Eng ine Performance Section 2
1
Geometric Properti es VC TC y
Piston displacement: y = l + a - s
B
s = a cosθ
L
BC
Wrist pin
l
s
Connecting rod
+ (l 2 − a 2 sin 2 θ )
1/ 2
When the piston is at TC (s= l+a ) the cylinder volume equals the clearance volume V c The cylinder volume at any crank angle is:
V
= Vc + Ac y = Vc +
πB 2 4
(l + a − s )
Maximum displacement, or swept, volume: a
Vd
=
πB 2 4
L
Compression ratio: For most engines B ~ L (“square engine”)
rc
=
VBC VTC
=
Vc
+ Vd Vc
2
Geometric Properti es
VC
TC B
Average piston velocity:
U p = 2 LN L
BC
l
s
stroke m rev rev stroke s
Average piston speed for standard auto engine is about 15 m/s. Ultimately limited by material strength. Therefore engines with large strokes run at lower speeds “Over square engines” (B > L) with light pistons have higher rev limits
s = a cosθ a
+ (l 2 − a 2 sin 2 θ )
1/ 2
Instantaneous piston velocity:
Up Up
=
π 2
sin θ 1 +
cosθ
((l / a )
2
=
Up
− sin 2 θ )
1/ 2
ds dt
3
Piston Velocity vs Crank Ang le
R = l/a R=3
TC
BC
4
Piston Accelera tion 1/ 2
Piston displacement is:
a 2 2 s = a cosθ + l 1 − sin θ l
For most modern engines (a/l)2 ~ 1/9 Using series expansion approximate (1- ε)1/2 ~ 1-(ε/2) and subst θ = ωt 2
So
s = a cos ωt + l − a sin 2 ωt 2l
Substituting
sin 2 ωt
= (1 − cos 2ωt ) / 2
a s = a cos ωt + l − (1 − cos 2ωt ) 4l 2
yields differentiating
d 2s dt 2
a = aω 2 cos ωt + cos 2ωt l 5
Pisto n Inerti a Force
The inertia force is simply the piston mass multiplied by the acceleration Inertia Force = − m
d 2s dt 2
a = −amω 2 cos ωt + cos 2ωt l Primary term
• The maximum force occurs at TC,
θ = ωt = 0
Secondary term
F ~ amω2
• The primary term varies at the same speed as the crankshaft and the secondary term varies at twice the crank shaft speed • For a very long connecting rod (a/l) << 1 secondary term vanishes and the force is harmonic • Complete cancellation of the forces is possible for in-line 6 and 8 as well as for V-12 and V-16 6
Tor que and Power
Torque is measured off the output shaft using a dynamometer. b Force F
Stator Rotor N
Load cell
The torque exerted by the engine is T: T
= F ⋅b
units : Nm = J
The power W delivered by the engine turning at a speed N and absorbed by the dynamometer is: W
= ω ⋅ T = (2π ⋅ N ) ⋅ T
units :
rad rev ( J ) = W (1 kW = 1.341 hp) rev s
Note: ω is the shaft angular velocity in units rad/s
7
Br ake Power
Torque is a measure of an engine’sability to do work and power is the rate at which work is done Note torque is independent of crank speed. The term brake power , Wb , is used to specify that the power is measured at the output shaft, this is the usable power delivered by the engine to the load. The brake power is less than the power generated by the gas in the cylinders due to mechanical friction and parasitic loads (oil pump, air conditioner compressor, supercharger, etc…). The power produced in the cylinder is termed the indicated power ,Wi . 8
Indic ated Wor k per Cycle
Given the cylinder pressure data over the operating cycle of the engine one can calculate the work done by the gas on the piston. This data is typically given as P vs V The indicated work per cycle is given by Wi
Compression W<0
Power W>0
Exhaust W<0
= ∫ PdV
Intake W>0
9
Indic ated Wor k per Cycle
Given the cylinder pressure data over the operating cycle of the engine one can calculate the work done by the gas on the piston. This data is typically given as P vs V The indicated work per cycle is given by Wi
= ∫ PdV
WA > 0
A
C
WB < 0
Compression W<0
Power W>0
Exhaust W<0
Intake W>0
10
Wor k per Cycle Gross i ndicate d w ork per cycle – net work delivered to the piston over the compression and expansion strokes only: Wi,g = area A + area C (>0)
Pump w ork – net work delivered to the gas over the intake and exhaust
strokes: Wp = area B + area C (<0) Net ind icated work per cycle – work delivered over all strokes: Wi,n = Wi,g – Wp = (area A + area C) – (area B + area C) = area A – area B 11
Indicated Power
Indicated power: Wi
=
Wi N nR
(kJ cycle)(rev s )
rev cycle
where N – crankshaft speed in rev/s nR – number of crank revolutions per cycle = 2 for 4-stroke
= 1 for 2-stroke Power can be increased by increasing: • the engine size, Vd • compression ratio, rc • engine speed, N
12
Indic ated W ork at WOT
The pressure at the intake port is just below atmospheric pressure
Po
Pintake
Pintake
The pump work (area B+C) is small compared to the gross indicated work (area A+C) Wi,n = Wi,g - Wp = area A - area B 13
Indic ated Work at P art Throt tl e
The pressure at the intake port is significantly lower than atmospheric pressure
Pintake
The pump work (area B+C) can be significant compared to gross indicated work (area A+C) Wi,n = Wi,g - Wp = area A - area B 14
Indicate d Work wi th Supercharging
Engines with superchargers or turbochargers have intake pressures greater than the exhaust pressure, yielding a positive pump work
Compressor
Pintake
Wi,n = area A + area B
Supercharge increases the net indicated work but is a parasitic load since it is driven by the crankshaft 15
Mecha nical Efficiency
Some of the power generated in the cylinder is used to overcome engine friction and to pump gas into and out of the engine. The term frictio n powe r , W f , is used to describe collectively these power losses, such that: W f
= Wi , g − Wb
Friction power can be measured by motoring the engine. The mechanical efficiency is defined as: ηm
W
= b = Wi , g
Wi , g
− W f
Wi , g
W
= 1− f Wi , g
16
Mechanical Effi ciency , con t’d
Mechanical efficiency depends on pumping losses (throttle position) and frictional losses (engine design and engine speed). Typical values for automobile engines at WOT are: 90% @2000 RPM and 75% @ max speed. Throttling increases pumping power and thus the mechanical efficiency decreases, at idle the mechanical efficiency approaches zero.
17
Power and Torq ue versu s Engin e Speed at WOT W
∝ N ⋅T
Rated brake power
1 kW = 1.341 hp
and W
∝ N ⋅W
cycle
so
T
∝W
cycle
There is a maximum in the brake power versus engine speed called the rated brake power (RBP). At higher speeds brake power decreases as friction power becomes significant compared to the indicated power Wb = Wi , g − W f
Max br ake torque
There is a maximum in the torque versus speed called maximum brake torqu e (MBT). Brake torque drops off: • at lower speeds due to heat losses • at higher speeds it becomes more difficult to ingest a full charge of air. 18
Indi cated Mea n Eff ecti ve Pressu re (IMEP) imep is a fictitious constant pressure that would produce the same work per cycle if it acted on the piston during the power stroke. imep =
Wi Vd
=
Wi ⋅ nR Vd ⋅ N
→ Wi =
imep ⋅ Vd ⋅ N nR
=
imep ⋅ A p ⋅ U p 2 ⋅ nR
Brake me an eff ective pressu re (bmep) is defined as: bmep =
Wb Vd
=
2π ⋅ T ⋅ nR
Vd
→ T=
bmep ⋅ Vd 2π ⋅ nR
imep is a better parameter than torque to compare engines for design and output because it is independent of engine size, Vd. 19
The maximum bmep of a good engine design is well established: Four stroke engines: SI engines: bmep= 850-1050 kPa* CI engines: bmep= 700 -900 kPa Turbocharged SI engines: bmep= 1250 -1700 kPa Turbocharged CI engines: bmep= 1000 - 1200 kPa Two stroke engines: Standard CI engines comparable bmep to four stroke Large slow CI engines: 1600 kPa *Values are at maximum brake torque and WOT Note, at the rated (maximum) brake power the bmep is 10 - 15% less
Can use the maximum bmep in design calculations to estimate engine displacement required to provide a given torque or power at a specified engine speed. 20
Maximum BMEP bmep =
Wb Vd
=
2π ⋅ T ⋅ nR
Vd
The maximum bmep is obtained at WOT at a particular engine speed Closing the throttle decreases the bmep For a given displacement, a higher maximum bmep means more torque For a given torque, a higher maximum bmep means smaller engine Higher maximum bmep means higher stresses and temperatures in the engine hence shorter engine life, or bulkier engine. For the same bmep 2-strokes have almost twice the power of 4-stroke 21
Typi cal 199 8 Passenger Car Engi ne Characterist ics Vehicle
Engine
Displ.
Max Power
Max Torque
BMEP at
BMEP at
type
(L)
(hp@rpm)
(lb-ft@rpm)
Max BT
Rated BP
(bar)
(bar)
I4
1.84
122@6000
117@4000
10.8
9.9
I4
2.25
150@5700
152@4900
11.4
10.4
Mazda Millenia S
I4 Turbo
2.26
210@5300
210@3500
15.9
15.7
BMW
I6
2.80
190@5300
206@3950
12.6
11.5
V8
3.50
375@8250
268@6000
13.1
11.6
V12
5.47
436@6250
398@4500
12.4
11.4
V12
5.71
492@7000
427@5200
12.7
11.0
Mazda Protégé LX Honda Accord EX
328i Ferrari F355 GTS Ferrari 456 GT Lamborghini Diablo VT
22
Road- Lo ad Power
A part-load power level used for testing engines (fuel economy, emissions) is the power required to drive a vehicle on a level road at a steady speed. The road-loa d p ower , Pr , is the engine power needed to overcome rolling resistance and the aerodynamic drag of the vehicle. Pr
ρ a C D Av S v2 ) ⋅ S v
= (C R M v g + 1 2
where CR = coefficient of rolling resistance (0.012 - 0.015) Mv = mass of vehicle g = gravitational acceleration ρa = ambient air density CD = drag coefficient (for cars: 0.3 - 0.5) Av = frontal area of the vehicle Sv = vehicle speed *Modern midsize aerodynamic cars only need 5-6 kW (7-8 HP) to cruise at 90 km/hr, hence the attraction of hybrid cars!
23
Drag Forc e Parameters
Auto manufacturers can improve the drag force by reducing: Vehicle frontal area: 2005 Corvette is 0.57 m 2 Most cars around 0.8 m2 2006 Hummer H3 is 1.56 m2 Drag coefficient CD: 2004 Toyota Prius – 0.26 2005 Porsche Boxster – 0.29 1983 Audi 100 – 0.3 2006 Dodge Challanger – 0.33 2003 Hummer – 0.57 Formula 1 car – 0.7 to 1.1 (DRS)
24
Engin e power requireme nts
α
F=Mvg
For a 1500 kg car moving at 50 km/hr up a 10o incline: PC
= ( M g sin α ) ⋅ S = 35 kW = 47 hp v
v
F = Mva
For a 1500 kg car accelerating (constant) from 0-100 km/hr in 8 s: Pa
= M ( S / t ) ⋅ S = 145 kW = 207 hp v
v
v
Ferrari 430 Scuderia has 508 hp at 8500 rpm 4.3L V8 engine 1270 kg and 0-100 km/hr in 3.6 s requires 272 kW (389 hp)
25
Automobile transmission
Engine operates between 600 – 7000 rpm whereas car wheels rotate at 0 -1800 rpm Highest torque is obtained in the mid engine speed range while the greatest torque is often required at the lowest wheel speed
Transmission produces torqueoperating at low carinspeeds andspeed also operates at highway speeds with high the engine the same range transmission changes the gear ratio as the car speeds up Automatic transmission – gears shift automatically based on input data from the sensors on the engine and the transmission (e.g., engine speed, vehicle speed, throttle position, brake pedal position) Manual transmission – operator shifts gears 26
Automobile transmission
Differe ntial pr ovides Further gear ratio (3:1)
Torque converter provides fluid dynamic coupling with torque multiplication during idle and acceleration
27
Gears
Gears change the speed of rotation and torque transmitted between shafts Consider a simple gear set consisting of two gears:
ωo
ωi R
Ro
GR =
Ro Ri
i
Vc, Fc
Gear ratio (GR) is the number of turns of the input shaft required to give one revolution of the output shaft
Vc = ωi Ri = ωo Ro
Fc
= Ti / Ri = To / Ro
R ω ωo = i ωi = i GR Ro
To
R = o Ti = GR ⋅ Ti Ri 28
Automobile Transmission
An automobile is more complicated because you need several gear ratios so the car can accelerate smoothly (shift for power or fuel economy?) Automatic transmission uses two sets of planetary gears to give three or four forward gear ratios and one reverse Manual transmission typically has five forward gears and a reverse Gear
GR
ωo/ωi
To/Ti
1
3:1
1/3
3
2
2.5:1
2/5
5/2
3
1.5:1
2/3
3/2
4
1:1
1
1
0.75:1
4/3
3/4
5 (OD)
29
To h e l p p r o te c tyo u r p r i va c y, Po we r Po i n tp r e ve n te d th i s e xte r n a lp i c tu r e fr o mb e i n g a u to ma ti c a l l yd o wn l o a d e d .To d o wn l o a d a n d d i s p l a yth i s p i c tu r e ,c l i c k O p ti o n s i n th e Me s s a g e B a r , a n d th e n c l i c k En a b l e e xte r n a lc o n te n t.
Manual tr ansmissi on (two gears)
Flywheel Bearings
Splined shaft/collar
TO WHEELS CRANKSHAFT
Clutch
Synchronizer
Starter
30
Automatic transmission planetary gear sys tem
A reduction, B
overdrive, C reverse
Locking any two locks up the whole device at a 1:1 gear
Input
A
B
Sun ( S)
Planet Carrier (C)
C Sun ( S)
Out
Stationary
i:
o
Planet Carrier (C)
Ring (R)
1 + R/S
3.4:1
Ring (R)
Sun (S)
1 / (1 + S/R)
0.71:1
Ring (R)
Planet Carrier (C)
-R/S
-2.4:1
Planet gears linked by carrier
ring gear = 72 teeth and sun gear has 30 teeth.
31
500 1999 Neon DOHC
ω2 nd =
Engine
450
1st gear (GR=3.54) 2nd gear (GR=2.13)
400
GR2 GR1
ω1st
3rd gear (GR=1.36) 4th gear (GR=1.03)
350 ) lb t-
F ( e u q r o T
5th gear (GR=0.72)
300 250
4210 rpm 3610 rpm
200 150 100 50 0 0
1000
2000
3000
4000
5000
6000
7000
8000
Engine speed (RPM)
32
Specific Fue l Consump tio n
For transportation vehicles fuel economy is generally given as mpg, or L/100 km (in the future gm of CO2/km). In engine testing the fuel consumption is measured in terms of the fuel mass flow rate m f. specific fuel consumption , sfc, The is a measure of how efficiently the fuel supplied to the engine is used to produce power, bsfc =
m f
Wb
isfc =
m f
Wi
units :
g kW ⋅ hr
Clearly a low value for sfc is desirable since for a given power level the lesser the fuel consumed the better it is.
33
Brake S pecific Fuel Consu mpti on vs Engin e Size
Bsfc decreases with engine size due to reduced heat losses from gas to cylinder wall.
Note cylinder surface to volume ratio increases with bore diameter. heat loss energy released
=
cylinder surface area cylinder volume
=
πBL 2
πB L
∝
1
B
34
Brake S pecific Fuel Consu mpti on vs Engine S peed
There is a minimum in the bsfc versus engine speed curve
bsfc =
m f
Wb
At high speeds the bsfc increases due to increased friction i.e. smaller Wb At lower speeds the bsfc increases due to increased time for heat losses from the gas to the cylinder and piston wall, and thus a smaller Wi Bsfc decreases with compression ratio due to higher thermal efficiency
35
Perf or mance Ma ps
Performance map is used to display the bsfc over the engines full load and speed range. Using a dynamometer to measure the torque and fuel mass flow rate for different throttle positions you can calculate: bmep =
2π ⋅ T ⋅ nR
Vd
bsfc =
m f
Wb
Wb
= (2π ⋅ N ) ⋅ T
bmep@WOT
Constant bsfc contours from a 2 litre four cylinder SI engine 36
Engin e Effic iencies
The time for combustion in the cylinder is very short, especially at high speeds, so not all the fuel may be consumed A small fraction of the fuel may not react and exits with the exhaust gas The combusti on efficie ncy is defined as: ηc
=
actual heat input theoretical heat input
=
Qin m f ⋅ QHV
=
Q in m f ⋅ QHV
where Qin = heat added by combustion per cycle mf = mass of fuel added to cylinder per cycle QHV = heating value of the fuel (chemical energy per unit mass)
37
Engin e Eff ici encies (2 )
The thermal e ffic iency is defined as: ηth
=
work per cycle heat input per cycle
=
W Qin
=
W ηc ⋅ m f ⋅ QHV
or in terms of rates ηth
=
power out rate of heat input
= QW in
= η ⋅ mW⋅ Q c f HV
Thermal efficiencies can be given in terms of brake or indicated values Indicated thermal efficiencies are typically 50% to 60% and brake thermal efficiencies are usually about 30%
38
Engin e Eff ici encies (3 ) Fuel conversion effic iency is defined as: ηf
=
W m f ⋅ QHV
=
W m f ⋅ QHV
Note: ηf is very similar to ηth, difference is ηth takes into account actual fuel combusted. sfc =
Recall:
m f
W
Therefore, the fuel conversion efficiency can also be obtained from: ηf
=
1 ( sfc ) ⋅ QHV 39
Volum etric Effi ciency
Due to the short cycle time at high engine speeds and flow restrictions through the intake valve less than ideal amount of air enters the cylinder. The effectiveness of an engine to induct air into the cylinders is measured by the volumetric efficiency: ηv
=
actual air inducted theoretical air
=
ma ρ a ⋅ Vd
=
nR ⋅ m a ρ a ⋅ Vd ⋅ N
where ρa is the density of air at atmospheric conditions Po, To and for an ideal gas ρa =Po / RaTo and Ra = 0.287 kJ/kg-K (at standard conditions ρa= 1.181 kg/m3) Typical values for WOT are in the range 75%-90%, and lower when the throttle is closed
40
Air-Fuel Ratio
For combustion to take place the proper relative amounts of air and fuel must be present in the cylinder. The air-fuel ratio is defined as AF
=
ma mf
=
m a m f
For gasoline fuel the19. ideal AF is about 15:1, with combustion possible in the range of 6 to For a SI engine the AF is in the range of 12 to 18 depending on the operating conditions. For a CI engine, where the mixture is highly non-homogeneous, the AF is in the range of 18 to 70.
41
Relatio ns hip s Between P erfor mance P arame ters
By combining equations presented in this section the following additional working equations are obtained: W =
T=
η f ⋅η v ⋅ N ⋅ Vd ⋅ QHV ⋅ ρ a ⋅ (1 / AF )
nR
η f ⋅ηv ⋅ Vd ⋅ QHV ⋅ ρ a ⋅ (1 / AF ) 2π ⋅ nR
mep = η f ⋅ηv ⋅ QHV ⋅ ρ a ⋅ (1 / AF )
42
