1). For a defined driving cycle, calculate the energy required for braking.
ans)
For a defined driving cycle let us take FTP75.
So we require some amount of power which is used for acceleration and also braking.
(Kinetic energy) K = 0.5*m*v^2.
where
m=mass of vehicle=1000Kg
v=Vf-Vi=final speed - initial speed=75 - 0 =75 Kmph=20.8333m/s^2
t=time from 75 to 0 speed is 10sec
Now power is required.
P=K/t
So, K = 0.5*1000*(20.833-0)^2
=217.01KJ
Now P=217.01/10=21.701KW
So energy is required for braking of the vehicle from 75 to 0 in 10 sec.
P=k/t------>K=P*t=21.701*10/3600=0.06028 KWhr
2). Why electric motor can’t develop braking torque at high speed similar to starting? How electric and mechanical brakes are coordinated?
ans)
Why electric motor can’t develop braking torque at a high speed similar to starting?
In this curve, we have two regions constant torque and constant power. At the start, the motor reaches peak torque till it reaches a certain motor speed range from where it rapidly drops. The reason for the drop is motor shaft reaches critical speed and the area where the peak torque is constant is known as the constant torque region of the motor. The region where the torque is decreased is known as the constant power region. Here power generated by the motor is at its peak.
Since the torque produced at the constant power region is very low that's why it is not possible to develop enough braking in this region.So that's why electric motor cannot develop braking torque at a high speed similar to starting.
How electric and mechanical brakes are coordinated?
Braking systems are necessary for the correct and safe operation of mechatronic systems. Most functional braking is achieved using electrical braking, which is made possible by the addition of a variable speed drive system (e.g., a variable frequency drive system). Electrical braking is efficient and ensures smooth operation and allows the designer to recover the mechanical energy. However electrical braking cannot be used as a safety braking system in the cases of emergency braking and parking braking. It is for these reasons that mechanical braking systems are still employed today, in order to act as emergency and parking brakes.
The Rationale for Braking systems in any motion control system, braking is necessary for the following reasons:
1. In order to stop the system in cases of emergency.
2. In order to ensure that the system remains stationary when not in operation.
Two types of braking systems are employed in most drive systems: electrical and mechanical. In this note, a general overview is presented of the two types of braking systems, and it concludes with a comparison of the two types.
Types of Electrical Braking Systems
There are six methods of electrical braking in drive systems, listed below:
1. Plugging:
The three-phase voltage sequence is reversed at the stator terminals of the induction motor. The equivalent form of plugging in dc drive systems is the reversal of the polarity of the armature voltage of a dc motor.
2. DC Injection Braking:
A d.c. voltage is applied to any two of the three terminals of the induction motor.
3. Eddy Current Braking:
A separate eddy current braking device is coupled to the shaft of the motor. Eddy current braking results when electromagnets are activated and they induce eddy currents in the rotating metallic disk. The eddy currents produce magnetic fluxes that oppose the flux produced by the electromagnets and thus result in a braking torque on the motor. The energy is dissipated as heat in the rotating metallic disk.
4. Dynamic Resistor Braking:
In variable frequency drive systems, when the motor is braking and acting as an induction generator, it returns power back to the d.c. link capacitor and the voltage of the d.c. link increases. A resistor can thus be used to dissipate the excessive charge on the capacitor, whereby an IGBT is controlled to proportionately switch the resistor in parallel with the
capacitor (using pulse width modulation switching). The excessive energy is thus dissipated in the resistor.
5. Regenerative Braking:
Rather than dissipating the excessive charge in the resistor, it is possible to return the excess energy back to the three-phase supply by using a dedicated inverter, placed in parallel with the uncontrolled rectifier that is used in the variable frequency drive system.
6. Sharing the DC Bus:
Another possible method of removing the excessive charge on the capacitor is to share the d.c. between a number of different variable frequency drive systems. The drive that is braking can supply energy to the drive that is motoring. In such a case, there is no need to dissipate the excess energy into a resistor or to return it to the three-phase power source.
In the first two methods, the excess energy is dissipated as heat in the motor itself. In the Eddy current method, the excess energy is dissipated as heat in the metallic disk. In the fourth method, the excess energy is dissipated as heat in the resistor. In the fifth method, the energy is returned to the ac supply. In the sixth method, the energy circulates between the different motors (from the one that is acting as a generator to the one that is acting as a motor).
In cases where it is possible to recover the energy, the following can be done:
1. It can be stored in batteries.
2. It can be stored in a flywheel (a flywheel that has an integrated permanent
magnet dc alternator).
3. It can be returned to the ac supply using a regenerative drive.
Mechanical Braking Systems:
There are four types of mechanical brakes used in drive systems:
1. Drum brakes.
2. Disk brakes.
3. Band brakes.
4. Pawl and Ratchet brakes:
The ratchet is linked to a spring-loaded friction plate. The ratchet is lifted using a solenoid.
The drum and disk brakes are either electromagnetically lifted or hydraulically lifted, but always spring applied. They must be spring applied for safety reasons, so that when the power supply is lost, the brake will apply and safely bring the load to a standstill.
Comparison between Electrical and Mechanical Braking Systems:
The following is a comparison between electrical and mechanical braking systems:
a. Electrical braking systems are used for functional braking (controlling the speed; bringing the load to a standstill). Mechanical braking systems are used for emergency stopping or for parking.
b. Electrical braking is usually very smooth and comfortable. Mechanical braking is usually rough and uncomfortable.
c. No wear results from electrical braking. Mechanical braking on the other hand causes wear on the braking components and requires regular maintenance.
d. It is possible (but not always feasible) in electrical braking systems to return the regenerated energy back to the main supply. This is not possible in
mechanical braking systems and the energy is always lost as heat, noise, and wear.
e. The electrical braking system cannot be used as a safety device. Most systems will require a mechanical braking system as a backup safety device.
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.
ans)
Matlab Code:
clc;
x=linspace(1,1000);%x axis for speed w
y=linspace(1,250);%yy axis for torque T
%now define motor losses
Kc=0.2;% it is for copper loss constant
Ki=0.0075;% Iron loss constant
Kw=0.00001;% Windage loss constant
Const=20;% Constant loss
[X, Y]= meshgrid(x,y);%meshgrid for 2-D or 3-D grids
OutputPower = (X .* Y);% it is Torque .* speed= power
TotalLoss = (Y.^2 )* Kc + Ki*X + (X.^3)*Kw + Const;%here X
% is speed and Y is torque
%Now eqution for inputpower
InputPower= OutputPower + TotalLoss;
Efficiency=OutputPower./InputPower;%define efficiency
%Now we set efficiency for which contour
Z=[0.60 0.70 0.75 0.80 0.85 0.90 0.925 0.95 0.975];
box off;%it displays outlines
grid off;%it displays grid lines
contour(X,Y,Efficiency,Z);
xlabel('Speed');
ylabel('Torque');
hold on;
Z=[8000 10000];
contour(X,Y,OutputPower,Z);
In MATLAB script First i defined 4 losses
1)Copper losses:
it is I^2 * R = Kc * T^2;
because torque is proportional to current.
2)Iron losses:
it is based on the magnetic field effect.
it is Ki * W.
3)Windage losses:
it is a mechanical loss that depends on the size or shape of the motor.
it is Kw * W^3.
4)constant losses:
it is Constant
So, total loss = Kc * T^2 + Ki * W + Kw * W^3+ Constant.
Contour Plot:
This plot is a group of efficiencies where innermost will have the highest efficiency and the height is maximum as we keep on going outside or outward the height reduces and efficiency lowest.