Energy Consumption & Power Requirements of A Vehicle
P M V SubbaraoProfessor
Mechanical Engineering Department
Know the Requirements Before You develop an Engine…..
Resistance Force : Ra
• The major components of the resisting forces to motion are comprised of :
• Aerodynamic loads (Faero) • Acceleration forces (Faccel = ma & I forces)• Gradeability requirements (Fgrade)• Chassis losses (Froll resist ).
grraero FFFmaF
Aerodynamic Force : Flow Past A Bluff Body
Composed of:1. Turbulent air flow around vehicle body (85%)2. Friction of air over vehicle body (12%)3. Vehicle component resistance, from radiators and air
vents (3%)
Aerodynamic Resistance on Vehicle
2
21 VPd
(Re)21 2 fAVFd
ACVF dd2
21
20, )()2.1(
21 VVACF ddesignd
VF = P designd ,
Dynamic Pressure:
Drag Force:
Aero Power
Cd =
coefficient of drag
=
air density 1.2 kg/m3
A =
projected frontal area (m2)
f(Re)
= Reynolds number
v =
vehicle velocity (m/sec)
V0 =
head wind velocity)(862 0
2VV V A C )10 .(1 = P d-6
aero
P
= power (kw)
A = area (m2)V
= velocity (KpH)
V0 = headwind velocityCd
= drag coefficient
= 1.2 kg/m3
Purpose, Shape & Drag
Shape & Components of Drag
Some examples of Cd:
• The typical modern automobile achieves a drag coefficient of between 0.30 and 0.35.
• SUVs, with their flatter shapes, typically achieve a Cd of 0.35–0.45. • Notably, certain cars can achieve figures of 0.25-0.30, although sometimes
designers deliberately increase drag in order to reduce lift.• 0.7 to 1.1 - typical values for a Formula 1 car (downforce settings change for
each circuit) • 0.7 - Caterham Seven • at least 0.6 - a typical truck • 0.57 - Hummer H2, 2003 • 0.51 - Citroën 2CV • over 0.5 - Dodge Viper • 0.44 - Toyota Truck, 1990-1995
• 0.42 - Lamborghini Countach, 1974 • 0.42 - Triumph Spitfire Mk IV, 1971-1980 • 0.42 - Plymouth Duster, 1994 • 0.39 - Dodge Durango, 2004 • 0.39 - Triumph Spitfire, 1964-1970 • 0.38 - Volkswagen Beetle • 0.38 - Mazda Miata, 1989 • 0.374 - Ford Capri Mk III, 1978-1986 • 0.372 - Ferrari F50, 1996 • 0.36 - Eagle Talon, mid-1990s • 0.36 - Citroën DS, 1955 • 0.36 - Ferrari Testarossa, 1986 • 0.36 - Opel GT, 1969 • 0.36 - Honda Civic, 2001 • 0.36 - Citroën CX, 1974 (the car was named after the term for drag
coefficient) • 0.355 - NSU Ro 80, 1967
• 0.34 - Ford Sierra, 1982 • 0.34 - Ferrari F40, 1987 • 0.34 - Chevrolet Caprice, 1994-1996 • 0.34 - Chevrolet Corvette Z06, 2006 • 0.338 - Chevrolet Camaro, 1995 • 0.33 - Dodge Charger, 2006 • 0.33 - Audi A3, 2006 • 0.33 - Subaru Impreza WRX STi, 2004 • 0.33 - Mazda RX-7 FC3C, 1987-91 • 0.33 - Citroen SM, 1970 • 0.32064 - Volkswagen GTI Mk V, 2006 (0.3216 with ground effects) • 0.32 - Toyota Celica,1995-2005 • 0.31 - Citroën AX, 1986 • 0.31 - Citroën GS, 1970 • 0.31 - Eagle Vision • 0.31 - Ford Falcon, 1995-1998 • 0.31 - Mazda RX-7 FC3S, 1986-91 • 0.31 - Renault 25, 1984 • 0.31 - Saab Sonett III, 1970 • 0.30 - Audi 100, 1983 • 0.30 - BMW E90, 2006 • 0.30 - Porsche 996, 1997 • 0.30 - Saab 92, 1947
• 0.195 - General Motors EV1, 1996 • 0.19 - Alfa Romeo BAT Concept, 1953 • 0.19 - Dodge Intrepid ESX Concept , 1995 • 0.19 - Mercedes-Benz "Bionic Car" Concept, 2005 ([2]
mercedes_bionic.htm) (based on the boxfish) • 0.16 - Daihatsu UFEIII Concept, 2005 • 0.16 - General Motors Precept Concept, 2000 • 0.14 - Fiat Turbina Concept, 1954 • 0.137 - Ford Probe V prototype, 1985
Rolling Resistance
Composed primarily of 1. Resistance from tire deformation (90%)2. Tire penetration and surface compression ( 4%)3. Tire slippage and air circulation around wheel ( 6%)4. Wide range of factors affect total rolling resistance5. Simplifying approximation:
WCF rrrr
ROLLING RESISTANCE
V M C )10 (2.72 = P
V M C 36009.81 = P
rr3-
rr
rrrr
where:
P
= power (kW)
Crr
= coefficient of rolling resistance
M
= mass (kg)
V
= velocity (KpH)
Rolling resistance of a body is proportional to the weight ofthe body normal to surface of travel.
MgFrr
147101.0 VCrr
Contact Type Crr
Steel wheel on rail 0.0002...0.0010
Car tire on road 0.010...0.035
Car tire energy safe 0.006...0.009
Tube 22mm, 8 bar 0.002
Race tyre 23 mm, 7 bar 0.003
Touring 32 mm, 5 bar 0.005
Tyre with leak protection 37 mm, 5 bar / 3 bar 0.007 / 0.01
Rolling Resistance And Drag Forces Versus Velocity
Grade Resistance
Composed of – Gravitational force acting on the vehicle
gg WF sin
gg tansin
gg WF tanGg tan
WGFg
For small angles,
θg W
θg
Fg
Inertial or Transient Forces
• Transient forces are primarily comprised of acceleration related forces where a change in velocity is required.
• These include:• The rotational inertia requirements (FI ) and • the translational mass (Fma). • If rotational mass is added it adds not only rotational
inertia but also translational inertia.
ra = k m = I =
dtd I = T
tire
vehiclewheelwheel
2wheeli
a r
k m = r
a k m = rT = F 2
tire
222
2tire
2
tire
ii
= angular acceleration k = radius of gyration t = time T = Torque
m = mass = ratio between rotating component and the tire
Transient Force due to Rotational Mass
Therefore if the mass rotates on a vehicle which has translation,
a m + mr
k = F tr2tire
22
i t&r
m +
r
k m a + Slope% + C gm + V A C = F t2tire
22
rrrt2
dtire
2
Resistance power, Presistance V FP tireceresis tan
P resi
stan
ce
Vehicle Speed
Power Demand Curve
Gr F = T tiretire
PE
)( r G
RPM = hkm tirePE 377.0/
The Powering Engine Torque is:
The speed of the vehicle in km/h is:
rtire = Tire Rolling Radius (meters)
G = Numerical Ratio between P.E. and Tire
Ideal capacity of Powering Engine: kWNTP PEPE
600002
Ideal Engine Powering Torque
Drive System Efficiency
• Drive train inefficiencies further reduce the power available to produce the tractive forces.
• These losses are typically a function of the system design and the torque being delivered through the system.
actual
PEdrivemech P
PEfficiencyMechanical
nredredreddrivemech ......21
Actual Capacity of A Powering engine
kWNTPPmech
PE
mech
PEactual
600002
auxtyre
tyretyrePE PkWN
rFP
600002
Correction for Auxiliary power requirements:
MATLAB for Vehicle Torque Requirement
MATLAB Model for Transmission System
MATLAB Model for Engine Performance
Engine Characteristic Surface
Requirements of Vehicle on Road & Engine Power
Urban Driving Cycle
Engine RPM during Urban Driving Cycle
Engine Fuel Consumption During Urban Driving Cycle
Inverse of Carnot’s Question
• How much fuel is required to generate required power?• Is it specific to the fuel?• A Thermodynamic model is required to predict the fuel
requirements.• Carnot Model• Otto Model• Diesel Model• A Geometric Model is required to implement the
thermodynamic model.