Week-11 Challenge: Braking
I am ARUNKUMAR P,
Student of live Ev batch-16
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Question.No.1:For a defined driving cycle, calculate the energy required for braking.
Braking:
The function of the braking system is to retard the speed of the moving vehicle or bring it to rest in the shortest possible distance whenever required. The vehicle can be held on an inclined surface against the pull of gravity by the application of a brake. Brakes are mechanical devices for increasing the frictional resistance that retards the turning motion of the vehicle wheels. It absorbs either kinetic energy or potential energy or both while remaining in action and this absorbed energy appears in the form of heat. While moving down a steep gradient the vehicle is controlled by the application of brakes. In this case, brakes remain in action for a longer period making it imperative to dissipate the braking heat to the atmosphere as rapidly as possible.
The brakes must be capable of decelerating a vehicle at a faster rate than the engine is able to accelerate it. Normally brakes have to absorb three times the amount of engine horsepower energy in its equivalent form. Kinetic energy is the force that keeps the vehicle moving. This energy is provided by the engine in order to accelerate the vehicle from a standstill to desired speed. Kinetic energy is dissipated as heat by the brakes during application of breaks (Fig. 28.1). The kinetic energy of a vehicle during braking is given by
`KE=1/2mV^2`
where, m=mass of the vehicle
V=velocity of vehicle
`KE=0.5*m*(u^2-v^2)`
u=initial velocity
v=final velocity
Procedure:
- create an excel spreadsheet containing data of drive cycle .which has first column as time and second column as velocity.
- Now open an mfile and run the following code
-
clear all clc mass = 1000; % mass of the vehicle in kg drivecycle=xlsread("drivecycle.xlsx"); t = drivecycle(:,1); % time period v = drivecycle(:,2); % velocity data in m/s for i = 1:(length(drivecycle)-1) if v(i)>v(i+1) BE(i) = 0.5*mass*(v(i) - v(i+1))^2; % BE stands for brake energy end end fprintf('Total Brake Energy is %d',sum(BE)) fprintf('Average Brake Energy is %d',mean(BE)) % plotting the results figure(1) subplot(2,1,1) plot(t,v,'linewidth',3,'color','g') xlabel('Time(s)') ylabel('Velocity(m/s)') title('Drivecycle') subplot(2,1,2) plot(BE,'linewidth',3,'color','r') xlabel('Time(s)') ylabel('Braking Energy') - after running the above code in command window it will give the average and total brake energy applied throughout the all instant of values that we have given.
- in the above graph we can see the brake that applied according to the drive cycle is showned.
Result:
Hence the brake energy is calculated and ploted in a graph according to drive cycle.
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Question.No.2:Why electric motor can’t develop braking torque at high speed similar to starting? How electric and mechanical brakes are coordinated?
Answer:
The electric motor can't develop braking torque at high speed similiar to starting because once the electric supply is pulled off from the motor,it acts as a generator and due to the flux weakening effect of the electric generator,the generator produces low torque as expected.
Regerative Braking:
- Regenerative braking is an energy recovery mechanism that slows down a moving vehicle or object by converting its kinetic energy into a form that can be either used immediately or stored until needed.
- In this mechanism, the electric traction motor uses the vehicle's momentum to recover energy that would otherwise be lost to the brake discs as heat. This contrasts with conventional braking systems, where the excess kinetic energy is converted to unwanted and wasted heat due to friction in the brakes, or with dynamic brakes, where the energy is recovered by using electric motors as generators but is immediately dissipated as heat in resistors. In addition to improving the overall efficiency of the vehicle, regeneration can significantly extend the life of the braking system as the mechanical parts will not wear out very quickly.
- The most common form of regenerative brake involves using an electric motor as an electric generator. The working of the regenerative braking system depends upon the working principle of an electric motor, which is an important component of the system.
- The electric motor gets activated when some electric current is passed through it. But, when some external force is applied to activate the motor (during the braking), then it behaves as a generator and generates electricity. This means that whenever the motor runs in one direction, the electric energy gets converted into mechanical energy, which is then used to accelerate the vehicle, and whenever the motor runs in the opposite direction, it performs functions of a generator, which then converts mechanical energy into electrical energy, which makes it possible to utilize the rotational force of the driving axle to turn the electric motors, which results in regenerating electric energy for storage in the battery and simultaneously reducing the speed of the car with the regenerative resistance of the electric motors. This electricity is then used for recharging the battery.
- Coordination of Electric and Mechanical Brake :
In an electric Motor, the Total Brake distribution is in two aspects which are Electrical Brake which is called Regenerative Brake, and Mechanical Brake or Classical Brake with a 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 the regenerative capacity of the machine we must apply Mechanical brakes as well.
- In a practical case, the combination of these two is applied. So usually it is a combined response of Mechanical brake and Regenerative brake. The reason for how this braking force is combined, that particular phenomena are 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 Regeneratively or apply Mechanical brake. The Series Brake Control Strategy has a further classification, that is based on whether we want the comfort of driver or Brake behavior compared to energy efficiency.
- Parallel - In the case of Parallel, we apply both brakes simultaneously. This Particular Parallel braking is going to happen both at a time. Electric Driven Intelligent Brake (EDIB) is a
combination of Hydraulic brake or Mechanical brake along with a Regenerative brake. It is a Parallel braking technique. - The Control criteria such that at low speed use regenerative braking higher and at high speed or during the initial braking period we actually
combine the both.
Result:
Hence regenerative braking can't used when vehicle moves in a high speed. Because, in the coasting mode the torque produced by the motor will be low to the torque that we expected. in this case the user has to apply brake in parallel coordination to stop the vehicle.
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Question.No.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
Solution:
To plot contour line for a given Motor. A Motor have various types of losses happening inside. The various types of losses are:
Copper Losses - `I^2R` , where R is the Armature resistance and we say that Torque is proportional to Current, this copper loss can be
consider as .where is to take care of resistance of brushes and also take care of ux and its effect.
Copper losses =`K_c*T^2`
Iron Losses - Iron Losses are nothing but Magnetic losses, which is denoted as , where factor is based on the Magnetic field effect. For Permanent Magnets have constant Magnetic field and this Iron losses is based on the value of and this are going to change according to the Speed. At Various Speed have different value of Back EMF and that is going to affect the Iron Losses.
Iron Losses =`K_i*omega^2`
Windage Losses - This is a Mechanical or friction of Windage losses. This is based on size of shape of Motor and the amount of windage losses that the motor is going to experience.
Windage Losses =`K_w*omega^3`
Constant Losses - This is denoted as C.
Total Losses - The Total Losses are considering the above four losses and denoted as,
Total losses=`K_c*T^2`+`K_i*omega^2`+`K_w*omega^3`
All Motors are model as per the above equation of losses. This above equation is not only applicable for Permanent Magnet DC Motor,
but other motors as well like induction motor.
MATLAB program for calculating the losses of the EV motor:
M.file:
clear all
close all
clc
w = linspace(0,1000);
T = linspace(0,250);
% motor constants
kc = 0.2; % for copper loss
ki = 0.008; % for iron loss
kw = 0.00001; % for windage loss
conL = 20; % for constant motor losses
% making mesh
[X,Y] = meshgrid(w,T);
OutputPower = (X.*Y); % Torque x Speed = Power
B = (Y.^2)*kc; % copper loss
C = (X.^3)*kw; % winding loss
D = X*ki; % Iron loss
InputPower = OutputPower + B + C + D + conL;
Z = OutputPower./InputPower;
% Set the efficiencies for which the contours will be plottted
V = [0.70 , 0.75 , 0.80 , 0.82 , 0.85 , 0.87 , 0.89 , 0.90 , 0.92];
box off
grid off
contour(X,Y,Z,V);
xlabel('Speed(rad/s)')
ylabel('Torque(N.m)')
hold on
V = [8000,10000];
contour(X,Y,OutputPower,V);
graphical representation:
Result:
Thus the contour plot for motor efficiency was ploted successfully.
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