Internal Combustion Engines
Dale R. Tree
Brigham Young University
Permission for use
This packet has been assembled for students in MeEn 425, Internal Combustion Engines,
at Brigham Young University. It is for educational purposes only. This material should not
be reproduced or distributed for any purpose outside the context of this course.
Fall 2024
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Table of Contents
1.
2.
Introduction ................................................................................................................. 5
Common Engine Nomenclature .................................................................................. 7
2.1. The Five Processes of and Internal Combustion Engine...................................... 7
2.2. Piston-Cylinder Nomenclature ............................................................................. 9
2.3. Engine Load ....................................................................................................... 10
2.4. Common Reciprocating Engine Parts ................................................................ 11
2.5. Summary ............................................................................................................ 16
3. Engine Classification ................................................................................................ 19
3.1. Number of Strokes ............................................................................................. 19
3.2. The Rotary or Wankel Engine............................................................................ 20
3.3. Fuel Type............................................................................................................ 21
3.4. Ignition source .................................................................................................... 22
3.5. Combustion Processes ........................................................................................ 22
3.6. Engine Cooling................................................................................................... 23
3.7. Piston Configuration .......................................................................................... 23
3.8. Air Delivery........................................................................................................ 24
3.9. Fuel Delivery ...................................................................................................... 26
4. Engine Performance Parameters ............................................................................... 29
4.1. Indicated Work ................................................................................................... 29
4.2. Engine Load ....................................................................................................... 31
4.3. Brake Work ........................................................................................................ 32
4.4. Power.................................................................................................................. 32
4.5. Mean Effective Pressure..................................................................................... 33
4.6. Indicated or Brake – Single or Multi-cylinder ................................................... 33
4.7. Mechanical Efficiency........................................................................................ 34
4.8. Volumetric Efficiency ........................................................................................ 35
4.9. Thermodynamic and Fuel Conversion Efficiencies ........................................... 37
4.10.
Specific Fuel Consumption ............................................................................ 38
4.11.
Piston Cylinder Geometry Motion ................................................................. 39
4.12.
The Power Equation ....................................................................................... 41
4.13.
Fuel - Air Ratio ............................................................................................... 43
4.14.
Indicated and Brake, Gross and Net - Revisited ............................................. 44
5. Engine Models .......................................................................................................... 49
5.1. Thermodynamics Review................................................................................... 49
5.1.1. Conservation of Energy .............................................................................. 49
5.1.2. Entropy Equation ........................................................................................ 50
5.1.3. Ideal Gas Law ............................................................................................. 50
5.1.4. Mixtures ...................................................................................................... 50
5.1.5. Ideal Mixtures ............................................................................................. 51
5.1.6. Ideal Gas Properties .................................................................................... 51
5.2. Overview of a real combustion cycle ................................................................. 51
5.3. Ideal Cycles ........................................................................................................ 53
5.3.1. Ideal Otto Cycle .......................................................................................... 54
5.3.2. Ideal Diesel Cycle ....................................................................................... 55
5.4. Ideal Cycle with Variable Heat Release ............................................................. 57
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5.4.1. Modeling a variable heat release................................................................. 57
5.5. The Four Stoke Cycle ......................................................................................... 58
5.5.1. Closed Portion of the Cycle ........................................................................ 58
5.5.2. Blow Down ................................................................................................. 58
5.5.3. The Exhaust Processes ................................................................................ 59
5.5.4. The intake Process ...................................................................................... 59
5.5.5. Residual Mass Calculation.......................................................................... 61
5.6. The Miller Cycle ................................................................................................ 67
6. Combustion ............................................................................................................... 71
6.1. Combustion Processes ........................................................................................ 71
6.1.1. Premixed Deflagration Wave ...................................................................... 71
6.1.2. Premixed Detonation Wave ........................................................................ 72
6.1.3. Mixing Controlled Jet ................................................................................. 72
6.1.4. Volumetric Reaction ................................................................................... 73
6.1.5. Multi-Mode Combustion ............................................................................ 73
6.2. Combustion Thermodynamics ........................................................................... 74
6.2.1. The Enthalpy of Formation ......................................................................... 74
6.2.2. Heat of Combustion .................................................................................... 76
6.2.3. Heat of Combustion for fuel containing only hydrogen and carbon .......... 77
6.2.4. Enthalpy of Formation for Hydrocarbon Fuels ........................................... 79
6.2.5. Finding Adiabatic Flame Temperature ....................................................... 80
6.2.6. Constant Volume Combustion Reactions ................................................... 82
6.2.7. Thermodynamic Equilibrium ...................................................................... 83
6.2.8. Equilibrium Calculation .............................................................................. 87
6.2.9. The Water-Gas Shift Reaction .................................................................... 88
6.2.10.
Equilibrium with Multiple Species ......................................................... 89
6.3. Zero Dimensional Thermodynamic Engine Models .......................................... 89
7. Engine Losses, Friction, Accessories and Pump Work ............................................ 95
7.1. Accessory Work ................................................................................................. 95
7.2. Friction ............................................................................................................... 95
7.2.1. Modes .......................................................................................................... 95
7.3. Pumping Mean Effective Pressure ..................................................................... 97
7.4. Hydrodynamic Bearing Friction......................................................................... 98
7.5. Piston Friction and Forces .................................................................................. 99
8. Air, Fuel and Exhaust Flow .................................................................................... 101
8.1. Flow through an Orifice ................................................................................... 101
8.2. Factors Influencing Volumetric Efficiency ...................................................... 103
8.3. Evaporative Cooling During Intake ................................................................. 105
8.4. Two Stroke Breathing Parameters.................................................................... 107
8.5. Fuel Flow.......................................................................................................... 109
8.5.1. Carburation ............................................................................................... 109
8.5.2. Port Fuel Injection..................................................................................... 111
8.5.3. Direct Fuel Injection ................................................................................. 112
9. Engine Measurements and Controls ....................................................................... 117
9.1. Dynamometers ................................................................................................. 117
9.1.1. Water Brake .............................................................................................. 117
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9.1.2. Eddy Current ............................................................................................. 117
9.1.3. Motoring AC or DC .................................................................................. 118
9.2. Fuel System ...................................................................................................... 118
9.2.1. Volumetric Displacement ......................................................................... 118
9.2.2. Pressure and Orifice .................................................................................. 118
9.2.3. Coriolis Flow Meter .................................................................................. 119
9.2.4. Gravimetric Device ................................................................................... 119
9.3. Air Flow Measurement..................................................................................... 119
9.3.1. Hot Wire or thin film anemometer............................................................ 119
9.3.2. Turbine Flow Meter .................................................................................. 119
9.3.3. Calibrated Orifice...................................................................................... 119
9.4. Exhaust Gas Measurement and Analysis ......................................................... 120
9.4.1. Non dispersive Infrared (NDIR) absorption, CO2 CO and NO ................ 120
9.4.2. Unburned Hydrocarbon (FID) .................................................................. 121
9.4.3. Chemiluminescence .................................................................................. 122
9.5. Calculating Equivalence Ratio from Exhaust Gas ........................................... 122
9.6. In-Cylinder Pressure Measurement .................................................................. 126
9.6.1. Calculating Heat Release Rate from Cylinder Pressure ........................... 126
9.7. Engine Sensors and Actuators .......................................................................... 128
9.8. Engine Control ................................................................................................. 128
10.
Spark Ignition Combustion .................................................................................. 131
10.1.
SI Combustion Processes.............................................................................. 131
10.1.1.
Spark Ignition ........................................................................................ 131
10.1.2.
Early Flame Development..................................................................... 131
10.1.3.
Flame Propagation................................................................................. 133
10.1.4.
Flame Termination ................................................................................ 134
10.2.
Two Zone Model of SI Combustion ............................................................. 135
10.3.
Temperature Distribution in a SI Engine ...................................................... 139
10.4.
Burn Duration and Location ......................................................................... 140
10.4.1.
Parameters Influencing Burn Duration ................................................. 142
10.5.
Abnormal Combustion ................................................................................. 145
10.5.1.
Lean Misfire .......................................................................................... 145
10.5.2.
Partial Combustion ................................................................................ 145
10.5.3.
Fuel Rich Combustion........................................................................... 146
10.5.4.
Surface Ignition ..................................................................................... 146
10.5.5.
Knock .................................................................................................... 146
10.5.6.
Fuel Considerations for Knock ............................................................. 148
10.6.
Pollutant Formation for SI Engines .............................................................. 149
10.6.1.
Emission Standards ............................................................................... 149
11.
Compression Ignition Combustion and Emissions .............................................. 153
11.1.
Conceptual Model of Compression Ignition Combustion ............................ 153
11.2.
Stages of Compression Ignition Combustion ............................................... 154
11.2.1.
Compression Ignition Emissions........................................................... 155
11.2.2.
NOx Particulate Trad-off Curve............................................................ 157
Appendix A Thermodynamic Data ................................................................................ 161
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1. Introduction
Cars, trucks, motorcycles, generators, lawn mowers, grass trimmers, chain saws and many
other devices used or experienced in everyday life are powered by internal combustion
engines. The objective of this book is to study the fundamental processes and principles
occurring in these engines in a way that enables an understanding of their operating
principles and design considerations while reviewing fundamental concepts of Mechanical
Engineering. Turbine engines or external combustion engines such as a steam engine or
Sterling cycle engine are not covered.
When a fuel is burned, the bonds between molecules in the fuel (for example gasoline) and
air are rearranged producing bonds between combustion products (mostly water and carbon
dioxide). The energy level of the products is much lower than the energy level of the
reactants producing a release of energy known as a combustion process. The energy
released is transformed from bond energy to rotational, translational, and vibrational
energy in the product molecules. When confined, this higher energy results in increased
pressure. Combustion is therefore a process whereby relatively slowly moving molecules
release their chemical energy resulting in highly disordered molecular motion. An internal
combustion engine converts this random molecular motion into a highly ordered motion of
molecules (a piston) moving in the same direction capable of mechanical work.
The process of conversion of chemical energy to work is illustrated in Figure 1.1 where
fuel and air represented by oxygen (O2) and nitrogen (N2) is converted to products of CO2
and H2O during the combustion process. Before combustion the fuel and air have chemical
energy but their molecular motion is low. After combustion the molecules have high
motion energy in random directions as illustrated. The molecular motion can be measured
as increased temperature and pressure. When molecules impact the piston surface and
exchange momentum, the piston moves downward and thereby the random molecular
motion is converted to bulk motion of the piston mass all moving in the same direction.
This produces a force over a distance or work.
N2
O2
Fuel
N2
O2
Combustion
H2O
H2O
Fuel
N2
Fuel
O2
Fuel
N2
N2
N2
N2
N2
N2
N2
N2
CO2
H2O
N2
CO2
H2O
N2
CO2
N2
H2O
N2
Figure 1.1 Diagram of the energy conversion process in an internal combustion process.
Cummins [1] suggests that the internal combustion engine began with the work of Christian
Huygens (1629-95) when he designed a device that would lift a ball in a cylinder using gun
powder. After lifting the ball, it could be captured and then tied to a pulley or weight to
perform work. It wasn’t until 1876 that Nicholas Otto’s four stroke engine was designed
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and built to conversion of fuel energy directly to work in way that was commercial
successful [1].
Today we have numerous internal combustion engines employing various types of
hardware, but all internal combustion engines achieve the following processes:
1. Fuel and air are delivered by some means into the combustion chamber.
2. A compression process of some working fluid such as air or a mixture of fuel and air
proceeds combustion
3. The fuel is burned converting chemical energy into sensible energy.
4. Energy is extracted from hot gases by an expansion process.
5. Product gases are removed from the combustion chamber in order to provide space for
fresh fuel and air delivery.
We begin the study of internal combustion engines by learning the parts, processes, and
performance measures associated with the most common internal combustion engine
currently in use today; a spark ignition, port fuel-injected, naturally-aspirated engine. A
brief overview of numerous engine designs will then follow to provide a vocabulary for
understanding engine designs and components.
Following this overview: The
thermodynamic, fluid mechanics, solid mechanics, and combustion processes of spark
ignition and Diesel engines will be considered in some detail. The final topic will be
pollutant formation and reduction.
Chapter 1 Homework
References
1. Cummins, Lyle, “Internal Fire,” Society of Automotive Engineers, 1989.
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2. Common Engine Nomenclature
2.1. The Five Processes of and Internal Combustion Engine
There are five processes present in all internal combustion engines including the fourstroke, spark-ignition, naturally aspirated engine and a direct injection Diesel engine. The
five processes are illustrated in Figure 2.1 and include:
1. Intake
2. Compression
3. Combustion
4. Expansion
5. Exhaust
Figure 2.1 Four of the five processes occurring in an internal combustion engine. The
combustion process occurs at the end of compression (b) and start of expansion (c). [1]
Intake is the process whereby air, or a mixture of fuel and air is added to the combustion
chamber. It begins when a valve opens (top left valve in (a)) and the piston moves
downward pulling air into the cylinder. Intake ends when the intake valve closes or when
the piston moves upward and flow changes direction out of the cylinder.
The compression process begins when the intake valve is closed and the piston moves
upward. The pressure of the trapped fuel and air (charge gas) increases as the specific
volume of the gas mixture decreases. This process requires external work to be done on the
fuel-air mixture.
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Combustion occurs near the end of compression and during the start of expansion. The
figure does not show the combustion process except for the lines extending from the spark
plug (Figure 2.1b). During combustion, the chemical energy in the fuel is converted to
sensible or thermal energy in the product gas resulting in a rapid increase in the temperature
and pressure within the combustion chamber.
The expansion process shown in Figure 2.1c begins when the piston is at the top of its
travel even if combustion is not complete. During expansion the specific volume of the
gases in the cylinder increases resulting in a decrease in pressure and temperature. The
thermal energy of the gas decreases as it is converted to the mechanical energy of piston
motion.
The final process of exhaust is illustrated in Figure 2.1(d). The process begins when the
exhaust valve opens and continues until the exhaust valve closes. With the completion of
the exhaust process, the cycle is complete and ready to start over. The mass remaining in
the cylinder at the end of exhaust is called the residual mass. It is left over to be a part of
the cylinder mass for the next cycle.
These five processes are common to all internal combustion engines although the hardware
used to achieve the processes may be different. An illustration of a 2-stroke engine is shown
in Figure 2.2. The piston on the left is shown to be nearing the end of the expansion process
when a port on the left side of the engine is opened and the exhaust process is occurring.
The pressurized cylinder contents are exiting the exhaust port because of the pressure
within the cylinder. This exhaust process is sometimes referred to as blow-down as the
pressure drops from the high cylinder pressure to the near atmospheric exhaust pressure.
Shortly after the piston moves below the position shown in (a) the intake port on the
opposite side of the engine is opened and fuel and air can move in from the crankcase
cavity below the piston. Since the pressure in the cylinder is reduced to near atmospheric
pressure, the air and fuel compressed in the crankcase cavity by the downward motion of
the piston can travel into the cylinder on the right side. This pressurized fuel and air from
below the piston are pushed into the combustion chamber for a short time until the intake
port is again covered and the compression process can continue. Combustion occurs near
the end of compression and extends through the beginning of expansion.
Additional insights on the five processes that occur in an I.C. engine for both 4-strok and
2-stroke engines can be obtained by observing on-line animations. An internet search will
typically yield a large number of results.
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(a)
(b)
Figure 2.2 An example of a two-stroke engine incorporating the five processes: intake,
compression, combustion, expansion, exhaust. (a) near end of expansion, middle of exhaust
and intake is about to begin. (b) the end of intake and the beginning of compression. [1]
2.2. Piston-Cylinder Nomenclature
The most common hardware employed in an internal combustion engine is a reciprocating
piston and cylinder. An illustration identifying several commonly named geometric
components of a reciprocating engine is shown in Figure 2.3. The piston motion is created
using a slider-crank mechanism which converts the rotational motion of the crank to a
translating motion. The connecting rod connects the rotating crank to the sliding piston. As
the rotating crank travels through one revolution or 360 degrees of motion the piston moves
through two strokes (one up and one down). When the piston is at the top position where
the volume above the piston is the smallest, the piston is said to be located at Top Dead
Center or TDC. This location is most commonly referred to as 0 degrees. The volume at
this location is called the clearance volume or Vc. As the crank turns through 180 degrees
and the piston reaches the bottom of its travel, the position is referred to as being at bottom
dead center or BDC. Here the crank is at 180 degrees after top dead center (ATDC) or 180
ATDC. The volume at this position is called the total volume, Vt.
As the piston moves from TDC to BDC it displaces a volume known as the displacement
volume (Vd). The displacement volume (not the total volume) of all of the cylinders in an
engine is the volume that is used commercially to describe an engine. Thus, a 5.9 L
Cummins Diesel is an engine that has a total displacement in all cylinders of 5.9 L.
The distance traveled by a point on the piston from TDC to BDC is called the stroke. The
diameter of the cylinder containing the piston is called the bore. A simple relationship
exists between the bore, stoke and displacement volume for a single cylinder as shown by
Equation 2.1, where B is the bore and S is the stroke.
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Figure 2.3 Important nomenclature for reciprocating engine geometry. [2]
Vd = π
B2
S
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2.1
When 0 ATDC is used to indicate the rotational piston at the start of the expansion process,
the compression process occurs from -180 to 0 ATDC. Negative crank angles are often
referred to as positive crank angles before top dead center or BTDC. Thus, -15 ATDC and
15 BTDC are the same crank position.
2.3. Engine Load
The power produced by an engine is determined by Equation 2.2 where 𝑊𝑊̇ is the power,
N is the engine speed and T is the torque. This equation shows that power can be increased
by increasing either the engine speed or the engine torque. The engine torque is directly
proportional to the work produced by the engine in a given cycle. The engine will reach
maximum torque when the maximum amount of fuel is burned, but the amount of fuel that
can be burned in a cycle is limited by the amount of air that can fit within the engine within
a cycle. The maximum amount of air is obtained in an engine when the throttle or butterfly
valve on the intake plenum is all the way open. The throttle position in a spark ignition
engine is controlled by the pedal position. Thus, the torque is directly correlated to the
pedal position, the throttle position, the induced air and the injected fuel. When one is
maximized, they are all maximized.
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𝑊𝑊̇ = 2𝜋𝜋𝜋𝜋𝜋𝜋
2.2
Engine load is a word that represents any of these variables: torque, air induced, fuel
burned, throttle position, and pedal position. Load is expressed as a percentage of
maximum. Thus 100% load indicates maximum torque, maximum pedal position, fully
opened throttle, etc. Engine load is not the percentage of maximum engine power.
As an example of how engine load is changed, imagine driving a car on a flat road in 4th
gear at 60 mph and approaching a hill. In order to maintain the car speed up the hill the
engine will have to produce more power. If the car remains in 4th gear at 60 mph, the engine
speed will remain constant. According to Equation 2.2, in order to obtain more power while
N is constant, the torque will need to increase. This means an increase in work produced
per cycle which means more fuel will need to be burned requiring more air, requiring the
throttle be opened more, which requires the driver to increase the pedal position. When the
driver depresses the pedal, they increase the engine load. This can be done until the pedal
is at the floor, 100% pedal position and 100 % load.
If the pedal is all the way to the floor and 100% load is achieved but the car begins to slow
down, it is an indication that the power required to maintain the vehicle speed is too high.
In this case, a look at Equation 2.2 suggests that if more power is needed but torque (T) is
at a maximum (100% load), then the only way to achieve more power is to increase the
engine speed. This can be achieved by shifting the car from 4th to 3rd gear increasing N.
Thus, there are two ways to increase engine power, by increasing engine speed, or by
increasing engine torque (or load).
2.4. Common Reciprocating Engine Parts
An engine begins with a block wherein a cylinder is bored and honed in order to produce
a smooth surface. A rendering of a cylinder block with four cylinders is shown in
Figure 2.4.
Engine Block
Figure 2.4 Deutz engine block with four cyclinders. [3]
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A crankshaft for the four-cylinder block shown in Figure 2.4 is shown in Figure 2.5. This
crankshft has five cylindrical components along the centerline and four cylindrical
components that are offset from the center. As the shaft turns, the offset compoents travel
in a circular path around the centerline of the crank. The distance from the center of the
crankshaft to the center of the offset cylinder is called the crank radius and is equal to one
half the stroke.
The piston which moves up and down in the cylinder is connected to the crankshaft by the
connecting rod. A piston pin or sometimes referred to as a wrist pin is used to connect the
piston and the connecting rod while on the other end of the connecting rod, a bearing is
placed between the connecting rod and the crankshaft. The bearing that is between the
connecting rod and the crankshaft is called the rod bearing. There are also bearings
between the crankshaft at the block. These bearings (not shown in the figure) are called
main bearings. Holes in each of the cylindrical parts of the crank are used to allow oil to
flow through the crank and onto the bearings for lubrication.
In order to seal gases from leaking from the combustion chamber past the piston walls a
piston ring is used (not shown in figure). There are typically three rings on each piston.
The two on top are used to seal gases from leaking and the third ring on the bottom is used
to spread oil across the cylinder. The rings ride on a film of oil to reduce friction and wear.
When a piston reverses direction, accelerations or declerations produce large dynamic
forces on the block. This would cause the block to move up and down or shake and vibrate.
To reduce these dynamic loads, weights are added to the crankshaft on the opposite side of
the offset for each connecting rod. This causes an opposite dynamic force on the block that
offsets the dynamic forces of the engine. When dynamic forces are equal on a cylinder
block, the engine is said to be balanced.
Bearing
Connecting
Rod
Piston
Piston Pin
Crank
Shaft
Counterweight
Figure 2.5 Parts used to convert translational to rotational motion. [3]
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The crankshaft is mounted in the block with the shaft protruding from each end. On one
end, a flywheel is attached to the crankshaft as shown in Figure 2.6. The flywheel contains
gear teeth along the outer diameter. In order to start the engine, teeth attached to the shaft
of a starter motor engage with the teeth on the flywheel and spin the engine. The flywheel
provides angular momementum or inertia that is used to smooth out the periodic pulses
obtained when each cylinder fires and it stores inertial energy from the previous cylcel to
provide the work required for the compression process.
Flywheel
Figure 2.6 Flywheel attached to crankshaft on the outside of the block. [3]
Figure 2.7. shows a cylinder head or head which is placed on top of the cylinder block to
seal off the combustion chamber formed in the cylinder above the piston. The four openings
on the side of the head are intake ports where the fuel-air mixture enters the engine. The
head is the housing for the poppet valves, valve seats, valve guides and valve springs as
shown in Figure 2.8. This engine has a total of 16 valves, 4 valves per cylinder with each
cylinder having two valves for intake and two valves for exhaust. When looking at an
engine with an even number of valves per cylinder (2 or 4) the larger valves will be used
for intake while the smaller valves will be used for exhaust. This is because getting air into
and engine is a bigger challenge than getting exhaust out. In engines with 3 valves, two of
the valves will generally be intake and one exhaust; allowing for the greater total flow area
of the intake to be larger than the exhaust.
Figure 2.7 Cylinder head. [3]
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Valve Spring
Valve Guide
Valve Seat
Valve
Figure 2.8 Intake and exhaust valves and associated parts located in the head. [3]
Figure 2.9. shows two camshafts suspended directly above the intake and exhaust valves.
The cams rotate causing the lobes on the cam to push down on the intake and exhaust
valves. When a camshaft is located directly above the valves it is referred to as an overhead
cam. Since there are two cams, the arrangement is called a dual overhead cam (DOHC).
Each cam has a gear or sprocket on the end. A timing chain or timing belt is used to connect
the rotation of the crankshaft with the rotation of the camshaft. In a 4-stroke engine, the
number of teeth on the crankshaft is double the number of teeth on the camshaft creating
two rotations of the crank for each camshaft rotation. A mechanical linkage exists between
the crankshaft to valve which goes in sequence: crankshaft – timing chain – camshaft –
follower – valve.
In some cases, cams are located outside of the head and mechanical levers called pushrods
(no a part of the engine shown) are used to transmit the motion of the cam to the valves.
When the camshaft is located below the valve a rocker arm is used to transmit the upward
motion of the pushrod to a downward motion on the valve. The mechanical linkage for an
under head camshaft is: crankshaft – timing chain – camshaft – pushrod – rocker arm –
follower – valve.
A mechanical linkage can be noisy or experience high forces when parts with little
elasticity push on each other. As a result, it is often the case that a piece of the linkage has
a hydraulic (fluid) component that can expand when pressurized to fill up the gaps between
linkage parts and to cushion the forces of acceleration. This linkage is called a hydraulic
lifter or lifter (not shown for this engine).
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Camshaft
Follower
Valve Seat
Valve
Valve Guide
Figure 2.9 Dual overhead cam shafts located in the head above the valves [3]
The head gasket is a thin piece of metal or composite material that is crushed between the
head and the block as shown in Figure 2.10. The head gasket contains a large hole for each
of the cylinders and smaller holes that allow the flow of coolant and oil between the block
and the head.
Head Gasket
Figure 2.10 Head gasket for a four-cylinder engine. [3]
From the outside of the engine the head and block are somewhat hidden by external
components. Figure 2.11 shows the intake side of an engine. The throttle contains a circular
butterfly valve that opens to control airflow into the engine. After passing through the
throttle valve, the air is distrbuted to each of the four cylinders in the intake plenum. Four
“runners” can be seen in the intake manifold or plunum, one for each cylinder. The fuel
injectors are mounted in the intake ports just upstream of where the intake plenum meets
the head. Fuel injectors are housed in a plastic case with a fuel line and electrical wire
attached to each injector. Spark plugs must screw directly through the head into the
combustion chamber of each cyclinder. In Figure 2.11, one of the four spark plugs is shown
directly above the hole where it is mounted in the engine. A relatively large diameter wire
(not shown) is connected on one end to the top of each spark plug and on the other end to
the distributor (not shown). The distributor is used to send a high voltage pulse to each
spark plug at the correct time in the cycle in order to initiate combustion. When combusiton
products leave the engine they exit through the exhuast manifold (not shown). The exhaust
manifold may look similar to the intake manifold with a runner for each cylinder. While
the intake manifold passes cool air and can be made of plastic materials, the exhaust
15
manifold is very hot during operation and must be made of metal. The metal rapidly
discolors with use and becomes rusty making it easy to identify.
On the top of the engine head is a valve cover used to keep oil (lubricant) in the camshaft
and valve area. On the bottom of the engine block is the oil pan (not shown). Oil drains by
gravity to the bottom of the engine where it collects in the oil pan. A pump in the oil pan
distributes oil throughout the engine.
Spark
Plug
Fuel
Injector
Valve
Cover
Intake
Plenum
Throttle
Flywheel
Starter Motor
Figure 2.11 Intake side of the engine. [3]
2.5. Summary
All internal combustion engines utilize five processes to convert chemical energy to work.
The five processes are Intake, Compression, Combustion, Expansion, and Exhaust. Four
of the five processes are mechanical in nature (Intake, Compression, Expansion, Exhaust)
and are typically achieved using a slider crank mechanism with poppet valves and an offset
crankshaft. A nomenclature for parts and processes common to most internal combustion
engines has been introduced. The parts and names of parts typically used to accomplish the
four mechanical processes have also been introduced and discussed. The next chapter will
discuss general details of the combustion process.
References
[1] Rogowski , A. R., Elements of Internal Combustion Engines, McGraw-Hill, 1953. Cited
by Heywood, J. B., Internal Combustion Engine Fundamental, McGraw-Hill, 1988.
[2] Heywood, J. B., Internal Combustion Engine Fundamental, McGraw-Hill, 1988.
[3] Deutz Screensaver, 2016. http://deutz-engine-screensaver.software.informer.com/1.0/
16
Chapter 2. Homework Questions
2.1 The text contains italicized name of engine parts which are common to most internal
combustion engines. Identify the parts listed below with the number pointing to the part on
the diagram.
____ Cylinder head
____ Piston and pin
____ Connecting rod
____ Valve spring
____ Rocker arm
____ Camshaft
____ Crankshaft
____ Piston Rings
____ Main bearing
____ Rod Bearings
____ Head Gasket
____ Oil pan
____ Poppet Valve
____ Valve Cover
http://www.justanswer.com/uploads/tracker40/2009-05-04_173558_pumppp.gif
17
2.2 There are several parts referred to in the text that are not shown in the figures in the
text. These parts are listed below. Complete an internet search on each of the parts to learn
what it looks like and where it fits. Describe the part and where it fits in your own words.
A. Piston Ring
B. Lifter or Hydraulic link
C. Main Bearing
D. Distributor
E. Oil Pan
F. Rocker Arm
2.3 Write the name of at least one part that is in contact with each of the following parts.
A. Engine block
B. Crankshaft
C. Spark plug
D. Fuel injector
E. Head Gasket
F. Piston
G. Oil Pan
H. Main Bearing
2.4 If you are driving up a hill at 30 mph and the engine produces a constant power output
but shift gears from a high gear (4th gear) to a lower gear (3rd) gear causing the engine to
turn and a higher engine speed. What will happen to the engine load, will it increase or
decrease.
2.5 A car is driving along a road at 40 mph with an engine speed of 2000 rpm and is
producing 0.5 kJ or work per cycle. The road grade changes and the car is going uphill.
The driver pushes on the gas pedal all the way to the floor (wide open throttle) and is just
barely able to still maintain the speed of 2000 rpm but is now producing 1.2 kJ of work
every cycle. At what % load was the engine at while driving on the flat road.
2.6 Write a mathematical equation relating the total volume in the cylinder with the
clearance volume and the displacement volume.
2.7 Using a clearance volume, Vc = 5 cm3, and a displacement volume Vd = 50 cm3, what
is the total volume of the cylinder at BDC.
2.8 If the compression ratio, rc, is defined as the maximum volume (Vbdc) divided by the
minimum volume (Vtdc), and the clearance volume is 5 cm3 and the displacement volume
is 50 cm3. What is the compression ratio, rc.
2.9 Write a mathematical expression for the stroke, S, in terms of the crank radius, R.
2.10 Write a mathematical expression for the clearance volume, Vc, in terms of the
compression ratio, rc, and the displacement volume, Vd.
18
3. Engine Classification
A list of various parameters used to classify engines is provided in Table 3.1. The
characteristics listed are by no means comprehensive and the discussion which follows is
brief and intended only to provide a basic literacy to understand the literature and to
facilitate discussions related to engines.
Table 3.1 Some of the most common characteristics used to classify I.C. engines.
Characteristic
Possible Configurations
Number of Strokes
4-Stroke
2 Stroke
Compression –
Expansion Hardware
Reciprocating
Rotary
Fuel Type
Gasoline
Diesel
Ignition Type
Spark
Compression
Combustion Process
Propagating
Mixing Limited
Compressed
Natural Gas (CNG)
Volumetric
Type of Engine Cooling Water Cooled
Air Cooled
Piston Configuration
In-line
V
Boxer – Flat
Air Delivery
Naturally
Aspirated
Turbocharged
Supercharged
Fuel Delivery
Port Injection
Direct Injection
Carbureted
3.1. Number of Strokes
A reciprocating engine can be used to accomplish the five processes common to all IC
engines using 2, 4, or even 6 strokes. The 2-stroke cycle is common among smaller engines
or engines looking to maximize power to weight ratio. The 4-stroke cycle is the most
common and has been described in Section 2.1.
A diagram of one method of producing a 2-stroke cycle is shown in Figure 3.1. Beginning
with the combustion process and the piston near top dead center (TDC), the piston moves
downward to produce the expansion stroke. Before the piston reaches bottom dead center,
an exhaust port is uncovered on the left side of the cylinder as shown in Figure 3.1b.
Exhaust leaves the combustion chamber because the cylinder is still pressurized and then
rapidly reaches the exhaust or atmospheric pressure. Shortly after the exhaust port is
opened, the intake port is uncovered on the opposite side of the engine. The downward
movement of the piston creates a positive pressure in the crankcase (volume beneath the
piston). This helps to push the fuel air mixture into the cylinder. The piston proceeds to
bottom dead center and then returns upward. After passing the intake and exhaust ports the
mixture in the cylinder is compressed as the piston moves toward TDC. Near TDC, the
spark ignites the combustion process starting the cycle over again.
19
(a) Combustion
(c) Intake
(b) Expansion/Exhaust
(d) Compression
Figure 3.1 The five combustion processes occurring in a two-stroke engine. [1]
The 2-stroke cycle produces one combustion event for every revolution of the crankshaft
and thereby produces a higher power to weight and power to volume ratio. Because the
exhaust and intake ports are open at the same time, it is difficult to fill the entire volume of
the combustion chamber with a fresh charge of fuel and air. This reduces the amount of
fuel that can be burned for a given displacement volume. The expansion portion of the
cycle is also reduced which decreases efficiency. These and other issues to be discussed
in more detail in later chapters create an advantage for two-strokes where power to weight
ratio is essential, but when power to weight ratio is not of highest priority the four-stroke
cycle is more fuel efficient and durable.
3.2. The Rotary or Wankel Engine
There are numerous means whereby engineers and inventors have come up with the
hardware to produce the five processes common to all IC engines. The two that have
succeeded commercially are the reciprocating and rotary engines. The reciprocating engine
is described in Chapter 2.
The rotary engine design was invented by Felix Wankel who developed a prototype in 1957
[2]. A diagram showing the hardware used to produce the 5 processes is shown in
Figure 3.2. The triangular shaped rotor moves in the clockwise direction inside and
elliptically shaped outer casing. The intake process can be seen to occur between the
20
volume produced by points A and C and the outer wall. This volume is seen to expand
through the sequence of Figure 3.2a, b, c, d. The expanding volume produces a vacuum
which draws the mixture into the engine (intake). At the same time, the mixture between
A and B is being compressed in the sequence between 4.2a and 4.2b and then burned
(combustion) between Figure 3.2b and c. Expansion begins in the volume confined
between points A and B in Figure 3.2c and extends through Figure 3.2d and continues until
point B reaches the exhaust port. Exhaust occurs in the volume defined by points B and C
on the rotor as they rotate between the points shown in Figure 3.2c, d, and a.
(a)
(b)
(c)
(d)
Figure 3.2 Basic operation of a rotary or Wankel engine [3].
It can be seen by the diagram that all five processes are occurring during a period where
the rotor moves 1/3 of a rotation. Each time the rotor makes one rotation, three cycles are
completed. This provides an advantage of higher power per unit weight or volume for this
type of engine. Also, the moving parts rotate in the same direction at all times and are not
required to change direction. This means the forces (dynamic forces) generated by the
acceleration and deceleration of a piston and cylinder are not present. The disadvantages
of the Wankel configuration are the difficulty in providing lubrication where the rotor seals
against the outer casing and the high surface area to volume ration produced by the
combustion chamber which leads to efficiency and emissions problems.
3.3. Fuel Type
Internal combustion engines have been operated on many different types of fuel. Rudolf
Diesel initially ran engines using coal dust while gas derived from heating coal was later
used in spark ignition engines. An internal combustion engine can operate on virtually any
kind of fuel. The most common fuels in use today are gasoline and diesel. A small but
significant number of engines are run on compressed natural gas.
Gasoline is a fuel that has been developed from petroleum to be: 1. Easily evaporated
(highly volatile) and 2. Resist ignition. This matches the requirements for a spark ignition,
deflagration wave or propagating flame combustion. Contrary to most people’s intuition,
gasoline is actually designed to resist ignition in order to allow an engine to compress the
fuel air mixture as much as possible before ignition is initiated by a spark. The ability of a
21
fuel to resist ignition is measured by the Octane number of the fuel. Octane number is not
a measure of fuel energy (heating value), fuel cleanliness or any other characteristic.
Therefore, two fuels may have completely different octane numbers but share the same
heating value. The octane number is not a measure of the amount of octane in a fuel. It is
a measure that compares the tendency of a fuel to autoignite with the tendency of octane
to autoignite.
Diesel fuel is designed to be the opposite of gasoline. It is easily ignited, not particularly
volatile and easy to inject (high lubricity). Ignition quality is measured by the cetane
number of the fuel. Fuels with a high cetane number ignite easily. This allows them to
ignite when injected into a hot oxidizer. As with octane number, the cetane number is not
an indication of the heating value, lubricity, cleanliness or any other characteristic of the
fuel. It is also not a measured of the amount of cetane in the fuel. The cetane number is a
comparison of how a fuel autoignites in comparison to cetane.
Natural gas is an excellent fuel for internal combustion engines. It has a high volatility and
high octane number making it ideal for spark ignition engines. The disadvantage of natural
gas is that it is a gas at normal atmospheric temperature and pressure and therefore the
energy density is much lower than liquid fuels. To overcome this issue, natural gas is
typically stored in high pressure tanks (about 200 psi) which are larger and heavier that
liquid fuel tanks. A spark ignition engine can be made to run on either gasoline or natural
gas with a switch to control which fuel source and fuel hardware deliver fuel to the engine.
Many other fuels have been and will continue to be burned in internal combustion engines
including liquid petroleum gas (LPG), hydrogen, biodiesel, ethanol, methanol, and
kerosene’s. Each has advantages and disadvantages. Fuels and engines are designed in an
evolutionary process. Fuels can be designed for a given engine or vis-versa.
3.4. Ignition source
There are basically two ways to initiate combustion in an IC engine: spark ignition or
compression ignition. The spark allows much greater control and repeatability for the start
of combustion. Compression ignition is typically controlled by heating the oxidizer well
above the ignition temperature and then ignition begins shortly after injection. Thus
injection timing is used to control ignition timing. In the past, diesel engines often
employed a heated glow plug that was used to create a local hot spot and ensure ignition
when fuel was injected in the vicinity of the glow plug. This practice was commercially
ended when injection pressures and higher compression ratios made the glow plug
unnecessary. Other ignition sources such as plasma ignitors or laser ignition have also been
investigated but are not commercially produced.
3.5. Combustion Processes
The three combustion processes encountered in IC engines are premixed deflagration,
mixing limited combustion, and volumetric reaction. Spark ignition engines normally
utilize premixed deflagration where the spark ignites a premixed mixture of fuel and air
and a flame propagates through the mixture at the flame speed. Diesel engines typically
22
utilize both volumetric and mixing limited combustion. As fuel is injected into the engine,
it begins to mix with the air, evaporate, and heat up. A small fraction of the volume inside
the engine containing hot fuel and air reaches the ignition temperature and reacts rapidly
throughout the mixed volume. This is volumetric combustion is typically referred to as a
premixed burn and is also the ignition source for the engine. As the remainder of the fuel
is injected, it burns at the rate the fuel is evaporated and mixes with the surrounding air in
the cylinder.
It is very common to hear or read of a fuel “exploding” in an engine. This general term for
combustion does not designate which of the three combustion processes are being used in
the engine and is therefore not helpful as an I. C. engine classification. An explosion is also
more commonly associated with a fourth mode of combustion called detonation. A
detonation occurs when a flame propagates through a mixture faster than the speed of
sound producing a shock wave. This type of combustion does not occur in I.C. engines and
therefore the term, explosion, is misleading. The processes of combustion are discussed
more completely in Chapter 6.1.
3.6. Engine Cooling
There are two methods for cooling IC engines; water and air cooled. Water cooling is
actually achieved by a water / coolant (antifreeze) mix that extends the boiling and freezing
point. Air cooling is achieved by adding fins to the engine block and head to increase
convection. A fan or blower attached to the drive shaft can also be employed to increase
convective heat transfer. Air cooled engines are used to reduce weight, decrease cost and
increase simplicity.
3.7. Piston Configuration
Numerous configurations for locating pistons relative to the crank shaft have been used
with various advantages and disadvantages. A diagram of the three most common
configurations is shown in Figure 3.3. The first configuration is, in-line, where a single
crankshaft is located directly below each cylinder. The number of cylinders is normally
specified such as; an in-line-4, for an engine with four cylinders located adjacent to each
other in a line. This type is easy to manufacture and packages nicely. Balancing of dynamic
loads is improved by using counterweights on the crankshaft and alternating the timing of
each cylinder.
The second configuration is a “V” where pistons share a single crankshaft but are moving
up and down at an angle to the vertical. The number of pistons is also typically given after
the letter V such as a V-8 engine signifying an engine with 8 cylinders in the “V”
configuration. This configuration has advantages for balancing dynamic loads. In addition
to dynamic vertical forces caused by the piston moving up and down, horizontal forces are
created by the angular position of the connecting rod. In V-8 engines, both the vertical and
horizontal forces can be offset by having an adjacent piston rotating such that when one is
moving up the other is moving down and while one is moving left, the other is moving
right. Engine balancing is not a topic covered in detail in this book but it is reported that a
23
V-8 engine can be designed for perfect dynamic load balancing, unlike any other engine
configuration.
The final configuration shown is a boxer or flat engine. This configuration again places
multiple pistons on a single crankshaft but provides the most space around the cylinder and
head for each piston. This design is advantageous for providing cooling fins to the engine
block and head and is often found in aircraft and other air-cooled engines.
There are many other innovative and interesting engine configurations that have been
devised for use with reciprocating pistons. Some offer unique advantages and
disadvantages. In the end, the three configurations shown in Figure 3.3 dominate the
market of currently available commercially manufactured IC engines.
(a) In-line
(b) V
(c) Boxer or Flat
Figure 3.3 Various piston location arrangements common in I.C. engines
3.8. Air Delivery
There are three common types of air delivery for an IC engine: naturally aspirated,
turbocharged, and supercharged.
A naturally aspirated engine utilizes the piston motion alone to draw air into the cylinder
from atmospheric conditions.
A turbocharged engine uses a turbocharger to compress air in the intake plenum. A
turbocharger consists of a turbine and compressor as shown in Figure 3.4. Hot exhaust gas
enters the turbine and moves radially inward and through the vanes causing the turbine to
spin. The exhaust exits in the axial direction. The spinning shaft is connected to a
compressor. Air enters the compressor axially and exits radially where is it delivered to the
intake plenum. Pictures and video of how a turbocharger is mounted and used on an IC
engine are readily available on the internet. Turbochargers utilize waste energy that would
otherwise be discarded in the exhaust to increase intake pressure. The turbocharger alone
does not increase engine efficiency but it does make a smaller engine more powerful by
increasing the amount of air delivered; enabling more fuel to be burned. This allows a
24
smaller engine to be used at a higher load. Because engines are more efficient at high load
than they are at low load, the turbocharger enables a more efficient use of the engine.
Figure 3.4 Picture of a cutaway turbocharger. [4]
Turbochargers require energy from the exhaust and therefore may not work well at low
engine speeds or loads when exhaust flows are low or at lower temperature. A supercharger
on the other hand receives power from the output shaft of the engine and can drive a
compressor at any engine speed. Superchargers are popular in applications where high
power is required (racing). An example of a supercharger is shown in Figure 3.5. In this
example the supercharger looks like the compressor half of a turbocharger but instead of a
turbine, a pulley placed on the compressor shaft must be rotated by the engine to provide
the power needed.
Figure 3.5 Picture of a supercharger with a compressor and pulley. [5]
25
3.9. Fuel Delivery
There are three types of fuel delivery systems that are most commonly found in IC engines:
carburetion, port fuel injection, and direct fuel injection. Each of these will be explained in
greater detail in a later chapter but are introduced conceptually here.
Carburetors use air passing through a venturi to create a low pressure. Piston motion creates
the air flow. The low pressure produced by the air flow sucks fuel into the venture which
mixes with the air. The ratio of fuel to air remains relatively constant so that increased air
flow automatically increases fuel flow as is generally desirable for an engine. The amount
of air flow is therefore the controlling factor in how much fuel is burned and how much
work is done in a cycle.
Port fuel injection is a design where fuel is injected into the intake port upstream of the
intake valve or valves. The injection system consists of a pressurized fuel line with a
computer controlled orifice that opens to inject the fuel. The amount of fuel injected is
determined by the amount of time the orifice is opened. The computer or engine control
unit (ECU) received information from sensors on the engine regarding the amount of air
flowing in order to determine how long the orifice should stay open. Port fuel injected
engines typically maintain a constant fuel to air ratio so once again, the air flow determines
how much fuel is injected and determines how much work is done in a cycle. Common
injection pressures are on the order of 100 psig.
Direct fuel injectors are located so they will inject fuel directly into the combustion
chamber. In compression ignition engines, fuel is injected near the end of the compression
process requiring a higher pressure for two reasons. First to overcome the pressure in the
cylinder and second to inject the fuel in a short period of time. Common injection pressures
are on the order of 10,000 – 25,000 psig. In these engines, the air flow rate remains
relatively constant while the amount of fuel increases when more work / power is desired.
Fuel can be injected in increasing amounts until the fuel can no long mix with enough air
to produce complete combustion. At this point additional fuel injection for more power
will produce large amounts of smoke. Thus, direct fuel injection engines are said to be
smoke limit; meaning the amount of fuel that can be injected and therefore the amount of
work that can be done in a cycle is limited by the production of smoke.
References
[1] http://www.roymech.co.uk/images/Thermo_22_Two_stroke.gif
[2] Ansdale, R.F., The Wankel RC Engine Design and Performance, Iliffe Books, London,
1968.
[3]
http://assets.blog.hemmings.com/wp-content/uploads//2013/12/SE47-dR1-NSUWankel-Engine-Cycle.jpg
[4] https://upload.wikimedia.org/wikipedia/commons/7/76/Turbocharger.jpg
[5] http://www.stage3motorsports.com/assets/images/pc-1ft202-sci20112014mustangv6prochargerintercooledsuperchargertunerkit0.jpg
26
Chapter 3 Homework Questions
3.1 Identify a car with an internal combustion engine that you can obtain visual access. For
this car, classify it according to the nine characteristics in Table 3.1.
3.2 Given two engines with the same displacement volume, which engine would produce
more work per cycle; a naturally aspirated engine or a turbocharged engine? Why?
3.3 Consider the engine on a push lawnmower. Classify the engine using the nine elements
in Table 3.1
3.4 Why do fuel injectors for direct injection engine cost more than port fuel injectors?
27
28
4. Engine Performance Parameters
The objective of this chapter is to learn the definitions of various engine performance
parameters and how they are calculated. Most of the performance parameters are common
to all engine types and the nomenclature is general for most of the manufacturers in the
engine industry.
4.1. Indicated Work
A precise nomenclature has been developed by engineers in order to make performance
comparisons between engines. In order to understand this nomenclature, we begin with a
P-V diagram produced by measuring the cylinder pressure as a function of piston position
or volume. A portion of a four stroke cycle is shown in Figure 4.1.
This diagram was produced from data collected with a rapid response piezoelectric
pressure transducer which produces a charge in proportion to the pressure. The charge is
converted to a digital signal and recorded on a computer. Before piezoelectric transducers
were available, cylinder pressures were recorded using a mechanical device which
responded to fluctuations in the cylinder pressure called and “indicator”. The term
“indicated” has remained in the engineering nomenclature to identify work produced by
the cylinder gases on the piston.
The area on the figure marked “closed portion” is the area enclosed by the pressure curve
between bottom dead center before compression to bottom dead center after expansion.
During this period, the intake and exhaust valves are mostly closed and therefore this is
referred to as the closed portion of the cycle. The closed area represents what is referred to
as the indicated gross work for the cycle. Work is positive out for this portion of the cycle
when the cycle proceeds in a clockwise rotation on a P-V diagram.
2
Pressure (MPa)
1.5
(BDC)
End Of
Expansio
1
Closed
Portion
TDC
0.5
0
Open
Portion
-0.5
0
0.2
0.4
(BDC)
Start of Compression
0.6
0.8
1
1.2
Volume (L)
Figure 4.1. A portion of a P-V diagram from a four-stroke cycle.
29
The area of the diagram denoted as the open portion represent the work done during the
exhaust and intake processes. The integral of pressure from BDC at the end of exhaust to
BDC prior to compression. The area in this portion of the P-v diagram is called the
indicated pump work. In this diagram, this portion of the cycle represents negative work
out of the engine or work input is required to keep the piston moving. In this case, the cycle
proceeds in a counter clockwise direction on the diagram with the exhaust stroke preceding
the intake stroke.
The net indicated work for the cycle is the total area integral of the entire cycle.
Mathematically these work terms can be defined by Equations 4.1 and 4.2 as follows
where: the “i” subscript is used to represent indicated, “g” – gross and “p” – pumping. The
pressure is given by P and the volume, V.
180
∫
Wi , g ,out = PdV
−180
∫
540
W p ,out = PdV
180
4.1
4.2
The indicted net (subscript, i,n) work out for the cycle is given as the sum of the indicated
gross and indicated pumping work as shown in Equation 4.3.
Wi ,n ,out = Wi ,g ,out + Wi ,p ,out = Wi ,g ,out − Wi ,p ,in
4.3
Because the gross and net work are always positive out of the cylinder and because
pumping work is normally negative out of the cylinder or positive work done on the
cylinder gasses, the subscripts in and out are normally eliminated or dropped producing
Equation 4.4
Wi ,n = Wi ,g − Wi ,p
4.4
This equation states that the net indicated work is the gross indicated work minus the
pumping work. Pumping work is seen in this expression to decrease the work coming out
of the piston. The importance of pumping work in naturally aspirated, SI engines cannot
be overstated. Negative pump work increases when the throttle is closed which is how the
output of a port injected, SI engine is controlled. Thus, the efficiency of all throttle
controlled engines is lowered because of pumping work. At idle, the pumping work is large
because the throttle is closed and the indicated work is small. At idle, the indicated and
pumping work may be almost equal.
Equation 4.4 may be confusing when a turbo or supercharger is involved which is typically
the case with a diesel engine. When turbo and supercharging are applied to an engine, the
30
intake pressure can be higher than the exhaust pressure which produces positive work out
during the pumping process. Thus, the more complete expression in Equation 4.3 with
subscripts should be kept in mind and is the more rigorous, albeit tedious approach.
Example: Gross, Indicated and Net Work
Given: In the P-V diagram below. Area “A and B” represent the closed portion of the cycle and areas
(B and C) represent the open portion of the cycle.
Area A+B = 0.214 kPa-m3
Area C+B = 0.020 kPa-m3
Solution:
The area on a P-V diagram represent the
work done. The cycle begins at the
bottom right and proceeds through
compression then expansion. Thus, the
area represented by A and B represents
the closed portion of the cycle which is
proceeding in a clockwise direction
making positive work out of the cycle.
500
450
400
Pressure (kPa)
Find: The indicated gross work, Wi,g, The
indicated pumping work, Wi,p, and the net
indicated work, Wi,n.
350
300
250
A
200
150
100
B
C
50
0
0
0.0002
0.0004
0.00
3
Volume (m )
Wi,g = Area A + Area B = 0.214 kPa-m3
= 0.214 kJ
The area C + B represents the closed portion of the cycle which is proceeding in a counterclockwise
direction. This means the work out is negative during this portion of the cycle.
Wi,p,out = -(Area B + Area C) = -0.020 kPa-m3 = -0.020 kJ or
Wi,p,in = Wp = 0.020 kJ
The net work is the difference between the indicated gross and the pumping work
Wi,n = 0.214 kJ – 0.020 kJ = 0.194 kJ
Note: The pumping work results in a direct loss in engine output.
4.2. Engine Load
The term “engine load” or “load” was introduced in Section 2.3 and is equal to the work
done in a cycle. When an engine is at “full load” the engine is producing the maximum
possible work per cycle. With the use of the P-V diagram, the impact of throttling the
intake air and pumping work on load can be more clearly understood. Maximum or peak
load is accomplished by opening the throttle completely in a throttled engine and by
producing maximum pressure from a turbocharger or supercharger. The maximum air
would then be matched by the maximum amount of fuel in order to produce the largest
indicated gross work and the smallest pumping work. Engine load should not be confused
with power. Maximum power is produced by the combination of load and speed that
31
produces the most work per unit of time. Each engine speed has a maximum load but
only one speed load combination produces maximum power.
4.3. Brake Work
The power produced by an engine is typically measured by connecting the output shaft to
a dynamometer. The dynamometer applies a force or “bakes the engine” to maintain a
desired speed. Without the dynamometer or some other force on the shaft, the rotational
speed of the engine would accelerate until dynamic forces destroyed the engine. The force
applied to the engine to slow it down is referred to as the brake force which can be used to
calculate a brake work and brake power for the engine.
The brake work of an engine is equal to the net indicated work minus the work done by
friction. Friction work is always a negative output and thus the subscript identifying work
as out or in is dropped resulting in Equation 4.5.
Wb = Wi , n − W f
4.5
By substituting for net indicated work, the brake work can also be determined by
Equation 4.6.
Wb = Wi , g − W p − W f
4.6
Where again the subscripts of “in” and “out” have been dropped assuming that pumping
work and friction work are negative work out. This equation states that the work coming
out of an engine is equal to the work produced during the closed portion of the cycle minus
the work required to pump air into the cylinder and minus the work lost to friction.
4.4. Power
Power is the rate at which work is being done. When work for a cycle is known, power is
calculated by dividing by the work for one cycle by the time for one cycle as shown in
Equation 4.7. Typical units are shown in brackets to illustrate the use of a new parameter,
nR, which is the number of revolutions required to produce a cycle.
W =
Wcycle
t cycle
=
Wcycle
[kJ / cycle ]
WN
[kW ]
=
n R / N [ rev / cycle ] /[ rev / s ] n R
4.7
When engine shaft output is measured, the brake power is normally determined by
measuring the torque and rotational engine speed. An equation for power output is thus
derived as shown in Equation 4.8.
Force Dis tan ce F 2πd T [kJ ] 2π [ rad / rev ]
W =
=
=
= 2π NT [kW ]
Time
t cycle
1 / N [ rev / s ]
4.8
32
It should be noted that the equations for power above are homogenous, and do not require
that the same units be used as are given as examples above. Any set of units can be used
with these equations and then later converted to the desired output units. The key is to
remember the units associated with nR [rev/cycle] and 2π [rad/rev] as these are not
explicitly stated in the equation. The most common units for power output are kW and
horsepower (hp) and all results should be converted to one of these two units unless
otherwise stated.
4.5. Mean Effective Pressure
The amount of work per cycle an engine produces should to the first order scale
proportional to engine size; therefore, dividing by the engine size provides a means of
comparing two engines of different size. The most common parameter used to scale
engines is the displacement volume, although other parameters such as engine weight or
total engine volume are occasionally used. The work output of an engine divided by the
engine displacement volume results in a value that has the units of pressure as shown in
Equation 4.9, and is called the Mean Effective Pressure (MEP).
mep =
W[kN − m]
= mep [kPa ]
Vd [m 3 ]
4.9
The mean effective pressure represents the average pressure acting on the expansion stroke
that will produce the same amount of work as is produced by the cycle.
Substituting the work from Equation 4.7 into Equation 4.9 produces the mean effective
pressure as a function of power, engine speed, and displacement volume as shown in
Equation 4.10.
mep =
W n R
N Vd
4.10
4.6. Indicated or Brake – Single or Multi-cylinder
Each of the above parameters, Work, Power, and mep and many other engine parameters
can be calculated for brake (engine out) or indicated (in-cylinder) conditions as shown in
Equation 4.11. This is simply done by using brake or indicated parameters consistently
throughout the equation. For example brake power can be determined from the brake work
while indicated gross power is found from the indicated gross work.
W N
W b = b [kW ]
nR
or
Wi , g N
Wi , g =
[kW ]
nR
4.11
Results from a cycle simulation or a cylinder pressure measurement usually represent on a
single cylinder. Calculations can be made for the entire engine or one cylinder of an engine
by using values consistently throughout the equations. For example, when calculating the
33
power produced by a single cylinder, use work produced by a single cylinder or when
calculating the power produced for the entire engine, use the work produced by a single
cylinder multiplied by the total number or cylinders as shown in Equation 4.12.
W sin gle =
Wsin gle N
nR
or
W total =
4.12
Wsin gle n cyl N
nR
The mean effective pressure is independent of the number of cylinders and care should be
taken to use either total engine or single cylinder values consistently in the numerator and
denominator of Equations 4.9 and 4.10 to obtain correct results.
Example: Calculating engine performance parameters
Given:
An Audi A-4 uses a 2.0 liter, 4 cylinder, 4-stroke engine with a compression ratio of 10.5:1 that
produces 200 hp at 5000 rpm.
Find:
A) The brake torque produced by the engine at this speed.
B) The brake work produced by a single cycle for a single cylinder.
C) The mean effective pressure of the engine at this speed and load.
Solution:
Engine manufacturers report brake outputs so the power reported is brake power. The volume
reported for and engine is the displacement volume.
A. Tb =
W b
200 hp 33,000 ft − lbf / hp − min
=
= 210 ft − lbf
2π N (2)(3.14159 rad / rev) 5,000 rev / min
B. Wb,all = Wb nR = 200 hp 2 rev / cycle 33,000 ft − lbf / hp − min = 2,640 ft − lbf
5,000 rev / min
N
Wb,sin gle =
Wb,all
n cyl

=
3.39 Btu
= 0.848 Btu / cyl
4 cyl
C. bmep = Wb n R =
N Vd
1Btu
= 3.39 Btu
778 ft − lbf
200 hp 2 rev / cycle 33,000 ft − lbf / hp − min
5,000 rev / min 2.0 L / cycle 0.035315 ft 3 / L 144 in 2 / ft 2
= 259 psi
Comments: The conversion of 33,000 ft-lbf/hp-min proves useful in a lot of engine calculations using
English units. Note also the units on the displacement volume for the mep calculation is L/cycle. This
engine produces and unusually high bmep which can be attributed to turbocharging.
4.7. Mechanical Efficiency
The mechanical efficiency is the ratio of work out of the engine to work done on the piston
before losses as shown in Equation 4.13. Because pumping work is normally a loss, the
work done on the piston is taken to be the gross indicated work. Mechanical efficiency as
defined here includes losses due to pumping work and to friction. Engine friction is a
function of engine speed and load and will be discussed in Chapter 8.
34
ηm =
Wb Wi ,g − W p − W friction
=
Wi ,g
Wi ,g
4.13
4.8. Volumetric Efficiency
The mass of fuel burned in a given engine cycle and therefore the work produced in that
cycle is limited by the mass of air or oxygen that is available to burn the fuel. An engine
designer therefore seeks to maximize the air flow trapped in the cylinder in order to
maximize power.
Using the ideal gas law, the amount of air drawn into the engine during the intake stroke is
given be Equation 4.14. In order to get more air in for a given displacement volume one
must either increase the pressure or decrease the temperature of the incoming air. The
typical choice is to increase the pressure. A throttle can be used to decrease the pressure
below atmospheric. A turbocharger can be used to increase the pressure above atmospheric.
m air =
PV d
R air T
4.14
The volumetric efficiency is a comparison to the air actually drawn into the engine in
comparison to the amount that could be drawn in as calculated by the ideal gas law as
shown in Equation 4.15.
ev =
m air , act
4.15
m air ,ideal , gas
A challenge arises in determining how to calculate the ideal amount of air that can fit in
the engine. Consider the two locations shown on the diagram of the intake path shown in
Figure 4.2 the could be used as a reference. The first is at atmospheric pressure (P0) and
temperature (T0) before air enters the engine and the second is the pressure (PI) and
temperature (TI) in the intake plenum downstream of the throttling valve.
If the first position is used to calculate the mass that can fit in the cylinder, the actual
amount will be much higher if the intake is turbocharged and much lower if the throttling
valve is closed or partially closed. Therefore, the volumetric efficiency will mostly indicate
how much air is entering compared to an open throttle atmospheric engine.
On the other hand, if the intake plenum pressure and temperature are used, the volumetric
efficiency is only influenced by flow characteristic of the intake value and intake port
downstream of the throttle. It becomes an indicator of flow friction losses.
Both definitions can be valuable. As a result, Equations 4.16 and 4.17 define two different
volumetric efficiencies each with a different reference or ideal air mass flow. The
atmospheric volumetric efficiency, ev,0, will typically vary from 0.2 for a naturally
aspirated engine at idle (throttle closed) to 3 atm. for a highly turbocharged diesel engine.
35
The volumetric efficiency based on intake plenum conditions will typically vary over a
narrower range between 0.6 and 1.1. The intake based volumetric efficiency can be greater
than one if favorable pressure waves are present in the intake plenum as will be discussed
in Chapter 9.
P0 T0
Pi Ti
Throttle
Figure 4.2 Diagram of an engine showing three locations where pressure and temperature
may differ; atmospheric, intake and in-cylinder.
𝑒𝑒𝑣𝑣,0 =
e v ,i =
𝑚𝑚𝑖𝑖𝑖𝑖 𝑅𝑅𝑇𝑇0
𝑚𝑚𝑖𝑖𝑖𝑖
=
𝑃𝑃0 𝑉𝑉𝑑𝑑
𝜌𝜌0 𝑉𝑉𝑑𝑑
4.16
m air ,in RTi
4.17
Pi V d
=
min
ρ iVd
Air entering an engine is normally measured as a flow rate. The volumetric efficiency can
be written in terms of flow rates as shown in Equations 4.18 and 4.19.
ev ,0 =
m air n R
Vd ρ 0 N
4.18
e v ,i =
m air n R
Vd ρ i N
4.19
36
It should be noted that the volumetric efficiency is a measure of the air entering the engine
but is not necessarily the air that is trapped or remains in the engine. It is possible that air
can come in the intake and exit through the exhaust without becoming trapped in the
cylinder. This is particularly true for 2-stroke engines. For these cases additional care must
be taken to describe the effectiveness of the intake system. Additional information will be
included in Chapter 9.
4.9. Thermodynamic and Fuel Conversion Efficiencies
The efficiency of many devices can be determined by applying a definition of desired
output over required input. Thus, the thermodynamic efficiency of an internal combustion
engine is given by Equation 4.20.
η th =
desired output Wout
=
required input
Qin
4.20
In internal combustion engines we know that heat is not really added but rather chemical
energy is converted to thermal energy during combustion. Additional information on the
way to describe combustion thermodynamically will be given in subsequent chapters. For
now, consider the heat that is released during a steady flow combustion process. If the fuel
and air reactants begin at the ambient temperature (T0 = 298 K), and then return to the
ambient temperature (T0 = 298 K) the heat released in this process is called the fuel heating
value. The processes can be written mathematically as shown in Equation 4.21 where: QHV
is the heating value, mf is the mass of the fuel, and H represents the enthalpy of a mixture
of reactants or products. The specific heating value of combustion (qHV) for gasoline is
44,000 kJ/kg.
Qideal = Q HV = m f q c = H reac tan ts (T0 ) − H ideal , prod (T0 )
4.21
The actual heat produced by a combustion process is given by Equation 4.22. In this
equation the heat generated in combustion is the difference of the enthalpies if the intake
and exhaust at the ambient temperature (Ta). This may be less than the heat of combustion
if the combustion process is incomplete. For example, the exhaust may contain unburned
fuel or partially burned fuel which still has chemical energy and therefore the actual energy
released and converted to heat is smaller than the total energy released.
Q act = H int ake (Ta ) − H exhaust (Ta )
4.22
The possibility of using two different values for Q in Equation 4.20 leads to two different
definitions of efficiency. The thermodynamic efficiency is defined in Equation 4.23 as the
ratio of work out to actual heat added during the combustion process. The fuel conversion
efficiency is given by Equation 4.24 as the ratio of the work produced to the ideal amount
of heat that can be produced by the fuel in a complete combustion process.
η th =
Wout
Q act
4.23
37
ηf =
Wout
Wout
=
Qideal m f q HV
4.24
The ratio of actual heat produced in a combustion process in comparison to the ideal
amount of heat that could be produced in a combustion process is shown in Equation 4.25
and is known as the combustion efficiency. The combustion efficiency in a properly
operating, modern spark ignition engines is 97-99%. There are however operating
conditions such as occur during start-up or acceleration when the combustion efficiency is
well below 90%.
ηc =
Q act
H
(T ) − H exhaust (Ta )
= int ake a
Qideal
m f q HV
4.25
The fuel conversion efficiency is therefore the product of the thermodynamic efficiency
and the combustion efficiency as shown in Equation 4.26.
4.26
η f = η thη c
Each of the relationships developed in this section can also be written in terms of rates. For
example the fuel conversion efficiency can be written in terms of the power output of the
engine, the fuel flow rate, and the fuel heating value as shown in Equation 4.27.
ηf =
4.27
W out
m f q HV
4.10. Specific Fuel Consumption
It is common practice in the engine manufacturing community to report fuel consumption
instead of efficiency. The specific fuel consumption for an engine is found by dividing the
fuel burned by the work produced in a cycle or alternatively, the rate of fuel burned
normalized by the power produced as shown in Equation 4.28. Common units for specific
fuel consumption are g/kW-hr or in some cases g/hp-hr.
sfc =
mf
Wout
=
m f
W
4.28
out
The specific fuel consumption can be related to fuel conversion efficiency by:
sfc =
1
η f q HV
4.29
which shows that specific fuel consumptions are inversely proportional to fuel conversion
efficiency if the fuel heating value remains constant. Because all diesel and gasoline fuels
have similar heating values, specific fuel consumption provides a good means of
comparing most commercial IC engines but cannot be used to compare engines burning
38
petroleum-based fuels and alcohols. As can be seen in Equation 4.29, when heating values
are different between fuels, the sfc can no longer be attributed solely to the fuel conversion
efficiency.
4.11. Piston Cylinder Geometry Motion
The compression ratio for an engine is the ratio of maximum volume to minimum volume
in the cylinder as shown in Equation 4.30; where: Vc is the clearance volume and Vd is the
displacement volume.
rc =
4.30
V max Vc + V d
=
W min
Vc
For the slider crank mechanism shown in Figure 4.3, the crank radius is “r” and the
connecting rod is length “c”.
The stroke is:
4.31
S = 2r
Vc
y
n
m
c
θ
r
Figure 4.3 Geometry of the piston, cylinder, and connecting rod.
The displacement volume is:
Vd =
SπB 2
2rπB 2
=
4
4
4.32
The total volume at a given crank angle θ is and distance, y, the piston has traveled from
TDC as shown in Figure 4.3 is given by Equation 4.33.
39
V (θ ) = Vc +
For:
A=
4.33
yπB 2
4
4.34
c
r
the volume as a function of crank angle theta is given by:
1
V (θ )
= 1 + (rc − 1)( A + 1 − cos θ − ( A 2 − sin 2 θ )1 / 2 )
2
Vc
4.35
Proof of this equation is left as an exercise for the reader. It is often difficult to find data
on the connecting rod length “c” while bore and stroke of an engine are normally readily
available. If no data are available, the connecting rod is often approximately four times the
crank radius or A = 4.
This final expression can be used to determine the volume as a function of crank angle,
compression ratio, connecting rod length and stroke. The volume can be seen to take on a
periodic, somewhat sinusoidal shape which changes very slowly near the top and bottom
dead center crank positions. The piston velocity can be found by writing the volume in
terms of the vertical directional component “y” and differentiating with respect to time.
The acceleration which is proportional to the dynamic forces produced by the piston can
be found by differentiating twice.
The mean piston speed can be found by dividing the distance the piston travels in one
rotation divided by the time for a single rotation or:
Up =
dis tan ce 2[ strokes / rev] L[m / stroke]
=
= 2N L
1 / N [rev / s ]
time
4.36
Example: Mean Piston Speed
Given:
Automotive News lists the Chevrolet Aveo to have a bore and stroke of 3.11 and 3.21 inches
respectively with a peak power of 103 hp at 5800 rpm. A large marine engine has a stroke of 36 inches
and a speed of 300 rpm.
Find the mean piston speed of these two engines.
For the Aveo
U p = 2 NL = 2[ strokes / rev] 5800 [rev / min] 3.21 [in / stroke]
For the Marine Engine
U p = 2 NL = 2[ strokes / rev] 300 [rev / min] 36 [in / stroke]
1 [min]
= 620.6 in / s
60 [sec]
1 [min]
= 360 in / s
60 [sec]
40
Although larger engines tend to have lower rational and mean piston speeds, mean piston
speed varies less between engines than engine rotational speed or bore and stroke. In this
example, the ratio of engine speeds is 19.3:1 and the ratio of engine strokes is 11.2:1 but
the ratio of mean piston speeds is only 1.7:1. For the commercial automotive market, bore,
stroke, and mean piston speed are relatively constant and vary less than a factor of 2
indicating that for automobiles torque is increased by adding cylinders rather than
increasing cylinder size.
4.12. The Power Equation
An equation can be derived which provides insights into the parameters that produce power
in an IC engine.
The derivation begins by writing power as a function of work and engine speed as shown
in Equation 4.37.
4.37
WN
W =
nR
Where, nR is the number of revolutions per cycle.
The work done per cycle is the product of the specific fuel energy, fuel mass, and the fuel
conversion efficiency.
q fuel m fuelη f N
4.38

W=
nR
The fuel mass flow is the product of the air mass flow and the fuel air ratio (F/A).
Substituting for fuel mass flow gives:
F
4.39
m air
η f q fuel N
A
W =
nR
The air mass flow is replaced with an expression for volumetric efficiency (see
Equation 4.19) to produce the final result.
F
4.40
W = e ρ V
N /n
η q
v ,i
i
d
A
f
fuel
R
This relatively simple expression is a powerful tool for explaining the performance of an
internal combustion engine. Each term in the equation should be considered carefully to
determine the possible influence on the power output.
The first term in the equation is the intake based volumetric efficiency. This parameter
describes how well air flows through the intake port and intake valve area. The intake based
41
volumetric efficiency typically has values in the range of 0.5 to 1.2 and will be discussed
in more detail in Chapter 9 where it will be seen that volumetric efficiency is a strong
function of engine speed. It is relatively easy to produce high volumetric efficiencies over
a narrow range of speeds but difficult to achieve high values at low, medium, and high
speeds. The wide range of possible values for volumetric efficiencies makes this a critical
parameter to consider when attempting to maximize torque and power in an engine.
The second term in the equation is intake air density. This variable can range from near
zero at idle to a value of two to three atmospheres using a turbocharger. The wide range of
possible values makes this the most important parameter for producing engine power. It is
this parameter which is used to control a spark ignition engine output using a throttle.
Turbocharging is a direct effort to increase the intake air density which then increases
power proportionally.
The next term in the equation is displacement volume. The equation shows that power
output is directly proportional to displacement volume. All things equal, an engine of twice
the displacement volume should produce twice the power. Dividing by the rate of engine
displacement produces the MEP equation as shown in Equation 4.41. This equation allows
the comparison of engines of different sizes running at the same speed. The mep equation
can also be considered a torque equation and provides insights into the terms that will
increase engine torque.
mep = ev ,i ρ i
F
η f q fuel
A
4.41
The fourth parameter in the power equation is the fuel air ratio. The fuel air ratio is normally
fixed to a value near stoichiometric for a spark ignition engine and therefore can have little
influence on the power output. Different fuels have different stoichiometric F/A ratio
values that can range over a factor of two. However, the product of F/A ratio and heating
value (qHV) is similar for all practical fuels in use today, including methanol, ethanol, and
other oxygenated fuels. In a diesel engine, the F/A ratio is not fixed and is the parameter
used to control the engine power. The F/A ratio in a diesel engine can be increased until
the engine reaches the “smoke limit” or point at which there is insufficient air to burn the
fuel at which time the exhaust becomes filled with smoke. Diesel engines are regulated to
operate well below the smoke limit to prevent unnecessary pollution, but aftermarket
devices are often sold that enable higher fuel air ratios and therefore produce more power
with the cost of polluting the air and create a bad public image for diesel engines.
The engine fuel conversion efficiency is the next term in the power equation. This
parameter can be the brake or indicated fuel conversion efficiency depending on whether
the brake in indicated power is the desired output. The indicated thermodynamic efficiency
is typically between 25 and 40 % and scales with engine size. While increasing efficiency
has many beneficially implications, it is not typically a design parameter capable of
providing significant increases in power.
42
The heating value of the fuel is next in the power equation. This value is relatively fixed
for hydrocarbon fuels and typically has little influence when seeking to increase power.
Alcohols, as mentioned above tend to have approximately half the energy content but burn
at twice the F/A ratio of gasoline making the change in power relatively small. There are
some fuels like nitro-methane that have substantially higher heating values and therefore
can produce significantly more power.
The last parameter in the numerator is engine speed. The power output is directly
proportional to engine speed and therefore, high engine speeds are used in racing to
produce high power output. Most automotive engine run below 6,000 rpm but numerous
race car and motorcycle engines run up to 15,000 rpm to access the extra power available.
The engine speed is the second most influential parameter on power output next to intake
pressure.
The final parameter is nR, the cycle parameter. This parameter points out that when all other
parameters are constant, the four stroke engine (nR = 2) will produce half as much power
as a two stroke engine (nR = 1). In practice, two-stroke engines do not produce twice the
power because volumetric efficiency is reduced for two-stroke engines but nevertheless, 2stroke engines do produce substantially more (usually about 50%) power for a given
displacement.
4.13. Fuel - Air Ratio
The amount of air entering an engine controls how much fuel can be burned. For a given
amount of air, the fuel can be increased until the mixture is stoichiometric, after which
the fuel can no longer be completely burned. When the amount of air in a mixture
matches the amount required to completely burn a fuel the mixture is said to be
stoichiometric. When the mixture has more fuel than the amount that can be burned the
mixture is said to be “fuel rich” or simply “rich” and when the mixture has less fuel than
stoichiometric, the mixture is “fuel lean” or lean.
The equivalence ratio (ϕ) defined in Equation 4.42 is fuel air ratio of a mixture divided by
the stoichiometric fuel air ratio. Table 4.1 summarizes the nomenclature of the
equivalence ratio with respect to stoichiometric. For an equivalence ratio less than 1.0 the
mixture is lean. In this condition the engine could produce more power by increasing the
amount of fuel and the fuel can theoretically still be burned and produce additional
energy release. For an equivalence ratio greater than 1.0, the fuel air mixture is rich.
Some fuel will be unburned or only partially burned and therefore wasted in a fuel rich
mixture. The fuel conversion efficiency of the engine will be low.
φ=
(F A)
(F A)
4.42
s
Table 4.1 Description of the equivalence ratio and associated terminology.
Equivalence
Ratio
Name
Description
43
φ < 1.0
Fuel lean
φ = 1.0
Stoichiometric
φ > 1.0
Fuel rich
The available oxygen could burn additional
fuel.
Maximum amount of fuel that can be burned
with the given amount of oxygen.
Not enough oxygen to complete combustion.
4.14. Indicated and Brake, Gross and Net - Revisited
At the beginning of this chapter, the terms Indicated, Brake, Gross and Net were introduced
with respect to work. Having introduced the numerous parameters that are used to describe
engines it is important to point out that for each occurrence of work in an equation, the type
of work must be specified, and the dependent parameter will change accordingly. For
example, if the thermodynamic efficiency is calculated using the indicated gross work then
the thermodynamic efficiency will be the gross indicated thermodynamic efficiency.
Table 4.2 provides a guide for naming the parameters based on the type of work that is
used in their calculation.
Table 4.2 Use of indicated, brake, gross and net prefixes for engine parameters.
Thermodynamic
Efficiency
Fuel Conversion
Efficiency
Specific Fuel
Consumption
Power
Mean Effective
Pressure
Indicated Gross
η th,ig
Indicated Net
η th,in
Brake
η f ,ig
η f ,in
η f ,b
igsfg
insfc
bsfc
W i , g
W i ,n
W b
igmep
inmep
bmep
η th,b
44
Chapter 4 Homework Problems
4.1 A cylinder pressure transducer and encoder are used to produce a P-V diagram of a
naturally aspirated, four-stroke cycle. The net indicated work from the P-V diagram is 0.18
kJ and the pumping work is 0.025 kJ. What is the gross indicated work.
4.2 The torque measured from a 4-stroke, 4-cylinder engine using a dynamometer is 150
ft-lbf at a dynamometer rotational speed of 2500 rpm. What is the brake work in Btu being
generated per cylinder, per cycle.
4.3 The torque measured from a 4-cyclinder engine using a dynamometer is 283 ft-lbf at a
rotational speed of 3000 rpm. The engine has a mechanical efficiency of 0.9. What is the
indicated gross work per cylinder, per cycle (Btu).
4.4 The torque measured by a dynamometer is 350 ft-lbf at a rotational speed of 1800 rpm.
What is the brake power produced by the engine in hp and kW.
4.5 Based on the exhaust composition and temperature it is determined that 3% of the fuel
energy was left unburned. If the brake thermodynamic efficiency is 28%, what is the
combustion efficiency and the brake fuel conversion efficiency.
4.6 A 6-cylinder engine is measured to produce 25 kW at the output shaft with a fuel flow
rate of 12 lbm per hour. What is the brake specific fuel consumption for this engine in
lbm/hp-hr.? What is the brake fuel conversion efficiency of the engine if the fuel heating
value is 44 MJ/kg.
4.7 From the list of typical engine components, identify one that you are not familiar with
or less familiar with. Find a picture of that component on the internet. Print the picture and
the URL used to find the picture and hand it in with your homework.
4.8 Does the P-V diagram plotted on a log scale below represent a naturally aspirated or
turbocharged engine? Why?
100000
Pressure (kPa)
10000
1000
100
10
0.00001
0.0001
0.001
Volume (m3)
45
4.9 The power measured at the shaft of a mini cooper, 4-cylinder, 4-stroke spark ignition
engine is 120 hp at 6000 rpm. If the mechanical efficiency is 0.9, what is the indicated
gross work per cycle for a single cylinder?
4.10 The intake based volumetric efficiency, ev,i, of a 4-stroke, single-cylinder, fuelinjected engine with a bore of 150 mm and stroke of 130 mm is 0.8. The intake pressure is
150 kPa and inlet temperature is 315 K. The engine speed is 2600 rpm. What is the air flow
rate to the engine?
4.11 A 3.8 L four-stroke fuel injected automobile engine has a brake power output of 88
kW at 4000 rpm and an atmospheric based volumetric efficiency of 0.85. Atmospheric
conditions are 298K 101 kPa. The bsfc is 0.35 kg/kW-hr. If the fuel has a heating value of
42 MJ/kg, what are the bmep, brake fuel conversion efficiency, and air to fuel ratio?
4.12 A six-cylinder, two-stroke engine produces a torque of 1100 Nm at a speed of 2100
rpm. It has a bore of 123 mm and a stroke of 127 mm. What is the bmep and mean piston
speed?
4.13 A two-stroke, 380 cc single cylinder motorcycle engine is operating at 5500 rpm. The
engine has a bore of 82 mm and a stroke of 72 mm. Performance testing gives a bmep of
6.81 bar, bsfc of 0.49 kg/kw-hr and an atmospheric based volumetric efficiency of 0.748.
A. What is the fuel to air ratio? B. What is the air flow rate (kg/s)?
4.14 A 4.0 liter, 4-stroke engine with a compression ratio of 9 is measured to produce 350
kW of brake power at 6000 rpm. Assuming the volumetric efficiency of the engine based
on standard atmospheric conditions is 1.0 and the mechanical efficiency is 0.9. The fuel air
ratio can be assumed to be 1:14.6 with a fuel heat of combustion of 44 MJ/kg. Find A. The
gross imep for this engine? B. The bsfc C. The gross indicated fuel conversion efficiency.
4.15 Assume that a diesel engine has an indicated thermodynamic efficiency of 0.4 and a
spark ignition engine has an indicated thermodynamic efficiency of 0.35 and that both are
naturally aspirated. It is a good assumption that both will have the same volumetric
efficiency and that both fuel have the same heating value but the diesel engine runs overall
lean at an A/F ratio of 20 while the spark ignition engine runs at stoichiometric (A/F =
14.6). Which engine has the higher MEP?
Homework - Modeling Problems
4.16 Using and Excel spreadsheet calculate the cylinder volume as a function of crank
angle. Place crank angle in the first column and volume (m3) in the second column. Use
1.0 crank angle increments beginning with -180 and continuing to +540 crank angles. Make
the spreadsheet such that the volume can be calculated for any combination of inputs for
bore, stroke, compression ratio, and connecting rod length. Using the following inputs
calculate the volume as a function of crank angle and print out the volume at the ten crank
angles from -90 to -80 crank angle degrees.
Bore (m)
Stroke (m)
Compression Ratio
0.0875
0.083
10
46
Con Rod (m)
0.166
4.17 Set up an EES calculation using the following inputs to calculate the listed outputs.
Use the program to check your answers on the problems above. Specifically, use your
program to solve problem 5.9. Print out the equation page and the solution block for 5.9.
Inputs
"Constants"
q_c = 44000 [kJ/kg]
R = 0.287 [kJ/kg-K]
AFratio = 14.6
"Normal Inputs"
bore = 0.0875 [m]
stroke = 2*0.0832 [m]
conrod = 0.175 [m]
r_c = 9
N = 6000 [rpm]
e_v0 = 1.0
P_0 = 101 [kPa]
T_0 = 298 [K]
P_intake = 101 [kPa]
T_intake = 298 [K]
“Mechanical Efficiency“
eta_m = 0.9
“Indicated Gross Fuel Conversion Efficiency”
eta_fig = 0.343
n_R = 2 [rev/cycle]
“Number of Cylinders”
n_cyl = 4
Outputs
V_d displacement volume {m^3}
V_c clearance volume {m^3}
rho_0 atmospheric density {kg/m^3}
QQ_c Rate of heat addition {kJ/s}
rho_i Intake Port Density {kg/m^3}
e_vi Intake based volumetric efficiency
m_dotair mass flow rate of air {kg/s}
m_dotfuel mass flow rate of fuel {kg/s}
P_ig Gross Indicated Power {kJ/s}
P_b Brake Power {kJ/s}
P_pumpfric Pumping Power + Friction Power {kJ/s}
igmep Indicated Gross Mean Effective Pressure {kPa}
bmep Brake Mean Effective Pressure {kPa}
pmepfmep Pumping + Friction Mean Effective Pressure {kPa}
W_ig Indicated Gross Work per cycle {kJ/cycle}
W_b Brake Work per cycle {kJ/cycle}
W_pumpfric Pumping + Friction Work per cycle {kJ/cycle}
igsfc Indicated gross specific fuel consumption {kg/kw-hr}
bsfc brake specific fuel consumption {kg/kw-hr}
T_b Brake Torque {N-m}
47
48
5. Engine Models
Producing a thermodynamic model of an IC engine can prove very insightful for learning
how various design parameters will impact engine performance. Since the implementation
of computing, engineers have developed numerous models ranging from very simple
spreadsheet type models to full 3-D comprehensive flow and combustion computational
models. In order to appreciate the accuracy and responsibly utilize more complicated
models it is important to understand the basic principles and assumptions that are required.
This chapter reviews the simplest of engine models introduced in most introductory
thermodynamics courses and extends these models to a slightly more sophisticated
spreadsheet model. Detailed comprehensive models are beyond the scope of this text.
5.1. Thermodynamics Review
Prior to attempting a thermodynamic model of an IC engine it is important to review and
establish basic principles of thermodynamics. This section reviews the conservation of
energy equation, the entropy equation, the ideal gas law and how to characterize a mixture
of ideal gases. The presentation is brief and the reader should consult a thermodynamics
text for a more detailed explanation if needed.
5.1.1. Conservation of Energy
The conservation of energy equation for a system is given by Equation 5.1 where; U is the
internal energy, KE is the kinetic energy, PE is the potential energy, h is enthalpy, Q is
rate of heat transfer, and W is rate of work or power.
d (U + KE + PE ) CV
= m i ( h + ke + pe) = m e ( h + ke + pe) + Q in − W out
dt
5.1
For a closed system where 1) no mass enters or leaves the system; 2) the only type of work
is boundary work and 3) changes in kinetic and potential energy are negligible; the energy
equation simplifies to:
d (U ) CV
PdV
= Q in −
dt
dt
5.2
dU = ∂Qin − PdV
5.3
or
Where: P is the pressure and V is the volume.
49
5.1.2. Entropy Equation
The entropy equation for a system is given by Equation 5.4. In an IC engine, entropy can
be generated by combustion, friction, viscous dissipation, unrestrained expansions, heat
transfer through a temperature gradient, and mixing.
Q
∂ in 
T 
dS CV

= m i s − m e s +
+Sgen
dt
dt
5.4
For processes which are reversible (do not generate entropy) and adiabatic, the entropy
equation reduces to:
dS CV = 0
5.5
5.1.3. Ideal Gas Law
The ideal gas law is given by Equation 5.6 where R is the ideal gas constant for the specific
gas or mixture being considered and T is the absolute temperature. The ideal gas constant
can be determined by dividing the universal ideal gas constant Ru = 8.314 J/mol-K by the
molecular weight of the gas: R = Ru/MW.
PV = mRT
5.6
5.1.4. Mixtures
Ideal gases often occur as a mixture of several pure component gases. For example, air is
a mixture of oxygen, nitrogen, argon, carbon dioxide and potentially many other gaseous
species. Mass fractions and mole fractions defined by Equations 5.7 and 5.8 are used to
quantify the composition of mixtures. The mass fraction of component “i” is represented
by mfi and the mole fraction by yi. The variables m and N represent total mass and total
moles respectively.
mf i =
mi
m
5.7
yi =
Ni
N
5.8
These definitions lead to the following relationships:
∑ mf = ∑ y = 1
i
i
5.9
50
The components of standard air that are near or greater than 1% are: y N 2 = 0.78, y O 2 = 0.21,
y Ar = 0.01. For simplicity, air is often modeled as y N 2 = 0.79 and y O 2 = 0.21 or the molar
percentages are 79% N2 and 21% O2.
5.1.5. Ideal Mixtures
Properties for ideal gases and ideal mixtures can be found by summing the contributions
of each component in the mixture. The following relationships may be applied.
Mass Based
Molar Based
∑m
h = ∑ mf h
u = ∑ mf u
m=
∑N
h =∑y h
u =∑y u
MW = ∑ y MW
N=
i
i i
i
i i
i i
i i
i
The ideal gas law for a mixture may be written
PV =
P=
∑ N RT
N i NRT
5.10
i
∑ N V =∑ y P =∑P
i
i
i
5.11
Where Pi is the partial pressure of a component of a mixture. For example, if a room is
filled with air that is 21% oxygen and 79% nitrogen and the total pressure in the room is
100 kPa, then the partial pressure of oxygen in the room is 21 kPa and the partial pressure
of nitrogen is 79 kPa.
5.1.6. Ideal Gas Properties
Because air is a mixture of ideal gases at normal atmospheric conditions, the properties of
air are often calculated and presented as a mixture. Properties for air as an ideal gas are
found in Table A.1 in the appendix. Temperature dependent properties for CP, Cv, and
k=CP/Cv for several gases commonly found in IC engines are given in Table A.2.
5.2. Overview of a real combustion cycle
Before selecting a thermodynamic model for the processes occurring in an internal
combustion engine, let us first consider the process that is actually occurring. A plot of
pressure vs. volume from an in-cylinder pressure transducer for a single cylinder directinjection Diesel engine is shown in Figure 5.1. Periods of compression, combustion,
expansion, exhaust, and intake are indicated in the figure.
51
Pressure (MPa)
15
10
End
Combustion
End
Compression
Start
Compression
Start Exhaust
End Intake
Start
Combustion
5
End Exhaust
Start Intake
0
0.05
0.25
0.45
0.65
0.85
1.05
Volume (L)
Figure 5.1 Pressure as a function of volume for a Cummins 5.9 L Diesel Engine
The five processes of the 4-stroke cycle are more easily identified on a LogP vs LogV
diagram as shown for the same data in Figure 5.2. On the logP vs. LogV diagram the
compression and expansion processes are clearly identified as straight lines indicating they
are polytropic processes. Combustion is a curved line occurring and the end of compression
and the start of combustion and the intake and exhaust processes are relatively constant
pressure processes made visible by the log scale.
Beginning in the bottom right corner, the piston is at BDC and the cylinder is full of fresh
air mixed with a small amount of residual gas left over from the previous cycle. As
compression proceeds the volume decreases and pressure increases and the diagram
follows a nearly straight line. Just before the end of compression, the line begins to curve
indicating heat addition. Fuel injected into the cylinder prior to heat release is begins to
burn at the point where the slope of the LogP vs LogV line changes.
During combustion the mixture is changing from fresh air and residual gas to products of
combustion. The slope of the logP vs LogV curve is continuously changing. The end of the
curve portion and the start of the straight line during expansion indicates the end of the
combustion process.
The straight line indicating a polytropic expansion ends when the exhaust valve is opened.
At this point blowdown occurs as the high pressure combustion products exit the exhaust
valve and seek equilibrium pressure with the exhaust port. A large fraction of the mass
exits during this process. Most of the remaining mass exits as the piston displaces the
cylinder contents during the exhaust stroke which ends at TDC or minimum volume. A
small amount of product gas remains in the cylinder at this point called the residual mass.
In this engine, the intake air is turbocharged and so the pressure increases when the intake
valve is opened. The intake process is seen to occur at a higher pressure than the exhaust
52
process which proceeds through the intake stroke until the piston reaches BDC, the intake
valve closes, and the process starts all over again.
10
Pressure (MPa)
Expansion
1
Blow
Down
Combustion
Compression
Intake
0.1
Exhaust
0.01
0.05
0.5
Volume (L)
Figure 5.2 Log P vs Log V diagram for the same data shown in Figure 5.1.
5.3. Ideal Cycles
The ideal Otto and ideal Diesel Cycles are simplifications of the real processes occurring
in an engine and are utilized to provide a basic understanding and prediction of what is
occurring thermodynamically in the engine. Both cycles employ the following simplifying
assumptions.
Ideal Cycle Assumptions
1. The working fluid throughout the entire cycle is air.
2. All processes are internally reversible and the compression and expansion processes are
also adiabatic.
3. The combustion process are replaced by a heat addition process
4. The exhaust and intake processes are replaced by a heat rejection process
As can be seen in the P-V data from an actual diesel engine, the combustion process takes
place over a short period of the cycle but the period is neither instantaneous, constant
volume or constant pressure. The Otto and Diesel Cycles provide two approximations
which bracket actual cycles. The Otto Cycle assumes heat is added at constant volume
while the Diesel cycle assumes heat is added at constant pressure. These assumptions allow
trends in engine performance to be explored even though they are seriously flawed in their
ability to predict accurate magnitudes for work and efficiency. Generally speaking, the
ideal Otto cycle will over predict work and efficiency estimations by a factor of 2.
53
5.3.1. Ideal Otto Cycle
A P-V diagram for an ideal Otto cycle is shown in Figure 5.3. The simplified energy and
entropy equations for the four processes are given in Table 5.1.
3
P
2
4
1
V
Figure 5.3 Pressure vs Volume diagram for an Otto Cycle
Table 5.1 Processes, assumptions and the resulting simplified energy and entropy equations
for the ideal Otto cycle.
Process
1-2
2-3
3-4
4-1
Assumptions
Adiabatic
Reversible
Constant Volume
Adiabatic
Reversible
Constant Volume
-
–
Energy
Entropy
U 2 - U 1 = Win,1- 2
S 2 = S1
U 3 - U 2 = Q in,2-3
dS =
U 4 - U 3 = − Wout,3- 4
S 4 = S3
U 1 - U 4 = Q in,4-1
dS =
∂Q
T
∂Q
T
Using the additional assumption of constant specific heats, the following relations can be
applied.
Table 5.2. Processes, assumptions and the resulting energy and entropy equations for the
ideal Otto cycle with constant specific heats.
Process
1-2
2-3
3-4
4-1
Assumptions
Adiabatic
Reversible
Constant Volume
-
Adiabatic
–
Reversible
Constant Volume
Energy
Entropy
Win,1- 2 = mC v (T2 - T1 )
T2  V1 
= 
T1  V 2 
k −1
mC v (T3 - T2 ) = Q in,2-3
mCv (T4 - T3 ) = − Wout,3-4
T3  V 4 
= 
T4  V3 
k −1
mC v (T1 - T4 ) = Q in,4-1
54
Using these relationships and the ideal gas law, the temperature and pressure can be
determined at each point in the cycle and the work and heat for each process can be
calculated.
For the ideal Otto cycle with constant specific heats, the overall thermodynamic efficiency
can be related to the compression ratio and ratio of specific heats with a relatively simple
result. The derivation begins by calculating the desired output or net work out and divided
by the required input or heat added as shown in Equation 5.12.
η th =
Wnet, out
Q in
=
Wout ,1− 2 + Wout ,3− 4
5.12
Qin, 2 −3
After substituting results from Table 5.2 it can be shown that:
η th = 1 −
1
rck -1
where k =
Cp
5.13
Cv
This results demonstrates that in order to obtain a high thermodynamic efficiency, the
compression ratio and/or the ratio of specific heats of the working gas should be increased.
Although the magnitude of this predicted efficiency is typically off by a factor of 2 or more,
the trend predicted by this equation is validated by engine performance. Compression ratios
are made as high as possible in an IC engine to improve efficiency. If gasoline is present
in the fuel air mixture, it will autoignite or knock when the compression ratio of an engine
approaches approximately 10:1. Thus gasoline engines are knock limited in their
efficiency. Compression ignition engines do not contain fuel in the mixture during
compression. According to Equation 5.13, efficiency would always increase with
increasing compression ratio and therefore an infinite compression ratio would be best. In
practice this is not the case because the adiabatic assumption during compression and
expansion used to obtain Equation 5.13 become increasingly erroneous as the compression
ratio increased. In practice, a maximum efficiency typically occurs around a compression
ratio of 20:1. Higher compression ratios typically produce lower efficiencies because if
increased heat loss during compression and expansion.
The other parameter impacting efficiency in Equation 5.13 is the ratio of specific heats. It
is also important and often overlooked. Air has a higher value of k than fuel or fuel
products. Therefore, engines which run on high air to fuel ratios can have higher efficiency
if the fuel burning processes is not significantly slowed by the higher air to fuel ratio.
5.3.2. Ideal Diesel Cycle
The ideal diesel cycle assumes a combustion process where the fuel is burned such that the
combination of fuel burning and expansion produces a constant pressure as shown in
55
Figure 5.4. Clearly this imaginary cycle does not match a real diesel cycle as is shown in
Figure 5.1. The real Diesel cycle does not produce a constant pressure heat release process.
As a matter of fact, spark ignition and Diesel engine P-V diagrams look very similar. With
regard to the combustion process, the real heat release for both engine types is slower than
and Otto cycle and more rapid than a Diesel cycle. The Otto and Diesel cycles are therefore
useful in bracketing the real cycles.
All of the processes in the Diesel cycle are the same as those of the Otto cycle except for
the combustion or heat addition process. For this process, all three terms in the energy
equation: internal energy, heat, and work are nonzero as shown in Equation 5.14
dU = δ Q in - δWout = δ Qin − PdV
2
5.14
3
P
4
1
V
Figure 5.4. Pressure – Volume diagram for an ideal Diesel cycle.
For a constant pressure process, dH = dU + PdV and therefore, Equation 5.14 can be
rearranged and integrated to produce:
H 3 - H 2 = Q in
5.15
If specific heats are assumed constant, the temperatures can be related to heat addition as
shown in Equation 5.16.
mC p ( T3 - T2 ) = Q in
5.16
For the diesel cycle the same isentropic relations apply as are shown for the Otto cycle in
Table 5.2, but care should be taken that the volume ratio V3/V4 is no longer the same as the
56
compression ratio. The volume ratio V3/V4 can be found by first using Equation 5.16 to
find T3 and then using the ideal gas law and the fact that P3 = P2 to obtain:
V3 P2 T 3
=
V2 P3 T2
5.17
5.4. Ideal Cycle with Variable Heat Release
One of the largest errors of the ideal Otto Cycle is the assumption of constant volume heat
release. While it is complicated and beyond our scope at this point to predict the rate of
heat release from first principles, an empirical approach is introduced here to allow a more
reasonable prediction to be implemented.
It has been shown, that heat release from a spark ignition engine takes on the shape of the
Weibe function which is given by Equation 5.18.
Where:
θ = crank angle
  θ − θ n 
s
 
xb ( θ ) = 1 − exp − a
  θ d  
5.18
θ s = crank angle at start of heat release
𝜃𝜃𝑑𝑑 = duration of heat release
n = Weibe form factor
a = Weibe efficiency factor
The shape of this function is an S-shaped curve with a small slope at the beginning and end
and a steep slope in the middle. Parameters that have been found to fit well to measured
burn rate data for spark ignition engines are a = 5.0, n= 3.0
5.4.1. Modeling a variable heat release
Having a defined function for the fraction of fuel burned makes it relatively easy to produce
a model that accounts for variable heat release. Beginning at the start of compression with
a known pressure, temperature and volume, the ideal gas law can be used to determine the
mass which completely defines the initial condition. For each step through the cycle, the
gas undergoes both a compression or expansion process and a heat addition process.
Combining an isentropic compression with a constant volume heat addition process
between each crank angle produces a temperature for each crank angle as shown in
Equation 5.19. Now for each crank angle θ, volume can be calculated using Equation 4.35,
the fraction of fuel burned by Equation 5.18 and the temperature using Equation 5.19.
Once T and V are known, the initial mass which is constant can used with T and V and the
ideal gas law to obtain pressure.
57
V 
Ti +1 = Ti  i 
 Vi +1 
k −1
+ (xb i +1 − xb i ) QHV / mCV , i
5.19
5.5. The Four Stoke Cycle
The ideal four stroke cycle is the simplest attempt to model both the closed and open
processes occurring in the cycle. A P-V diagram of an ideal 4-stroke process is shown in
Figure 5.5. The initial condition in a 4-stroke cycle is not readily available. The incoming
fresh air and perhaps fuel are mixed with the residual mass in the cylinder from the previous
cycle to produce a mixture that is higher in temperature than the intake but yet at the same
pressure. As will be seen, an iterative approach is required to obtain an initial temperature.
3
2
P
4
Pe
Pi
6
1
7
5
V
Figure 5.5 P-V diagram of an Ideal 4-Stroke Otto Cycle
5.5.1. Closed Portion of the Cycle
The processes from 1-4 of the 4-stroke ideal Otto cycle are the same as the ideal Otto cycle.
The calculation of temperature, pressure, work and heat should proceed in the same way
as the ideal Otto cycle with an initial guess for the value of T1 with P1 equal to the intake
pressure.
5.5.2. Blow Down
The process from state 4 to state 5 is known as blow-down and represent the opening of
the exhaust valve and decrease in pressure from the cylinder pressure to the exhaust
pressure. From state 4 to 5, the contents of the cylinder can be assumed to undergo and
isentropic expansion from the cylinder pressure P4 to the exhaust pressure P5. For the
assumption of constant specific heats this allows the temperature to be found using
Equation 5.20
k −1
P  k
T5 = T4  5 
 P4 
5.20
Example Problem –Constant Specific Heat, Otto Cycle
Find the pressure and temperature at the end of the blow-down process for an ideal Otto cycle with
the follow inputs: T1 = 320 K, P1 = 85 kPa, rc =9.5, QHV/m = 2830. Pexhaust = 85 kPa, k=1.4
58
V 
T2 = T1  1 
 V2 
k −1
= 320(9.5)1.4 −1 = 787.5 K
 T  V 
787.5 9.5
= 1987.2 kPa
P2 = P1  2  1  = 85
320 1
 T1  V2 
T3 = T2 +
P3 = P2
Q
2830 kJ / kg
= 787.5 K +
mCv
0.718 kJ / kgK
= 4729.0 K
T3
 4729.0 
= 1987.2
 = 11,933 kPa
T2
 787.5 
V 
T4 = T3  3 
 V4 
k −1
= 4729 (
1 1.4 −1
= 1921.7 K
)
9.5
 V  T 
1 1921.7
P4 = P3  3  4  = 11,933
= 510.4 kPa
9.5 4729.0
 V4  T3 
k −1
1.4 −1
P  k
 85  1.4
T5 = T4  5 
= 1151.5 K
= 1921.7 

 1510.4 
 P4 
5.5.3. The Exhaust Processes
The process from state 5 to state 6 is a constant pressure uniform flow of mass out of the
cylinder. Neglecting changes in kinetic and potential energy and assuming there is no mass
entering and no heat transfer, the energy Equation simplifies to:
5.21
dU CV
dV
= −m e he − P CV
dt
dt
For constant pressure:
5.22
d ( U + PV )CV = − heδme
d ( mh )CV = − heδme
mCV dhCV + hCV dmCV = −heδme
But the specific enthalpy of the control volume (hcv) must be equal to the specific enthalpy
of the gas exiting the control volume (he) and the change in mass of the control volume
(dmcv) is equal and opposite in sign to the mass exiting the control volume (-δme).
Therefore, the second term on the LHS is equal to the RHS and cancels leaving dhCV = 0,
or T6 = T5.
Combining all of the expression for the closed portion and the exhaust portion of the cycle;
for an ideal gas at constant specific heat, the temperature at the end of the exhaust process,
T6, can be related to T1 by Equation 5.23.
T6 = T1 +
 P5 
Q
 
( k −1) 
mCv rc
 P4 
k −1
k
5.23
5.5.4. The intake Process
In the 4-stroke cycle, point 6 on the P-V diagram represents the location where the
minimum exhaust or residual mass is present in the cylinder.
59
From state 6 to 7, the exhaust valve closes and the intake valve opens. During this process,
products flow in and out of the cylinder depending on the relative pressures of the intake
and exhaust. For naturally aspirated IC engine, the pressure during intake is lower than
during exhaust because the intake is throttled. This is the case shown in Figure 5.5. For this
case, exhaust will flow from the cylinder into the intake when the intake valve is opened
and then flow back into the cylinder as the piston moves down and draws in the intake
mixture. Just like blow-down, the temperature drops in an isentropic process within the
cylinder when the intake valve is opened. The temperature at T7 can be estimated by
Equation 5.24.
k −1
P  k
T7 = T6  7 
 P6 
5.24
It is difficult to solve the energy equation for the process between 7 and 1 because the
energy that exited the cylinder into the intake plenum changed the temperature and
composition of the intake and therefore hi is no longer known. A simpler approach is to
back up and consider the combined processes from 6 to 1 and assume that any mass that
left the cylinder from process 6-7 reenters during process 7-1. Thus, no mass is exiting and
the air entering is at the state of the intake air. The energy equation can be written for the
entire process from state 6 back to state 1 as shown in Equation 5.25.
dmi δW
dU
= hi
−
dt
dt
dt
5.25
Integrating and simplifying the energy equation results in:
m1u1 + P1V1 − m6u6 − P6V6 = mi hi + P1V6 − P6V6
5.26
Which can be further modified by assuming constant specific heats and solving for T1:
T1' =
miTi + m6T6
V
+ 6 ( P1 − P6 )
m1
m1C p
5.27
Where Ti is the temperature of the mixture in the intake and all other subscripts refer to
states on the P-V diagram in Figure 5.5. Note that we have now gone all the way around
the cycle ending with the initial temperature. AS can be seen, T1 is not equal to the intake
temperature. This is because the residual mass in the cylinder is higher than the intake
temperature and when the fresh mixture entering is mixed with the residual mass, the
temperature goes up. The temperature found by Equation 5.27 is therefore labeled T1’
because it is not the same temperature used at the start of the cycle. T1 is therefore rarely if
60
ever a known value. It is difficult to measure and must therefore be approximated. One
method is to iterate using the sequence of equation presented above. Beginning with a guess
for T1 the cycle is calculated to find T1’ and then T1’ is used as the new T1 to calculate the
cycle again. Rapidly, the value for T1 converges. This iterative process can be reduced to
two equations as will be shown in Section 5.5.5.
5.5.5. Residual Mass Calculation
The residual mass is an important quantity that impacts performance in IC engines. The
residual mass is composed of primarily combustion products, CO2, H2O and N2 and a little
oxygen. A large amount of residual mass reduces the capacity of the engine to burn fuel
and also reduces the peak combustion temperature by diluting the mixture. The residual
mass therefore has a strong impact on power and emissions.
If they are known, the residual mass can be calculated from the density and volume at state
6 or at TDC prior to intake valve opening as shown in Equation 5.28. The temperature in
the cylinder is difficult to measure and therefore is often found using a model of the cycle.
m res = ρ 6V6 =
P6
V6
RT6
5.28
As discussed in the section above, the temperature and pressure at each point in the cycle
can be estimated by guessing an initial temperature to start the cycle and then iterating until
the final temperature in the cycle is equal to the initial.
Instead of calculating each temperature and pressure in the cycle, it can be useful to
approximate the residual mass by iterating on a more condensed set of equations. We first
assume that all processes are ideal as in the ideal Otto cycle and all processes occur at
constant specific heats. Thus all the relationships developed above for the ideal 4-stroke
Otto cycle can be applied.
The residual mass fraction is defined as the ratio of the mass at state 6 divided by the total
mass or the mass at state 1 as shown in Equation 5.29 . Substituting as shown provides the
residual fraction in terms of the compression ratio, a temperature ratio and a pressure ratio.
f =
m6 m6 V6 / v 6 V6 v 4 1 T4 P6
=
=
=
=
m1 m 4 V4 / v 4 V4 v 6
r T6 P4
5.29
The temperature ratio can be related to a pressure ratio as follow in Equation 5.30.
k −1
k −1
P  k
T4 T4  P4  k
=
=  
=  4 
T6 T5  P5 
 P6 
5.30
And then substituting Equation 5.30 into 5.29 gives.
61
5.31
1
1  P k
f =  6 
r  P4 
Now the relationships for a 4-stroke Otto cycle can be used to relate the pressure at state 4
to the pressure at state 1 as shown in Equation 5.32 and when this is substituted into
Equation 5.31the result is Equation 5.33 which gives the residual fraction in terms of the
initial temperature, pressure, compression ratio, and heating value of the fuel. The
pressures at P5 and P1 represent the exhaust pressure (Pe) and intake pressure (Pi)
respectively.

Qr 1−k 
P4 = P1 1 +

 m1C vT1 
f =
(Pe / Pi )
1
1/ k
r 
Qr 1−k 

1 +
 m1CvT1 
1/ k
5.32
5.33
The final step to simplifying this expression is to recognize that m1 is the sum of the mass
of the fuel (mf), air (ma) and residual mass (mr) as shown in Equation 5.34.
m1 = ma + m f + mr
5.34
Dividing Equation 5.34 by m1 and solving for m1 in terms of the residual mass and air-fuel
ratio (A/F = ma/mf) gives:
A

m f  + 1
F

m1 =
(1 − f )
5.35
Substituting this result into Equation 5.33 gives:
f =
(Pe / Pi )1 / k
1
=
1/ k
rc 



1− k
 (1 − f )q c r

1 +  A 

  + 1C v T1 

 F

(Pe / Pi )1 / k
1
rc  (1 − f )q r 1− k 1 / k
mix
1 +



C v T1


5.36
Where qmix is the heating value of the fuel divided by the mass of the mixture.
This equation is clearly iterative as it contains the residual fraction on both sides of the
equation. Beginning with a guessed value for the residual fraction (a good initial guess is
zero) and T1 (a good guess is the intake temperature) and new value for residual fraction
can be found. The only problem is that T1 is not known and also requires iteration.
62
The Equation to determine T1 is developed as follows. Beginning with Equation 5.26, and
the definition of H = U + PV gives Equation 5.37.
m1h1 − m6 h6 = mi hi + V6 ( P1 − P6 )
5.37
Recognizing that mi = m1 - m6 and rearranging gives:
m1 h1 − m6 h6 − V6 ( P1 − P6 ) = mi hi = hi ( m1 − m6 )
5.38
Dividing by m1 and remembering the definition of “f” gives:
h1 −
m6
V
h6 − 6 ( P1 − P6 ) = mi hi = hi (1 − f )
m1
m1
5.39
Substituting the ideal gas law and making the assumption of constant specific heats (h =
CpT) gives:
C PT1 −
P6V6 RT1
V RT1
( P1 − P6 ) = C PTi (1 − f )
C PT6 − 6
RT6 P1V1
P1V1
5.40
Recognizing that P1=Pi and P6=Pe, Rearranging, factoring and canceling gives:
 P V
P 
R V6
T1 1 − e 6 −
(1 − e ) = Ti (1 − f )
Pi 
 Pi V1 C P V1
5.41
Now substituting: rc=V1/V6, R = Cp-Cv, k = Cp/Cv and simplifying gives:
T1 =
Ti (1 − f )

1  Pe
 + k − 1
1−
rc k  Pi

5.42
Equations 5.36 and 5.42 now constitute a set that can be used to iteratively determine the
initial temperature of the cycle T1 and the residual mass fraction f. It should be remembered
that these equations were developed assuming the ideal Otto cycle and constant specific
heats and therefore the results are only estimates arrived at using those assumptions.
Example: Residual Mass Fraction
Given: A naturally aspirated, Honda F4i, 600 cc, 4-stroke motor cycle engine has a compression ratio
of 12:1, runs on gasoline at an A/F ratio of 12.5 at 12000 rpm. The engine is being run on a hot day,
T0=310 K, at altitude, P0=85 kPa at full throttle (Pi = 85 kPa)
63
Find: A. The residual mass fraction, f. B. The temperature prior to compression, T1.
Solution: Begin by guessing T1 = Ti = T0 = 310 K, f = 0 and find an improved value for f using the
first equation below:
f =
(Pe / Pi )
1
1/ k
rc 


1− k 
1 + (1 − f )q c r 


A


 + 1C v T1 
F




1/ k
T1 =
Ti (1 − f )

1 P
1 −  e + k − 1
rk  Pi

We are given: r = 12, A/F=12.5, Pe = 85 kPa (naturally aspirated), Pi = 85 kPa, q = 44,000 kJ/kg (for
gasoline)
We assume constant specific heats at 300 K: Cv = 0.718, k = 1.4
f =
(85 / 85)
1
= 0.0221
1 / 1.4
12 



1−1.4

1 + (1 − 0)44,000 kJ / kg 12

  12.5 
+ 10.718 kJ / kgK 310 K 
 
1




1 / 1.4
Now use this result in the equation to find a new T1
T1 =
310 K ( 1 − 0.0221 )
= 330.7 K
 85 kPa

1

+ 1.4 − 1
1−
( 12 ) ( 1.4 )  85 kPa

Iterate until T1 is not changing:
Using T1 = 330.7 gives f = 0.0232
Using f = 0.0232 gives T1 = 330.3 K
Which is close enough to the previous values to stop the iteration.
T1 = 330.3
f = 0.0232
Example – Finding the mass of each component in a mixture: fuel, air, residual
Given: The residual mass fraction of a four-stroke engine has been determined to be
2%. A mixture of air and gasoline with a ratio of 14.6:1 are being burned in and
engine with a displacement of 2.5 L, and an atmospheric based volumetric efficiency
of ev,0 = 0.95. P0 = 101 kPa, T0 = 298 K.
Find: A. The fraction of fuel and air in the mixture and B. The total amount of fuel
delivered in a single cycle of all engine cylinders. C. The amount of heat released in
one cylinder for each cycle if this were a four-cylinder engine
Solution: The sum of all the mass fractions must equal 1.
mf
m
+
ma mres
+
=1
m
m
The residual mass fraction is given and the A/F ratio is known to be 14.6
64
A.
mf
mf
A
F + 0.02 = 1
+
m
m
mf 
A
1 +  = 0.98
m  F
mf
0.98
=
= 0.063
m 1 + 14.6
ma
= 1 − 0.02 − 0.063 = 0.917
m
B.
Using the definition of volumetric efficiency
ma = ρ 0Vd ev ,0 =
P0
101 kPa
Vd ev ,0 =
2.5 x10 −3 m 3 0.95 = 2.805 x10 −3 kg
RT0
0.287 kJ / kgK 298 K
Using the fuel air ratio we obtain.
m f = ma
1
F
= 2.805 x10 −3
= 1.921x10 − 4 kg
14.6
A
C.
Note that this would produce (1.921x10-4 kgfuel)(44,000 kJ/kgfuel) = 8.45 kJ of heat
per cycle when burned. If this were a four-cylinder engine it would be 2.11
kJ/cylinder.
Example: Finding the temperature at the start of compression
Given: A cycle simulation has been set up to produce the temperature and pressure
at every crank angle from -180 to +360 crank angle. The inputs for the simulation
are as follows:
Bore (m)
Stroke (m)
Con Rod (m)
r_c
A/F
q_LHV (kJ/kgfuel)
C_p (kJ/kg-K)
P_in (kPa)
P_atm
P_ex (kPa)
T_atm (K)
T_in (K)
0.0875
0.083
0.166
8
14.7
44,000
1.005
40
100
100
300
300
The resulting temperature at the end of blow-down is T181 = 1541.38 K.
Find: The temperature in the cylinder at the end of the intake (T540) stroke or the
start of compression.
Solution:
65
Begin by calculating same basic parameters:
Vd =
πB 2 L
4
V c = V360 =
=
π * 0.0875 2 * 0.083
4
= 0.000499 m 3
Vd
0.000499
=
= 7.123 x10 −5 m 3
rc − 1
8 −1
V −180 = Vc + Vd = 7.123x10 −5 + 0.000499 = 0.0005704 m 3
The temperature remains constant throughout the exhaust stroke, therefore, T_360
= T181 = 1541.38 K.
Between crank angles 360 and 540, the energy equation can be solved to produce
Equation 5.27which has been rewritten in terms of crank angles.
T 540=
miTi + m360T360
V360
+
( P−180 − P360 )
m−180
m−180 C p ,540
Or substituting gives:
The challenge to using this equation is the mass entering the cylinder mi depends on
the temperature we are trying to find, T540. Thus, the solution requires iteration. For
the first iteration find the mass entering assuming T540 is equal to T-180 at the start
of the cycle and use the ideal gas equation to find the masses before and after intake.
 40 kPa * 0.0005704 m 3   180 kPa * 7.123 x 10 -5 m 3 
 PV 
 PV 
−

−
m i = (m−180 − m360 ) = 
 = 

 RT  −180  RT  360  0.287kJ / kgK 300 K   0.287kJ / kgK 300 K 
(
)
mi = 2.650 x10 −4 − 1.612 x10 −5 = 0.000249 kg
m−180 = 2.650 x10 −4 kg
The temperature, T540 can then be estimated calculated:
0.000249 m 3 * 300 K + 1.612 x10 −5 m 3 *1541.38 K
7.123x10 −5 m 3
+
(40 − 100)
2.650 x10 −4 m 3
0.0002650 m 3 *1.005 kJ/kg - K
T 540 = 359.60 K
T 540 =
This temperature now becomes the new T-180 and the entire cycle calculation is
repeated until T540 of the current cycle matches T-180
Note that CP,540 was found at 300 K for this iteration but if variable specific heats
are used, Cp can be updated each cycle to the value that is correct for the current
temperature at 540 crank angles.
66
5.6. The Miller Cycle
The 4-stroke Otto cycle shown in Figure 5.5 shows a throttled engine where the intake
pressure is lower than the exhaust pressure. This is the most common method of controlling
the amount of air and fuel entering a spark ignition engine. This method is however very
inefficient. The lower the intake pressure, the higher the amount the amount of pump work
required to draw the mixture into the engine. An alternative method for controlling the
amount of intake air is early or late intake valve closing. Such a cycle is referred to as a
Miller Cycle named after R. H. Miller, who is credited with first presenting the idea of
creating a larger expansion ratio than compression. The cycle is shown schematically in
Figure 5.6.
3
2
P
4
6
1
5
7
V
Figure 5.6 Ideal Miller cycle or ideal cycle with early intake valve closing.
The Miller cycle proceeds just at the Otto cycle through the first five processes (numbers
1-6). Note that process 1-2 is compression and has a smaller compression ratio than 3-4
which is the expansion process. Beginning at state 6 and BDC after exhaust, the piston
moves down and brings in fresh fuel and air. Before reaching maximum volume the intake
valve closes and stops the intake of more fuel and air mixture. This causes the pressure to
decrease in an isentropic expansion from point 1 to point 7 on the diagram. At point 7 the
piston changes direction and the volume decreases returning to point 1 before compression
begins. Points 5, 6, and 1 are at atmospheric pressure.
The change in pumping work can be considered graphically by comparing the area under
the curve for process 5-6-7-1 in the ideal 4-Stoke Otto cycle shown in Figure 5.5and the
area under the curve 6-4-1-7 for the ideal Miller cycle shown in Figure 5.6. In order for the
two cycles to produce the same indicated gross work, the pressure at point 7 in the Miller
cycle must equal the pressure at point 1 in the ideal Otto cycle.
67
Chapter 5 Homework
5.1 Using Equation 5.13 which was derived for an ideal Otto cycle; calculate the (a) thermal
efficiency using the thermal efficiency and the information below calculate the (b) net
work, and (c) imep.
P1 = 100 kPa T1 = 300 K
r = 10
V1 = 0.5 L
k = 1.3
Q = 1.5 kJ
5.2 Given the engine characteristics listed below and assuming an ideal Otto cycle with
constant specific heats, find: (a) the pressure and temperature at each of the four states in
the cycle (b) the work for each process (c) the heat transfer for each process (d) the thermal
efficiency using net work and heat (Eq 5.12) and (d) the imep.
P1 = 100 kPa T1 = 300 K
r = 10
V1 = 0.5 L
1
k = 1.3
Q = 1.5 kJ
5.3 Derive the Otto Cycle efficiency equation. η = 1 − k −1
r
5.4 Given the engine characteristics listed below and assuming an ideal Diesel cycle with
constant specific heats use the energy equation and analyze each process to produce: (a)
the pressure and temperature at each of the four states in the cycle (b) net work. (c) thermal
efficiency (d) imep.
P1 = 100 kPa T1 = 300 K
r = 20
V1 = 0.5 L
k = 1.3
Q = 1.5 kJ
5.5 A Miller cycle is to be used on a spark ignition engine in place of a throttle. In order to
achieve medium load, the intake valve closes early such that the volume at intake closing
is 80% of the total volume at BDC. Compare the pumping work for this engine to the
pumping work of a throttled engine achieving the same load (the same pressure at BDC).
You may assume the intake and exhaust pressure of the Miller cycle are 100 kPa, Vd = 1L,
k =1.4, and r = 9.0.
5.6 If the temperature of stoichiometric exhaust products (MW approx. equal MWair)
leaving the cylinder at the end of the exhaust process is known to be 1000 K. Find the
residual mass for a 0.5 L displacement single cylinder engine with a compression ratio of
10. Assume the exhaust pressure is one atmosphere 101 kPa.
5.7 For an engine operating under the following conditions. Pi=59.5 kPa, T0=Ti = 300 K,
Pe = 101 kPa, φ=1.0, gasoline, k =1.4, Vd = 2.0 liter, rc = 9.5.
A. Estimate the initial temperature in the cylinder prior to compression.
B. Determine the mass of fuel in the cylinder.
5.8 Given and intake temperature of 276 K, Pi = 60 kPa, P0 =Pe= 101 kPa. Estimate the
residual mass fraction, and temperature T1. Use: r = 9.0 , φ=1.0, fuel = gasoline, Vd = 0.25
liters, k = 1.4
68
Homework - Modeling Problems
5.9 Produce an excel spreadsheet model of an ideal Otto Cycle with constant specific heats.
The model should produce the following at each crank angle from -180 to + 180: crank
angle (degrees), volume (m3), pressure (kPa), temperature (K), Cp (kJ/kg-K), CV (kJ/kgK), k, mass (kg) and work (kJ). The work column should produce the work out from the
initial crank angle to that point in the cycle. All of the heat should be added between 0 and
1 crank angle degrees and should be assumed to be added at constant volume. Inputs and
outputs listed below should be listed on a separate worksheet labeled “inputs-outputs”,
each in a single column. Two P-V plots should be in the input-output worksheet, one on a
log-log scale, the other on a linear scale.
Turn in: The table of outputs, the P-V diagram on linear coordinates and the crank angle
resolved results from -10 to +10 cad.
Inputs
Bore (m)
Stroke (m)
r_c
P_1 (kPa)
T_1 (K)
Con Rod (m)
Q_in (kJ)
C_p (kJ/kg-K)
0.0875
0.083
9
100
300
0.166
1.50
1.005
Outputs
V_d (m3)
V_crank angle 180 (m3)
V_crank angle 0 (m3)
A
Wi,g,out (kJ)
igmep (kPa)
T_crank angle 0
T_crank angle 1
T_crank angle 180
5.10 Make a model of an Otto cycle with constant specific heats as is defined in the
previous problem, but now add the heat release over a period of crank angles as defined by
the Weibe function. Add columns to your spreadsheet for the Wiebe function, total heat
added, and heat added in each crank angle. Adjust the temperature to increase according to
the heat released during each crank angle. Use the following inputs to produce the
following outputs as well as the P-V diagrams listed above. Turn in the table of outputs,
the P-V diagram on a linear scale, and all the results from -10 to +10 crank angles.
Inputs
Bore (m)
Stroke (m)
r_c
P_1 (kPa)
T_1 (K)
Con Rod (m)
Q_in (kJ)
C_p (kJ/kg-K)
Wiebe Function Parameters
Theta Start
Theta Duration
n, form factor
a, efficiency factor
0.0875
0.083
9
100
300
0.166
1.50
1.005
Outputs
V_d (m3)
V_crank angle 180 (m3)
V_crank angle 0 (m3)
A
Wi,g,out (kJ)
igmep (kPa)
T_crank angle 0
T_crank angle 1
T_crank angle 180
-20
60
3
5
69
5.11 Extend the constant specific heat, variable heat release model made in Excel to a 4stroke model which includes the exhaust and intake strokes. The P-V worksheet should
include all of the same output columns as before but should now begin at -180 crank angles
and proceed through an entire four stroke cycle to +540 crank angles. On the input and
output worksheet include following:
Inputs
Outputs
Bore (m)
Stroke (m)
Con Rod (m)
r_c
A/F
q_LHV (kJ/kgfuel)
C_p (kJ/kg-K)
P_in (kPa)
P_atm
P_1
P_ex (kPa)
T_atm (K)
T_in (K)
0.0875
0.083
0.166
8
14.6
44,000
1.004
80
100
80
100
300
300
T_1
f, res, frac
300.00
0.0414
Wiebe Function Parameters
Theta Start
Theta Duration
n, form factor
a, efficiency factor
0
1
3
5
V_d (m3)
V_1 (m3)
V_2 (m3)
A
m_1 (kg)
m_fuel (kg)
T_exhaust
T_intake
T_361
T_540
Wi,g,out (kJ)
Wi,p,in (kJ)
Wi,net
ηi,g
ηi,n
f, res. Mass frac.
rho_0*V_d, atm
rho_i*V_d
ev,0
ev,i
igmep (kPa)
inmep (kPa)
pmep
Using the solve function, produce and the corrected result and print them out.
Turn in the input-output results, the P-V plot and the P-V-T data from -10 to + 10 crank
angle degrees.
70
6. Combustion
6.1. Combustion Processes
In the combustion process, fuel and oxidizer react to produce combustion products. The
process is illustrated in Figure 6.1 where fuel and oxidizer at a high chemical and low
sensible energy are converted to products at a high sensible and low chemical energy. The
total energy remains constant until there is some type of heat loss or conversion of sensible
energy to work. A common misconception about the combustion process is that it occurs
instantaneously. In terms of the engine cycle, the process typically occurs over a period of
30 to 90 crank angles.
Products
Sensible Energy
Chemical Energy
Energy
Fuel
Oxidizer (air)
Time
Figure 6.1 Diagram of the combustion process which changes chemical energy in the fuel
into sensible energy in the products.
There are four mechanisms whereby combustion can occur. They are: 1. Premixed
deflagration wave (typical of a classic spark ignition engine), 2. premixed detonation wave,
3. mixing controlled jets (typical of a classic diesel engine or compression ignition), and
volumetric reaction. Because understanding the differences in these processes is important
for understanding how engines are classified, they will be described here for reference. A
more detailed discussion of combustion will be given in subsequent chapters.
6.1.1. Premixed Deflagration Wave
Premixed deflagration is a combustion process that proceeds by the propagation of a flame
through a mixture of fuel and oxidizer. The process begins by mixing fuel and oxidizer
71
(usually the oxidizer is mostly air) in a chamber such that the mixture is fairly uniform in
composition. The mixture is at a temperature too low to produce a reaction. A spark or hot
spot is introduced into the mixture creating a small region where the temperature is high
enough to cause a rapid chemical reaction. The reaction releases energy producing hot
products in the small region near the spark. The adjacent fuel and air then become heated
and react producing more high temperature products which in turn react with the adjacent
fuel and oxidizer. The subsequent reaction caused by hot products causes the reaction to
propagate through the mixture moving outward in a spherical shape until meeting a wall or
void in the fuel air mixture. The result of this combustion process is a relatively slow
moving reaction zone or flame front through a mixture. The reaction is limited by the
propagation speed or flame speed through the mixture.
This is the type of combustion process utilized by spark ignition engines. The spark initiates
a reaction within the combustion chamber. The fuel is typically gasoline but many other
fuels are possible. Because many fuels can be used, it is more general to refer to an engine
using this type of combustion process as a spark ignition engine. The fuel mixes with the
air during the intake process producing a relatively uniform mixture. During compression,
the fuel air mixture is heated but the mixture does not ignite until the spark is fired. The
flame then propagates from the spark to the combustion chamber walls.
6.1.2. Premixed Detonation Wave
As a flame moves through a mixture, the rate of the reaction is determined in part by the
temperature of the reactants. When the reaction occurs rapidly enough that the flame moves
at the speed of sound, the flame will create a compression wave in front of the reaction
which further heats the reactants before the arrival of the flame and thereby greatly
increases the flame speed. The result is a supersonic flame front proceeded by a shock
wave. If this occurs in an engine the pressure in the cylinder is not uniform and does not
push uniformly on the piston resulting in poor transformation of gas energy to piston
energy. The shock wave can also produce enhanced heat transfer causing melting and
focused pressure forces on specific locations that lead to material failure. Detonation is
therefore not a desirable combustion process in an IC engine and is avoided.
6.1.3. Mixing Controlled Jet
In a mixing controlled jet, the fuel is injected directly into the hot oxidizer. The oxidizer is
hot enough to cause a reaction with the fuel at the boundary of the fuel and oxidizer. The
fuel then burns as it mixes from within the jet and move to the surrounding oxidizer.
Because the fuel cannot burn without oxygen, the reaction rate is limited by the rate at
which oxygen is mixed into the fuel jet; thus the term mixing limited combustion.
This is the type of combustion process typically employed in a “diesel engine”. The intake
process involves only the ingestion of air. The compression process heats the air to a
temperature above the ignition temperature of the fuel. The fuel is injected utilizing very
high pressures at the time when combustion is desired. Fuels which ignite easily are
desirable for this type of combustion process. Because many fuels besides diesel can be
72
used for this type of combustion, it is more general to name engines that use this process
as compression ignition engines because the ignition processes is initiated by compression
rather than a spark. Gasoline is actually a poor fuel for this type of combustion process
because it is designed to resist compression ignition. Instead of igniting shortly after
injection, gasoline tends to evaporate, mix with the oxidizer, and then burn volumetrically
as will be describe in section 6.1.4.
6.1.4. Volumetric Reaction
When a mixture of fuel and air is heated uniformly, such as occurs in a compression
process, a temperature can be reached which causes the mixture to react. The rate of
reaction is not limited by mixing or the propagation of energy but by the kinetic reaction
rate which is exponential with temperature. For example, a deflagration wave typically last
for 30 to 90 crank angles while volumetric combustion can release the majority of the fuel
chemical energy in 0.5 crank angles or less.
It is very difficult if not impossible to produce a perfectly uniform temperature and mixture
concentration in an engine and therefore the volume that reacts is typically only a fraction
of the total volume. When a small volume reacts very rapidly, it produces a pressure wave
which reflects back and forth across the combustion chamber at the speed of sound. The
reflections result in a noise which is called engine knock. Like detonation, this extremely
rapid combustion process is not desirable. It causes engine damage due to extreme heat
transfer on specific surfaces and the rapid pressure rise can exceed the mechanical capacity
of components such as the head gasket, causing failure. The non-uniform pressure also
causes lower efficiency in the conversion of sensible energy to work.
Since the early 1990s, engine manufacturers have worked to control the volumetric reaction
of fuels by slowing down the reaction rate and controlling the engine temperature during
compression to produce the reaction at the desired time in the cycle. Two techniques are
commonly used to make this type of combustion process useful in an engine. The first is
to dilute the fuel air mixture with exhaust gas to produce a lower concentration of oxygen
in the mixture. This slows the kinetic rates of reaction and can also allow control of the
temperature. Second, the fraction of the engine volume filled with the mixture is reduced.
As a result, volumetric reactions have been slowed to useful rates and the combustion
process is now employed for certain operating conditions, particularly in engines utilizing
diesel fuel. These combustion processes have several names in the literature including:
Homogenous Charge Compression Ignitions (HCCI) and Partially Premixed Combustion
(PPC). The goal of these combustion processes is to release fuel chemical energy in a
premixed mode in order to avoid the formation of soot and in a lower temperature mode in
order to avoid the formation of nitrogen oxides.
6.1.5. Multi-Mode Combustion
The development of computer controlled electronic fuel injectors has allowed engineers
increased control over the mode of combustion utilized by internal combustion engines.
Fuel can be delivered at any time and location (in-cylinder or intake port) and in varying
amounts. Multiple injections of fuel can be delivered within a single cycle. This is enabling
73
a single engine to be operated in any one of the three acceptable combustion modes and to
be able to switch fuels or utilize more than one fuel on the same cycle and engine.
Traditional classifications of “gasoline” or “diesel” or “spark ignition” and “compression
ignition” are no longer good enough to completely describe an engine and the associated
technology.
6.2. Combustion Thermodynamics
The internal energy of a substance is the sum of sensible, latent, and chemical energies.
The three types of energy are described as:
Sensible Energy is the energy associated with the motion of molecules and includes the
vibrational, rotational, and translational energy. This is the form of energy that is most
closely associated with temperature (sense).
Latent Energy is the energy associated with bonds between molecules. Ice and liquid water
at the freezing point have the same temperature but ice molecules are bonded together and
it requires a fixed about of energy (latent heat) to separate them.
Chemical Energy is the energy associated with bonds within a molecule. Water consists of
two hydrogens and one oxygen atom bonded together. When hydrogen and oxygen
combine from separated elements to form H2O, a significant amount of energy is released.
Similarly, when water is separated from H2O into hydrogen and oxygen it takes the same
amount of energy to separate them as was obtained when H2O was formed.
The total enthalpy of a molecule is shown in Equation 6.1 where h is the sum of the
sensible, and latent enthalpy, and h 0f is the heat of formation or chemical energy. For nonreacting mixtures or substances, the heat of formation of all components in the mixture are
constants and therefore need not be considered when finding a change in enthalpy. In this
text the subscript “tot” will not be used and the difference between total enthalpy (htot) and
sensible plus latent enthalpy (h) will be determined by context.
htot = h 0f + h
6.1
In this chapter the method for determining chemical energies is introduced and then applied
to fuel air mixture to determine the amount of energy that is released and the temperature
produced when a mixture reacts or burns.
6.2.1. The Enthalpy of Formation
The chemical energy of a molecule is found by first recognizing that energy is a relative
and not an absolute value. It makes little sense to seek an absolute value for the elevation
of the city you live in. Elevation is always relative to something, normally sea level. So it
is with energy. There is no substance that has zero energy, not even at zero Kelvin.
74
Therefore, in order to assign a chemical energy to a substance, a reference point is selected
where energy is defined to be zero. All energies are then determined relative to that
reference. For chemical energy, each element needs a zero reference state. The default
value is to define the most naturally occurring molecule for each element to be zero. Several
important elements related to combustion are listed in Table 6.1. The term chemical energy
is given the name, heat of formation, or enthalpy of formation. This term comes from the
experiments that were done to measure the amount of heat produced when molecules
combine or form a new molecule. A general rule is that an element that is most commonly
found in nature is the form that is assigned to be zero. For example, oxygen can exist as O,
O2, or O3 (ozone) but the most common form of oxygen found in nature is O2 and therefore
O2 is assigned and energy of zero as a reference for all other forms of oxygen.
Table 6.1. Elements and the molecule for which the elements chemical energy is zero.
Element
Molecule
Enthalpy of Formation h 0f
Hydrogen, H
Oxygen, O
Nitrogen, N
Carbon, C
H2
O2
N2
Graphite Solid
0
0
0
0
The second reference used if for sensible enthalpy. For combustion, it is convenient to
assign zero enthalpy to that of an ideal gas at 298.15 K . This is written mathematically in
Equation 6.2.
h (T = 298) = 0 or h = h (T ) − h (T = 298)
6.2
Using these two reference points, the total enthalpy is given in Equation 6.3. The equation
shows that the total enthalpy is the sum of the heat of formation plus the sensible enthalpy.
Values for sensible enthalpy for molecules frequently occurring in combustion reactions
are found in Appendix A.
htot = h f0 + h (T ) − h (T = 298) = h f0 + ∆h (T )
6.3
Now consider an experiment when hydrogen and oxygen are burned to produce water H2O
as shown in Figure 6.2. The energy equation for this process can be written as shown in
Equation 6.4. The left side of the equation is zero because the reaction is steady-flow. The
inflows and outflows are found as the sum of the molar flow rates multiplied by the total
molar specific enthalpies.
75
Qin
½ O2
H2O
H2
T=298 K
T=298 K
Figure 6.2 A schematic diagram of an experiment that can be done to measure the enthalpy
of formation for H2O
(
(
)
)
0 = ∑ N i h 0f + ∆h (T ) i − ∑ N i h 0f + ∆h (T ) i + Q in
Re act .
6.4
Pr od .
With the experiment set up to produce a temperature of 298 K for both the inflows
(reactants) and outflows (products), all of the sensible enthalpies ( ∆h (T ) ) are zero. Also,
the heats of formation for hydrogen and oxygen are zero. Therefore, the energy equation
simplifies to:
h f0, H 2O = Q in / N H 2O
6.5
Which shows that the heat added during the reaction is the enthalpy of formation for H2O.
When this experiment is done, the result is 285,800 kJ of heat is produced or leaves the
control volume for every kmol of H2O formed. Instead of heat being required to form H2O,
heat is produced. Thus, the heat of formation of liquid H2O is negative or -285,800 kJ/kmol.
The heat of formation for vapor H2O is the heat of formation of liquid H2O plus the
enthalpy of vaporization or -241,826 kJ/kmol. The heats of formation of several gas species
can be found in Table A.3.
6.2.2. Heat of Combustion
Using the thermodynamic data, the heat produced from a combustion reaction with a given
starting temperature and final exhaust temperature can be found. The heat produced is
referred to as the heat of combustion or heat of reaction.
The first step to calculating the heat of a reaction is to write the reaction equation. This
involves writing an equation which defines the reactant and product species and
temperatures. The reaction equation contains the fuel and oxidizer on the left side and
products on the right. The equation can be balanced if the reaction is assumed to go to
completion or in other words the products are at their lowest possible heats of formation.
Complete or ideal products of combustion include: CO2, H2O, N2, O2 and SO2. A generic
reaction equation for a fuel containing carbon, hydrogen, nitrogen, and oxygen is shown
reacting with air to form complete products of combustion in Equation 6.6. If all of the
76
coefficients on the left side of the equation are given, the right side coefficients can be
determined by balancing each element as shown in Equations 6.7.
6.6
C a H b Od N e + A(O2 + 3.76 N 2 ) = xCO2 + yH 2 O + zN 2 + wO2
6.7
C: a = x
H : b = 2y
O : d + 2 A = 2 x + y + 2w
N : e + 2 A = 2z
Once the balance is complete the coefficients are used along with the energy equation
(Equation 6.4) to calculate the heat produced from the reaction as shown in Equation 6.8.
(
)
(
)
)
(
)
(
)
Qout = 1 h 0f + ∆h (T ) Ca H bOd N e S f + A h 0f + ∆h (T ) O2 +3.76 A h 0f + ∆h (T ) N 2 − x h 0f + ∆h (T ) CO2
(
)
(
(
)
6.8
− y h 0f + ∆h (T ) H 2O − z h 0f + ∆h (T ) N 2 − w h 0f + ∆h (T ) O2
6.2.3. Heat of Combustion for fuel containing only hydrogen and carbon
A chemical reaction equation for a fuel containing only carbon and hydrogen with
unknown amounts of complete products of combustion and an unknown amount of air is
shown in Equation 6.9.
C a H b+ A(O2 + 3.76 N 2 ) = xCO2 + yH 2O + zN 2 + wO2
6.9
Consider first the case where the amount of air is stoichiometric or just enough to burn the
fuel completely but without any air left over. In this case the amount of air is called the
theoretical stoichiometric air or stoichiometric air (As). The Reaction equation becomes:
C a H b+ A s (O2 + 3.76 N 2 ) = xCO2 + yH 2O + zN 2
6.10
The equation can now be balanced or solved by balancing the number of moles of each
element (C, H, O, and N):
C:a = x
H : b = 2y
O : 2 As = 2 x + y
N : 3.76 * 2 * As = 2 z
The resulting stoichiometric equation is:
77
b
b
b


C a H b+ a + (O2 + 3.76 N 2 ) = aCO2 + H 2 O + 3.76 a +  N 2
4
2
4



6.11
It is generally reasonable to assume complete products of combustion when the fuel air
mixture is lean or the equivalence ratio is less than one. In this case the combustion reaction
equation becomes.
b

a + 
6.12
 1 
b
3.76 
b
b
4
C a H b+ 
(O2 + 3.76 N 2 ) = aCO2 + H 2 O +
 a +  N 2 +  − 1 a + O2
φ
φ 
2
4
4
 φ 
Once the reaction equation is complete, the energy equation can be applied to the reaction
and the coefficients for each gas species, reactants and products, are found from the
reaction equation. The enthalpies of formation and sensible enthalpies are found in Tables
A.3 - A.4 in Appendix A.
Example 7.1 Finding the heat of combustion for Methane
Given: Natural gas is assumed to consist of 100% methane and is burned to produce heat
for a water heater. The incoming fuel-air mixture is at a temperature of 298.15 K with an
equivalence ratio of 0.9. The mixture is assumed to produce complete products of
combustion and after leaving the water heater the products are at 400 K.
Find: The amount of heat delivered to the water per kg of fuel burned.
A reaction equation is first written for methane burning in air at an equivalence ratio of 0.9.
 4
1 + 
3.76  4 
4
4
 1

CH 4+ 
− 1O2
(O2 + 3.76 N 2 ) = 1CO2 + H 2 O +
1 +  N 2 + 
0.9
2
0.9  4 
 0.9 
CH 4+2.22(O2 + 3.76 N 2 ) = 1CO2 + 2 H 2 O + 8.36 N 2 + 0.22O2
Next substitute into the energy equation.
(
)
(
)
(
)
(
)
Qout = 1 h 0f + ∆h (T ) CH 4 +2.22 h 0f + ∆h (T ) O2 +3.76(2.22) h 0f + ∆h (T ) N 2 −1 h 0f + ∆h (T ) CO2
(
)
(
)
(
)
− 2 h 0f + ∆h (T ) H 2O −8.36 h 0f + ∆h (T ) N 2 −0.22 h 0f + ∆h (T ) O2
Which gives:
Qout = 1 kmol CH 4 (− 74,900 + 0)kJ/kmol CH 4 + 2.22(0 + 0) + 3.76(2.22)(0 + 0)
− 1 kmol CO 2 (− 393,500 + 4010.2)kJ/kmol CO 2 − 2 kmol H 2 O (− 241,820 + 3453.5)H 2O
− 8.36 kmol N 2 (0 + 2972.1)kJ/kmol N 2 − 0.22 kmol O 2 (0 + 3030.7 )kJ/kmol O 2
Qout = 765,809.3 KJ
78
For the reaction used, 1 kmol of CH4 was burned or 16.04 kg. The heat of combustion per
kg of CH4 burned is found by dividing Qout by the molecular weight of methane (16.04
kg/kmol). The final result is:
Qout = 765,809.3 KJ or Q out = 765,809.3/16.04 = 47,743 kJ/kg CH 4
6.2.4. Enthalpy of Formation for Hydrocarbon Fuels
When the heat of combustion of a fuel is measured, it can be used to find the enthalpy of
formation. This has already been done for most common fuels but oil, coal and many other
practical fuels are a mixture of pure component fuels. The enthalpy of formation for these
fuels must be found by first measuring the heat of combustion and the relative amounts of
hydrogen and carbon in the fuel. Various laboratories can be hired to measure what is
referred to as the “heating value” and the “ultimate analysis” of a fuel. The heating value
provides the energy released when reactants at 25 oC are burned completely and then cooled
to 25 oC. The ultimate analysis measures the percentage of carbon, hydrogen, nitrogen and
oxygen in a fuel. There are two types of heating values that can be reported. One is the heat
release when the products are cooled so that the water is in the liquid phase. This is the
higher heating value. The other assumes the water vapor remains in the gas phase. This is
called the lower heating value.
Example – Finding the Enthalpy or Heat of Formation for a sample of gasoline
Given: A laboratory analysis of a sample of gasoline returned the following values.
Lower Heating Value: 48,980 kJ/kg
Ultimate Analysis (mass %): Carbon: 84.21
Hydrogen: 15.79
Nitrogen: 0.0
Oxygen: 0.0
Find: The enthalpy of formation for this gasoline:
Solution: Begin by assuming a 100 g basis for the fuel and calculate the ratio of
Hydrogen to Carbon for the fuel
Element
Mass (grams) Moles (gmol) Mole Ratio
Carbon
84.0
7.0175
1
Hydrogen
15.7
15.79
2.250
The fuel can therefore be represented by C1H2.25 remembering that this represents the
carbon to hydrogen ratio of the fuel but not the actual average fuel molecule which
might be larger in size.
Now use this fuel to write a stoichiometric reaction equation for the fuel producing
complete products with the water in the products in the liquid phase to match the
higher heating value that was measured.
C1 H 2.25 + 1.5625(O2 + 3.76 N 2 ) = 1CO2 + 1.125H 2 O(l ) + 5.875 N 2
79
The measured heating value is found by subtracting the enthalpy of the reactants from
the enthalpy of the products. For the reaction written, the mass of fuel burned is: 12
g/mol C * 1 mole C + 1 g/mol H * 2.25 mole H = 14.25 g
0 = H R − H P − Qout
Qout = 12.432 g * 48.980 kJ/g
(
H R = 1 mol fuel * h of, fuel + 1.5625 mol O 2 * 0 kJ/mol + 1.5625 * 3.76 mole N 2 * 0 kJ/mol
)
H p = 1 mol CO 2 * ( - 393.52 kJ/mol ) + 1.125 mol H 2 O * - 241.820 kJ/kmol + 5.875 mol N 2 * 0
Substituting Gives:
697.965 kJ = l mol fuel * h of, fuel − (−665.567 kJ)
h of, fuel = 32.398 kJ/mol CH 2.25
6.2.5. Finding Adiabatic Flame Temperature
The adiabatic flame temperature for a steady flow system is found by solving the energy
equation with reactants turning into products in an adiabatic process or for Q = 0. The
energy equation for a steady-flow adiabatic combustion process is shown in Equation 6.13.
(
)
(
)
0 = ∑ N i h 0f + ∆h (T ) i − ∑ N i h 0f + ∆h (T ) i
R
6.13
P
Because the incoming flow is made of reactants and the outgoing flow of products; the
adiabatic energy equation states that the enthalpy of the reactants must equal the enthalpy
of the products. A shorthand notation is given in Equation 6.14.
HR =
∑ N (h + ∆h (T )) = ∑ N (h + ∆h (T )) = H
R
i
0
f
i
P
i
0
f
i
P
6.14
An example of finding the adiabatic flame temperature which follows illustrates and
iterative technique to solving the energy equation. Commercial programs exist that can be
used to calculate the adiabatic flame temperature for a given fuel air mixture and starting
temperature and pressure.
Example: Finding an adiabatic flame temperature for ideal products of combustion
Find the constant pressure or steady-flow adiabatic flame temperature of methane which begins with a
stoichiometric mixture of fuel and air at 25oC.
First we write the combustion reaction equation for stoichiometric methane to obtain the number of moles of
reactants and products.
80
CH4 + 2(O2 + 3.76N2)  CO2 + 2H2O + 7.52N2
Next we write the energy equation in terms of moles of products and reactants for a steady flow process.
(
)
(
)
0 = ∑ h f0,i + hi (T ) − hi (T0 ) P − ∑ h f0,i + hi (T ) − hi (T0 ) R
P
R
Substituting from the reaction equation into the energy equation give.
1kmolCO (h f0,CO + hs ,CO (T ) − hs ,CO (298)) +

2
2
2
2


0 = 2kmol H 2O, g (h f0, H 2O, g + hs , H 2O, g (T ) − hs , H 2O, g (298)) + 


7.52kmol N (h f0, N + hs , N (T ) − hs , N (298))

2
2
2
2

 products
1kmolCH (h f0,CH + hs ,CH (298) − hs ,CH (298)) + 
4
4
4
4


− 2kmolO2 , g (h f0,O2 + hs ,O2 (298) − hs ,O2 (298)) + 


7.52kmol N (h f0, N + hs , N (298) − hs , N (298)) 
2
2
2
2
 reac tan ts

Because the reactants are at 298 K which is equal to T0, the h(T) – h(T0) terms for the reactants are all zero.
Also, the heats of formation for N2 and O2 are zero. Removing these terms gives.
(
) (
)
1 h f0,CO + hs ,CO (T ) − hs ,CO (298) + 2 h f0,H O ,g + hs ,H O ,g (T ) − hs ,H O ,g (298) + 
2
2
2
2
2
2
 kJ
0 = 
0


 7.52 hs ,N 2 (T ) − hs ,N 2 (298) − 1 h f ,CH 4

(
)
Using data from the appendix allows substitution for all of the heats of formation.
(
) (
)
1 − 393520 + hs ,CO (T ) − hs ,CO (298) + 2 − 241820 + hs , H O , g (T ) − hs , H O , g (298) + 
2
2
2
2
 kJ
0 = 

 7.52 hs , N (T ) − hs , N (298) − 1 (− 74,850 )
2
2


(
)
Which can be rewritten as:
[( (
) (
)
(
))] kJ
802310 kJ = 1 hs ,CO2 (T ) − hs ,CO2 (298) + 2 hs , H 2O , g (T ) − hs , H 2O , g (298) + 7.52 hs , N 2 (T ) − hs , N 2 (298)
We now guess a value for the temperature of the products. Adiabatic flame temperatures are normally
between 2000 and 3000 K. Since the reactant temperature is low, 298 K, then we will guess toward the lower
end.
Guess T= 2200 K
802310 kJ ≠ 1 (103624 ) + 2(83075) + 7.52(63401) kJ = 746549 kJ
Since the right side of the equation is smaller than the left, the temperature guess is too low.
Guess T=2400 K
802310kJ ≠ 1 (115843) + 2(93649) + 7.52(70685) kJ = 834692 kJ
This temperature guess is too high.
Interpolation gives
T − 2200
802310 − 746549
=
2400 − 2200 834692 − 746549
or
T = 2326.5
81
Normally, if the guesses are within 200 k of the interpolated temperature, this first iteration on temperature
will be with 2 K of the final result.
6.2.6. Constant Volume Combustion Reactions
For constant volume problems, the adiabatic flame temperature can be found by solving
the energy equation for a closed system. The energy equation becomes:
∑ N (u ) − ∑ N (u ) = 0
i
i
P
i
6.15
i
R
Or the internal energy of the products equals the internal energy of the reactants. Rather
than measure an internal energy of formation, chemical energy is defined using the
enthalpy of formation. Enthalpy is defined as given in Equation 6.16
6.16
h = u + Pv = h f0 + h (T ) − h (T0 )
The internal energy is then found to be:
u = h f0 + h (T ) − h (T0 ) − Pv
6.17
u = h f0 + h (T ) − h (T0 ) − Ru T
6.18
And for and ideal gas:
The value for internal energy can therefore be determined in terms of the enthalpy.
Example: Adiabatic Flame Temperature for a constant volume process with ideal products of
combustion
Find the constant pressure or steady-flow adiabatic flame temperature of methane which begins with a
stoichiometric mixture of fuel and air at 25oC.
First we write the combustion reaction equation for stoichiometric methane to obtain the number of moles of
reactants and products.
CH4 + 2(O2 + 3.76N2)  CO2 + 2H2O + 7.52N2
Next we write the energy equation in terms of moles of products and reactants for a closed system. Note this
is the same as the steady flow equation except for the RuT terms.
(
(
)
)
0 = ∑ h f0,i + hi (T ) − hi (T0 ) − Ru T P − ∑ h f0,i + hi (T ) − hi (T0 ) − Ru T R
P
R
Substituting from the reaction equation into the energy equation give.
82

1kmolCO (h f0,CO + hs,CO (T ) − hs,CO (298) − Ru (T )) +
2
2
2
2


0 =  2kmol H 2O, g (h f0, H 2O, g + hs , H 2O, g (T ) − hs , H 2O, g (298) − Ru (T )) + 



7.52kmol N (h f0, N + hs , N (T ) − hs , N (298) − Ru (T ))
2
2
2
2
 products

1kmolCH (h f0,CH + hs ,CH (298) − hs ,CH (298) − Ru (298)) + 
4
4
4
4


0

− 2kmolO2 , g (h f ,O2 + hs ,O2 (298) − hs ,O2 (298) − Ru (298)) + 


7.52kmol N (h f0, N + hs , N (298) − hs , N (298) − Ru (298)) 
2
2
2
2
 reac tan ts

Because the reactants are at 298 K which is equal to T0, the h(T) – h(T0) terms in the for the reactants are all
zero. Also, the heats of formation for N2 and O2 are zero. Removing these terms gives.
(
) (
)
1 h f0,CO + hs ,CO (T ) − hs ,CO (298) − Ru T + 2 h f0,H O ,g + hs ,H O ,g (T ) − hs ,H O ,g (298) − Ru T + 
2
2
2
2
2
2
 kJ
0 = 
 7.52 hs ,N 2 (T ) − hs ,N 2 (298) − Ru (T ) − 1 h f0,CH 4 − Ru * 298 − 2(Ru * 298) − 7.52(Ru * 298) 
) (
(
)
Using the tables in from the appendix allows substitution for all of the heats of formation.
(
) (
)
1 − 393520 + hs ,CO (T ) − hs ,CO (298) − 8.314T + 2 − 241820 + hs , H O , g (T ) − hs , H O , g (298) − 8.314T + 
2
2
2
2
 kJ
0 = 
 7.52 hs , N (T ) − hs , N (298) − 8.314T − 1(− 74850 − 8.314 * 298) − 2(− 8.314 * 298) − 7.52(− 8.314 * 298)
2
2


(
)
Which is rewritten as:
[( (
) (
)
(
]
))
776209 kJ = 1 hs ,CO2 (T ) − hs ,CO2 (298) + 2 hs ,H 2O ,g (T ) − hs ,H 2O ,g (298) + 7.52 hs ,N 2 (T ) − hs ,N 2 (298) − 10.52 * 8.314T kJ
We now guess a value for the temperature of the products. Adiabatic flame temperatures are normally
between 2000 and 3000 K. Since the constant volume process will produce a higher temperature than constant
pressure.
Guess T= 2800 K
776209.9 kJ ≠ 1 (140511) + 2(115349 ) + 7.52(85386 ) - 87.463 * 2800 kJ = 768093.6 kJ
Since the right side of the equation is smaller than the left, the temperature guess is too low.
Guess T=3000 K
776209.9 kJ ≠ 1 (152935) + 2(126421) + 7.52(92782 ) - 87.463 * 3000 kJ = 840776.8 kJ
This temperature guess is too high.
Interpolation gives
𝑇𝑇 − 2800
776209.9 − 768093.56
=
3000 − 2800 840776.8 − 768093.56
𝑜𝑜𝑜𝑜
𝑇𝑇 = 2822.3
Note that in comparison to the steady flow adiabatic flame temperature the value is considerably higher. This
is because there is no energy removed by expanding the gas during the reaction.
6.2.7. Thermodynamic Equilibrium
In the previous discussion we have assumed that reactions have produced complete or ideal
products of combustion. This assumption breaks down under two common conditions in
IC engines: 1) When the temperature of the products is very high (above 1500 K) and 2)
when there is an insufficient amount of oxygen to completely burn the fuel. In both of these
83
cases, the products are no longer limited to CO2, H2O, N2 and O2. At high temperatures,
CO2 and H2O will dissociate or break apart and form CO and O2. Water, H2O will dissociate
into H2, O2 and OH. Under fuel rich conditions, there is not enough oxygen to fully oxidize
carbon and hydrogen resulting in partially burned species such as CO, OH, and H2.
Consider the reaction equation written for a generic fuel containing carbon, hydrogen,
oxygen, and nitrogen. Now instead of limiting the products to complete products of
combustion, the products list consists of 10 that have been found to be important related to
energy and emissions including CO, H2, H, O, OH, and NO in addition to the ideal products
as shown in Equation 6.19. Now there are 10 unknowns on the right side of the equation
and only four elements from which to write a balance equation. Thus, using the previous
method for balancing we can write 4 equations but need 6 more equations to solve for the
10 unknowns. Fortunately, the distribution of the four elements on the left side among the
products can be determined if the products are in thermodynamic equilibrium.
C a H b Od N e + A(O2 + 3.76 N 2 ) = n1CO2 + n 2 H 2 O + n3 N 2 + n4 O2 + n5 CO + n6 H 2 + n7 H + n8 O + n9 OH + n10 NO
6.19
Thermodynamic equilibrium is reached when the elements are distributed among the
products in a way that maximizes entropy. The elements may keep changing the molecules
they form as long as entropy is increasing but once maximized, the composition can no
longer change or it would produce a decrease in total entropy. The equilibrium condition
is described by Equation 6.20.
6.20
dS prod = 0
It turns out that there is a more convenient parameter than entropy for determining
equilibrium. Introducing the Gibbs function which is define as:
G = H − TS
6.21
dG = dH − TdS − SdT
6.22
Then:
Now for a fixed temperature and pressure the elements may move around between
molecules but the total enthalpy cannot change. Therefore:
dH = 0 and dT = 0 and dS = 0 gives dG = 0
6.23
Using the relationships for a mixture:
dG = 0 = d
∑ N g = ∑ N dg + g dN
i
i
i
i
i
i
6.24
84
Again if the temperature and pressure are fixed and we seek the equilibrium at that
condition then dg i = 0 which leads to the equilibrium condition of:
∑ g dN
6.25
This states that in equilibrium the elements can move around to change the amount of Ni,
the moles of each compound but they will reach a composition of Ni where changing Ni
will increase entropy and at that composition, the Ni can no longer change.
For a generic reaction:
0=
i
i
6.26
aA + bB = cC + dD
The values of dNi are the stoichiometric coefficients a, b, c, and d. As written, the reactants
A and B are being destroyed to form C and D. Thus dNA = -a or is the negative of the
coefficient of A.
The Gibbs property can be written in terms of enthalpy and entropy as:
(
g i = h 0f ,i + hi (T ) − hi (T0 ) − T s i 0 − R ln( y i P)
)
6.27
Separating the terms that are only temperature dependent gives:
6.28
g i = h f0,i + hi (T ) − hi (T0 ) − Ts i 0 − TR ln( y i P) = g i0 (T ) − RT ln( y i P )
Where: g i0 = h f0,i + hi (T ) − hi (T0 ) − Ts i 0
Now substituting back into Equation 6.25 for the reaction in Equation 6.26gives:
(
) (
) (
) (
)
0 = −a g a0 − RT ln(PA ) − b g b0 − RT ln(PB ) + c g C0 − RT ln(PC ) + d g D0 − RT ln(PD )
6.29
Which can be rearranged to give :
 − ∆G 0  PCc PDd N Cc N Dd  P 
=

= a b 
exp
a b

N A N B  P0 N tot 
 RT  PA PB
(c + d − a −b )
6.30
Where: ∆G 0 = cg C0 + dg D0 − ag a0 − bg b0
Each of the terms in Equation 6.30 can be used to define the equilibrium constant, KP
where:
 − ∆G 0  PCc PDd
=
K P = exp
a b

 RT  PA PB
6.31
The first term in Equation 6.31 shows that the equilibrium constant is only a function of
temperature and is a function of a thermodynamic property, g, the Gibbs energy. The Gibbs
energy for an ideal gas can be readily found in thermodynamic tables. The second term
illustrates that the equilibrium constant is related to the partial pressures of the products
85
and reactants. When the equilibrium constant is large, the partial pressures of the products
are large in comparison to that of the reactants and equilibrium favors products. When the
equilibrium constant is small (<< 1.0) then equilibrium favors the reactants.
The partial pressures can be rewritten in terms of moles or mole fractions and the total
pressure as shown in Equation 6.32. This equation provides a mathematical relationship
between the mass fractions of each species involved in a reaction and thereby allows for a
solution to be determined regarding how much of each species will exist at a given
temperature and pressure.
KP =
N Cc N Dd  P 


N Aa N Bb  P0 N tot 
( c + d − a −b )
=
y Cc y Dd  P 
 
y Aa y Bb  P0 
6.32
( c + d − a −b )
The equilibrium coefficient is helpful for understanding semi-quantitative amounts or
trends. Take for example the reaction CH 4 + O2 = CO2 + 2 H 2 O . If Kp for this reaction is
KP = 108, then it suggests that the mole fractions of the products are much larger than the
mole fractions of the reactants, or that equilibrium dictates that the reaction will produce
almost all CO2 and H2O and the will be very little CH4 or O2 remaining.
Rather than calculate the change in Gibbs function at each temperature, curve fits have
been made that describe KP as a function of temperature for various reactions. A curve fit
commonly used for KP takes on the form of Equation 6.33 where the coefficients for the
reactions shown in Table 6.2 can be found in Table 6.3.
log 10 K p (T ) = A ln(T / 1000) +
B
T
6.33
+C + D T + E T 2
Table 6.2 Reactions for which KP has been found using a curve fit.
Reaction
1
H2 ↔ H
2
1
O2 ↔ O
2
1
1
H 2 + O2 ↔ OH
2
2
Number
1
2
3
Reaction
1
1
O2 + N 2 ↔ NO
2
2
1
H 2 + O2 ↔ H 2 O
2
1
CO + O 2 ↔ CO 2
2
Number
4
5
6
Table 6.3. Coefficients to be used for finding KP for reactions in Table 6.2 and according
to Equation 6.33.
Reaction
1
2
3
4
5
6
A
0.432168 E00
0.310805 E00
-0.141784 E00
0.150879 E-01
-0.752364 E00
-0.415302 E-02
B
-0.112464 E05
-0.129540 E05
-0.213308 E04
-0.470959 E04
0.124210 E05
0.148627 E05
C
0.267269 E01
0.312779 E01
0.853461 E00
0.646096 E00
-0.260286 E01
-0.475746 E01
D
-0.745744 E-04
-0.738336 E-04
0.355015 E-04
0.272805 E-05
0.259556 E-03
0.124699 E-03
E
0.242484 E-08
0.344645 E-08
-0.310227 E-08
-0.154444 E-08
-0.162687 E-07
-0.900227 E-08
86
Reactions 5 and 6 are important for understanding a combustion process in an internal
combustion engine. They concern the formation of H2O and CO2. At low temperatures the
KP values for these two reactions are very large, for example at 300 K they are 6.06E39
and 6.7E44 respectively. This indicates that at low temperature, CO2 and H2O are very
stable species and that at equilibrium there will be virtually no O2, CO, H2 and almost all
CO2 and H2O. At high temperature however, for example 3000 K, KP for the two reactions
are 22 and 3.06 respectively. Thus at high temperatures, H2O and CO2 tend to break apart
and form significant amounts of H2 and CO. This breaking apart of molecules at high
temperature is called dissociation and has important implications in engines. Dissociation
is an endothermic process and therefore results in a reduction in temperature and storage
of chemical energy which is released as gas temperatures reduce during expansion.
Dissociation is the reason that ideal or complete products of combustion are a poor
assumption at high temperatures.
6.2.8. Equilibrium Calculation
The composition of products in equilibrium can be determined through a combination of
the conservation of elemental mass and the equilibrium equation. Consider the dissociation
reaction in Equation 6.34. The quantity of reactants CO2 is known but amounts of O2, CO2
and CO products are unknown.
6.34
CO 2 ↔ xCO2 + yCO + zO 2
A solution can be found at a fixed temperature and pressure for the products by first using
an elemental balance to produce two equations balancing carbon and oxygen and a third
equation can be written using KP for Reaction 6 listed in Table 6.2.
Example - Simple Equilibrium Problem.
Given a container starting with 1 mole or CO2 3000 K and 1 atm. What will the equilibrium concentration of this mixture
considering possible products of CO, CO2, and O2.
Step 1. Write the actual equation
CO2  xCO2 + yCO + zO2
Step 2 – Write the elemental balance equations
C: 1 = x + y
O: 2 = 2x + y + 2z
Step 3. Write the stoichiometric equation for a reaction containing only the species of interest
CO + ½ O2 = CO2
Step 4 Find KP for this reaction. This is done by using Eq. 6.33 and coefficients from Table 7.2.
KP = 3.057
Step 5 – Write the KP equations in terms of unknowns
K P = 3.06 =
1−1−1 / 2


P


N CO N O2 1 / 2  P0 ( N CO2 + N CO + N O2 ) 
N CO2
1−1−1 / 2
=

x 
1


yz1 / 2  1( x + y + z ) 
87
Step 6 – Solve the set of equations
x = 0.564
y = 0.436
z = 0.218
Find the mole fractions for each product specie
yCO2 = x/(x+y+z) = 0.463 = 46.3%
yCO = 0.358 = 35.8%
yO2 = 0.218 = 17.9%
Note that this is a lot of CO. Ideal or complete combustion is not a good assumption at high temperatures.
6.2.9. The Water-Gas Shift Reaction
One of the most important reactions in the area of combustion and gasification is the water
gas shift reaction shown in Equation 6.35. The forward reaction shows hydrogen can be
used to separate CO2 into CO and H2O. The reverse reaction is extremely important
because it shows that CO and water can be used to make hydrogen. An equation for the
equilibrium coefficient is given in Equation 6.36. The amount of each species produced by
this reaction is dependent on the temperature. At low temperature, the left side is favored
while at high temperature equilibrium shifts to the right. Any fuel containing carbon can
be partially burned to produce carbon monoxide. When water is added, equilibrium
demands that hydrogen be produced. A catalyst can help the reaction proceed more rapidly
at lower temperatures.
H 2 + CO 2 ↔ CO + H 2 O
ln K P (T ) = 2.743 −
1761 1.611x10 6 2.803 x10 8
−
−
T
T2
T3
6.35
6.36
The water gas shift can be used to produce the simplest possible prediction of product
species for a fuel rich process. Consider reaction Equation 6.19 but now with only six
products of combustion; the ideal products plus CO and H2 as shown in Equation 6.37. For
the case when the mixture is fuel lean, it is reasonable to assume that CO and H2 are
negligible, and the equation is balanced using only the elemental balance equations. When
the mixture is fuel rich, it is reasonable to assume that O2 is negligible leading to 5 product
species. A solution can be found using an elemental balance to produce 4 equations and
the water gas shift equilibrium equation as the fifth equation to solve for the five unknowns.
The equations are shown in
Table 6.4.
C a H b O d N e + A(O 2 + 3.76 N 2 ) = n1CO 2 + n 2 H 2 O + n3 N 2 + n 4 O 2 + n5 CO + n 6 H 2
6.37
88
Table 6.4 Equations used to solve a combustion reaction using the water gas shift reaction
to determine equilibrium of fuel rich species.
Fuel Lean ϕ ≤ 1.0
Fuel Rich ϕ >1.0
n6 = 0
a = n1 + n5
n5 = 0
a = n1
b = 2n 2
d + 2 A = 2n1 + n 2 + 2n 4
e + (3.76)(2) A = 2n 3
n4 = 0
b = 2n 2 + 2n 6
d + 2 A = 2n1 + n 2 + n5
e + (3.76)(2) A = 2n 3
KP =
n5 n 2
n1 n 6
6.2.10.
Equilibrium with Multiple Species
Returning to the initial combustion reaction shown in Equation 6.19, the equilibrium
products in this equation can now be determined using a combination of elemental balances
and KP equations. There are four elements in the equation, C, H, N, O allowing an equation
for each element or four equations. The right side has 10 unknowns suggesting six more
equations are required to solve for the unknown products. These six equations can be
produced by writing a KP equation for the six reactions listed in Table 6.2. The result is a
set of 10 equations and 10 unknowns. The equations can be solved simultaneously using a
numerical solver.
6.3. Zero Dimensional Thermodynamic Engine Models
When a model has no spatial resolution, for example the temperature of an entire volume
is represented by a single value, it is said to be zero-dimensional. The simple Otto and
Diesel cycle models as well as the 4-stroke Otto cycle are all zero-dimensional. They also
assume the properties of air are constant or fixed. Complexity and improved accuracy can
be added by incorporating thermodynamic principles taught in this chapter such as treating
the working fluid as a mixture of fuel and air, determining specific heats as a function of
temperature, and calculating mixture compositions based on equilibrium calculations. The
development of such a model is beyond the scope of this text but it is interesting to consider
results from such a model and compare them to simpler air-standard models.
Figure 6.3 shows that the molecular weight of does not change greatly with the type of fuel
mixture or if the mixture is a burned or unburned gas. Molecular weight is typically very
close to that of air, 28.95.
Figure 6.4 shows the heat capacity increases significantly with temperature at all fuel /air
mixtures. When a specific heat at 300 K is used and held constant, the temperatures in the
model will be greatly over predicted because the actual CP is much higher than the value
being used.
Figure 6.5 shows the ratio of specific heats is decreasing with temperature and is also lower
when fuel is added to the mixture. When the ratio of specific heats is higher, more energy
is extracted during expansion and converted to work. These data show why a lean burn
engine can produce a higher efficiency than a stoichiometric engine. If air is assumed as a
89
working fluid, the value for the ratio of specific heats will be overestimated and the cycle
efficiency will be too high.
Figure 6.3 Molecular weight of iso-octane/air mixtures. (xb is the residual mass fraction) [1]
Figure 6.4 Dependence of Cp for various equivalence ratios (air-iso-octane) as a function
of temperature. (xb is the residual mass fraction) [1]
90
Figure 6.5 The ratio of specific heats for various equivalence ratios as a function of
temperature. (xb is the residual mass fraction) [1]
Overall engine efficiency in a constant specific heats model is compared to a real engine
in Figure 6.6. The figure shows that the model predicts the same trend of increasing
efficiency with increasing compression ratio as does an actual engine but the modeled
efficiency is double that of actual engines. Making the engine model contain a fuel-air
mixture which increases CP and lowers CP/Cv makes the model closer to that of a real
engine.
Figure 6.6 Comparison of thermodynamic efficiency with a fuel-air model and engine
measurements. [1]
The real engine hits a maximum efficiency around a compression ratio of 20 after which
higher compression ratios make no improvement and can even decrease efficiency. This is
because heat transfer or heat loss from combustion products to the walls increases with
increasing compression ratio. The loss of heat transfer becomes greater than the gain
produced by higher compression ratio above 20:1. None of the models used in the figure
include heat transfer and therefore continue to predict an increase in efficiency with
increasing compression ratio.
References
1. Heywood, J.B., Internal Combustion Engine Fundamentals, McGraw Hill, 1988.
91
Chapter 6 Homework
6.1 Name three types of acceptable combustion processes and one unacceptable
combustion process for and internal combustion engine.
6.2 What type of combustion process is found in a car with a spark and a fuel injector in
the intake port?
6.3 What is a typical duration in crank angles for a combustion process?
6.4 If it is true that faster burning fuel produces higher efficiency, why is detonation which
has a very fast burn rate a problem in an internal combustion engine?
6.5 What is knock and what type of combustion processes produces it?
6.6 Look up three vehicles and the engines that are used in those vehicles. List the type of
combustion process used in those engines.
6.7 What are the molecular weight (kg/kmol) and specific enthalpy (kJ/kg) of a gas mixture
at P = 1000 kPa and T = 500 K, if the mixture contains the following species and mole
fractions.
Species
CO2
H2O
N2
CO
yi
0.10
0.15
0.70
0.05
6.8 If a lean mixture of methane and air (φ=0.8) is burned and produces ideal combustion
products at a temperature of 1500 K and a pressure of 500 kPa. What are the mole fractions
and the specific molar enthalpy (kJ/kmol) of the products?
6.9 Compute the higher and lower heating values in kJ/kg for methanol CH3OH (l).
6.10 Find the adiabatic flame temperature of Iso-Octane vapor burning in air at an
equivalence ratio of 0.9 in a steady state process. The initial temperature of the reactants is
25°C.
6.11 Find the adiabatic flame temperature of Iso-Octane vapor burning in air at an
equivalence ratio of 0.9 at constant volume. The initial temperature of the reactants is 25
°C.
92
6.12 Use EES to find the constant volume and constant pressure adiabatic flame
temperatures of Iso-octane vapor burning in air at an equivalence ratio of 0.9. Print out the
Equation Page and the Solution page for each result. (note EES does not have a lookup
value for the enthalpy of iso-octane so you will need to input the value into EES yourself)
6.13 If one mole of CO2 is raised to a temperature of 3000 K at a pressure of 1 atm. what
will be the equilibrium concentrations of CO2, CO, and O2.
6.14 Nitrogen (N2) can react with O2 at high temperature to produce NO. Assuming one
mole of N2 and one mole of O2 are initially available, what is the equilibrium composition
of N2, O2, and NO at 2500 K, 1 atm.
6.15 Using Excel, write a program using the “solve” function which calculates the
equilibrium composition of any initial composition of N2 and O2 as a function of
temperature from 2500 to 3500 K, P= 101 kPa. Print out a table of the mole fraction of NO
as a function of temperature at 100 K increments between these two temperatures for an
initial concentration of .21 moles of oxygen and .79 moles nitrogen.
6.16 Write and excel program which calculates the products of combustion for the reaction
in Equation 6.19 but considering only the first 6 products (n1 – n6). The fuel to be used is
iso-octane (C8H18) at an equivalence ratio of 1.2 at T = 1000 K, φ = 1.2 and T = 1500 K.
Because this is a fuel rich condition, it can be assumed that there is no O2 in the products
(i.e. n4 = nO2 = 0). This leaves five unknowns. Four can be found with an elemental balance.
The water gas shift reaction can be used to write a fifth, KP equation.
93
94
7. Engine Losses, Friction, Accessories and Pump Work
The brake work out of an engine may be characterized by Equation 7.1 where the brake
work is the indicted work minus losses produced by friction (Wf), accessories (Wa), and
pumping work (Wp). Friction and accessory work are always negative contribution or a loss
while pumping work can be positive in the case of a turbocharged or supercharged engine
(i.e. pump work in is negative or pump work out is positive for a turbocharged engine). If
not stated explicitly, the pumping work can be assumed to be a loss.
Wb,out = Wi ,out − W f ,in − Wa ,in − W p ,in
7.1
In some references, accessory work and pumping work are lumped together with friction
work and the sum of all three is called friction work. This can be confusing but is
convenient when the focus is on naturally aspirated gasoline engines where pump work out
is always negative. In this text, the three will be considered separately.
It is often convenient to normalize work by the engine displacement volume and consider
the mean effective pressures of each component and shown in Equation 7.2.
bmep =
Wb,out
Vd
= imep − fmep − amep − pmep
7.2
7.1. Accessory Work
Various accessories can be attached to an engine as either essential for operation or for the
use of powering non-essential equipment. By essential it is meant that the engine could nor
run without it. Potential accessories include: a cooling fan, water pump, and alternator.
Non-essential axillaries might include and air conditioner compressor, power steering
pump, auxiliary power supply. At full load the accessory work is typically not a large
fraction of the maximum possible work produced by an engine and is on the order of 110%. At idle or low load accessory work can become a large fraction of the total work
being produced by the engine.
7.2. Friction
Friction occurs in engines between moving surfaces which include: the piston and cylinder;
the camshaft and head; the main bearings and the block; and the connecting rod and the
block. At low load or idle, friction consumes a large fraction of the work produced by the
engine. At full load, friction becomes a much smaller fraction.
7.2.1. Modes
There are three modes of friction: contact or boundary, intermediate, and hydrodynamic.
95
Contact friction is a mode where two solid parts are in contact as one moves relative to the
other. Resistance to motion is characterized by Equation 7.3 where the force of friction Ff
is proportional to a friction coefficient (μf) multiplied by the normal force (FN).
Ff,c = μf FN
7.3
Hydrodynamic friction is characterized by the viscosity, μ, and the velocity gradient du/dy
of a fluid as shown in Equation 7.4. The friction force produced by hydrodynamics is about
two orders of magnitude smaller than boundary friction. In this type of friction, the two
parts are not touching but riding on a film of oil. This mode is essential in order to produce
long life for the parts and low friction resistance for reduced losses.
Ff,h = Aμ du/dy
7.4
Mixed friction is a combination of boundary and hydrodynamic friction. It occurs during
the transition from one to the other.
Using these two equations, the effect of load and speed on friction force can be understood.
Load increases the normal force between the moving surfaces and increases contact friction
proportional to the load. For example; a higher load in-cylinder (higher pressure and more
work per stroke) will produce more force on the connecting rod. This produces a larger
normal force on the bearing between the connecting rod and the crankshaft. On the other
hand, a higher engine speed produces a greater velocity gradient (du/dy) between a bearing
and stationary surface also creating a larger friction force.
F f ,c ∝ load
7.5
F f ,h ∝ N
Results for measured values of friction vs engine speed are shown in Figure 7.1 for a 4cylinder spark ignition engine. The figure shows what happens to the friction loss in a
motored engine as various components are removed. The largest contributors to friction
are seen to be in the pistons, rings, pins, and connecting rods. At low speeds the friction
increases rapidly but from 2500 – 4000 rpms the engine appears to agree with the result
from hydrodynamic friction that the fmep is linear with increasing speed.
96
Figure 7.1 Measured engine friction as a function of engine speed. [1]
Results for engine friction at a fixed speed with varying load are shown in Figure 7.2 for a
four-cylinder spark ignition engine. The figure shows rubbing friction (“rfmep”) increases
with increasing load at a fixed engine speed of 1600 rpm. The higher compression ratio
engine has a greater increase in friction with load. The pumping mean effective pressure
(pmep) is seen to decrease with engine load which is expected because the intake value
will be open further to allow the mixture to enter the cylinder with less resistance.
Figure 7.2 Mean effective pressure due to rubbing friction (rfmep) and pumping mean
effective pressure are shown as a function of bmep at fixed engine speed (1600rpm). [2]
7.3. Pumping Mean Effective Pressure
The idea of pumping work was introduced in Equation 4.2. As the piston moves down
during the intake stroke, the pressure in the cylinder is normally lower than atmospheric
97
pressure creating a situation where work is required to pull the air through the intake valve.
This is normally caused primarily by the throttle which is closed to reduce air flow into the
engine and thereby control engine output or load. It is therefore expected that pmep will be
inversely proportional to the load. The results is shown experimentally is Figure 7.2. The
addition of turbocharging can cause a positive pressure in the intake during intake and
produce positive work out of the engine during intake. A more complete discussion of the
intake process is given in Chapter 8.
7.4. Hydrodynamic Bearing Friction
Figure 7.3 shows the hydrodynamic oil film gap between a bearing and a rotating shaft
inside the bearing. Between the bearing a gap of distance “c” is filled with oil that has a
velocity of u = πDN at the shaft and zero at the bearing. The velocity gradient in fluid is
du/dr = u/c = πDN/c.
Substituting this into Equation 7.4 gives:
F f = πDLµ
7.6
πDN
c
Where: L is the width of the bearing and D the bearing diameter.
Bearing
Shaft
Shaft
2
Oil Film
u
r
c
Figure 7.3 Cross sectional view of a shaft rotating in a bearing with oil film thickness “c”
and velocity gradient, du/dr = u/c.
The result provides insight into the factors that increase friction in an engine. An increase
in bearing size (L and D), and increase in engine speed (N) and an increase in viscosity
will all increase bearing friction. A decrease in the gap or oil film thickness will also
increase friction. Some of the variables are inter-related. For example, a low viscosity oil
may produce a thinner oil film. Adding a load to the bearing surface increases the
pressure on the oil film and decreases film thickness.
98
The friction mean effective pressure can be found by multiplying friction force by the
distance to obtain friction work and then dividing by the displacement volume. The
resulting theoretical fmep for a bearing is found in Equation 7.7, where: nb is the number
of bearings, nc is the number of cylinders, B is the cylinder bore, S is the stroke, and Cb is
a constant.
fmep = π DLµ
nb LD 3 N
πDN nbπD
=
C
b
c ncπB 2 S
nc B 2 S
4
7.7
The actual friction on a bearing will differ from that predicted by Equation 7.7 because
the load on the bearing and therefore the thickness of the oil film will not be uniformly
distributed or uniformly applied over time. Patton et al. 1989 [1] suggested an empirical
constant of Cb = 3.03x10-4 kPa-min/rev-mm) which provides an estimate for bearing
friction.
7.5. Piston Friction and Forces
A slider-crank mechanism undergoes various forces during the processes of a 4-stroke
cycle. A simple free body diagram of a piston in a cylinder is shown in Figure 7.4.
Summing forces in the x and y directions produces the results shown in Equations 7.8 and
7.9.
∑ F =0 = F − F sin(θ )
7.8
∑ Fy = ma = F cos(θ ) + F − PA − mg
7.9
x
y
T
R
R
f
PA
y
x
FT
mg
Ff
θ FR
Figure 7.4 Free body diagram of forces acting on a piston during expansion.
99
The thrust force, FT, is the horizontal force imparted by the cylinder walls on the piston.
This force is what is largely responsible for wear on the cylinder liner. It switches from
the left side to the right side depending on the position of the crank and the acceleration
of the piston. The largest thrust force occurs during the expansion stroke when the force
produced by pressure on the top of the piston is greatest. Because the force is greatest on
this side, it is known as the thrust side of the engine.
References
1. Millington, B.W. and Hartles, E.R., “Friction Losses in Diesel Engines,” SAE Paper
680590, SAE Trans. Vol. 77, 1968.
2. Gish, R.E. McCullough, J.D., Retzloff, J.B., and Mueller, H.T., “Determination of True
Engine Friction,” SAE Trans. Vol. 66, pp 649-661, 1958.
3. Patton, K. J., Nitchke, R.G., and Heywood, J.B, “Development and Evaluation of a
Friction Model for IC Engines,” SAE Paper 890836.
Chapter 7 Homework Problems
7.1 When an engine is at high load, what variable in Equation 7.6 is impacted making the
friction force on the engine to increase.
7.2 Use Patten’s equation (Eqn. 7.7) to predict the bearing friction mean effective pressure
produced by a 2.0 L, 4 cylinder, square (bore = stroke) engine. Assume the engine has six
bearings each 50 mm in diameter and 20 mm wide at a speed of 4000 rpm.
7.3. If the engine described in problem 7.2 produces 92 indicated hp at 4000 rpm. What
fraction of the engine indicated power is consumed by bearing friction?
7.4 Using a free body diagram of the forces on a piston, derive an equation that shows the
dependence of the thrust force on crank angle position (θ). Is it possible for the thrust force
to be zero when the piston is not at top or bottom dead center?
100
8. Air, Fuel and Exhaust Flow
Intake and exhaust exchange are central to the operation of an internal combustion engine.
The largest influence on engine torque and mean effective pressure comes through the
intake mixture density and the volumetric efficiency. It will be seen in this chapter that the
gas exchange process is responsible for defining the torque and power curves of an engine
and thereby how and why engines perform as they do. Understanding air flow into the
engine begins with a study of air flow through small passages and a basic understanding of
compressible flow. A complete understanding of flow in engines requires a sophisticated
understanding of flow dynamics and wave behavior which is beyond the scope of this text.
8.1. Flow through an Orifice
An intake or exhaust valve may be considered a restriction or an orifice and therefore flow
equations for an orifice may be applied. For an isentropic flow or a flow without viscous
effects, the mass flow rate can be derived as a function of the upstream (intake plenum)
and downstream (valve) pressures and other geometric and fluid properties as shown in
Equation 8.1.
1/ 2
 2   P  2 / k  P  ( k +1) / k  
 v  −  v 

m = ρ 0 A f c0 
P 
 
 k − 1   P0 
 0



8.1
The subscript “0” is used to signify a locations far upstream of the valve and “v” denotes a
property at the valve throat. Other symbols include: the density, ρ, the area through which
the flow travels, Af, the speed of sound, c0, pressure, P, and the ratio of specific heats, k.
The speed of sound is found from Equation 8.2.
8.2
c0 = kRT0
For the intake process it can be assumed that the pressure at the valve is equal to the
pressure in the cylinder. Thus, the movement of the piston produces a vacuum or low
pressure in the cylinder that induces flow through the valves. The flow rate increases with
decreasing pressure in the cylinder until the pressure in the cylinder reaches the critical
pressure when the velocity of the mixture at the valve throat area reaches the speed of
sound. At this point the velocity can go no higher and the flow is said to be choked.
The critical pressure ratio for chocked flow occurs when the pressure ratio is equal to or
above that shown in Equation 8.3. For k=1.4 the critical pressure ratio is 0.529.
k
 Pv 
2  ( k −1)
  = 

 P0  cr  k + 1 
8.3
101
The mass flow rate reaches a maximum at the critical pressure ratio and is designated the
critical flow rate. The critical flow rate can be found by substituting the result of Equation
8.3 into Equation 8.1 to produce Equation 8.4.
8.4
k +1
 2  2( k −1)
m cr = ρ 0 A f c 0 

 k +1
Or for k=1.4.
8.5
m cr = 0.578 ρ 0 A f c 0
This equation provides the maximum flow rate that can pass through an area Af. The area,
Af, is the area through which isentropic streamlines would flow or the effective area of the
flow. This is typically a smaller area than the actual valve opening or valve area. The
effective flow area is related to the actual valve area through a discharge or flow coefficient.
For poppet valves, the valve area changes as a function of valve position. Figure 8.1 shows
a poppet valve in an open position with two potential areas denoted. Area 1 is the valve
seat area defined by the area of the valve port minus the valve stem area. Area 2 is the area
defined by the valve curtain or the circumference of the valve times the valve lift.
d
A1
l
D
Figure 8.1 Diagram of a poppet valve and seat and associated valve areas.
The effective flow area can be defined by a discharge (Cd) or flow (Cf) coefficient as
defined by Equation 8.6.
A f = C f A1 = C f
(
)
π 2
D − d 2 or A f = C d A2 = C d πDl
4
8.6
The condition of choked flow defines an upper limit for flow into an engine. Consider the
volumetric efficiency using choked flow as the actual flow into the engine. The volumetric
efficiency is the mass entering the cylinder divided by the mass that could fit into the
displacement volume. The numerator of Equation 8.7 shows the mass into the cylinder as
102
the product of the choked flow rate times the time the valve is open (Δθ /N). The
denominator is the air density times displacement volume where the displacement volume
is found as the product of mean piston speed (UP), piston area (AP), and time (π/N) for the
piston to move from TDC to BDC.
ev ,i =
8.7
0.578 ρ i A f c 0 ∆θ / N
ρ iU P AP π / N
After simplification, the volumetric efficiency is seen to be 0.578 divided by Z, the Mach
index. Experimentally when the mach index is below 0.6, the volumetric efficiency is
relatively constant and in the range of 0.8 – 1.0. This indicates that the flow is not choked
and the volumetric efficiency is not controlled by choked flow at the valve throat. However,
when Z > 0.6, the volumetric efficiency drops linearly with increasing Z. This indicates
that above Z = 0.6 the flow at the valve is choked and that increasing the piston speed will
cause a dramatic decrease in volumetric efficiency and thereby engine torque and power.
The Mach index is therefore useful for determining the operating speed at which power
drops because the flow rate through the valve cannot keep up with the flow demanded by
the moving piston.
𝑒𝑒𝑣𝑣,𝑖𝑖 =
0.578
𝑍𝑍
𝑈𝑈 𝐴𝐴
where; Z = 𝑐𝑐𝑃𝑃𝐴𝐴 𝑃𝑃
8.8
0 𝑓𝑓
8.2. Factors Influencing Volumetric Efficiency
When flow is not choked, there are several factors that influence the volumetric efficiency.
These include: valve timing, flow friction, heat transfer, and flow dynamics. Modeling of
the volumetric efficiency has been done where the model can include or exclude individual
effects in order to demonstrate their impact on volumetric efficiency. Figure 8.2 shows a
plot of volumetric efficiency as a function of engine speed. The figure begins with the
dashed line showing a volumetric efficiency of 1.0 for a case where the model includes no
losses or resistance to flow. Each successive line numbered 1 through 7 shows the added
effect of a physical phenomenon added to the model which impacts volumetric efficiency.
Beginning with the dashed line, the volumetric efficiency for a quasi-static engine is
shown. This value should be less than 100% if residual gas already in the cylinder were
included in the model. The first line, line 1, models flow friction in the intake hardware
(induction system) upstream of the valve. The volumetric efficiency is seen to decrease
with increasing engine speed. Higher engine speeds are seen to produce increased flow
friction which causes lower volumetric efficiency as speed is increases.
Adding the effects of heat transfer to the induction system causes the intake air density to
decrease and lowers volumetric efficiency further resulting in line 2. At slow engine speeds
there is more time for heat transfer causing the decrease in volumetric efficiency to be
greater at low speed but enabling better volumetric efficiency at high speed.
103
The next curve (line 3) adds the effect of late intake valve closing. At low speeds this is
detrimental to vol. eff. because the valves is still open when the piston returns and pushes
air back out of the cylinder (backflow). At high speeds however, the air entering the engine
has a high velocity and momentum and continues to flow for a period of time even when
the piston changes direction (ram effect). This improves volumetric efficiency at high
speed.
The next effect (line 4) is flow friction caused by the valve. This effect is the same as the
induction system friction. At low speeds the impact is lower than at high speeds.
Line 5 adds the effect of choking flow. This has an impact only when the engine speed is
high enough to cause choked flow in the valves, but once reached, the volumetric efficiency
is seen to drop off rapidly.
Effect 6 is heat transfer to the incoming air within the cylinder. This impact is the same as
effect 2, heat transfer in the induction system.
The final effect, 7, is intake tuning. This is the impact of pressure waves on the intake
system caused by rapid valve closure. When a valve closes a pressure wave is produces
that travels through the intake plenum and intake runners and reflects off of surfaces back
toward the valve. If the pressure wave produces a positive pressure at the intake when the
valve closes it can have a positive effect on volumetric efficiency. In this model, the
pressure waves are the most beneficial at two speeds, about 3200 and 4300 rpm. A similar
impact can be produced when a negative pressure results at exhaust valve closing. A
negative pressure reduces residual mass enabling more intake mixture.
Figure 8.2 Modeling results for volumetric efficiency as a function of engine speed. Source
unknown.
104
The accumulated impact of all of the parameters on volumetric efficiency produces a curve
that describes the amount of air ingested into the engine. Since the air inducted dictates the
amount of fuel that can be burned, the volumetric efficiency curve also represents the fuel
burned curve and thereby the torque curve. The reasons for a torque curve taking on a
specific shape are the same as those for the shape of the volumetric efficiency curve.
IC engines typically have low torque at low speeds because of backflow, and heat transfer.
Torque decreases at high speeds because of flow friction and choking. Engines have “sweet
spots” at intermediate speeds where torque is high due to valve timing and dynamic flow
effects producing high volumetric efficiencies at specific engine speeds.
The power curve for an engine follows from the torque curve. Power is the product of
torque and engine speed. Shapes for a typical torque curve and power curve are seen in
Figure 8.3. The power curve peaks to the right of the torque curve because power continues
to increase until the torque drops at a higher rate than the speed increases.
Power
Torque
Engine Speed (rpm)
Figure 8.3 Example of the shape of a torque and power curve for an IC engine
8.3. Evaporative Cooling During Intake
For port injected and carbureted engines, fuel evaporation in the intake port can cause a
significant change in the temperature and density of the incoming air. The impact of
evaporation can be studied using a thermodynamic analysis of the intake port. A control
volume beginning with atmospheric air and extending into the cylinder at the boundary of
the intake valve is shown in Figure 8.4. A simplification of the energy equation for this
control volume shown in Figure 8.4 is shown in Equation 8.9. The subscripts “a” represents
air, “l” liquid phase, “f” fuel, “0” atmospheric conditions, and “cyl” the boundary in the
cylinder at the intake valve seat. The variable, xe, represents the mass fraction of the fuel
that has vaporized at the boundary.
(
) (
)
0 = (m a ha )0 + m f h f l − m a ha + xe m f hv + (1 − xe )m f h f cyl +Q in
8.9
105
Treating the air and fuel vapor as an ideal gas with constant specific heat, h = CpT,
substituting and rearranging gives Equation 8.10. Factoring out the fuel flow terms leaves
a hv - hf which is equal to hfg or the enthalpy of vaporization.
(
) (
)
(
)
m a C p ,a Tcyl − m a C p ,a T0 = m f h f 0 − m f h f cyl − x e m f hv − x e m f h f cyl + Q in
8.10
Additional factoring and arranging gives Equation 8.11 which shows the temperature
change of the air in terms of the fuel evaporated, enthalpy of vaporization, and heat transfer.
Fuel Injector
P0 T0
Throttle
Pcyl Tcyl
Figure 8.4. Control volume used to analyze evaporation of the fuel during intake.

(Tcyl − T0 ) = m QCin − x+e mm f hCfg
a

p ,a
f
8.11
p, f
The intake runner and intake valve are typically hot and transfer heat into the intake air.
The fraction of fuel evaporated depends on the temperature of the air, intake valve, and
runners. If the intake parts are cold causing Qin to be low, the fraction of fuel evaporated
will be lower, if the intake components are hot then heat transfer will be higher but a greater
fraction of the fuel will be evaporated. The largest possible temperature drop will occur
when all fuel is evaporated and there is no heat transfer into the intake port. This limit can
be readily calculated to show the relative impact of the evaporation of different fuels on
volumetric efficiency.
Example: Find the maximum possible temperature drop for intake air when gasoline
is used as a fuel and the fuel air mixture is stoichiometric.
Solution: The maximum temperature drop will occur when Qin = 0 and xe = 1.0.
Substituting this result into Equation 8.11 gives.
(Tcyl − T0 ) = m C0 − m+ fmh fgC

a
p ,a
f
p, f
106
Dividing the numerator and denominator by m f gives:
(Tcyl − T0 ) = m
−h fg
a
m f
C p ,a + C p , f
=
− (350)kJ / kg
= −20.6 K
(14.6)1.0 + 1(2.4)kJ / kg − K
8.4. Two Stroke Breathing Parameters
The volumetric efficiency introduced in Chapter 4 is the ratio of air entering the engine
through the intake valve to the air that would fit within the displacement volume at some
reference temperature and pressures. It is often assumed that the air entering is all trapped
within the cylinder when the intake valve closes. This may not be true if the exhaust valve
is open during a portion of the intake valve opening period, but it is usually a reasonable
assumption for a 4-stroke engine. Two-stroke engines on the other hand very frequently
have an intake and exhaust valve or port open at the same time and therefore have a high
probability that gas will pass from the intake to the exhaust without being trapped in the
cylinder. When a mixture passes from intake to exhaust without being trapped in the
cylinder it is called short circuiting. Short circuiting gives rise to a variety of possible
parameters used to describe the effectiveness of two-stroke air exchange.
A list of parameters used to describe two-stroke gas exchange is given in Table 8.1. For all
of the terms, the mixture mass is used for a spark ignition engine because fuel and air are
entering the cylinder through the intake valve while for a diesel engine, only air is used.
The first term listed is the delivery ratio. This parameter is similar to the equivalence ratio,
but unlike equivalence ratio, the delivery ratio can easily be greater than unity when short
circuiting is occurring. In this case, it could be undesirable for a delivery ratio to be greater
than one while a volumetric efficiency of greater than one is desirable. The delivery ratio
alone is therefore incomplete in describing the effectiveness of the intake system.
Table 8.1 List of parameters used to describe the effectiveness of two-stroke intake
systems.
Parameter Name
Delivery Ratio
Charging Efficiency
Trapping Efficiency
Scavenging Efficiency
Purity
Equation
Dr =
mass of delivered air (or mixture) per cycle
ρ a,0 Vd
ec =
mass of delivered air (or mixture) reatained
ρ a,0 Vd
Γ=
mass of delivered air (or mixture) retained
mass of delivered air (or mixture)
es =
mass of delivered air (or mixture) retained
mass of trapped cylinder charge
P=
mass of air in trapped cylinder charge
mass of trapped cylinder charge
The second term in the table is charging efficiency. It considers only the fraction of air or
mixture that is retained or captured in the cylinder. This parameter is useful for determining
107
how well the cylinder is charged with a fresh fuel air mixture but does not provide any
information about how much fuel is short circuited.
The third parameter is trapping efficiency. This parameter describes how well the intake
system is capturing the fuel air mixture but it does not tell how much mixture is actually
being trapped. It is possible to trap 100% of the incoming mixture while only a small
fraction of the cylinder contains a fresh charge.
The scavenging efficiency is the ratio of the delivered mass to the mass in the cylinder.
This is similar to the delivery ratio except that the denominator is determined by the mass
that is actually in the cylinder instead of the mass that could fit in the cylinder.
The purity is the fraction of the total trapped mass of the cylinder that is air.
It is often useful when evaluating two-stroke engine parameters to draw a diagram with the
intake flow taking two possible paths as shown in Figure 8.5. One path represents the
mixture that is delivered and retained, md,r and the other for the mixture that is short
circuited, ms,c. The mass that is short circuited can be assumed to have the same
composition as the intake mixture while the mass of the mixture that is delivered and
retained can often be assumed to be burned completely. Making these assumptions allows
the mass delivered and retained to be calculated from exhaust gas measurement.
m d ,r
m in
Cylinder
m s ,c
m out
Figure 8.5 Schematic diagram of the cylinder as a control volume.
Example:
Given: The mass of a stoichiometric fuel and air mixture of gasoline entering a twostroke, S.I, engine is 2.0 kg/min. A hydrocarbon analyzer measures the equivalent of
15.13 g/min of fuel exiting in the exhaust. Given the engine and operating conditions
listed, and assuming that all fuel trapped within the cylinder is burned.
Find: The delivery ratio and charging efficiency
ρmix = 1.0 kg/m3
B = 0.1m
Pin = 1 atm
L = 0.12m
m f , sc = m f ,out = 0.01513 kg/min
N = 2000 rpm
Tin = 350 K
m sc = 0.01513(1 + 14.6) = 0.0236 kg/min
m sc = m f , sc + m air , sc = m f , sc +
A
A

m f , sc = m f , sc 1 + 
F
F


m sc + m dr = 2.0 kg / min
m dr = 2.0 − 0.236 = 1.764 kg / min
108
Dr =
ec =
d
m
2.0
2 kg/min 1 rev/cycle
=
=
= 1.06
3
2
3
ρVd N / n R (1.0 kg/m )(0.12)(0.1) π / 4 m / cycle 2000 rev / min 1.885
 d,r
m
ρVd N / n R
=
(m in − m sc ) (2 − 0.236) kg/min
=
1.885 kg/min
ρVd N / n R
=
1.764
= 0.936
1.885
8.5. Fuel Flow
There are primarily three methods that are currently used for the introduction of fuel into
an engine, they are: carburation, port injection, and direct injection. These three
technologies will be introduced in the following sections.
8.5.1. Carburation
Although the number of applications for carburetion has decreased significantly,
carburation is still used in numerous small engine applications including lawn mowers,
gas trimmers, and small off-road vehicles. Carburetors are purely mechanical devices that
require no electricity or sensors and can therefore less complicated and less expensive.
Carburetors do not require batteries or computer control technology.
A simple carburetor is shown schematically in Figure 8.6. The primary feature of a
carburetor is a narrow neck where the air accelerates. The high velocity at the neck
produces a vacuum or lower pressure at the throat of the venturi. The low pressure draws
fuel up-hill through an elevation change “h” before it enters the throat area. As the mass
flow rate of air increases, the pressure drop at the throat increases producing an
approximately proportional increase in fuel drawn up the fuel tube. Thus, a carburetor
maintains a relatively constant fuel air ratio as varying amounts of fuel pass through the
venturi. This creates an inexpensive mechanical device useful for introducing fuel into an
engine.
Air
h
Fuel
Figure 8.6 Diagram of a simple carburetor.
The simple carburetor suffers from several significant drawbacks which are listed below.
1. The fuel flow rate at low air flow rate is zero. Fuel enrichment is required at low speeds.
2. Under cold conditions, the fuel does not completely evaporate causing the mixture to be
too lean during start-up. Over-fueling during startup is not inherent in a simple carburetor.
3. Under maximum power requirements it is desirable to run fuel rich to provide a power
boost.
109
4. During times of low power demand, it is beneficial to operate fuel lean to improve fuel
consumption.
An advanced carburetor, as shown in Figure 8.7, can overcome many of the problems
associated with a simple carburetor. The following features are typically added to improve
performance.
1. A butterfly valve (not shown) is placed on the top of the carburetor to restrict the flow
of air during startup and place the entire path of the carburetor under a vacuum. This pulls
fuel through the main orifice (6) without the need for air flow enriching the mixture at
startup.
2. A second orifice is added (11) to allow additional fuel flow when the throttle is closed
and air is not flowing well at low engine speed. This also provides enrichment of the fuel
air mixture at low speeds and load (idle)
3. An air bleed orifice (10) allows additional air into the fuel air mixture causing a leaner
mixture at medium loads.
4. An ejector pump (not shown) is connected to the throttle plate. When the throttle position
is changed rapidly, fuel is injected by a mechanical pump into the Venturi by pressurizing
the fuel tank.
5. A sealed bellows in the float bowl changes the elevation difference between the float
bowl and engine thereby altering the fuel/air ratio for a given pressure. This allows the fuel
required for a stoichiometric mixture to change as needed with a change in altitude.
1. Main Venturi
2. Boost Venturi
3. Main metering spray tube or nozzle
4. Air bleed orifice
5. Emulsion tube or well
6. Main fuel metering orifice
7. Float chamber
8. Throttle plate
9. Idle air-bleed
10. Idle fuel orifice
11. Idle mixture orifice
12. Transition orifice
13. Idle mixture adjusting screw
14. Idle throttle setting adjusting screw
Figure 8.7 Schematic diagram of an advanced carburetor. [1]
110
Even with the features of an advanced carburetor, there are still undesirable issues related
to their use. The carburetor introduces fuel for all cylinders at a single source. Liquid fuel
is transported toward each cylinder but some larger drops may not turn a corner as easily
as smaller drops thereby accumulating and causing uneven fuel air ratios. Carburetors have
no way of sensing temperature and adjusting fuel flow accordingly. Over time a carburetor
jet (orifice) may be become partially plugged and the fuel flow rate changes with no way
of automatically adjusting. The phrase “engine tune-up” originated from the need to adjust
carburetors periodically for changes in weather or for cleaning.
8.5.2. Port Fuel Injection
Port fuel injection involves the introduction of fuel into the intake port just upstream of the
intake valve(s). There is typically one injector for each cylinder. Port fuel injection allows
for individual control of fuel to each cylinder and the ability to change the fuel flow rate
based on engine operating condition or ambient conditions. A common gasoline system
consists of a fuel pump, fuel line and fuel injector. The injector is an electronically
controlled orifice.
The pump which is often located in the gas tank feeds and pressurizes a fuel line which
transports the fuel to a location near the intake port. The fuel line is typically pressurized
to approximately 100 psig. The fuel injector is controlled by the electronic control unit
(ECU) which is a computer that tells the injector how long to open the nozzle and inject
the fuel. After a brief transition period, the amount of fuel injected is linearly proportional
to the opening time as shown in Equation 8.12.
m f = m f ∆t = m f
∆θ
N
8.12
The amount of fuel injected must increase when the throttle is opened or is proportional to
the engine load as suggested by Equation 8.13, where C1 is a constant.
m f ∝ Load = C1 Load
8.13
In a typical fuel injector, the rate of fuel injection ( m f ) remains constant while the amount
of time the injector is opened is used to control how much fuel is injected. Combining
Equations 8.12 and 8.13 and solving for the amount of engine rotation required to inject
fuel into the intake port (Δθ in Equation 8.14) shows that the injection crank angle time
increases proportional to the product of the engine speed (N) and the load. This means that
an engine at full load (100%) and the peak power engine speed (for example 6000 rpm)
will require 100 times as many crank angles to inject fuel in comparison to an engine at
idle (600 rpm, 10% load). This creates a challenge for port fuel injection because a nozzle
that allows injection in 5 crank angles at idle would require 500 crank angles at full speed
and load. The limit on crank angles allowed for an injection in a four-stroke cycle is 720
crank angles.
111
∆θ =∝ Load =
8.14
C1 Load N
m f
8.5.3. Direct Fuel Injection
Direct fuel inejection is now common in both spark ignition and compression ignition
engines. Spark ignition engines must deliver the fuel during the compression process
requiring a higher pressure than port injection but not as high as the compression ignition
engine. The compression ignition injector must deliver fuel into the pressurized cylinder
(3 – 5 MPa) within an acceptable combustion burn period (30 – 90 CAD). The solution has
been to produce fuel injectors that operate at relatively high pressures, typically in the range
of 55 – 172 MPa (8,000 – 25,000 psi). The result is higher cost and more complexity. There
are three types of fuel injectors that have been developed to meet these design
requirements: the in-line pump, the common rail and the unit fuel injector.
Each of these three systems are shown schematically in Figure 8.8. Each system requires a
pump or a mechanism to pressurizing the fuel, fuel line or method for delivering the fuel
to each cylinder, a timing mechanism or method for starting and ending the injection
process, and a nozzle with a small orifice where the fuel is injected. Each of these systems
will be described in more detail below but the figure shows that in the pump line nozzle,
the pressure and timing mechanisms are in the pump and the line carries the fuel to an
injector in each cylinder. The common rail has the timing mechanism (solenoid) separated
from the pump and the unit injector, with the exception of a small low pressure fuel pump,
has all of the system in one location for each cylinder.
Common Fuel Rail
Solenoids
Nozzles
Injectors
Lines
Pump
Engine
Pump
Pump-Line-Nozzle
Engine
Common Rail
Common Fuel Rail
Cam Shaft
Unit
Injectors
Pump
Engine
Unit Fuel Injectors
Figure 8.8. Schematic diagrams of three common fuel systems used for direct injection
engines.
112
The most common type of fuel injector prior to the turn of the century and still used in
larger, lower pressure compression ignition engines is the in-line pump or pump-linenozzle. As implied by the name, this system consists of a fuel pump that pressurizes the
fuel, a fuel line that transfers the fuel from the pump to the engine and a nozzle that is
mounted to spray the fuel into the cylinder. In this system, the pump as shown in Figure 8.9
accomplishes the three requirements of increasing the fuel pressure, controlling the
injection time and controlling the injection duration. These three features are accomplished
mechanically without the need for electronic controls.
The pump is driven by the engine crankshaft which rotates a cam shaft causing a plunger
to move up and down. There is one plunger for each cylinder in the engine. The plunger
creates the motion that compresses the fuel and pressurizes the fuel line, one fuel line for
each cylinder. The increased pressure moves through the line to the injector causing the
needle which is initially held in place by a spring to overcome the spring force and lift a
needle valve off its seat. There is one nozzle or injector for each cylinder. The amount of
fuel injected is controlled by the amount of time the pressure remains high in the fuel line.
A “rack” is used to rotate a helix on the plunger in the fuel pump. The helix location
determines when the fuel pressure is dumped by the plunger. The “gas pedal” on a “diesel
engine” is in actuality moving this fuel rack to change the amount of fuel injected and
thereby control how much work or torque the engine will produce.
Figure 8.9. Image of a fuel pump used in a pump-line-nozzle fuel system.
The common rail system also uses a pump, line and nozzles as shown in Figure 8.10 but
there is one common plenum or pressure collector for all cylinders. The pump can be driven
electronically or by the crankshaft. This system which is also used for port injected spark
ignition engines will have a much higher pressure in the rail (pressure collector in the
figure) for compression ignition applications than it does for spark ignition applications. In
this system, the fuel injector is designed to produce a constant high pressure (approximately
55 MPa) in the rail which is controlled by a regulator. The injection timing and duration is
electronically controlled by a computer electronic control unit (ECU) and solenoid
113
controlling flow to each cylinder. When the solenoid valve is opened, fuel flows into the
top of the injector. The injector consists of two pistons and two cylinders. The piston on
top where the fuel enters from the common rail has a larger area than the piston on the
bottom where the fuel exits to the engine. This area change multiples the pressure by a
factor of two or three over the rail pressure. As with the pump-line-nozzle injector, an
increase in pressure created by the solenoid lifts a needle valve producing injection.
The advantage of the common rail system over the pump-line-nozzle is electronic control
over fuel injection. This enables a wider range of injection timings and durations. It also
enables multiple injections within the same cycle. It is now common to have three separate
injection events within a cycle.
Figure 8.10. Common Rail Fuel System. https:// www.researchgate.net / figure / Diagram-of-the-
Common-Rail-CR-system_fig2_282448607
The final method is a unit injector. For a unit injector, the fuel is delivered to each
individual injector at relatively low pressure. The injector is however linked to the cam
which rotates to cause a plunger to move within the injector. The plunger pressurizes the
fuel liquid creating high pressures on the order of 20,000 psig. Unit injectors can be either
mechanically controlled or electronically controlled. The control comes by opening a valve
that dumps the fluid in the injector through a relatively large orifice and depressurizes the
system. Unit injectors tend to be used in larger engine while common rail injectors
dominate the smaller engine (gasoline and diesel) market.
The final type of fuel injector used in compression ignition engines in the unit fuel injector.
An image of a unit fuel injector is shown in Figure 8.11. The unit fuel injector places all
three functions of the fuel system (pressure, timing and duration) within one small unit.
114
The injector is mounted below the camshaft which provides the work needed to pressurize
the fuel. The camshaft moves a piston compressing fuel within the injector. A solenoid also
mounted on the injector opens and closes to create or relieve pressure within the cylinder
inside the injector. Two pistons and two cylinders are used to amplify the pressure as in
the common rail injector. The advantage is having all of the injector functions within a
single unit. The disadvantage is that there is a separate unit required for each cylinder and
it the injector size is enlarged because there is essentially a pump within each unit. The unit
injector is favorable when engine sizes are large and there is enough space to accommodate
the unit within the engine head.
Figure 8.11 Image of a unit fuel injector. https://en.wikipedia.org/wiki/Unit_injector
References
[1] Heywood, J.B., “Internal Combustion Engine Fundamentals,” McGraw Hill, 1988.
115
Chapter 8 Homework Problems
8.1 Find the maximum critical (assuming Cf = 1.0) flow rate of air (298 K, 101 kPa) through
an intake valve that is 50 mm in diameter and has a valve stem diameter of 6 mm.
8.2 An engine has a bore of 0.1 m, stroke of 0.08 m, inlet flow effective area 4.0x10-4 m2,
and inlet temperature of 320 K, what is the maximum speed for the engine to maintaining
good volumetric efficiency.
8.3 You will be given a spreadsheet with valve lift vs. crank angle data for a single intake
valve on a Honda F4i, 600 cc, 4-cylinder motorcycle engine. Calculate the engine speed
(in RPM) for which the Mach index, Z, is equal to 0.6 for this engine. The bore = 67 mm ,
S = 42.5 mm, and the valve diameter = 27.5 mm, valve stem diameter = 3.975 mm, Tintake
= 300 K, Pintake = 85 kPa. There are two intake valves per cylinder and four cylinders for
the engine.
8.4 Use the mach index criteria that Z < 0.6 to estimate displacement volume required of a
naturally aspirated engine running at sea level (T = 298 K, P = 101 kPa) to produce 300
hp. Assume: 1. Below the critical engine speed the volumetric efficiency is 1.0. 2. The
intake valve area is 1/3 of the piston area. 3. The engine is square (bore = stroke). 4. The
engine is burning a stoichiometric mixture of gasoline and air. 5. The engine has a fuel
conversion efficiency of 0.30.
8.5 Compare the amount of energy inducted (mf*QLHV) into one cycle of one cylinder for
a stoichiometric mixture of gasoline and air compared to stoichiometric mixture of
methanol and air. Assume the parameters for a 2.0 L, 4 cylinder, rc = 9.8, ev,i = 1.0, Pi = 101
kPa, T0=300 K Dodge Neon engine. You will need to calculate the air flow in and use the
F/A ratio to obtain mf. Include evaporative cooling effects but not residual gas effects. You
may assume there is no heat transfer in the intake process and all of the fuel evaporates.
For your answer write: 1) the energy inducted for the gasoline air mixture, 2) the energy
inducted for the methanol air mixture.
8.6 The air fuel ratio entering a 350 cm3 2-stroke, atmospheric, carbureted gasoline engine
is stoichiometric. If the engine is running at full load and 3800 rpm and has a delivery ratio
of 1.3 and a charging efficiency of 0.8, what is the trapping efficiency?
8.7 Given a diesel engine with a bore = 125 mm, stroke = 150 mm, compression ratio = 15.
The fuel flow rate at 1800 rev/min is 1.6 g/s per cylinder. The intake air flow rate is
measured to be 80 g/s at an intake temperature and pressure of 300 K, 1 atm. Optical
diagnostics are used to measure the CO2 and O2 concentration in-cylinder before the
exhaust valve opens and it is determined that the equivalence ratio of the trapped gases is
0.5. Find (a) the charging efficiency, (b) the delivery ratio, and (c) the trapping efficiency.
8.8 A small motorcycle engine is running at idle (1200 rpm) and the intake manifold
pressure is 2 psia (the atmospheric pressure is 12.5 psia). The electronic control unit (ECU)
is holding the fuel injectors open for 0.139 msec. If the throttle is fully opened (12.5 psia)
and the engine is running at 14,000 rpm. How long should the ECU be programed to open
the fuel injector? How many crank angles will the injector be open for each injection
event?
116
9. Engine Measurements and Controls
Engines have changes more significantly over the past 30 years than they did in a 70 year
span between 1920 and 1990. This change has been enabled primarily by sensors and
computer control. The miniaturization of the computer in size and cost has enable
electronic fuel injection, exhaust gas sensing, fuel and air flow measurement and adaptive
smart tuning of engines. Not only can engines be controlled through sensors and real time
feedback but systems are monitored to determine if they are functioning correctly and
changes are made automatically to adjust operating parameters over time in order to
compensate for wear, fouling and other forms of degradation. This chapter will first
describe how a engine test cell is set up to collect measurements on an engine and then
describe how measurements are now continuously made and used to control and improve
performance on an engine.
When an engine is removed from a vehicle, the fuel system, air and exhaust system, coolant
system and means of braking or loading the engine are typically also disturbed. Each of
theses systems must therefore be replaced in a the test cell to allow the engine to function.
The output shaft of the engine is typically connected to a dynamometer which places an
opposing torque on the engine. The dynamometer (dyno) often has a means of measuring
the torque and speed on the shaft connecting the engine and dyno. The radiator in the car
must be replaced by a radiator and fan in the test cell or some type of heat exchanger. The
air intake is often unchanged but the exhaust tailpipe, catalytic converter and tailpipe are
often removed and replaced with some other type of system that move exhaust safely from
the engine to outside. Finally, the fuel tank and fuel delivery system must be replaced
and/or modified.
9.1. Dynamometers
There are three types of dynamometers typically used for IC engines: Water Brake, Eddy
Current, and Motoring. Each has advantages and disadvantages which will be described
below.
9.1.1. Water Brake
The water brake dynamometer works by connecting the input shaft to veins which force
water through channels during rotation. As the water passes through narrow channels,
viscosity provides a resistance to flow and energy is dissipated through viscous dissipation.
The water heats up absorbing the energy. Very large torques can be absorbed with a
relatively small rotor. It is difficult to dissipate a lot of energy at low engine speeds. This
type of dynamometer is the least expensive and requires no electrical parts unless they are
used to control the direction of the vanes and therefore the amount of dissipation at a given
speed. A water brake dynamometer can only absorb energy.
9.1.2. Eddy Current
The eddy current dynamotor utilizes forces created by induced currents and the resulting
magnetic field. When a conducting material passes through a magnetic field, a circular
(eddy) electrical current is induced in the material. This current creates a magnetic field
that opposes the original magnetic field and thereby places a force on the moving object.
The force is proportional to the strength of the original magnetic field and the speed of the
117
object, and the resistance to electrical current. The induced current produces resistive
heating in the conductive material and thereby dissipates the kinetic energy of the moving
object.
An eddy current dynamometer connects a rotating conductive material to the shaft of the
engine and as the engine rotates, the eddy current is induced in the rotating member. The
resistance of a variable resistor is changed to impede the current. A higher resistance
produced a higher loss and higher resistance heating. This part is sprayed or passes through
water to keep the part from overheating due to the induced eddy currents. This type of
dynamometer is very popular in engine testing and is widely used in industry.
9.1.3. Motoring AC or DC
The motoring dynamometer is essentially a DC or AC motor output shaft connected to the
engine output shaft. It is both a power producing and a power generating (absorbine)
device. When it is desired to rotate the engine without combustion in order to start the
engine or to study friction or pumping losses, the engine can be turned by the motor. When
the engine is producing power, the motor can be used as a brake by generating current
which can be placed back onto the grid or used to charge a battery. In some cases, the
power produced is delivered to a resister bank and dissipated as heat.
9.2. Fuel System
For a test cell, the fuel tank must be supplied for the engine. Fuel tanks are typically smaller
for the test cell and may be rectangular or simplified shapes as they do not need to fit
beneath a vehicle. A fuel pump is needed to get the fuel from the tank to the engine or the
tank must be mounted higher than the engine to enable gravity to flow the fuel to the engine.
Fuel flow can be measured with the same device used on the engine or a more sophisticated
or accurate system. Four types of flow meters are typical in test cells.
9.2.1. Volumetric Displacement
The fuel flow sensor in most cars is a volumetric device. The volume flow is measured and
convert to a mass flow using the density of the fuel. This is typically done by passing the
fuel through a rotating turbine or some type of positive displacement device. A typical
accuracy might be on the order of ± 5% which can be attributed to changes in density or
the very small volumes associated with each individual rotation or displacement.
9.2.2. Pressure and Orifice
A second method, not common in a vehicle, but sometimes used in a test cell is pressure
and orifice flow meter. Here a pressure sensor is placed upstream and downstream of an
orifice. The velocity of the flow can be related to the change in pressure and the orifice
diameter. This method will also depend on the density of the fluid which is a slight function
of temperature.
118
9.2.3. Coriolis Flow Meter
One exception to the volumetric approach is the Coriolis flow meter produce by Flowtron.
This device works by measuring the vibrational frequency of a tube bent in a “U” shape.
As the flow moves along the U turn and changes direction it induces a rotational moment
on the tube that is proportional to the mass flow rate. Once this device is calibrated, liquids
of differing density can be measured accurately in the same device. This is an expensive
but perhaps the most accurate method for measuring liquid fuel flow.
9.2.4. Gravimetric Device
The final measurement of flow is an old fashion scale. The fuel tank is placed on a scale
and the change in mass is monitored as the fuel flows out of the tank. This is a less
expensive method of obtaining accurate flow rates.
The most common method used on vehicles to determine flow rate is not a measurement
at all. For fuel injected cars, the amount of fuel passing through the injector is a function
of the time the fuel injector orifice is open. The flow rate is measured as a function of the
injector opening time to create a calibration table or curve. This curve is placed into the
engine computer and the computer calculates the flow rate for the given injector opening
time. This works fairly well until the injector becomes clogged, damaged or worn and then
the calibration curve in the computer is no longer valid.
9.3. Air Flow Measurement
There are several different methods used to measured air flow in an engine. The hot wire
or hot film anemometer is typically used when the car is mounted under the hood but in a
test cell other options are used.
9.3.1. Hot Wire or thin film anemometer.
A hot wire anemometer places a current through a thin wire causing it to heat up. The
resistance of the wire changes with wire temperature. As the flow increases, the wire
temperature decreases changing the resistance. Thus, the wire electrical resistance is a
function of air flow rate. A calibration curve is produced to relate the two values.
9.3.2. Turbine Flow Meter
Air flow rated can also be measured with a turbine wheel rotating proportional to volume
flow rate or a spring loaded flapping device that cycles with volume flow rate. Both of
these techniques also require a density (temperature and pressure) measurement in addition
to the volume flow to convert the flow from volume to mass flow. Unlike water, air density
is a strong function of air temperature and pressure.
9.3.3. Calibrated Orifice
Pressure drop can also be related to flow rate. Measurement of the pressure upstream and
downstream of a flow orifice can be calibrated to the air flow. Once again, this flow rate is
dependent on the density of the air entering the orifice and therefore temperature must also
be measured. A special case of the calibrated orifice is the chocked flow orifice. As the
pressure difference upstream and downstream of the orifice increases, the velocity at the
orifice eventually reaches the speed of sound. At this velocity and above, the downstream
119
pressure no longer has any impact on the flow rate which can be determine solely from the
upstream temperature and pressure. The choked flow orifice eliminates the need for two
pressure measurements but is only applicable over a smaller range of flows where the gas
velocity is sonic.
9.4. Exhaust Gas Measurement and Analysis
Engines exhausts consists of primarily H2O, O2, N2, and CO2 (major species) with small
but significant amounts of CO, H2, NO, NO2, SO2, aldehydes (minor species), and particles.
Gases are always measured as molar basis, not on a mass basis. Major species are typically
measured in mole fractions or percent while minor species are normally measured in parts
per million (ppm). Minor species are typically unimportant for energy analysis or
understanding engine efficiency but are extremely important for understanding pollutants
and the combustion process. Common methods for measuring some of these gas species
are described below.
9.4.1. Non dispersive Infrared (NDIR) absorption, CO2 CO and NO
Some combustion products such as CO2, CO, and H2O are strong absorbers of infrared
radiation. Because water is a strong absorber and can condense on surfaces within an
analyzer, it is often removed before gases are analyzed. The bulk of the water in a product
gas can be removed by passing water through a heat exchanger of chilled water. The
product gas can then pass through a desiccant to remove remaining moisture.
Once water vapor is removed there are spectral regions where CO2, CO, and NO are known
to be strong absorbers. A non-dispersive infrared (NDIR) analyzer utilizes this property of
the gas to make a measurement by setting up a system as illustrated in Figure 9.1. An
infrared radiation source passes irradiation through two cells; one containing a nonabsorbing gas such as nitrogen and the other containing the gas to be measured. After
passing through the measurement or reference cell, the emission terminates in an absorbing
detector gas cell. The detector cell compares the pressure on each side of the cell with a
diaphragm. When both cells contain non absorbing gasses, both sides are heated by the
incoming emission from the IR source and the diaphragm is in a neutral position. If the
measurement cell contains an absorbing gas, emission from the IR source reaching the
measurement cell will be reduced causing the pressure on that side of the cell to be reduced
and allowing the diaphragm to move. Longer sample cell lengths allow for a smaller
concentration of gas to be measured.
The instrument is calibrated by placing a non-absorbing (zero gas) such as nitrogen in the
sample cell and utilizing the diaphragm location to indicate zero concentration. Then a
known sample gas concentration or span gas is placed in the path giving a known span
diaphragm location. The actual concentration can then be determined by interpolating
between the two known concentrations.
In order for an NDIR analyzer to function the water vapor must be removed because it
participates in the absorption of IR radiation in the same bands as CO2, CO, and NO. When
an analyzer requires water to be removed from the gas, the resulting measurement is called
a “dry measurement”.
120
Figure 9.1 A diagram of an non dispersive infrared measurement device. [1]
9.4.2. Unburned Hydrocarbon (FID)
When a hydrocarbon fuel is burned in an electric field, positive ions are produced
proportional to the number of carbon atoms in the hydrocarbon. A schematic drawing of a
flame ionization detector (FID) is shown Figure 9.2 which utilizes this principle to measure
unburned hydrocarbons in exhaust. A hydrogen flame which contains no carbon is used as
an ignition source for the unburned hydrocarbons. For a fixed flow of exhaust, the current
produced by the ions is proportional to the concentration of unburned carbon in the exhaust.
Figure 9.2 A diagram of a flame ionization detector (FID) used to measure unburned
hydrocarbons. [1]
121
The FID is not able to distinguish the type of hydrocarbon in the exhaust; it only counts
the number of carbons. For example; 20 ppm of C3H8 would produce the same results as
60 ppm of CH4 because both have the same number of carbon atoms available for
ionization.
9.4.3. Chemiluminescence
Although NO can be measured using NDIR absorption, it is more common and typically
more accurate to measure NO using a chemiluminescence analyzer as shown in Figure 9.3
This analyzer works by creating a reaction between ozone (O3) and NO. The reaction
produces and excited NO2* molecule which subsequently decays and produces a photon.
The photon or luminous signal is measured by a photomultiplier. As with the other gas
analyzers, the instrument is calibrated by flowing a span gas (gas of known concentration)
and a zero gas (gas free of hydrocarbons) into the measurement volume and then real gas
measurements are determined by interpolating between known measured points.
Figure 9.3 A diagram of a chemiluminescence analyzer. [1]
9.5. Calculating Equivalence Ratio from Exhaust Gas
The air fuel ratio of a mixture can be determined from exhaust gas measurements. While
fuel and air flow rates entering an engine can be measured, exhaust gas measurements can
often produce a lower uncertainty in determining the air fuel ratio. The analysis utilizes an
elemental balance to determine the amounts of carbon, hydrogen, and oxygen in the
system.
A powerful concept in combustion analysis is that of the mixture fraction. The mixture
fraction is the ratio of mass originating from the fuel to the total mass of the mixture and
is defined in Equation 9.1
xf =
mass originating in the fuel mo, f
=
total mass
mt
9.1
One of the uses for a mixture fraction is to determine the mixture equivalence ratio from a
measure of the product species. For a balanced reaction equation, the mass of products
originating from the fuel equals the mass of the fuel while the total mass is the mass of the
122
fuel plus the mass of the air. The mixture fraction can therefore be related to the fuel air
ratio by Equation .9.2
m fuel
mair
=F
A
=
m fuel
=
mt − m fuel
9.2
xf
1− x f
Now combining this with the definition of the equivalence ratio shows that if the mixture
fraction is known for a given fuel (the stoichiometric fuel air ratio is also known), then the
equivalence ratio can be determined.
9.3
xf
(F A) = 1 − x
φ=
(F A) (F A)
f
s
s
The mixture fraction is a mass based parameter while combustion measurements are all
volume or molar based. Finding the mixture fraction involves converting molar
measurements to a mass. For a 100 mole basis, the mass in the exhaust products originating
in the fuel can be determined by adding the carbon mass and the hydrogen mass in the
products as shown in Equation 9.4. In this case, only CO2, CO, and H2O species were
considered; however, if additional carbon or hydrogen containing species are present they
must be included.
(
mo, f = y CO2
MWC
MWH 2
MWC
) MW
+ ( y CO )
+ (y H O )
MW
MW
CO2
CO
2
9.4
H 2O
The total mass is determined from Equation 9.5, where “I” represents each of the species
present in the products.
mt =
∑ ( y )MW
i
9.5
i
Example: Finding Equivalence ratio from product gas composition.
Given: The exhaust gas composition shown in the table for propane.
Find: The mixture fraction and equivalence ratio.
Example of products of combustion for propane.
Gas
Species
Concentration
(%)
123
CO2
H2O
O2
N2
9.45
12.60
3.94
74.0
Because propane contains only carbon and hydrogen, the carbon in the CO2 and
hydrogen in the H2O originated in the fuel. All other mass in the products originated
in the air. On a 100 moles of products basis, the mass originating from the fuel can
be found.
mo, f = 9.45 moles CO2
2 g H2
12 g C
+ (12.60) moles H 2O
= 138.6 g
mole H 2O
mole CO 2
The total mass in the products is:
mt = 9.45 moles CO2
28g
32g
18 g H
44 g CO2
+ (12.60) moles H 2 0
+ 3.94moles O 2
+ 74.0moles N 2
= 2840.7 g
mole N 2
mole O 2
mole H 2O
mole CO2
The mixture fraction then becomes:
xf =
138.6
= 0.0488
2840.7
The stoichiometric F/A ratio can be found in the literature to be: 0.0638.
The equivalence ratio can then be found using Equation 9.3.
 0.0488 


1 − 0.0488 
= 0.8
φ=
0.0638
There are several complications encountered when trying to implement this technique with
real measurements. The first is that measurements are always incomplete. Water (H2O) and
nitrogen (N2) concentrations are rarely measured and therefore not available. Also, many
of the measurement techniques do not work with water vapor in the gas. The water vapor
is often removed before the measurement is taken. Such a measurement is referred to as a
dry measurement.
The following assumptions are typically made in order to complete an equivalence ratio
analysis from products.
1. If CO is not measured, and the mixture contains O2, it is assumed that combustion is
complete and CO and H2 concentrations are negligible.
2. If CO is measured and is significant (i.e. greater than 500 ppm). Then it can be assumed
that the H2 concentration is half the CO concentration.
3. It can be assumed that the product H/C ratio is the same as the fuel H/C ratio in order to
find the water concentration from the CO2 and CO concentrations.
4. The balance of the mixture concentration not measured or calculated from the above
assumptions is assumed to be nitrogen (N2)
124
Example: Finding the equivalence ratio from incomplete exhaust gas
measurements.
Given: Dry measurements of exhaust gas from propane (C3H8) are measured to be:
10.8% CO2 and 4.5% O2.
Find: What are the wet concentrations of these gases and what is the mixture
fraction.
Solution: Since there is a significant amount of O2, it is assumed that CO and H2
concentrations are negligible. We therefore assume the balance of the dry gas is
nitrogen.
y N 2 = 100 − yCO2 − yO2 = 1 − 0.108 − 0.045 = 0.847
Now for 100 moles of gas, there are 10.8 moles of CO2. The fuel is propane with a
H/C ratio of 8/3. So the moles of H atoms that would be present for 10.8 moles of
CO2 is:
N H = N CO 2
8
8
= 10.8 = 28.82
3
3
Or the moles of H2O = 14.41.
The total number before the water was removed to make 100 moles of dry gas was
114.41.
The wet concentrations can now be calculated by taking the moles of each gas
divided by the total number of moles.
10.81
= 0.0945
114.41
14.41
y H 2O =
= 0.1260
114.41
4.50
y O2 =
= 0.0394
114.41
84.68
y N2 =
= 0.740
114.41
y CO2 =
Note these are the same concentrations as in the previous example showing that the
dry concentrations from this example came from the same mixture used in the
previous example except the water has been removed.
Now, either the wet or the dry product concentrations can be used to find the
mixture fraction but the type used must be consistent. Since the wet concentrations
were used in the last example, we will use the dry concentrations to demonstrate
the results are the same.
The mass originating from the fuel is:
mo, f = 10.8 moles CO 2
The total mass is:
2 g H2
12g C
+ 14.41 moles H 2 O
= 158.55 g
mole CO 2
mole H 2
125
mt = 10.81 moles CO2
28g
32g
18 g H
44 g CO2
+ (14.41) moles H 2 0
+ 4.50 moles O 2
+ 84.68 mole N 2
= 3250.5.1g
mole N 2
mole O 2
mole H 2O
mole CO2
The mixture fraction is:
xf =
158.55
= 0.0488
3250.5
The equivalence ratio is:
 0.0488 


1 − 0.0488 
= 0.8
φ=
0.0638
9.6. In-Cylinder Pressure Measurement
The cylinder pressure is typically measured using a piezoelectric pressure transducer. The
transducer consists of a piezoelectric crystal that produces a small charge when
compressed. The charge is amplified and then recorded using an analog to digital converter
and data acquisition software. The pressure transducer must be calibrated to create a
voltage vs pressure relationship. Typically, the crystal response is linear with pressure and
a single constant is used to relate a change in pressure to an output voltage.
Although linear, the gage responds to changes in pressure but not to absolute pressure.
Therefore, a reference pressure is also needed. This is typically done by setting the pressure
at the end of intake valve closing equal to the intake plenum pressure.
An analogue to digital converter is used to covert the voltage signal to a digital number.
Rather than collect the data at specified time intervals, it is more useful to collect it at
specific crank angle positions. This is done using an encoder attached to the output shaft.
The encoder is made to provide a trigger signal at defined crank intervals; typically, 0.25,
0.5 or 1.0 crank angles. A second pulse is obtained each time the engine rotates. This is
called the TDC signal. The TDC signal may occur at any crank angle but remains constant
at that angular position and the difference between TDC and the crank angle where the
pulse is given must be known.
The final result of an in-cylinder pressure measurement is a pressures at each sampled
crank angle. It is common to average the pressure at each respective crank angle or
ensemble average the pressure. It is typical to utilize an average of 50 or more cycles
although it is also common to record each cycle and also calculate a variation between
cycles or a variance in the pressure data.
9.6.1. Calculating Heat Release Rate from Cylinder Pressure
The energy equation and piston-cylinder geometries can be used to estimate the apparent
rate of heat release from cylinder pressure data.
The energy equation is written in Equation 9.6 where Qin represents two terms: 1) actual
heat transfer entering the gases within the cylinder and 2) the sensible energy change
126
produced by burning the fuel. In reality, fuel energy is changing from high to low chemical
energy which is converted to sensible energy in the products. The heat transfer term is
therefore considered an apparent heat transfer which combines both physical effects.
9.6
dU δQin
dV
=
−P
dt
dt
dt
The infinitesimal rate of time change, dt, can be related to a change in crank angles dθ by
a constant. Such that dt = Cdθ. Also, dU = mCvdT. This allows the energy equation to be
written as shown in Equation 9.7.
9.7
δQin mC v dT
dV
=
+P
dθ
dθ
dθ
Now using the ideal gas law for constant mass:
9.8
dT
P dV
V dP
=
+
dθ mR dθ mR dθ
And substituting the ideal gas law (9.8) into the energy equation (9.7) gives.
δQin
V dP 
dV
 P dV
= mC v 
+
+P
dθ
mR
d
mR
d
θ
θ
dθ


9.9
Arranging variables of like terms and remembering that R = Cp-Cv and k=Cp/Cv leads to:
δQin  k  dV  1  dP
=
+
P
V
dθ
 k − 1  dθ  k − 1  dθ
This equation suggests that the apparent heat release rate
9.10
δQin
can be determined from a
dθ
knowledge of k, P, V and the rates of change of pressure and volume. A cylinder pressure
measurement provides both P and dP/dθ. The volume and rate of volume change can be
obtained from an encoder signal which provides the crank angle position and Equation 4.35
which relates crank angle to volume. The result assumes constant specific heat but when
combined with a model of the cylinder contents, a gas temperature and ratio of specific
heats can be calculated on a crank angle basis and used to produce a more accurate result.
127
9.7. Engine Sensors and Actuators
Today’s engines utilize numerous sensors that are read and interpreted by computers using
a manufacturer based control algorithm to produce engine operation.
In 1991 the California Air Resources Board (CARB) instituted a requirement that all cars
sold in California have some type of on-board diagnostic system to monitor emission
equipment. Later in 1994, CARB mandated a standard for the on-board diagnostic system
now called OBD II that would become mandatory on cars in 1996. The OBD II system has
grown in capability to the point that not only emissions but almost all of the cars sensors
and performance parameters can be monitored through the ODB II connector port.
The first part of the OBD-II standard is the connector. The 16-pin female interface
connector must be located in the vehicle's cabin within 2 feet of the steering wheel. For
most cars, this means in the driver's foot well or just below the steering wheel. Although,
the physical connection is always the same in OBD-II-compliant vehicles, not all of the 16
pins are always utilized and the data isn't always sent over those pins in exactly the same
manner, so there is some variation within the standard. Specifically, there are five major
signaling protocols for vehicles sold in the United States between 1996 and 2008 that can
usually be discerned by the configuration of pins used. New legislation has narrowed these
five variations down to one, ISO 15765 CAN, for all vehicles sold after 2008 [2].
9.8. Engine Control
Current engines utilize a variety of sensors and inputs to control the engine. When a fuel
injector is involved, the injector opening time controls the amount of fuel injected. The
engine control unit (ECU) contains a map of injector opening time as a function of
measured parameters, typically the intake manifold pressure and the engine speed. The
engine map is a look-up table correlating intake pressure and engine speed with fuel
injector opening time.
An oxygen sensor is used to modify the engine map based on operating conditions. If the
oxygen sensor detects oxygen, a signal is sent to increase injector opening time and inject
more fuel. If the O2 sensor detects a fuel rich condition, fuel injection opening time is
reduced. Almost all engines now contain “smart” ECUs that will adjust the base time in
the engine map permanently if the O2 sensor continually shows excess air or excess fuel
for a certain engine speed - intake pressure condition.
References
[1] Ferguson, C. R. and Kirkpatrick A. T,, Internal Combustion Engines, John Wiley and
Sons, ISBN 0-471-35617-42001.
[2] http://reviews.cnet.com/8301-13746_7-20002489-48.html
128
Chapter 9 Questions
9.1 The exhaust composition of a test engine is as follows:
CO2 = 11.5%
H2O = 7.11 %
N2 = 78.11 %
O2 = 3.19 %
CO = 0.06 %
H2 = 0.03 %
Find the following:
A. Dry concentrations for each gas.
B. The equivalence ratio.
9.2 A diesel engine operated on C14H27 produced exhaust gas of the following dry
composition: Find the equivalence ratio.
CO2 = 6.22%
CO = 0.024 %
O2 = 12.2 %
9.3 Using experimental P-V data calculate the apparent heat release for a diesel engine.
Assume gamma = 1.35. Note that heat release requires a numerical derivative of the
pressure data which is normally too noisy to produce a good result. First calculate the result
without smoothing the pressure date. Next smooth the pressure data and re-calculate the
heat release rate. Smoothing can be done by the using the following formula:
1
(Pi−4 + 3 pi−3 + 5 pi−2 + 7 Pi−1 + 9 Pi + 7 Pi+1 + 5Pi+2 + 3Pi+3 + Pi+4 ) .
41
Plot the heat release from -30 to +60 crank angles as a function of crank angle before and
after smoothing.
Pi =
Print out 10 crank angles of data from zero to 10 deg. atdc including crank angle, pressure,
volume, and Q_dot. Turn in the graph and the table of data.
Manifold Pressure (psia)
9.4 A map is being created to control the fuel flow rate of a port injected IC engine. The
map inputs are manifold absolute pressure (MAP) and engine speed. The map outputs the
orifice opening time in msec. The map is divided into 6 intake pressure ranges and 10 speed
ranges. If one of the points on the map is given below, fill in the rest of the map assuming
that air flow rate reduces by 2% for every 600 rpm of engine speed.
12
10
8
6
4
2
2.14
600
1200
1800
2400 3000 3600
Engine Speed (rpm)
4200
4800
5400
6000
129
130
10. Spark Ignition Combustion and Emissions
In Chapter 3, three modes of combustion were discussed which commonly occur in internal
combustion engines: premixed deflagration wave, mixing limited jet, and volumetric
combustion. This chapter will consider these processes in more detail with some basic
computational models that can be used to provide additional insights into the performance
characteristics of IC engines. Engines are evolving in an attempt to optimize the
combustion process for high fuel conversion efficiency and low emissions. Understanding
the basics of each process should enable an evaluation of new engine concepts as they
evolve.
10.1. SI Combustion Processes
From the early 1900s through 2010 almost all spark ignition engine implemented premixed
deflagration waves as the sole means for burning the fuel. It is useful to understand this
combustion process as it continues to be the most prevalent mode of combustion in
automotive IC engines. The combustion process can be broken down into four parts:
1. Spark Ignition
2. Early Flame Development
3. Flame Propagation
4. Flame Termination
10.1.1.
Spark Ignition
Spark ignition is the event whereby the combustion process is initiated. A high voltage (10
kV) is created between two electrodes separated by a small gap of approximately 1 mm.
The voltage begins to break down the gas and create an ionic flow between the electrodes.
Breakdown is followed by arcing and then glow discharge. Typically, on the order of 30 –
50 mJ are released in approximately 10 ns. Although a spark of 0.5 mJ is all that is typically
required to initiate exothermic reactions under ideal conditions, the higher energy helps to
ensure a more repeatable result and is still only a small fraction of the energy of the fuel in
the cylinder which is typically on the order of 1 kJ or 104 - 106 larger than the spark. The
spark creates a cylindrical plasma that produces the energy necessary to create exothermic
reactions. Once these reactions begin, the spark ignition process is complete, and the spark
has no additional influence on the combustion process.
10.1.2.
Early Flame Development
During early flame development a flame kernel is established surrounding the spark
plasma, the flame kernel expands at the laminar flame speed. The laminar flame speed is
the speed at which a flame propagates in a quiescent environment. The flame kernel itself
may be transported by the bulk motion of the flow. If the flame kernel loses heat more
rapidly than the exothermic reactions produce heat, the flame will go out.
The laminar flame speed is one of the fundamental properties of deflagration waves. The
laminar flame speed is surprisingly slow compared to intuitive expectations. Some typical
values are given in Table 10.1 for fuel air mixtures which are at one atmosphere and a
temperature of 25 oC. The standard state laminar flame speed of iso-octane which has a
similar laminar flame speed to gasoline is 33 cm/s. With a radius of 4.5 cm which would
be common for an automotive engine a flame would take 142 ms to travel from the spark
131
to the wall at the laminar flame speed. At an engine speed of 2500 rpm, this would require
2,135 crank angles to complete the combustion process. Clearly, the flame cannot continue
to propagate at this slow speed in order to produce a successful combustion event.
Understanding the laminar flame speed is however critical to understanding the actual
flame speed.
Table 10.1 Laminar Flame speeds for stoichiometric mixtures of fuel and air for several
fuels.
Fuel
Formula
Methane
Ethylene
Ethane
Propane
Iso-Octane
Hydrogen
CH4
C2H4
C2H4
C3H8
C8H18
H2
Laminar
Flame Speed
(cm/s)
40
136
43
44
33
210
The laminar flame speed has been shown to be the highest at a slightly rich equivalence
ratio of 1.08 – 1.13 and decrease above and below this optimum value. Metkalchi and Keck
[1] generated a correlation for laminar flame speed as shown in Equation 10.1 where SL is
the laminar flame speed, the subscript 0 indicates the reference state at which the reference
flame speed (SL,0) is measured, the subscript u indicates the unburned (reactants) gas
temperature, γ and β are empirical constants, and Ydil represents the mole fraction of
diluents in the mixture.
γ
β
 T   P
S L = S L,0  u    (1 − 2.1Ydil )
 Tu ,0  P0

  
10.1
Metgalichi and Keck [1] suggest γ and β be determined according to Equation 10.1. The
result suggests flame speed is increases significantly by an increase in temperature but is
influenced very little by pressure. Contrary to intuition, increasing pressure actually
produces a slight decrease in flame speed. It can also be notices that for this empirical
equation, flame speed peaks at an equivalence ratio of 1.0. Since an engine compresses the
fuel air mixture before the spark occurs, the temperature at the time of spark is
approximately double that of the standard state. Looking at Equations 10.1 and 10.2, a
doubling of the temperature will produce approximate a four-fold increase in flame speed.
Thus, in the example above, the flame could consume the fuel in 2135/4 = 533 crank angles.
Almost fast enough for the engine to work but still well above the 60 crank angles typical
for an engine. Clearly some other factor must increase the flame speed, and that factor is
turbulence.
132
10.2
γ = 2.18 − 0.8(φ − 1)
β = −0.16 + 0.22(φ − 1)
10.1.3.
Flame Propagation
As the flame expand further, the flame becomes wrinkled and begins to thicken from a
very thin reaction zone of approximately 1 mm thick to a somewhat thicker 2-3 mm
reaction zone. The wrinkling of the flame is caused by turbulence in the fluid motion.
Because turbulence is critical to flame speed and engine combustion, a brief discussion
will be provided.
10.1.3.1. Turbulence
Turbulence is the random motion of fluid structures within a velocity field. The fluid
motion can be characterized by a combination of the bulk average motion and a fluctuating
component. The total velocity is characterized by Equation 10.3 where the total velocity is
the sum the average velocity, u plus the turbulent velocity component, u ′ .
10.3
u = u + u′
One way of visualizing turbulence is to consider a fluid element translating in one direction
with superimposed rotating vortices. Actual turbulence need not be circular in motion, it
may be a back and fourth motion, but the circular motion allows the time and length scales
of turbulence to be visualized. A schematic diagram of a fluid eddy rotating at the edge of
a flame front is shown in Figure 10.1. The turbulent eddy shown is rotating in a clockwise
direction. The diameter of the eddy can be used to characterize the length scale (lt) of the
turbulence while the time to complete one rotation of the fluid around the eddy can be
considered the turbulence time scale (τt). The entire eddy is being transported at the mean
velocity of the flow.
Eddy
lt
lL
Flame Front
Figure 10.1 A schematic diagram of a flame front and a turbulent eddy.
133
A second time scale can be determined for a laminar flame propagating through a mixture.
If, for example the flame is moving from left to right in Figure 10.1 with reactants on the
right and products on the left. The time that it takes a mixture to move from one side of the
flame to the other is determined by the flame thickness “lL” divided by the laminar flame
speed SL, as shown in Equation 10.4. This is the laminar time scale. It represents the amount
of time required to transform the mixture from reactants to products.
τL =
lL
SL
10.4
The Damkohler number is a dimensionless parameter defined as the ratio of the turbulent
to laminar time scales as shown in Equation 10.5. The Damkohler number is very useful in
describing the relative importance of turbulence in a reacting mixture. When Da << 1 the
turbulent time scale is very short in comparison the laminar flow reaction time. This
suggest a turbulent eddy can transport the reacting mixture to a new location before the
reaction is complete. In such a case, the reaction occurring in the flame gets transported
out into the reactants. In the case of a very low Damkohler number the reaction becomes
distributed or does not form a flame at all but a reacting volume. At moderate Damkohler
numbers near 1.0, the flame shape will be distorted or wrinkled by the flame, but the flame
will be discernable. At high Damkohler numbers the flame will be very flat. The reaction
time will dominate (be much shorter) over the time required to mix the reactants and the
flame will be a smooth laminar surface.
Da =
τt
τL
10.5
Because turbulence can have the effect of wrinkling a flame, the laminar surface area (AL)
of the flame is increased to a turbulent area (At) and the reactant mixture is consumed more
rapidly. Turbulence results in an increase of the laminar flame speed to the turbulent flame
speed as shown in Equation 10.6. The turbulent flame speed can be many times the laminar
flame speed. It is the combination of the laminar flame speed increasing because of the
temperature after compression and turbulence that allows a flame to propagate fast enough
through a mixture to produce a successful combustion event in an engine.
ST =
At
SL
AL
10.6
10.1.4.
Flame Termination
As the flame approaches the wall, heat transfers from the reacting mixture to the wall and
reactions slow rapidly. A small amount of mixture located near the walls and in crevices
burns more slowly as they mix from locations too cold to produce reactions out into the
products where they will react. As the piston expands and the reactants cool, there is also
a portion of heat released due to reverse dissociation reactions. The result is a small amount
of heat released at a lower rate than during flame propagation.
134
Fuel Burned (% )
The four processes of spark ignition combustion are shown graphically in Figure 10.2
which shows the typical shape of fuel fraction burned as a function of crank angle. Spark
ignition is a very short event with very little fuel energy released. It takes place shortly
before the fuel burn fraction begins to rise. The slow increase in fuel burn fraction
consuming approximately the first 10% is early flame development when the flame area is
small and is propagating primarily at the laminar flame speed. Flame propagation is the
mostly linear portion of the curve between 10 and 90% fuel burned. The final 10% of fuel
burning is flame termination.
Early
Flame
Dev.
Spark
Ignition
Flame
Propagation
Flame
Termination
Crank Angle
Figure 10.2 Typical cumulative heat release curve for a spark ignition engine. [2]
10.2. Two Zone Model of SI Combustion
The presence of a flame that separates reactants from products within a cylinder leads to
the concept of a two zone model for spark ignition combustion: one zone for the unburned
and one zone for the burned gases. A simple constant specific heats model provides the
basis from which a more complex variable specific heats model could be developed and
provides valuable insights into the formation of pollutants, particularly nitrogen oxide
(NO).
Consider a top view looking down on the piston through the head with the spark in the
center of a circle projected from the piston as shown in Figure 10.3. For simplicity, assume
the combustion chamber is a flat cylinder. A flame propagating outward from the center of
the cylinder produces a circular cylinder enclosing the products inside and reactants outside
of the flame front. Generally, the flame propagates over a period when the piston is near
top dead center and therefore is not moving rapidly and constant volume can be assumed.
The flame front is not a physical boundary but a location where reactions are taking place.
The pressure is equal on each side of the flame. Therefore, the burned pressure Pb, is equal
to the unburned gas pressure (Pu).
135
Rf
Tb
Burned
Gas
Tu
R
Unburned
Gas
Figure 10.3 Top view of a cylindrical combustion chamber with the spark in the center
The temperature of the unburned gas prior to ignition, Tu,ig, is estimated by assuming an
isentropic compression as discussed in Chapter 6 which is repeated here in Equation 10.7.
V 
Tu ,ig = Tu ,0  bdc 
 Vig 


k −1
10.7
As the flame propagates, the burning process across the flame can be considered a quasisteady flow with the burned and unburned gas temperatures related by Equation 10.8. As
an estimate it can be assumed that the volume at ignition is the clearance volume. Because
of the sinusoidal motion of the piston, the volume remains relatively constant during the
crank rotation before and after TDC.
Tb − Tu =
QLHV
q
= mix
mmix C p
Cp
10.8
As the mixture burns, the flame becomes a boundary with unburned reactants on one side
and hot products on the other side. Although the temperature is very different on each side
of the flame, the pressure is not. The flame is like a bubble that expands until the pressure
is equal on each side. The heat release of the fuel and higher temperature on the burned
side produces an increasing pressure on both sides of the flame as fuel is burned.
The unburned mixture is being compressed as the fuel burns and the flame approaches. The
temperature of the unburned mixture can be estimated by assuming an isentropic
compression as shown in Equation 10.9. Where: ku is the ratio of specific heats of the
unburned gas, vu,0 is the specific volume of the unburned gas at the time of the spark, and
vu is the specific volumes of the unburned gas at the time when the flame has reached Rf.
136
Because the mass of the unburned gas is changing along with the volume the specific
volume must be used in the isentropic equation.
k −1
 vu ,0  u

Tu = Tu ,0 
 vu 
10.9
Rearranging the specific volumes gives:
k −1
or
V m  u
Tu = Tu ,0  u ,0 u 
 mu ,0Vu 


k −1
V m  u
= Tu ,0  u ,0 u 
 Vu mu ,0 


 V (m − mb )  u

= Tu ,0  u ,0 u ,0


V
m
,
0
u
u


k −1
 Vu ,0 (1 − xb )  u
 where xb = mb
Tu = Tu ,0 
V
mu ,0
u


10.10
k −1
10.11
This indicates that the unburned gas temperature can be found from an estimate of the
burned mass fraction xb.
The burned mass fraction can be estimated from the flame location using the following
sequence.
xb =
mb
mb
ρbVb
=
=
mu , 0 mb + mu ρbVb + ρuVu
10.12
Now using the fact that Vu = V-Vb gives:
 ρ V + ρ u (V − Vb ) 
xb =  b b

ρ bVb


And by defining yb =
−1
10.13
Vb
gives:
V


ρ  1
− 1
xb = 1 + u 
ρ b  yb


−1
10.14
A simple rule of thumb can be used to estimate the density of the unburned gas to the
burned gas to be 4:1. Using this rule, the burned mass fraction can be estimated.
Example: Fuel burned fraction
137
Video images show a circular flame located 75% of the distance between the center
of the engine and the wall. Estimate the fraction of fuel burned at this time.
From the given information, Rf /R=0.75
For a cylinder of length L and diameter, R, the burned volume divided by the total
volume is:
2
Vb πRf 2 L  .75R 
2
yb =
=
=
 = 0.75 = 0.5625
2
V
πR L  R 
Substituting into 11.12 gives
 4 1

xb = 1 + 
− 1 
 1  0.5625 
−1
= 0.243 = 24.3%
Or the burned fraction in the cylinder is 24.3%. This simple calculation demonstrates
that a large fraction of the fuel is compressed into a small volume before burning.
The flame location can be a misleading visual indicator of how much fuel has been
burned.
Example: Unburned gas temperature
If the unburned gas temperature of a stoichiometric mixture of gasoline and air at
bottom dead center is 330 K, the compression ratio is 9:1 and the flame at TDC is
located at Rf /R = 0.75. Estimate the temperature of the unburned gas and the
temperature of the next element of unburned gas after it is burned. Assume k = 1.30.
The temperature of the unburned gas after compression is:
V 
Tu ,ig = Tu ,0  bdc 
 Vig 


k −1
9
= 330  
1
1.30 −1
= 637.9 K
The values for yb=0.5625 and xb=0.243 are the same as in the previous example.
Using these values, the temperature of the unburned gas can be determined:
 V (1 − xb )  u

Tu = Tu ,0  u ,0
Vu


k −1
 (1 − 0.5625) 

Tu = 637.9
 (1 − 0.243) 
1− k u


Vu

= Tu ,0 
 Vu ,0 (1 − xb ) 


1− k u
 V − Vb 

= Tu ,0 
 V (1 − xb ) 
1− k u
 (1 − yb ) 

= Tu ,0 
 (1 − xb ) 
1−1.30
= 772.9 K
Now as this gas is burned the temperature changes to the burned gas temperature as
described by Equation 10.8
CP can be found using the given value for k =1.30
CP =
kR 1.30(0.287)
=
= 1.243 kJ/kg - K
k −1
1.3 − 1
Tb − Tu =
q mix
2830 kJ/kg
=
= 2275.5 K
CP
1.243 kJ/kg - K
Tb = 772.9 + 2275.5 K = 3048.4 K
138
Note that this is a very high temperature but consistent with the assumption that k
remains constant during the combustion process.
10.3. Temperature Distribution in a SI Engine
In a spark ignition engine, the first fuel to be burned is in the region near the spark plug
while the last fuel burned is in the region near the cylinder walls. This burning process can
produce an uneven temperature distribution in the cylinder that can be somewhat
understood by considering a simple thermodynamic model of the unburned and burned
gases.
Consider an ideal Otto Cycle as shown in Figure 10.4 consisting of sates 1-4 as shown. The
compression process (1-2) is followed by a constant volume heat addition process (2-3).
The piston motion is very small near the top of the stroke thereby allowing time for
combustion at a relatively constant volume. All of the unburned gases are compressed
during the compression process from 1 to 2.
Pressure
2’’
3’’ 3
2
3’
2’
4
1
Specific Volume
Figure 10.4 Pressure vs Specific Volume for an ideal Otto Cycle showing the first element
of fuel burned denoted by ’ and the last element of fuel burned denoted by ’’.
At point 2, consider an infinitesimal mass element burning near the center of the cylinder.
The temperature of this mass increases when it burns but the small amount burned does not
impact the cylinder pressure. Therefore, the temperature at the end of combustion can be
determined by a constant pressure process as shown in Equation 10.15. If this mass does
not mix with the surrounding products it is then compressed to the final pressure in the
cylinder ending at state P3. The temperature at TDC of this element is determined by an
isentropic compression from 2’ to 3’ according to Equation 10.16. The temperature at 3’ is
higher than the average gas temperature in the cylinder (T3) and therefore the specific
volume is higher.
139
T2' = T2 +
qmix
Cp
10.15
k −1
P  k
T3' = T2'  3 
 P2 
10.16
Now consider an infinitesimal element of mixture burned near the wall (the last element
burned). This gas is being compressed from P2 to P3 prior to the arrival of the flame and
the combustion process. The temperature of this element increases due to isentropic
compression. After compression it reacts at constant pressure to produce a burned gas
temperature T3’’ that is lower than the average gas temperature.
Of course the combustion process in an actual engine is not constant volume nor do the
product and reactant elements remain separate without mixing or diffusion but the trends
described in this exercise can be found in functional IC engines. The first elements of fuel
burned reaches the highest temperature and those temperatures tend to be located closer to
the spark plug where flame propagation begins. The temperature of gas burned last is
lower. The temperature in the cylinder shortly after combustion is complete is not uniform.
One of the important results of this temperature profile is that NO emissions increase
exponentially with temperature and therefore the first elements of fuel burned tends to the
locations where NO is produced the most. Because NO formation is non-linear with
temperature, the average temperature (T3) is not the best temperature to use to predict NO
formation.
10.4. Burn Duration and Location
The burn duration and the location of the heat release within a cycle have an impact on the
thermodynamic efficiency and pollutant formation. A study of the Otto cycle with variable
burn duration shows that the shorter the burn, the higher the thermodynamic efficiency.
This is generally true in real engines until the duration reaches approximately 5 crank
angles or shorter. At this duration the pressure rise becomes so rapid that several
undesirable effects occur. First, the instantaneous pressure may exceed the material limits
of the head gasket and produce forces that bend the connecting rod. Second, the high rate
of pressure rise produces undesirable noise. Third, the pressure rise may be non-uniform
within the cylinder which does not allow work to be done as efficiently during expansion
because the expansion process is not quasi steady.
A typical burn duration is on the order of 30 – 90 crank angles with 60 being a good rule
of thumb at a moderate speed and load. The burn duration in an SI engine is controlled by
the time it takes a flame to propagate from the spark to the furthest cylinder wall. A simple
model for burn duration with the spark plug in the center of the cylinder is given by
Equation 10.17, where B is the bore in meters, N is the engine speed in rpm, ST is the
turbulent flame speed in m/s, and 3 is a constant for unit conversions. The equation suggests
that burn duration increases linearly with engine speed, but fortunately this is not the case
or engines would not function over a wide range of speeds as they do. In reality, the
140
turbulent flame speed increases with increasing engine speed such that the duration in crank
angles increases only slightly with increasing engine speed.
θb =
3BN
(cad)
ST
10.17
Figure 10.5 shows 10 and 90% burn durations for a spark ignition engine over a speed
range from 800 to 3000 rpm. The 90% burn duration is seen to change from approximately
38 cad at 1000 rpm to approximately 55 cad at 3000 rpm. These data show that tripling
(300% increase) the engine speed produces a 44% increase in burn duration.
Figure 10.5 Experimental measurements of 10% and 90% burn durations in a SI engine.
[3]
The location of the burn is also of importance. Figure 10.6 shows simulated log P vs log V
data for three spark timings. The figure illustrates the need to center the heat release near
top dead center in order to produce the maximum work for a given amount of fuel burned.
The figure shows how either early or late heat release produces a smaller area on the P-V
diagram and therefore produces less work than the diagram where the fuel burn is centered
about TDC.
ln P
Ln P
ln P
ln V
Optimal
Too Early
Too Late
ln V
ln V
141
Figure 10.6. Ln P vs. Ln V diagrams illustrating the work (area) produced by various spark
timings.
Experiments show the optimal timing for an engine will produce a heat release or fuel
burned curve such that 50% of the fuel is burned 10 cad after top dead center as shown in
Figure 10.7. It would seem intuitive that the optimal efficiency would occur with 50% of
the fuel burned at TDC and therefore the heat release would be exactly centered (one half
before and one half after TDC). A model of the ideal Otto cycle that does not consider heat
transfer does indeed produce this result. However, when heat transfer from the cylinder gas
to the walls is considered it can be seen that there is a greater heat transfer penalty for
burning fuel early compared to burning fuel late. When the fuel is burned early, the early
temperature rise provides more time for heat transfer causing more energy to be lost and
less to be converted to work.
1.0
Xb
0.5
0
10 ATDC
Angular Position
Figure 10.7 For optimal efficiency, work, and torque, the fuel burn curve should be located
as shown in the figure with 50% of the fuel burned at 10 cad ATDC.
The location of the spark controls the burn location and to some degree the burn duration.
The spark location that produces the maximum torque is called Maximum Brake Torque
(MBT) timing. When the timing is changed to be earlier in the cycle it is said that the timing
is “advanced”. When timing is changed to be later in the cycle, the timing is “retarded”.
Because the burn duration changes with engine speed, there is an MBT timing for each
engine speed. When engine speed is increased from idle, the spark ignition timing must be
advanced in order to accommodate a longer burn duration and still produce MBT timing.
10.4.1.
Parameters Influencing Burn Duration
The burn duration is influenced by 1) the distance the flame must travel to complete
combustion 2) the laminar flame speed 3) the area of the flame front (chamber geometry
and turbulence).
142
The distance the flame travels is influenced by the location of the spark plug and the
geometry of the combustion chamber. The optimal spark location for burn duration is
equidistant from all points in the cylinder and thus a spark plug is often located in the center
of an engine. Another method of decreasing burn duration is to provide two spark locations.
An experiment allowing the visualization of a spark ignition flame is shown in Figure 10.8.
In the top row of images a spark is located in the center of the combustion chamber and the
flame can be seen to propagate outward toward the walls in a circular pattern. The flame is
almost to the wall at 20 cad and appears to have burned all of the fuel by 25 cad. The second
row of images shows a combustion process initiated by two spark plugs. In this sequence,
two flames propagate in a circular pattern around each spark. At 20 cad the two flames
have met in the center but regions near the wall on the left and right sides are not yet burned.
A simple geometric analysis shows that the distance between both spark plugs and points
on the left and right walls is greater than the radius of the cylinder and therefore all of the
fuel cannot be burned as quickly even though a large fraction of the fuel can be burned
more rapidly. The experiment leads to the conclusion that the two spark plugs will produce
a more rapid burn initially but that the centered spark will complete the entire combustion
process more rapidly. It also shows that simply drawing circles around the spark plug can
provide insight into combustion rates relative to spark location and engine geometry.
Figure 10.8. Images of propagating flames for the same conditions but different spark
locations. The top row shows a spark initiated in the center of the chamber. The bottom
row shows two sparks located half-way between the center of the cylinder and the wall. [4]
The second way of reducing burn duration is through the laminar flame speed. As has been
discussed earlier, the laminar flame speed increases with increasing temperature and is
maximum at a slightly rich fuel air mixture. Since fuel rich mixtures produce inefficiencies
and pollution, the practical maximum flame speed is stoichiometric and as the mixture is
leaned out, the flame speed drops rapidly. Pressure has little impact on laminar flame speed.
The final parameter influencing burn duration is turbulence. Turbulence has the effect of
wrinkling the flame and producing more surface area for reactions. Turbulence can be
increased by increasing the shearing forces in the cylinder. This is achieved by high inlet
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velocity (high mean piston speed), high swirl, strong squish and tumble. As the piston
increases speed, the air flow into the engine increases speed and thereby creates more
turbulence. Turbulence can be created during compression through swirling flow. Swirl
can be intentionally increased by adding shroud to the intake valve to direct the flow in a
radial motion. Near the end of compression, flow moving over the crown (squish) and into
the bowl (tumble) interacts with the swirling flow to create turbulence.
An experiment showing the flame propagation for three levels of swirl in a cylindrical
combustion chamber is shown in Figure 10.9. The top row shows the flame propagating
with no swirl. In the final frame (equivalent to 20 cad) the flame has propagated to about
the center of the combustion chamber covering less than half of the projected area. As swirl
is increased, the flame front can be seen to move in a clockwise flow and the area covered
by the flame increases. At high swirl, the final frame shows approximately 80% of the area
engulfed in flame. The high swirl turbulent flame is clearly progressing more rapidly than
the quiescent flame.
Figure 10.9A sequence of images showing flame propagation for three levels of swirl in a
cylindrical combustion chamber. [4]
Having reviewed the importance of various parameters on burn duration it is interesting to
consider the design trade-offs for the geometry of an IC engine. For the purposes of burn
duration, it would be optimal to:
1. Place the spark plug in the center of the cylinder on a flat head.
2. Provide a shroud to induce swirl into the intake flow.
3. Provide a piston with a bowl and crown. The motion of fluid from larger radius above
the crown into the bowl (squish) will produce an increase in swirl at the time of ignition. It
will also confine the fuel air mixture to a smaller volume where the flame does not need to
travel as far to complete combustion.
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4. Create a spherical shape to the bowl so that the flame can propagate to the walls more
rapidly.
Clearly not all spark ignition engines have this geometry due to the fact that other
performance parameters such as power, cost, or manufacturability produce a trade-off with
efficiency.
10.5. Abnormal Combustion
In a normal combustion event the four processes outlined above are completed and 99% or
more of the fuel is burned. In a typical well operating automobile greater than 98% of the
combustion events are normal. There are however many possible ways abnormal
combustion can occur. These include: lean misfire, partial combustion, fuel rich
combustion, surface ignition, and knock. Each of these will be briefly discussed with a
more in-depth discussion of knock.
10.5.1.
Lean Misfire
If a spark is provided to a mixture of fuel and air forming a flame kernel but the flame does
not propagate the mixture is said to be below the lean limit of combustion and the mixture
will not burn. The lean limit depends on the fuel air ratio and the temperature but is
typically somewhere near an equivalence ratio of 0.5 for gasoline. Engines typically
experience the lean limit during startup when the intake components are cold and the fuel
does not evaporate rapidly. This is overcome by introducing an extra amount of liquid fuel
during startup so that the evaporated component will be large enough to form an ignitable
mixture. As the engine warms, the fuel injection rate is reduced until it matches
stoichiometric. Carburetors achieve this initial over-fueling with a choke or a with a primer
pump.
10.5.2.
Partial Combustion
In some cases, a flame can propagate through a mixture but it propagates too slowly to
complete combustion before the exhaust valve opens and quenches reactions. Such a case
would be a partial combustion event. Partial combustion can be caused by a lean mixture
at startup, by a low mixture temperature, low turbulence or high engine speed. For most
engines partial combustion is most likely during startup.
Figure 10.10 shows a two-dimensional map of equivalence ratio and spark timing. Regions
of no combustion, partial combustion and normal or complete combustion are shown. The
spark timing is seen to impact the partial burn limit as an advanced timing enables a lower
partial burn limit. Also shown on this figure is the MBT timing. This is seen to advance as
the mixture becomes leaner.
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Figure 10.10 Figure showing lean limit regions where combustion processes can be
abnormal due to an ignition failure or partial combustion. [5]
10.5.3.
Fuel Rich Combustion
As with the lean limit, there is also a rich limit for flame propagation. Above this limit, a
flame will also not propagate. Below the rich limit and above equivalence ratio of 1.0 fuel
is wasted, but the flame will propagate. Fuel rich combustion is common late in the startup
process after the extra liquid fuel injected to produce an ignitable mixture evaporates. Fuel
rich combustion can also be temporarily advantageous when maximum power is desired.
10.5.4.
Surface Ignition
Engine surfaces can collect deposits which protrude into the combustion gas like a fin. As
with fins, the material can be very hot relative to the cooler surface on which they reside.
They can also heat the mixture in the immediate vicinity of deposit. The high temperature
of the deposit and the surrounding mixture can lead to ignition which is not initiated by the
spark. A flame can begin propagating before the spark or a second flame can begin to
propagate after the initial flame created by the spark. Early ignition causes reduced
efficiency and power. An early spark can also lead to knock as will be discussed in the next
section. In carbureted engines, surface ignition can lead to run-on which is a condition
where the engine continues to fire event though the engine ignition switch is turned off and
the spark is no longer being produced.
10.5.5.
Knock
Knock is a condition where the fuel air mixture ignites volumetrically either before or after
the spark. The combustion process proceeds very rapidly (less than 1 crank angle)
throughout the entire volume of the remaining mixture. Knock can be more or less severe
depending on the amount of fuel involved in the reaction. Severe knock creates a rapid
pressure rise and a pressure wave that reflects back and forth across the combustion
chamber. The pressure and heat impinging on a surface can cause severe damage. Light
knock can be tolerated but is undesirable because of reduced efficiency and noise.
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Understanding the process that produces knock can be helpful for developing strategies of
how to reduce or eliminate it. A schematic diagram is given in Figure 10.11 showing the
sequence leading up to knock. A fuel air mixture is compressed causing increased
temperature and density. When heated the fuel changes composition as bonds break within
molecules releasing smaller molecules and in some cases reactive ions or radical species
such as OH- and H+. These radicals are more reactive than the base fuel molecules from
which they came, but their formation does not release heat and the temperature is not
increased rapidly because of their formation. Over time, the radical pool builds until it is
large enough that the reaction rate between the radicals and oxygen is fast enough to
generate a temperature rise. The increased temperature leads to even faster reaction rates,
leading to faster heat release and so forth. The entire process spirals rapidly to consume the
radicals and the fuel. Thus, the buildup of the radical pool can take on the order of
milliseconds (many crank angles) but once the oxidation reactions start the process is
completed in a matter of microseconds (less than one crank angle).
Fuel
Fuel
Fuel
Fuel
Fuel Fuel
Fuel
Fuel
Fuel
Volumetric Chain Reaction
Radicals Form
Fuel
Compression
increases
temperature
Fuel
Fuel
Fuel
Fuel
Fuel
Fuel
Fuel
Fuel
Fuel
Premixed End
Gas Cooks over
a period of time
R+
Fuel
R+ Fuel
R+
Fuel R+
R+Fuel
Fuel
Fuel
Fuel
R+ Fuel
+
Fuel R
More time
Knock !!
Figure 10.11 Block diagram of the sequence of events leading to knock in and IC engine.
Engine knock can be reduced by first selecting a fuel that creates fewer radical or destroys
radicals after they are formed. Second, reduce the temperature in the combustion chamber
which is accomplished by reducing the compression ratio, reducing the intake temperature
(intercooling) or changing the spark timing. And third, reduce the induction time or the
time that the radicals exist before the flame arrives and they are consumed by normal
combustion. Increasing the flame speed is one way of achieving reduced induction time.
Finally, reduce the mixture density. This is difficult to do while maintaining power. Part
throttle or lower altitudes reduce density while turbocharging increases charge density. The
factors influencing knock are in order of importance: fuel composition, temperature,
induction time, and charge density.
Engine knock can be classified into three categories: pre-ignition knock, surface ignition
knock and spark knock. In pre-ignition knock, the fuel air mixture auto ignites prior to the
spark due to the mixture being heated during compression. Typically, an engine is designed
to have a compression ratio that will not allow this to happen and so this form of knock is
not common. Surface ignition knock occurs when a deposit begins flame propagation
creating a deflagration wave prior to ignition. The end gas or gases out in front of the flame
are heated and before the flame arrives they react. Spark knock is the same as surface
ignition knock except the flame propagation begins with a spark. Spark knock can be
147
controlled by changing the spark timing. If the spark timing is retarded, the flame
propagation occurs during expansion and the end gas does not reach as high a temperature.
10.5.6.
Fuel Considerations for Knock
A fuels tendency to produce knock is shown graphically in Figure 10.12 where the critical
compression ratio is plotted as a function of the number of carbon atoms in a fuel molecule.
The correlation between knocking compression ratio and number of carbon atoms is seen
to be strong for paraffin fuels or straight-chain, single-carbon bond hydrocarbons. For these
fuels the larger the molecule the more readily the fuel auto ignites. This is consistent with
the observation that the longer the molecule, the weaker the bonds between atoms within
the molecule and the more readily radicals are formed.
Improved knock resistance is seen to be associated with molecules containing double
bonds, isomers, and aromatics (ring structures). In these fuels the strength between bonds
is increased and radicals are formed more slowly.
In some cases, additives can be used to reduce knock. Additives work by either destroying
radicals, reducing radical formation, or by increasing flame speed. Alcohols are examples
of fuels that reduce knock by increasing flame speed. The laminar flame speed of methanol
is 25% faster than gasoline. Leaded gasoline is an example of a fuel that contains a radical
destroyer. When radicals interact with lead they recombine to form more stable molecules.
Figure 10.12 The critical compression ratio for various fuels. [6]
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10.6. Pollutant Formation for SI Engines
There are three primary pollutants in spark ignition car engines that utilize flame
propagation. They are nitrogen oxides (NOx), unburned hydrocarbons (UHC) and carbon
monoxide (CO). Because these pollutants impact the local air quality they will be
designated as local pollutants. More recently CO2 has also been designated as a pollutant.
CO2 has no local impact on air quality but the global increase in CO2 world-wide has been
attributed to combustion processes. In this section only the local pollutants will be
discussed.
Two forms of nitrogen are produced in significant quantities in IC engines: nitrogen oxide
(NO) and nitrogen dioxide (NO2). Combined the two pollutants are referred to as NOx.
The more abundant of the two is NO being typically 80 – 90 % of the NOx formed, but
NO2 formation is large enough to be of concern. The primary formation route for NOx in
SI engines is oxygen and nitrogen reacting at high temperature. Temperatures must be over
1600 K for these reactions to become significant, but this is typically the case for the
products of a stoichiometric flame. The strategy for reducing NOx emissions is to lower
the flame temperature by adding diluents to the combustion process.
Unburned hydrocarbons result from unreacted fuel. The primary source of UHC is fuel
trapped in crevices available to the combustion chamber such as the top land crevice above
the first piston ring, a crevice in an intake or exhaust valve. As the flame propagates toward
the wall, reactions are quenched by the colder wall temperature and the flame can no longer
consume the fuel in the crevice. Although these volumes are small, the mixture in the
volume has been compressed due to combustion in the cylinder and the temperature is
lower than the product temperature, therefore the mass in the crevice is much greater than
existed at the start of the compression process. Some of the UHC will move out of crevice
volumes after the flame has passed and will burn but the remainder exits with the exhaust
as UHC.
When fuel is burned the process involves many elementary steps or elementary chemical
reactions. During this process carbon monoxide is formed as an intermediate step. If the
combustion process is not completed due to quenching of a flame or the mixing of a
partially burned element with a cold element that stops reactions, then CO remains as an
intermediate or a sign of partial combustion. A flame quenching at the wall or unburned
hydrocarbons that exit crevice volumes and do not completely burn are examples of how
CO survives the combustion process and exits the engine in the exhaust.
10.6.1.
Emission Standards
Emissions are measured as mole fractions and reported as a percentage or in ppm of the
total gas. A gas concentration of 1% is equivalent to 10,000 ppm. It can be more useful to
report or compare emissions on a mass specific or energy specific basis. If for example two
engines produce 200 ppm of NOx but one engine produces 100 hp while the other produces
200 hp, then the latter engine could be considered better relative to NOx formation.
Emission standards are typically defined on an energy specific basis.
149
References
1. Metgalichi, M, and Keck, J., “Burning Velocities of Mixtures of Air with Methanol,
Isooctane and Indolene at High Pressure and Temperature,” Combustion and Flame, Vol.
48, No. 2, pp 191120, 1982.
2. Heywood, J. B., Internal Combustion Engine Fundamental, McGraw-Hill, 1988.
3. Hires, S.D., Tabaczynski, R. J. and Novak, J. M., The Prediction of Ignition Delay and
Combustion Intervals for a Homogenous Charge, Spark Ignition Engine, SAE Paper
780232, SAE Trans., Vol. 87, 1978. Cited by Heywood, J. B., Internal Combustion Engine
Fundamental, McGraw-Hill, 1988.
4. Witze, P. O., The Effect of Spark Location on Combustion in a Variable Swirl Engine,
SAE Paper 820044, SAE Trans. Vol. 91, 1982. Cited by Heywood, J. B., Internal
Combustion Engine Fundamental, McGraw-Hill, 1988.
5. Peters, B. D., Mass Burning Rates in a Spark Ignition Engine Operating in the PartialBurn Regime, paper C92/79, Conference Proceedings on Fuel Economy and Emissions of
Lean Burn Engines, Institute of Mechanical Engineers, London, June 12-14, 1979. Cited
by Heywood, J. B., Internal Combustion Engine Fundamental, McGraw-Hill, 1988.
6. Lovell, W. G., “Knocking Characteristics of Hydrocarbons, Ind. Engng. Chem., Vol. 40,
pp 2388-2438, 1948. Modified and Cited by Heywood, J. B., Internal Combustion Engine
Fundamental, McGraw-Hill, 1988.
150
Chapter 10 Homework Problems
10.1 If the temperature and pressure at BDC for a stoichiometric mixture of gasoline
(assume k = 1.35) are 300 K, 50 kPa, and the compression ratio is 8, find the unburned gas
temperature when the piston reaches -30 crank angles. Use; bore = 0.0875 m, Stroke =
0.083 m, ConRod = 0.1666 m. Note that you can do this by hand or use a spreadsheet to
calculate the volume change.
10.2 For the same engine as described in 11.1 it has been found that:
2.19
 Tu 
sl = 0.2525 P

 (1 − 2.1 f )
 298 
Where f is the residual mass, Tu is the unburned gas temperature and P is the pressure.
−0.13
And that ignition delay is inversely proportional to laminar flame speed: ∆θid = C/Sl
If the ignition delay is ∆θid =25 crank angles when the spark fires, θs at -25 crank angles
then find the ignition delay when the spark fires at different crank angles. Plot the ignition
delay vs spark timing from -50 to 0 cad.
In order to find the values needed to calculate the laminar flame speed, assume P = 50 kPa
at BDC, Tu = 300 K at BDC. In order to simplify, the residual mass, f, can be found by a
correlation f = 0.10(8/rc) where: rc is the compression ratio
10.3 An engine is assumed to compress intake air at 320 K, P = 0.8 atm to TDC with a
compression ratio of 12:1. A flame is then assumed to propagate at constant volume until
the flame is 90% of the way from the spark plug in the center of the cylinder to the cylinder
wall. Assume k=1.35. Estimate:
A. The temperature of the unburned gas.
B. The fraction of fuel that has burned.
C. The temperature of the most recently burned gas.
10.4 Consider an engine which compresses a stoichiometric gasoline fuel air mixture (k =
1.35) beginning at 300 K, P=100 kPa over a compression ratio of 8:1 and then the mixture
burns at constant volume.
A. Find the temperature of the first mixture element burned right after it has burned.
B. Find the temperature of the first mixture element burned at the end of the constant
volume combustion process.
C. Find the temperature of the last element of fuel after it has burned.
10.5 A 2.0 L, spark ignition, gasoline engine with a brake fuel efficiency of 0.32 and a
volumetric efficiency of ev,0 = 0.98 produces 187 ppm of NO as measured by an
chemiluminescence (wet concentrations) analyzer. Assume gasoline has a heating value of
44,000 kJ/kg, (A/F)s = 14.6, P0 = 101 kPa, T0 = 298 K.
151
(a) Emissions index – grams of nitric oxides per kilogram of fuel.
(b) Mass fraction – grams of nitric oxide per kilogram of exhaust.
(c) Energy specific – grams of nitric oxide per kW-hr of brake work
10.6 Explain the most important physical parameter influencing: NO, unburned
hydrocarbon (UHC), and CO formation in a spark ignition engine.
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11. Compression Ignition Combustion and Emissions
Compression ignition combustion is created by injecting an easily ignited fuel (usually
diesel fuel) into a compressed oxidizer (usually air). The oxidizer is typically compressed
in a piston-cylinder, slider-crank mechanism creating a high temperature (800 – 950 K)
and pressure (3.5 – 16.2 MPa). The fuel is injected at very high pressure (50 – 170 MPa)
through 5 – 8 small holes (100 – 200 µm in diameter) in the injector where it exits at a high
velocity. This liquid jet is atomized, evaporates, mixes with the air, and reacts.
11.1. Conceptual Model of Compression Ignition Combustion
Traditional compression ignition combustion involves the following processes: liquid fuel
injection, hot oxidizer entrainment, fuel evaporation and heating, a premixed or volumetric
reaction, and a mixing limited reaction. A conception model of diesel combustion was
created by Dec (1997) after collecting optical diagnostics of combustion events in an
optically accessible diesel engine. His conceptual model is shown in Figure 11.1 for a
single spray after the process has been fully developed and before the injection process
ends.
Figure 11.1 Conceptual model of compression ignition combustion.
The model shows liquid fuel (dark brown) exiting the injector and expanding in volume as
it moves from left to right. Because the jet is at a very high velocity relative the oxidizer,
the fluid shears at the boundary forming droplets and entrains hot gas into the liquid jet
causing an expansion of the liquid spray. The solid liquid core exiting the injector becomes
a mixture of hot entrained gas (usually air) and fine liquid drops of fuel. Simultaneously
liquid is being heated and transitions from liquid to vapor. The distance from the injector
to the end of the dark brown region where the liquid has all evaporated into a vapor is
called the liquid length. The amount of oxidizer entrained into the mixture at this point is
typically too small to burn all of the fuel, the mixture is overall fuel rich.
153
As this fuel rich jet of products proceeds from left to right (light brown region), the fuel
heats further until the temperature is high enough and a reaction occurs represented by the
small light blue region call the premixed flame. In this region the reaction releasing heat
and rapidly increases the temperature of the mixture creating fuel rich products. The high
temperature produced in the absence of oxygen causes unburned fuel to form soot. The fuel
rich products and soot continue to move from left to right, continue to form soot, and
continue to entrain oxidizer on their outer border.
All along the outer border of these hot products, oxidizer is available and reacts with the
hot products and soot to complete the combustion process. This reaction is called a
diffusion flame but is more properly described as a mixing limited flame because the rate
of reaction is controlled by the rate of mixing of oxidizer into the jet of fuel rich products.
The mode of transport is dominated by mixing rather than diffusion. The vast majority of
the soot will burn out as it passes through the diffusion flame, but some soot may remain
if the diffusion flame becomes stretched and is quenched allowing the rich mixture to exit
the encompassed volume without reacting. The products of the diffusion flame are at a very
high temperature, close to the adiabatic flame temperature. This high temperature creates
a region on the oxidizer side of the flame where oxygen and nitrogen are available at high
temperature and NO is formed. The green boundary on the outer edge of the diffusion flame
is shown as the location of NO formation.
Compression ignition engines introduce fuel directly into to the cylinder following the
compression process. This results in a challenging design requirement that the injector
delivers fuel into a significantly higher pressure in a much shorter period of time. The fuel
then burns in two different modes: volumetric and mixing limited combustion. The
resulting reactions produce a unique set of emission challenges and advantages. This
chapter will first present the types of injectors typically implemented in compression
ignition engines. It will then detail how the combustion process proceeds followed by the
resulting energy release and emissions produced.
11.2. Stages of Compression Ignition Combustion
The conceptual model of compression ignition combustion captures a portion of the
combustion process where the fuel jet and mixing limited flame have reached a fully
developed quasi steady-state. The image of the flame in this state can be used to imagine
the entire process including the time prior to and after the process captured in the model.
In order to facilitate an understanding of the complete process, consider the Apparent Heat
Release Rate (AHRR) diagram as shown in Figure 11.2. The AHRR is the rate at which
energy is being converted from chemical to thermal energy or the rate at which the fuel is
being burned. The five processes of combustion can be identified by changes in the AHRR
shown in the figure. The five processes are: ignition delay, premixed combustion, jet
growth, quasi-steady mixing limited, and burnout.
When the liquid fuel jet is first injected, the fuel entrains the hot gas in the combustion
chamber and evaporates. This has the effect of lowering the temperature and pressure and
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creates the appearance of heat being lost or a negative heat release rate. This can be seen
in the drop in AHRR between 347.5 and 352 CAD. Fuel is present but is not yet burning
causing this to be referred to as the ignition delay.
At some point, the evaporated fuel that is premixed with gas in the cylinder is hot enough
to burn volumetrically. This rapid rise and drop in AHRR is seen between 352 and 358
CAD indicating the premixed burn or a volumetric combustion event. At this point, the
liquid jet feeds vaporized fuel into the center of a jet surrounded by flame and the jet grows
in length until the rate of fuel burning around the jet becomes large enough to equal the
rate of fuel being injected into the jet and the burn rate reaches a brief steady-state. In the
data shown, this is a very brief period between 360- 361 CAD. At this point, each jet
exiting the fuel injector produces a flame resembling the schematic in Figure 11.1. As the
injection ends and there is no longer a source of fuel for the jet, the flame surrounds the
unburned fuel which then burns out. This period is represented by the period between 361
– 377 CAD.
Figure 11.2 Measured (black) and modeled (grey) Apparent Heat Release Rate and
cumulative total heat release from a compression ignition engine.
11.2.1. Compression Ignition Emissions
The dominant emissions problems associated with compression ignition combustion are
NOx and particulate. They are a natural result of the mixing limited combustion process
but can be controlled or reduced in quantity by understanding how they are formed and
designing the combustion process to limit their formation and enhance their destruction.
Particulate consists of soot particles that have absorbed unburned hydrocarbons. Soot is
formed when unburned hydrocarbon fuel is heated in a fuel rich environment. This
environment is created in the dark blue region shown in Figure 11.1 where a flame
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surrounds unburned evaporated fuel. Fossil fuel is a mixture of various hydrocarbon
molecules. The hydrocarbon molecules pyrolyzes at high temperature causing the lighter
hydrocarbon molecules to be released and sperate from more stable carbon dominant
molecules. The hydrogen tends to react first with any available oxygen leaving the carbon
molecules to coalesce and form solid soot particles. The soot particles agglomerated and
absorb remaining unburned hydrocarbons. After being formed, the soot passes through the
flame along with the surrounding unburned fuel and unless the flame is interfered with in
some way, the soot will burn out. The orange yellow glow of a candle is an example of
soot burning as it passes through a flame.
Unfortunately, soot can survive the combustion process if the flame is prematurely
quenched. This can occur when the flame interacts with a wall or of the fluid motion
stretches the flame causing to go out. The flame can also be quenched if the expansion
process cools the reactants too quickly, and the exhaust valve opens before the fuel can be
completely burned. The amount of soot exiting an engine can be thought of the difference
between the soot formed and the soot oxidized. The worst case for soot emissions is when
the amount of oxidizer is insufficient to burn all of the fuel injected.
Soot is controlled by either reducing soot formation or increasing soot oxidation. Soot
formation can be reduced by increasing the amount of oxidizer mixed into the fuel jet
before ignition or during entrainment. This creates a leaner mixture when the fuel is burned.
Soot oxidation is improved by increasing the time and temperature for burnout and by
avoiding impingement of fuel on surfaces that quench reactions or collect soot deposits
before they pass through the flame. Advancing the injection timing to earlier in the process
is an example of a design parameter that can decrease soot formation and increase
oxidation. When fuel is injected earlier in the compression process, the air in the cylinder
is colder allowing more time and more fuel to evaporate before ignition occurs. This leads
to a larger premixed burn where the mixture is leaner and less soot is formed. The
combustion process then ends earlier when expansion is less complete producing a higher
temperature and more time for the fuel to burn out. Advancing the timing is limited by the
amount of fuel that can be burned as a premixed fuel because this is a volumetric
combustion event that occurs rapidly and produces a large pressure spike. It is the same
combustion process as knock in a spark ignition engine and can damage engine
components. It will also be discussed below that advancing the timing increases NO
formation.
Another method for reducing soot formation is to reduce the jet diameter. This results in
increasing the fraction of air entrained into the fuel jet and produces a lower equivalence
ratio in the rich combustion products where soot is formed. Fuels can also impact soot
formation by either containing oxygen within the fuel itself or by resisting ignition leading
to more air entrainment and a lower equivalence ratio for the evaporated mixture where
soot is formed. Some fuels also contain molecules that are less stable and more likely to
burn instead of surviving and forming soot.
NOx is term used to describe the total concentration of NO and NO2. In compression
ignition engines, the primary component of NOX is NO. As mentioned above, NO is formed
156
on the oxidizer side of a flame where very high temperatures (1400 C and higher) cause
nitrogen (N2) and oxygen (O2) to react and produce NO. NO produced by this mechanism
is called thermal NO and it is very difficult to eliminate when a flame (deflagration or
mixing limited) is present. Note that the definition of a flame is a thin region (normally on
the order of 1 mm thick) that separates reactants from products. Products leaving a flame
are at or near the adiabatic flame temperature. Because oxygen and nitrogen are present in
air, the three ingredients necessary for NO formation are present (O2, N2 and high
temperature) where the hot products meet the reactants and NO is formed.
NO is controlled by reducing the temperature of the flame so that the reaction rate rates
converting O2 and N2 to NO are reduced. This can be done by increasing the amount of
non-reacting molecules in the oxidizer which then dilute the mixture and decrease the
adiabatic flame temperature. The atmospheric concentration of O2 in air is typically 21%.
By adding exhaust gas to the air, the concentration of O2 can be significantly reduced.
Exhaust gas contains primarily CO2 and H2O. These molecules dilute the mixture and
decrease the adiabatic flame temperature. This dilution is achieved using exhaust gas
recirculation (EGR). It is more effective when the exhaust gas is cooled using an
intercooler.
A second common method for reducing NO is to retard the injection timing. This design
technique allows and increased fraction of the fuel to be injected during the expansion
process when the cylinder gases are being cooled from expansion. The result is a lower
average flame temperature and lower formation rates of NO. Unfortunately, this technique
decreases the thermodynamic efficiency of the engine and has been discussed above
increases particulate emissions by reducing soot oxidation.
11.2.2. NOx Particulate Trad-off Curve
As discussed in Section 11.2.1 particulate emission can be reduced by advancing engine
injection timing and NO emissions can be reduced by retarding the injection timing. The
result is what is referred to as the NOx – Particulate trade-off curve which is shown in
Figure 11.3. The two curves on the graph represent measured NOx and particulate value
for a given engine when the timing is changed. As the timing is advanced, particulate
decreases and NOx increases. Improvements in emissions occur only when this curve is
moved closer to the origin. The second curve that is closer to the origin represents
emissions typical of the best 2005 engines. The boxes drawn on the graph represent the
federal emissions standards for on-highway diesel engines that took place in the year
indicated. Diesel engines have been redesigned and can now meet the 2010 standards
which are approximately 1% of the emissions of diesel engines in the early 1980s. The
emissions improvements were obtained by:
1) Higher pressure fuel injection which increased fuel air mixing and completed
combustion earlier to enable soot oxidation
2) Exhaust gas recirculation (EGR) which reduced flame temperature and decreased
NOx formation.
157
3) Multiple fuel injections per cycle. Multiple injections allowed multiple premixed
burns which are volumetric combustion events that do not create very little soot or
NOx
Figure 11.3. Example of the NOx – Particulate trade-off curve for a diesel engine and the
changing emissions standards over time.
The introduction of multiple fuel injections and the increased variation in combustion from
the conceptual model introduced by Dec (1997) have increased the formation of CO and
unburned hydrocarbons in modern compression ignition engines. These emissions are
becoming increasingly important in engines that utilize what amounts to a hybrid
combustion approach where volumetric and mixing limited combustion dominate at certain
speeds and loads.
References
1. Dec, J. E., “A Conceptual Model of DI Diesel Combustion based on Laser-Sheet
Imaging,” SAE Transactions, Vol. 106, Sec. 3, pp. 1319-1348, paper no. 970873, 1997.
158
Chapter 11 Homework Problems
11.1 What are the five processes of combustion for a typical compression ignition engine?
11.2 What type of combustion discussed in Chapter 6.1 occurs during the premixed burn
process?
11.3 What are the two most dominant emissions from a compression ignition engine?
11.4 What property is most important in creating and controlling NO formation?
11.5 Where are soot particles formed in a compression ignition engine?
11.6 Where are NO molecules formed in a compression ignition engine?
159
160
Appendix A Thermodynamic Data
Table A.1 Ideal Gas Properties for Air
T
(K)
240
260
280
300
320
340
360
380
400
420
440
460
480
500
520
540
560
580
600
620
640
660
680
700
720
740
760
780
800
820
840
860
880
900
h
(kJ/kg)
240.04
259.97
279.96
300.00
320.11
340.28
360.51
380.80
401.16
421.59
442.08
462.65
483.28
503.99
524.77
545.63
566.56
587.56
608.65
629.81
651.04
672.36
693.76
715.23
736.79
758.43
780.14
801.94
823.82
845.78
867.82
889.94
912.15
934.43
u
(kJ/kg)
171.16
185.35
199.60
213.91
228.27
242.70
257.19
271.74
286.36
301.05
315.80
330.63
345.53
360.49
375.54
390.65
405.84
421.11
436.45
451.87
467.37
482.95
498.60
514.34
530.15
546.05
562.03
578.09
594.23
610.45
626.75
643.13
659.59
676.14
s0
(kJ/kg-K)
1.47890
1.55872
1.63283
1.70203
1.76695
1.82812
1.88598
1.94088
1.99313
2.04299
2.09070
2.13644
2.18038
2.22267
2.26345
2.30284
2.34092
2.37780
2.41356
2.44828
2.48202
2.51484
2.54680
2.57794
2.60833
2.63799
2.66696
2.69529
2.72301
2.75014
2.77671
2.80276
2.82830
2.85336
T
(K)
920
940
960
980
1000
1050
1100
1150
1200
1250
1300
1350
1400
1450
1500
1550
1600
1650
1700
1850
1900
1950
2000
2050
2100
2150
2200
2250
2300
2350
2400
2450
2500
h
(kJ/kg)
240.04
259.97
279.96
300.00
320.11
340.28
360.51
380.80
401.16
421.59
442.08
462.65
483.28
503.99
524.77
545.63
566.56
587.56
608.65
629.81
651.04
672.36
693.76
715.23
736.79
758.43
780.14
801.94
823.82
845.78
867.82
889.94
912.15
U
(kJ/kg)
171.16
185.35
199.60
213.91
228.27
242.70
257.19
271.74
286.36
301.05
315.80
330.63
345.53
360.49
375.54
390.65
405.84
421.11
436.45
451.87
467.37
482.95
498.60
514.34
530.15
546.05
562.03
578.09
594.23
610.45
626.75
643.13
659.59
s0
(kJ/kg-K)
1.47890
1.55872
1.63283
1.70203
1.76695
1.82812
1.88598
1.94088
1.99313
2.04299
2.09070
2.13644
2.18038
2.22267
2.26345
2.30284
2.34092
2.37780
2.41356
2.44828
2.48202
2.51484
2.54680
2.57794
2.60833
2.63799
2.66696
2.69529
2.72301
2.75014
2.77671
2.80276
2.82830
Cp vs T data from B.G. Kyle, Chemical Process Thermodynamics (Englewood Cliffs, NJ Prentice Hall, 1984)
161
Table A.2 Ideal Gas Specific Heats
T (K)
250
300
350
400
450
500
550
600
650
700
750
800
900
1000
Air R = 0.2870 kJ/kg-K
CP
Cv
k
(kJ/kg-K)
(kJ/kg-K)
1.003
0.716
1.401
1.005
0.718
1.400
1.008
0.721
1.398
1.013
0.726
1.395
1.020
0.733
1.391
1.029
0.742
1.387
1.040
0.753
1.381
1.051
0.764
1.376
1.063
0.776
1.370
1.075
0.788
1.364
1.087
0.800
1.359
1.099
0.812
1.354
1.121
0.834
1.344
1.142
0.855
1.336
Hydrogen (H2) R= 4.124 kJ/kg-K
CP
Cv
k
(kJ/kg-K)
(kJ/kg-K)
14.051
9.927
1.416
14.307
10.183
1.405
14.427
10.302
1.400
14.476
10.352
1.398
14.501
10.377
1.398
14.513
10.389
1.397
14.530
10.405
1.396
14.546
10.422
1.396
14.571
10.447
1.395
14.604
10.480
1.394
14.645
10.521
1.392
14.695
10.570
1.390
14.822
10.698
1.385
14.983
10.859
1.380
Nitrogen (N2) R = 0.2968 kJ/kg-K
CP
Cv
k
(kJ/kg-K)
(kJ/kg-K)
1.039
0.742
1.400
1.039
0.743
1.400
1.041
0.744
1.399
1.044
0.747
1.397
1.049
0.752
1.395
1.056
0.759
1.391
1.065
0.768
1.387
1.075
0.778
1.382
1.086
0.789
1.376
1.098
0.801
1.371
1.110
0.813
1.365
1.121
0.825
1.360
1.145
0.849
1.349
1.167
0.870
1.341
Carbon Dioxide (CO2) R = 0.1889 kJ/kgK
Carbon Monoxide (CO) R = 0.2968 kJ/kg-K
Oxygen (O2) R = 0.2598 kJ/kg-K
CP
Cv
k
CP
Cv
k
CP
Cv
k
(kJ/kg-K)
(kJ/kg-K)
(kJ/kg-K)
(kJ/kg-K)
(kJ/kg-K)
(kJ/kg-K)
250
0.791
0.602
1.314
1.039
0.743
1.400
0.913
0.653
1.398
300
0.846
0.657
1.288
1.040
0.744
1.399
0.918
0.658
1.395
350
0.895
0.706
1.268
1.043
0.746
1.398
0.929
0.668
1.389
400
0.939
0.750
1.252
1.047
0.751
1.395
0.941
0.681
1.382
450
0.978
0.790
1.239
1.054
0.757
1.392
0.956
0.696
1.373
500
1.014
0.825
1.229
1.063
0.767
1.387
0.972
0.712
1.365
550
1.046
0.857
1.220
1.075
0.778
1.382
0.988
0.728
1.358
600
1.075
0.886
1.213
1.087
0.790
1.376
1.003
0.743
1.350
650
1.102
0.913
1.207
1.100
0.803
1.370
1.017
0.758
1.343
700
1.126
0.937
1.202
1.113
0.816
1.364
1.031
0.771
1.337
750
1.148
0.959
1.197
1.126
0.829
1.358
1.043
0.783
1.332
800
1.169
0.980
1.193
1.139
0.842
1.353
1.054
0.794
1.327
900
1.204
1.015
1.186
1.163
0.866
1.343
1.074
0.814
1.319
1000
1.234
1.045
1.181
1.185
0.888
1.335
1.090
0.830
1.313
th
Source Kenneth Wark. Thermodynamics 4 ed. New York: McGraw Hill, 1983. Originally published in Tables of Thermal Properties
of Gases, NBS Circular 564, 1955.
162
Table A.3 Enthalpy of formation for combustion related gas species at a reference state of
298.15K, 1 atm.
Species
Formula
State
h f0 (kJ/kmol)
g 0f (kJ/kmol)
C
O2
CO
Molar Mass
(kg/kmol)
12.01
32.00
28.01
Carbon
Oxygen
Carbon
Monoxide
Carbon
Dioxide
Nitrogen
Hydrogen
Water
(liquid)
Water
(gas)
Methane
Ethane
Propane
n-Butane
n-Octane
IsoOctane
IsoOctane
Cetane
Methanol
Ethanol
Solid
Gas
Gas
0
0
-110,530
0
0
-137,150
5.74
205.04
197.66
CO2
44.01
Gas
-393,520
-395,360
213.78
N2
H2
H2O(l)
28.01
2.02
18.02
Gas
Gas
Liquid
0
0
-285,830
0
0
-237,180
191.61
130.68
69.92
H2O(g)
18.02
Gas
-241,820
-228,590
188.83
CH4
C2H6
C3H8
C4H10
C8H18
C8H18
C8H18
C16H34
CH3OH
C2H5OH
16.03
30.07
44.10
58.12
114.23
114.23
114.23
226.44
32.04
46.07
Gas
Gas
Gas
Gas
Liquid
Gas
Liquid
Liquid
Liquid
Liquid
-74,850
-84,680
20,410
-126,150
-249,950
-224,094
-259,280
-454,500
-238,660
-277,690
-50,790
-32,890
62,720
-15,710
6.610
186.16
229.49
266.94
310.12
360.79
-166,360
-174,890
126.80
160.70
NIST Chemistry Web Book, http://webbook.nist.gov/chemistry/
s0 (kJ/kmol)
163
Table A.4 Sensible enthalpy for several combustion related ideal gases.
h(T) - h(T0) , kJ/kmol T0 = 298.15 K
T(K)
298
C
0.00
CO
0.00
CO2
0.00
H2
0.00
H2O
0.00
N2
0.00
O2
0.00
300
400
500
16.74
1,046.5
2,381.8
54.42
2,976.2
5,931.6
66.98
4,010.2
8,317.6
54.42
2,959.5
5,885.5
62.79
3,453.5
6,923.6
54.42
2,972.1
5,914.8
54.42
3,030.7
6,090.6
600
700
800
900
1000
3,964.1
5,743.2
7,664.6
9,703.1
11,821.3
8,945.5
12,026.4
15,182.6
18,405.8
21,696.0
12,922.2
17,769.6
22,826.3
28,054.6
33,421.0
8,815.7
11,754.3
14,295.2
17,690.0
20,695.6
10,502.7
14,190.5
17,999.8
21,934.6
25,990.9
8,895.3
11,942.7
15,052.9
18,230.0
21,470.0
9,251.1
12,507.8
15,848.2
19,255.6
22,717.4
1100
1200
1300
1400
1500
14,010.5
16,254.2
18,552.4
20,879.8
23,240.7
25,044.8
28,439.7
31,880.6
35,355.0
38,867.0
38,913.1
44,505.6
50,181.8
55,933.3
61,743.5
23,734.6
26,807.1
29,921.5
33,077.8
36,284.2
30,181.1
34,492.6
38,921.4
43,467.4
48,118.1
24,768.6
28,121.5
31,516.4
34,953.1
38,423.3
26,229.5
29,779.2
33,366.6
36,983.3
40,629.3
1600
1700
1800
1900
2000
25,626.7
28,029.5
30,453.2
32,889.4
35,338.2
42,404.2
45,962.3
49,545.5
53,149.6
56,766.3
67,612.3
73,527.1
79,479.6
85,469.7
91,493.4
39,541.0
42,835.3
46,171.6
49,545.5
52,957.1
52,869.2
57,712.4
62,639.3
67,645.8
72,723.4
41,922.8
45,451.6
49,005.5
52,576.2
56,167.7
44,300.4
47,992.5
51,713.8
55,460.3
59,227.7
2100
2200
2300
2400
2500
37,795.4
40,269.3
42,747.4
45,238.1
47,733.0
60,404.0
64,050.0
67,708.6
71,379.7
75,059.2
97,546.4
103,624.4
109,723.4
115,843.4
121,984.2
56,406.4
59,889.1
63,401.2
66,946.7
70,525.7
77,868.0
83,075.4
88,337.2
93,649.2
99,011.5
59,776.1
63,401.2
67,038.8
70,684.8
74,347.5
63,016.0
66,959.3
70,668.1
74,527.5
78,412.2
2600
2700
2800
2900
3000
50,240.4
52,752.0
55,267.8
57,796.1
60,328.6
78,751.2
82,447.5
86,156.3
89,869.2
93,586.4
128,146.0
134,320.4
140,511.5
146,715.1
152,935.5
74,125.7
77,755.0
81,409.3
85,084.6
88,785.1
104,419.8
109,865.8
115,349.4
120,870.8
126,421.4
78,018.7
81,698.2
85,386.0
89,078.1
92,782.7
82,313.5
86,240.0
90,187.4
94,155.7
98,145.0
Generated from Cp(T) data. Dale Tree
164
Table A.5 Properties of selected fuels
Fuel
Formula
Molar
Mass
Specific
Gravity
(g/cm3)
Lower
Heating
Value
(MJ/kg)
Lower
Heating
Value
Stioch.
Mixture
(MJ/kgmix)
Stoichiometric
Air Fuel Ratio
(A/F)s
Heat of
Vaporization
(kJ/kg)
Specific Heat
(CP)
Fuel Octane
Rating
Liquid
(kJ/kg)
Vapor
(kJ/kg)
RON
MON
Methane
CH4(g)
16.04
0.00072
50.0
2.72
17.23
509
0.63
2.2
120
120
IsoOctane
C8H18
114.23
0.692
44.3
2.75
15.13
308
2.1
1.63
100
100
Gasoline
CnH1.87n
(l)
~110
~ 0.75
44.0
2.83
14.6
350
2.4
1.7
9199
82-89
Diesel
CnH1.8n (l) ~170
~0.81
43.2
2.79
14.5
270
2.2
1.7
----
----
Methanol
CH3OH (l) 32.04
0.792
19.9
2.68
6.47
1103
2.6
1.72
106
92
Ethanol
C2H5OH (l) 46.07
0.785
26.9
2.69
9.00
840
2.5
1.93
107
89
Hydrogen
H2
0.000090
120.0
3.4
34.3
----
-----
1.44
----
----
2.01
Heywood, J.B. McGraw Hill, 1988.
165