Week-11 Challenge: Braking

Q1.For a defined driving cycle, calculate the energy required for braking.

Ans

The energy required for acceleration or deceleration is nothing but the kinetic energy required by the vehicle.

Energy required for Acceleration = Energy required for Deceleration = (m*v^2)/2

But with respect to calculation of braking energy of a vehicle, we have to consider the initial velocity and the final velocity of the vehicle to determine the kinetic energy required.

Therefore,

BE= m*(u^2 - V^2) / 2

BE = Braking Energy (J) = BE / (3600*1000) (kWh)

m = Mass of the vehicle (kg)

u = Initial Velocity (m/s)

v = Final Velocity (m/s)

Now in order to calculate the Braking Energy, we need to create a driving cycle using an Excel sheet as shown below:

The Driving cycle excle sheet and the Plot of the data are as shown below;

Now, From the data or the plot we can observe the deceleration is happining 3 times so, here we have to calculate the breaking power for all three deceleration period.

From data or plot we get,

In 1st braking condition:

Initial velocity = 40 m/s

Final velocity = 20 m/s

Assuming mass of vehicle including kerb and payload weight=1575 kg

Therefore, BE = (1575*(1600-400)/2) / (3600*1000) = 0.2625 KWh      

In 2nd decelerating or braking condition:

Initial velocity = 78 m/s

Final velocity = 2 m/s

Therefore, BE = (1575*(6084-4)/2) / (3600*1000) = 1.33 KWh

In 3st braking condition:

Initial velocity = 70 m/s

Final velocity = 0 m/s

Therefore, BE = (1575*(4900-0)/2) / (3600*1000) = 1.072 KWh

Hence, total Braking power required during above driving cycle = 2.6645 KWh

Q2.Why electric motor can’t develop braking torque at high speed similar to starting? How electric and mechanical brakes are coordinated?

Ans

Electric motor can't develop braking torque at high speed similar as Starting because:

• The torque developed by an electric motor during deceleration is very low compared to acceleration period because during acceleration the motor operates at the constant torque region where the starting torque is very high.

• Whereas during deceleration period the motor operates in the constant power region where the torque developed is very low.

• During deceleration i.e when the motor is at high speeds (i.e at speeds greater than base speed), due to the field weakening effect the torque developed will be low.

• Hence it is not possible to obtain maximum regenerative braking as acceleration efficiently due to  motor limitation.

• Another limitation with respect to regenerative braking is due to the limitation impose by the battery chemistry.The discharge rate of a battery is twice that of its charging rate.

• Hence during regenerative braking the energy or power developed by motor (which is acting as generator during deceleration) cannot be used to charge the battery at the desired rate (i.e the battey cannot charged at same rate at which it dischare).

• The final limitation concerning regenerative braking is associated with the power converter. To supply the energy back to battery during regeneration from the motor we need a bi-directional power converter, which in turn increases the complexity of the system and also losses during conversion.

Practically in EV and HEV both the Mechanical and Electrical brakinng are employed.

• Coordination between electric and mechanical brakes can be basically obtained using two methods :                          
•  Series strategy and

1. Parallel strategy.

• Under series strategy, both the electric and mechanical brakes are applied one after another.With series type a blending between friction brake and regenerative brake can be obtained which lead to the vehicle deceleration corresponding to the driver's braking demand. This type of braking offers a high regeneration efficiency and good braking feel, but required several modifications in the hydraulic brake line to mechanically decouple the connection between the brake pedal and the brake pressure.
• Under parallel strategy, both the electric and mechanical brakes are applied simultaneously.In parallel systems, the brake pressure is coupled to the brake pedal, and the regenerative braking is added directly to the friction brake operated by brake pedal, which results in a low regeneration efficiency and poor brake comfort.

Electric Driven Intelligent Brake (EDIB) developed by NISSAN is one of the most efficient type of coordinated braking established between electric and mechanical brakes.After stepping on the brake pedal, the system produces natural and adequate braking force that corresponds to the operation. The amount of energy regeneration also needs to be increased as much as possible.

EDIB controls the regenerative brake and friction brake to support both of these requirements. Further, it also controls the reactive force from the pedal in order to unify the feeling when stepping down on the pedal and the sense of deceleration.

EDIB is programmed such that during the initial phase of braking i.e at high speeds both the regenerative and friction brakes are applied in a pre-determined ratio.Whereas at lower speeds the regenerative brakes are applied pre-dominantly.This is because at lower speeds, large torque is present which increases the efficiency of regenerative braking.

The figure below shows the EDIB brake control developed by NISSAN

Q3.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.

ANS;

• To plot the efficiency contours of an electric motor, we have to consider the various losses the motor is subjected to.The various losses considered are as follows :

1.Copper Losses :  These losses are due to heat produced by electrical currents in the conductors of transformer windings and other electrical devices.

• Copper Loss  is proportional to I^2 * R
• where, R is the armature resistance. Since, current is proportional to torque and by considering a constant for copper loss, we get
• Therefore,copper loss= Kc * T^2

2.Iron Losses  : These losses are due to the magnetic field effect. Iron losses consist of three components namely eddy current loss, hysteresis loss and anomalous loss. Since, these losses vary with the speed of motor and by considering a constant for Iron Losses, we get

• Iron loss = Ki*W

3.Windage Losses: These are the mechanical and friction losses experienced by the motor. They depend on the size and shape of the motor. It is  determined by the below equation,

•  Windage loss = Kw * W^3

4.Constant Loss: These are determined by various design factors. It is represented by C.

Therefore, Total loss = Kc*T^2 + Ki*W + Kw*W^3 + C.

MATLAB PROGRAM TO PLOT CONTOUR FOR THE GIVEN MOTOR SPEED IS AS BELOW;

% Plotting Efficiency Contours for Electric Motors

clear all
close all
clc

% Determining the Speed Range (rad/s)
x = linspace(1,1000);

% Determing the Torque Range (Nm)
y = linspace(1,250);

% Allocating Motor Loss Constants 

% For Copper Losses
kc = 0.2;

% For Iron Losses
ki = 0.008;

% For Windage Losses
kw = 0.00001;

% For Constant Motor Losses
ConL = 20;

% Meshing Speed and Torque ranges to create a 2D Array
[X,Y] = meshgrid(x,y);

% Output Power 
Po = (X.*Y);

% Copper Losses
B = kc.*(Y.^2);

% Iron Losses 
C = ki.*(X);

% Windage Losses
D = kw.*(X.^3);

% Input Power 
Pi = Po + B + C + D + ConL;

% Efficiency 
Z = Po./Pi;

% Setting the Efficiencies for which a contour will be plotted
V = [0.70,0.75,0.80,0.85,0.90,0.91,0.92,0.93,0.94,0.95];

% Plotting the Efficiency Contour 
box off 
grid off
contour(X,Y,Z,V)
xlabel('Speed (rad/s)')
ylabel('Torque (Nm)')
hold on 
% Plotting contours of Output Power at 8kW and 10kW
V = [8000,10000];
contour(X,Y,Po,V)

The Contour Plot is shown below:-

Here, The starting 6 or 7 line represent the constant torque region or the starting torque which is actually high and the circle like egg is the constant torque region which is the operating region of motor.

The circle represents the operationg condition i.e this is the region where, dTL/dWm > dTm/dWm.