Week-11 Challenge: Braking

Aim #1 :   For a defined driving cycle, calculate the energy required for braking.

Theory :

For a defined driving cycle, calculate the energy required for braking. At first, the drive cycle is determined in a Table and a plot is made. Using the plot, the values from the braking regions of the vehicle are taken. The plot of the drive cycle is attached as .xlsx file

The braking performance is one of the most important concerns in vehicle safety. In urban area driving, a significant amount of energy is consumed in braking. Electrification of vehicle makes it feasible to recover some of the energy lost in braking.

A hybrid braking system was developed to ensure the efficient braking of vehicle and recovering the braking energy. The design and control objectives of such hybrid braking system is to ensure sufficient braking force  to quickly reduce the vehicle speed, properly distributing the brake force to front and rear axle, and also to recover as much braking energy as possible. So understanding of braking energy of a driving cycle is important.

 

GOVERNING EQUATION:

Braking Energy = 1/2 x  m x (Vi - Vf ) ^2

E = 0.5 * m * (V,  - V. )^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.

From the drive cycle 10 braking points were obtained. The mass is set at a constant 1150 Kg. The energy required for braking is given in the table below.

 

 

Result : Hence , the total energy required for braking is 129318/1000 = 129.318 kj

 

 

============================================================ ============================== 

 

Aim #2 : Why electric motor can’t develop braking torque at high speed similar to starting? 

How electric and mechanical brakes are coordinated? 

 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 P_hp / n_r                         (1)

where

T = full load torque (lb ft)

P_hp = rated horsepower

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

In metric units the rated torque can be expressed as

T = 9550 P_kW / n_r                              (2)

where

T = rated torque (Nm)

P_kW = rated power (kW)

n_r = 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:

T_fl = 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.

1. 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:

1. 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.
2. Electrical braking is usually very smooth and comfortable. Mechanical braking is usually rough and uncomfortable.
3. No wear results from electrical braking. Mechanical braking on the other hand causes wear in the braking components and requires regular maintenance.
4. 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.
5. 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.

 

 

 

============================================================ ============================== 

 

 

 

 

Aim  #3 : To Make a MATLAB program which plots contour of given motor speed, torque and efficiency values.

 

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.