Week-11 Challenge: Braking

1. For a defined driving cycle

Answer: 

The special feature of electric vehicles and hybrid vehicles is their ability to absorb, store and reuse the braking energy. A successfully designed braking system for a vehicle must always meet two distinct demands. While applying the sudden brake, the vehicle must come to rest in the shortest possible distance and at the same time, the vehicle must have control over the vehicle’s direction. The first one requires that the vehicle braking system be able to provide enough braking torque on all wheels. The second demand requires braking force to be distributed on all the wheels equally.

Here, the braking torque required is much larger than the torque that an electric motor can produce. So in order to bring the two demands success in EV and HEVs, both the mechanical braking system and electrical braking system should consider as major roles.

BRAKING :-

Braking is a process which applies energy to decrease the speed of a moving vehicle.When brake is applied in a moving vehicle the kinetic enegy possed by the moving vehicle get opposed by various factor like , air drag , rolling resistance etc , but it takes a lot time and much distance to come to rest which is not a practical scenario. So to stop the moving vehicle with lesser time braking is used which is opposit pheomena to accleration.In braking care the kinetic energy is converted either in the form of heat or we can use to recharge the battery back in case of  EV & HEV application. It includes rapid stopping of the electric motor, holding the stopping of the electric motor, holding the motor shaft to a specific position, maintaining the speed to the desired value, or preventing the motor from overspeeding.

A regenerative braking system is the electrical braking of a vehicle using motors as brakes. The total kinetic energy which is to be utilized by the wheels travel a certain distance was being stopped by for some reason by the braking system then the accumulated energy near the power train is directed to travel back to the energy storage through the motor. So here the motor acts like a generator that stores mechanical energy into electrical energy in the battery. This stored energy can be reused by the electric motor for vehicle propulsion. This regenerative system improves motor efficiency by simply converting kinetic energy into electrical energy which is to be generated as heat energy during frictional braking.

Ex:-

If a 1500kg vehicle traveling at 70km/h has 300kJ of kinetic energy, which drops to zero as the vehicle comes to rest. If we can utilize this recovered it would be enough to propel the same vehicle for 1.8km at 70km/h.

This will significantly recover energy which leads to increased vehicle range.

Whenever a force is applied to a vehicle's brakes, there is work done due to the friction between the brakes and the wheel. This reduces the vehicle's kinetic energy, slowing it down and causing the temperature of the brakes to increase.

Decelerating from very high speeds too quickly can cause several problems:

• The brakes could overheat and the material they are made from could shatter. This will lead to the driver losing control. 
• Very high temperatures near the tires could cause the tires to explode. This will lead to the driver losing control.
• A high deceleration needs a high force (Newton's second law), and these high forces can seriously injure the driver. 

calculate the energy required for braking.

We know that the equation of kinetic energy from breaking is given as

E = (1/2) *m *(Vi – Vf )^2

Where,

M – mass of the vehicle in Kg

Vi – initial speed of the vehicle in m/s

Vf – final speed of the vehicle in m/s

Form the drive cycle 15 breaking points were obtained. The mass is set at a constant 2000. The energy required for breaking is given below.

no.of breaking	initial speed (Vi) m/s	final speed (Vf) m/s	mass Kg	breaking energy
1	8.25	7.5	2000	562.5
2	10	7	2000	9000
3	9	5.8	2000	10240
4	11	10	2000	1000
5	11.5	9.2	2000	5290
6	10	7.7	2000	5290
7	12	8.3	2000	13690
8	12.5	0	2000	156250
9	13.5	10.5	2000	9000
10	15	11.2	2000	14440
11	11.4	9.2	2000	4840
12	15.3	11	2000	18490
13	11.2	10	2000	1440
14	13.4	12	2000	1960
15	15	0	2000	225000
		total energy in KW		476.4925

 The total energy used for 15 time braking is 476.495

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

Answer:  

 

 Explanation :  Discussing each term Starting torque and braking torque ..

 

Starting Torque is the torque an electrical motor develops when starting at zero speed.

A high Starting Torque is more important for application or machines hard to start - like positive displacement pumps, cranes etc. A lower Starting Torque can be accepted for centrifugal fans or pumps where the start load is low or close to zero.

Pull-up Torque

The Pull-up Torque is the minimum torque developed by an electrical motor when it runs from zero to full-load speed (before it reaches the break-down torque point).

When a motor starts and begins to accelerate the torque in general will decrease until it reach a low point at a certain speed - the pull-up torque - before the torque increases until it reach the highest torque at a higher speed - the break-down torque - point.

The pull-up torque may be critical for applications that needs power to go through some temporary barriers achieving working conditions.

 

Break-down Torque

The Break-down Torque is the highest torque available before the torque decreases when the machine continues to accelerate to working conditions.

 

Full-load (Rated) Torque or Braking Torque

The Full-load Torque is the torque required to produce the rated power of an electrical motor at full-load speed.

In imperial units the Full-load Torque can be expressed as

T =  5252 Php / nr                         (1)

where

T = full load torque (lb ft)

Php = rated horsepower

nr = rated rotational speed (rev/min, rpm)

In metric units the rated torque can be expressed as

T = 9550 PkW / nr                              (2)

where

T = rated torque (Nm)

PkW = rated power (kW)

nr = rated rotational speed (rpm)

 

Example - Electrical Motor and Braking Torque

The torque of a 60 hp motor rotating at 1725 rpm can be calculated as:

Tfl = 5252 (60 hp) / (1725 rpm)

= 182.7 lb ft

Accelerating Torque

Accelerating Torque =  ( Available Motor Torque - Load Torque )

 

Two types of braking systems are employed in most drive systems: Electrical and Mechanical.

I 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 excessive 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 excessive energy into a resistor or to return it to the three phase power source.

 

Heat dissipation comparision above methods :

In the first two methods, the excessive energy is dissipated as heat in the motor itself.

In the Eddy current method, the excessive energy is dissipated as heat in the metallic disk. I

n the fourth method, the excessive energy is dissipated as heat in the resistor.

In the fifth method the energy is returned to the ac suppy.

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 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 in 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 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.

 

Tabular form of difference : 

Sr No.	Mechanical Braking	Electrical braking
1	Low efficient	High efficient method.
2	The energy of the rotating parts is wasted as heat in friction.	The energy of the rotating parts can be converted to electrical energy which can be utilized or returned to the supply mains.
3	It require frequent maintenance like adjustment of brakes, replacement of brake linings. They are prone to tear & wear.	It require very little maintenance because of absence of mechanical equipments.
4	Depending upon the conditions the braking may not be very smooth.	Braking is very smooth, without snatching.
5	This braking is applied to hold the system at any position.	It cannot produce holding torque. It requires electrical energy for operation.
6	Brake drum, brake shoes and brake linings are needed.	Equipment of higher rating than the motor rating may. be needed in certain types of braking.

 

Electric and mechanical brakes are coordination:

There are two brake control strategies namely, SERIES & PARALLEL.

SERIES:

Brakes will be applied one after the another so which they are connected in series. Either one brake stragety at a
time either Regenerative braking or Mechanical braking at a time.

PARALLEL:

In case of parallel both the brakes are applied simultaneously, Both at a time.

 

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.

Answer:  

 

To plot contour line for a given Motor. A Motor have various types of losses happening inside. The various types of losses are

A : Copper Losses - ( I^2 *R) where R is the Armature resistance and we say that Torque is proportional to Current, this copper loss can be consider as (Kc. T^2) where kc is to take care of resistance of brushes and also take care of flux and its effect.

          Copper losses =Kc *T^2

B: Iron Losses - Iron Losses are nothing but Magnetic losses, which is denoted as ( Ki*w) , where k; 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 Ki 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 = Ki*w

C : 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 = Kw *W^3

D : Constant Losses - This is denoted as C.

         Total Losses - The Total Losses are considering the above four losses and denoted as,

F : Total Losses = Kc*T^2  + Ki - w + Kw *W^3 + C

 

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

 

% A MATLAB PROGRAMM FOR PLOTTING CONTOUR PLOT OF MOTOR SPEED, TORQUE & Efficiency

 

clear all

 

close all

 

clc

 

% SPEED ARRAY

 

w= linspace(0,1000);

 

% TORQUE ARRAY

 

T= linspace(0,300);

 

% LOSS COEFFICIENTS

 

kc= 0.20; % COPPER LOSSES

 

ki = 0.008; % IRON LOSSES

 

kw = 0.000020; % WINDAGE LOSSES

 

% MESHGRID

 

[X, Y] = meshgrid(w,T);

 

copper_loss = (Y.^2)*kc;

 

iron_loss = X*ki;

 

windage_loss = (X.^3)*kw;

 

K = 20;

 

 

output_power = X.*Y;

 

input_power = (copper_loss)+(iron_loss)+(windage_loss)+(output_power)+K;

 

Eff = (output_power)./(input_power);

 

n = linspace(0.7,0.95,10);

 

contourf (X, Y, Eff,n);

 

title('SPEED TOROUE CHARACTERISTICS' )

 

xlabel ( 'SPEED rad/s')

 

ylabel ( 'TORQUE N-m')

 

Explanation for above Program:

Step-1: Assign Input of speed and Torque in an array using 'linspace' Command and allocate Constant Values for four losses (Copper Loss, Iron Loss, Windage Loss and Constant Motor Loss)

Step-2 : Mesh the Speed and Torque values using 'meshgrid' command.

Step-3 : Calculate the Output Power by multiplying Speed and Torque values.

Step-4 : Calculate Copper Loss, Iron Loss, Windage Loss using those formulas.

Step-5 : Calculating Inputpower by adding Output Power and Four Losses(Copper Loss, Iron Loss, Windage Loss and Constant Motor Loss)

Step-6 : The ratio of Output Power and Input Power we can get the Efficiency.

Step-7 : Now Set the Efficiencies for which a contour will be plotted. The range of contour efficiency is to plot from 70% to 94% efficiency. V is the Mapping of those efficiency points.

Step-8 : Plot the Contour Efficiency using 'contour' command and plot Speed, Torque,Efficiency and V.

                           

                                            Contour : contour(z) creates a contour plot containing the isolines of matrix 2, where Z contains height values on the x-y plane. MATLAB automatically selects the contour lines to display. The column and row

indices of Z are the x and y  coordinates in the plane, respectively. contour(X,Y,)' specifies the x and y coordinates for the values in Z.

Step-9 : Now plot a contour of Power Output. Plotting contour of 8kw and 10kw of a Motor.

Step-10 : Run the Program and get the Motor Speed-Torque Efficiency Contour Plot.

 

 

OUTPUT

 The above Plot shows the Motor Speed-Torque Efficiency Contour Plot. A Point is denoted in innermost contour line is group of efficiencies which is highest value of the efficiency as 94.54%, because where the

height is maximum and another point is denoted outermost contour line is contain group of efficiencies which is the lowest value of efficiency as 70.97 %, because where the height reduces.

Increasing efficiency  order  as going inward and Decreasing efficiency order as going outward.