IC ENGINES TESTING AND SUPERCHARGING

2. Syllabus : IC Engines
UNIT-IV : TESTING OF IC ENGINES
Objective of testing, various performance parameters for
IC engines - indicated power, brake power, friction power, SFC,
AF ratio etc, methods to determine various performance
parameters, characteristic curves, heat balance sheet
SUPERCHARGING
Supercharging and turbo-charging methods and their limitations

3. Lecture No 23
Learning Objectives:
• To understand objectives of engine testing
• To learn about engine performance parameters

4. Objectives of Testing
• Engine performance during development
• Engine performance after development/
sample testing after production by
manufacturers
• Engine performance testing by Govt Testing
Agencies for certification like ARAI

5. Objectives of Testing
• Whether engine is performing as per design?
 Indicated Power (IP)
 Brake Power (BP)
 Frictional Power (FP)
 Mechanical Efficiency
 Thermal Efficiency
 Brake Specific Fuel Consumption (bsfc)
 Heat Balance Sheet
• Whether engine is meeting emission norms?
 CO
 HC
 NOx
 PM
 Soot

6. Objectives of Testing
• Leakage in engine?
• Smooth running of engine?
• Engine performance after overhaul/repairs?

7. Performance Parameters
Brake Power (BP)
Power available at output shaft/ crank shaft
Mechanical Losses/Frictional Power (FP)
Sum of frictional losses and pumping losses
incl power required to operate engine
accessories like water pump, dynamo etc
Indicated Power (IP)/ Theoretical Power
Power produced within eng cylinder
IP = BP + FP

8. Tests Performed on IC Engines
1. Indicated Power (IP)
2. Brake Power (BP)
3. Frictional Power (FP)
4. Mechanical Efficiency
5. Air-Fuel Ratio
6. Brake Specific Fuel Consumption (bsfc)
7. Thermal Efficiency
8. Working out Heat Balance Sheet

9. Engine/ Mechanical Indicator
Indicator Paper
Wrapped Drum
p-V diagram
Stylus
Weight
Coupling Nut
To Combustion Chamber
Piston
Rope connected
To Piston Rod
Pulleys

10. Measurement of IP on Mech/Eng Indicator
• To determine IP, p-V diagram is required, the area of
which represents work developed by engine per cycle
• Apparatus used for drawing actual p-V diagram is
called Mechanical/ Engine Indicator
• Eng indicator consists of a cylinder, piston, piston rod
coupling nut, straight line linkage with stylus, spring
of required stiffness, indicator card wrapped drum,
pulley, rope and weights .
• Vertical movement of stylus and horizontal movement
of the cord combines to produce a closed figure
called Indicator diagram
• Area of indicator diagram can be measured by
Planimeter to a definite scale giving work developed

11. Mean Effective Pressure
Indicated Mean Effective Pressure (imep)
• imep is the average pressure, which if acted over the
entire stroke length, would produce the same work
done by the piston as is actually produced by the
engine during a cycle
• Let ‘a’ be the net area of indicator diagram (cm2)
‘l’ be length of diagram (cm) and
‘k’ be spring stiffness N/cm2/cm
Hence, mean height of
diagram = a/l
2
/
)
( cm
N
xk
l
a
imep
pm 


12. Indicated Power (IP)
Let pm= imep (N/cm2);
A= Piston top area (cm2)=Лd2/4;
L= Length of stroke (m)
n= Power stroke /min(=N/2 for 4 S eng as one
power stroke per 2 rev & N for 2S eng)
N= RPM
Hence, Force on Piston=
WD per Cycle=
Pm x A (Newton)
pm x A x L (Nm)
Hence, IP = pm x A x L x n (Nm/min)/Cycle/Cylinder
)
/
/
(
60
cylinder
s
Nm
LAn
pm

)
(
000
,
60
kW
Cylinders
of
No
x
LAn
p
IP m



13. Lecture No 24
Learning Objectives:
• To learn working out/ measurement of Brake Power (BP) &
Friction Power (FP)

14. Measurement of BP
1. Rope Brake Friction Dynamometer
Spring Balance
Flywheel/Brake Drum
Rope
Weight
S
W

15. Rope Brake Friction Dynamometer (Contd.)
Let W=Dead Weight (mg) in Newton (N)
S=Spring Balance Reading (N)
Rb=Radius of Brake Drum (D+d)/2 (m)
D=Brake Drum dia and ‘d’ rope dia
N=Engine RPM
Hence, net Brake Load= (W – S)
Braking Torque = (W-S) x Rb
Hence, Brake Power (BP) =
 
)
(
000
,
60
2
kW
N
x
xR
S
W b 

Brake Mean Effective Pressure (bmep)
)
(
000
,
60
kW
Cylinders
of
xNo
LAn
p
BP
bmep


16. 2. Prony Brake Dynamometer
Brake
Shoes
Flywheel/
Brake Drum
Length, L
Weight
W
Load
Arm

17. Prony Brake Dynamometer (Contd.)
Let W (=mg) be the weight (N)
Let L be the distance from centre of brake drum to
hanger, called load arm (m)
Then, Torque=W x L (Nm)
  )
(
000
,
60
2
,
kW
N
L
x
W
BP
Hence


60
2
&
N
x
T
BP






18. Frictional Power (FP)
• Difference between IP and BP is called FP
• FP includes:
- Pumping losses due to intake & exhaust processes
- Frictional losses in bearings, rotary/sliding parts
- Power required to drive auxiliaries like governor,
water, lub oil, fuel pumps, alternator/dynamo,
valve operating mechanism etc
• FP increases as square of N but practically FP ∞ N1.6
• Higher FP results in:
- Reduced power output
- Decreased mech efficiency
- Increased bsfc
- Increased requirement of cooling

19. Methods of Measurement of FP
1. By measurement of IP and BP
2. Willan’s Line Method
3. Morse Test
4. Motoring Test

20. FP by Willan’s Line Method
( Fuel Rate Extrapolation Method )
- 8 - 4 0 4 8 12 16 20
4
3
2
1
Fuel Flow
Rate (kg/h)
BP (kW)
A
At Constant Eng Speed, say
1500 RPM

21. FP by Willan’s Line Method
• A graph between fuel consumption rate (kg/h) taken
on y-axis and BP (kW) on x-axis is drawn, while
engine is made to run at some constant speed, say
1500 RPM
• The graph is extrapolated back to zero fuel
consumption, which cuts on –ve x-axis at point ‘A’
• The –ve intercept on x-axis represents FP at that
speed of the engine
• Although when BP=0, some fuel consumption is there.
This fuel is consumed to overcome engine friction
• Only for CI engine to be run at constant speed as
Fuel consumption rate v/s BP plot is almost straight
line in case of diesel engine, hence can be extrapolated

22. FP by Morse Test
• Morse Test can be used for determining FP/IP of
multi-cylinder IC engines, generally 3 cyl and more
by cutting off each cylinder in turn
• In SI engines, each cylinder is rendered in-operative
by short-circuiting the SP or cutting off fuel supply
in MPFI systems. In CI engines, fuel supply is cut off
• Consider 4 stroke, 4 cylinder SI engine coupled with
dynamometer
• Engine is run at constant speed N throughout one
set of test parameters, as FP ∞ N2
• It is assumed that pumping & mech losses are same
whether a cylinder is working or not
• Throttle position is kept fixed, however, to attain
same speed N, load is decreased by dynamometer

23. FP by Morse Test
• Let B=BP of eng when all cylinders are working
B1=BP of eng when Cylinder No 1 is cut off
Similarly, B2=BP of eng when Cylinder No 2 is cut off
• Let I1, I2, I3 & I4 be the IPs developed by Cylinder
Nos 1, 2, 3 & 4 respectively and their corresponding
FPs be F1, F2, F3 & F4
B3=BP of eng when Cylinder No 3 is cut off
B4=BP of eng when Cylinder No 4 is cut off
• Total BP(B) = (I1+I2+I3+I4) - (F1+F2+F3+F4)

24. BP(B) = (I1+I2+I3+I4) - (F1+F2+F3+F4)
FP by Morse Test
Hence, B1=(I2+I3+I4) – (F1+F2+F3+F4)
On subtracting; B – B1 = I1
Similarly, B – B2 = I2
B – B3 = I3
B – B4 = I4
On adding; IP = I1 + I2 + I3 + I4
= 4B – (B1+B2+B3+B4)
B, B1, B2, B3 & B4 can be measured by Dynamometer,
Hence IP can be calculated
Therefore, FP = IP - BP

25. Lecture No 25
Learning Objectives:
• To understand working out of heat balance sheet
• To learn measurement of air/fuel consumption

26. Theoretical/ Air Std Efficiencies
Otto Cycle: 1
1
1 

 

r
Diesel Cycle:
 








 
1
1
1
1 1






r
Dual Cycle:
 
   










 
1
.
1
1
.
1
1 1









r

27. Some Definitions
Thermal Efficiencies:
i) Indicated Thermal Efficiency
CV
x
m
IP
f
i 

ii) Brake/Overall Thermal Efficiency
CV
x
m
BP
f
b 

- Where mf is fuel consumed in kg/sec
- CV is Calorific Value of fuel in kJ/kg
- IP/BP is in kW

28. Some Definitions
Mechanical Efficiency:
100
x
IP
BP
mech 

Relative Efficiency : Defined as the ratio of Brake
Thermal Efficiency to Air Standard Efficiency at same
Compression Ratio(CR)
100
x
a
b
r


 

29. Some Definitions
Volumetric Efficiency:
Temp
Atm
at
vol
swept
to
ing
correspond
e
ch
of
Mass
inducted
e
ch
actual
of
Mass
v
&
Pr
arg
arg


• Reduced Volumetric Efficiency causes reduction in
Power Output
Volume
Swept
conditions
suction
at
inhaled
e
Ch
of
Vol
Actual
Also v
arg
, 

• Ratio of actual mass of charge inducted during
suction stroke to mass of charge corresponding to
swept volume of the engine at atm pr & temp
• Volumetric Efficiency puts a limit on the amt of fuel
that can be burnt and hence on its power, since
output of eng depends on amt of air inducted

30. Some Definitions
(Brake) Specific Fuel Consumption (bsfc/sfc):
)
/
(
3600
kWh
kg
BP
m
bsfc
x
f


Specific Output:
• BP per unit of piston displacement
AxL
BP
Output
Specific 
• bsfc is defined as the amount of fuel required to be
supplied to eng to develop 1kW of power per hour at
crankshaft

31. Heat Balance Sheet
• Heat Balance Sheet is an account of heat released
on combustion of fuel in the combustion chamber
and its utilization in the engine
• To draw heat balance sheet, tests are carried out
on engine, while it is run at some constant speed
Heat Supplied:
Heat Supplied = mf x CV (kJ/min)
Where mf = mass flow rate of fuel (kg/min)
CV = Calorific Value of fuel (kJ/kg)

32. Heat Balance Sheet
Heat Expenditure/Utilization:
1) Heat Equivalent to BP:
3) Heat carried away by Exhaust gases
Heat carried by exh gases=mgCpg(Tge – Tsa) kJ/min
Where mg=(ma+mf) =flue gases flow rate (kg/min)
Heat Equivalent to BP = BP x 60 (kJ/min)
2) Heat Rejected to Cooling Water:
Heat carried away by water =mwCpw(Two – Twi) kJ/min
Where mw=cooling water circulation kg/min
& Cpw=4.187 kJ/kgK
4) Unaccounted Heat:
By difference

33. Heat Balance Sheet
Heat
Supplied
kJ/
min
% Heat Utilization kJ/
min
%
Heat
Supplied
by comb
of
Fuel
=mf x CV
100
Total 100
a) Heat to BP=BPx60
b) Heat to water
=mwxCpw(Two – Twi)
c) Heat carried
away by exhaust
gases
=mgxCpg(Tge – Tsa)
d) Heat Unaccounted
(By difference)
100

34. 100cc 200cc
3-Way Cock
3-Way Cock
Fuel to Eng
Fuel from Tank
Start
Stop
Start
Stop
Volumetric Fuel Flow Meter(Burette Type)

35. Fuel Measurement
• Time required to supply given volume of fuel is noted
• Mass Flow Rate of Fuel Supply:
Fuel
of
xDensity
Time
Volume
mf 
• Density of Fuel = Sp Gravity of fuel x Density of water
• This method does not give very accurate mass flow
rate due to variation in density with temp
Water
of
Density
material
of
Density
Gravity
Specific 

36. Gravimetric Fuel Flow Meter
Weighing Machine
Flask
Fuel to Engine
Fuel Tank
Valves
A
B

37. Air Flow Meter
ΔH
Thermometer
Orifice Plate (A, Cd)
Manometer
Air
Surge
Tank
Air Intake to Eng

38. Measurement of Air Consumption by
Air Flowmeter
• Surge tank is connected to intake side of the engine
• Manometer measures the pressure difference
• Vol Flow Rate w
d H
g
x
A
x
C 
 .
2
-Cd – Coeff of discharge for given orifice
-A – Orifice cross sectional area
-ΔHw-Head of water to be converted to air head

39. Lecture No 26
Learning Objectives:
• To learn about engine characteristic curves

40. SI Engine Characteristic Curves
POWER
(kW)
Speed
IP
BP
FP

41. SI Engine Characteristic Curves
• Lab tests carried out to determine eng performance
• During tests, throttle is kept full (full /rated load,
max fuel consumption) and speed is varied by
adjusting the brake load
• IP, BP, FP, bsfc, mechanical & volumetric efficiencies
etc are worked out
• Same tests can be repeated at half load
• At rated output, max p-V diagram area, hence max
imep; For given torque; power ∞ N
kW
N
Tx
hp
000
,
60
2


42. SI Engine Characteristic Curves
• IP increases when imep or speed or both increase
• IP initially increases faster with speed, if inlet
conditions are kept constant
• However, after certain limit, rate of increase of IP
reduces with speed due to reduction in vol efficiency
as air/charge velocity increase results in inlet pr drop
• Mech losses increase with
increase in speed(FP∞ N2)
due to which increase in
IP is off-set by steep
increase in FP

43. SI Engine Characteristic Curves
IP
BP
bsfc
Mech Eff
Speed
IP
BP
bsfc
Mech Efficiency
x

44. SI Engine Characteristic Curves
• As FP ∞ N2, mech efficiency reduces due to steep
increase in FP
• At lower speeds, due to lower charge velocity because
of low piston speed, bsfc reduces since volumetric
efficiency increases and mech efficiency also increases
• After certain speed, bsfc
increases due to
reduction in volumetric
efficiency and increase
in mech losses
x
• Point x represents
economical speed of eng
for min fuel consumption

45. SI Engine Characteristic Curve
Vol
Efficiency
Speed

46. SI Engine Characteristic Curve
• Volumetric Efficiency reduces with increase in speed
due to increase in intake velocity resulting in drop of
suction pressure
• Higher the speed, lesser the time available for
induction of charge
• Suction valve fully opens
only when pressure inside
cylinder slightly below
the surrounding pressure,
thus reducing effective
suction stroke

47. CI Engine Characteristic Curves
Power
bsfc
Speed
BP
bsfc

48. CI Engine Characteristic Curves
• IP and BP increase with speed but due to steep
increase in FP, IP and BP start coming down
• For bsfc curve, same reasons as in SI engine

49. Engine Characteristic Curves
bsfc
BP
SI
CI

50. CI Engine Characteristic Curve
Brake/
Overall
Efficiency
A/F Ratio
Stoichiometric Mixture
Lean
Mixture
Rich
Mixture

51. Lecture No 27
Learning Objectives:
• To understand working out of engine parameters through
numerical problems

52. Q1. Obtain cylinder dimensions of a twin-cylinder, 2-S
IC engine from the following data:
Engine speed=4000RPM; Volumetric efficiency=77%;
Mech Efficiency=75%; Fuel consumption=10 lit/hr;
Sp. Gr. Of fuel=0.73; A/F ratio=18;
Piston speed= 600m/min; imep=5 bar.
Also, determine power output at STP conditions
(p=101325 N/m2; Ta=25˚C; R for air=0.287 kJ/kgK)
Solution:
Cylinder Dimensions=? D & L
Piston speed=2LN Since speed & N are given, L=?
Now, to find out D=? Cylinder
stroke
one
for
L
D
Vs /
4
2


 
engine
S
for
N
S
for
N
Strokes
Power
strokes
power
of
xNo
cyl
of
xNo
Vs
Rate
Flow
Vol
Total
4
2
/
&
2
min
/


53. Solution(Contd):
v
v
Va
Vs
Vs
Va

 


Ta
R
ma
Va
pa
Va
Rate
Flow
Vol
actual
For .
.
. 

)
(
18
/ given
Ratio
F
A
m
m
ma
For
f
a



f
f x
h
m
x
h
lit
m 














 
3
3
10
10
w
x
fuel
of
gr
Sp
x
h
m
x 
.
10
10
3
3


h
kg
h
kg
x
x
x
/
3
.
7
/
1000
73
.
0
10
10 3

 

54. Solution(Contd):
h
m
Va
Vs
Vs
Va
Since
v
v
/
144
77
.
0
9
.
110 3







 
h
m
x
x
Va
Rate
Flow
Hence
Ta
R
ma
Va
pa
/
9
.
110
101325
273
25
287
4
.
131
,
.
.
.
3




h
kg
h
kg
x
m
m
m
a
f
a
/
4
.
131
/
3
.
7
18
18 




55. Solution(Contd):
2
2
600
4
min
60
144 2
3
xNx
N
x
D
m 









Cylinders
x
N
x
L
D
Vs
Rate
Flow
Vol 2
4
2


2
3
2
10
094
.
5 m
x
D 


mm
m
D 4
.
71
0714
.
0 


mm
m
x
L
And
75
075
.
0
4000
2
600




56. Q2. A 6 cylinder gasoline engine operates on 4 stroke
cycle. Bore of cylinder is 80mm and stroke 100mm.
Clearance volume per cylinder is 70CC. At 4000 RPM,
Fuel consumption is 20kg/hr and the torque developed
is 150Nm. Calculate:-
(a)BP (b) Brake mean effective pressure (c) Brake
thermal efficiency
If CV of the fuel is 43000kJ/kg, find relative efficiency
on brake power basis, assuming engine works on
Constant volume cycle and gamma for air =1.4.
Solution:
kW
N
Tx
BP
000
,
60
2
 kW
x
x
83
.
62
60000
4000
2
150




57. ?

bmep
cyliders
of
no
x
AxLxn
x
bmep
BP
60000

 
bar
m
N
x
x
x
x
x
bmep
25
.
6
/
10
25
.
6
6
60000
2
4000
1
.
0
08
.
0
4
83
.
62
2
5
2





58. CV
x
m
BP
f
b 

;
a
b
r


 
%
3
.
26
263
.
0
43000
3600
20
83
.
62



x
1
1
1 

 

r
a
;
c
c
s
V
V
V
r

 CC
x
Vs 6
.
502
10
8
.
0
4
2



;
18
.
8
70
70
6
.
502



r 5686
.
0
18
.
8
1
1 1
4
.
1


 
a

%
25
.
46
4625
.
0
5686
.
0
263
.
0



r


59. Lecture No 28
Learning Objectives:
• To understand working out of engine parameters and heat
balance sheet through numerical problems

60. Q3. During trial of a single cylinder, 4 stroke oil engine
the following results were obtained:
Cyl bore=200mm, Stroke=400mm, mep=6 bar,
Torque=407Nm, speed=250 RPM,
Oil consumption=4kg/hr, CV of fuel=43MJ/kg,
Cooling water rate=4.5kg/min, Air used per kg of fuel=
30kg, Rise in cooling water temp=45°C, Temp of
Exhaust gases=420°C, Room temp=20°C, mean sp.
heat of exhaust gases=1kJ/kgK, Sp. Heat of water=
4.18kJ/kgK, Barometric pressure=1.01325 bar
Find IP, BP and draw up heat balance sheet in kJ/hr.
Solution:
)
(
000
,
60
kW
cyl
of
no
x
AxLxn
x
imep
IP 

61. 60000
2 N
x
T
BP


 
kW
x
x
x
x
x
IP 7
.
15
1
000
,
60
2
250
4
.
0
2
.
0
4
10
6
2
5



kW
x
x
65
.
10
60000
250
2
407



Heat Balance Sheet
1. Heat supplied by fuel to eng =mfxCV
=4x43000kJ/hr=172,000kJ/h

62. 2. Heat utilized
(ii) Heat to cooling water=mw x Cpw x ∆Tw
=4.5x60x4.18x45 = 50,787 kJ/h
(iii) Heat to exhaust gases=mg x Cpg x (Te-Ta )
To find mg : ma=30kg/kg of fuel; hence mg=31kg
Since fuel consumption is 4kg/h; mg=31x4kg/h
Hence, Heat to exhaust gases=31x4x1(420-20)
=49,600kJ/h
(iv) Unaccounted Heat=33,255kJ/h (by difference)
(i) Heat to Power Output=BPx3600kJ/h
=10.65x3600=38,358kJ/h

63. Heat
Supplied
kJ/hr % Heat Utilized kJ/hr %
Heat
supplied by
fuel= mfxCV
172,000 100 To BP=
BPx3600
38,358 22.3
To Cooling
Water
=mw.Cpw.∆Tw
50,787 29.5
To exhaust
gases
=mg.Cpg.∆Tg
49,600 28.8
Unaccounted
Heat
33,255 19.3
Total 172,000 100 Total 172,000 100
Heat Balance Sheet

64. Q4. During a test on a 4 stroke oil engine,
the following data were obtained:
Mean height of indicator diagram = 21mm
Indicator spring number/stiffness=27kN/m2/mm
Swept volume=14 lit, effective brake load=77kg,
Effective brake radius= 0.7m, speed=6.6 rev/s,
fuel consumption=0.002kg/s, CV of fuel=44MJ/kg,
Cooling water rate=0.15kg/s, water inlet temp=38°C,
cooling water outlet temp=71°C, Sp. heat of water=
4.18kJ/kgK, energy carried by exhaust gases=33.6kJ/s
Determine IP, BP and mech efficiency and draw up
heat balance sheet in kJ/s and %.
Solution:
)
(
000
,
60
kW
cyl
of
no
x
AxLxn
x
imep
IP 

65. )
(kW
cyl
of
no
x
n
x
L
x
A
x
imep
IP 
mm
m
kN
x
mm
k
x
l
a
imep .
/
27
)
(
21 2


2
2
/
567
.
/
27
)
(
21 m
kN
mm
m
kN
x
mm 

kW
m
x
l
m
kN
IP 2
.
26
2
6
.
6
).
(
10
)
(
14
).
/
(
567 3
3
2

 
 
000
,
1
2 N
x
R
S
W
BP b 


kW
x
x
x
x
92
.
21
000
,
1
6
.
6
2
7
.
0
81
.
9
77



%
65
.
83
100
2
.
26
92
.
21

 x
mech


66. Heat
Supplied
kJ/s % Heat Utilized kJ/s %
Heat supplied
by fuel=
mf x CV =
0.002x44000
(=88kJ/s)
88 100 To BP 21.92 24.9
To Cooling
Water
=mw.Cpw.∆Tw
0.15x4.18x
(71-38)
20.69 23.5
To exhaust
gases (given)
33.6 38.2
Unaccounted
Heat
11.79 13.4
Total 88 100 Total 88 100
Heat Balance Sheet

67. Q5. A 4 cylinder 4 stroke SI engine has a bore of
5.7cm and stroke 9cm. Its rated speed is 2800 RPM
and it is tested at this speed against a brake which
has a torque arm of 356mm. The net brake load is
155N and fuel consumption is 6.74 lit/hr. Sp gravity of
petrol is 0.735 and CV is 44200kJ/kg. A Morse test is
carried out and cylinders are cut off in order of 1, 2, 3
& 4 with corresponding brake loads of 111, 106.5,
104.2 and 111N. Determine engine torque, bmep,
brake thermal efficiency, sfc, mech efficiency and imep.
Solution:
Engine torque= (W-S).Rb=WxL
= 155x0.356=55.18Nm

68. )
(
60000
2
000
,
60
kW
N
Tx
cyl
of
no
x
AxLxn
x
bmep
BP



kW
x
x
x
x
bmep
17
.
16
4
000
,
60
2
2800
09
.
0
)
057
.
0
(
4
2


kW
x
BP 17
.
16
60000
2800
.
2
18
.
55



xCV
m
BP
f
b 

bar
bmep 55
.
7



69. s
kg
x
x
x
m
x
h
lit
mf
/
10
376
.
1
3600
1000
735
.
0
)
(
10
)
/
(
74
.
6
3
3
3




kWh
kg
x
x
BP
m
bsfc
f
/
306
.
0
17
.
16
3600
10
376
.
1 3




%
58
.
26
2658
.
0
44200
10
376
.
1
17
.
16
3
or
x
x
xCV
m
BP
f
b


 


70. We know that IP = 4BP - (BP1+BP2+BP3+BP4)
60000
2 N
x
WxR
BP b 

 
 
60000
2
4 4
3
2
1
N
x
R
W
W
W
W
W
IP b 





 
 
60000
2800
2
356
.
0
111
2
.
104
5
.
106
111
155
4
x
x
x
IP






kW
54
.
19

%
75
.
82
100
54
.
19
17
.
16


 x
IP
BP
m

8275
.
0
55
.
7
.
.



imep
LAn
imep
LAn
bmep
m

bar
imep 12
.
9



71. Q6. During trial of a 4 cylinder 4 stroke SI engine
running at 50 rev/s, the brake load was 267N when all
cylinders were working. When each cyl was cut off in
turn and speed returned to same 50 rev/s, brake
readings were 178N, 187N, 182N and 182N.
Determine BP, IP and mech efficiency of the engine.
For brake, BP=F.N/455(kW), where F is brake load in
Newtons and N rev/s. The following results were
obtained: Fuel consumption=0.568lit in 30 seconds,
SG of fuel=0.72, CV=43000kJ/kg, A/F ratio=14:1,
Exh temp=760°C, Cpg=1.015kJ/kg, Water inlet temp=
18°C and outlet temp=56°C, water flow rate=0.28kg/s,
Ambient temp=21°C. Draw heat balance sheet in kJ/s.
Solution:
kW
x
N
F
BP 34
.
29
455
50
267
455
.




72. We know that IP = 4BP - (BP1+BP2+BP3+BP4)
kW
x
N
F
BP 56
.
19
455
50
178
455
.
1 


kW
x
N
F
BP 55
.
20
455
50
187
455
.
2 


4
3 20
455
50
182
455
.
BP
kW
x
N
F
BP 



Therefore IP= 4x29.34 - (19.56+20.55+20+20)
=37.25kW
%
76
.
78
100
25
.
37
34
.
29


 x
IP
BP
mech


73. Heat utilized
(ii) Heat to cooling water=mw x Cpw x ∆Tw
=0.28x4.187x(56-18) =44.55 kJ/s (7.6%)
(iii) Heat to exhaust gases=mg x Cpg x (Te-Ta )
To find mg :(ma +1)xmf=(14+1)x0.01363=0.204kg/s
Hence, Heat to exhaust gases=0.204x1.015(760-21)
=153kJ/s (26.1%)
(iv) Unaccounted Heat=356kJ/s (61%) (by difference)
(i) Heat to BP=BP= 29.34kJ/s (5%)
Heat Balance Sheet
Heat supplied =mfxCV
s
kJ
x
x
x
x
x
/
2
.
586
43000
01363
.
0
43000
30
1000
72
.
0
10
568
.
0 3





75. Syllabus : IC Engines
SUPERCHARGING
Supercharging and turbo-charging methods and their limitations

76. Lecture No 29
Learning Objectives:
• To learn effects of supercharging /turbo-charging and
limitations

77. • Increasing Eng speed (BP= T x 2πN)
(FP ∞ N2 & Volumetric η ↓ )
How can engine power be increased?
• Higher CR (Peak Pr increases; Thermal Load
increases; Weight to Power ratio increases)
(HUCR limited due to knocking/detonation in SI
engines and heat load in CI engines)
• Utilization of exh energy in gas turbine, thus ↑ BP
• Use of 2-stroke cycle; but cooling, emission
problems, lower volumetric & thermal efficiency
• Increasing charge density by
- Lowering charge temp (Cooling) and / or
- Increasing induction pressure

78. Objectives of Supercharging
• To increase power output
• To increase power to weight ratio
• To compensate loss of power at high altitude
• To reduce bulk size of engine

79. Supercharging
• Supplying air /Air-Fuel mixture at higher pressure
than the pressure, at which the engine naturally
aspirates, by a boosting device is called
supercharging
• The device which boosts the pressure is called
supercharger.
• Purpose of supercharging to have small displacement
engines but developing more power and to meet
emission legislation on fuel consumption for emission
control
• More power is achieved by raising density of charge,
thus more mass of air making available more oxygen
for combustion

80. Supercharging & Turbo Charging Systems

82. Effects of Supercharging
• Increased Eng Output (p-V diagrams)
• Turbulence Effect (Higher BP)
• Power required to drive supercharger, thus ↓ BP
• Mech Efficiency increases
• SI Engines→ Knocking tendency as ign delay ↓
• bsfc ↑ for SI (due to reduced delay) but ↓ for CI
engines due to better combustion & higher
mechanical efficiency
• Better scavenging; Increase in power output
• For CI Engines→ Smoother running, low F/A
ratio, ↑ durability & reliability and lower bsfc

83. Effects of Supercharging
• Better atomization
• Better mixing of air and fuel
• Reduced exhaust smoke
• Better torque characteristic over whole speed range
• Increased gas load
• Better and smoother combustion
• Increased thermal stresses
• Increased valve overlap period of 60° to 160° of
crank angle
• Increased cooling requirements of piston and valves

84. -(b)
p
patm
V
+(a)
7
1
5
6
2
4
3
+(d)
+(c)
patm
p
V
4
3
7
1
5
6
2
Naturally Aspirated Engine Supercharged Engine
P-V Diagrams of Naturally Aspirated &
Supercharged Engines

85. Limitations of Supercharging
• Power o/p limited by knock, thermal & mech loads
• For SI engines, knocking reached earlier
• In CI Engs, thermal & mech loads reached earlier
• Increase in peak pr, increases bearing loads
• Increase in intake pr increases peak pr leading to
increase in weight of cylinder (limitation on peak pr)
• Increase in peak pr→ ↑ tendency to detonate (SI)
• Increase in peak pr increases friction losses
• ↑ peak pr → ↑ peak T →Reqmt of better cooling sys
• ↑T → ↑ exh gas temp →overheating of exh valves
Due to the above reasons, supercharging
generally limited to 2.5 bar

86. Limitations of Supercharging In SI Engs.
• Detonation is the limitation as it increases with ↑ pr,
↑ T, ↑ CR, ↑ density of charge
• Strongest detonation at stoichiometric A/F ratio
• CR limited due to detonation for given Octane
Rating of fuel used
• Detonation can be reduced by reducing CR but
BP & thermal efficiency decreases & bsfc increases
For the above reasons, SI Engines are normally
NOT supercharged except for aircraft, high
altitude compensation or higher power of aero
engines required at the time of take off of aircraft
at the expense of higher fuel consumption

87. Limitations of Supercharging In CI Engs.
• Increased induction pr helps in suppressing
knocking tendency, improve combustion,
higher power output & thermal efficiency and
hence can use lower Cetane fuels
• Supercharging is limited by:
- peak pressure →mechanical loading
- peak temp →thermal loading
- thermal stresses developed
- mean temp of cylinder walls
- loads on bearings

88. Lecture No 30
Learning Objectives:
• To understand methods of supercharging
• To learn various methods of turbocharging

89. Types of Superchargers
• Centrifugal Type Supercharger
• Root’s Type Supercharger
• Vane Type Supercharger

90. Centrifugal Type Supercharger

91. Root’s Type Supercharger

92. Vane Type Supercharger

93. Vane Type Supercharger

94. Arrangements of Supercharging
Engine Load
Compressor
Air inlet to
Compressor
Gears
Exhaust from
Engine
Inlet to Engine
Air outlet from
Compressor
After Cooler

95. Arrangements of Turbocharging
Engine Load
Compressor
Air inlet to
Compressor
Exhaust from
Engine
Inlet to Engine
Air outlet from
Compressor
After Cooler
Turbine
Exhaust from
Turbine

96. Method of Super/Turbocharging
Engine
Load
Compressor
Air inlet to
Compressor
Exhaust from Engine
Inlet to Engine
Air outlet from
Compressor
After Cooler
Turbine
Exhaust from
Turbine

97. Method of Super/Turbocharging
Engine Load
Compressor
Air inlet to
Compressor
Gears
Exhaust from
Engine
Inlet to Engine
Air outlet from
Compressor
After Cooler
Exhaust from
Turbine
Turbine Load

98. Turbochargers
• Exhaust gases carry about 1/3 of the total energy
generated in the eng cylinder
• In order to utilize this energy, hot gases can be
allowed to expand further in a gas turbine and its
work output can be utilized to drive a supercharger.
This system of supercharger coupled to Turbine is
called Turbocharger
• Due to cyclic fluctuations of the pressure in exhaust
pipe, turbo charging is not employed in single
cylinder eng, however, system is suitable for
engines having 4 or more cylinders

99. Turbo Charger
• Turbocharger does not consume eng power
• No gearing required between turbine and
compressor as both are connected by single shaft
• Gain in power at nominal cost
• Exhaust energy, which is 1/3 of total energy
generated in the engine, is gainfully utilized
• Exhaust noise level reduces
• Suitable for high speed engines
Advantages
Disadvantages
• Increase in fuel consumption at low power output
• Total cost of unit increases