant 7.8 (b)

V= constant
CH+2Op
16 kg
CO, +2H,O
64kg
1 64
1kgkg inimum num
Process 1-2: Reversible adiabatic compression So. to burn 1 kg of CH,
Process 2-3: Constant pressure heat addition Kilograms of oxygen needed is equa
Process 34: Reversible adiabatic expansion
54
Process 4-1: Constant volume heat rejection Okg 4 kg
16
 Thermodynamics 527
MADEEA S Y
7.12 (a)
Amuffler (silencer in British language) is a device
ofiintake air will
for reducing the noise which is emitted by the
t e m p e r a t u r e

in
efficiency a s
the density of air
i n c r e volumetric
a n overall
exhaust of an internal combustion engine.
cause
This wIill
e s s
down.
intake
G0es
g o e s

d e c r e a s ei n e f f i c i e n c y 7.13 (b)
Compression ratio,

=
Surtace area of cylinder r= 5.5
heat loss Vc
l area for Tx2x2 = 47T
= ndl =

V = 5.5 V
surtace area is
UPSC only
Vote: As p e r
cooled IC engine.
For air W WV
nsidered for wate water
oiston head area is also mep VV-Ve
IC engine pist
oled
considered if mention. P

AVOID MISTAKE

combustion
chamber
where
side
.As atthe top take place, so
heat
heat loss c a n n o t
is there, a r e a of piston.
the s u r f a c e V2 Ve
lost from
23.625x 10 x Ve
A rDL+D
5.5V- V

= Ttx2x 2x (2)
23.625 x10Vc 5.25x 105 Pa
4.5V
= 4Tt + t = 5rt = 5.250 bar

the option.
which is also given in 7.14 (a)
1
POINTS T O REMEMBER = 1- 8.5)4-1
0.5751 = 57.51%
TDL+2xD =

cylinder
Surface area ofthe
where we calulate
But in of IC engines
case 7.15 (b) actual fuel air
of defined a s the
heat loss then the area
of bottom and top Equivalence
ratio is
fuel air ratio
the cylinder is not considered.
stoichiometric
ratio to
considered
is
A S per UPSC, only surface a r e a
cooled IC
Tor water cooled IC engines. For air
is also considered
Actual
engines, piston head a r e a
f mention. Stoich
Fuel-air
condition.
power
and peak
11 6 For both idling stoichiometric
fuel-
than the
is m o r e
ratio required

(AFR)altitude Paltitude air ratio.

AFRgroundlevel VPground level 7.16 (c ratio and

the same heat

- AFR)gu roundlevel compression

and diesel
(AFRat itude
So with ference to he ground level, the air-fuel
For same

supplied,
Otto cycle
is most
efficient

efficient.
least
cycle is
atio
at this altitude will be
2
 GATE Previous Years Solved Papers:
ME
528
compression
ratio of the 7.19 (c MADEL
in practice, however, the
between 14 and
25 whereas Method I:
D i e s e l engine ranges

ranges
between 6 (B.P.)1,2.3,4 3037 kW
that of the Otto cycle engine
has higher
and 12. Because of
diesel engine (.P)1,2.3,3,4= (6.,2.3,4++(FP),
Number 1 cylinder not firing,
efficiency than the petrol engine.
(B.P.)23,4 =
2102 kW
7.17 (1.P.)2,3,4 (8.P2,3,4+ (FP),,.
=

Volumetric efficiencyY
Eq.(i)-Eq. (i), we get
Actual volume
Swept volume
=0.9 (1.P.)1,2 . 3 , 4 - ( P 2 3 , 4 = ( 6 . 2 3 . 4 ( B . P

(I.P.), 3037-2102 3.4

V= 0.9V =
=
935 kW
Mass of air, Similarly, number 2 cylinder not firing.
maPai =0.9 (B.P.)1,3,4 2102 kW
m= 0.05 x 0.9V, 0.045V
=

(.P.) (B.P)1, 2,3,4(BP)34

PmepXLA
Ithermal 3037-2102 935 kW
m,xC.V.
Number 3 oylinder not firing,
PnxV
0.3
0.045 V, x 45 x 10 (B.P.)1,2 4= 2100kW
P 0.6075 x 105 Pa = 6.075 bar (1.P.) (B.P)1, 2, 3, 4 (6.P,24
= 3037 -2100 937 kW
7.18 (d
Number 4 cylinder not firing,
V 10%V, = 0.10V,
(B.P.)12.a = 2098 kW
O 1
0.10 1 0
(1.P.)=(B.P.)1, 2,3,4 -(B.P,23
Y 1.4 3037 -2098 939kW
=

PA Total I.P
(.P)1234 (1L.P), +(1.P),+(L.P)}h+(1P},
= 935 +935+937 +939
= 3746 kW
Mechanical efficiency.
3037
V m (B.P.h234
.P.)h234 3746
= 0.8107 = 81.07%
Compression ratio, Method l:
Given:
V Brake power with 4-cylinder, 4B 3037K
Brake power with 3-cylinder,
= 1+ 10 = 11
2102+2102+2100+209821005
3B=
4

oto 1--1 1114-1 Indicated power, I.P.

4(4B-38) = 4(3037-2100

- 1 - , 4 2 . 6 019
= 4x 936.5 3746 kW
11 Mechanicakl efficiency,
=1-0.3832 = 0.6168 = 61.67%
BP
IP3746 3037 = 0.8107=8
 Thermodynamics 529
MOEEASY
n=1 1 (8.85)14-1
1--8.850.4
0.5819 =58.2%

722 (6Assumption of air-standard Otto cycle

both internally as well as
V
1. All process are

V externally reversible.
2. Air behaves as ideal gas.
3. Specific heat remains constant
G i v e nd a t a :
Y 1.4 volume heat addition
= 15 cm; 4. Intake process is constant
cm;
10 cm3 exhaust process is constant
d- process and
=
196.3
cC
196.3
V= volume heat rejection process.
4=1800k/kg
S w e p tv o l u m e ,

V,= (10x15 723(d

Given data for Otto cycle:
= 1177.5 cm3

Compression ratio,

r . + 1 9 6 . 3 - 1 1 7 7 . 5
6.99
196.3
V
1
Tlotto (6.99)14 -V V

1-0.4594 = 0.5406 Compression ratio,

=

W r = 10
V2
also lotogs P 100kPa
27°C (27+ 273)K
W =

0.5406 1800 T, =

= 300 K
W= 973.08 kJ/kg
9 1500 kJ/kg
qp 700 kJ/kg
R 0.287 kJ/kgk
Stroke length, PV=RT
0.25 m
I= 250 mm
=
= 0.287 x 300
200 =0.2 m 100 x v,
Boredia: d mm
= 0.861 mkg
Clearance volume, Of. V = 10 V
also
V0.001m3 0.861 10V
Y 1.4 =
0.0861 m'/kg
V
Displacement volume, work done per
cycle,
Specific
x0.25 W g-92
V xl =

4
x(2) 1500-700
= 800 kJkg
=

effective p r e s s u r e ,
= 7.85 x 10 m3 that mean
We know
otal volume in the cylinder, W
800
7.85 x 103
= V+ V.=0.001 + PmV
Vs V-V2
= 8.85 x 10 ms
800 1032.39 kPa
Compression ratio, 0.861-0.0861

8.85x10 = 8.85
0.001
Air-standard cycle efficiency.
530 GATE Previous Years olved Papers
ME
7.24 (a)
Cy 288.8
Given data: y-1 04 /222 Jikgk
N
n for four-stroke engine
Stroke volume,
V = 0.0259 m3
(8)04
Power, P= 950 kW T 308 x 80
Speed, N= 2200 rpm
=
707.6 K
Q mo, AT =1x
=

1409.6 kJ/kg 722(2660-707 e

660-701 6
We know that power output,

P- mAln W= n
6 0 kW= Pnn
r
60 7.27 Sol
where Pis in kW; pm is in kPa; Vis in m p
N
n rpm
2 pVi4 =C
x = 1, number of cylinder

950Pm X0.0259 N
x1
60

950 PmX0.0259x2200
120
or N
Pn 2000 kPa =2 MPa
Compression ratio,
7.25 (d)

Crank radius, r= 60 mm V2
Diameter of cylinder, d= 80 mmm Ratio of specific heats,
Swept volume = ?
I= 2r 2x 60 mm 120 mm
=
= y= 1.4
v
Swept volume, Cut-off = 0.1 x stroke volume
V-V2 = 0.1 x V
V= d=x(80 x120x103 V-V = 0.1 x(V, -V)
= 602.88 cm
7.26 Sol.

Va-1 0.1
or 3= 0.1 16 +1 26x
2
Cut-off ratio,

==26
Given:
308 K;
TDiesel 1 -C-
p, =
T,
0.1 MPa; =
y= 1.4
R= 288.8 J/kgK; r= 8 = 1- 264-1
170.4 1.4(2.6-0J
T2660 K
a, = mc,(T-7)
 Thermodynamics 531

MaDE
EASY So, it is important to note
that compression
13.81-1) ratio is equal to the multiplicationof cut-off ratio
11704 1.4x1.6 and expansion ratio. And the value of cut-off
=
59.6%
=
0.596 ratio, expansion ratio and compression ratio
are always greaterthan 1.

Sol

7.29 Sol.

P 1 kg/m3
I 21
m 30 x 10 kg/s
W= 15 kW
V
J0X10 m°/s
= 32.42bar V ==
1
bar, p2 P
P,1
Given,
= 30 x 10 m/s
y P = 1.4

Cy V- 30x 10
21
8 = 1.428 x 10 m/s
V
ForprocesS 1-2, V, =V-Ve 10-3
30 x 10-3-1.428
x
=

28.572 x 10-3 m/s

Mean effective pressure,
P3242
W 15
Pm V28.572 x10
12
a= (32.42)1/14=11.999 524.99 kPa =525 kPa
V r= 12
ratio, 7.30 Sol.
: Compression
Brake torque x angular velocity
Cut-off ratio, Brake power
=

P To
15 10,000

O T 260 XN
Desel
6060

1 154-1
1000025
400
Nm
:1224 1.4x0.5 7.31 (b)
= 0.596 = 59.6%

POINTS TO REMEMBER

Here, inthis problem, cut-off ratio,

V

So, cut-off ratio.r, =

*
 532 GATE Previous Years Solve
Papers: ME
number of moles of O, MADE
As it is burnt with 10% required=5
PC CO. deficient O. t wgeneg
That is, the new
equation is
CH+0.9x 50,oCO, +BCO 4 ,
Balancing carbon
3 =+B
atoms, we get 0+O
Efficiency of diesel cycle, Balancing hydrogen atoms, we
get
8 2
or 4
1 e a t rejected
Heat supplied Balancing oxygen atoms, we get
Heat is added in the constant 0.9x 10 2a +B
pressure process, +Y
2s h-h 9 20 +B+4
cT-T) 5 20+ B
Heat is reject in the constant volume
process, Eq.(i)-Eq. (), we get
2 a
a1ha-h, =o{T-11 or 2
efficiency, n Q2-3 Substituting a = 2 in Eg. (i), we get
5 2x2+B
1C4 T) O B 1
CpT-T2) .CH+4.5 0,2CO + CO+ 4H,O
The volume percentage of CO in the product
=
1-1-7)
Y(T-T2)
- x100
7.32 (b)

For same
compression ratio and heat addition x 100 =14.28
lotto dual diesel 7.34 Sol.
For same compression ratio and heat rejection,
otto dual ldiesel Stoichiometric reaction
For same peak temperatur and heat rejection
diesel dualotto CH,+2 0 2,0 C0-2
For same peak pressure and heat 50% excess air:
rejection
ldiesel 'ldual otto
7.33 Sol.
CH+3 2H,O+CO,+3xx,-0
CHg+ a0, bCO, + cH,O 3x79
Balancing carbon atoms, we get 21 100
3 = b
%N2
2+1+1+3x
or b= 3
= 73.83%
Balancing hydrogen atoms, we get
8 2c
POINTS TO REMEMBER
or C 4
Balancing oxygen atoms, we get eacton
If student don't know stoichiometric reabu
2a 2b+C
2a 2 x 3+4 =10 CH, then they can easily find by
a = 5
79. CH,0+0
For stoichiometric burning (i.e., chemically correct). CH +a(Op+N) bCO,
 Thermodynamics
533

uNDEEASY P
C b a l a n c e :1 = b

H-balance:4=
2C: C = 2 (MPa)
:4 =20 =
C: 2x1+2 =
4a =2
=
2b 4+

O-balance:
2a
72
N-balance:
ax
dd-2x 21
21
imetric reaction willbe:
S ostoichime
For process 1-2
+240+2xN

+20+)
0% P = PV
H 21
0.1x ()4
Y 1.4
1.8379 MPa
B.P. 70kW Pa =

10.3 kg/hr and

m =

T
CV= 44000 kJ/kg
1.8379Xx 308
B.P
70 20.1x8
m, xCv 10.344000 T 707.6 K
3600 3
Forprocess 2
= 0.556
a = ,(T-T) = 500 kJ/ kg
0.718(T,-707.6) = 500
=0.556 1403.97 K
T =

.4-70.556
7.37 Sol.
1- As per given data,
0.5
Totto
1 -0.4439 0.287x 300
0 4 RI
r= 7.61
V P 100
= 0.861 m/kg
36 S0l 1

35°C =308 K otto 1-T

P=0.1 MPa, T
=

r 8 0.5 1-
V2
(r04= 2
Q 500 kJ/kg
r= 5.65
1.005 kJ/kgK
C, = 0.718 kJ/kgk V = 5.65
V2
R 0.287 kJ/kgk
0.861

P=1.3991.40 V25.65
Cy = 0.1524 m/kg
2
T max ? Specific work output meps
= 1000 kPa x (V, - V2)
=

0.1524)
1000 x (0.861
= 708.6 kJ/kg
 Papers: M
GATE Previous Years
Solved
ME MADE
534| 7.40 Sol.

7.38 Sol.
As given compression atio (CR)
0.8 diese
r =
8, Q, = 10 kW, r = 1.4, mech r = 1.3, cut off ratio r= 2 15t0
0.5647
1-
n=1- = 1R04 8.4
=

Na, r=21
Y(p-1 4.87

nd,15 1- p-
T(P-1).08
d, 15 = (54.87 50.08)%
d, r=21
W n x a
48
W= 10x564/ = 5.647 kW
1000

5.647
B mech X W = 0.8 x

= 4.5175 kW
 MISC (Internal
Combustion Engine)
termining the ignition quality of
o

engine fuels the eference fuels

compression 8.4 At the
ynition used are time of
al lsooctane and n-heptane starting, idling
operation, the carburettor and low
b)
Cetane and o-methyinapthalene can be termed as
supplies a mixturespeed
which
Hexadecane and n-heptane (a) Lean
id) Cetane
and isooctane (6) slightly leaner than
c) stoichiometric stoichiometric
[1991:2 Marks] (d) rich
f air fuel
ratio of the mixture in petrol engine
2 more than 15:1 is
8.5
[2004: 1 Mark]
(a) NO,is reduced (b) CO,is reduced For a simple
Tare specific compressible system, v, s, p and
c)HCisreduced (d) CO reduced volume,
and specific entropy, pressure
[1991 2 Marks] temperature, respectively. As per Maxwell's
relations, is equal to
4 Knocking tendency in a sl
engine reduces with
increasing (T
a) Compression ratio
(b) Wall temperature
(a)aps (dV)T
ds
(c)Supercharging
(d) Engine speed
(CaT (d)

[1993 2 Marks] [2019: 1

Mark, Set-2]
8.6 For an ideal gas, the value of the
Joule-Thomson
coefficient is
(a) positive (6) zero
(C) negative (d) indeterminate
[2020:1 Mark, Set-1]
Answers Miscellaneous of I.C. Engine
8.1 (6) 8.2 (a) 8.3 (d) 8.4 (d) 8.5 (a)8.6 (b)
Eplanations Miscellaneous of I.C. Engine
2 a) 8.3 (d)
in Sl engine reduces with
"dil-uel ratio of the mixture in Knocking tendency a
petrol engine is
re than 15:1 it means that the mixture is lean, increasing
earner AFR Turbulence
(air-fuel ratio) results in higher
temperatures as the mixture is combusted. As Engine speed
We
know, Self ignition temperature of fuel
imum NOx is produced at AFR

is
tween 14:1 to 15:1, SO when the air-fuel ratio Octane rating
more than 15: 1 then the NOx formation
reduces.
536 GATE Previous Years Solved
Papers: M
A (d)
At the time of starting, idling and low speed
operation, mixture should be rich for smooth engine
operation.
The richening of mixture increases the probability
of contact between fuel and air particles and thus
improves combustion.

As per Maxwell's relation,

- ap
8.6 (b)

Value of Joule-Thomson coefficient for ideal gas,

= 0
3P )h