Week-11 Challenge: Braking

Aim: To calculate energy required for braking and to explain the Motor efficiency at required speeds by using MATLAB program.

 Objective:

1. For a defined driving cycle, calculate the energy required for braking.
2. Why electric motor can’t develop braking torque at high speed similar to starting? How electric and mechanical brakes are coordinated? 
3. Make a MATLAB program which plots contour of given motor speed, torque and efficiency values. Attach the code as a .m file attach a screenshot of all the plots.

1- For a defined driving cycle, calculate the energy required for braking.

The energy required for vehicle braking is nothing but to neutralize the Kinetic energy of the vehicle which it acquired during acceleration phase.

The kinetic energy of the body is KE = (½) * m* v^2.

We all know that the power is equal to Ratio of Work done with respect to time.

P = Work done/Time.

Where Work done = Kinetic energy of the body with respect to the speed.

The below shown Excel sheet is the driving cycle data which tells about the speed of the vehicle with respect to time.

The Below shown is the amount of energy required by the vehicle to stop/reach the required speed.

2 -Why electric motor can’t develop braking torque at high speed similar to starting? How electric and mechanical brakes are coordinated?

Answer:

Electric Motor Can’t Develop Braking Torque at High Speed Similar to Starting:

Before going to discuss why Motor isn’t developing the braking torque similar to the starting, we have to know how the starting torque is developing by the Motor (DC/AC).

Motor Operating Regions:

The above image talks about the motor operating regions like constant Toque and constant power, not every motor works on same operating regions where it depends on the power of the Motor. We can see that the Motor below the critical speed works under the constant torque, above critical speed it works under the constant power condition.

P = T*W

• T = Torque
• W = Speed
• P = Power

By increasing the speed of the motor torque is going to reduce, this condition is to maintain the constant power. We know that by supplying the variable frequency to the AC motor we can get variation in the speed and torque of the Motor. When we supply the voltage more than the rated voltage the motor doesn’t take to run at higher torque, but the motor will take the higher frequencies that leads to be decrease in the torque increase in the speed.

Decreasing in Torque and increase in the speed results the constant power condition. When the supply is given to the motor it starts under high torque during acceleration phase it reaches to the constant power phase. When the driver releases the throttle, the vehicle moves under deceleration phase, during that phase the direction of current become reverse and the Motor acts like the Generator, in this phase Negative torque is generated, and the generated torque is in counter to the direction of rotation, because of this the Generated torque is not similar to the starting torque because the motor is running in the constant power condition while the negative torque generation.

How electric and mechanical brakes are coordinated?

While applying the brakes for EV’s we need to consider so many conditions like charging of the battery during deceleration/braking phase, and simultaneously to apply mechanical brakes after battery is charged/ during charging.

To avoid massive braking of the vehicle because of generated negative torque while the motor is acting like a generator, we need a strategy to apply both electric and mechanical brakes, to avoid this kind of Problems we need to obtain the coordination between electric and mechanical braking strategy.

This can be done by using two strategies they are.

• Series strategy.
• Parallel strategy.

Series strategy:

Both electric and mechanical brakes are applied one after another. The Regenerative braking is applied first to charge the battery and then the Mechanical brakes/Friction brakes are applied. This type of control strategy is totally depending on the Energy recovery to the battery, by applying this series type we can recover/Store more energy to the battery, and offers Good Braking feel, but need some modifications in hydraulic brake line.

Parallel Strategy:

Both the electric and mechanical brakes are applied at a time simultaneously. In this system the brake pressure is applied on the brake pedal, the brakes are applied according to the movement of brake pedal. Regenerative braking system is directly coupled to the Mechanical brakes which are operated through the brake pedal. This results in low regenerative power and poor brake comfort.

By undertaking the parallel strategy Nissan developed the EDIB (Electronic Driven Intelligent Brake) system to increase the coordination between both Electric and Mechanical brakes, this leads to improve braking comfort and High Regenerative power.

During starting phase of braking movement of brake pedal make controller to create the operating percentage of both braking systems. By observing the above image, the we can say the application of friction brakes are more during high speeds and slowly decreases with respect to time, in the case of Regenerative braking system the Negative torque is increases with respect to time which leads to increase charge in the battery as well as it increases the comfort.

3- Make a MATLAB program which plots contour of given motor speed, torque and efficiency values. Attach the code as a .m file attach a screenshot of all the plots.

Before we move to make a MATLAB program, we need to know about various losses, which occur in the motor during Acceleration Phase, decides the performance of the motor. Increase in the performance can increase the efficiency of the motor.

Various losses are:

Copper losses:

Copper losses are totally depending on number of armature windings and other electrical parts which exhibits heat during running.

Copper loss = I²R

Copper loss = Kc * T²

R = Armature resistance

I = current passing through it

Iron losses:

This is due to variation in magnetic field which results in formation of eddy currents and other losses like hysteresis loss and anomalous loss. These losses are increased with respect to speed of the motor.

Ki = iron loss constant

Iron loss = Ki * W

Windage losses:

These are the losses experienced by motor due to its size and shape. Mainly this loss is due to friction.

Kw = windage constant

Windage losses = Kw * W³

Constant losses: losses other than the above explained losses.

Total loss = Copper loss + Iron loss+ Windage Loss + Constant.

Total power input = power Output + total loss

 Matlab Program:

clc
%to plot the Contour lines
% windage losses constant
Kw = 0.0002;
%copper losses constant
Kc = 0.2;
%iron losses constant
Ki = 0.001;
%constant losses
Cl = 20;
% x axis will represents the speed(rad/sec)
X = linspace(1,1000);
% y axis will be torque (Nm)
Y = linspace(1,200);
%meshing the speed and torque valuesin to 2d array
[x,y] = meshgrid(X,Y);
% copper loss
A = Kc.*(y.^2);
%windage loss
B = Kw.*(x.^3);
%iron loss
C = Ki.*(x);
%output power
Op = (y.* x);
%input power 
Ip = Op + A + B + C + Cl;
% efficiency
Z = Op./Ip;
% Efficiencies will be plotted with the references of v values
v = [0.70,0.72,0.75,0.80,0.82,0.85,0.90,0.92,0.95];
box off
grid off
contour(x,y,Z,v)
xlabel('speed rad/sec')
ylabel('Torque Nm')
hold on
v = [8000 ,9000];
contour(x,y,Op,v)

 Result: