engine power for agricultural machines

288 Pages·2015·14.36 MB·English

Selections from Ajit K. Srivastava, Carroll E. Goering, Roger P. Rohrbach, and Dennis R. Buckmaster. 2006.
Engineering Principles of Agricultural Machines, 2nd ed. St. Joseph, Michigan: ASABE, Copyright American Society of
Agricultural and Biological Engineers. (Corrected reprint 2012).

ENGINE POWER
FOR
AGRICULTURAL

MACHINES
INTRODUCTION
The earliest farm equipment made use of human power and, for a period in the 19th
and 20th centuries, animals supplied the power needs of farm equipment. Modern
agricultural equipment, however, is powered by internal combustion (IC) engines and,
since the 1970s, nearly all new agricultural engines have been compression ignition
(CI) engines that burn diesel fuel. The engine can be a part of the machine itself, as on
a self-propelled combine, or can provide the power for an agricultural tractor.
Engines consume fuel to produce power. The power is delivered to some load
through the crankshaft and flywheel of the engine. Much of the energy in the fuel is
lost before it is converted to useful power. The purpose of this chapter is to clarify the
processes by which an IC engine produces power and to provide insights into how
engines may be made to operate efficiently. By reading this chapter, you will also
learn the important terminology of diesel engines.
2.1 THE POWER IN FUEL
Liquid fuels are a highly concentrated form of chemical energy storage. Burning
the fuel at even a modest rate releases a large amount of energy that can be calculated
using Equation 2.1:
Hgm&f
P = (2.1)
fe 3600
where P = fuel equivalent power, kW
fe
H = gross heating value of the fuel, kJ/kg
g
m&f = fuel consumption rate, kg/h
The heating values are measured by burning a sample of fuel in a calorimeter. The
heating values are defined as gross (H ) or net (H ) depending on whether the water
g n
created in combustion is recovered as liquid or vapor, respectively. The terms higher
16 CHAPTER 2 ENGINE POWER FOR AGRICULTURAL MACHINES

Figure 2.1 – Energy flows through an engine.
and lower are sometimes used instead of gross and net, respectively. Heating values
tabulated in books (see Table 2.1) are gross values unless otherwise indicated. Less
than half of the fuel equivalent power is available for useful work at the flywheel of an
engine (see Figure 2.1). In the remainder of this chapter, the various power losses are
identified.
2.2 COMBUSTION
Combustion is a very complex process, particularly in a CI engine. The fuel must
vaporize and mix with air to form a combustible mixture. Burning of the fuel-air
mixture generates exhaust emissions, but also generates increased pressure to drive the
pistons. The rate of pressure rise affects engine performance and durability.
2.2.1 Combustion chemistry
Insights that are very useful in understanding engines can be obtained by making
two simplifying assumptions regarding combustion chemistry. The first is that all of
the hydrogen in the fuel links with oxygen to form water. The second is that all of the
carbon in the fuel is converted to carbon dioxide (CO ) and carbon monoxide (CO), so
2
that no free carbon appears in the combustion products. Most conventional, petroleum-
based engine fuels are mixtures of a variety of hydrocarbon molecules, but
representative molecules are given in Table 2.1 for each of the common petroleum-
based fuels. Alcohols, which may become engine fuels of the future, are also listed.
Atomic weights of 12 for carbon, 1 for hydrogen, 16 for oxygen and 14 for nitrogen
may be used in the combustion calculations. Although various gases are in the earth’s
atmosphere, it is common practice in combustion calculations to neglect all gases
except oxygen and nitrogen. The composition of earth’s atmosphere is such that 3.76
molecules of nitrogen (N ) accompany every molecule of oxygen (O ). Combustion
2 2
chemistry then becomes a simple matter of counting atoms, as indicated in Example
Problem 2.1.
ENGINEERING PRINCIPLES OF AGRICULTURAL MACHINES 17
Table 2.1. Comparison of properties of several fuels.
Higher Stoichio-
API Heating Research Boiling metric
Gravity, Density, Value, Octane Range, Air-Fuel
Fuel degrees kg/m3 kJ/kg Number °C Formula Ratio
Butane 112 580 49.500 98 0 C H 15.5
4 10
Propane 146 509 50,300 111 – 42 C H 15.7
3 8
Reg.gasoline 61 735 47,600 93 30 - 230 C H 15.2
6 18
No. 1 diesel 40 823 45,700 40[a] 160 - 260 C H26 15.0
12
No. 2 diesel 38 834 45,500 40[a] 200 - 370 C H 15.0
16 34
Methanol --- 792 22,700 110 65 CH O 6.49
4
Ethanol --- 785 29,700 110 78 C H O 9.03
2 6
Methyl --- 885 38,379 51[b] C H O 12.5
19 36 2
soyate
[a] Minimum cetane rating for diesel fuel
[b] Cetane rating

Example Problem 2.1

Calculate the stoichiometric (chemically correct) air/fuel ratio when diesel fuel is
burned with air. Also analyze the products of combustion when No. 2 diesel is burned.
Solution
From Table 2.1, the cetane molecule (C H ) is used to represent diesel fuel. Under
16 34
the standard simplifying assumptions, the complete combustion reaction becomes:
C H + 24.5 O + 92.12 N → 92.12 N + 16 C0 + 17 H O
16 34 2 2 2 2 2
The reaction is balanced on the basis of one molecule of fuel. The hydrogen balance
determines the amount of water in the combustion products, while the carbon balance
determines the amount of CO . Then enough O must be supplied to form the CO and
2 2 2
H O; each mole of O is accompanied by 3.76 moles of N. The nitrogen is nearly inert
2 2 2
and simply appears in the combustion products. The stoichiometric air/fuel ratio is:
A/F = (24.5 × 32 + 92.12 × 28) / 226 = 14.9
Note that 17 moles of water appear in the exhaust for each mole of fuel burned or, on a
mass basis, 1.35 kg of water appear per kilogram of fuel burned. The difference
between the gross and net heating values of the fuel is exactly equal to the latent
energy of the water produced by combustion, i.e., the energy needed to convert that
liquid water to vapor. A major reason why quick warm up of engines is important is to
cause the combustion water to exit the engine as vapor rather than liquid. If the fuel
contains sulphur impurities, the sulphur compounds created in combustion can react
with liquid water to form sulfuric acid and corrode the engine.
18 CHAPTER 2 ENGINE POWER FOR AGRICULTURAL MACHINES
Engine exhaust gases are normally analyzed on a dry, volume basis. Since the exhaust
gases are intermingled at the same temperature and pressure, each molecule occupies
the same volume according to Avogadro’s Law. Thus, the analysis of the dry exhaust
gases in Example Problem 2.1 is:
92.12 / (92.12 + 16) = 0.852 volume fraction (85.2%) is occupied by N ,
2
and 16 / (92.12 + 16) = 0.148 volume fraction (14.8%) is occupied by CO .
2

The equivalence ratio, φ, is a measure of mixture richness. It is defined as follows:
(F/A)
φ= actual (2.2a)
(F/A)
stoichiometric
(A/F)
or φ= stoichiometric (2.2b)
(A/F)
actual
Note that the F/A ratio is just the inverse of the A/F ratio. Thus, in Example
Problem 2.1, the stoichiometric ratios were A/F = 14.9 or F/A = 0.0671. An air-fuel
mixture is rich if φ >1, stoichiometric if φ =1, or lean if φ <1. A rich mixture contains
more fuel than the available oxygen can combust, while a lean mixture contains more
oxygen than is theoretically needed to combust all the fuel. When φ >1, not enough
oxygen is available to convert all the carbon in the fuel to CO ; consequently, CO
2
appears in the exhaust. When φ <1, not all of the oxygen is needed in combustion and
free oxygen appears in the exhaust products. For a spark ignition engine, φ must be
reasonably close to unity to sustain combustion. For a compression ignition engine, φ
should not exceed 0.7 to prevent engine damage.
The following generalized combustion reaction is valid for any air-fuel mixture
under the two simplifying assumptions given earlier:
ψ ψ
C H O + 1O +3.76 1 N →
x y z 2 2
φ φ
(2.3)
ψ y
3.76 1 N +ψ CO +ψ CO+ψ O + H O
φ 2 2 2 3 4 2 2 2
where x = number of carbon atoms in fuel molecule
y = number of hydrogen atoms in fuel molecule
z = number of oxygen atoms in fuel
Ψ = x + y/4 – z/2
1
Ψ = x for φ < 1
2
x – 2Ψ (1 – 1/ φ) for φ > 1
1
Ψ = 0 for φ < 1
3
2Ψ (1 – 1/ φ) for φ > 1
1
Ψ = Ψ (1 / φ – 1) for φ < 1
4 1
0 for φ > 1
ENGINEERING PRINCIPLES OF AGRICULTURAL MACHINES 19
Note that the Combustion Reaction 2.3 accommodates oxygenated fuels, such as the
alcohols in Table 2.1. The number of carbon, hydrogen, and oxygen atoms need not be
integer numbers. The stoichiometric air/fuel ratio for the combustion is:
137.3ψ
A/F= 1 (2.4)
φ(12x+y+16z)
The theoretical concentrations of the dry exhaust products on a volume basis are:
Conc. N = 3.76 Ψ /(φ T) (2.5a)
2 1
Conc. C0 = Ψ /T (2.5b)
2 2
Conc. CO = Ψ /T (2.5c)
3
Conc. O = Ψ /T (2.5d)
2 4
where T = Ψ + Ψ + Ψ + 3.76Ψ /φ.
2 3 4 1
Equations 2.5a through 2.5d give good approximations to actual exhaust emissions,
except that minute amounts of other gases also appear. A small amount of oxygen and
nitrogen react with each other to form oxides of nitrogen, i.e., NO and NO . The
2
combined NO and NO gases are commonly referred to as NO . Also, φ is typically
2 x
not uniform throughout all of the combustion chambers of an actual engine. Thus,
small amounts of CO and O may appear in the exhaust whether the overall φ is less
2
than or greater than one. Some free carbon may also appear, as well as trace amounts
unburned hydrocarbons (HC), hydrogen, and other gases.
2.2.2 Energy release in combustion
The purpose of the combustion reaction is to release energy to drive the pistons. A
cross section of a typical diesel engine is shown in Figure 2.2. The combustion process
can be carried out in either two or four strokes of the piston, but the four-stroke cycle
is most common. Unless otherwise indicated, all engines discussed in this book will be
assumed to use the four-stroke cycle.
Through a combined experimental and analytical technique, it is possible to infer
the rate of energy release throughout the combustion process. The technique relies on
measurement of combustion chamber pressures in a running engine while simultane-
ously measuring the crankshaft rotation, and computing the volume within the
combustion chamber. The spatially averaged temperature in the combustion chamber
can be calculated from the pressure and volume. Then, from changes in pressure,
volume, and temperature, the heat loss through the chamber walls, work done on the
piston, and changes in internal energy of the mixture in the combustion chamber can
be calculated. The energy released from the fuel is equal to the sum of the heat loss,
work, and increases in internal energy. Figure 2.3 shows a typical energy release
diagram for a diesel engine; the rate of energy release is plotted versus crankshaft
position.
20 CHAPTER 2 ENGINE POWER FOR AGRICULTURAL MACHINES

Figure 2.2 – Cross section of a typical diesel engine.

Figure 2.3 – Rate of energy release from fuels in a compression ignition engine.
ENGINEERING PRINCIPLES OF AGRICULTURAL MACHINES 21
In a diesel engine, air without fuel is taken in during the intake stroke and com-
pressed. Late in the compression stroke, at approximately 20° before HDC (Head
Dead Center), injection of fuel into the combustion chamber begins. An apparent
negative energy release rate appears initially as energy is withdrawn from the chamber
to evaporate the injected fuel. The evaporated fuel mixes with air and undergoes
certain pre-reactions during an ignition delay period. Then ignition occurs and all of
the air-fuel mixture prepared during the ignition delay burns suddenly to produce a
sharp, triangular-shaped energy release pattern identified as premixed combustion. For
combustion to continue, fuel vapor and air must diffuse toward each other across the
regions burned out in the premixed combustion. The rate of diffusion limits the latter
combustion, which is identified as diffusion combustion. The total energy release is the
sum of the premixed and diffusion combustion. Premixed combustion is efficient and,
except for the production of NO , is also clean combustion. However, the rapid release
x
of energy produces the greatest stress on the engine and also most of the combustion
noise. The slower diffusion burning is quieter and less stressful on the engine, but
produces exhaust smoke and most of the CO emissions and is less efficient. Using
fuels of higher cetane rating and less-advanced injection timing shifts more of the
combustion from the premixed to the diffusion mode; the converse is also true.
In a diesel engine, the air supply is never throttled to control the engine speed.
Rather, control is achieved by controlling only the fuel delivery rate. Consequently, φ
is close to zero when the engine is idling without load and increases as more fuel is
injected with increasing load. To limit smoke emissions and avoid excessive engine
temperatures, it is necessary to operate a diesel engine with φ below approximately
0.7. As Reaction 2.3 and Equation 2.5d would show, considerable free oxygen appears
in the exhaust when φ = 0.7 or less. Engine users sometimes increase the fueling rate
to diesel engines to take advantage of the extra oxygen and boost the power output of
the engine, but at the cost of reduced engine life. For their own protection, engine
manufacturers put a seal on the injector pumps of their engines; if the seal is broken to
increase the fueling rate, the engine warranty is automatically voided.
2.3 THERMODYNAMIC LIMITS
TO ENGINE PERFORMANCE
The effective pressure that can be obtained from fuel to drive the pistons and also
the combustion efficiency have thermodynamic limits which are defined in this
section. The engine is designed to carry out the combustion cycle in four strokes of the
piston. As is required in an engine with a four-stroke cycle, the timing gears in Figure
2.2 are arranged such that the crankshaft makes two revolutions for each revolution of
the camshaft. Valve timing in a four-stroke cycle is shown on a valve-timing spiral, as
illustrated in Figure 2.4. Valve timing is designed to maximize airflow through the
engine and may differ somewhat from that shown in Figure 2.4. The cycle begins just
before HDC (Head Dead Center) with the opening of the intake valve, and the air
intake process ends well after CDC (Crank Dead Center) with the closing of the
intake valve. As the piston approaches HDC on the compression stoke, fuel is injected
and, after a short delay, ignites and forces the piston down on the power stroke. The
22 CHAPTER 2 ENGINE POWER FOR AGRICULTURAL MACHINES
exhaust process begins with the opening of the intake valve late in the power stroke
and ends with the closing of the exhaust valve soon after HDC. Thus, the four strokes
of the cycle are intake, compression, power, and exhaust. Note that there is valve
overlap, i.e., both valves are open simultaneously during a brief part of the cycle.
Alternate terms used in the literature are TDC (Top Dead Center) instead of HDC and
BDC (Bottom Dead Center) instead of CDC.
The dual cycle of Figure 2.5 is the best thermodynamic model of the modern diesel
engine. It illustrates the theoretical variations in combustion gas pressure and cylinder
volume during an engine cycle. The dual cycle is a combination of the Otto cycle,
which represents spark ignition engines, and the original Diesel cycle that Dr. Rudoph
Diesel proposed to represent his engine. Parameter γ defines the relative proportion of
energy input to the dual cycle at constant pressure, i.e.:
q
p
γ= (2.6)
q +q
p v
where q = energy input at constant pressure
p
q = energy input at constant volume
v

Figure 2.4 – Valve timing spiral showing typical valve timing.
ENGINEERING PRINCIPLES OF AGRICULTURAL MACHINES 23

Figure 2.5 – The theoretical dual cycle.
When γ = 0, the dual cycle becomes the Otto cycle with points 2a and 3 becoming
coincident (Figure 2.5). When γ = 1, the dual cycle becomes the original Diesel cycle
with points 2 and 2a becoming coincident.
In the dual cycle, 0-1 is the intake process, followed by compression, 1-2. Process
2-2a is energy input to the cycle at constant volume and 2a-3 is constant-pressure
energy input. Work is extracted from the cycle between points 2a and 3, followed by
heat rejection, 4-1. Process 1-0 is exhaust, at which point the cycle starts over. The
cylinder volume at CDC, V , is the maximum gas volume. The cylinder volume at
1
HDC, V , is called the clearance volume. The displacement of a single cylinder is:
2
V =V −V (2.7)
c 1 2
The displacement, V, of a multi-cylinder engine is simply V times the number of
e c
cylinders. The compression ratio of the engine is:
V
r= 1 (2.8)
V
2
The cycle mean effective pressure is the net area within the p-v diagram of Figure
2.5 divided by V. Multiplying the cycle mean effective pressure by the piston area
c
and stroke length gives the actual work performed by each power stroke. The cycle
mean effective pressure can be calculated from:
24 CHAPTER 2 ENGINE POWER FOR AGRICULTURAL MACHINES
p r−rk +Θ [r−r2−krk−1+r r−1(k−1)(r −1)]
cme = r co co co (2.9)
p (k−1)(r−1)
1
where p = cycle mean effective pressure, kPa
cme
p = absolute pressure at beginning of compression, kPa
1
Θ = Θ /Θ
r 3 1
λ = k(γ -1 – 1)
r = (λ +1)/( λ + (Θ /Θ )rk-1) = fuel cutoff ratio
co 3 1
k = 1.4 for air standard cycle
The pressure, p , is very nearly equal to the atmospheric pressure unless the engine
1
is turbocharged. The fuel cutoff ratio is defined as the proportion of the power stroke
during which energy is being released into the cycle from the burning fuel. The cycle
efficiency is defined as:
q +q −q
v p out
η =
cy q +q
v p
where each heat transfer quantity is calculated from the mass (M) in the combustion
chamber, multiplied by the appropriate specific heat and temperature difference, i.e.:
q = Mc (T −T )
v v 2a 2
q =Mc (T −T )
p p 2a 3
q =Mc (T −T )
out v 4 1
where c = specific heat at constant volume, J/kg °K
v
c = specific heat at constant pressure, J/kg °K
p
k = c /c
p v
The temperatures are those at the corresponding points in the cycle of Figure 2.5.
Note that the mass (M) cancels out in the efficiency equation. Through use of the
definition of k, the specific heats also cancel out of the efficiency equation. Then, in a
lengthy derivation making use of the ideal gas law, the temperatures can be reduced to
volume ratios (r or r ) and the inlet pressure, p . The result is the following equation
co 1
for cycle efficiency:
γ(rk −1)+k(r −1)(1−γ)r1−k
η =1− co co (2.10)
cy k(r −1)
co
The theoretical values, p and η , cannot be achieved in practice, but are
cme cy
thermodynamic upper limits and targets against which practical designs can be
compared.