BRAKING ENERGY,BRAKING STRATEGY AND MOTOR LOSSES
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.
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
and,
,
j = deceleration in m/sec^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
M_v = Vehicle mass
Once, the braking forces are known we can easily calculate the Braking Energy for given braking distance.
MATLAB CODE FOR CALCULATING BRAKING ENERGY.
%% Calculate the braking energy of a defined Drive cycle.
% Rahul Ramakrishnan
clear all
close all
clc
%% Input Parameters of the vehicle
M_veh = 1500; %Mass of vehicle in Kg
L = 2.7; % Wheel base of the vehicle
L_front = 0.4*L; %Horizontal distance between the vehicle centre of gravity and the front wheel
L_rear = 0.6*L; %Horizontal distance between the vehicle centre of gravity and the rear wheel
H_g = 0.55; % Height of the vehicle centre of gravity from the ground.
g = 9.8; % Acc due to Gravity
% Velocity matrix of IM240 drive cycle in(m/sec)
Velocity = [0;0;0;0;0;1.34111666666667;2.63752944444444;3.84453444444444;...
5.14094722222222;6.39265611111111;7.55495722222222;7.73377277777778;...
8.09140388888889;9.25370500000000;9.70074388888889;10.0136711111111;...
10.0583750000000;9.87955944444444;9.61133611111111;9.34311277777778;...
9.11959333333333;8.85137000000000;7.59966111111111;6.66087944444444;...
6.66087944444444;6.79499111111111;6.92910277777778;7.15262222222222;...
7.64436500000000;8.53844277777778;9.43252055555556;10.1477827777778;...
10.2371905555556;10.1477827777778;10.1030788888889;9.52192833333333;...
8.49373888888889;7.64436500000000;7.06321444444444;7.06321444444444;...
7.91258833333333;8.85137000000000;9.65604000000000;10.3713022222222;...
10.8183411111111;10.9971566666667;11.1312683333333;11.1759722222222;...
11.4888994444444;11.6677150000000;11.9359383333333;12.2935694444444;...
12.7853122222222;13.0982394444444;13.3217588888889;13.4558705555556;...
13.5899822222222;13.7240938888889;13.7240938888889;13.6346861111111;...
13.5899822222222;13.5452783333333;13.5899822222222;13.7687977777778;...
13.5899822222222;13.3664627777778;13.1876472222222;13.3217588888889;...
13.5452783333333;13.7240938888889;13.8135016666667;13.8582055555556;...
13.8135016666667;13.5899822222222;13.3217588888889;13.3664627777778;...
13.5005744444444;13.7240938888889;13.9476133333333;14.2158366666667;...
14.3946522222222;14.4840600000000;14.3946522222222;14.1711327777778;...
12.7853122222222;11.2206761111111;9.65604000000000;8.09140388888889;...
6.52676777777778;4.96213166666667;3.39749555555556;1.83285944444444;...
0.268223333333333;0;0;0;0;0;1.47522833333333;2.95045666666667;4.4256850...
;5.90091333333333;7.37614166666667;8.85137000000000;9.92426333333333...
;10.8630450000000;11.5336033333333;11.8018266666667;11.4888994444444...
;11.2206761111111;11.0418605555556;11.2653800000000;11.3547877777778;...
12.1594577777778;11.8465305555556;10.7289333333333;10.1477827777778;...
8.67255444444444;7.91258833333333;7.68906888888889;8.09140388888889;...
8.31492333333333;8.94077777777778;9.25370500000000;9.70074388888889;...
10.0136711111111;10.0583750000000;9.87955944444444;9.61133611111111;...
9.34311277777778;9.11959333333333;8.85137000000000;7.59966111111111;...
7.64436500000000;7.06321444444444;7.06321444444444;7.91258833333333;...
8.85137000000000;9.65604000000000;9.92426333333333;10.9524527777778;...
11.0418605555556;11.0865644444444;11.0418605555556;10.9971566666667;...
10.9971566666667;11.2206761111111;11.4441955555556;11.4888994444444;...
11.3547877777778;11.1312683333333;11.1759722222222;11.3547877777778;...
11.6230111111111;11.6230111111111;11.4888994444444;11.6677150000000;...
11.9359383333333;12.2041616666667;13.6346861111111;14.9758027777778;...
16.1828077777778;16.6745505555556;17.5686283333333;18.1050750000000;...
18.8203372222222;19.4461916666667;20.1614538888889;20.5637888888889;...
20.9214200000000;21.2343472222222;21.2343472222222;21.1449394444444;...
21.1002355555556;21.1002355555556;21.1896433333333;21.4131627777778;...
21.6813861111111;21.9496094444444;22.1284250000000;22.3519444444444;...
22.6201677777778;22.7989833333333;23.0225027777778;23.3354300000000;...
23.7824688888889;24.1848038888889;24.4083233333333;24.5424350000000;...
24.5871388888889;24.5424350000000;24.4083233333333;24.4083233333333;...
24.4977311111111;24.6318427777778;24.8106583333333;24.9000661111111;...
25.0788816666667;25.1682894444444;25.3024011111111;25.3471050000000;...
25.3471050000000;25.1682894444444;25.0341777777778;24.5871388888889;...
23.8718766666667;23.0672066666667;23.1566144444444;23.2907261111111;...
23.4695416666667;23.6930611111111;23.9165805555556;24.1401000000000;...
24.5424350000000;24.7659544444444;24.8553622222222;25.0341777777778;...
25.0341777777778;24.9447700000000;24.6765466666667;24.3636194444444;...
23.9612844444444;23.4695416666667;23.0225027777778;22.5754638888889;...
21.4578666666667;19.8932305555556;18.3285944444444;16.7639583333333;...
15.1993222222222;13.6346861111111;12.0700500000000;10.5054138888889;...
8.94077777777778;7.37614166666667;5.81150555555556;4.24686944444444;...
2.68223333333333;1.11759722222222;0;0];
Time = (linspace(1,241,241))'; % Time Matrix
% Energy calculation
Energy = zeros(1,241);
Del_v = zeros(1,241);
Del_T = zeros(1,241);
j = zeros(1,241);
F_br = zeros(1,241);
F_bf = zeros(1,241);
Energy_kwh = zeros(1,241);
Brake_energy_f = zeros(1,241);
Brake_energy_r = zeros(1,241);
Total_Energy = zeros(1,241);
Dx = zeros(1,241);
for i = 1:(length(Velocity)-1)
% Condition for deceleration
if Velocity(i)>Velocity(i+1)
% Calculation of deceleration
Del_v(i) = Velocity(i+1) - Velocity(i);
Del_T(i) = Time(i+1) - Time(i);
j(i) = Del_v(i)/Del_T(i); % Deceleration in m/sec^2
% Force on the front and rear axels in KN
F_bf(i) = ((j(i)*M_veh)/L*(L_rear + (H_g/g)*j(i)))/1000;
F_br(i) = ((j(i)*M_veh)/L*(L_front - (H_g/g)*j(i)))/1000;
% Small distance dx RETARTED during the braking operation
Dx(i) = Del_v(i)*Del_T(i);
% Braking energy in KwH
Brake_energy_f(i) = (F_bf(i)*Dx(i))/3600; % Braking energy of front wheels
Brake_energy_r(i) = (F_br(i)*Dx(i))/3600; % Braking energy of rear wheels
Total_Energy(i) = Brake_energy_f(i) + Brake_energy_r(i); % Total energy in KwH
end
end
Brake_energy_f(length(Velocity)) = 0;
Brake_energy_r(length(Velocity)) = 0;
% FORCE CALCULATION
Force_Front_axle = sum(F_bf); % Total force exerted on the front axle during the drive cycle
Force_Rear_axle = sum(F_br); % Total force exerted on the rear axle during the drive cycle.
Total_Force = Force_Front_axle + Force_Rear_axle ;
% ENERGY CALCULATION
Total_Brake_energy_front = sum(Brake_energy_f); % Total energy in joules to KwH
Total_Brake_energy_rear = sum(Brake_energy_r);
Total_energy = sum(Total_Energy); % Total energy in KwH
%% Result Display
disp('BREAKING ENERGY')
disp(' ')
disp(['Total braking energy@Rear = ' num2str(Total_Brake_energy_rear) ' KwH'])
disp(['Total braking energy@Front = ' num2str(Total_Brake_energy_front) ' KwH'])
disp(['Total Braking Energy = ' num2str(Total_energy) ' KwH'])
disp('----------------------------------------------------------------------------')
disp('BREAKING FORCE')
disp(['Total braking Force@Rear = ' num2str(abs(Force_Rear_axle)) ' KN'])
disp(['Total braking Force@Front = ' num2str(abs(Force_Front_axle)) ' KN'])
disp(['Total Braking Force = ' num2str(abs(Total_Force)) ' KN'])
% Drive cycle plot
figure(1)
subplot(2,1,1)
plot(Time,Velocity)
xlabel('Time(sec)')
ylabel('Velcity(KmpH)')
title('Drive cycle IM240')
subplot(2,1,2)
plot(Time,j)
xlabel('Time(sec)')
ylabel('Deceleration(m/sec)^2')
title('Deceleration')
%Force plot
figure(2)
subplot(3,1,1)
bar(Time,abs(F_bf))
xlabel('Time(sec)')
ylabel('Brake Force Front axle(kN)')
title('Front Axle')
subplot(3,1,2)
bar(Time,abs(F_br))
xlabel('Time(sec)')
ylabel('Brake Force Rear Axle(kN)')
title('Rear Axle')
subplot(3,1,3)
bar(Time,abs(F_br))
xlabel('Time(sec)')
ylabel('Total Brake Force(kN)')
title('Net Force')
% Energy Plot
figure(3)
subplot(3,1,1)
bar(Time,Brake_energy_f)
xlabel('Time(sec)')
ylabel('Brake Energy')
title('Brake Energy - Front Axle')
subplot(3,1,2)
bar(Time,Brake_energy_r)
xlabel('Time(sec)')
ylabel('Brake Energy')
title('Brake Energy - rear Axle')
subplot(3,1,3)
bar(Time,Total_Energy)
xlabel('Time(sec)')
ylabel('Brake Energy')
title('Total Brake Energy ')
the drive cycle used is IM240.
RESULTS
We can observe that the braking force and the braking energy on the front axle are greater than that of the rear axle.
This is necessary to prevent the rear wheels becoming locked before the front wheels, resulting in unstable braking.
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 maximum 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 mechanical brake must be controlled to meet the braking force command from the driver.
In order to meet these requirements, we have three control strategies
a) series braking with optimal braking feel
b) series braking with optimal energy recovery
c) parallel braking
a) SERIES BRAKING WITH OPTIMUM BRAKING FEEL
The control objective is to minimize 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.
The above figure illustrates the principle of this braking strategy.
When the commanded deceleration (Governed by the Brake pedal Position ) is less than 0.2g, only regenerative braking on the front wheels is applied.
which emulates the engine retarding function in general vehicles.
When the commanded deceleration is greater than 0.2 g, the braking forces on the front and rear wheels follow the ideal braking forces distribution
curve I, as shown in Figure.
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 is less than the maximum braking force that the electric motor can produce, only electrically regenerative braking
will apply. When the commanded braking force 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
b) SERIES BRAKE WITH OPTIMAL ENERGY RECOVERY
The principle of the series braking system with optimal recovery is to recover the braking energy as much as possible in the condition of meeting
the total braking force demanded the given deceleration.
When the vehicle is braked with an acceleration rate of j/g<u, the braking forces on the front and rear wheels can be varied in a certain range, as long as
the F_bf + F_br = M_v*j is satisfied.
This variation in the range of the front and rear axles is shown by the line ab,
where, u = 0.9 and j/g = 0.7.
In this case, regenerative braking should be used in priority.
If the available regenerative braking force (maximum braking force produced by the electric motor) is in this range, the braking force on the rear wheels, represented by point e, should be developed in order to meet
the total braking requirement.
If on the same road, the available regenerative braking force is less than the value corresponding to point a , the 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, 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, j/g, is equal to the road adhesive coefficient µ, the operating point of the front and rear braking forces must be on
the 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, j/g, is greater than the road adhesive coefficient µ, this commanded deceleration rate will never be reached due to the limitation of the road adhesion.
The maximum deceleration 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 both optimal feel and energy recovery needed active control of both electric regenerative braking and mechanical braking forces on the front and rear wheels.
c) PARALLEL BRAKE
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 figure.
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 forces on the front and rear axles are proportional to the hydraulic pressure in the master cylinder.
The regenerative 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 balance.
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 braking of both 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) CONTOUR PLOT OF MOTOR SPEED, TORQUE AND EFFICIENCY
The Dc motor has various losses associated with it.
a) COPPER LOSSES
The copper losses or the i^2r losses can be expressed as a function of the torque.
b) IRON LOSSES
Since at various speeds the back EMF will be different the iron losses will be a function of the rotational speed.
c) WINDAGE LOSSES
Windage or mechanical losses
d) CONSTANT LOSSES
The constant losses comprise all other losses other than those described above.
Therefore, given this information, we can now calculate the Total Losses.
TOTAL LOSSES = COPPER LOSS + IRON LOSS + WINDAGE LOSS + CONSTANT LOSS
The values of K_c, K_i and 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.
The following MATLAB script plots the CONTOUR characteristics of the motor.
% A program for plotting the efficiency contours of the electric motors.
% The x-axis corresponds to motor speed(omega),
% and the y-axis to Torque T.
% Rahul Ramakrishnan
% Motor Speed and Torque
x = linspace(1,1000); % rad/sec
y = linspace(1,250); % 0 to 40 N.M torque.
% Motor loss constants.
K_c = 0.2; %For Copper losses
K_i = 0.008; %For Iron losses
K_w = 0.00001; %For windage losses
ConL = 20; %Constant motor losses
% MeshGrid
[X,Y] = meshgrid(x,y);
Outpower = (X.*Y); % power
B = (Y.^2)*K_c; % copper losses
C = X*K_i; % Iron losses
D =(X.^3)*K_w; %Windage losses
Inpower = Outpower + B + C + D + ConL; % Total input power
Z = Outpower./Inpower; % Efficiency
% We can now set the efficiency for contour plots.
V = linspace(0.7,0.95,15); % Mapping of efficiency points
box off
grid off
contour(X,Y,Z,V);
xlabel('Speed/rad.s^-^1'),
ylabel('Torque/N.m')
hold on
% Now plot a contour of the power output
% The array Output Power has already been calculated. We draw contours at
% 3Kw and 5Kw.
V = [8000,10000];
contour(X,Y,Outpower,V);
title('MOTOR EFFICIENCY,SPEED and TORQUE CONTOURS')
legend('Contour lines with the given Motor constants','8Kw and 10Kw Motor characteristics')RESULT
It's quite evident from the contour plot that the efficiency increases as we move inside.
The yellow lines represent the contour lines for 8Kw and 10Kw motors.