Week-11 Challenge: Braking

1) Braking energy

A significant amount of energy is consumed in braking. To stop a car entirely, the kinetic energy of the moving vehicle has to be absorbed.

K.E = 1/2 . M. V²

Braking power and braking energy consumed by the front and rear wheels are closely related to the braking forces on the front and rear wheels. Ignoring the drags, the braking forces on the front and rear wheels can be expressed as

j = deceleration in m/s^2

La, Lb = The horizontal distance between the vehicle centre of gravity and o the centre of the front and rear wheels respectively.

h_g = height between the vehicle centre of gravity to the ground

I used the nedc driving cycle :

the BRAKING ENERGY FOR THE DRIVING CYCLE IS:

braking energy=0.5*(m+0.05*m)*(v_i^2-v_f^2)
let m=500 kg
so total mass= 525kg
constant K=0.5*525=262.5
in NEDC driving cycle :
at t=4
B.E.1= K*(15^2-0)=59062.5
at t=9
B.E.2=K*(32^2-0)=268800
at t=15
B.E.3=K*(50^2-35^2)=334687.5
at t=17
B.E.4=K*(35^2-0)=321562.5
BE=(B.E.1+B.E.2+B.E.3+B.E.4)*4=3936450
till t=72

at t=79
B.E.5=K*(70^2-50^2)=630000
at t=87
B.E.6=K*(120^2-80^2)=2100000
at t=88
B.E.7=K*(80^2-50^2)=1023750
at t=89
B.E.8=K*(50^2-0)=656250

TOTAL BRAKING ENERGY= BE+B.E.5+B.E.6+B.E.7+B.E.8= 8346450=8346.45kJ

2.) Variation of Braking force/Torque with vehicle/motor speed

It should be noted that the maximum regenerative braking force produced by an electric motor is closely related to the electric motor’s speed.

At low speed (lower than its base speed), the maximum torque is constant. However, at high speed (higher than its base speed), the maxing torque decreases hyperbolically with its speed. Therefore, the mechanical brake torque at a given vehicle deceleration varies with vehicle speed.

REGENERATIVE AND MECHANICAL BRAKES

Regenerative braking in EVs and HEVs adds some complexity to the braking system design.

Two basic questions arise:

1) one is how to distribute the total braking forces required between the regenerative brake and the mechanical friction brake so as to recover the kinetic energy of the vehicle as much as possible.

2) the second is how to distribute the total braking forces on the front and rear axles so as to achieve steady-state braking.

Usually, regenerative braking is effective only for the driven axle. The traction motor must be controlled to produce the proper amount of braking force for recovering the kinetic energy as much as possible.

The machine must be controlled to meet the braking force command from the driver.

In order to meet these requirements, we have three controlled strategies

a) series braking with optimal braking feel

b) series braking with optimal energy recovery

c) parallel braking

a) series braking with optimal braking feel

The control objective is to minimise the stopping distance and optimize the driver’s feel.

The shortest braking distance and good braking feel require the braking forces on the front and rear wheels to follow the ideal braking force distribution curve I.

Which emulates the engine retarding function in general vehicles. When the commanded deceleration is greater than 0.2g, the braking force on the front and rear wheels follow the ideal braking forces distribution curve I, as in fig. The braking force on the front wheels (driven axle) is divided into two parts: regenerative braking force and frictional braking force. When the braking force demanded less than the maximum braking force that the electric motor can produce, only electrically regenerative braking will apply. When the commanded deceleration is greater than the available regenerative braking force, the electric motor will operate to produce its maximum braking torque, and the remaining braking force is met by the mechanical brake system.

series braking with optimal energy recovery

The principal of the series braking with optimal energy recovery is to recover the braking energy as much as possible.

If on the same road, the available regenerative braking force is less than the value corresponding to point a electric motor should be controlled to produce its maximum regenerative braking force. The front and rear braking forces should be controlled at point f, so as to optimize the driver feel and reduce braking distance. In this case, the additional braking force on the front wheels must be developed by mechanical braking by the amount represented by F_bf-mech, and the braking force on the rear axle is represented by point h. When the commanded deceleration rate is, j/g, is much smaller than the (e.g at point j) road adhesive coefficient (j/g = 0.3), and the regenerative braking force can meet the total braking force demand, only regenerative braking is used without mechanical braking on the front and rear wheels. When the commanded deceleration rate is, j/g, is equal road adhesive coefficient µ, the operating point of the front and rear braking forces must be on curve I. On a road with a high adhesive coefficient (µ = 0.7, the operating point f), the maximum regenerative braking force is applied and the remaining is supplied by the mechanical brake. When the commanded deceleration rate is, j/g is greater than the road adhesive coefficient µ, the commanded deceleration rate will never be reached due to the limitation of the road adhesion. The maximum deceleration rate that the vehicle can obtain is (a/g)max=µ.

The operating point of the front and rear braking forces is on the curve I, corresponding to µ (µ = 0.4 and j/g = 0.4 in). The operating point is point k and the maximum deceleration rate is j/g = 0.4.

A series brake with optimal feel and energy needed active control of both electric regenerative braking and mechanical braking forces of the front and rear wheels.

parallel braking

The parallel brake system includes both an electrical (regenerative) brake and a mechanical brake, which produce braking forces in parallel and simultaneously.

The operating system is shown in the fig.

The parallel brake system has a conventional mechanical brake which has a fixed ratio of braking force distribution on the front and rear wheels. Regenerative braking adds additional braking force on the front wheels, resulting in total braking force distribution curve. The mechanical braking force developed by the electric motor is a function of the hydraulic pressure of the master cylinder, and therefore a function of vehicle deceleration. Because the regenerative braking force available is a function of motor speed and because almost no kinetic energy can be recovered at low motor speed, the regenerative braking force at high vehicle deceleration (e.g a/g = 0.9) is designed to be zero so as to maintain braking distance. When the demanded deceleration is less than this deceleration, regenerative braking is effective. When the braking deceleration commanded is less than a given value, say 0.15 g, only regenerative braking is applied.

This emulates the engine retarding in a conventional vehicle.

The parallel braking system does not need an electronically controlled mechanical brake system. A pressure sensor senses the hydraulic pressure in the master cylinder, which represents the deceleration demand. The pressure signal is regulated and sent to the electric motor controller to control the electric motor to produce the demanded braking torque. Compared with the series brake with optimal feel and energy recovery, the parallel braking system has a much simpler construction and control system. However, the driver’s feeling and the amount of energy recovered is compromised.

3.)

The dc motor has various losses associated with it.

1. a) COPPER LOSSES

The copper losses or the i^2r can be expressed as a function of the torque.

K_c . T²

1. b) IRON LOSSES

Since at various speeds the back EMF will be different the iron losses will be a function of the rotational speed.

K_i . ɷ

1. c) WINDAGE LOSSES

K_w . ɷ³

1. d) CONSTANT LOSSES

It comprises all other losses other than those described above.

Therefore, given this information, we can calculate the total losses.

TOTAL LOSSES = COPPER LOSS + IRON LOSS + WINDAGE LOSS + CONSTANT LOSS

The values of K_c, K_i, K_w can be determined experimentally and for a given value of Torque and Speed the losses can be computed, and hence the output power and efficiency can be found.

CODE:

% A program for plotting efficiency contours
% for electric motors
% X-axis - W (speed)
% Y-axis - T

x = linspace (1, 1000); % speed -> rad/s
y = linspace (1, 250); % 0-40 Nm

Kc = 0.2; % Cu loss
Ki = 0.008; % Iron loss
Kw = 0.00001; % windage loss
L = 20; % constant motor loss
% Now make mesh,
[X, Y] = meshgrid (x, y);
Outputpower = (X.*Y); % T*W=P
B = (Y.^2) * Kc; % Copper loss
C = X * Ki; % Iron loss
D = (X.^3) * Kw; % Windage loss

Inputpower = Outputpower + B + C + D + L;
	Z = Outputpower./Inputpower;

% We know efficiencies for which a contour will be plotted

V = [0.7, 0.8, 0.85, 0.9, 0.91, 0.92, 0.925, 0.93];

box off

grid off
contour (X, Y, Z, V);

xlabel ('speed /rad.s');
ylabel ('Torque /N.m');

hold on

% Now plot a contour of power output
% The array Output Power has already been calculated

% We draw contours at 3 & 5hZ

	V = [8000, 10000];
	contour (X, Y, Outputpower, V);

output: