Week-11 Challenge: Braking

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

Drive Cycle:

• Drive cycle is a test cycle for a vehicle with various input conditions which will be done for all the newly produced and about to launch vehicles.
• This kind of drive cycle is done to test the emission, fuel efficiency, with various input of throttle and various hill climbing angles in case of an IC engine.
• In case of an electric vehicle the test is done with various input of throttle and various hill climbing angles to analyze power efficiency, range of vehicle and the state of charge that is left after running the test in battery.
• There are many defined Drive Cycles but, we have taken the 'Urban Dynamometer Driving Schedule(UDDS)' as our reference drve cycle.

Urban Dynamometer Driving Schedule(UDDS):

UDDS stands for Urban Dynamometer Driving Schedule,and refers to a United States Environmental Protection Agency mandated dynamometer test on fuel economy that represents city driving conditions which is used for light duty vehicle testing.

It is also known as U.S. FTP-72 (Federal Test Procedure) cycle, LA-4 cycle, in Sweden as A10 or CVS (Constant Volume Sampler) cycle and in Australia as the ADR 27 (Australian Design Rules) cycle.

Though it was originally created as a reference point for fossil fueled vehicles, it is also used to give estimates on how many miles an electric vehicle will travel on a single charge.

Braking Energy:

In acceleration the velocity changes from initial to final velocity value, While braking the velocity change from final to initial value or 0.

So, in this case of braking, The kinetic energy is given by,

Braking Energy = 1/2* M * (Vf - Vi)^2

where, M = mass of vehicle

Vi = Initial velocity

Vf = Final velocity

To calculate the braking energy required in a difined driving cycle, we use the above formula in UDDS drive cycle data.

The mass of vehicle is assumed is 2000 kg and total time period of this UDDS drive cycle is 1370 sec

MATLAB code to calculate the braking energy:

clear all 
close all
clc

mass = 2000;
g = 9.8;
time_period = 1370;
filename = 'UDDS.xlsx';
num = xlsread(filename);
time = num(1:time_period,1);
velocity = num(1:time_period,3);
instantaneous_braking_energy = zeros(1,time_period);

for i = 1:(length(velocity)-1)
    if velocity(i)>velocity(i+1)
        instantaneous_braking_energy(i) = 0.5*mass*0.001*((velocity(i))^2-(velocity(i+1))^2)/3600;
    end
end

figure
subplot(1,1,1)
plot(time,velocity)
xlabel('Time(sec)')
ylabel('velocity(m/s)')
title('UDDS drive cycle data')

figure
plot(time,instantaneous_braking_energy,'r')
xlabel('Time(sec)')
ylabel('Braking energy(kwh)')
title('Braking energy values for UDDS drive cycle data')

fprintf('the total braking energy for the UDDS drive cycle of %d seconds is %.3f Kwh',time_period,sum(instantaneous_braking_energy))

Output:

Result:

The total braking energy for the UDDS drive cycle of 1370 seconds is 15.109 Kwh

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

Electric Motor Braking:

• The electric motor braking is also known as Regenerative Braking, which is an energy recovery mechanism that slows down a moving vehicle orobject by converting its kinetic energy into a form that can be either used immediately or stored until needed.
• When we press the brake pedal, we cutoff the power supply to the motor and now the motor will be in rotation due to its inertia
• As this rotation is happening inside the magnetic field a Back Emf is generated which produces a negative Torque and that is used as the brakes to reduce the vehicle speed.
• Also the current flows in reverse direction, Using a Bi-directional Inverter we can store the current back into the battery.
• This process of Braking by the effect of negative Torque produced and where the motor acts as a generator in producing the current out of the motor's rotation and storing it back into the battery is called Regenerative braking.

Speed and Torque Characteristics of a DC motor:

• From the above figure we can see the speed and torque behaviour of a electric motor
• During the initial low speed the Torque is high which is the constant Torque region. As speed builds up, after the Critical speed point the torque gradually decreases which is known as constant Power region.
• So during high speeds the motor will beat constant power region and while braking, due to high speed of the motor the negative Torque produced will be less.
• But while starting the motor, the speed will be very less and at this stage the motor will be operating in constant Torque region and this is the reason for High accelerating torque during the start of motor
• From this we can say that the Braking torque produced at High speed motor operations will be lesser than the starting Torque of an Electric motor.
• Also there are other reasons like Battery Limitations due to which we cannot get High braking torque.
• The rate of discharge is always higher than the rate of charging due to the battery's chemical compositions.

Combination of Electric and Mechanical brakes:

• Motor of electric vehicle is able to be used to brake regeneratively, so braking energy can be recycled.
• Braking stability of electric vehicles with electro-hydraulic hybrid braking system can be influenced by braking force distribution between hydraulic braking force and regenerative braking force
• Yes electric cars have regenerative braking and this does substantially reduce the need for hydraulic braking.
• But there are some limitations like, the maximum deceleration with the engine alone is far smaller than with brake assistance. So you still need a nordinary Frictional brake system to make the car stop instantly if required.
• When the regenerative capacity is less than the actual braking required, mechanical brakes are applied at those times.

Brake control Strategy:

• When the brake pedal is pressed the control signal is sent to both the Hydraulic cylinder and power converters for producing both Frictional Braking and Regenerative braking.
• There are two ways of applying the combination of brakes and they are classified as,

1. Series-The Regenerative braking and Frictional braking are applied one after another.

2. Parallel-The Regenerative braking and Frictional braking are applied at the same time

• The series broking and parallel braking is chosen based on two parameters, one is Driver's comfort and other one is based on the Efficiency.
• This parallel combined braking Technology is further developed as EDIB, which is Electric Driven Intelligent Braking.

Electric Driven Intelligent Braking System.

• The above figure shows the Electric driven intelligent braking system, where we see that both the Regenerative Brakes and the Frictional brakes are applied at the sametime.
• After braking the speed reduces from higher speed values and at this moment the % of composition of the combined braking changes.
• So after brake application the speed is reduced to low speeds. Here Torque will be high and the motor will be operating at constant Torque region.
• Now at this point the % of Frictional brake application is reduced and Regenative Braking is Increased because at this point more Negative Torque can be generated and thus higher electric power generated out of this braking can also be fed back to the battery.
• Thus using this technology in Braking an electric driven vehicle certainly increases the stopping power and also contributes to higher efficiency figures.

Question 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.

Contour plot

• A contour plot is a plot which represents heights in Two-dimensions figure.
• A matlab program is done to plot the contour of the Electric DC Motor's efficiency at given speed and torque ranges of the motor.

• The losses that happens is taken into consideration while calculating the efficency and they are,
  1. Copper Loss
  2. Iron loss
  3. Windage Loss
  4. Constant motor Loss

1. Copper Loss:

• Copper loss is given by ((i)^2*R), where I-Current ; R-Resistance
• As the Torque is proportional to the current,

((i)*2*R) = KC*(T)^2

• Where Kc - Copper loss constant that takes care of the Resistance in Brushes and Magnetic Flux Effects.

2. Iron Loss:

• Iron losses are simply due to the Magnetic Losses that is happening and is given by

Iron loss = (Ki*w) where, Ki-Iron loss constant; w - speed of the motor

• Ki depends on the effect of magnetic field. The Magnetic field changes with change in the speed values and at various speeds the Back Emf changes accordingly.

3. Windage Loss:

• The Windage loss is based on the Size & Shape of the motor and the amount of windage that the motor experiences and is given by

Windage loss = (Kw*(w)^3) Where, Kw - Windage loss constant; w - speed of the motor.

4.Contant motor loss:

• In all the motor there are some constant losses happening and is given as C.
• The Total loss = copper Loss + iron loss + windage loss + Constant motor loss.
• Output Power = Input Power - The Total losses.
• Finally the Efficiency of the DC Motor is given by, Efficiency = Output Power/Input Power.

MATLAB Code:

clear all
close all
clc

%Input quantities
V = linspace(1,1200); % velocity array (assumed)
T = linspace(1,500);  % Torque array (assumed)
Kc = 0.1;             % Copper loss constant
Ki = 0.006;           % Iron loss constant
Kw = 0.0001 ;         % Windage loss constant
Cl = 20;              % motor loss constant

% calculations
[A,B] = meshgrid(V,T); % array to matrix

% power calculations
P_out = (A.*B);        % motor output power
P_c = (B.^2)*Kc;       % Copper loss power
P_i = A.*Ki;           % Iron loss power
P_w = (A.^3)*Kw;       % windage loss power
P_in = P_out + P_c + P_i + P_w + Cl; % motor input power

% Efficiency Calculation
Efficiency = P_out./P_in;

% Contour Plotting
N = [0.75,0.8,0.85,0.86,0.87,0.88,0.89,0.9,0.91,0.92,0.93,0.94,0.95]; % Contour Mapping

figure(1)
box off
grid off
contourf(A,B,Efficiency,N);
colorbar
title('Torque-Speed-Efficiency contour of an induction motor')
xlabel('Rotor Speed in RPM')
ylabel('Induced Torque in N-m')
zlabel('Motor Efficiency')

figure(2)
surf(A,B,Efficiency)
colorbar
title('Torque-Speed-Efficiency Surface contour of an induction motor')
xlabel('Rotor Speed in RPM')
ylabel('Induced Torque in N-m')
zlabel('Motor Efficiency')

Output: