Week-11 Challenge: Braking

1. For a defined driving cycle, calculate the energy required for braking.
2. Why electric motor can’t develop braking torque at high speed similar to starting? How electric and mechanical brakes are coordinated?
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.

Q1

Braking:

Braking is the process of controlling the velocity of an object by inhibiting its motion. An object in motion possesses kinetic energy and to bring the object to a stop this kinetic energy must be removed. Removing the kinetic energy can be accomplished by dissipating the energy to the atmosphere through friction or by converting it into another form of energy. The most common type of braking is a mechanical brake which inhibits motion through friction brake pads. A mechanical brake applies a friction force to convert the kinetic energy of the vehicle into thermal energy which then dissipates into the atmosphere. Braking systems which don't use friction are referred to as Regenerative Braking System (RBS). In RBS, the kinetic energy is converted into other forms of useful energy, which can be stored for later use, increasing fuel efficiency.

As with any system, the process of braking must follow the principal of conservation of energy. Energy cannot be created or destroyed but only converted from one form to another,

ΣE_IN=ΣB_OUT

The energy present in an object in motion is given by the following equation:

where,

m-mass of the object in Kilogram(Kg).

v-velocity of the object in meters per second(m/s).

E_KINETIC - Kinetic energy in joules(1)

From this above equation, and assuming the mass of the object is constant, it is clear the in order to remove the kinetic energy from the system, the velocity must be brought to zero.

Braking Energy Calculation:

For Calculating the Energy required for Braking, Drivecycle data is prepared in the Excel sheet, which Drive cycle data includes Time and Speed

Mass =1000kg

V1=20mps

V2=5mps

K.E.= 0.5*M*[v₁²-v₂²]

K.E.=0.5*1000*[20²-5²]

K.E.=1,87,500 joule

K.E.=187.5KJ

Q2

Electric Motor Braking:

Eectric motor can't develop braking torque at high speed similar to starting, because Braking torque is applied in the opposite direction of the rotation. We can use a mechanical brake to produce a braking torque. We can also do it electrically.

Regenerative Braking: It is a form of braking in which the kinetic energy of the motor is returned to the power supply system.

Regenerative braking takes place whenever the speed of the motor exceeds the synchronous speed. This baking method is called regenerative baking because here the motor works as generator and supply itself is given power from the load, i.e. motors. The main criteria for regenerative braking is that the rotor has to rotate at a speed higher than synchronous speed, only then the motor will act as a generator and the direction of current flow through the circuit and direction of the torque reverses and braking takes place.

Sometimes after accelerating you can't reach the maximum torque. The reason is that no matter how much the current loop wants to increase current (proportional to torque) by increasing the PWM duty cycle of the power stage, the Vmotor is too close to Vbemf and therefore current can't increase more .This situation does not occur at the beginning of an acceleration because most of the torque is used to increase the kinetical energy of the load (and motor). But when the speed is already high and no extra load is applied it is not possible to increase the torque.

Peak Acceleration Torque and Peak Deceleration torque are not equal for Electric Vehicle Motor. So the Peak Torque during acceleration and the Peak torque during deceleration or braking are not equal. Acceleration is usually higher, at low speed we get higher Torque but high speed the value of the torque gets lower. So we cannot say that the motor which is producing 7Nm Torque for Acceleration can also produce -7Nm of Braking Torque. So beacuse we Operating the Motor in Field Weakening region or we are operating the Motor above base speed we must know that the torque is compromised or the torque value is lower. So Electric motor can't develop braking torque at high speed similar to starting, because Braking Torque of an EV Motor is not as high as the Acceleration Torque. Therefore Regenerative braking is cannot happen so effectively that we can get all the power back to the battery.

Coordination of Electric and Mechanical Brake:

In a 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 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 regenerative capacity of the machine we must apply Mechanical brakes as well. In practical case the combination of these two is applied. So usually it is combined response of Mechanical brake and Regenerative brake. The reason for how these brake force are combined, that particular phenomena is 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 Regenerative or apply Mechanical brake.

The Series Brake Control Strategy has further classification, that is based on whether we want comfort of driver or Brake behaviour compared to energy efficiency.

Parallel - In 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 initial braking period we actually combine the both.

Q3

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 - - R. where R is the Armature resistance and we say that Torque is proportional to Current, this copper loss can be consider ask where is to take care of resistance of

brushes and also take care of flux and its effect.

Copper Losses - IR, where R is the Armature resistance and we say that Torque is proportional to Current, this copper loss can be consider ask, 1² where , is to take care of resistance of brushes and also take care of flux and its effect.

Copper losses = ₁72

Iron Losses - Iron Losses are nothing but Magnetic losses, which is denoted as w, 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 = A

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

Constant Losses - This is denoted as C

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

Total Losses = k-1²+k₁+k₁-²² +C

as,

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

Codings are

close all

clc

%defining speed and torque

speed= linspace(0,1000,500);

torque= linspace(0,1000,500);

%motor constants

kc=0.1; % copper loss constant

ki=0.001; % iron loss constant

kw=0.0001; % windage loss constant

cm=25; % motor loss constant

%2-D grid for speed and torque

[x,y]=meshgrid(speed,torque);

powerout = x.*y;

total_loss=((y.^2).*kc)+(ki.*x)+(x.^3).*kw+cm;

inputpower = powerout + total_loss;

efficiency= powerout./inputpower;

%efficiency level for contouring

El=[0.70,0.72,0.75,0.77,0.79,0.80,0.81,0.82,0.83,0.84,0.85,0.86,0.87,0.88,0.89,0.90,0.91,0.92,0.93,0.95];

%plotting the efficiency w.r.t speed and torque

contourf(x,y,efficiency,El);

colorbar

xlabel("speed(rpm)")

ylabel("torque(N-m)")

title("Efficiency contour")