Week-11 Challenge: Braking

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

First, a drive cycle is created in the Excel sheet and then the braking energy is calculated using the formula,

Braking energy = 0.5 * m *((V2)^2 -(V1)^2)

where m = mass of the vehicle = 500 kg

V2=final velocity

V1=initial velocity

The below drive cycle is a self-created drive cycle for a time period of 34 seconds

The braking energy required for the above-created drive cycle is = 200 kJ

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

when the motor is at rest and it has to operate from a standstill position to reach the moderate speed the higher torque is required to bring the motor to operate. whereas when the vehicle is already at high speed the capping or the limit reached by the motor operation and thus any further higher torque could not be obtained while braking. This braking operation happens in the motion which is distributed as per proportion in the brake energy and the regenerative braking.

The coordination between hydraulic and regenerative braking involves various reasons such as electromagnetic torque oscillation at low speed, the requested braking torque is higher than the power absorption capacity of the electric source(including conductors) and emergency situation. Also below a certain speed, the kinetic energy is very low, there is no enough regenerative braking, and it is proposed to use the mechanical braking system with the help of brake pedals. This coordination between the electric and mechanical braking system in an electric vehicle helps in the efficient running of the vehicle.

In mechanical braking which we usually see and experience in all vehicles, the brake pedal in the vehicle helps us to engage or disengage the brakes. Currently, braking technology has done wonders by inventing life-saving braking technologies such as ABS, EBD, BA, and many others. All these technologies play a vital role in stopping the vehicle and coming back to rest from high speed. In all these vehicles when the driver applies the brakes the brake pads come closer to the disc connected with the driving disc. The brake pads usually help inefficient braking.

There are two generic methods by which the vehicle could be braked,

• series- The regenerative braking and the mechanical braking both are applied one after the other.
• parallel-Both the regenerative braking and the mechanical braking are applied parallelly.

When the brake pedal is applied it supplies the hydraulic fluid to the disc to slow down the tires at the same time when the acceleration pedal is lifted a certain amount of electric brakes are applied. At high speed, since the braking force required is a high, large amount of hydraulic brakes are applied and once the speed is reduced and the torque value is also high so the electric brakes become predominant so the battery charging occurs efficiently. The ECU system with a feedback system to find the torque region and regenerative braking predominate in the high torque region.

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.

Motor speed= 0 to 1000 rpm

Motor torque = 0 to 700 N-m

Motor constants:

kc=0.1 (copper loss constant)
ki=0.001(iron loss constant)
kw=0.0001(windage loss constant)
cm=25(motor loss constant)

Matlab code:

clear all
close all
clc

%defining speed and torque
speed= linspace(0,1000,300);
torque= linspace(0,1000,300);

%motor constants
kc=0.1;            % copper loss constant
ki=0.001;          % iron loss constant
kw=0.0001;         % windage loss constant
cm=25;             % motor loss constant

%2-D grid for speed and torque
[x,y]=meshgrid(speed,torque);

powerout = x.*y;
 
total_loss=((y.^2).*kc)+(ki.*x)+(x.^3).*kw+cm;

inputpower = powerout + total_loss;

efficiency= powerout./inputpower;

%efficiency level for contouring
El=[0.70,0.72,0.75,0.77,0.79,0.80,0.81,0.82,0.83,0.84,0.85,0.86,0.87,0.88,0.89,0.90,0.91,0.92,0.93,0.95];

%plotting the efficiency w.r.t speed and torque
contourf(x,y,efficiency,El);
colorbar
xlabel("speed(rpm)")
ylabel("torque(N-m)")
title("Efficiency contour")

Results:

The contour plot for motor with defined constants is plotted in the defined speed vs torque region based on efficiency. It can be seen that for increasing speed and torque, the efficiency of the motor gradually decreases.The level of efficiency is depicted using colorbar.