Week-11 Challenge: Braking

1. ANSWER

Suppose we have UDDS cycle: -

We can observe in this test cycle vehicle needs to stop in given time at some distance also for this we will apply brakes. We will calculate the power require for braking operation.

the drive cycle has been attached below

• To reduce speed from 80 to 40 Kmph in 20 sec = 292 KJ
• To reduce speed from 80 to 0 Kmph in 10 sec = 560 Kj

2. ANSWER

Electric Motor Braking :
Electric motor can’t develop braking torque at high speed similar to starting, because Braking torque is applied in the opposite direction of the rotation. We can use a mechanical brake to produce a braking torque. We can also do it electrically.
Regenerative Braking:

It is a form of braking in which the kinetic energy of the motor is returned to the power supply system.Regenerative braking takes place whenever the speed of the motor exceeds the synchronous speed. This baking method is called regenerative baking because here the motor works as generator. The main criteria for regenerative braking is that the rotor has to rotate at a speed higher than synchronous speed, only then the motor will act as a generator and the direction of current flow through the circuit and direction of the torque reverses and braking takes place.

Sometimes after accelerating you can't reach the maximum torque. The reason is that no matter how much the current loop wants to increase current by increasing the PWM duty cycle of the power stage, the Vmotor is too close to Vbemf and therefore current can't increase more. This situation does not occur at the beginning of an acceleration because most of the torque is used to increase the kinetical energy of the load (and motor). But when the speed is already high and no extra load is applied it is not possible to increase the torque.

Peak Acceleration Torque and Peak Deceleration torque are not equal for Electric Vehicle Motor. Acceleration is usually higher, at low speed we get higher Torque but high speed the value of the torque gets lower. So we cannot say that the motor which is producing 7Nm Torque for Acceleration can also produce -7Nm of Braking Torque.Because we are operating the Motor above base speed so we must know that the torque value is lower. So Electric motor can’t develop braking torque at high speed similar to starting, because Braking Torque of an EV Motor is not as high as the Acceleration Torque. Therefore Regenerative braking is cannot happen so effectively that we can get all the power back to the battery.

Coordination of Electric and Mechanical Brake :
In a electric Motor the Total Brake distribution is in two aspects first one is regenerative Brake and Mechanical Brake or Classical Brake with hydraulic cylindrical system. Brake Force distribution must be uniform, If the Regenerative braking power requirement is more than the charging limit of the battery or more than regenerative capacity of the machine we must apply Mechanical brakes as well. In practical case the combination of these two is applied. So usually it is combined response of Mechanical brake and Regenerative brake. The reason for how these brake force are combined, that particular phenomena is called Brake Control Strategies. There are two Brake Control Strategies, which are

Series - In case of Series we apply Mechanical brake and Regenerative electric brake one after another. We can apply only one brake strategy at a time either apply Regenerative or apply Mechanical.
Parallel - In case of Parallel we apply both brakes simultaneously. This Particular Parallel braking is going to happen both at a time.


3. ANSWER

%A program for plotting efficiency contours for electric motors
clc
close all
clear all
x=linspace (1, 1000); % speed rad/sec
y=linspace (1,250); % torque N-m
Kc=0.2; % copper losses
Ki=0.008; % iron losses
Kw=0.00001; % windage losses
Conl=20; % constant motor losses
% Now making mesh
[X, Y]=meshgrid (x, y) ;
outputpower= (X.*Y) : % power = torque x speed
B= (Y.^2) *Kc; % copper losses
C= (x*Ki); % iron losses
D= ((X. ^3) . *Kw); % windage losses
inputpower=outputpower+B+C+D+ConL;
Z= outputpower./inputpower;
% Now setting the etticiences for which the concours will be plotted
V=[0.70,0.80,0.85,0.90,0.91,0.92,0.925,0.93];
grid on
contour (X, Y, Z, V) :
colorbar
xlabel ('Speed rad/sec')
ylabel ('Torque N-m')
title ('Plotting of efficiency Contour')
hold on

Result: