Week-11 Challenge: Braking

Ans.1) For a defined driving cycle, calculate the energy required for braking.

The Energy of Traction of wheels mainly compromises of Drag and Braking Energy. Where Braking Energy contributes the highest. In Fig.1.1. We can notice that when a standard driving cycle undergoes changes in speed, acceleration and deceleration. The vehicle at many points is brought to rest or decelerated (i.e. 80 Km/hr - 0 Km/hr) within 10- 15 sec. Hence a high braking energy is required. Hence the Braking energy also increases with deceleration of vehicle.

Fig.1.1.

As per the above Fig.1.1, Acceleration and deceleration at certain points are considered and the Brake power is calculated.

The related spreadsheet is attached.

From the calculations, we can say that the Braking Energy requirement decreases as the difference in the deceleration speed decreases for a given set of time.

Case 1: Deceleration from 80Km/hr – 0 Km/ hr in 10 sec – Energy required – 247 KJ

Case 2: Deceleration from 45Km/hr – 23 Km/ hr in 10 sec – Energy required – 19 KJ

Case 3: Deceleration from 35Km/hr – 30 Km/ hr in 10 sec – Energy required – 0.96 KJ

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

Fig.1.2.

The fig.1.2. Represents the two quadrants of the motor operation where in the first quadrant both torque and speed are +ve (Forward motoring) and in the second quadrant the speed is +ve and the torque is –ve. (Braking)

At High speed the torque will be less & operates at constant power region hence it is difficult to bring the vehicle to rest or decelerate immediately. The peak acceleration torque is greater than the peak deceleration torque. Hence Braking torque of EV motor is not as high as acceleration Torque.

The term braking comes from the term brake. We know that brake is an equipment to reduce the speed of any moving or rotating equipment, like vehicles, locomotives. The process of applying brakes can be termed as braking. Now coming to the term or question what is braking. First of all we can classify the term braking in two parts

1. Mechanical Braking -

   Mechanical braking mostly used in scooters, motor vehicles, and motorcycles where small power is required. It is essential in manufacturing power transmission applications, material handling, etc. It delivers forces to the axle or a wheel in order to stop motion. It helps to reduce the speed of the system slowly by the mechanical process when compared to electrical braking. The working of a mechanical brake depends on the pedal. When the pedal is pressed, the brake shoes are pushed outwards and rotates against the drum which is connected to the wheels. Hence the machine or vehicle gets slow down and stopped. And when the pedal is released, it goes to normal position due to the pullback action of spring shoes.

• Drum braking
• Disk braking
• Band braking
• Pawl and Ratchet braking

1. Electrical Braking - Electrical braking is used to reduce the speed of the machine depending on flux and torque. This type of braking is mainly used for functional braking to control the speed of the machine. It is easy to handle and comfortable. But it cannot be used for emergency braking and parking braking.

• Plugging type braking
• DC injection type braking
• Eddy current braking
• Dynamic resistor type braking
• Regenerative braking
• Sharing DC bus type braking

Electrical and mechanical braking systems can are coordinated by using Brake stratergy subsystem

1) Serial Braking Stratergy

2) Parallel Braking Stratergy

Ans.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

Matlab Code :

clear all

close all

clc

x = linspace(1,1000,250);

y = linspace(1,250,250);

kc = 0.3; %Copper Loss constant

ki = 0.009; %Iron Loss Constant

kw = 0.00005; %Windage Loss Constant

conL = 30;

[X,Y] = meshgrid(x,y);

pout = X.*Y;

B = kc*(Y.^2);

C = ki*X;

D = kw*(X.^3);

pin = pout+B+C+D+conL;

eta = pout./pin; %Efficiency

V = [0.5,0.55,0.6,0.65,0.7,0.75,0.8,0.85,0.9,0.95];

box off;

grid off;

contour(X,Y,eta,V);

xlabel('Speed in rad/s');

ylabel('Torque in N.m');

OUTPUT :

From the output graph you could notice that only at particular speed and torque the efficiency is at maximum i.e. 90 %. where Speed x = 157 rad/s @ 40 Nm Torque. As the Speed increases the efficiencies of the motor will be reduced.

Hence, to attain maximum efficiency the motor has to constantly run at around 157 rad/s speed @ 40 Nm.