Internal Combustion Engines (IC Engines)
The internal combustion engine (IC Engine) is a heat engine that converts heat energy
(chemical energy of a fuel) into mechanical energy (usually made available on a rotating output
shaft).
Applications of IC Engines:
Mainly used as ‘prime movers’, e.g. for be the propulsion of a vehicle i.e., car, bus, truck,
locomotive, marine vessel, or airplane. Other applications include stationary saws, lawn mowers,
bull-dozers, cranes, electric generators, etc.
Classifications of IC Engines:
IC engines can be classified according to:
1. Number of cylinders – 1, 2, 3, 4, 5, 6 to 16 cylinder engines.
2. Arrangement of cylinders – Inline, V-type, Flat type, etc.
3. Arrangement of valves and valve trains – In-block camshaft, OHC, DOHC, etc.
4. Type of cooling – Air-cooled, Water-cooled, etc.
5. Number of strokes per cycle – 2-stroke, 4-stroke engines.
6. Type of fuel burned – Petrol, diesel, CNG, etc.
7. Method of ignition – Spark Ignition (SI), Compression Ignition (CI).
8. Firing order – 1-3-4-2, 1-2-4-3, etc.
9. Primary mechanical motion – Reciprocating, rotary.
Fig. 1: Major Components of a reciprocating single cylinder Petrol Engine.
Page 1 of 15
Problem-1: A four cylinder car engine has
bore x stroke = 79 mm x 77 mm.
What is the capacity of the engine in cc?
Solution: Capacity in cc = N.(π/4).B2. S
Here, N= number of cylinders
B= bore diameter in cm
S= stroke length in cm
Therefore, Engine Capacity in cc
= 4 x (π/4) x (7.9)2 x (7.7) = 1509 ≈ 1500 cc Ans.
Fig. 2: Main Geometric parameters of a reciprocating IC Engine.
Four-stroke Petrol Engine.:
A 4-stroke petrol engine operates on air
standard Otto cycle. It completes the Otto cycle
in 4 strokes (4 TDC to BDC movements of the
piston), namely, (1) Suction Stroke, (2)
Compression Stroke, (3) Power Stroke, (4)
Exhaust Stroke.
Otto cycle shown below consists of four
processes:
1 – 2 : Isentropic Compression Process
2 – 3 : Constant Volume Combustion
3 – 4 : Isentropic Expansion Process
4 – 1 : Constant Volume Blowdown
Fig. 3: Four-strokes of an IC (Petrol) Engine.
Page 2 of 15
Fig. 4: Air standard Otto cycle for petrol engine.
For a petrol engine working on air standard Otto cycle,
Compression ratio,
r=
vd + vc v1 v4
= =
vc
v2 v3
For an isentropic process, x → y ,
Tx ⎛ px ⎞
=⎜ ⎟
Ty ⎜⎝ p y ⎟⎠
k −1
k
⎛ vy ⎞
=⎜ ⎟
⎝ vx ⎠
k −1
Otto cycle Efficiency = Work Output / Heat Supplied
= (Heat Supplied – Heat Rejected) / Heat Supplied,
or,
η=
Wout Qin − Qout
=
Qin
Qin
or,
η=
ma cv (T3 − T2 ) − ma cv (T4 − T1 )
T −T
= 1− 4 1
ma cv (T3 − T2 )
T3 − T2
So,
η = 1−
T4 − T1
T3 − T2
Now for isentropic compression process, 1 → 2 :
or,
(1)
T1 ⎛ v2 ⎞
=⎜ ⎟
T2 ⎝ v1 ⎠
T2 = T1r k −1
(2)
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k −1
⎛1⎞
=⎜ ⎟
⎝r⎠
k −1
T3 ⎛ v4 ⎞
=⎜ ⎟
T4 ⎝ v3 ⎠
Again, for isentropic expansion process, 3 → 4 :
T4 =
or,
T3
r k −1
Putting (2) and (3) into (1), we get,
k −1
= (r )
k −1
(3)
T3
− T1
k −1
r
η = 1−
T3 − T1r k −1
T3 − T1r k −1
k −1
η = 1 − r k −1
T3 − T1r
or,
Therefore,
ηOtto = 1 −
1
r k −1
where,
k=
cp
cv
and k = 1.4 for air.
So we see, the efficiency of the petrol engine depends only on the compression ratio r.
Diesel Engine working on diesel cycle
Diesel cycle shown above consists of four
processes:
1 – 2 : Isentropic Compression Process
2 – 3 : Constant Pressure Combustion Process
3 – 4 : Isentropic Expansion Process
4 – 1 : Constant Volume Blowdown
Fig. 5: Air standard Diesel Cycle for Diesel Engine
For a diesel engine working on air standard Diesel cycle,
Compression ratio,
r=
vd + vc v1
=
vc
v2
and
Fuel cut-off ratio,
ρ=
vcut + vc v3
=
vc
v2
Diesel cycle Efficiency = Work Output / Heat Supplied
= (Heat Supplied – Heat Rejected) / Heat Supplied,
or,
η=
Wout Qin − Qout
=
Qin
Qin
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or,
η=
ma c p (T3 − T2 ) − ma cv (T4 − T1 )
ma c p (T3 − T2 )
η = 1−
So,
= 1−
cv T4 − T1
1 T4 − T1
= 1−
c p T3 − T2
k T3 − T2
1 T4 − T1
k T3 − T2
(1)
Now for isentropic compression process, 1 → 2 :
T1 ⎛ v2 ⎞
=⎜ ⎟
T2 ⎝ v1 ⎠
T2 = T1r k −1
or,
⎛1⎞
=⎜ ⎟
⎝r⎠
k −1
⎛r⎞
=⎜ ⎟
⎝ρ⎠
k −1
(2)
For constant pressure combustion process, 2 → 3 :
T2 v2 1
= =
T3 v3 ρ
T3 = T2 ρ = T1r k −1 ρ
or,
k −1
Again, for isentropic expansion process, 3 → 4 :
(3)
T3 ⎛ v4 ⎞
=⎜ ⎟
T4 ⎝ v3 ⎠
k −1
⎡
v4 v4 v3 v1
1
r⎤
since,
=
⋅
=
⋅
=
⎢
⎥
v3 v2 v2 v2 (v3 / v2 ) ρ ⎦
⎣
T3 ρ k −1 T1r k −1 ρ .ρ k −1
T4 = k −1 =
= T1 ρ k
k −1
r
r
or,
Putting (2) and (3) into (1), we get,
1
T1 ρ k − T1
η = 1 − ⋅ k −1
k T1r ρ − T1r k −1
1 T1 ( ρ k − 1)
η = 1 − ⋅ k −1
k T1r ( ρ − 1)
or,
Therefore,
(4)
η Diesel
1 ⎡ ρ k −1 ⎤
= 1 − k −1 ⎢
r ⎣ k ( ρ − 1) ⎥⎦
where,
k=
cp
cv
and k = 1.4 for air.
So we see, the efficiency of the diesel engine depends not only on the compression ratio but
also on the fuel cut-off ratio.
Page 5 of 15
Comments: The fuel cut-off ratio ρ of the diesel engine is usually greater than 1 and k is 1.4 for
air so the expression in square bracket is always greater than 1. So for the same compression
ratio, petrol engine is more efficient than the diesel engine. But usually the compression ratio of
diesel engine is much higher than that of petrol engine (from 8 to 12). Therefore, the diesel
engine is more efficient due to it higher compression ratio (from 15 to 23).
Problem-2: Calculate the air standard cycle efficiencies of an Otto cycle engine and a Diesel
cycle with a compression ratio of 10:1. Take ρ = 1.5 for the diesel engine.
1
Solution: Otto cycle efficiency, ηotto = 1 – k −1
r
=1–
1
1
=1–
= 0.6019 = 60.19% Ans.
1.4 −1
0.4
10
10
⎡ k
⎤
1 ⎢ ρ −1 ⎥
Diesel cycle efficiency, ηdiesel = 1 – k −1
r ⎢⎣ k ( ρ − 1) ⎥⎦
1.4
⎤
⎡
1 ⎢ 1 .5 − 1 ⎥
= 0.5654 = 56.54% Ans.
= 1–
1.4 −1
⎢1.4(1.5 − 1) ⎥
10
⎦
⎣
Problem-3: Calculate the air standard Diesel cycle efficiency of the engine with a compression
ratio of 11:1; if the fuel supply is cut-off at 6% of the stroke (or swept or displacement volume).
Solution:
Given,
vcut
= 0.06
vstroke
Now from Fig.5, we can write,
v3 − v2
= 0.06
v4 − v2
or,
or,
v3 − v2
= 0.06
v1 − v2
v2 (v3 / v2 − 1)
= 0.06
v2 (v1 / v2 − 1)
or,
or,
ρ −1
= 0.06
r −1
ρ −1
= 0.06
11 − 1
ρ = 1.6
or,
Diesel cycle efficiency,
⎤
⎡ k
1 ⎢ ρ −1 ⎥
ηdiesel = 1 – k −1
r ⎢⎣ k ( ρ − 1) ⎥⎦
1.4
⎡
⎤
1 ⎢ 1.6 − 1 ⎥
= 0.5753
η = 1 − 1.4−1
11 ⎢⎣1.4(1.6 − 1) ⎥⎦
η = 57.53 % Ans.
Page 6 of 15
Actual Indicator Diagram of a 4-stroke Petrol Engine
Valve Timing Diagram for a 4-stroke Petrol Engine
Valve overlap: The duration (10o + 10o = 20o) when both the inlet and exhaust valves
remain open is called valve overlap.
Spark Advance: The ignition is initiated 20 – 30o before TDC. This is called spark
advance.
Differences between Petrol and Diesel engines
Page 7 of 15
drilling machines, etc.
Two-Stroke Engines:
Two strokes engines (both petrol and diesel engines) completes the cycle (both Otto and
Diesel cycle) in just two strokes of the piston. In moving from BDC to TDC, both the
suction and compression occur whereas, during TDC to BDC movement, both power and
exhaust occur thus completing the total cycle.
In Brief:
(a) and (b): Up-stroke of the piston : Suction and Compression.
(c) and (d) : Down stroke of the piston : Expansion and Exhaust.
Page 8 of 15
(a) Suction
(b) Compression
(c) Expansion (Power)
(d) Exhaust
Fig. 6: Two-stroke IC (Petrol) Engines.
Scavenging in 2-Stroke Cycle Engine:
At the end of the expansion stroke, the exhaust port opens and allows the exhaust gas to
exit. As the piston starts moving down, air-pressure in the crank case increases and air,
through the scavenging port, enters the cylinder (combustion chamber) and pushes out
the burnt gas clear from the cylinder. This is referred to as scavenging.
Differences between 2-stroke and 4-stroke engines
4-stroke Engine
1. The intake, compression, combustion
and exhaust occur in two upward and two
downward strokes of the piston.
2. Needs complicated valve train
arrangement for intake and exhaust strokes.
2-stroke Engine
1. All four events are accomplished in one
downward stroke, and one upward stroke.
2. Intake and exhaust are both integrated
into the compression and combustion
movement of the piston, eliminating the
need for valves.
3. Outputs power once in every two
3. The engine delivers power on every
revolutions of the crankshaft.
revolution.
4. The engine is heavier for the same power 4. Higher power-to-weight ratio because it
rating, i.e., low power to weight ratio.
is much lighter.
5. More expensive than the 2-stroke
5. Less expensive because of its simpler
engines.
design.
6. It has limited orientation if oil is to be
6. It can be operated in any orientation
retained in the sump.
because it lacks the oil sump
7. More fuel efficient, less noisy, less
7. Less fuel-efficient because of the simpler
polluting and longer lifespan.
design, resulting in poorer mileage than a
four stroke engine.
8. Less noisy.
8. Twice as much noisy.
9. Less polluting.
9. Very much polluting.
10. Usually lasts longer.
10. Does not last very long.
Page 9 of 15
Engine Knocking:
Engine knocking (also called detonation) is a sudden blow on the piston just like a
hammering. Knocking occurs due to localized ignition inside the combustion chamber.
This can be explained thus: at the end of the compression stroke the sparkplug gives
electric spark to initiate
ignition of the air fuel charge.
Ignition takes place and very
quickly advances like a heat
wave to all corners of the
combustion
chamber.
Consequently,
localized
ignition starts before the flame
reaches it. Therefore, knocking
is a post ignition phenomenon.
Detonation or knocking is
harmful for the engine and
causes the engine-running
shaky.
Both
high
combustibility of fuel and the
Fig. 7: Engine Knocking
high compression ratio are responsible for knocking. To stop engine knocking generally a
special fuel or a chemical (tetraethyl lead) is mixed with gasoline. Mixing of a small
amount satisfactorily stops knocking. High Octane rating also prevents engine knocking.
Octane Number
The property that describes how well petrol will or will not self-ignite is called the octane
number of petrol or just octane. The higher the octane number of petrol, the less likely it
will self-ignite. Engines with low compression ratios can use petrol with lower octane
numbers, but high-compression engines must use high-octane petrol to avoid self-ignition
and knock.
Common octane numbers (anti-knock index) for petrol used in cars range from 87 to 95,
with higher values available for special high-performance and racing engines. A 93octane petrol is more knock resistant than an 89-octane petrol. Reciprocating SI aircraft
engines usually use low-lead fuels with octane numbers in the 85 to 100 range.
Cetane Number
In a compression ignition engine, self-ignition of the air-fuel mixture is a necessity. The
correct fuel must be chosen which will self-ignite at the precise proper time in the engine
cycle. It is therefore necessary to have knowledge and control of the ignition delay time
of the fuel. The property that quantifies this is called the cetane number. The larger the
cetane number, the shorter is the ID and the quicker the fuel will self-ignite in the
combustion chamber environment. A low cetane number means the fuel will have a long
ID. Normal cetane number range is about 40 to 60.
Page 10 of 15
Engine Subsystem:
The main engine subsystems are:
1.
2.
3.
4.
5.
Fuel System ---- carburetor, Electronic Fuel Injection (EFI), etc.
Lubrication System ---- splashed lubrication, pump-forced lubrication, etc.
Cooling System ---- air-cooling system, water-cooling system.
Ignition System ----Spark Plug Ignition (SI), Compresion Ignition (CI).
Starting System ---- Battery/Starting motor, manual cranking, compressed air
motor, etc.
Carburetor:
Carburetor is a vital component of the fuel system of a conventional petrol engine.
It composed of 4 main parts:(i) Air horn, (ii) Venturi, (iii) Fuel nozzle and (iv) Throttle
valve
Petrol pumped
from tank
Float bowl
Fig. 8: Engine Carburetor
The above figure shows how petrol is practically atomized while passing through the
carburetor venture. At the venture, the passage is the smallest, resulting in high velocity
of the air and high kinetic energy as well. Since total energy must be unchanged, the rise
in kinetic energy is balanced by the fall in pressure (vacuum) at the venture. Therefore,
the air flowing through the venture creates a sort of vacuum (lower than atmospheric
pressure) at the other end (downstream) of the narrow portion of the passage, which
causes suction (air pushes the petrol) of the liquid petrol from the float bowl.
Page 11 of 15
Fuel Injection System
A fuel injection system is used usually in diesel engines to inject diesel fuel at the end of
compression stroke at a very high pressure.
A fuel injection pump and its operation are shown below:
Fig. 9: Fuel injection pump and its control unit
The plunger is operated by a cam as shown. As the cam pushes, the plunger moves
upward against a heavy spring and the fuel is delivered towards injection nozzle. The fuel
then enters from the nozzle into the combustion chamber under high pressure. It will be
seen that there is a spiral groove on the body of the plunger to control the amount of fuel
injection. There is a hole (feed hole) on the body of the plunger connected to its central
hole on top. The plunger can be rotated right way or left way (clock wise or anti
clockwise) with the help of a rack and pinion mechanism fitted with the body of the
plunger. Thus the actual amount of fuel per stroke as desired by the operator can be
injected. The rack and pinion again is operated by the engine governor.
Page 12 of 15
Engine Performance
Efficiency:
Mechanical Efficiency = Brake Power (bhp)/ Engine Power (hp)
Thermal Efficiency = Engine Power (hp) / Fuel Power (hp)
Overall Efficiency = Mechancal x Thermal
Brake Specific Fuel Consumption (bsfc): Fuel consumed per unit power generation per
unit time. Unit kg/kW-hr.
Bsfc = brake power/ fuel consumed per unit time
An Engine is not 100% efficient (typical efficiency is about 30%).
Because it has some losses.
Heat balance is as follows:
Heat loss to cooling water
30%
Heat loss to exhaust gases
30%
Heat loss to lubricating oil
5%
Heat loss due to friction of mechanical components
5%
Useful brake power available from engine
Total Heat Input to the engine
30%
100%
Mean effective pressure
Generally, the mean effective pressure is the ratio of the net work done to the
displacement volume of the piston. It is a valuable measure of an engine's capacity to do
work that is independent of engine displacement.
It is defined as,
Here, pmep = mean effective pressure, Pa
T = torque, N-m
Vd = displacement volume, m3
nc = number of revolutions per cycle (for a 4-stroke engine nc = 2, for a 2-stroke
engine nc = 1 )
Page 13 of 15
This is useful for comparing engines of different displacements (a specific torque of sorts,
i.e. torque per unit displacement). Mean effective pressure is also useful for initial design
calculations; that is, given a torque, we can use standard mep values to estimate the
required engine displacement.
Brake Mean Effective Pressure (bmep) is, calculated by putting the measured
dynamometer torque into the above equation.
For spark-ignition engines : maximum values are in the range 8.5 to 10.5 bar (850 to
1050 kPa; 125 to 150 psi), at the engine speed where maximum torque is obtained. At
rated power, bmep values are typically 10 to 15% lower.
For four-stroke diesels: the maximum bmep is in the 7 to 9 bar range (700 to 900 kPa;
100 to 130 psi).
Problem-4: A four-stroke engine producing 160 N·m from 2 litres of displacement. What
will be it brake mean effective pressure (bmep) ?
Solution: bmep is given by,
So, bmep = (160 N·m) (4π)/(0.002 m³)
Here, T = 160 N-m
Vd = 2 liters = 2 x 10-3 m3
nc = 2 for a 4-stroke
engine
= 1,005,000 N/m2 =1005 kPa (10.05 bar).
Problem-5: If the same engine (i.e., four-stroke, 2 liters) as above produces 76 kW at
5400 rpm (90 Hz), Find its bmep.
Here, P = 76 x 103 W
ω=2πN/60=2π(5400)/60
=565.5 rad/s
We have,
Power, P = T ω
T=P/ ω=(76 x 103)/565.5=134.4 N.m
Where,
Now,
Vd = 2 liters = 2 x 10-3 m3
nc = 2 for a 4-stroke
engine
So, bmep = (134.4 N·m) (4π)/(0.002 m³)
= 844460 N/m2 =844.5 kPa (8.34 bar).
Page 14 of 15
Problem-6: A 4-Cylinder, 2-stroke IC engine has the following particulars: engine speed
= 3000 rpm, bore = 120 mm, crank radius = 60 mm, mechanical efficiency = 90% and the
engine develops 75 bhp. Calculate the swept volume and mean effective pressure (MEP).
Mechanical efficiency, η =
Or,
0.9 =
75
,
P
BrakePower (bhp )
EnginePower (ihp )
i.e.,
P = 83.33 hp
Now, Engine Power, P = T ω
Here, P = 83.33 hp = 83.33 x 746 W = 62166.67 W
ω=2πN/60=2π(3000)/60
=314.16 rad/s
T=P/ ω=(62166.67)/314.16=197.88 N.m
We get, Mean Effective Pressure (MEP or Pmep) as follows:
where,
Vd = N.(π/4).B2. S
Here Stroke, S = 2 x crank radius
= 2 x 0.06 m = 0.12 m
Vd = 4.(π/4).(0.12)2(0.12)
= 5.43 x 10-3 m3
= 5.43 liter
nc = 1 for a 2-stroke engine
Therefore,
MEP = (197.88 N·m) (2π)/(0.00543 m³)
= 228971.77 N/m2 =228.97 kPa
Page 15 of 15