220 likes | 242 Views
Learn about the aerodynamic & rolling resistance forces affecting a vehicle's power needs. Discover how grade & shape influence energy usage.
E N D
Energy Consumption & Power Requirements of A Vehicle P M V Subbarao Professor 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 ).
Aerodynamic Resistance on Vehicle 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 P = power (kw) A = area (m2) V = velocity (KpH) V0 = headwind velocity Cd = drag coefficient = 1.2 kg/m3
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.
ROLLING RESISTANCE Rolling resistance of a body is proportional to the weight of the body normal to surface of travel. where: P = power (kW) Crr = coefficient of rolling resistance M = mass (kg) V = velocity (KpH)
Grade Resistance Composed of • Gravitational force acting on the vehicle For small angles, θg Fg θg W
Total Resistance Force on A Vehicle Instaneous total force on an accelerating vehicle F = Faero + Froll resist + Fgrade + Faccel Steady state force are equal to the summation of Faero + Froll resist + Fgrade
Power To be Developed by the engine @ Steady Speed The Powering Engine Torque is: rtire = Tire Rolling Radius (meters) G = Numerical Ratio between P.E. and Tire Ideal capacity of Powering Engine:
Actual Capacity of A Powering engine Correction for Auxiliary power requirements:
Inertial Resistance where: FIR = inertia resistance [N] meff-vehicle = Vehicle mass + Equivalent mass of rotating parts [kg] a = car acceleration [m/s2], (from 0 to 100 km/h in: 6 s (4.63 m/s2), 18 s (1.543 m/s2)) mvehicle = Vehicle mass [kg] meq = Equivalent mass of rotating parts [kg]
Industrial Relations for Equivalent Rotating Mass • Experience has shown the total equivalent mass of the typical car can be estimated with following engineering rule of thumb: