History of IC Engines
1860 Lenoir’s engine (a converted steam engine) combusted natural gas in a double acting piston, using electric ignition. Efficiency = 5% Internal combustion Engines: History, engine types and operation
of 2 & 4 stroke engines
Dr. Primal Fernando
[email protected]
Ph: (081) 2393608
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History if internal combustion (IC) engines
History ‐ continued
• Both power generation and refrigeration are usually accomplished by systems that operate on a thermodynamic cycle po er cycles
by systems that operate on a thermodynamic cycle: power cycles and refrigeration cycles.
• 1876 Nikolaus Otto patented the 4 cycle engine, it used gaseous fuel
• 1882 Gottlieb Daimler, an engineer for Daimler, left to work on his own engine. His 1889 twin cylinder V was the first engine to be produced in quantities. Used liquid fuel and Venturi type carburetor, engine was named “Mercedes” after the daughter of his major distributor
• 1893 Rudolf Diesel built successful CI engine which was 26% efficient (double the efficiency of any other engine of its time)
efficient (double the efficiency of any other engine of its time) • Power producing devises: engines
• Refrigeration producing devices: refrigerators, air‐conditioners and heat pumps
and heat pumps.
• Steam engine ‐ 1700 (external combustion engines)
2
4
1
V Engine
Classification of Engines
•
•
•
•
•
External vs Internal Combustion
E
l I
lC b i
Spark Ignition SI or Compression Ignition CI
Configuration
Valve Location
2 Stroke or 4 Stroke
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7
Wankel Rotary Piston Engine
Engine Configurations
In Line
(Automobile)
(A
bil )
Horizontally
Opposed (Subaru)
Radial (Aircraft)
V
(Automobile)
Opposed Piston
(crankshafts geared
together)
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8
2
Basic considerations in the analysis of power cycles
Rotary “Wankel” Engine
• Cycles encountered in actual devices y
are difficult to analyze because of the presence of complicating effects such as friction etc. • Consider a cycle that resembles the actual cycle closely but it made up totally of internally reversible process (id l
(ideal cycle) l )
Thermal efficiency,  th 
Wnet
Qin
or
wnet
qin
Ref. Internal combustion engines and air pollution, E. F. Obert
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11
Idealizations and simplifications • Cycle does not involve any friction: no pressure drop in the working fluid.
ki fl id
• Expansion and compression process: quasi equilibrium.
• Pipes connecting various p
components are well insulated.
• Neglecting changers in KE and PE
10
12
3
Net work of the cycle
Air‐standard assumption
• Gas power cycles (working fluid gas): spark ignition engines, diesel engines, conventional gas turbines, etc.
• All
All these engines energy is provided by burning a fuel within the system these engines energy is provided by burning a fuel within the system
boundary.
• Working fluid (air) mainly contains nitrogen and hardly undergoing any chemical reactions in the combustion chamber and can be closely resembles to air at all times in the chamber. – Assumptions:
p
working fluid as air, behaves as ideal gas, internally g
,
g ,
y
reversible cycle, combustion process replace by heat addition process by a external source, exhaust process replace by heat rejection process that re‐stores initial state of working fluid, specific heat values determines at room temperatures (call cold‐air‐standard assumptions).
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Reciprocating Engines
Carnot cycle Top Dead Center (TDC) : Upper most position
Bottom Dead Center (BDC) : Lower most position
Exhaust
valve
Intake
valve
Stroke : Length of piston travel
TDC
Stroke
Bore
BDC
Bore : Diameter of the cylinder
Clearance Volume (Vc) : V where piston is at TDC
Displacement Volume (Vd) :Swept Volume (Vmax‐Vmin)
Compression Ratio (rv) = (Vmax/Vmin) = (VBDC/VTDC)
Mean Effective Pressure (MEP) :
Wnet = (MEP) x (Displacement Volume)
• The
The Carnot cycle is the most efficient cycle Carnot cycle is the most efficient cycle
that can be executed between heat a source and a heat sink.  th,Carnot  1 
Reciprocating Engine is INTERNAL COMBUSTION ENGINE, and is Classified into 2 types:
1.
Spark Ignition (SI): Gasoline Engine, Mixing air‐fuel outside cylinder, ignites by a spark plug (Auto cycle)
2.
Compression Ignition (CI): Diesel engine, fuel is injected into the cylinder, self ignited as a result of compression (Diesel cycle).
TL
TH
รศ.ดร.สมหมาย ปรีเปรม
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Four Stroke Engine – spark ignition engine
Parts of an engine
Intake
Compression
Power
Exhaust
1. Intake Stroke piston moves from TDC to BDC,
drawing in fresh air-fuel mixture.
2. Compression Stroke piston moves from BDC to
TDC, compress air-fuel mixture.
3. Power Stroke piston at TDC, spark plug ignite
the air-fuel mixture. the combustion occur
very fast that, in theory, the piston still at
TDC. After that the piston is pushed to BDC.
4. Exhaust Stroke piston moves from BDC to TDC,
pushes the combustion gases out.
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Mean Effective Pressure, MEP Concept
Actual and ideal cycle in spark ignition engine
Actual Processes
P
P
19
Equivalent by MEP
Equivalent
Wnet
MEP
Wnet
vmin
TDC
vmax v
BDC
vmin
vmax v
Wnet = (MEP) x (Displacement Volume)
= (MEP) x (Vmax-Vmin)
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Two Stroke Engine
Intake &
Exhaust
Power
Compression
Energy balance – Otto cycle (I) T
3
qin
Neglecting changes in KE and PE
2
(qin  qout )  ( win  wout )  u (kJ / kg )
4
qout
1
1. Compression Stroke piston moves from BDC to TDC, compress air‐fuel mixture.
2. Power Stroke piston at TDC, spark plug ignite the air‐fuel mixture. After the piston is pushed to BDC. Meanwhile, about half way, combustion gases are b t h lf
b ti
discharged and fresh air‐fuel mixture is drawing in .
Heat transfer to/from the system is under constant volume (no work)
qin  u 3  u 2  cv (T3  T2 )
q out  u 4  u 1  cv (T4  T1 )
 th ,Otto
s1=s2
P
Evaluate at room tem: called cold air standard assumption
s
s3=s4
3
wout
T

T1  4  1 
 T1

 T3

  1 
2
T
 2

2
4
T T
w
q
 net  1  out  1  T4  T1  1 
3
2
qin
qin
T
2‐stroke engines generally less efficient than 4‐stroke engines since partial expulsion of unburned mixture with exhaust gas. It has higher power/weight ratio. win
1
v
v1=v4
v2=v3
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Air Standard Otto Cycle (Nikolaus A. Otto 1876)
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Energy balance – Otto cycle (II)T
Ideal cycle of spark ignition engine, comprises of 4- Process:
Process 1-2 Isentropic Compression (piston moves from BDC to TDC)
Process 2-3 v = constant, heat added (piston stays still, represents combustion)
Process 3-4 Isentropic expansion (piston moves from TDC to BDC gives POWER)
Process 4-1 v = constant, heat rejection (piston stays still, represents EXHAUST and INTAKE stroke)
T
 th ,Otto 
P
T
3
Processes 1‐2 and 3‐4 are isentropic and v2=v3
and v4=v1 (Pvk=constant)
wout
2
2
win
v2=v3
TDC
1
1
v1=v4
v
T1  v 2

T2  v1
4
4



k 1
v
  3
 v4



k 1

4
qout
1
s1=s2
P
wout
2
s3=s4
s
r
Vmax V1 v1


Vmin V2 v 2
BDC
 th ,Otto  1 
What is the different of Otto cycle from Carnot cycle, the most efficient cycle
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s
s3=s4
3
T4
T3
qout
Compression ratio
s1=s2
2
T2   1 
 T2

3
qin

T1  4  1 
wnet
q
 1  out  1  T4  T1  1   T1 
qin
qin
T3  T2
 T3

There are only 2-stroke of all 4-processes,
3
qin
1
4
win
v2=v3
1
v1=v4
v
r k 1
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Thermal efficiency of a Otto cycle (I)
 th ,Otto  1 
Compression Issues
1
r
k 1
• High compression ratios: temperature of air/fuel mixture rises above auto ignition temperature (premature ignition)‐produces audible noise is called engine knock.
• Improvement of thermal efficiency was obtained utilizing higher compression ratios (up to 12) gasoline i
ti ( t 12)
li
blend with tetraethyl lead (improving octane rating) but it has been prohibited to use since the hazardous of lead to health. • Problems can occur during a cycle if there is:
– Lack of Compression caused from gasses leaking past the piston, a hole in the piston, or the intake or exhaust valves are not sealing properly.
k=1.4
– Lack of Spark caused by malfunctioning spark plugs, dirty spark plugs, mistimed firings, or bad connections between plugs and the battery.
l
d th b tt
Octane rating = measure of fuel quality (measure of engines knock resistance)
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Thermal efficiency of a Otto cycle (II)
How Fuel is Handled
Monatomic gas (He, Ar)
• Most compression ratios are around 10:1, meaning that the gas let into the cylinder is
meaning that the gas let into the cylinder is compressed to 1/10 times its original size.
air
CO2
k=1.2
• Efficiency is better with a higher compression ratio but only to the limits of the fuel quality.
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ethane
Molecular weight of the working fluid increases
• Thermal efficiency of actual spark ignition engine ~ 25‐30% 26
• Structure of Gasoline
– Is mostly comprised of hydrocarbon molecules having from six to ten carbon atoms.
o si to te ca bo ato s.
– Octane is a measure of the resistance to detonation. The octane number assigned to gasoline (87,89, 93, 100, 114, 120) represents the ratio of heptane, which easily detonates, to isooctane, which does not want to detonate (better to say octane number above 100 as “performance number”. It is calculated by different way. Often itʹs done y
y
by pure extrapolation. ) . Eighty‐seven‐octane gasoline is gasoline that contains 87‐percent octane and 13‐percent heptane (or some other combination of fuels that has the same performance of the 87/13 combination of octane/heptane). 28
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Diesel cycle: The ideal cycle for compression ignition (CI) engine (Rudolph Diesel 1890)
Chemical Energy of Gasoline
• The chemical energy of one gallon of gasoline is, on the average, 125 000 BTU per gallon (132×106 J per 3.8 L). 125,000 BTU per gallon (132×10
J per 3 8 L)
• Similar to spark ignition engine differing mainly in the method of initiating combustion.
• In spark ignition (SI) engines (gasoline engines), air fuel mixture compressed below auto ignition temperature of the air/fuel mixture and combustion starts by firing spark plugs. • Only about 25% of chemical energy in gasoline is converted to mechanical energy.
• In combustion ignition (CI) engines (diesel engines) air compressed above the auto ignition temperature of the air fuel p
g
p
mixture and then fuel inject into the air. • Basically out of a one dollar gallon of gasoline, 75 cents is wasted.
• SI engines has a carburetor and diesel engine has a fuel pump.
• The compression ratio of diesel engines typically higher (12 ‐24)
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Diesel engine
Cylinder Configurations
• The fuel injection starts when the piston reaches to TDC.
• Combustion process takes place over longer interval.
Straight Configuration
• Because of this longer period the heat addition process can be approximated as constant pressure heat addition process.
pressure heat addition process. V Configuration
Flat
Configuration
Displacement refers to
the volume inside each
piston chamber. For
example: a 3.0 Liter
engine with 6 cylinders
will have 0.5 liters per
cylinder.
• Other parts are common for both SI and CI engines. 30
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Otto vs. Diesel
Energy balance – Diesel engine (I)
 th ,Otto  1 
(qin  qout )  ( win  wout )  u (kJ / kg )
q in  P2 (v 3  v 2 )  (u 3  u 2 )  h3  h2  c p (T3  T2 )
r k 1
 th , Diesel  1 
• Limiting value of rc=1; when efficiencies of both Otto and Diesel cycles are identical.
wnet
q
 1  out
qin
qin
• Diesel cycle operates much higher compression ratios, therefore thermal efficiency of Diesel engines are usually higher than SI eengines (35 to 40%). gi e (
o %)
T

T1  4  1
T1
(T4  T1 )


 1
 1
k (T3  T2 )

 T3
kT2   1

 T2
• Diesel engines burns fuels more completely than gasoline engines.
Energy content of 1 gallon of diesel on average, 147,000 BTU per gallon (155×106 J per 3.8 L).
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Energy balance – Diesel engine (II)
th , Diesel 
wnet
q
 1  out
qin
qin
• More realistic way to model:
Combination of heat transfer
Combination of heat transfer processes in gasoline and diesel cycles.
V3 v3

V2 v 2
Utilizing definition of isentropic ideal‐gas relations
 thh , Diesel
Di l  1 
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Dual cycle

T
T1  4  1
T1
(T4  T1 )


 1
 1
k (T3  T2 )

 T3
kT2   1

 T2
Define new quantity; cutoff ratio rc 
1  rck  1 


r k 1  k (rc  1) 
 th ,Otto   th , Diesel (when both cycles operate on the same compressio n ratio)
q out  u 4  u 1  cv (T4  T1 )
th , Diesel 
1
• The relative amount of heat transfer during each process can be adjusted to approximate j
pp
actual cycle more closely.
1  r 1 


r k 1  k (rc  1) 
k
c
r is the compression ratio
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9
Brayton Cycle: The ideal cycle for gas turbine engines (George Brayton 1870)
Simple Ideal Brayton Cycle
• Used in aircraft propulsion and electric power generation.
• Gas turbines usually operate on an open cycle.
• Fresh air ambient condition compressor (temperature and pressure raised) combustion chamber (fuel burn) turbine (work produce) expands to atmospheric pressure.
Modeled as a closed cycle. Air standard assumptions are applied. Air is the working fluid.
1‐2: isentropic compression
1‐2: isentropic compression • However, open gas turbine cycle can be modeled as a closed cycle. 2‐3: constant pressure heat addition 3‐4: isentropic expansion
4‐1: constant pressure heat rejection 37
Open and closed cycle gas turbine
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Energy balance for steady flow process
(qin  qout )  ( win  wout )  h (kJ / kg )
q out  h4  h1  c p (T4  T1 )
q in  h3  h2  c p (T3  T2 )
th , Brayton
open‐cycle gas turbine
closed cycle gas turbine
closed‐cycle gas turbine
w
q
(T  T )
 net  1  out  1  4 1  1 
qin
qin
(T3  T2 )
T

T1  4  1 
T
 1

T

T2  3  1 
 T2

P
Processes 1‐2 and 3‐4 are isentropic, and P
1 2 d3 4
i t
i
d P2=P
P3 and P
d P4=P
P1
• Combustion and exhaust processes are constant pressure processes.
• Exhaust propels craft or used to generate steam.
T2  P2 
 
T1  P1 
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 k 1 


 k 
P
  3
 P4



 k 1 


 k 

T3
T4
 th , Brayton  1 
1
 k 1 

k 
rp 
Pressure ratio
rp 
P2
P1
40
10
Typical pressure ratios and highest temperatures
• Common pressure ratios 11 to 16.
• Highest temperature occurs at the end p
of the combustion process and limited maximum temperature of turbine blades can withstand. • For a fixed inlet temperature, the net work output increases with pressure ratio, reaches to a maximum value and then decreases. • Air/fuel ratio above 50 is not Ai /f l ti b
50 i
t
uncommon.
• Higher back work (usually more than one‐half of turbine work output).
Gas Power Cycle ‐ Jet Propulsion Technology
k=1.4
In jet propulsion, gas turbine produces little power and the high velocity exhaust gas responsible for producing essay thrust for moving. 41
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Gas Turbine Improvements
Advantageous and disadvantageous
• Increase the gas combustion temperature (T3) before it enters the turbine.
 Gas turbine engines have a great power‐to‐weight ratio compared to reciprocating engines. That is, the amount of power you get out of the engine compared to the weight of the engine itself is very good. d t th
i ht f th
i it lf i
d
Gas turbine engines are smaller than their reciprocating counterparts of the same power. The main disadvantage of gas turbines is that, compared to a reciprocating engine of the same size, they are expensive. Because they spin at such high speeds and because of the high operating temperatures, designing and manufacturing gas turbines is a tough problem from both the engineering and materials standpoint
turbines is a tough problem from both the engineering and materials standpoint. Gas turbines also tend to use more fuel when they are idling, and they prefer a constant rather than a fluctuating load. That makes gas turbines great for things like transcontinental jet aircraft and power plants, but explains why you donʹt have one under the hood of your car. 42
Limited by metallurgical restriction: ceramic coating over the turbine Limited
by metallurgical restriction: ceramic coating over the turbine
blades
 Improved intercooling technology: blow cool air over the surface of the blades (film cooling), steam cooling inside the blades.
• Modifications to the basic thermodynamic cycle: intercooling, reheating, regeneration
• Improve design of turbomachinery components: multi‐stage compressor and turbine configuration. Better aerodynamic design on blades (reduce stall). 44
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Deviation of actual gas‐turbine cycles from idealized ones. Draw diagram
P
qin
2
3
Isentropic efficiency of the compressor
c 
ws h2 s  h1

wa h2 a  h1
1
Isentropic efficiency of the turbine
T 
4
q out
wa h3  h4 a

ws h3  h4 s
v
45
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Start analysis
EXAMPLE PROBLEM
The pressure ratio of an air standard Brayton cycle is 4.5 and the inlet conditions to the compressor are 100
4.5 and the inlet conditions to the compressor are 100 kPa and 27C. The turbine is limited to a temperature of 827C and mass flow is 5 kg/s. Determine
Let’s get the efficiency:
  1
1
rpk 1 k
From problem statement, we know rp = 4.5
a) the thermal efficiency
b) the net power output in kW
c) the BWR (back work ratio)
  1
Assume constant specific heats.
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1
4.5 1.4 1 1.4
 0.349
48
12
Net power output:
Solving for temperatures:
Net Power:

W net  m wnet  m wturb  wcomp

T2:
T2  300 4 .5 
T4:
T4  1100 0 .222 
 461 K
0 .4 1.4
0 .4 1.4
 715.7 K
Net power is then:
Substituting for work terms:
W net  m (h3  h4 )  (h2  h1 )

W
kg/ s)(1.0035 kJ / (kg K )) 
net  (5
(1100
Applying constant specific heats:
W net  m c p (T3  T4 )  (T2  T1 )
 715.7) 
(461 300)
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Back Work Ratio
Need to get T2 and T4
Use isentropic relationships:
T2  p2 
 
T1  p1 
;
K
W net  1120 kW
49
k 1
k

T4  p4 
 
T3  p3 
BWR 
k 1
k
wcomp
wturb

h2  h1
h3  h4
Applying constant specific heats:
T1 and T3 are known along with the
pressure ratios:
BWR 
50
T2  T1
T3  T4

461  300
 0 .42
1100  715.7
52
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The Brayton cycle with regeneration (III)
The Brayton cycle with regeneration (I)
• Therefore, the thermal efficiency of Brayton cycle with regeneration is depends on minimum/maximum temperature ratio as well as ll
pressure ratio.
• In gas turbine engines the temperature of the exhausted gas leaving the turbine is often considerably higher than the temperature of the air leaving the compressor.
f h i l
i
h
• Therefore, counter‐flow (call regenerator) heat exchanger can be used to capture portion of heat from the exhaust gas.
 T1   k 1 
rp  k 
 T3 
th ,regen  1  
• Re generation also most effective at lower pressures.
• This improves the efficiency of the cycle. 53
55
The Brayton cycle with regeneration (II)
q regen ,act  h5  h2
qregen ,max  h5 '  h2
 h4  h2

q regen, act
q regen , max

h5  h2
h4  h2
With cold‐air standard assumption

T5  T2
 0.85 (in practice )
T4  T2
Under cold air assumption with regeneration
 
T1  k 1 
rp  k 
 T3 
th,regen  1  
54
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