Class Notes on Thermal Energy Conversion SystemClass Notes on Thermal Energy Conversion System 1 Chapter 1. Gas Power Cycles and Reversibility 1.1 Introduction In the study of thermodynamics,

Class Notes on Thermal

Conversion System

Ramesh Khanal

Assistant Professor

Nepal Engineering College

Bhaktapur, Nepal

For the students of Civil & Rural

3rd semester

Class Notes on Thermal Energy

Conversion System

Assistant Professor

Nepal Engineering College

the students of Civil & Rural

Energy

2015

Class Notes on Thermal Energy Conversion System

i

Course Structure

MEC 209.3: Thermal Energy Conversion System (3-1-2)

Theory Practical Total

Sessional 30 20 50

Final 50 - 50

Total 80 20 100

Course Objective

The objective of this course is to make the students familiar with air standard cycles,

principle and systems of internal combustion engines and fuels and their combustion

properties.

Course Contents:

1. Gas Power Cycles and Reversibility (6 hours)

Introduction; Air Standard Efficiency of a Cycle; Thermodynamic Reversibility;

Carnot’s Cycle; Otto Cycle and actual pV diagram for Otto Cycle; Diesel Cycle and

actual pV diagram for Diesel Cycle; Dual Combustion Cycle; Comparison of Otto,

Diesel and Dual Combustion Cycles; Brayton Cycle; Stirling Cycle.

2. Reciprocating Steam Engine: (5 hours)

The Rankine Cycle; Comparison of Rankine and Carnot’s Cycle; Rankine Cycle applied

to steam engine plant; Rankine Cycle on TS diagram; Steam Engine Indicators;

Hypothetical and Actual Indicator Diagram of Steam Engine.

3. Internal Combustion Engines (7 hours)

Introduction; Classification of Internal Combustion Engines; Working Cycles of

Internal Combustion Engines; Two Stroke Cycle Petrol and Diesel Engines; Four

Stroke Cycle Petrol and Diesel Engines; Basic Parameters of Internal Combustion

Engines; Components of Internal Combustion Engines.

4. Performance of Internal Combustion Engines (12 hours)

Indicated Power, Brake Power, Mean Effective Pressure; Engine Efficiency; Heat

Balance for Internal Combustion Engines; Power, Torque, Speed Relationship;

Specific Fuel Consumption; Gasoline and Gaseous Fuel Systems: carburation system,

temperature and altitude effects, gaseous fuel storage, air-mixing, relative efficiency;

Ignition Systems: spark ignition engines, conventional system- primary and

secondary circuits, electronic ignition, comparison and effects on engine

performance, compression ignition engines, fuel injector characteristics, engine

output and efficiency; Cooling Systems: liquid (water/anti-freeze) coolant, dry and

wet liners corrosion inhibitors, air cooling, temperature limitations, fan power

requirement, comparative advantage of liquid and air cooling system; Lubrication


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System: lubrication requirements of spark ignition and compression ignition engines,

low pressure splash lubrication system, high pressure gear pumps and distribution

system.

5. Reciprocating Air Compressor (8 hours)

Primary Components of a Reciprocating Air Compressor; Reciprocating Compression,

Clearance Volume Effects, p-V diagram and work done; Volumetric and Adiabatic

Efficiencies, Compression Process on T-s Diagram; Multi-stage compression,

Intercooling, Optimum pressure distribution Work done, Representation of p-V and

T-s diagram; Positive Displacement Compressor Types, Axial flow compressors,

Roots Blower, Rotary compressors.

6. Fuels and Combustion (7 hours)

Introduction; Classification of Fuels; Solid Fuels; Liquid Fuels; Gaseous Fuels;

Calorific and Heating Values of Fuels; Determination of Calorific Values for Solid,

Liquid and Gaseous Fuels; Combustion Equations for Hydrocarbon Fuels;

Combustion in Spark Ignition and Compression Ignition Engines; Pre-Ignition and

Ignition Delays; Detonation and Effects of Operating Variables on Detonation.

Total Lectures 45 hours

Laboratories:

Five laboratory Exercises to be performed in this course, as stated hereunder:

i. Studies on Parallel Flow and Counter Flow Heat Exchangers

ii. Performance Evaluation of Reciprocating Air-Compressor

iii. Study of Systems and Components of Internal Combustion Engines

iv. Performance Testing of Spark Ignition Engine: Ignition Timing, Fuel Combustion,

Losses, Mechanical Efficiency, Air-Fuel Ratio, Volumetric Efficiency, Compression

Ratio, Exhaust Emission, Energy Balance

v. Performance Testing of Compression Ignition Engine: BMEP, Injection Timing, Fuel

Consumption, Losses, Mechanical Efficiency, Air-Fuel Ratio, Volumetric Efficiency,

Compression Ratio, Exhaust Emission, Energy Balance

Tutorial:

Problem Solving and Assignments

Text Books and References:

i. J.B. Heywood. Internal Combustion Engine Fundamentals. McGraw Hill Book Co.

ii. C.R. Ferguson. Internal Combustion Engine. Wiley Publishers (latest edition).

iii. P.L. Ballaney. Thermal Engineering (Heat Engines). Khanna Publishers, New Delhi.


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Content Chapter 1. Gas Power Cycles and Reversibility ..................................................................................... 1

1.1 Introduction ................................................................................................................................... 1

1.2 Air standard efficiency of a cycle.................................................................................................. 1

1.3 Thermodynamic reversibility ....................................................................................................... 2

1.4 Carnot cycle.................................................................................................................................... 2

1.5 Brief introduction to petrol and diesel engine operation .......................................................... 4

1.6 Approximation of real cycle with air standard cycle .................................................................. 5

1.7 Otto cycle ........................................................................................................................................ 6

1.8 Diesel cycle ..................................................................................................................................... 8

1.9 Dual cycle ..................................................................................................................................... 11

1.10 Comparison of Otto, Diesel and Dual cycle ................................................................................ 13

1.11 Brayton cycle ............................................................................................................................... 17

1.12 Stirling Cycle ................................................................................................................................ 20

Chapter 2. Reciprocating Steam Engine ............................................................................................... 32

2.1 Rankine cycle ............................................................................................................................... 32

2.2 Mean temperature of heat addition ........................................................................................... 33

2.3 Comparison of Rankine and Carnot cycle .................................................................................. 34

2.4 Effect of pressure and temperature on Rankine cycle ............................................................. 35

2.5 Working principle of reciprocating steam engine .................................................................... 36

2.6 Steam engine indicators .............................................................................................................. 37

2.7 Hypothetical indicator diagram ................................................................................................. 37

2.8 Hypothetical and actual indicator diagram ............................................................................... 39

Chapter 3. Internal Combustion Engine ............................................................................................... 45

3.1 Introduction ................................................................................................................................. 45

3.2 Classification of IC engines ......................................................................................................... 46

3.3 Working cycle of IC engines ........................................................................................................ 46

3.4 Valve timing diagram of four stroke IC engine ......................................................................... 49

3.5 Components of IC engines .......................................................................................................... 51

3.6 Parts common to SI engine ......................................................................................................... 54

3.7 Parts common to CI engine ......................................................................................................... 55

3.8 Basic engine parameters ............................................................................................................. 55

Chapter 4. Performance of Internal Combustion Engine .................................................................... 57

4.1 Mean effective pressure (MEP) .................................................................................................. 57

4.2 Indicated power (IP) ................................................................................................................... 57


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4.3 Brake power (BP) ........................................................................................................................ 58

4.4 Engine efficiencies ....................................................................................................................... 59

4.5 Heat balance for IC engine .......................................................................................................... 60

4.6 Torque-power-speed relationship ............................................................................................. 61

4.7 Gasoline and gaseous fuel system .............................................................................................. 62

4.8 Ignition system ............................................................................................................................ 68

4.9 Engine cooling system ................................................................................................................. 71

4.10 Advantage of liquid cooling ........................................................................................................ 73

4.11 Disadvantage of liquid cooling ................................................................................................... 74

4.12 Advantage of air cooling ............................................................................................................. 74

4.13 Disadvantage of air cooling ........................................................................................................ 74

4.14 Engine lubricating system .......................................................................................................... 74

Chapter 5. Reciprocating air compressor ............................................................................................ 77

5.1 Reciprocating compression ........................................................................................................ 77

5.2 Work done in compressor .......................................................................................................... 78

5.3 Isothermal compression ............................................................................................................. 80

5.4 Effect of polytropic index on work required ............................................................................. 80

5.5 Compression process in T-s diagram ......................................................................................... 81

5.6 Clearance volume effects, P-v diagram and work done ........................................................... 82

5.7 Work done in single stage compressor with clearance volume .............................................. 83

5.8 Volumetric and adiabatic efficiencies ........................................................................................ 85

5.9 Effect of delivery pressure on the performance of air compressor......................................... 86

5.10 Multi-stage compression ............................................................................................................ 87

5.11 Representation of P-v and T-s diagram ..................................................................................... 88

5.12 Optimum pressure distribution work done .............................................................................. 89

5.13 Positive displacement compressor ............................................................................................ 91

Chapter 6. Fuels and Combustion ......................................................................................................... 99

6.1 Introduction ................................................................................................................................. 99

6.2 Classification of fuels ................................................................................................................... 99

6.3 Calorific and heating values of fuels .......................................................................................... 99

6.4 Determination of caloric values ................................................................................................. 99

6.5 Combustion equation for hydrocarbon fuels ............................................................................ 99

6.6 Combustion in SI and CI engine ................................................................................................ 103

6.7 Abnormal combustion in SI engine .......................................................................................... 104

6.8 Abnormal combustion in CI engine .......................................................................................... 105


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1

Chapter 1. Gas Power Cycles and Reversibility

1.1 Introduction

In the study of thermodynamics, a cycle is defined as a set of processes involving transfer of

heat and work. The cycle starts from one state and after completion of set of processes

return back to the initial state. The cycle is capable of continuing same process repeatedly. If

the purpose of the cycle is to receive heat from surrounding and transfer work to the

surrounding, it is called power cycle. We also know that some kind of medium is required

for a cycle to transfer heat and work. If a device that operates in thermodynamics power

cycle uses gas as working medium then it is called gas power cycle. We also know that

according to the second law of thermodynamics, heat can be converted into work and while

doing so some part of heat received from the source must be rejected to the surrounding.

The difference in heat received from source and heat rejected to surrounding is work done

by the cycle to the surrounding. Generally, four processes are required to complete a gas

power cycle and they are:

a. Compression of gas

b. Head addition from source

c. Expansion of gas

d. Heat rejection to the surrounding

1.2 Air standard efficiency of a cycle

We know that air is freely and abundantly available in our atmosphere. For this reason air is

commonly used as working medium in gas power cycle. Though the composition of air in

some power cycle may change during the cycle (fuel is burned in presence of air in petrol

engine and diesel engine), working medium is always treated as ideal gas. All power-

producing engines such as petrol engine, diesel engine, gas turbine; which use air as

working medium are approximated as air standard cycle for analysis. One of the parameters

used to measure the performance of a power cycle is thermal efficiency. Thermal efficiency

of power cycle working with air is also known as air standard efficiency. From the second

law of thermodynamics, we know that efficiency of a power cycle can be expressed as,

�� = �� ℎ�� Eq. 1.1

Let us consider that qin amount of heat is supplied to an air standard cycle to produce wout

amount of work and in the process qout amount heat is rejected.

wout = qin – qout Eq. 1.2

So, the equation of air standard efficiency of a cycle can be written as


2

�� = �� − ��

�� = 1 − �� Eq. 1.3

Air standard efficiency of a cycle is an ideal efficiency and no real cycle can have higher

efficiency than air standard efficiency.

1.3 Thermodynamic reversibility

Thermodynamics reversibility of a process is defined as the capability of the process to

return back to original state without changing the condition of either system or

surrounding.

Factors causing irreversibility in a process:

a. Friction

b. Free expansion

c. Heat transfer through a finite temperature difference

d. Mixing of two different substances

1.4 Carnot cycle

We know that the efficiency of a heat engine is always less than 100%. Then there should be

an engine that is most efficient among all others. For an engine to be most efficient, all the

processes forming the cycle must be reversible. The engine which works on cycle with

reversible compression, reversible heat addition, reversible expansion and reversible heat

rejection is the most efficient. For the heat transfer (either addition or rejection) to be

reversible it should be carried out across infinitesimally small temperature difference or at

constant temperature. Therefore, reversible heat transfer is possible in isothermal process.

For the expansion/compression process, it should be reversible adiabatic or isentropic

process. Therefore, the most efficient reversible cycle has the following four processes

a. Reversible isothermal heat addition from high temperature source

b. Reversible adiabatic (isentropic) expansion

c. Reversible isotheral heat rejection to low temperature sink

d. Reversible adiabatic compression

The cycle which operates based on above four processes is called Carnot cycle. Carnot cycle

is a heat engine cycle which operates between the given high-temperature and low-

temperature reservoirs such that every process is reversible. If every process is reversible,

the cycle is also reversible. This cycle was proposed by a French engineer Leonard Sadi

Carnot (1796-1832).


3

1.4.1 Assumptions for Carnot cycle

For the reversibility of the processes following assumptions are made

1. No friction is involved.

2. The transfer of heat does not affect the temperature of source or sink.

3. Compression and expansion are reversible.

4. Perfectly reversible adiabatic conditions are achieved.

Fig. 1-1 Cylinder arrangement for Carnot Cycle

Process 1-2

It is a reversible isothermal process in which heat is transferred from the high-temperature

reservoir at constant temperature (T1 or TH). The source at a high temperature is brought in

contact with the bottom of the cylinder so that air expands isothermally from V1 to V2.

For an isothermal process:

)ln()ln()ln( 1

1

21

1

21121 crmRT

V

VmRT

V

VVPQ ===−

Fig. 1-2 P-v and T-s diagram of Carnot Cycle

Process 2-3

It is a reversible adiabatic expansion process in which the temperature of the air decreases

from T1 to T2. The source is removed and the bottom of the cylinder is insulated so that the

air is allowed to expand adiabatically. The volume increases from V2 to V3.

032 =−Q


4

Process 3-4

It is a reversible isothermal process in which heat is rejected to the low temperature

reservoir at T2. The bottom of the cylinder is brought in contact with the sink at temperature

T2 so that air is compressed isothermally from V3 to V4.

)ln()ln()ln( 3

3

43

3

43343 crmRT

V

VmRT

V

VVPQ ===−

Process 4-1

It is a reversible adiabatic process in which temperature of working fluid is increased from

T1 to T2 by compressing it reversibly and adiabatically. The sink is removed and the bottom

of the cylinder is insulated so that the air is allowed to compress adiabatically so that the

volume decreases from V4 to V1.

014 =−Q

Now,

)ln()ln(

31

4321

cc rmRTrmRT

QQ

SuppliedHeatNetDoneWorkNet

−=

−=

=

−−

And, the efficiency of a Carnot cycle is:

1

3

1

31

1

31

1

)ln(

))(ln(

T

T

T

TT

rmRT

TTrmR

SuppliedHeat

doneWork

c

c

−=

−=

−=

=η

�� = 1 − �� Eq. 1.4

1.5 Brief introduction to petrol and diesel engine operation1

In petrol engine, mixture of air and petrol is enters into the cylinder through the inlet valve

in intake stroke. In compression stroke, inlet valve is closed and air fuel mixture is

compressed inside the cylinder. At the end of the compression, the fuel is burned with the

help of spark plug. Because of high pressure after the combustion of fuel, the burned gases

expand inside the cylinder. In exhaust stroke, exhaust valve opens and all the burned gases

escape outside the cylinder. In this way the cycle is repeated.

1 Detail description of operation principle of petrol and diesel engine is given in chapter 3


5

Fig. 1-3 Working mechanism of IC engine

In a diesel engine, only air is drawn inside the cylinder and fuel is burned by spraying diesel

over the compressed air. The method of compression and expansion and exhaust of diesel

engine are similar to that of petrol engine.

1.6 Approximation of real cycle with air standard cycle

The cycle that undergoes inside the cylinder of petrol and diesel engine is very complex. To

make the analysis of the engine cycle easier, real cycle is approximated with ideal air

standard cycle. Petrol engine cycle is approximated with air standard Otto cycle and diesel

engine cycle is approximated with air standard Diesel cycle. The difference and

approximation in ideal and real cycles are:

1. The gas mixture in the cylinder is treated as air for the entire cycle, and property values

of air are used in the analysis. In real cycle, fuel is mixed with air (in SI engine) in first

part of cycle. In second part of the cycle the fuel is burned in presence of air and the

composition of gas then after is of mixture of mostly N2, CO2 and H2O. As the fraction of

fuel is very less compared to air and mixture of N2, CO2 and H2O can be approximated as

air, approximation with air does not result large errors.

2. The real open cycle is changed into a closed cycle by assuming that the gases being

exhausted are fed back into the intake system. This works with ideal air standard cycles,

as both intake gases and exhaust gases are air. Closing the cycle simplifies the analysis.

3. The combustion process is replaced with a heat addition term Qin of equal energy value.

4. The open exhaust process, which carries a large amount of enthalpy out of the system, is

replaced with a closed system heat rejection process Qout of equal energy value.

5. The almost-constant-pressure intake and exhaust strokes are assumed to be constant

pressure.

6. Compression strokes and expansion strokes are approximated by isentropic processes.

To be truly isentropic would require these strokes to be reversible and adiabatic. There


6

is some friction between the piston and cylinder walls but, because the surfaces are

highly polished and lubricated, this friction is kept to a minimum and the processes are

close to frictionless and reversible.

7. The combustion process is idealized by a constant-volume process (petrol cycle), a

constant-pressure process (diesel cycle), or a combination of both (CI Dual cycle).

8. Exhaust blow down is approximated by a constant-volume process.

1.7 Otto cycle

Working cycle of petrol engine is approximated with Otto cycle, which is described by four

processes.

Process 1-2: isentropic compression: Fuel-air mixture is taken into the cylinder through

suction and the mixture inside the cylinder is compressed until the piston reaches the Top

Dead Center (TDC). This process is reversible and adiabatic. The pressure and temperature

of the air increase.

Process 2-3: constant volume heat addition: The compressed air-fuel mixture is burned

with a spark which makes the pressure and temperature of the combustion gases to rise.

This process is assumed to occur at constant volume.

Process 3-4: isentropic expansion: The piston begins to move until it reaches Bottom

Dead Center (BDC) so the expansion of gas occurs adiabatically and reversibly. This process

generates work output and the pressure and temperature of air decreases consequently.

Process 4-1: constant volume heat rejection: The exhaust valve opens and the pressure

and temperature of gas decreases at constant volume. Then the gas is removed from the

cylinder by movement of piston and the cycle is completed.

Above processes can be shown in P-v and T-s diagram as in Fig. 1-4.

Fig. 1-4 P-v and T-s diagram of Otto Cycle


7

The thermal efficiency of the cycle is given as:

−

−

−=

−

−−=

−=

−=

1

1

1

)(

)(1

1

2

3

2

1

41

23

14

T

TT

T

TT

TTmC

TTmC

Q

Q

Q

QQ

V

V

H

L

H

LHη

Since process 1-2 and 3-4 are isentropic, we can write,

( ) 1

1

2

1

1

2 −

−

=

=

γ

γ

rV

V

T

T

Where,

=

2

1

V

Vr is called compression ratio

And,

1

2

1

2

1

1

3

4

4

3

T

T

V

V

V

V

T

T=

=

=

−− γγ

So,

( ) 1

2

1 111

−−=−=

γη

rT

T

�� = 1 − 1!�"#$% Eq. 1.5

Mean Effective Pressure (mep) is defined as the pressure that, if it acted on the piston

during the entire power stroke, would do an amount of work equal to that actually done on

the piston.

i.e. &'( = )�*�+% − +, Eq. 1.6


8

The Fig. 1-4 represents the idealized or theoretical P-v diagram of petrol engine but the

actual P-v diagram of petrol engine closely resembles to the diagram shown in Fig. 1-5.

Fig. 1-5 Actual P-v diagram for Petrol Engine

1.8 Diesel cycle

Working cycle of diesel engine can be theoretically approximated with diesel cycle which is

completed by four processes.

Process 1-2: isentropic compression: Air is taken into the cylinder through suction and

air inside the cylinder is compressed until the piston reaches the Top Dead Center (TDC).

This process is reversible and adiabatic. The pressure and temperature of the air increase.

Fig. 1-6 P-v and T-s diagram of Diesel Cycle

Process 2-3: constant pressure heat addition: Fuel is injected in the cylinder so that

compressed air and fuel is burned spontaneously which makes the pressure and


9

temperature of the combustion gases to rise. Simultaneously, the piston moves maintaining

the pressure inside the cylinder. This process is assumed to occur at constant pressure.

Process 3-4 isentropic expansion: The piston begins to move until it reaches Bottom Dead

Center (BDC) so the expansion of gas occurs adiabatically and reversibly. This process


Process 4-1 constant volume heat rejection: The exhaust valve opens and the pressure



All the four processes can be shown in P-v and T-s diagram as in Fig. 1-6

The thermal efficiency of diesel cycle is given by,

HQ

W=η

H

LH

Q

QQ −=

H

L

Q

Q−=1

)(

)(1

23

14

TTmC

TTmC

P

V

−

−−=

� = 1 − !�- − �%".!�/ − �," Eq. 1.7

For a Diesel cycle,

Compression Ratio,

=

2

1

V

Vr

Cut-off Ratio,

=

2

3

V

Vρ

i.e. �0 = +%+, × +,+/ = +%+/ = +-+/ Eq. 1.8

For isentropic process 1-2 as shown in Fig. 1-6

�,+,#$% = �%+%#$%

�, = �% 2+%+,3#$% = �%!�"#$% Eq. 1.9

Similarly for isobaric process 2-3


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+,�, = +/�/

�/ = �, +/+, = �,0

Substituting the value of T2 from Eq. 1.9

�/ = �%0�#$% Eq. 1.10

Process 3-4 is isentropic expansion,

�-+-#$% = �/+/#$%

�- = �/ 2+/+-3#$%

From Eq. 1.8 and Eq. 1.10

�- = �%0�#$% 40�5#$%

�- = �%0# Eq. 1.11

Now, from Eq. 1.7, Eq. 1.9, Eq. 1.10 and Eq. 1.11

� = 1 − !�%0# − �%".!�%0�#$% − �%�#$%"

�6�*7*8 = 1 − 1�#$% 9!0# − 1".!0 − 1": Eq. 1.12

The Fig. 1-6 represents the idealized or theoretical P-v diagram of petrol engine but the

actual P-v diagram of petrol engine closely resembles to the diagram shown in Fig. 1-7.

Fig. 1-7 Actual P-v diagram for Diesel Engine


11

1.9 Dual cycle

The dual cycle (limited pressure cycle) is the ideal cycle for modern Compression Ignition

(CI) engines. The dual cycle differs from Otto cycle and Diesel cycle, in that heat is supplied

at both constant volume and constant pressure. There are five processes involved in dual

cycle which are explained with the help of P-v and T-s diagram as shown.

Process 1-2: isentropic compression: Air is taken into the cylinder through suction and

air inside the cylinder is compressed until the piston reaches the Top Dead Center (TDC).

This process is reversible and adiabatic. The pressure and temperature of the air increase.

Process 2-3: constant volume heat addition: Fuel is injected in the cylinder before the

piston reaches the TDC during compression stroke so some of the combustion occurs at

almost constant volume at TDC as in Otto cycle.

Process 3-4: constant pressure heat addition: The fuel is continued to be injected at TDC

so the combustion continues and the piston moves maintaining the pressure inside the

cylinder. This process is assumed to occur at constant pressure.

Process 4-5: isentropic expansion: The piston begins to move until it reaches Bottom

Dead Center (BDC) so the expansion of gas occurs adiabatically and reversibly. This process


Process 5-1: constant volume heat rejection: The exhaust valve opens and the pressure



All the processes involved in theoretical dual cycle can be represented in P-v and T-s

diagram as shown in Fig. 1-8.

Fig. 1-8 P-v and T-s diagram of Dual Cycle

Some parameters used in dual cycle based on the P-v and T-s diagram

Heat added at constant volume during process 2-3 (Qin1) = mcv(T3 – T2)


12

Heat added at constant pressure (during process 3-4) (Qin2) = mcp(T4 – T3)

Total heat added per cycle (Qin)= mcv(T3 – T2) + mcp(T4 – T3)

Heat rejected at constant volume during process 5-1 (Qout) = mcv(T5 – T1)

Compression ratio (rc) = v1/v2

Cut-off ratio (ρ) = v4/v3

Pressure ratio (rp) = P3/P2

Now air standard thermal efficiency of the dual cycle is

� = �� ;�ℎ�� ; = <�� − <��<�� = 1 − <��<��

� = 1 − =>?!�@ − �%"=>?!�/ − �," + =>B!�- − �/"

� = 1 − !�@ − �%"!�/ − �," + .!�- − �/" Eq. 1.13

Process 1-2: Isentropic compression of air

�%C%#$% = �,C,#$%

�, = �% 2C,C%3#$% = �%�D#$%

�, = �%�D#$% Eq. 1.14

Process 2-3: Constant volume heat addition (,�, = (/�/

�/ = �, (/(, = �,�B

�/ = �%�D#$%�B Eq. 1.15

Process 3-4: Constant pressure heat addition C/�/ = C-�-

�- = �/ C-C/ = �/0

�- = �%�D#$%�B0

�- = �%�D#$%�B0 Eq. 1.16

Process 4-5: Isentropic expansion of air


13

�-C-#$% = �@C@#$%

�@ = �- 2C-C@3#$%

9∵ C-C@ = C-C% = C,C% × C-C, = C,C% × C-C/ = 0�D:

�@ = �%�D#$%�B0 20�D3#$%

�@ = �%�B0# Eq. 1.17

Now from Eq. 1.13, Eq. 1.14, Eq. 1.15, Eq. 1.16 and Eq. 1.17

� = 1 − F�%�B0# − �%GF�%�D#$%�B − �%�D#$%G + .F�%�D#$%�B0 − �%�D#$%�BG

�6��8 = 1 − F�B0# − 1G�D#$%HF�B − 1G + �B.!0 − 1"I Eq. 1.18

The actual P-v diagram of modern diesel engine something like as shown in Fig. 1-9.

Fig. 1-9 Actual P-v diagram of modern Diesel engine

1.10 Comparison of Otto, Diesel and Dual cycle

The important factors, which are used as the basis for comparison of different cycles are

compression ratio, peak pressure, heat addition, heat rejection and the net work. In order to

compare the performance of the Otto, Diesel and Dual combustion cycles, some of the

variable factors must be fixed. In this section, a comparison of these three cycles is made for

the same compression ratio, same heat addition, constant maximum pressure and

temperature, same heat rejection and net work output. This analysis will show which cycle

is more efficient for a given set of operating conditions. We know that efficiency of any heat

engine can be expressed as


14

� = 1 − <��<�� Eq. 1.19

1.10.1 Same compression ratio and heat addition

Fig. 1-10 Same compression ratio and heat addition

The P-v and T-s diagram of Otto cycle 1-2-3-4-1, the Diesel cycle 1-2-3”-4”-1 and the dual

cycle 1-2-2a-3’-4’-1 are shown in the Fig. 1-10

In P-v diagram, v1 and v2 is same for all cycles which means the compression ratio is same for

all. The T-s diagram is plotted in such a way that area under the curve 2-3 (for Otto cycle),

curve 2-2a-3’ (for dual cycle) and 2-3” (for diesel cycle) is same. We also know that area

under the curve in T-s diagram indicates the heat transfer. So heat addition (qin) is same for

all cycles. Again from T-s diagram, it is clear that,

Area under curve 4-1 < Area under the curve 4’-1 < Area under the curve 4”-1

i.e. (qout)otto < (qout)dual < (qout)diesel

So with the help of Eq. 1.19, it can be concluded that

ηotto > ηdual > ηdiesel for same compression ratio and same heat addition.

1.10.2 Same compression ratio and heat rejection

Fig. 1-11 Same compression ratio and heat rejection

1-2-3-4-1 Otto cycle

1-2-2a-3’-4’-1 dual cycle

1-2-3”-4”-1 Diesel cycle


1-2-2a-3’-4-1 dual cycle

1-2-3”-4-1 Diesel cycle


15

As in the previous case, compression ratio is same for all three cycles (Fig. 1-11). The cycle

in T-s diagram is drawn in such a way that heat rejection (area under the cure 4-1) is same

for the all cycles. From the T-s diagram it is also clear that

(qin)otto > (qin)dual > (qin)diesel

(area under the curve 2-3 for Otto cycle, area under the curve 2-2a-3’ for dual cycle and area

under the curve 2-3” for Diesel cycle in T-s diagram)

So for same compression ratio and heat rejection.

ηotto > ηdual > ηdiesel

1.10.3 Same peak temperature, peak pressure and heat rejection

Fig. 1-12 Same peak temperature and peak pressure

From the Fig. 1-12, it is clear that peak pressure (P3) is same for each cycle. Similarly same

heat rejection (area under curve 4-1) is also same for all cycles.

But,

(qin)diesel > (qin)dual > (qin)otto

(Area under the cure 2”-3 for diesel, area under the curve 2’-2a-3 for dual and area under

the curve 2-3 for otto in T-s diagram)

So,

ηdiesel > ηdual > ηotto


1-2’-2a-3-4-1 dual cycle

1-2”-3-4-1 Diesel cycle


16

1.10.4 Same peak pressure and heat addition

Fig. 1-13 For same peak pressure and heat addition

For the same peak pressure (P3) and same heat addition (area under the curve 2”-3” for

diesel, 2’-2a-3’ for dual and 2-3 for otto)

(qout)diesel < (qout)dual < (qout)otto

(area under the curve 4”-1 for diesel, 4’-1 for dual and 4-1 for otto in T-s diagram)

So, for same peak pressure and heat addition


1.10.5 Same maximum pressure and work output

Fig. 1-14 Same maximum pressure and work output

We know that,

� = �� = �� + �� Eq. 1.20

For same maximum pressure (P3) and same work output (same area of the cycle 1-2-3-4-1

Otto, 1-2’-2a-3’-4’-1 dual, 1-2”-3”-4”-1 Diesel in either of T-s and P-v diagram)

(qout)diesel < (qout)dual < (qout)otto


1-2’-2a-3’-4’-1 dual cycle

1-2”-3”-4”-1 Diesel cycle


1-2’-2a-3’-4’-1 dual cycle

1-2”-3”-4”-1 Diesel cycle


17

(area under the curve 4”-1 for diesel, 4’-1 for dual and 4-1 for otto in T-s diagram)

So, from the Eq. 1.20, for same peak pressure and work output


1.11 Brayton cycle

The Brayton Cycle, also called the Joule Cycle, was developed originally for use in a piston

engine with fuel injection. This cycle is the ideal cycle for the simple gas turbine. The air

standard Brayton cycle is composed of constant pressure heat transfer processes separated

by isentropic expansion and compression processes.

a. Closed Brayton Cycle b. Open Brayton Cycle

Fig. 1-15 Closed and open Brayton Cycle

The closed-cycle Brayton engine is shown in Fig. 1-15a. The working fluid air enters the

compressor in state 1, where it is compressed isentropically until state 2 is reached and

enters high temperature heat exchanger. In the heat exchanger heat will be added to the

fluid at constant pressure until state 3 is reached. Now high temperature air enters the

turbine, where an isentropic expansion occurs, producing mechanical work. The working

fluid (air) leaves the turbine at state 4 and enters low temperature heat exchanger, where

heat will be rejected from the fluid until state 1 is reached. After completing a cycle previous

processes will be repeated in same order.

Compression in compressor and expansion in turbine are assumed to be isentropic

processes in the ideal closed cycle air standard Brayton cycle. It is easier to construct

compressor and turbine which operates nearly adiabatic. However, it is difficult to approach

reversibility.

The P-v and T-s diagram of closed Brayton cycle can be represented as in Fig. 1-16


18

Fig. 1-16 P-v and T-s diagram of Brayton Cycle

The efficiency of Brayton cycle is given by,

)(

)(1

1

23

14

TTmC

TTmC

Q

Q

P

P

H

L

B

−

−−=

−=η

)1(

)1(

1

2

3

2

1

4

1

−

−

−=

T

TT

T

TT

Bη

Here, process 2-3 and 4-1 are constant pressure process so P2 = P3 and P1 = P4

For isentropic process 1-2,

γ

γ 1

1

2

1

2

−

=

P

P

T

T

For isentropic process 3-4,

So,

2

3

1

4

T

T

T

T= Eq. 1.21

Now, using Eq. 1.21 the equation of efficiency can be written as

γ

γη

1

1

2

3

4

2

1 1111

−

−=−=−=

P

PT

T

T

TB


19

( ) γ

γη

1

11

−−=

p

B

r

Eq. 1.22

Where, 4

3

1

2

P

P

P

Prp == is the pressure ratio

1.11.1 Effect of isentropic efficiency of pump and turbine

In the previous section, we have assumed the turbine and pump are isentropic. But, in real

case, they cannot be isentropic. If isentropic efficiency of turbine and pump are considered,

the P-v and T-s diagram of Brayton cycle will be as shown in the Fig. 1-17. In the figure, the

cycle 1-2-3-4-1 is the ideal cycle and the cycle 1-2’-3-4’-1 is for non-isentropic turbine and

pump.

Fig. 1-17 Isentropic efficiency of turbine and pump

The head addition in the cycle occurs at process 2’-3 while heat rejection occurs at 4’-1

process.

The turbine efficiency can be defined as

�J = !)J"/$-K!)J"/$- = ℎ/ − ℎ-Kℎ/ − ℎ-

�J = �/ − �-L�/ − �- Eq. 1.23

Similarly the pump efficiency can be defined as

�M = !)M"%$,!)M"%$,L = ℎ, − ℎ%ℎ,L − ℎ%

�M = �, − �%�,L − �% Eq. 1.24

The equation for efficiency of the cycle can be written as

� = 1 − ��


20

� = 1 − NB!�-L − �%"NB!�/ − �,L"

� = 1 − !�-L − �%"!�/ − �,L" Eq. 1.25

1.12 Stirling Cycle

Ideal Stirling cycle consists of two isothermal and two constant volume processes as follow

Fig. 1-18 P-v and T-s diagram of Stirling cycle

Stirling Cycle Processes:

a) The air is compressed isothermally from state 1 to 2.

b) The air at state-2 is passed into the regenerator from the top at a temperature T1. The air

passing through the regenerator matrix gets heated from TL to TH.

c) The air at state-3 expands isothermally in the cylinder until it reaches state-4.

d) The air coming out of the engine at temperature TH (condition 4) enters into regenerator

from the bottom and gets cooled while passing through the regenerator matrix at

constant volume and it comes out at a temperature TL, at condition 1 and the cycle is

repeated.

e) It can be shown that the heat absorbed by the air from the regenerator matrix during the

process 2-3 is equal to the heat given by the air to the regenerator matrix during the

process 4-1, then the exchange of heat with external source will be only during the

isothermal processes.

So we can write net work done as

)�*� = <�� − < − <��

For isothermal heat addition during process 3-4

<�� = =O� ln!+-+/"

For isothermal heat rejection during process 1-2


21

<�� = =O�� ln 2+,+%3

So thermal efficiency of the cycle is

�� = )�*�<�� = 1 − <��<��

�� = 1 − =O�� ln 4+,+%5=O� ln 4+-+/5

�� = 1 − �� Eq. 1.26


22

Solved Numerical Problems

1. In a Carnot cycle, the maximum pressure and temperature are limited to 18 bar and

410°C. The ratio of isentropic compression is 6 and isothermal expansion is 1.5.

Assuming the volume of the air at the beginning of isothermal expansion as 0.18 m3,

determine:

a. The temperature and pressure at main points in the cycle

b. Change in entropy during isothermal expansion

c. Mean thermal efficiency of the cycle

d. Mean effective pressure of the cycle

e. Theoretical power if there are 210 working cycles per minute.

Maximum pressure (P2) = 18 bar

Maximum temperature (T2) = 410+273 = 683 K

V1/V2 = 6

V3/V2 = 1.5

V2 = 0.18 m3

a. Temperature and pressure at main points

Process 1-2 is isentropic

(%+%# = (,+,#

(% = (, 2+,+%3# = 18 × 10@ × 2163%.-

VW = X. WYZ [V\

�%+%#$% = �,+,#$%

�% = �, 2+,+%3#$% = 683 × 2163%.-$%

^W = ___. `` a

Process 2-3 is isothermal

(,+, = (/+/

(/ = (, 2+,+/3 = 18 × 10@ × 2 11.53 = 1.2 &(�

V_ = W. d [V\

^_ = ^d = Ze_ a

Process 3-4 is isentropic

(/+/# = (-+-# �/+/#$% = �-+-#$%

^Y = ^W = ___. ` a


23

(- = (/ 2+/+-3# = (/ 2+/+-3# = (/ 2+,+%3#

= 1.2 × 10f × 2163%.-

P4 = 97.67 kPa

b. Change in entropy during isothermal expansion

For mass of the working substance

= = (,+,O�, = 18 × 10@ × 0.18287 × 683 = 1.65 �h

∵ !i/ − i," = =O ln 2+/+,3 = 1.65 × 287 × ln 1.5 = X. Wjd kl/a

c. Mean thermal efficiency of the cycle

�D�� = 1 − �� = 1 − 333.5683 = `W. Wn%

d. Mean effective pressure of the cycle

=�� = )+- − +, = �D��<��+- − +, = �D�� × =O�, ln!+//+,"=O!�-/(- − �,/(,"

=�� = �D�� × �, ln!+//+,"!�-/(- − �,/(,"

= 0.5117 × 683 ln 1.54 333.597.67 × 10/ − 68318 × 10@5 = YZ. Zj kV\

e. Theoretical power if there are 210 working cycles per minute

) = �D��<�� = �D�� × =O�, ln!+//+,"

= 0.5117 × 1.65 × 287 × 683 × ln 1.5 = 67.10 �q

(�� = )��>r>�� × >r>��>�;� = 67.10 × 10/ × 21060

= d_Y. e` ks


24

2. A reversible engine converts one-sixth of the heat input into work. When the

temperature of the sink is reduced by 70°C, its efficiency is doubled. Find the

temperature of the source and the sink. )<�� = 16

��, = ��% − 70

� = )<�� = 16

We also know that

� = 1 − ��%� = 16

��%� = 56 ⇒ ��% = 56 �

��% = 56 �

Again,

2� = 13 = 1 − ��,�

��,� = ��% − 70� = 23 ⇒ ��% = 23 � + 70

��% = 23 � + 70

Solving for TL1 and TH

TH = 420K

TL1 = 350 K

3. An engine working on Otto cycle has a volume of 0.45 m3, pressure 1 bar and

temperature 30°C at the beginning of compression stroke. At the end of compression

stroke, the pressure is 11 bar. 210 kJ of heat is added at constant volume. Determine:

a. Pressure, temperature and volume at salient points in the cycle

b. Percentage clearance

c. Efficiency

d. Net work per cycle

e. Mean effective pressure

f. Ideal power developed by the engine if the number of working per minute is 210.


25

P1 = 1 bar,

V1 = 0.45 m3

T1 = 30+273 = 303 K

P2 = 11 bar

Qin = 210 kJ

a. Pressure, temperature and volume at salient points

For mass of the air

= = (%+%O�% = 1 × 10@ × 0.45287 × 303 = 0.517 �h

For isentropic compression process 1-2

(%+%# = (,+,# � = +%+, = 2(,(%3# = 2111 3 %%.- = 5.54 +, = +%� = 0.455.54 = X. XeW v_

�, = �% 2+%+,3#$% = 303 × 5.54%.-$% = ZXX. jZW a

For constant volume heat addition process 2-3

<�� = =N?!�/ − �,"

210 × 10/ = 0.517 × 718 × !�/ − 600.961"

^_ = WWZZ. Ze` a

(/�/ = (,�, ⇒ (/ = (,�/�, = 11 × 10@ × 1166.685600.961

(/ = 21.355 × 10@ = dW. _`` w\x

+/ = +, = 0.081 =/

For isentropic expansion process 3-4

(- = (/ 2+/+-3# = 21.355 × 10@ × 2 15.543%.- = 1.94 × 10@ = W. jY w\x

�- = �/ 2+/+-3#$% = 1166.685 × 2 15.543%.-$% = èe. d__ a

+- = +% = X. Y` v_


26

b. Percentage clearance

��>�;��h� >��;>� = +,+% − +, = 0.0810.45 − 0.081 = dd%

c. Efficiency

� = 1 − 1!�"#$% = 1 − 1!5.51"y.- = Yj. Yn%

d. Net work per cycle

) = � × <�� = 0.4947 × 210 × 10/ = 103.887 �q

e. Mean effective pressure

(z = )+% − +, = 103.887 × 10/0.45 − 0.081 = 2.815 × 10@ = d. eW` w\x

f. Power if revolution per minute is 210

(�� = )��>r>�� × ��C��; �� >�;� = 103.887 × 10/ × 21060 = _Z_. ZX ks

4. An engine with 200 mm cylinder diameter and 300 mm stroke works on theoretical

Diesel cycle. The initial pressure and temperature of air used are 1 bar and 27°C. The cut

off is 8% of the stroke. Determine if compression ratio is 15 and working fluid is air

a. Pressure and temperature at all salient points

b. Theoretical air standard efficiency

c. Mean effective pressure

d. Power of the engine if the working cycles per minutes are 380

Diameter (D) = 200 mm,

Stroke (L) = 300 mm

P1 = 1 bar, T1 = 300 K

Cutoff = 8% of stroke,

Compression ratio (r)= 15

i�� C��=� !+7" = {|,}4 = { × 0.2, × 0.34 = 0.00942 =/

N��~~ = 8% �~ ��

+/ − +, = 8%!+% − +," +/+, − 1 = 0.08 × 2+%+, − 13


27

0 − 1 = 0.08 × !� − 1"

0 − 1 = 0.08 × !15 − 1"

0 = 2.12

a. Pressure and temperature at all silent points

For isentropic compression 1-2

(, = (% 2+%+,3# = 1 × 10@ × 15%.- = 44.312 × 10@ = YY. _Wd w\x

�, = �% 2+%+,3#$% = 300 × !15"%.-$% = eeZ. d`_ a

For constant pressure heat addition 2-3 +,�, = +/�/ ⇒ �/ = �,+/+,

�/ = �,0 = 886.253 × 2.12 = Wene. e` a

(/ = (, = YY. _Wd w\x


(- = (/ 2+/+-3# = (/ 40�5# = 44.312 × 10@ × 22.1215 3%.- = d. `Z w\x

�- = �/ 2+/+-3#$% = 1878.85 × 22.1215 3y.- = eè. jj a

b. Theoretical efficiency

� = 1 − 0# − 1�#$%.!0 − 1" = 1 − 2.12%.- − 115y.- × 1.4 × !2.12 − 1" = `j. nn%


+7 = +% − +, = 0.00942 =/ +%+, = 15

Solving for V1 and V2

V1 = 0.0101 m3, V2 = 0.000673 m3

= = (%+%O�% = 1 × 10@ × 0.0101287 × 300 = 0.0117 �h

<�� = =NB!�/ − �," = 0.0117 × 1005 × !1878.85 − 886.253" = 11.671 �q

) = �<�� = 0.5977 × 11.671 = 6.975 �q


28

(z = )+7 = 6.975 × 10/0.00942 = 7.404 × 10@ = n. YXY w\x

d. Power at 380 rpm

(�� = )��>r>�� × >r>��>�;� = 6.975 × 10/ × 38060 = YY. Wn` ks

5. In an engine working on Dual cycle, the temperature and pressure at the beginning of the

cycle are 90°C and 1 bar respectively. The compression ratio is 9. The maximum

pressure is limited to 68 bar and total heat supplied per kg of air is 1750 kJ. Determine:


b. Air standard efficiency


T1 = 90+273 = 363 K

P1 = 1 bar

Compression ratio (rc) = 9

P3 = 68 bar

qin,total = 1750 kJ/kg


For the isentropic compression process 1-2

(, = (% 2+%+,3# = 1 × 10@ × 9%.- = 21.67 × 10@ = dW. Zn w\x �, = �% 2+%+,3#$% = 363 × 9y.- = enY. We a

For constant volume heat addition process 2-3 (,�, = (/�/

�/ = (/�,(, = 68 × 10@ × 874.1821.67 × 10@ = dnY_. Wn a

Heat supplied at constant volume

�,$/ = N?!�/ − �," = 718 × !2743.17 − 874.18" = 1341.93 �q/�h

For constant pressure heat addition process 3-4

Heat added at constant pressure

�/$- = ��8 − �,$/ = 1750 − 1341.93 = 408.07 �q/�h


29

Also,

�/$- = NB!�- − �/"

�- = �/$-NB + �/ = 408.069 × 10/1005 + 2743.17 = _WYj. dW a

(- = (/ = Ze w\x

>��~~ �� !0" = +-+/ = �-�/ = 3149.212743.17 = 1.15


(@ = (- 2+-+@3# = (- 20�D3# = 68 × 10@ × 21.159 3%.- = 3.81 × 10@ = _. eW w\x

�@ = �- 2+-+@3#$% = 3149.21 × 21.159 3y.- = W_ed. ee a

b. Air standard efficiency

Heat rejection at constant volume

�� = N?!�@ − �%" = 718 × !1382.88 − 363" = 732.27 �q/�h

� = 1 − �� = 1 − 732.271750 = è. W`%


(z = �� ;�� C��=� = �� − ��C% − C, = �� − ��O 4�%(% − �,(,5

(z = 1750 × 10/ − 732.27 × 10/287 × 4 3631 × 10@ − 874.1821.67 × 10@5 = 10.99 × 10@ = WX. jj w\x

6. Calculate the efficiency and specific work output of a simple gas turbine working on the

Brayton cycle. The maximum and minimum temperatures of the cycle are 1000 K and

288 K respectively, the pressure ratio is 6, and the isentropic efficiencies of the

compressor and turbine are 85 and 90 percent respectively.

T3 = 1000 K

T1 = 288 K

rp = 6

ηT = 90%

ηC = 85%

a. Efficiency of the cycle


30

For isentropic compression process 1-2

�%(%%$## = �,(,%$## �, = �% 2(%(,3%$## = 288 × 2163$y.-%.- = 480.53 �

�- = �/ 2(/(-3%$## = 1000 × !6"$y.-%.- = 599.34 � For the turbine efficiency

�J = �/ − �-L�/ − �-

0.9 = 1000 − �-L1000 − 599.34 ⇒ �-L = 639.406 �

For compressor efficiency

�� = �, − �%�,L − �%

0.85 = 480.53 − 288�,L − 288 ⇒ �,L = 514.51 �

The cycle efficiency

� = 1 − !�-L − �%"!�/ − �,L" = 1 − 639.406 − 2881000 − 514.51 = dn. Zd%

b. Specific work output

�� = NB!�/ − �,L" = 1005 × !1000 − 514.51" = 487.917 �q

�� = �� = 0.2762 × 487.917 = W_Y. nZ kl

1. A Carnot engine working between 400°C and 40°C produces 30 kJ of work. Determine:

a. The engine thermal efficiency

b. The heat added

c. The entropy changes during heat rejection process

2. An inventor claims that a new heat cycle will develop 0.4 kW for a heat addition of 32.5

kJ/min. Temperature of heat source is 1990 K and that of sink is 850 K. Is his claim

possible?


31

3. An engine of 250 mm bore and 375 mm stroke works on Otto cycle. The clearance

volume is 0.00263 m3. The initial pressure and temperature are 1 bar and 50°C. If the

maximum pressure is limited to 25 bar, find the following:

a. The air standard efficiency of the cycle [56.5%]

b. The mean effective pressure [1.334 bar]

4. In a constant volume Otto cycle, the pressure at the end of compression is 15 times that

at the start, the temperature of air at the beginning of compression is 38° C and

maximum temperature attained in the cycle is 1950°C. Determine:

a. Compression ratio [6.9]

b. Thermal efficiency [53.8%]

c. Work done [597.5 kJ]

5. A diesel engine has a compression ratio of 15 and heat addition at constant pressure

takes place at 6% of stroke. Find the air standard efficiency of the engine. [61.2%]

6. The stroke and cylinder diameter of CI engine are 250 mm and 150 mm respectively. If

the clearance volume is 0.0004 m3 and fuel injection takes place at constant pressure for

5% of the stroke determine the efficiency of the engine. [59.3%]

7. Calculate the percentage loss in the ideal efficiency of a diesel engine with compression

ratio 14 if the cut off is delayed form 5% to 8%. [2.1%]

8. The mean effective pressure of a Diesel cycle is 7.5 bar and compression ratio is 12.5.

Find the percentage cut off of the cycle if its initial pressure is 1 bar.

9. An oil engine working on the dual combustion cycle has a compression ratio 14 and the

explosion ratio obtained form an indicator card is 1.4. If the cut off occurs at 6% of

stroke, find the ideal efficiency. Take γ for air = 1.4. [61.4%]

10. The compression ratio for a single cylinder engine operating on dual cycle is 9. The

maximum pressure in the cylinder is limited to 60 bar. The pressure and temperature of

the air at the beginning of the cycle are 1 bar and 30°C. Heat added during constant

pressure process upto 4% of the stroke. Assuming the cylinder diameter and stroke

length as 250 mm and 300 respectively, determine:

a. the air standard efficiency [57.56%]

b. the power developed if the number of working cycles are 3 per second [51 kW]


32

Chapter 2. Reciprocating Steam Engine

2.1 Rankine cycle

Rankine Cycle is the theoretical cycle on which steam power plant (such as nuclear power

plant Fig. 2-1) works. In Rankine cycle has water as working fluid which is used to handle the

phase change between liquid and vapor.

Fig. 2-1 Rankine cycle when used in steam power plant

The processes involved in a Rankine cycle are:

Process 1-2: Reversible adiabatic compression in pump

Process 2-3: Constant pressure transfer of heat in the boiler/steam generator

Process 3-4: Reversible adiabatic expansion in the turbine

Process 4-1: Constant pressure transfer of heat in the condenser

In an ideal Rankine Cycle as shown in Fig. 2-2, the process 1-2, a pump is used to increase the

pressure of the working fluid. The working fluid enters the pump as a saturated liquid (State

1, x1 = 0) and exits the pump as a sub-cooled liquid (state 2). The fluid entered into boiler

has relatively low temperature is heated at constant pressure and leaves the boiler as a

saturated vapor (state 3, x3 = 1). This saturated vapor is expanded isentropically through

turbine to produce the work and leave as state 4. The low pressure saturated mixture is

again condensed at constant pressure in the condenser leaving as low pressure low

temperature saturated liquid (state 1). Following condensation, the liquid enters the pump.

The working fluid is returned to the high pressure for heat addition at the higher boiler

temperature, and the cycle is repeated. If we apply SFEE for four components of the Rankine

cycle, we have

For Boiler: Heat addition qin = h3 – h2

For Pump: Work Absorbed wP = h2 – h1

For Turbine: work produced wT = h3 – h4


33

For Condenser: Heat loss qout = h4 – h1

Fig. 2-2 Ideal Rankine Cycle

So, efficiency of Rankine Cycle is:

)(

)()(

23

1243

hh

hhhh

−

−−−=η Eq. 2.1

As compared to wT, wP is very small so in many cases we can write,

)(

)(

23

43

hh

hh

−

−=η

Eq. 2.2

Fig. 2-3 P-v diagram of Rankine Cycle

2.2 Mean temperature of heat addition

In Rankine cycle, heat is added reversibly at a constant pressure, but at variable

temperature. If Tm is the mean temperature of heat addition, as shown in Fig. 2-4 so that the

area under 2-3 is equal to the area under 2’-3’, then heat added

qin = h3 – h2 = Tm(s3 – s2)


34

Tm = (h3 – h2)/(s3 – s2)

qout = h4 - h1

qout = T2(s4 - s1)

qout = T2(s3 – s2)

Fig. 2-4 Mean temperature of heat addition

Now, efficiency of Rankine cycle

��* = 1 − �� = 1 − �,!�/ − �,"�z!�/ − �,"

��* = 1 − �,�z Eq. 2.3

2.3 Comparison of Rankine and Carnot cycle

We know that Carnot cycle is the most efficient engine for any temperature range but it is

not preferred in vapor power cycle. When Carnot cycle is used in vapor power cycle, there

are some practical difficulties. In the following section we will compare Carnot cycle and

Rankine cycle when used as vapour power cycle.

Fig. 2-5 Comparison between Rankine and Carnot cycle

Fig. 2-5 considers three different cases for the comparison between Carnot and Rankine

cycle. In each case operating temperature range is same for both Carnot and Rankine cycle.

We know that the efficiency of Rankine and Carnot cycle depends on the operating

temperature range such that (Eq. 2.3 and Eq. 1.4).

��* = 1 − ��z

�� = 1 − ��


35

Case a: TL is same for both cycle whereas (TH)Carnot is more than (Tm)Rankine. Therefore, ηCarnot

is greater than ηRankine

Case b: TL is same for both cycle whereas (TH)Carnot is more than (Tm)Rankine. Therefore, ηCarnot


Case c: TL is same for both cycle whereas (TH)Carnot is more than (Tm)Rankine. Therefore, ηCarnot


It shows that theoretically Carnot cycle is more efficient than Rankine cycle. But the Carnot

cycle cannot be realized in practice because the pump work to increase the pressure in all

cases is very large. Heat addition in case a and case b should be carried out at variable

pressure which is difficult to achieve. In case c, the problem of heat addition at constant

pressure is eliminated but to achieve the condition, it is difficult to control the quality at 1’.

2.4 Effect of pressure and temperature on Rankine cycle

2.4.1 Decreasing condenser pressure

Let us consider the exhaust pressure (condenser pressure) is dropped from P4 to P4’ as

shown in the Fig. 2-6 Because of the drop in pressure, heat rejection temperature also drops

to TL’ from TL. There is no significant change in mean temperature of heat addition (Tm).

Therefore, the net result is an increase in efficiency.

Fig. 2-6 Decreasing condenser

pressure

Fig. 2-7 Increasing boiler

pressure

Fig. 2-8 Superheating

steam

2.4.2 Increasing boiler pressure

If pressure is increased to P2 from P1 keeping the maximum temperature same, the mean

temperature of heat addition (Tm) is also increased while the temperature heat rejection

(TL) remains same. Hence the efficiency also increases.

2.4.3 Superheating steam in boiler:

When superheating of steam is increased, the mean temperature of heat addition (Tm)

increases where as temperature of heat rejection (TL) remains the same which finally leads

to the increase in cycle efficiency


36

2.5 Working principle of reciprocating steam engine

The working principle of a reciprocating steam engine is shown in Fig. 2-9. In a reciprocating

engine, the front port is exposed to steam chest pressure allowing the steam to enter the

cylinder and start to drive the piston. The valve then starts to travel forward and will soon

cut off the steam admission from the front port and expose the rear port. The piston will

continue on to the end of its stroke at the rear of the cylinder. The valve traveled forward,

exposing the rear port to the pressurized steam and allowed steam to enter the backside of

the piston, which will drive it back again. The valve then moves back in the original direction

as the piston continues forward. The valve will soon cover the rear port and expose the front

port to begin the process all over again.

The left port is open

allowing steam chest

pressure into the left side

of the cylinder which

pushes the piston towards

the right.

The valve is starting to

travel left and will soon

cut off steam through the

left port. The piston will

continue to the end of its

stroke due to flywheel

momentum.

Fig. 2-9 Working mechanism of reciprocating steam engine

The valve has traveled

fully left exposing the

right port to pressurized

steam.The piston will be

driven towards the left.

The left side of the

cylinder is now exposed to

the exhaust port allowing

old steam to vent.

As the piston continues its

stroke the valve again

begins moving towards

the right. It will soon close

the right port and open

the left port driving the

system towards the

condition in diagram A.


37

Valve action in relation to piston action is shown in Fig. 2-9. Steam chest is filled with high

pressure steam at all times. A valve admits the steam into the cylinder through ports in the

cylinder wall. The same valve allows steam to escape through a center exhaust port. Since

steam force is used on both sides of the piston the engine is called “Double Action.” This

design allows very low rpm and high torque. “Single acting” engines operate by admitting

steam to only one side of a piston. The other side of the piston is exposed to a much lower

atmospheric pressure. They produce power for only 50% of the revolution.

2.6 Steam engine indicators

Engine indicator is a mechanical device, which is used to analyze the change in pressure

with respect to the piston movement. It can plot the graph showing the pressure inside the

cylinder at every point of the stroke.

Fig. 2-10 Engine indicator

The device is fitted with a plotting mechanism, pencil and paper on which the diagram is

drawn. As the piston moves back and forth and as the pressure rises and falls, the pencil will

draw a diagram indicating the pressure in the cylinder at every point of the stroke. A

provision is provided in such a way that the paper move together with the piston. The

vertical and horizontal movement of pencil is reduced by appropriate mechanism so that the

diagram will be in reduced scale. The diagram obtained from indicator is also called

indicator diagram.

2.7 Hypothetical indicator diagram

If we consider the single acting steam engine with zero clearance volume, the operations can

be divided into following processes:

Process 1-2: Constant boiler pressure (Pa) admission of steam

Process 2-3: Expansion of steam which is assumed to follow the relation Pv = constant

Process 3-4: Constant volume release of steam

Process 4-5: Constant pressure release of steam at back pressure (Pb)


38

Fig. 2-11 P-v diagram of steam engine

The total work done can be calculated as:

) = )%$, + ),$/ + )/$- + )-$@

) = (�!+, − +%" + (�+, log 2+/+,3 + 0 + (�!+@ − +-"

) = (�+, + (�+, log 2+/+,3 − (�+-

If expansion ratio r = V3/V2

) = (� +7� + (� +7� log!�" − (�+7 Eq. 2.4

We know that,

&��; �~~�>��C� �� !(z" = )+7

(z = (� +7� + (� +7� log!�" − (�+7+7

(z = (�� + (�� log!�" − (�

(z = (�� !1 + log �" − (� Eq. 2.5

If clearance volume is not considered zero, the P-v diagram of steam engine will be as shown

in Fig. 2-12. As in previous case, work done can be calculated as:


39

Fig. 2-12 P-v diagram of engine with clearance volume

) = )%$, + ),$/ + )/$- + )-$@ + +)@$%

) = (�!+, − +%" + (�+, log 2+/+,3 + 0 + (�!+@ − +-" + 0

) = (�!+, − +D" + (�+, log!�" − (�+7 Eq. 2.6

Where, V3/V2 = r

Similarly

(z = (�!+, − +D" + (�+, log!�" − (�+7+7 Eq. 2.7

2.8 Hypothetical and actual indicator diagram

The hypothetical and actual indicator diagram is shown in the Fig. 2-13. The solid line

represents the hypothetical diagram and dotted line represents actual diagram.

Fig. 2-13 Hypothetical and actual indicator diagram


40

The various points of actual indicator diagram as shown in the figure are:

A ⇒ Admission of steam

B ⇒ Cutoff of steam

C ⇒ Release of steam

D ⇒ The exhaust port closes and trapped steam in clearance volume is compressed

E ⇒ Admission port opens and the trapped steam is mixed with admission steam.

There is some deviation in actual indicator diagram compared with hypothetical indicator

diagram. The admission of steam in actual is not at constant pressure. The pressure

gradually drops which is because of condensation of steam and pressure loss during steam

flow. The expansion process is not hyperbolic as assumed in hypothetical case. The opening

and closing of admission and exhaust port take some time so the region around cutoff,

release and admission of steam is not sharp. The steam left in clearance volume is

compressed at D since exhaust port is closed. The compression continues until the

admission port opens at A. The low pressure steam left in clearance volume is mixed with

high pressure admission steam which drops the admission pressure to A.

The area of actual indicator diagram is less than area of hypothetical indicator diagram. The

ratio of the area of actual indicator diagram to the area of hypothetical indicator diagram is

called Diagram Factor (DF).

|� = �� ~ �>�� ;��>�� h��=�� ~ ℎr��ℎ��>�� ;��>�� h��=

|� = =��; �~~�>��C� �� ~ �>�� ;��>�� h��= × +7=��; �~~�>��C� �� ~ ℎr��ℎ��>�� ;��>�� h��= × +7

|� = !(z"�D��8!(z"��B��*��D�8 Eq. 2.8


41

Solved Numerical Problems

1. In a Rankine cycle steam leaves the boiler and enters the turbine at 4 MPa, 400°C. The

condenser pressure is 10 kPa. Determine,

a. The pump work per kg

b. The work output per kg

c. The cycle efficiency

d. The condenser heat flow per kg

Boiler pressure = 4 MPa

Temperature T3 = 400°C

Condenser pressure = 10 kPa

From steam table, at 4 MPa and 400°C the steam is in superheated state. The following

properties can be noted from the steam table for 4 MPa and 400°C.

h3 = 3213.4 kJ/kg

s3 = 6.7688 kJ/kgK

Since process 3-4 is isentropic, the entropy at state 3 and 4 are same.

s4 = 6.7688 kJ/kgK

Since state 4 is in two-phase region, from steam table (saturated water), for 10 kPa

sl = 0.6493 kJ/kgK and slg = 7.4989 kJ/kgK

hl = 191.83 kJ/kg and hlg = 2392.0 kJ/kg

�- = �8 + ��8�

6.7688 = 0.6493 + 7.4989�

� = 0.82

Similarly for h4

ℎ- = ℎ8 + �ℎ8�

ℎ- = 191.83 + 0.82 × 2392.0

ℎ- = 2153.27 �q/�h

The state 1 is in saturated liquid line and has pressure of 10 kPa

h1 = hl =191.83 kJ/kg

The process 1-2 is an isentropic pumping of water. The water is incompressible substance

so the work required to increase the pressure of water from pressure P1 to P2 can be written

as:


42

�B = C%!(, − (%"

v1 is the specific volume of saturated water at pressure P1. So from steam table, v1 =

0.001010 m3/kg. So,

�B = 0.001010 × !4 × 10f − 10 × 10/" = YXdj. jX l/k�

Pump work can also be expressed as

�B = ℎ, − ℎ%

ℎ, = �B + ℎ% = 4029.9 + 191.83 × 10/ = 195.86 �q/�h

Work output per kg

�J = ℎ/ − ℎ- = 3213.4 × 10/ − 2153.27 × 10/ = WXZX. W_ kl/k�

The cycle efficiency

� = !ℎ/ − ℎ-" − !ℎ, − ℎ%"ℎ/ − ℎ, = !3213.4 − 2153.27" × 10/ − !195.86 − 191.83" × 10/!3213.4 − 195.86" × 10/

� = _`. X%

Condenser heat flow per kg

�� = ℎ- − ℎ% = !2153.27 − 191.83" × 10/ = WjZW. YY kl/k�

2. Conventional indicator card without clearance shows the percentage cut-off as 33%.

Intake pressure is 9.85 bar and exhaust pressure is 1.125 bar. For a 50×60 cm double

acting single cylinder steam engine with a mean piston speed of 18 m/s, find the

Indicated Power for a diagram factor of 0.78. Neglect the piston rod area.

Pa = 9.85 bar

Pb = 1.125 bar

Cutoff = 33% of stroke

Bore (D) × stroke (L) = 50 × 60 cm

Mean piston speed = 18 m/s

Diagram factor = 0.78

The stroke/swept volume is,

+- = +7 = {|,}4 = { × 0.5, × 0.64 = 0.118 =/ Since, the cutoff is 33% of the stroke

+, = 0.33 × +7 = 0.33 × 0.118 = 0.0389 =/

Theoretical mean effective pressure for single acting cylinder


43

!(z"�� = )+7 = (�+, + (�+, ln 4+/+,5 − (�+-+7

!(z"�� = �9.85 × 0.0389 + 9.85 × 0.0389 ln 4 0.1180.03895 − 1.125 × 0.1180.118 � × 10@

!(z"�� = 5.72 ��

!(z"�D� = |� × !(z"�� = 0.78 × 5.72 = 4.46 ��

In one revolution, piston covers 2L distance. So in N rpm, the mean piston velocity can be

expressed as,

+ = 2}�60 ⇒ � = 60+2} = 60 × 182 × 0.6 = 900 ��=

Actual indicated power for double acting cylinder can be calculated as

�( = 2!(z"�D� }��60 = 2!(z"�D� +7�60 = 2 × 4.46 × 10@ × 0.118 × 90060

�V = W`ne. eY ks

3. Determine the actual mean effective pressure for a steam engine receiving steam at a

pressure of 6 bar; cut-off takes place when the piston has travelled 0.4 of the stroke in

which clearance volume is 10% of the stroke volume. Back pressure is 1.03 bar and the

diagram factor is 0.7.

Pa = 6 bar

Pb = 1.03 bar

Cutoff = 0.4 of stroke

Clearance volume = 10% of stroke volume

Diagram factor = 0.7

+D = +% = 0.1+7

+, − +% = 0.4 +7 ⇒ +, = 0.1+7 + 0.4+7 = 0.5+7

+/ = +- = +7 + +D = +7 + 0.1+7 = 1.1+7

Theoretical mean effective pressure is

!(z"�� = )+7 = (�!+, − +%" + (�+, ln 4+/+,5 + (�!+@ − +-"+7


44

!(z"�� = (� × 0.4+7 + (� × 0.5+7 × ln 21.1+70.5+73 − (�+7+7

!(z"�� = (� × 0.4 + (� × 0.5 × ln 21.10.53 − (�

!(z"�� = 26 × 0.4 + 6 × 0.5 × ln 21.10.53 − 1.033 × 10@ = 3.73 ��

!(z"�D� = |� × !(z"�� = 0.7 × 3.73 = d. ZW w\x


45

Chapter 3. Internal Combustion Engine

3.1 Introduction

Any type of engine or machine which derives heat energy form the combustion of fuel or any

other source and converts this energy into mechanical work is termed as heat engine. Heat

engine may be classified as follow:

Fig. 3-1 Classification of heat engine

• External combustion engine: In external combustion engine, the fuel that runs the engine

is burned outside the engine. Example: Steam turbine

• Internal combustion engine (IC engine): In IC engine, the fuel that runs the engine is

burned inside the cylinder of the engine. Example: Automobile

In reciprocating engine, piston, connecting rod and crank-shaft are used to convert

reciprocating motion into rotary motion. There are two types of reciprocating engines:

Spark-Ignition engine (SI engine) and Compression-Ignition engine (CI engine). The

differences between the two are:

• The types of fuel used

• The way the fuel gets into the engine cylinder

• The way the fuel is ignited

The spark-ignition engine uses a highly volatile fuel which turns to vapor easily, such as

gasoline. The fuel is mixed with air before it enters the engine cylinder. The fuel turns into a

vapor and mixes with the air to form a combustible air-fuel mixture. This mixture then

enters the cylinders and is compressed. Next, an electric spark produced by the ignition

system sets fire to, or ignites, the compressed air fuel mixture.

In the compression-ignition engine, the fuel is mixed with the air after the air enters the

engine cylinder. Air alone is taken into the cylinder of the diesel engine. The air is then

compressed as the piston moves up. The air is compressed so much that its temperature

goes up to 550°C or higher. Then the diesel-engine fuel is injected into the engine cylinder.

The hot air, or heat of compression, ignites the fuel.


46

3.2 Classification of IC engines

Internal-combustion engine may be classified according to:

• Number of cylinder:

o Single cylinder or multi cylinder

• Arrangement of cylinder:

o Horizontal engine

o Vertical engine

o V-type engine

o Radial engine

Fig. 3-2 Cylinder arrangement

• Arrangement of valves:

o Overhead type

o T-head type

o L-head type

o F-head type

• Types of cooling

o Air cooled or liquid cooled

• Number of strokes per cycle

o Two stroke or four stroke

• Type of fuel used

o Gasoline (Petrol) or Diesel

• Method of ignition

o Compression ignition or spark ignition

• Reciprocating or rotary

3.3 Working cycle of IC engines

An internal combustion engine can work on any one of the following cycles:

1. Constant volume or Otto cycle

2. Constant pressure or Diesel cycle

3. Dual combustion cycle


47

3.3.1 Engine operation- four stroke SI cycle

1. Intake stroke: During the intake stroke (Suction stroke/induction stroke), the piston

moves downward form TDC (Top Dead Centre) to BDC (Bottom Dead Centre). The intake

valve is open. As the piston moves down, vacuum is created inside the cylinder. So air-

fuel mixture enters the cylinder through the inlet valve. As the piston passes through

BDC, the intake valve closes. Since the outlet valve is still closed, the upper end of the

cylinder is sealed off.

2. Compression stroke: After the piston passes through BDC, it starts moving up. The air

fuel mixture is compressed. The degree of compression is based on the compression

ratio of the engine. The average compression ratio of SI engine is 8:1.

Fig. 3-3 Four stroke SI engine operating cycle

3. Power stroke (Expansion stroke): As the piston nears TDC on the compression stroke,

an electric spark is produced by spark plug in the cylinder. The spark sets fire to the

compressed air-fuel mixture. The temperature of burning mixture rises up to 3300°C

and the pressure inside the cylinder also increases. The work is obtained during this

stroke. The high pressure in the cylinder pushes the piston down. The downward push is

carried through the connecting rod to the crankshaft. The flywheel is mounted on the

engine shaft to store energy so that it can be used during remaining three ideal strokes.

4. Exhaust stroke: As the piston approaches BDC on the power stroke, the exhaust valve

opens. Now, as the piston moves up again after BDC, the burned gases escape through

the exhaust valve. Then, as the piston nears TDC, the intake valve opens. As the piston

passes through TDC and start down again, the exhaust valve closes. Now another intake

stroke takes place.


48

3.3.2 Engine operation- four stroke CI cycle

As in four stroke SI engine, four stroke CI engine is also completed in four strokes: intake

stroke, compression stroke, power stroke and exhaust stroke. The working principle of each

stroke is almost similar as it is in SI cycle. During intake stroke, only air is pumped in the

cylinder unlike air-fuel mixture in SI engine. The air is compressed with compression ratio

of around 22:1. As the air is compressed the temperature of air rises up to 538°C. As piston

nears TDC on compression stroke, the fuel injector sprays diesel into combustion chamber.

The high temperature-air ignites the fuel and the power stroke and then exhaust stroke

follows.

Fig. 3-4 Four stroke CI engine operating cycle

3.3.3 Engine operation-two stroke cycle

The Fig. 3-5 shows a two stroke petrol engine. During upward stroke of the piston, the gas in

the piston is compressed and at the same time fresh air-fuel mixture enters the crank

chamber through the valve V. At the end of the compression, spark plug ignites the air-fuel

mixture burns. The increased pressure in the cylinder pushes the piston and near the end of

the stroke, the piston uncovers the exhaust gas port and exhaust gas escape through the

port. During downward movement, the air-fuel mixture in the crank case (CC) also gets

compressed. As soon as the transfer port (TP) is uncovered by the down moving piston, the

air-fuel mixture form the crank case enters to the cylinder. The special shape in the piston

top helps air-fuel mixture to flow upward and helps the exhaust gas to escape out of the

exhaust port. During exhaust, some of the unburned air-fuel mixture also escapes through

exhaust port along with the burned gas. As the piston again moves upward, the air-fuel

mixture gets compressed and another cycle is repeated.


49

Fig. 3-5 Two stroke engine

3.4 Valve timing diagram of four stroke IC engine

In describing the operation of the four stroke cycle engine, it was explained that the inlet

valve is open during the suction stroke and the exhaust valve during the exhaust stroke. But

in real practice the valve opening timing is different.

Fig. 3-6 Actual valve timing for four stroke engine

The Fig. 3-6 shows the typical valve timing diagram. Intake valve opens about 20° before

TDC. This allows the valve to be open completely when suction strokes starts. The intake

valve remains open well after the piston reaches the BDC during suction stroke. This gives


50

additional time for air-fuel mixture or air to flow into the cylinder. The delivery of adequate

amount or charge to the cylinder is the critical item in the engine operation. Actually, the

cylinders are never quite filled up when intake valve closes especially during high speed.

This period varies from 30 to 70 degree in modern automotive engines.

In similar manner, exhaust valve starts to open about 35° before piston reaches BDC on the

power stroke and stays open until 10° after TDC on the intake stroke. This gives more time

for the exhaust gas to leave the cylinder. By the time the piston reaches 35° before BDC on

the power stroke, the combustion pressure has dropped considerably. Little power is lost by

giving the exhaust gas this extra time to leave the cylinder. Closing the exhaust valve after

TDC on suction stroke allows exhaust gases to move rapidly for the cylinder into exhaust

port thus prevents the exhaust gas to be retained in the clearance space of the cylinder.

The spark plug produces spark about 30 – 40 degree before TDC on power stroke thus fuel

gets more time to burn.

Exercise for you

• List the major differences between SI engine and CI engine.

• What are the differences between two stroke and four stroke engine


51

3.5 Components of IC engines

Fig. 3-7 Engine components

The working principle of CI and SI engine is almost same so the construction of both engines

is same. There are numbers of components common to both engines.

1. Cylinder block/cylinder: The cylinder block is the foundation of the engine. Everything

else is put inside of or attached to the block. Most cylinder blocks are cast in one piece

form grey iron or iron alloy or aluminum alloy. It contains large holes for cylinder bore

where the gas is compressed (Fig. 3-7)


52

2. Piston and piston rings: Piston is usually made of aluminum alloy. Piston can slide up

and down in the cylinder easily. Piston undergoes great stress. During power stroke, up

to 1814 kg is suddenly applied to the piston head. This happens 30 to 40 times a second

at high speed. The temperature above the piston head can reach 2200°C. The piston

must be strong. But it must be light, too, to reduce inertia loads on the bearing. Modern

automotive engines use aluminum alloy piston. Since the diameter of piston is smaller

than that of cylinder, the gap must be sealed to avoid the leak of compressed air-fuel

mixture (in SI engine)/air (in CI engine) and high pressure burned gas. The escape of

unburned and burned gas from the combustion chamber, past the pistons, and into the

crankcase is called “blow by”. “Blow by” would greatly reduce the efficiency of the engine.

To avoid this, piston rings are installed on the piston. The piston rings are split at one

point so they can be expanded slightly and slipped over the piston and into the grooves

cut in the piston. Piston rings are of two types and they do two jobs:

a. Compression rings: The compression rings seal in the unburned gas as it is

compresses. They also seal in the combustion pressure as the mixture burns.

b. Oil-control rings: Oil-control rings scrape excessive oil that is splashed on the wall so

that it does not get up into the combustion chamber where it would burn.

Fig. 3-8 Piston and piston rings

3. Cylinder head: The cylinder head encloses one end of the engine cylinders and forms

the upper end of the combustion chambers. The cylinder head includes water jackets

and passages from the valves ports to the openings in the manifolds.


53

Fig. 3-9 Combustion chamber

4. Inlet and exhaust manifold: The intake manifold is a series of tubes that carry air to the

engine cylinders. The exhaust manifold is a set of tubes that carry the exhaust gases form

the engine cylinder.

5. Connecting rod: The connecting rod is attached at one end to a crankpin on the

crankshaft. It is attached at other end to a piston, through a piston pin. The connecting

rod must be very strong and rigid, and as light as possible. The connecting rod carries

the power thrust form the piston to the crankshaft.

6. Crankshaft: The crankshaft is a strong one-piece casting or forging of heat-treated alloy

steel. The crankshaft must be strong enough to take the downward push of the piston

during the power strokes without excessive twisting. In addition, the crankshaft must be

carefully balanced to eliminate undue vibration resulting form the weight of the offset

cranks. To provide balance, crankshafts have counterweight opposite to the cranks.

Crankshafts have drilled oil passage through which oil flows form the main to the

connecting-rod end.

Fig. 3-10 Crankshaft


54

7. The valves: Most engines have two holes in the enclosed upper end of the cylinder. One

of two holes is the intake valve. It allows the air or air-fuel mixture or air to enter the

cylinder. The other is exhaust valve. It allows the burned gases to exhaust form the

cylinder.

Fig. 3-11 Valve mechanism

8. Engine bearing: Bearings are placed in the engine wherever there is rotary motion

between engine parts. These engine bearings are called sleeve bearing because they are

shaped like sleeves that fit around the rotating shaft. The part of the shaft that rotates in

the bearing is called a journal. Connecting-rod and crankshaft bearings are of the split, or

half, type. Typical bearing half is made up of a steel or bronze back. The bearing material

is soft. Therefore the bearing wears with more ease compared with the other mechanical

part of the engine. The cost of bearing is less than that of engine part so the replacement

cost of bearing is less.

9. Crank case: The bottom of the cylinder block and the oil pan form crank case. The oil

pan holds lubricating oil. When the engine is running, the oil pump sends oil form the oil

pan up to the moving engine parts.

10. Fly wheel: Fly wheel attached with one end of crank shaft performs the following

functions:

a. Stores energy required to rotate the shaft during ideal strokes (intake,

compression and exhaust strokes)

b. Makes crank shaft rotation more uniform.

c. Facilitates the starting of the engine.

3.6 Parts common to SI engine

1. Spark plug: The main function of spark plug is to conduct the high potential form the

ignition system into the combustion chamber. The spark plug is a metal shell in which a

porcelain insulator is fastened. An electrode extends through the centre of of the


55

insulator. The metal shell has a short electrode attached to one side. The outer electrode

is bent inward to produce the proper gap between it and the centre electrode. High

tension current jumping from the supply electrode produces the necessary spark. The

correct type of plug with correct gap between the electrodes is important factor.

Fig. 3-12 Spark plug

2. Carburetor: Carburetion is the mixing of gasoline fuel with air to get a combustible

mixture. The function of carburetor is to supply a combustible atomized fuel and air

mixture of varying degree of richness to suit the engine operation conditions. The

mixture must be rich for starting, acceleration and high speed operation. The lean

mixture is desirable at intermediate speed with a warm engine.

3.7 Parts common to CI engine

1. Fuel injector: Fuel injector injects the fuel inside the cylinder of CI engine at the start

of the expansion stroke.

2. Fuel pump: The fuel form fuel tank is pumped to the fuel injector by fuel pump.

3.8 Basic engine parameters

1. Top dead centre (TDC): The top most position that piston can reach.

2. Bottom dead centre (BDC): The lowest position that piston can reach.

3. Stroke length: The distance between TDC and BDC or length covered by piston during

one stroke

4. Bore: Diameter of the cylinder

5. Swept volume (Vs): It is the product of cross-section area of the cylinder and stroke

length. It can also be defined as a volume covered by the piston during one stroke.

Engine capacity is defined in terms of swept volume expressed in cc. A four cylinder

engine marked as 1.2 ltr has four cylinders each having swept volume of 300 cc.


56

6. Clearance volume (Vc): Volume left above the piston when the piston is in TDC.

7. Compression ratio (r): r = (Vs + Vc)/Vc

Fig. 3-13 Bore stroke


57

Chapter 4. Performance of Internal Combustion Engine

4.1 Mean effective pressure (MEP)

When work done per cycle is divided by swept volume per cycle, the parameter obtained is

called mean effective pressure. Mean Effective Pressure (MEP) can also be defined as the

pressure that, if it acted on the piston during the entire power stroke, would do an amount

of work equal to that actually done on the piston (Fig. 4-1).

Fig. 4-1 Mean effective pressure

&'( = )�*�+7��* Eq. 4.1

4.2 Indicated power (IP)

Fig. 4-2 Pressure in the cylinder during each stroke

Indicated power is the total power that the engine develops inside the combustion chamber

during the combustion process. A special device is required to measure IP. It measures the

pressure in the engine cylinder. As shown in the Fig. 4-2 pressure is highest during power


58

stroke. The average of pressure in four strokes is also called indicated mean effective

pressure.

The relation between indicated mean effective pressure and IP can be derived as follow:

From Eq. 4.1

) = (z+7

) = (z}�

[L = Length of stroke, A = Cross-section area of the cylinder]

∵ (�� = )��>r>�� × >r>��>�;�

�( = )�60 = (z}�60

For n cylinder engine

�( = ;(z}�60

In four stroke engine, power is generated once in two revolution of the crankshaft but in two

stroke engine power is developed in every revolution of engine. So equation of IP can be

written in general terms as,

�( = �;(z}�60 × 1000 Eq. 4.2

Where, IP = Indicated power (kW)

n = Number of cylinder

Pm = Indicated mean effective pressure (N/m2)

L = Length of stroke (m)

A = Cross-section area of cylinder (m2)

N = rpm of crank shaft.

k = 1 for two stroke engine and 1/2 for four stroke engine.

4.3 Brake power (BP)

The power developed by an engine at the output shaft is called the brake power. Not all the

IP developed by the engine is available as useful power to do work. Part of is lost in the

engine as friction power (FP). FP is the power required to overcome the friction of the

moving parts in the engine. One of the major causes of friction loss is piston-ring friction. In

addition to the friction loss, FP also includes the work of charging absorbed during exhaust

and suction stroke. Resistance of air to flywheel rotation and power required to drive the

auxiliaries.


59

BP = IP – FP Eq. 4.3

100060

knLAN

BPbmep

××=

Eq. 4.4

Where, BP = Brake Power (kW)

n = Number of cylinder

bmep = Break mean effective pressure (N/m2)

L = Length of stroke (m)

A = Cross-section area of cylinder (m2)

N = rpm of crank shaft.

4.4 Engine efficiencies

Mechanical efficiency IP

BPmechanical =η

Eq. 4.5

Brake thermal efficiency f

thermalBmC

BP= ,η

Eq. 4.6

Indicated thermal efficiency f

thermalImC

IP= ,η

Eq. 4.7

Where, BP = Brake Power, IP = Indicated power, m = mass of fuel, Cf = Calorific value or

heating value of the fuel.

• Volumetric efficiency: During intake stroke, the passage of air is restricted by filter,

carburetor and throttle plate. The parameter used to measure the effectiveness of an

engine induction process is called volumetric efficiency. It is the ratio of the volume

Va of air admitted or air-fuel mixture admitted to the engine cylinder during suction

referred to NTP to the suction volume of the piston.

p

a

volumetricV

V=η

Eq. 4.8

• Relative efficiency

ration compressio samefor efficiency standardair Ideal

efficiency thermalindicated actual=relativeη

Eq. 4.9

• Air-fuel ratio (Fuel-air ratio): It is the ratio of mass of air to the mass of fuel in air-

fuel mixture.

• Stoichiometric air-fuel ratio: Ratio of amount of air (oxygen in air) required to

combust a certain amount of fuel in chemically balanced condition. According to

chemical reaction, 14.7 kg of air is required to burn 1 kg of gasoline. So the

stoichiometric A/F ratio for gasoline is 14.7:1. Similarly that of diesel is 14.6:1.


60

• Relative air-fuel mixture: Ratio of actual A/F ratio to stoichiometric A/F ratio. It is

also known as lamda.

( )( )

Stoich

actual

FA

FA

/

/=λ

Eq. 4.10

Lamda of 1 is stoich mixture, rich mixtures are less than 1 and lean mixtures are

more than 1.

• Equivalence ratio (φ): Ratio of actual F/A ratio to stoichiometric F/A ratio. It is an

inverse of λ

(((( ))))(((( ))))

Stoich

actual

AF

AF

/

/====φφφφ

Eq. 4.11

Specific fuel consumption (kg/kWh) BP

msfc

f= Eq. 4.12

Where, mf= mass of fuel used per hour

4.5 Heat balance for IC engine

First law of thermodynamics for a steady flow open system can be well applied for the

analysis of the overall heat (energy) balance for the IC engine. According the principle, total

energy leaving the system is equal to the total energy incoming to the system. In case of an

IC engine, the relation can be expressed as

LHVf

.

e

.

misc

.

cool

.

b QP mHQQ ====++++++++++++ Eq. 4.13

In the above Eq. 4.13, f

.

m is the mass flow rate of fuel entering the cylinder. It combined

with lower heating value of fuel (QLHV), represents the total energy incoming the cylinder in

the form of fuel. The fraction of heat produced upon burning of fuel is used for brake power

(Pb). Remaining heat is lost through the cooling water (cool

.

Q ), exhaust gas enthalpy ( e

.

H )

and miscellaneous loss (misc

.

Q ). Miscellaneous heat loss includes the friction power lost in

bearings, valve mechanism and heat lost during cooling by lubricating oil. Typical values of

each of each of these terms with reference to percentage of fuel heating value are given in

the table below.

Pb cool

.

Q misc

.

Q e

.

H

Percentage of fuel heating value

SI engine 25-28 17-26 3-10 36-50

CI engine 34-38 16-35 2-6 23-37


61

Energy balance can be illustrated through the Fig. 4-3below.

Fig. 4-3 Energy balance in IC engine

4.6 Torque-power-speed relationship

The torque that an engine can develop changes with engine rpm. During intermediate

speeds, volumetric efficiency is high. There is sufficient time for the cylinder to become

fairly well filled up. This means that with a fairly full charge of air-fuel mixture, higher

combustion pressure will develop. With higher combustion pressure, the engine torque is

higher. But, at higher speed, volumetric efficiency drops off. There is not enough time for the

cylinders to become filled up with air-fuel mixture. Since there is less air-fuel mixture to

burn, the combustion pressure is not as high. There is less push on the pistons. Therefore,

engine torque is lower. The Fig. 4-4 shows how the torque drops off as engine speed

increases. The decrease in BP is due to reduced torque at higher speed and to increased FP

at the higher speed. Note that the curves are for one particular engine only. Different

engines have different torque and BP. Peaks may be at higher or lower speed.

Fig. 4-4 Torque-power-speed relationship


62

The empirical equation that defines the relation between shaft power and the engine speed

is

−+=

2

maxmaxmax

max 09.156.153.0N

N

N

N

N

NPP

Eq. 4.14

Where, P = Brake Power

Pmax = maximum power an engine can produce (obtained in engine specification)

Nmax= shaft rpm at which engine can produce maximum power

Engine shaft power and shaft torque that engine can develop can be related as

60

2 NTP

π=

Eq. 4.15

Where, P = Brake Power, N = engine rpm, T = torque at engine shaft

Figure 4-1 Torque-sfc and Power-sfc relationship

4.7 Gasoline and gaseous fuel system

4.7.1 Gasoline

Gasoline is a hydrocarbon (abbreviated HC), made up largely of hydrogen and carbon

compounds. The resulting exhaust gas contains water vapor, carbon dioxide, and the

pollutants carbon monoxide (partly burned gasoline) and hydrocarbons (unburned

gasoline).

Gasoline is made from crude oil by a refining process that also produce engine lubricating

oil, diesel fuel oil, and other products. During the refining process, several additives are put

into the gasoline to improve its characteristics. Good gasoline should have:

1. Proper volatility, which determines how quickly gasoline vaporizes

2. Resistance to spark knock, or detonation

3. Oxidation inhibitors, which prevent formation of gum in the fuel system

4. Antirust agents, which prevent rusting of metal parts in the fuel system

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 1000 2000 3000 4000 5000 6000 7000

RPN

sfc

kg

/kW

h

0

10

20

30

40

50

60

70

Po

we

r kW

sfc

Power

0

20

40

60

80

100

120

140

160

180

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 1000 2000 3000 4000 5000 6000 7000

Torque

N-m

sfc

kg/k

Wh

r pm

s fc Torque


63

5. Anti-icers, which retard carburetor icing and fuel-line freezing

6. Detergents, which help keep the carburetor clean

7. Dye for identification

After gasoline is mixed with air in the carburetor, the gasoline must vaporize quickly, before

it enters the engine cylinder. If the gasoline is slow to vaporize, tiny drops of liquid gasoline

will enter the cylinders. Because these drops do not burn, some of the fuel is wasted. It goes

out the tail pipe and helps create atmospheric pollution. Also, the gasoline drops tend to

wash the lubricating oil off the cylinder walls. This increases the wear on the cylinder walls,

piston rings, and pistons.

The ease with which gasoline vaporizes is called its volatility. A high-volatility gasoline

vaporizes very quickly. A good gasoline should have just the right volatility for the climate in

which the gasoline is to be used. If the gasoline is too volatile, it will vaporize in the fuel

system. The result will be a condition called vapor lock. It prevents the flow of gasoline to

the carburetor. Vapor lock causes the engine to stall from lack of fuel.

4.7.2 Carburetion system:

Carburetion is the mixing of gasoline fuel with air to get a combustible mixture. The function

of carburetor is to supply a combustible atomized fuel and air mixture of varying degree of

richness to suit the engine operation conditions. Atomization is breaking of fuel into fine

droplets. Each droplet is then exposed to air so that it vaporized quickly. It would be

desirable to pass on the fuel in completely vaporized phase mixed with air but complete

vaporization is not possible. The mixture must be rich for starting, acceleration and high

speed operation. The lean mixture is desirable at intermediate speed with a warm engine.

Fig. 4-5 Basic carburetor construction

Basic components of carburetor are venturi tube, fuel reservoir, throttle valve and float

system. The carburetor works on Bernoulli’s principle. As air passes the venture tube, some

of static head of air is converted to velocity head. The high velocity air through the venture

creates partial vacuum or suction in it. The suction draws the gasoline form the discharge

jet. The fuel gets mixed with the air flowing through the venture. The level of fuel in fuel

reservoir and in the discharge tube is maintained by the float system in the reservoir. As the

fuel level in the reservoir falls, the float moves down and the needle valve opens thus the


64

fuel flows in the reservoir. When fuel level rises to certain level, the float moves up and

pushes needle valve into the valve seat.

4.7.3 Air mixing

As discussed in the previous section, air is mixed with fuel in carburetor and combustible

mixture is sent to the combustion chamber in intake stroke. The ideal mixture of air and fuel

is called stoichiometric mixture (λ or φ = 1). But mixture requirements are different

depending on operating condition. Different operating conditions and their respective air

mixing process are discussed as follow.

Idle system: When throttle valve is closed or only slightly opened, only a small amount of air

can pass through the air horn. The air speed is low, and very little vacuum develops in the

venture. This means fuel nozzle does not discharge fuel. Therefore the carburetor must have

another system to supply fuel when the throttle is closed or slightly open. This system,

called idle system, is shown in the Fig. 4-6. It includes passage through which air and fuel

can flow. The air passage is called air bleed. With the throttle valve closed there is a high

vacuum below the throttle valve form the intake manifold. Atmospheric pressure pushed air

and fuel through the idle mixture screw. The mixture has high proportion of fuel.

Fig. 4-6 Idle system

Fig. 4-7 Low speed operation

Low-speed operation: When the throttle valve is open slightly, as shown in Fig. 4-7, the

edge of the throttle valve moves past the low speed-operation port in the side of the air

horn. Therefore additional fuel is fed into the intake manifold through the low-speed port.

This fuel mixes with the additional air moving past the slightly opened throttle valve. The

additional mixture fuel provides sufficient mixture richness for part-throttle low-speed

operation.

Main-metering system: When the throttle valve is opened enough, sufficient air moves

through the air horn to produce a vacuum in the venture. As a result, the fuel nozzle begins

to discharge fuel. The wider the throttle is opened and the faster the air flows through the

air horn, the greater the vacuum in the venture. This means that additional fuel will be


65

discharged from the main nozzle. As the result nearly constant air-fuel ratio is maintained

by the main-metering system form part to wide opened throttle.

Fig. 4-8 Main metering system

Power enrichment system: For high-speed full power wide-open-throttle operation, the

air-fuel mixture must be enriched so engine can deliver its maximum power but utilizing all

the oxygen present in induced air. The rich mixture ensures the consumption of all the

oxygen. Additional device are incorporated in the carburetor to provide the enriched

mixture during high-speed-full-power operation.

Accelerator pump: When the throttle plate is opened rapidly, the fuel-air mixture flowing

into the engine cylinder leans out temporarily. The primary reason for this is time lag

between fuel flow into the air stream at the carburetor and fuel flow past the intake valve.

While much of the fuel flow into the cylinder is fuel vapor or small fuel droplets carried by

the air stream, a fraction of the fuel flows into the manifold and port walls and form a liquid

film. The fuel which impacts on the walls evaporates more slowly than fuel carried by the air

stream and introduces a lag between the air/fuel ratio produced at the carburetor and

air/fuel ratio delivered to the cylinder. An accelerator pump is used as the throttle plate is

opened rapidly to supply additional fuel into the air stream at the carburetor to compensate

for this leaning effect.

4.7.4 Temperature effect

When the temperature of the surrounding drops considerably than normal, only part of the

fuel vaporizes. Therefore extra fuel must be delivered to get the combustible air fuel mixture

that permits the engine to start. So during cold weather, the carburetor must supply rich

mixture. The elementary carburetor cannot enrich the mixture as required for cold start up.

For this purpose, choke system is supplied with elementary carburetor. Primary element of

choke system is a plate kept upstream of the carburetor.

Choke system

The working of choke can be explained with the help with the help of the Fig. 4-9. During

cranking, air speed through the carburetor air horn is very low. Vacuum form the venturi


66

action and vacuum below the throttle valve would be insufficient to produce adequate fuel

flow for starting. To produce enough fuel flow during cranking, the carburetor has a choke.

When the choke is closed, it is almost horizontal. Only a small amount of air can get past it.

The valve has “choked off” the air flow. Then, when the engine is cranked with choke closed,

a fairly high vacuum develops in the air horn. This vacuum causes the main nozzle to

discharge a heavy stream of fuel. The quantity delivered is sufficient to produce the air-fuel

mixture needed for starting the engine.

Fig. 4-9 Choke system

4.7.5 Altitude effect

An inherent characteristic of the conventional carburetor is that it meters fuel flow in

proportion to the air volume flow rate. Air density changes with ambient pressure and

temperature but change in pressure with altitude is most significant. To reduce the impact

of changes in altitude, modern carburetors are fitted with provisions for altitude

compensation.

4.7.6 Gasoline fuel injection system

Fig. 4-10 Distribution of air fuel mixture in intake manifold

One of the most difficult problems in a carbureted system is to get the same amount and

richness of air-fuel mixture to each cylinder. The problem is that the intake manifold acts as


67

sorting device, sending a richer air-fuel mixture to the end cylinders (Fig. 4-10) since the air

flow readily around corners and through variously shaped passages. However, the fuel is

unable to travel as easily around the bends in the intake manifold. As a result, some of the

fuel particles continue to move to the end of the intake manifold, accumulating or puddling

there. This enriches the mixture going to the end cylinders. The centre cylinder, closest to

the carburetor, get the leanest mixture.

The port fuel-injection system (Fig. 4-11) solves this problem because the same amount of

fuel is injected at each intake-valve port. Each cylinder gets the same amount of air-fuel

mixture of the same mixture richness. In port fuel-injection system, the fuel is injected into

the intake manifold through fuel-injection valves. Injection valve is positioned in the intake

port near the intake valve.

Fig. 4-11 Port injection system

4.7.7 Electronic fuel injection

The development of solid-state electronic devices such as diodes and transistors made

electronic fuel injection possible. In this system, fuel metering is controlled primarily by

engine speed and the amount of air that actually enters the engine.

Fig. 4-12 Basic block diagram of Electronic Fuel Injection system


68

The airflow sensor in the air intake measures the amount of air that enters. Modern

electronic fuel-injection systems include an oxygen sensor to measure the amount of oxygen

in the exhaust gas and send this information to the electronic control unit. If there is too

much oxygen, the mixture is too lean. If there is too little, the mixture is too rich. In either

case, the ECU adjusts air-fuel ratio by changing the amount of fuel injected.

4.8 Ignition system

The ignition system supplies high voltage surges to the spark plugs in the cylinders. These

surges produce electric sparks across the spark-plug’s electrode gap. The spark ignites the

air-fuel mixture. It must create this spark at the appropriate time during the compression

stroke. Usually spark timing is set to give maximum brake torque for the specific operating

condition. For a given engine design, this optimum spark timing varies as engine speed, inlet

manifold pressure and mixture composition vary. Thus, in most applications, and especially

the automotive application, the system must have means for automatically changing the

spark timing as engine speed and load vary. When the engine is idling, the spark appears at

the plug gap just as the piston nears TDC on the compression stroke. When the engine is

operating at higher speed, or with part throttle, the spark is advanced. It is moved ahead and

occurs earlier in the compression stroke. This gives the compressed mixture more time to

burn and deliver its energy to the piston.

4.8.1 Contact point ignition system (Conventional system)

The basic circuit of contact point ignition system is shown in the Fig. 4-13. It includes the

battery, the primary and secondary coil, the ignition distributor, the spark plugs and the

wires and cables that connect them. The distributor has a shaft that is driven by gear on the

cam shaft. There are two separate circuits through the distributor. One is primary circuit

and the other is secondary circuit.

The primary circuit includes the battery, the ignition switch, primary resistance, primary

winding (consists few hundred turns of relatively heavy wires) in the ignition coil, contact

point circuit breaker, and primary wiring. As the distributor cam rotates, the cam lobe

pushes the arm out so that the contact point is moved away from the stationary point and

primary circuit breaks. As soon as the lobe passes out, the spring bring back the arm the

primary circuit is connected. There are same numbers of lobes on the cam as there are

cylinders in the engine.

The secondary circuit consists of the cables that connect the secondary winding (consists of

thousands of turns of a very fine wire) to the cap centre terminal and the outer cap terminal

to the spark plugs.


69

Fig. 4-13 Circuit diagram of contact point ignition system

When the contact points are closed, they connect the primary winding of the ignition coil

with the battery. Current flows through the coil, causing a magnetic field to build up around

it. Then, when the points are opened, this disconnects the primary winding form the battery.

The resulting decay of magnetic flux in primary coil induces the short pulse of high voltage

in the secondary winding of the coil. The high-voltage surge (equal to break down potential

of spark plug) flows through the distributor cap and rotor to the spark plug in the cylinder

in which piston is nearing the end of compression stroke.

Centrifugal advance: As engine speed increases, the spark must occur in the combustion

chamber earlier in the cycle. If the engine is idling, the spark must occur just before the

piston reaches TDC. At low speed, the spark has enough time to ignite the compressed air-

fuel mixture. However, at higher speeds, the spark must occur earlier. The spark must be

given enough time to ignite the mixture. Without this spark advance, the piston would be up

over TDC and moving down before the combustion pressure reaches a maximum. As a result

the piston would keep ahead of the pressure rise. A weak power stroke would result. The

action of centrifugal-advance mechanism causes the contact points to open and close earlier.

Therefore, the ignition coil produces high voltage surge earlier and and the spark appear

earlier in the combustion chamber.

4.8.2 Limitations of conventional ignition system

The major limitations of the breaker-operated induction-coil system are the decrease in

available voltage as engine speed increases due to decreasing time available to build up the

primary coil stored energy. In addition to this, the breaker points are subjected electrical

and mechanical wear, which results in short maintenance interval.


70

4.8.3 Electronic ignition system

To eliminate the limitations of conventional ignition system, electronic ignition system is

used electronic ignition system is widely used. The basic difference between the contact-

point and electronic ignition system is in the primary circuit. The primary circuit in the

contact-point is opened and closed by contact points. In electronic system, the primary

circuit is opened and closed by electronic control unit (ECU). For example, the distributor

points and cam assembly of the conventional ignition system are replaced by a magnetic

pulse generating system which detects the distributor shaft position and sends electrical

pulses to an electronic control unit. The unit switches off the flow of current to the coil

primary windings, inducing the high voltage in the secondary windings which is distributed

to the spark plugs as in the conventional breaker system.

With the help of electronic ignition system, maintenance of ignition system is reduced; spark

plug life is extended and ignition of lean and dilute mixtures is improved.

4.8.4 Compression ignition engine

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.

Fuel is injected by the fuel-injection system into the engine cylinder toward the end of the

compression stroke, just before the desired start of combustion. The liquid fuel, usually

injected at high velocity as one or more jets through small orifices or nozzles in the injector

tip, atomizes into small drops and penetrates into the combustion chamber. The fuel

vaporizes and mixes with the high-temperature high-pressure cylinder air. Since the air

temperature and pressure are above the fuel's ignition point, spontaneous ignition of

portions of the already-mixed fuel and air occurs. The cylinder pressure increases as

combustion of the fuel-air mixture occurs. The consequent compression of the unburned

portion of the charge shortens the delay before ignition for the fuel and air which has mixed

to within combustible limits, which then bums rapidly. It also reduces the evaporation time

of the remaining liquid fuel. Injection continues until the desired amount of fuel has entered

the cylinder.

4.8.5 Types of diesel combustion system

Based on combustion chamber design, diesel engine is divided into two categories, direct

injection (DI) engine which have single open combustion chamber into which fuel is

injected directly; and indirect injection (IDI) engine in which combustion chamber is

divided into two regions and the fuel is injected into the “pre-chamber” which is connected

to main chamber via nozzle. IDI design engines are only used in small engine sizes.


71

4.8.6 Fuel injector characteristics

Fuel is introduced into the cylinder of the diesel engine through the nozzle in fuel injector.

There is very high pressure differential across the nozzle which enables the fuel to penetrate

through the pressurized air. The cylinder pressure at injection is generally in the range of 50

-100 atm. Fuel injection pressure is in the range of 200 – 1700 atm are used depending on

the engine size and type. These large pressure differences across the injector nozzle are

required so that injected liquid fuel jet will enter the chamber at sufficiently high velocity to

(i) atomize into small sized droplets to enable rapid evaporation and (ii) traverse the

combustion chamber in the time available and fully utilize the air charge. The task of fuel

injector is to meter appropriate quantity of fuel for the given speed and load to each cylinder

and inject fuel at the appropriate time. To accomplish this task, fuel is usually drawn from

the fuel tank by a supply pump, and forced through a filter to the injection pump. The

injection pump sends fuel under pressure to the nozzle pipes which carry fuel to the injector

nozzles located in each cylinder head. Excess fuel goes back to the fuel tank.

4.9 Engine cooling system

The purpose of the cooling system is to keep the engine at its most efficient operating

temperature at all speeds and under all operating conditions. During the combustion of the

air-fuel mixture in the cylinder, temperature of 2200°C or higher may be reached by the

burning gases. At such high temperature the metals will their characteristics and piston will

expand considerably and seize the liner. Of course theoretically thermal efficiency of the

engine will improve without cooling but actually the engine will seize to run. Cylinder wall

temperature must not go higher than about 200 to 260°C. Temperature higher than this

causes the lubricating oil film to break down and lose its lubricating properties. The cooling

system removes excess heat when the engine is hot and slowly or not at all when the engine

is cold or warming up. Generally, cooling systems are designed to remove about 30 to 35%

of heat produced in the combustion chamber by the burning of air-fuel mixture.

Two general types of cooling systems are used in heat engines: air cooled and liquid cooled.

4.9.1 Air cooled

In this method, heat is carried away by the air flowing over and around the engine cylinder.

Fins and flanges on the outer surface of the cylinder are used to increase the exposed area to

cooling air which raises the cooling rate. The common applications of air cooled engine are

motorcycles and airplanes. In this type of cooling, the cylinders are semi-independent. They

are not grouped in a block so the cylinder design is simpler and light. The main disadvantage

of air cooling is non uniform cooling of cylinder especially for multiple cylinders.

4.9.2 Liquid cooled

In the liquid cooling system, a liquid (generally water) is circulated around the cylinders to

absorb heat form the cylinder walls. The liquid water to which an antifreeze solution and

corrosion inhibitor is added is called coolant. The coolant absorbs heat as it passes through


72

the engine. Then the hot coolant is passed on to air that flowing through the radiator. The

cooled coolant is then passed to the engine. The circulation of coolant continuously removes

heat form the engine. The major components of liquid cooling are:

Figure 4-2 Liquid cooling in engine

Water Jacket: Water jacket is designed to keep the cylinder block and cylinder head cool.

The water jacket is open space between the outside wall of the cylinder and the inside of the

cylinder block and head.

Water pump: Water pump is impeller-type centrifugal pumps. The driving force for the

pump is taken from the engine shaft. Coolant from the radiator is drawn into the pump and

sent to the engine block.

Engine fan: The fan is driven by the same belt that drives pump. The purpose of fan is to

pull air through radiator. This improves cooling at slow speeds and idle. At higher speeds,

the air rammed through the radiator by the forward motion of the vehicle provides al the

cooling air that is needed.

Radiator: In the cooling system, the radiator is a heat exchanger that removes heat form

coolant passing through it. The radiator holds a large volume of coolant in close contact with

a large volume of air so that heat will transfer from the coolant to the air.


73

Figure 4-3 Radiator

Thermostat: Thermostat is placed in the coolant passage between the cylinder head and the

top of the radiator. Its purpose is to close off this passage when the engine is cold.

Radiator pressure cap: The cooling systems on engines today are sealed pressurized by a

radiator pressure cap. There are two advantages to sealing and pressurizing the cooling

system. First, the increased pressure raises the boiling point of the coolant. Second, sealing

the cooling system reduces coolant looses form evaporation.

Antifreeze solution: If water freezes in the engine cooling system, it stops coolant

circulation thus some part of the engine will overheat. On the other hand, water expands

while freezing. Water freezing in the cylinder block and head could expand enough to crack

the block or head. Water freezing in the radiator could split the radiator tubes. To prevent

freezing of the water in the cooling system, antifreeze is added to form the coolant. Some

antifreeze compounds plug small leaks in the cooling system.

Corrosion inhibitor: Corrosion shortens the life of metal parts. Also, corrosion forms an

insulating layer which reduces the amount of heat transferred from the metal part to the

coolant. So corrosion inhibitor is also added in anti-freeze solution.

4.10 Advantage of liquid cooling

• Compact design of engine

• The fuel consumption of high compression liquid cooled engine is rather than for air

cooled one.

• Uniform cooling


74

• Engine can be installed anywhere since cooling system can be conveniently located

wherever required where as air cooled engine requires it to be installed in the frontal

area.

• The size of the engine does not involve serious problem as far as design of cooling

system is concerned. In case of air cooled engines particularly in high power range

difficulty is encountered in circulation of required quantity of air for cooling purpose.

4.11 Disadvantage of liquid cooling

• This is dependent system in which supply of water for circulation in the jacket is

required.

• Power is absorbed by the pump and radiator fan.

• In the event of failure of cooling system serious damage may be caused to the engine.

• Cost of system is considerably high.

• Requires continuous maintenance.

4.12 Advantage of air cooling

• Simpler design due to absence of water jacket

• Simpler cooling system due to absence of pipes, fan and radiator assembly

• Easier installation

• No problem of freezing

4.13 Disadvantage of air cooling

• Non-uniform cooling

• The output of air cooled engine is less than that of a liquid cooled engine

• Smaller useful compression ratio

4.14 Engine lubricating system

Engine lubricating system supplies lubricating oil to all engine moving parts. The lubricant

and lubricating system function can be summarized as follow:

1. Protection of the engine against the wear.

2. Minimization of the frictional resistance of the engine to a minimum to ensure maximum

mechanical efficiency.

3. Cooling the piston and region of the engine where friction work is dissipated.

4. The clearance between bearings and rotating journals are filled with oil. When heavy

loads are suddenly imposed on the bearings, the oil helps to cushion the load. This

reduces the bearing wear.


75

5. Forming a gas tight seal between piston rings and cylinder walls.

6. Removing all impurities form lubricated regions.

4.14.1 Desired property of lubricants

Oxidation Stability: The lubricating oil subjected to atmospheric oxygen and high

temperature, through oxidation, contribute to deposit formation. These deposits would

eventually lead to ring sticking. Anti-oxidant and anti corrosive additives are used to control

these problem.

Detergency/Dispersion: The function of detergent additive is to reduce the amount of

deposits formed and make their removal. It loosens and detaches particles of carbon and

grit form engine parts and carry then down to the oil pan as oil circulates.

Wear Reduction: Wear is due to the individual and combined effects of corrosion, adhesive

and abrasion. Corrosive wear is effectively prevented by the use of detergent oils which

neutralize the corrosive acids as they form. Oils with anti-wear additives and low viscosity

at low temperature provide a remedy. Abrasion result form the presence of atmospheric

dust, and metallic derbies from corrosive and adhesive wear, in the lubricating oil. Efficient

air filtration is therefore most important. Elimination of abrasive particle impurities form

the oil system by filtration and periodic oil change is essential.

Viscosity: Oil viscosity is the most important parameter of lubricating oil. There are several

grade of viscosity oil. They are rated for winter, for other than winter and for both purpose.

For example SAE 20W is rated as winter use oil. SAE stands for Society for Automotive

Engineers and W stands for winter. For other than winter use, the rating is given as SAE20.

Higher the number higher is the viscosity of the oil. Now day multiple-viscosity oil is used.

For example multiple viscosity oil or multi grade oil is graded as SAE 10W-30. This means

that oil is same as SAE 10W when cold and SAE 30 when hot.

Sludge formation

Sludge is a thick, creamy, black substance that sometimes forms in the crankcase. It clogs oil

screens and oil lines, preventing oil circulation.

Water is collected in crankcase by combustion byproduct and the air. During cold operation

the moisture condensates into water and gets deposited into the oil. The crankcase mixes

the water, oil and carbon deposits continuously to form black thick substance called sludge.

If the engine is run for long time, the water gets evaporated and leaves through the crank

case ventilation but when the engine is used for short trip the water does not get enough

time to evaporate and hence sludge forms.

4.14.2 Low pressure splash lubricating system

The crankcase is used as the oil sump (reservoir) in a splash system, and the crankshaft

rotating at high speed in the oil distributes it to the various moving parts by splash; no oil

pump is used. All components, including the valve train and camshaft, must be open to the


76

crankcase. Oil is splashed into the cylinders behind the pistons acting both as a lubricant

and a coolant. Many small four-stroke cycle engines (lawn mowers, golf carts, etc.) use

splash distribution of oil.

4.14.3 High pressure distribution system

An engine with a pressurized oil distribution system uses an oil pump to supply lubrication

to the moving parts through passages built into the components. A typical automobile

engine has oil passages built into the connecting rods, valve stems, push rods, rocker arms,

valve seats, engine block, and many other moving components. These make up a circulation

network through which oil is distributed by the oil pump. In addition, oil is sprayed under

pressure onto the cylinder walls and onto the back of the piston crowns. Most automobiles

actually use dual distribution systems, relying on splash within the crankcase in addition to

the pressurized flow from the oil pump.


77

Chapter 5. Reciprocating air compressor

5.1 Reciprocating compression

The air compression process in reciprocating air compressor (clearance volume is neglected

in this case) involves the following steps (Fig. 5-1):

a. Induction of air

When piston moves from TBC to BDC, partial vacuum is created in the space above the

piston. The pressure at inlet manifold is at atmospheric level so the inlet valve is pushed by

higher pressure outside the valve. The pressure at delivery manifold is high because of the

high pressure in storage tank. The high delivery manifold pressure closes the delivery valve.

When the piston reaches BDC, the cylinder space is filled with air. It is assumed that the

induction of the air occurs at constant pressure.

TDC BDC

Piston moves from TDC

to BDC. Air enters the

cylinder at atm pressure

P

v

TDC BDC

The air is compressed

until the delivery valve

is open

P

v

TDC BDC

Once the pressure inside the cylinder

reaches the delivery pressure, delivery

valve gets open and compressed air is

delivered to the storage tank

P

v

P1 P1 P1

P2

Fig. 5-1 Working principle of reciprocating air compressor

b. Compression of air

The piston moves from BDC to TDC. The compression of the air inside the cylinder starts.

The volume decreases and the pressure increases. The increased pressure closes the inlet

valve but it is still not sufficient to open the delivery valve since delivery manifold pressure

is still higher. With the both valves closed, the air is compressed polytropically inside the

cylinder


78

c. Delivery of air

As the pressure inside the cylinder increases and reaches the delivery pressure, it is

sufficient to open the delivery valve. Now the piston forces the compressed air out of the

compressor to the storage tank.

5.2 Work done in compressor

P-v diagram of the compressor without clearance is shown in Fig. 5-2. We can express the

total work transfer during the air compressor cycle as the sum of work transfer in each

process.

Fig. 5-2 P-v diagram of reciprocating air compressor without clearance volume

Total work done in the cycle

��*� = �-$% + �%$, + �,$/ + �/$-

= (%!C% − C-" + M�?�$M�?�%$� + (,!C/ − C," + 0

Since, v4 and v3 = 0 for zero clearance volume

= (%C% + M�?�$M�?�%$� − (,C,

= !(,C, − (%C%" 4 %%$� − 15

= 4 �%$�5 !(,C, − (%C%"

��*� = 4 ;1 − ;5 (%C% 2(,C,(%C% − 13 Eq. 5.1

Since the process 1-2 is polytropic

(%C%� = (,C,�

C,C% = 2(%(,3%/�


79

C,C% = 2(,(%3$%/�

Substituting Eq. 5.1

��*� = 4 ;1 − ;5 (%C% �(,(% 2(,(%3$%/� − 1�

= 4 �%$�5 (%C% 94M�M�5!�$%"/� − 1:

��*� = − 4 ;; − 15 (%C% �2(,(%3!�$%"/� − 1� Eq. 5.2

We know that an air compressor is driven by energy from external means. The sign

convention that we adopt for work input in the system is negative. Therefore, the work

transfer is negative in the air compressor. However, our interest is only on the amount of

work/energy required in the compressor, we can eliminate the negative sign from our

equation. If we only consider the magnitude of the work, we can express the work required

for a reciprocating air compressor with zero clearance volume as,

� = 4 ;; − 15 (%C% �2(,(%3!�$%"/� − 1� Eq. 5.3

Where,

n = polytropic index of compression

P1 = Intake pressure of air (Suction pressure)

P2 = Delivery pressure of compressor

v1 = Volume of air drawn inside the compressor during suction (swept volume)

If swept volume is termed as vs, we can further express the equation for the work required

in compressor as:

� = 4 ;; − 15 (%C7 �2(,(%3!�$%"/� − 1� Eq. 5.4

OR

� = 4 ;; − 15 O�% �2(,(%3!�$%"/� − 1� Eq. 5.5

OR (in extensive term)

) = 4 ;; − 15 =O�% �2(,(%3!�$%"/� − 1� Eq. 5.6


80

5.3 Isothermal compression

We cannot use above equation when compression process is isothermal (n = 1). If the

compression process follows isothermal law (n = 1), work required for the compressor can

be expressed as

��*� = �-$% + �%$, + �,$/ + �/$-

= (%!C% − C-" + (%C% ln ?�?� + (,!C/ − C," + 0

= (%C% + (%C% ln ?�?� − (,C, (since P1v1 = P2v2)

= (%C% + (%C% ln C,C% − (%C%

= (%C% ln C,C%

= (%C% ln (%(,

= −(%C% ln (,(%

If only magnitude of the work required is considered

� = (%C% ln (,(% Eq. 5.7

OR

� = O�% ln (,(% Eq. 5.8

OR (in extensive term)

) = =O�% ln (,(% Eq. 5.9

5.4 Effect of polytropic index on work required

We have already discussed that the compression of air in a reciprocating air compressor is a

polytropic process. From the Eq. 5.3, we can say that the work required also depends on the

polytropic index. If the suction pressure, the delivery pressure and the swept volume are

same and only the value of polytropic index is different, we can have the following cases:

• n > γ (=1.4 for air)

• n = γ (Adiabatic compression)

• 1<n < γ

• n = 1(Isothermal compression)

We also know that inclination of curve Pvn = constant increases as the value of n increases.

The P-v diagram for above four cases will be as shown in Fig. 5-3.


81

Fig. 5-3 Effect of polytropic index

Process 1-2 for n > γ (=1.4 for air)

Process 1-2’ for n = γ (Adiabatic compression)

Process 1-2”for 1<n < γ

Process 1-2’’’ for n = 1(Isothermal compression)

We also know that area bounded by a cycle in P-v coordinate system represents the work

transfer. In Error! Reference source not found., the area bounded by the cycle is the work

required for the reciprocating air compressor. When we compare the area enclosed by the

cycle for different value of n, we come to know that work required is the least when the

compression is isothermal (area enclosed between 1 - 2’’’ – 3 – 4 - 1). As the value of n

increases, the area enclosed increases and thus the work required also increases. In actual

practice, it is impossible to achieve isothermal compression. But an attempt is made to

achieve the value of n as close to 1 as possible. To get the compression as close to

isothermal, the compressor is cooled by air or by any other cooling medium like water.

5.5 Compression process in T-s diagram 5.5 Compression process in T-s diagram

The four different compression processes can be shown in T-s diagram as shown in the

figure.

Fig. 5-4 Compression process in T-s diagram


82

In the Fig. 5-4 path -2’” represents isothermal compression (n = 1), path 1-2” represents

polytropic compression (1 < n < γ), path 1-2’ represents isentropic compression (n = γ) and

finally path 1-2 represents polytropic compression with polytropic index n> γ.

5.6 Clearance volume effects, P-v diagram and work done

So far, in the previous section we discussed about the air compressor without clearance

volume. In actual scenario, it is impossible to have zero clearance space. In this section, the

effects of clearance volume in operating parameters of reciprocating air compressor will be

discussed.

Fig. 5-5 Compression with clearance volume

As shown in the Fig. 5-5 the compressor has clearance volume vc above the TDC. The piston

cannot move past TDC and only moves between TDC and BDC. The volume contained

between TDC and BDC is called displacement volume (vd). Displacement volume is the

volume covered by the piston while moving between TDC and BDC. The fundamental

working principle of compressor with or without clearance volume is almost similar. In

former case, during the delivery of air (process 2-3) some amount of air is left in the

clearance volume since piston cannot move beyond TDC. The residual air in the clearance

volume has pressure P2 (delivery pressure). When the piston returns back to BDC from

TDC, the air left in clearance volume expands polytropically (process 3-4). During the

expansion process, the pressure inside the cylinder is higher than atmospheric pressure.

The higher inside pressure prevents the inlet valve from getting open. As the inlet value is

still closed, the air from atmosphere cannot be drawn inside the cylinder. When the

pressure finally drops to P1 (suction pressure), the inlet valve opens and air is drawn inside

the compressor. So suction of air only starts from point 4 and ends at point 1. Here we


83

should note that the displacement volume of the compressor is vd while the volume of air

drawn in the cylinder is vs (often called effective swept). Now we can conclude that when

clearance volume of the compressor is not zero, the volume of air drawn in the compressor

is less than the actual cylinder capacity of the compressor.

5.7 Work done in single stage compressor with clearance volume

The P-V diagram as shown in Fig. 5-5can be divided as in Fig. 5-2.

Fig. 5-6 P-v diagram of compressor with clearance volume

P-v diagram in Fig. 5-6 is same as that of compressor when clearance volume is neglected as

discussed in section 5.4.

So work done in compressor with clearance volume can be written as

w = w1 – w2

Where, w1 and w2 can be written in the form of Eq. 5.3

� = 4 ;; − 15 (%C% �2(,(%3!�$%"/� − 1� − 4 ;; − 15 (%C- �2(/(-3!�$%"/� − 1�

= 4 ;; − 15 (%C% �2(,(%3!�$%"/� − 1� − 4 ;; − 15 (%C- �2(,(%3!�$%"/� − 1�

= 4 ;; − 15 (%!C% − C-" �2(,(%3!�$%"/� − 1�

∵ C% − C- = C7 !�� C�=�"

� = 4 ;; − 15 (%C7 �2(,(%3!�$%"/� − 1� Eq. 5.10

Alternatively, the above expression can be derived as follow:

From the Fig. 5-6, work can be calculated as

� = �%$, + �,$/ + �/$- + �-$%

� = (,C, − (%C%1 − ; + (,!C/ − C," + (-C- − (/C/1 − ; + (%!C% − C-"


84

= (,C,1 − ; − (%C%1 − ; + (,C/ − (,C, + (-C-1 − ; − (/C/1 − ; + (%C% − (%C-

Since, P2= P3 and P1 = P4

= (,C, 2 11 − ; − 13 − (%C% 2 11 − ; − 13 + (,C/ + (%C-1 − ; − (,C/1 − ; − (%C-

= (,C, 4 ;1 − ;5 − (%C% 4 ;1 − ;5 + (,C/ 21 − 11 − ;3 + (%C- 2 11 − ; − 13

= (,C, 4 ;1 − ;5 − (%C% 4 ;1 − ;5 − (,C/ 4 ;1 − ;5 + (%C- 4 ;1 − ;5

= 4 ;1 − ;5 !(,C, − (%C% − (,C/ + (%C-"

= 4 ;1 − ;5 �(,!C, − C/" + (%!C% − C-"�

� = 4 ;1 − ;5 (%!C% − C-" 9(,!C, − C/" (%!C% − C-" − 1: Eq. 5.11

From the Fig. 5-6 we know that v1 - v4 = vs

And, for compression and expansion process (,(% = 2C%C,3� = 2C-C/3�

C%C, = C-C/

C,C/ = C%C-

C,C/ − 1 = C%C- − 1

C, − C/C/ = C% − C-C-

C, − C/C% − C- = C/C- = 2(%(,3%/�

C, − C/C% − C- = 2(,(%3$%/�

Substituting in Eq. 5.11

� = 4 ;1 − ;5 (%!C% − C-" �(,(% 2(,(%3$%/� − 1�

� = 4 ;1 − ;5 (%!C% − C-" �2(,(%3!�$%" �⁄ − 1�


85

� = 4 ;; − 15 (%C7 �2(,(%3!�$%"/� − 1�

Which is same as Eq. 5.10

If indices of compression and expansion are different, the equation for work required can be

written as

� = 4 ;; − 15 (%C% �2(,(%3!�$%"/� − 1� − 4 == − 15 (%C- �2(/(-3!z$%"/z − 1� Eq. 5.12

Where, m is index of expansion and n is index of compression

5.8 Volumetric and adiabatic efficiencies

For a compressor with clearance volume, the volume of air induced during the suction is

less than the displacement volume of the compressor. To establish the relation between

volume of cylinder and the actual volume of air coming in the compressor, volumetric

efficiency of the compressor is established. The volumetric efficiency of an air compressor is

defined as the ratio of volume of air induced in the cylinder to the displacement volume of

the cylinder.

Mathematically,

+��=��> �~~�>��;>r !�?�8" = C��=� �~ �� ;�>�� ; �ℎ� >r��;��>�=�;� C��=� �~ �ℎ� >��;��

Let us define another term called clearance factor (C) such that C = vc/vd

Since,

�?�8 = C7C� Eq. 5.13

�?�8 = C% − C-C�

�?�8 = !CD + C�" − C-C� Eq. 5.14

For polytropic expansion process 3-4

(-C-� = (/C/�

C-� = (/(- C/�

C- = C/ 2(/(-3% �⁄

C- = CD 2(,(%3% �⁄

Substituting in Eq. 5.14


86

�?�8 = !CD + C�" − CD 4(,(%5% �⁄C�

�?�8 = 1 + N − N 2(,(%3% �⁄ Eq. 5.15

Adiabatic efficiency of an air compressor is defined as

�� = 1 + N − N 2(,(%3% �⁄ Eq. 5.16

5.9 Effect of delivery pressure on the performance of air compressor

The performance of a reciprocating air compressor varies significantly with the increase in

delivery pressure. The effect of delivery pressure on the performance of an air compressor

can be explained with the help of Fig. 5-7

Let us take a compressor with clearance volume vc and displacement volume vd. The suction

pressure is P1. Initially the compressor is run to deliver the air at pressure P2. In this case,

we can draw a cycle as 1-2-3-4-1 (Fig. 5-7). The volume of air delivered for delivery

pressure P2 is vs1 (v1 – v4).

Fig. 5-7 Effect of delivery pressure


87

If we increase the delivery pressure to P3, the cycle will be 1-2’-3’-4’-1. Since clearance

volume is fixed for a given compressor size, the volume of air trapped in clearance volume

will be same even we increase the pressure. For the delivery pressure P3, the volume of air

delivered is vs2 (v1-v4’). If we further increase delivery pressure to P4, from the Fig. 5-7 we

can say that all the air is compressed in clearance space at the end of compression. As the air

left in clearance volume cannot get out of outlet valve, there is no delivery of air when

delivery pressure is P4. When the piston moves back to BDC, all the air left in clearance

volume expands. The expansion process follows same path followed by compression

process (1-2”-1). At the end of expansion (state 1), the air left in clearance volume occupies

the cylinder which prevents additional air to enter from atmosphere to the cylinder.

From above three cases, we can say that, for same intake condition, if delivery pressure is

increased, the volumetric efficiency of the compressor will be decreased. If we keep on

increasing the delivery pressure, at some point there will be a case where volumetric

efficiency of the compressor will be zero i.e., no air is delivered.

5.10 Multi-stage compression

Multistage compression is a serial arrangement of compressor cylinders in which

compressed air from the one cylinder is fed to the next cylinder for the further compression.

An intercooler is connected between two compressors, which cools the compressed air

coming from low pressure compressor and supplies to high pressure compressor.

Fig. 5-8 Two stage compression

5.10.1 Intercooling

Intercooler is a type of heat exchanger which is fitted between low pressure compressor and

high pressure compressor. It cools the compressed air delivered by low pressure

compressor. The cooled air is then passed to the high pressure compressor for the further


88

compression. The ideal intercooler (or perfect intercooler) should satisfy the flowing

condition.

• The air is cooled at constant pressure.

• The temperature of the cooled air is same as the temperature of the air at the start of

compression in low pressure compressor.

5.11 Representation of P-v and T-s diagram

The P-v and T-s diagram of double stage compressor fitted with perfect intercooler is shown

in the Fig. 5-9. The clearance volume is neglected in this case. The low pressure compressor

compress the air from P1 (state 1) to P2 (state 2). The compressed air is then cooled in

perfect intercooler from state 2 to state 2’. During the cooling process, the pressure is

constant at P2. The temperature of cooled air is equal to the temperature of the air at the

start of compression (state 1 and 2’ are in the isothermal curve). The air received from the

intercooler is further compressed to P3 (state 3) in the high pressure compressor.

P-v diagram of double stage compression

Compression process in T-s diagram

Fig. 5-9 P-v and T-s diagram of double stage compressor

From the P-v diagram show in the Fig. 5-9, we can say that 1-2-2’-3 is the path followed by

compression process in two stage air compressor fitted with perfect intercooler. If the

intercooler is not included, the path followed by the compressor process will be 1-2-3’. The

shift of compression from process 2-3’ to process 2’-3 is only due to the intercooling of air.

From the P-v diagram in the Fig. 5-9, we can say that due to the shift in the compression

process, work as represented by the shaded area is saved. In other word, with intercooler,

we need less work to compress the air compared to without intercooler. So, we can say that

the multistage compression process is efficient when intercoolers are used in between the

compressors.

The work required to compress in above case will be


89

� = ��M + � M

From Eq. 5.4, we have

� = ;; − 1 (%C% �2(,(%3�$%� − 1� + ;; − 1 (,C,L �2(/(,3�$%� − 1�

Since, 1 and 2’ state are in same isothermal line for a perfect intercooling

(%C% = (,C,K So we have,

� = ;; − 1 (%C% �2(,(%3�$%� − 1� + ;; − 1 (%C% �2(/(,3�$%� − 1�

� = ;; − 1 (%C% �2(,(%3�$%� + 2(/(,3�$%� − 2� Eq. 5.17

5.12 Optimum pressure distribution work done

In the previous section, we discussed about the two stage compressor where low pressure

compressor rises the pressure of air up to certain level and then high pressure compressor

further compresses the air up to delivery pressure. In the following section, we will discuss

about how the pressure should be distributed between the low pressure compressor and

the high pressure compressor so that minimum required work is achieved.

Total work required in two stage compressor with perfect intercooler is (from Eq. 5.17)

� = 4 ;; − 15 (%C% �2(,(%3!�$%"/� + 2(/(,3!�$%"/� − 2�

Since the induction pressure P1 and delivery pressure P3 are fixed, we can only change

intermediate pressure P2 to obtain the minimum work required. For the work to be

minimum, the first derivative of the expression of work with respect to P2 must be zero. ��(, = 0

��(, �4 ;; − 15 (%C% �2(,(%3!�$%"/� + 2(/(,3!�$%"/� − 2�� = 0

��(, �4 ;; − 15 (%C% �2(,(%3!�$%"/� + 2(,(/3!%$�"/� − 2�� = 0

2; − 1; 3 2 1(%3!�$%"/� !(,"!$%/�" + 21 − ;; 3 2 1(/3!%$�"/� !(,"!%$,�" �⁄ = 0


90

2; − 1; 3 2 1(%3!�$%"/� !(,"!$%/�" = − 21 − ;; 3 2 1(/3!%$�"/� !(,"!%$,�" �⁄

2; − 1; 3 2 1(%3!�$%"/� !(,"!$%/�" = 2; − 1; 3 2 1(/3!%$�"/� !(,"!%$,�" �⁄

2 1(%3!�$%"/� !(,"!$%/�" = 2 1(/3!%$�"/� !(,"!%$,�" �⁄

2 1(%3!�$%"/� !(,"!$%/�" = !(/"!�$%"/�!(,"!%$,�" �⁄

!(,"!$%/�"!(,"!%$,�" �⁄ = !(%(/"!�$%"/�

!(,",!�$%" �⁄ = !(%(/"!�$%"/�

(,, = (%(/

(/(, = (,(% Eq. 5.18

Eq. 5.18 is the condition for optimum pressure distribution in two stage compressor.

Substituting in Eq. 5.17, we have

� = 4 ;; − 15 (%C% �2(,(%3!�$%"/� + 2(/(,3!�$%"/� − 2�

= 4 ;; − 15 (%C% ��(%(/(% !�$%"/� + � (/�(%(/ !�$%"/� − 2�

= 4 ;; − 15 (%C% �2(/(%3!�$%"/,� + 2(/(%3!�$%"/,� − 2�

� = 2 2;; − 13 (%C% �2(/(%3!�$%"/,� − 1�

� = 2 2;; − 13 (%C7 �2(/(%3!�$%"/,� − 1� Eq. 5.19

This is the required expression for work required in two stage air compressor with perfect

intercooler and optimum pressure distribution.

For the pressure distribution in multistage (N-stage) compression process, Eq. 5.18 can be

written as

(¡(¡$% = … … … = (-(/ = (/(, = (,(% Eq. 5.20

Similarly work required in multi stage (N-stage) compressor is:


91

� = 2 �;; − 13 (%C7 �2(¡£%(% 3!�$%"/¡� − 1� Eq. 5.21

5.13 Positive displacement compressor

Positive displacement compressors are those compressors in which air is compressed by

being trapped in the reduced space formed by two sets of engaging surfaces. This type of

compressor deliver fixed volume of air at high pressure. Positive displacement machines

ensure positive admission and delivery preventing undesired reversal of flow. For example,

reciprocating compressor, root blower, rotary screw compressor

5.13.1 Axial flow compressor

An axial flow compressor is one in which the flow enters the compressor in an axial

direction (parallel with the axis of rotation), and exits in an axial direction. The axial-flow

compressor compresses its working fluid by first accelerating the fluid and then diffusing it

to obtain a pressure increase. The air is accelerated by a row of rotating airfoils (blades)

called the rotor, and then diffused in a row of stationary blades (the stator). The diffusion in

the stator converts the velocity increase gained in the rotor to a pressure increase. Axial

flow compressors produce a continuous flow of compressed air, and have the benefits of

high efficiencies and large mass flow capacity. Axial compressors are widely used in gas

turbines, such as jet engines, high speed ship engines, and small scale power stations. In an

axial flow compressor, air passes from one stage to the next, each stage raising the pressure

slightly. By producing low pressure increases on the order of 1.1:1 to 1.4:1, very high

efficiencies can be obtained. The use of multiple stages permits overall pressure increases

of up to 40:1 in some aerospace applications and a pressure ratio of 30:1 in some Industrial

applications.

Front part of axial compressor

Position of rotor and stator

Fig. 5-10 Axial flow compressor

5.13.2 Roots blower

The Roots type compressor (supercharger) or Roots blower is a positive displacement lobe

pump which operates by pumping air with a pair of meshing vanes (lobes). The roots type

compressor is two counter-rotating meshed lobed rotors. The two rotors trap air in the gaps

between rotors and push it against the compressor housing as they rotate towards the


92

outlet/discharge port. During each rotation, a specific fixed amount of air is trapped and

moved to the outlet port where it is compressed, which is why the roots type compressor

falls under the broader category of fixed-displacement superchargers. Roots blowers are

typically used in applications where a large volume of air must be moved across a relatively

small pressure differential. This includes low vacuum applications, with the Roots blower

acting alone, or use as part of a high vacuum system, in combination with other pumps. Two

lobe and three lobe type roots blower are shown in the Fig. 5-11.

Three lobe root blower

Two lobe root blower

Fig. 5-11 Root blower

5.13.3 Rotary compressor

A rotary compressor, or rotary vane compressor, is a device used for moving a fluid through

a system. A rotary vane compressor has a central, spinning rotor and a number of vanes. The

rotor assembly is encased inside of a sealed container with an inlet and outlet valve. The

vanes are spring-loaded so that they constantly push against the outside of the case, creating

a tight seal.

Fig. 5-12 Rotary compressor

The rotor is offset to one side. When a rotary blade spins past the inlet valve, it creates a

vacuum. The air flows out of the valve behind it, filling the vacuum. As it approaches the

outlet valve, the chamber shrinks. This creates more pressure on the air. The air has

nowhere to go but out of the outlet valve, so it shoots out of it. Then the vane continues on to

draw more air at the inlet valve. Common uses of vane pumps include high

pressure hydraulic pumps and automotive uses


93

Solved Examples

1. A single stage, single acting air compressor 30 cm bore and 40 cm stroke runs at 200

rpm. The suction pressure is 1 bar at 15°C and delivery pressure 5 bar. Neglecting the

clearance, determine the indicated mean effective pressure and the ideal power required

to run it, when

a. Compression is isothermal

b. Compression follows the law PV1.25=C

c. Compression is reversible adiabatic

d. Compression is irreversible adiabatic (n=1.5)

Mean effective pressure (mep) = work(w) / displacement volume(vd)

For single stage compressor with no clearance volume (vs = vd for zero clearance volume)

+7 = {|,}4 = { × 0.3, × 0.44 = 0.0283 =/ a. When the compression is isothermal

) = (%+7 ln (,(% = 1 × 10@ × 0.0283 × ln 5 × 10@1 × 10@

W =4.55 kJ

=�� = )+7 = 4.55 × 10/0.0283

mep = 1.609×105 Pa

(�� = ¤�� *D�D8* × ��=/60 = 4.55 × 10/ × 200/60

Power = 15.167 kW

b. When compression is polytropic with n = 1.25

) = 4 ��$%5 (%+7 �4M�M�5!¥¦�"¥ − 1�

= 2 1.251.25 − 13 1 × 10@ × 0.0283 ��5 × 10@1 × 10@ !%.,@$%"%.,@ − 1�


94

W = 5.373 kW

=�� = )+7 = 5.373 × 10/0.0283

mep = 1.898 ×105 Pa

(�� = ¤�� *D�D8* × ��=/60 = 5.373 × 10/ × 200/60

Power = 17.91 kW

c. When compression is reversible adiabatic ie, n = 1.4

) = 4 ��$%5 (%+7 �4M�M�5!¥¦�"¥ − 1�

= 2 1.41.4 − 13 1 × 10@ × 0.0283 ��5 × 10@1 × 10@ !%.-$%"%.- − 1�

W = 5.782 kW

=�� = )+7 = 5.782 × 10/0.0283

mep = 2.04×105 Pa

(�� = ¤�� *D�D8* × ��=/60 = 5.782 × 10/ × 200/60

Power = 19.27 kW

d. When the compression is irreversible adiabatic with n = 1.5

) = 4 ��$%5 (%+7 �4M�M�5!¥¦�"¥ − 1�

= 2 1.51.5 − 13 1 × 10@ × 0.0283 ��5 × 10@1 × 10@ !%.@$%"%.@ − 1�

W = 6.027 kW

=�� = )+7 = 6.027 × 10/0.0283

mep = 2.12×105 Pa

(�� = ¤�� *D�D8* × ��=/60 = 6.027 × 10/ × 200/60

Power = 20.09 kW


95

2. Determine the size of the cylinder for a double acting compressor of 37 kW, in which air

is drawn in at 1 bar and 15°C and compressed, according to law PV1.2=constant to 6 bar.

The compressor runs at 100 rpm with average speed of 152.5 m/min. Neglect clearance.

We can write the equation of power in terms of rate of work done as

)§ = 4 ��$%5 (%+§7 �4M�M�5!¥¦�"¥ − 1�

Since it is double acting compressor, work required for single side of the compressor is half

of 37 kW

Substituting the known values

18.5 × 10/ = 2 1.21.2 − 13 × 1 × 10@ × +§7 ��6 × 10@1 × 10@ !%.,$%"%., − 1�

+7§ = 18.5 × 10/208.80 × 10/ = 0.0885 =//��>

Swept Vs volume per cycle

+7 = 0.0885 !=/ ��>⁄ "100 60⁄ !>r>��/��>" = 0.053 =//>r>��

We also know that compressor (piston of the compressor) runs 2L (L=length of stroke)

distance in one revolution.

Distance covered by the compressor in N rpm = N×2L

Average velocity of the compressor = 2LN/60

Since the velocity of the compressor is 152.5 m/min (=2.542 m/sec)

2.542 = 2LN/60

L = 60×2.542/(2×100) = 0.7625 m

Length of stroke L = 76.25 cm


96

Swept volume per cycle Vs = πD2L/4

D2 = 0.053×4/(π×0.7625)

Bore D = 29.74 cm

3. A single stage, double acting air compressor is required to deliver 14 m3 of air per

minute measured at 1.013 bar and 15°C. The delivery pressure is 7 bar and the speed

300 rpm. Take the clearance volume as 5% of the displacement volume with the

compression and expansion index of n=1.3. Calculate:

a. Displacement volume

b. The delivery temperature

c. Indicated Power

d. Volumetric efficiency

If only single side of the compressor is considered, the delivery volume (swept volume) is

half of that of double acting compressor. So,

vs = 14/2 = 7 m3/min

The clearance volume is 5% of displacement volume

vc/vd = 0.05

Clearance factor (C) = vc/vd = 0.05

Volumetric efficiency (ηvol) = 1 + C - C(P2/P1)1/n

ηvol = 1 + 0.05 – 0.05(7/1.013)1/1.3

ηvol = 82.9%

We also know that

ηvol = vs/vd

vd = vs/ ηvol = 7(m3/min)/0.829

vd = 8.444 m3/min

Since the compressor runs at 300 rpm (or cycle/min), displacement volume per cycle can be

written as

C� = 8.444 � =/=�; × 1300 2 =�;>r>��3 vd = 0.028 m3

Delivery temperature

For process 1-2

�%(%!%$�" �⁄ = �,(,!%$�" �⁄


97

�, = �% 4M�M�5!%$�" �⁄ = 288 4%.y%/¨ 5$y./ %./⁄ = 550 K

T2 = 550 K

Indicated power of the compressor

)§ = 4 ;; − 15 (%+§7 �2(,(%3!�$%"� − 1�

= 2 1.31.3 − 13 1.013 × 10@ × 1460 �2 71.0133y./%./ − 1�

=57.58 kW

IP = 57.58 kW

4. Air at 103 kPa and 27°C is drawn in low pressure cylinder of two stage air compressor

and is isentropically compressed to 700 kPa. The air is then cooled at constant pressure

to 37°C in an intercooler and is then again compressed isentropically to 4 MPa in the

high pressure cylinder, and is delivered at this pressure. Determine the power required

to run the compressor if it has to deliver 30 m3 of air per hour measured at inlet

conditions. [4.194 kW]

The volume of air delivered at inlet condition (P =103 kPa and T = 27°C) is 30 m3 per hour.

For the simplification of the calculation, the volume of air is changed into mass that is same

at any condition.

=�� ~�� ~ �ℎ� �� !=§ " = (%+%O�% = 103 × 10/ × 30287 × 300 = 35.889 �h/ℎ�

Power required in low pressure compressor

)§ � = 2 1.41.4 − 13 35.8893600 × 287 × 300 × �27001033!%.-$%"%.- − 1� = 2.2 �)

Power required in high pressure compressor

)§ = 2 1.41.4 − 13 35.8893600 × 287 × 310 × �24000700 3!%.-$%"%.- − 1� = 2.2 �)


98

5. A single acting two stage compressor with complete inter cooling delivers 6 kg/min of

air at 16 bar. Assuming an intake at 1 bar and 150C and compression and expansion with

law PV1.3=constant. Calculate:

a. Power required to run the compressor at 420 rpm

b. Isothermal efficiency

c. Free air delivered per sec

d. If clearance ratios for LP and HP cylinder are 0.04 and 0.06, calculate

volumetric efficiency and swept volume for each cylinder.

e. Net heat transferred in LP and HP cylinder during compression and also in the

intercooler


99

Chapter 6. Fuels and Combustion

6.1 Introduction

The fuel can be defined as a substance, which can deliver heat energy. Some fuels like

petroleum releases energy due to combustion in presence of oxygen while other fuels such

as uranium release heat because of the nuclear fission. In this chapter, we will discuss only

the fuels that liberate energy because of oxidation process.

6.2 Classification of fuels

Most fuel which is used for combustion fall into three categories-Solid fuel, Liquid fuel and

Gaseous fuel.

6.2.1 Solid fuel

The common solid fuels are wood, peat, lignite, coal, anthracite, coke and briquettes. The

advantage of solid fuel over other fuel is that it is easily available.

6.2.2 Liquid fuel

The common liquid fuels are liquid petroleum like Petrol, Diesel and Kerosene. There are

some other liquid fuels such as ethanol and biodiesel.

6.2.3 Gaseous fuel

6.3 Calorific and heating values of fuels

Calorific value of a fuel is the amount energy produced when 1 kg of fuel is burned

completely.

a. Higher calorific value

It is the total heat released in kJ/kg or kJ/m3 of a fuel.

6.4 Determination of caloric values

6.5 Combustion equation for hydrocarbon fuels

When hydrocarbon fuel is burned in presence of oxygen, theoretically it is converted into

CO2 and H2O.

For Hydrogen

2H2 + O2 ⇒ 2H2O

Molecular wt 4 32 36

4 kg of hydrogen needs 32 kg of oxygen for the combustion

1 kg of hydrogen needs 8 kg of oxygen

For Carbon


100

The complete combustion of carbon results in the emission of CO2 while partial combustion

of carbon results in the emission of CO

For complete combustion

C + O2 ⇒ CO2


12 kg of carbon needs 32 kg of oxygen for the combustion

1 kg of carbon needs 8/3 kg of oxygen

For partial combustion

2C + O2 ⇒ 2CO


24 kg of carbon needs 32 kg of oxygen for the partial combustion

1 kg of carbon needs 4/3 kg of oxygen

For Carbon monoxide

2CO + O2 ⇒ 2CO2


56 kg of carbon monoxide needs 32 kg of oxygen for complete combustion

1 kg of carbon monoxide needs 4/7 kg of oxygen

For Sulphur

2S + O2 ⇒ SO2


32 kg of sulphur needs 32 kg of oxygen for the combustion

1 kg of sulphur needs 1 kg of oxygen

For Methane

CH4 + 2O2 ⇒ CO2 + 2H2O

Molecular wt 16 64 44 36

16 kg of methane needs 64 kg of oxygen for complete combustion

1 kg of methane needs 4 kg of oxygen for complete combustion


101

The result of above combustion equation is summarized in the table below

Constituent Mass of oxygen per kg of

constituent By product and its mass

Hydrogen 8 kg H2O (36/4 = 9 kg)

Carbon (Complete

combustion) 8/3 kg CO2 (44/12 = 11/3kg)

Carbon (Partial combustion) 4/3 kg CO (56/24 = 7/3kg)

Carbon monoxide 4/7 kg CO2 (88/56 = 11/7 kg)

Sulfur 1 kg SO2 (64/32 = 2 kg)

Methane 4 kg CO2 (44/16 =11/4 kg)

H2O (36/16 = 9/4 kg)

6.5.1 Minimum air required

In the following section, we will discuss about the amount of air required for theoretical

combustion of a fuel. Theoretical combustion of fuel is the oxidation of fuel as represented

by the chemical equation. The theoretical air required is also called stoichiometric quantity

of air. For the calculation, we adopt the composition of air as follow:

By weight (Nitrogen 77% and Oxygen 23%)

By volume (Nitrogen 71% and Oxygen 29%)

If we calculate the amount of air required with the help of chemical equation, we can easily

calculate the amount of air required.

Example

The analysis of coal used in boiler trial is as follows: C-81%, H2-4%, O2-4%, S-2.5% and

remainder incombustible. If 80% of the carbon is burned to CO2 and remainder into CO,

determine the minimum air required for combustion of 1 kg of coal.

Constituent Mass per kg of fuel (kg) O2 required (kg)

Carbon (complete combustion) 0.8×0.81 = 0.648 0.648×8/3 = 1.728

Carbon (incomplete combustion) 0.2×0.81 = 0.162 0.162×4/3 = 0.216

Hydrogen 0.04 0.04×8 = 0.32

Sulphur 0.025 0.025×1= 0.025

Total oxygen required = 2.289

The mass of oxygen present in fuel = 0.04

Net oxygen required = 2.249

So minimum air required = 2.249×100/23 = 9.778 kg


102

6.5.2 Excess air

The analysis of coal used in boiler trial is as follows: C-82%, H2-5%, O2-3%, S-3.5% and

remainder incombustible. Determine the minimum air required for combustion of 1 kg of

coal. Also find the volumetric composition of exhaust gas if 30% excess air is supplied.

Constituent Mass per kg of fuel (kg) O2 required (kg) Byproduct (kg)

Carbon 0.82 0.082×8/3 = 2.87 CO2 = 0.82×11/3 = 3.01

Hydrogen 0.05 0.05×8 = 0.40 H2O = 0.05×9 = 0.45

Sulphur 0.035 0.035×1= 0.035 SO2 = 0.035×2 = 0.07

Total oxygen required = 3.62

The mass of oxygen present in fuel = 0.03 (subtracted)

Net oxygen required = 3.59

Minimum air required = 3.59×100/23 = 15.61 kg

Excess air supplied = 15.61×30% = 4.68 kg

Total air supplied = 15.61 + 4.68 = 20.29

Byproduct Mass (kg) (A) Mol wt (B) C = A/B % by volume

= C/ ƩC

CO2 3.01 44 0.0684 10.35%

SO2 0.07 64 0.0011 0.17%

N2 (from total air

supplied)

20.29×77% =

15.62 28 0.5578 84.38%

O2 (from excess

air supplied) 4.68×23% = 1.08 32 0.0337 5.10%

ƩC = 0.661

6.5.3 Stoichiometric air requirement

Calculate the stoichiometric air for the complete combustion of 1 kg of LPG. Assume the

composition of the LPG as 70% butane and 30% propane.

Chemical equation for oxidation of butane

C4H10 + 6.5O2 ⇒ 4CO2 + 5H2O



103

Chemical equation for oxidation of propane

C3H8 + 5O2 ⇒ 3CO2 + 4H2O


From the equation we have,

58 kg of butane requires 208 kg of oxygen for complete combustion

i.e. 0.7 kg of butane requires ,y©@© × 0.7 = 2.51 kg of oxygen

44 kg of propane requires 160 kg of oxygen for complete combustion

i.e. 0.3 kg of propane requires %fy-- × 0.3 = 1.09 kg of oxygen

So, for 1 kg of LPG (2.51+1.09) requires 3.6 kg of oxygen

Amount of air required= /.f,/ × 100 = 15.65 kg

6.6 Combustion in SI and CI engine

6.6.1 Normal combustion in SI engine

Under normal operating conditions, combustion in SI engine is initiated towards the end of

the compression stroke at the spark plug by an electric discharge. Following inflammation, a

turbulent flame develops, propagates through this essentially premixed fuel, air, burned gas

mixture until it reaches the combustion chamber walls, and then extinguishes. The

combustion process of SI engines can be divided into three broad regions: (1) ignition and

flame development, (2) flame propagation, and (3) flame termination. Flame development is

generally considered the consumption of about the first 5% of the air-fuel mixture. During

the flame development period, ignition occurs and the combustion process starts, but very

little pressure rise is noticeable and little or no useful work is produced. Just about all useful

work produced in an engine cycle is the result of the flame propagation period of the

combustion process. This is the period when the bulk of the fuel and air mass is burned.

During this time, pressure in the cylinder is greatly increased, and this provides the force to

produce work in the expansion stroke. Nearly the final 5% of the air-fuel mass which burns

is classified as flame termination. During this time, pressure quickly decreases and

combustion stops.

6.6.2 Normal combustion in CI engine

Combustion in a compression ignition engine is quite different from that in an SI engine.

Engine torque and power output controlled by the amount of fuel injected per cycle. Fuel is

injected into the cylinders late in the compression stroke by one or more injectors located in

each cylinder combustion chamber. In addition to the swirl and turbulence of the air, a high

injection velocity is needed to spread the fuel throughout the cylinder and cause it to mix

with the air. After injection, the fuel must go through a series of events to assure the proper

combustion process:


104

1. Atomization: Fuel drops break into very small droplets. The smaller the original drop

size emitted by the injector, the quicker and more efficient will be this atomization

process.

2. Vaporization: The small droplets of liquid fuel evaporate to vapor. This occurs very

quickly due to the hot air temperatures created by the high compression of CI engines.

High air temperature needed for this vaporization process requires a minimum

compression ratio in CI engines.

3. Mixing: After vaporization, the fuel vapor must mix with air to form a mixture within the

AF range which is combustible. This mixing comes about because of the high fuel

injection velocity added to the swirl and turbulence in the cylinder air.

4. Self-Ignition: After the mixing the air-fuel mixture starts to self-ignite. Actual combustion

is preceded by secondary reactions, including breakdown of large hydrocarbon

molecules into smaller species and some oxidation. These reactions, caused by the high-

temperature air, are exothermic and further raise the air temperature in the immediate

local vicinity. This finally leads to an actual sustained combustion process.

5. Combustion: Combustion starts from self-ignition simultaneously at many locations.

When combustion starts, multiple flame fronts spreading from the many self-ignition

sites quickly consume all the gas mixture which is in a correct combustible air-fuel ratio,

even where self-ignition wouldn't occur. This gives a very quick rise in temperature and

pressure within the cylinder. The higher temperature and pressure reduce the

vaporization time and ignition delay time for additional fuel particles and cause more

self-ignition points to further increase the combustion process.

6.7 Abnormal combustion in SI engine

Fuel composition, certain engine design operating parameters and combustion chamber

deposits-may prevent the normal combustion process from occurring. The abnormal

combustion in the engine is harmful since it can harm the engine and produces undesired

noise while engine in operation. Two major types of abnormal combustion are: knock and

surface ignition.

6.7.1 Surface-ignition

Surface ignition is ignition of the fuel/air mixture by hot spot on the combustion chamber

walls such as an overheated valve or spark plug, or glowing carbonaceous deposits in

combustion chamber, i.e., any means other than the normal spark discharge. It can occur

before the occurrence of the spark (pre-ignition) or after (post-ignition). Between pre-

ignition and post-ignition phenomenon, pre-ignition is potentially the most damaging. Any

process that advances the normal start of combustion will cause higher heat rejection

because of increasing burned gas pressure and temperature that result. Higher heat

rejection causes high temperature components which, in turn, can advance the pre-ignition

point even further until critical components fail. Causes of pre-ignition include the following:


105

• Carbon deposits, an overheated spark plug. Glowing carbon deposits on a hot exhaust

valve

• A sharp edge in the combustion chamber or on top of a piston and valves.

• A lean fuel mixture.

• Ineffective cooling system.

6.7.2 Knocking (Detonation)

Knocking is the noise associated with auto ignition of a portion of the fuel-air mixture ahead

of the advancing flame front. Auto ignition is the spontaneous ignition and the resulting very

rapid reaction of a portion or all of the fuel-air mixture. It is generally agreed that knock

originates in the extremely rapid release of much of the energy contained in the end-gas

ahead of the propagating turbulent flame, resulting high local pressure. Knocking is

governed by different operating variables like temperature and pressure of combustion

chamber. Knock primarily occurs under wide open throttle operating conditions. It also

constrains engine efficiency by limiting temperature and pressure of the end gas; it limits

the engine compression ratio. The occurrence and severity of knock depend on the knock

resistance of the fuel and on the antiknock characteristics of the engine. The ability of a fuel

to resist knock is measured by its octane number. Higher the octane number greater is the

resistance to knock. The octane number (ON) scale is based on two hydrocarbons which

define the ends of the scale. By definition, normal heptane (n-C7H16) has a value of zero and

isooctane (C8,H16) has an octane number of 100.

6.8 Abnormal combustion in CI engine

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. The period between the start of fuel injection into the combustion chamber and the

start of combustion is called ignition delay. At normal engine conditions (low to medium

speed, full warmed engine) the minimum delay occurs with the start of injection at about 1

to 15° BTDC. The increase in the delay with earlier or later injection timing occurs because

the air temperature and pressure change significantly close to TDC. If injection starts earlier,

the initial air temperature and pressure are lower so delay will increase. If injection starts

later (closer to TC) the temperature pressure are initially slightly higher but then decrease

as the delay proceeds. The most favorable conditions for ignition lie in between.

The property that quantifies Ignition Delay 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.

Like octane number rating, cetane numbers are established by comparing the test fuel to

two standard reference fuels. The fuel component n-cetane, (C16H34), is given the cetane

number value of 100, while heptamethylnonane (C12H34), is given the value of 15. The

cetane number (CN) of other fuels is then obtained by comparing the ID of that fuel to the ID

of a mixture

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