Braking

Braking

Objective –

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

Case 1 – For a defined driving cycle, calculate the energy required for braking.

Solution – In this case we have defined the drive cycle which is a 100 seconds long.

In the drive cycle there are five instances when the vehicle decelerates. Now the energy required for the vehicle to accelerate is the same energy required for deceleration.

E = 0.5* m*v^2

And power required for the deceleration is P = E/ɗt

In the excel sheet the calculations are carried out

https://drive.google.com/file/d/1I6b9ztxHt-iTWiiuZG4Kov4vxXOiTSd7/view?usp=sharing

Total Energy - 2514.429 kilo joules

Total Power - 222.574 Kw

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

Solution - a) The image below shows the 1^st and 4^th quadrant of operation for a electric motor which are also called forward motoring and reverse Braking quadrants

When a electric motor operates in 1^st quadrant at the start of drive cycle the torque produced by the is very high at low speed as we can see.

When a motor operates usually there are two regions one is constant torque region (at low speed) and the other is constant power region (at high speed).

During Braking the vehicle speed is high and vehicle/motor is operating in the constant power region.

The peak charging rate is not as high as peak discharging rate of the battery due to convertors and chemistry.

Acceleration Torque > Braking Torque

So during the start of motor the torque is high with respect to braking.

B) Regenerative Braking

In Electric vehicle we need both electrical as well as mechanical(frictional) brakes

When a vehicle is running at cruising speed and we need to apply brakes we need to consider in how much time and how much distance is required to reduce the speed or to even make it zero.

If at cruising speed we remove the foot from paddle the vehicle need some time and some distance after which it will come to rest. This is because of various resistances acting on the vehicle

With the help of control strategies we can obtain the best characteristics from both mechanical as well as electric brakes.

Types of Control strategies

Series brakes - In this type mechanical and electrical brakes are applied one after another.

Parallel Brakes – In this type both mechanical and electrical brakes are applied simultaneously.

Further these are divided in to optimum braking feel and optimum energy recovery.

Let see a example, suppose if the initial speed is 105 Km/h and we need to reduce it to 90 Km/h and time is 4 seconds then regenerative brakes are enough to decrease the speed.

But if we want to reduce the speed for 100 km/h to 0 km/h in 7 seconds and 200m distance electrical brakes are not enough and frictional brakes must be applied.

Also for regenerative braking energy recovered while braking is not as same as for discharging/ accelerating. This is because the limitation of power convertors.

Case 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

Solution – Code

We have Kc as constant of copper losses

Kw as constant of windage losses

and Ki as constant of iron losses

Now the losses depend on the speed and torque produced by motor

output Power = speed X torque

Iron losses =speed* Ki

copper losses = torque^2*Kc

Windage losses =speed^3* Kw

Inpurt power = output power + Iron losses +Copper Losses + Windage Losses + constant losses

efficiency = output/input power

also we have constant motor loss or correction factor to the duty cycle

clear all
close all
clc

s= linspace(1,1000);    %motor speed in rad/s
t = linspace(1,250);    %Torque in N-M

Kc = 0.1;            %copper losses constant
Ki = 0.009;          %Iron losses constant
kw= 0.0001;         %windage losses constant

con_l = 20;  %constant motor losses

[x,y] = meshgrid(s,t);
Pout = (x.*y);       %Power = torque X speed
B = (y.^2)*Kc;       %copper losses = T^2.*Kc
C = (x.*Ki);         %iron losses   = speed * Ki
D = (x.^3)*kw;       %windage losses  speed^3*Kw
Pin =Pout + B + C + D + con_l;
Z =Pout./Pin;
% we set now efficiencies for contour
V = [0.70,0.75,0.80,0.84,0.88,0.90,0.92];
grid off
contour(x,y,Z,V)
xlabel('speed _ rad/s')
ylabel('torque_ N-m')

Conclusion - We have calculated the energy needed for braking in a drive cycle

Studied about types of control strategies in braking of Electric vehicle

Plotted contour for various efficiencies of motor.