Week-11 Challenge: Braking

Ans 1 : Key parameters required for calculating braking energy

Vehicle Mass ($m$: The total mass of the vehicle in kg.

Initial Speed ($v_{\text{initial}}$): The speed of the vehicle before braking in meters per second (m/s).

Final Speed ($v_{\text{final}}$): The speed of the vehicle after braking in meters per second (m/s).

Energy required for Braking (E) = 1/2*m*(Vinitial - Vfinal)^2

Drive cycle : UDDS ( urban dynamometer drive schedule)

No. of Braking	Initial Velocity (m/s)	Final Velocity (m/s)	Mass of Vehicle (kg)	Energy required for braking (J)
1	8.85	6.65	1500	3630
2	10.1	7		7207.5
3	11.17	10.95		36.3
4	11.08	11		4.8
5	11.48	11.13		91.875
6	11.62	11.48		14.7
7	13.72	13.54		24.3
8	13.7	13.18		202.8
9	13.85	13.32		210.675
10	14.48	0		157252.8
			Total Energy required for braking (J)	168675.75

Ans 2 : Refer "Ans 2" graph image

From above curve , we see the 3 regions first one is constant torque region , constant second is power region , third is breakdown torque region. So at start of motor, the motor achieved the peak torque till the moment at it reaches the certain speed and after that it start rapidly drop.

The reason of dropping the torque of motor because the motor shaft reaching the critical speed. The area motor achieve the peak torque and constant is called constant torque region.

The back EMF increases with speed. At very high speeds, the back EMF can become nearly equal to the supply voltage, making it difficult for the motor to generate additional braking torque.

Electric Brakes and Mechanical brakes are coordinated in the following manner :

Refer Ans 2 (B) for graphical representation of Electrical and Mechanical braking

In an Electric vehicle, at start of the vehicle the friction braking gives the 60 % contribution and regenerative braking give 40 % contribution and speed of vehicle start reduce. After some of the braking time the regenerative braking in role and provide the some of power supply back to the source . As we know the braking torque is very low at high speed at start so friction braking apply brake and slow the speed of vehicle and after that the regenerative braking in which motor act as generator and supply toward the source by producing negative torque

Ans 3 : Please refer the .m file for MATLAB code

clear all
close all
clc

%% arrays
% speed arrays
w = linspace(1,1000);  % in Rad/s
% torque arrays
t = linspace(1,250);   % in N-m

%% 3D array create
% Meshgrid
[W T] = meshgrid(w,t);

%% losses coefficients
% copper losses
Kc = 0.23;
% iron losses
Ki = 0.009;
% windage losses
Kw = 0.000020;
% constant motor losses
K = 20;

%% Losses Calculations
Cu_L = (T.^2)*Kc; %copper losses = (Torque^2*copper loss coeff)
I_L = W.*Ki;  % iron losses = (speed*iron loss coeff)
W_L = (W.^3)*Kw  % windage losses = (speed^3*windage loss coeff)

%% calculation efficiency
O_P = W.*T;  % output power = (speed*torque)
I_P = (Cu_L)+(I_L)+(W_L)+(O_P)+K;  % Input power = (copper losses+iron losses+windage losses+output power)*Constant motor losse
Eff = O_P./I_P;  %efficiency = (output power/input power)

% Mapping contour of efficiency
%n = [0.7,0.8,0.85,0.9,0.91,0.92,0.925,0.93];
n = linspace(0.7,0.95,10);
contourf(W,T,Eff,n);
title('Speed Torque Characteristics','color','r')
xlabel('Speed [Rad/S]')
ylabel('Torque [Nm]')
contourcbar

code explanation : 

close all: Closes all figure windows.
clc: Clears the command window.
w = linspace(1,1000): Creates a vector w with values linearly spaced from 1 to 1000 (speed in rad/s).
t = linspace(1,250): Creates a vector t with values linearly spaced from 1 to 250 (torque in N·m).
meshgrid(w, t) : Creates two 2D arrays W and T . 
Kc: Copper loss coefficient (related to the resistance of the motor windings).
Ki: Iron loss coefficient (related to magnetic losses in the motor core).
Kw: Windage loss coefficient (related to losses due to air resistance in the motor).
K: A constant term representing other motor losses.
Calculating Losses with specified formulas 
Cu_L = (T.^2)*Kc: Copper losses are proportional to the square of torque, multiplied by the copper loss coefficient.
I_L = W.*Ki: Iron losses are proportional to speed, multiplied by the iron loss coefficient.
W_L = (W.^3)*Kw:Windage losses are proportional to the cube of speed, multiplied by the windage loss coefficient.
O_P = W.*T: Output power is the product of speed and torque.
I_P = (Cu_L) + (I_L) + (W_L) + (O_P) + K: Input power is the sum of the losses (copper, iron, windage), output power, and constant motor losses.
Eff = O_P./I_P: Efficiency is calculated as the ratio of output power to input power.
n = linspace(0.7,0.95,10): Creates a vector n with 10 linearly spaced values between 0.7 and 0.95. These are the contour levels for the efficiency plot.
contourf(W,T,Eff,n): Plots a filled contour plot of efficiency, where W is the speed axis, T is the torque axis, and Eff is the efficiency at each (speed, torque) pair.
title('Speed Torque Characteristics','color','r'): Adds a title to the plot with red color.
xlabel('Speed [Rad/S]'): Labels the x-axis as “Speed [Rad/S]”.
ylabel('Torque [Nm]'): Labels the y-axis as “Torque [Nm]”.
contourcbar: Adds a color bar to the contour plot to indicate the efficiency levels.

Please refer image for graph 

explanation : From the plot it can be observed that the speed and torque varies as efficiency increase. This is occur due to motor losses. As we see the dark blue region represent the 70% efficiency and the yellow region represent 93% efficiency. Ex.- when it is at 70 % efficiency the torque reaches its peak value 250 Nm. it means the when the vehicle is at very low speed the torque decrease as the speed increase for same efficiency. and at 93% efficiency the speed and torque required much less compared to an efficiency 70%. Greater the motor efficiency lower speed torque combination is required to overcome the motor losses.