Week-11 Challenge: Braking

1)

Aim:

To calculate the energy required for braking.

Principle:

Braking:

• Under the braking condition vehicle starts decelerating i.e. the speed of the vehicles reduces.
• The function of the brake system is to stop the vehicle within the smallest possible distance.
• Hence it is done by converting the kinetic energy of the vehicle into the heat energy which is dissipated into the atmosphere.

E=(1/2)*m*v^2
Where,
m= mass of the vehicle
v= velocity of the vehicle
E= energy

Driving cycle:

• A driving cycle is a series of data points representing the speed of a vehicle versus time.
• It is used to assess the performance of fuel consumption, vehicle autonomy and pollutants emission etc.

Excel Sheet:

Duty cycle:

• The driving cycle is created for 40 intervals of time.
• The velocity from km/h is converted into m/s by multiplying it to 5/18.

• The driving cycle curve obtained is plotted in graph.
• Energy required for braking is 0.5*m*v^2.
• The 1000 kg is considered as the vehicle mass.
• The initial velocity period is the time at which the vehicle starts to decelerate and reaches to zero.
• So in the above driving cycle twice the vehicle is decelerated.

Energy =0.5*m*(vi-vf)^2
where,
m=mass
vi=initial velocity
vf=final velocity

Braking1:0.5*1000*(13.85-0)^2
=95911.25J
Braking2:0.5*1000*(20.498-0)^2
=210084J
Total energy= 95911.25+210084
=305995.3 J

• Hence for the driving cycle ,the energy required for braking is 305995.3 J.

Result:

Thus the energy required for braking for a defined driving cycle is done.

2)

Aim:

To find how the electric and mechanical brakes are coordinated.

Explaination:

• The above image represents the speed-torque characteristics of an electric motor.
• At low speed, the torque maintained in the motor is constant leading to higher acceleration.
• As the speed increases the torque reduces , which is constant power region.
• And the value of torque obtained in deceleration is negative.
• So the peak acceleration torque and peak deceleration torque are not equal.
• From this it is clear that electric motor cannot develop braking torque at high speed similar to starting.

Brake control stratergies:

• Series brake control stratergies: In this the both mechanical and regenerative electric brake are operated one after the another.
• Parallel brake control stratergies:In this both the mechanical and regenerative electric brake are applied at a time.

Electric Driven Intelligent Brake:

• The conventional power brake system derive vaccum from the engine intake manifold and use this energy to apply brake.
• This reduces drivers pedal force and it becomes highly responsive with easiness to operate.
• But the major disadvantage of the conventional power brake is that it puts load on the engine since the braking energy is extracted from the engine.
• Also the energy lost in stopping the vehicle goes waste.
• But in the Electric Driven Intelligent Braking System, the braking force is provided with the help of a ECU controlled motor and ball screw arrangement.
• So there is no direct load on the engine while applying brakes.
• A more controlled and uniform braking force is applied due to the usage of ECU controller.
• Thus helps to recover thr braking losses when a regenerative braking system is used.

• It is the combination of regenrative braking and the friction brake.
• Both the regenerative braking and the friction brake are operated parallelly for the high speed of vehicle.
• At low speed the percentage of regenerative braking is high.
• So the torque produced is high when the vehicle is at low speed.
• Control signal must detect the either regenerative or friction brake from the pedal.
• And these control signal should be diverted to the brake cyclinder or regenerative brake to power converter.

Conclusion:

Thus the regenerative braking are efficient but it requires a customized 3 phase bidirectional power flow inverter to take power from the motor and feed it back to the battery.

3)

Aim:

To make a MATLAB program which plots contour of given motor speed, torque and efficiency values.

Principle:

• A contour line of a function of two variables is a curve along which the fucntion has a constant value, so that the curve joins points of equal value.
• It is a plane section of three dimensional graph of the function parallel to the plane.
• The contour line is represented for the motor speed, torque and efficiency.

Efficiency contour plot:

Copper losses= I^2*R= kc*T^2
Iron losses= ki*w
Windage loss= kw*w^3
Constant losses= C
Total losses= (kc*T^2)+(ki*w)+(kw*w^3)+C

MATLab Script:

• The x axis is declared as speed upto 1000 rad/s.
• In the y axis torque is specified upto 250 N.m.
• The value of 0.2 is given as kc.
• The value of 0.008 is given as ki.
• 0.00001 for kw.
• 20 as constant motor losses.
• The product of torque and speed is done to obtain power.
• As per the formula the copper loss, iron loss and windage losses are calculated.
• The sum of output power and the losses gives input power.
• Z is the ratio of output power to the input power.
• To plot the efficiency , the contour values are given from 0.70 percent upto 93 percent.
• The graph is plotted along the X,Y,Z and V axis.

Output:

• The above image represents the contour plotting of the motor speed, torque and effciency.
• It X axis represents the speed upto 1000 rad/s.
• The Y axis represents the torque upto 250N.m.
• In the above image we can find the values of speed , torque and efficiency of the motor by clicking the desired curve.
• As per the point in the above image, the speed is 182.636 rad/s, 96.5758 N.m as the torque and 90.05 percentage as the efficiency.

Result:

Thus a MATLAB program is compiled in which the contour plots of motor speed, torque and efficiency values are plotted.

Attachment:

https://drive.google.com/file/d/1BoCyaMz5hIQ__CQ-sYUGUECA-yt9z7ik/view?usp=sharing