Week-11 Challenge: Braking

QUESTION 1:- The driving cycle taken into consideration is the UDDS cycle which stands for Urban Dynamometer Driving Schedule. Although the data is taken for a total driving time for 1370s, for simplicity sake on 150s of driving time is considered. The speed which is in miles per hour has been converted to meters per second.

The equation for kinetic energy from braking is given as,

Where,

m = Mass of the vehicle in Kg.

Vi = Initial speed of the vehicle in m/s.

Vf = Final speed of the vehicle in m/s.

From the drive cycle 10 braking points were obtained. The mass is set at a constant 1500 Kg.

Therefore the total energy required for braking is 16.8 KW.

The excel sheet file of the drive cycle and braking data is provided below.

QUESTION:2-

Let us consider the torque speed characteristics.

From this curve we have two region, constant torque and constant power region. At the start the motor achieves peak torque till it reaches a certain motor speed range from where it will rapidly drop. The reason due to the drop in torque is due to the motor shaft reaching critical speed and the area where the peak torque is constant is known as the constant torque region of the motor. The region where the torque decreases is known as the constant power region as the power generated by the motor will be at it's peak.

Since the torque produced at the constant power region is very low, it is not possible to develop enough braking torque in this region. The braking torque produced increases as the speed decreases.

How are electric brakes and mechanical brakes co-ordinated?

In electric vehicles, two types of braking systems are used electric brake system and mechanical brake system.

Electric Braking System: In electrical braking, the winding of the motor is used to produce a braking torque. A braking torque is developed during the braking operation. This braking torque opposes the motion of the rotating member or shaft. This is achieved by suitably changing the electrical connections of the motor. The motor operates on a speed–torque characteristic, depending upon the method of braking employed.

Mechanical Braking System: In mechanical braking, the frictional force between the rotating parts and the brake drums produces the required braking. To achieve this, mechanical equipment, such as brake linings and brake drums, are required. Whether electrical or mechanical, the braking of the drive should stop the motor at the specified point of time and position.

The electric braking system used in EVs and HEVs is regenerative braking. This method uses the motor as a generator during braking, developing a retarding torque, which acts on the running motor and halts it. The kinetic energy and potential energy, minus the losses of the motor are returned to the mains and the motor runs at a constant speed. Thus, regenerative braking eliminates the tendency of the load to accelerate the motor.

Mechanical braking systems used can be either disc or drum brake system. Where in case of a disc brake system friction pads rub against a disc when the brake pedal is applied and the kinetic energy is converted to heat and slows the vehicle down. In a drum brake system a set of brake shoes which are the friction materials press against a drum.

Since the braking torque is very less at high speeds, mechanical brakes are applied to slow down the vehicle. As the motor reaches the constant torque region the regenerative braking system takes effect where the motor acts as a generator and the negative torque produced charges the battery pack.

QUESTION:3-PROGRAM

%A program for plotting efficency for electric motors.
%The x-axis corresponds to motor speed(W).
%The y-axis corresponds to torque(T).
% copper loss=kcT^2 ,iron loss=kiW,windage loss=kwW^3,other motor contant
% loss=c,(total loss=kcT^2+kiW+kwW^3+c)
%first ,set up arrays for range
x=linspace(1,1000);%speed,N.B.rad/sec not RPM
y=linspace(1,250);%0 to 40 N.m
%allocate motor loss constt
kc=0.2;%for copper loss
ki=0.008;%for iron loss
kw=0.00001% for windage loss
c=20;%for constt motor loss

%now make mesh
[x,y]=meshgrid(x,y);
outputpower=(x.*y);%power=T*W
B=(y.^2)*kc;%copper loss
C=x*ki;%iron loss
D=(x.^3)*kw;%windage loss

inputpower=outputpower+B+C+D+c;
z=outputpower./inputpower;
%we now set the efficencies for which the contour will be plotted 
V=[0.70,0.80,0.85,0.90,0.91,0.92,0.92,0.925,0.93];
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 power output 
%the array output power has already been calculated.
%we draw contours at 3 and 5kw
V=[8000,10000];
contour(x,y,outputpower,V);

EXPLANATION

• The speed and torque are defined as 2-Dimensional linearly spaced arrays using the linspace command.
• The constants of the motor losses are defined.
• The 2-D arrays of the speed and torque are then defined as 3-Dimensional arrays using the meshgrid command.
• The output power equation is defined as the product of speed and torque.
• The equations of the motor losses based on the speed and torque are defined.
• The input power is defined as the sum of output power plus motor losses.
• The efficiency is then defined as the ratio of output power to input power.
• In order to map the contour for different efficiency ranges an array is defined which consists of the efficiencies from 0.7 to 0.93.
• The graph is plotted using the contour command to plot the speed, torque, and efficiency.

PLOTTING

THEORY

An efficiency map of an electric motor is a contour plot of torque and speed. It is used to describe the speed and torque combinations required at different efficiency ranges. In order to model the efficiency contour we must consider the losses that occur in electric motors and these losses are,

• Iron or Core Losses: Iron losses are made of two components which are hysterisis and eddy currents. Hysteresis losses occur as we switch the electromagnets in our motor on and off, and depends on the magnetic properties of the iron material used in the motor. Eddy current losses depend on many variables, including the thickness of the laminations in the stator, the frequency at which the motor operates, the flux density and the material's resistivity. The iron losses are proportional to the square of the lamination thickness.
• Copper or Variable Losses: Copper losses are due to the resistance of the copper conductors and the current that we ask them to carry, and they lead to heating of the conductors. Of course, because copper is heavy, we want to use as little of it as possible, minimising cross-sectional area. This increases resistance and therefore increases losses and heating.
• Windage Losses: The loss caused by air friction between rotor and surrounding air is called windage losses. It occours due to air resistance to the armature. Higher the speed of the motor more wil be the windage losses.