Week-11 Challenge: Braking

TASK 1.

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

AIM:

To calculate the total energy required for braking in a Drive cyele.

Ans:

Under the condition of braking,some amount of energy is required to do the same as that of energy required for accelerating the vehicle.

Acceleration energy equation:

Ek=1/2mv^2

Ek=Kinetic energy (J)

m= Mass of object (Kg)

v= Speed of the object (ms-1)

Calculation the energy required for a defined drive cycle.

From the drive cycle,there 2 different situations of deceleration of which energy has to be calculated.

First Braking : 24 - 15 km/hr in 4 sec.

Second Braking : 45 - 23 km/hr in 5 sec.

SOLUTION:

Total Braking Energy Required=0.0346kwh=346KJ

TASK 2.

Why electric motor can't develop braking torque at high speed similar to starting ? How electric and mechanical braks are coordinated ?

AIM:

To study the characteristics of electric motor braking.

ANS:

The braking torque in a electric motor is not as high as that of the acceleration torque at heigher speed.This can be attributed to many different reasons.The rate of discharge of the battery is higher than the charging rate it can be said that the discharge rate can betwice as that rate of charging rate.There are other limitations to themotors in the form of power converters too.This power converters in customised 3 phase inverters which are bi directional as it needs to convert regenerative braking energy back to dc current for the battery and this reduces the efficiency of the power converter at low speeds regenerative braking is very high as speed increases.The regenerative braking reduces as the effictiveness of inverter plays an important role and the overall efficiency reduces resulting in in reduced braking torque.hence for effective braking at high speeds,we use a combination of hydraulic brake and regenerative brake.

There are different ways of coordinating between the hydraulic brake and regenerative braks.One of the brake control startegies involves series connection where both brakes are connected togerther which implies that in this scenario both brakes are used together and the next onr in the parallel connection where either of the two brakes is operated individually whoteout relying on one another depending on the braking requirement.

Types of control strategy:

1.Series:

The braking is applied one after the other either of regenerative and mechanical braking is applied alternatively.

2.Parallel:

Both the brakes are applied the regenerative and mechanical are applied in parallel.

when the brake pedal is applied it supplies the hydraulic fluid to the disc to slow the tires at the same time when the acceleration pedal is lifted a certain in amount of electricbrakes are applied at high speed ,since the braking force is required is a high large amount of hydraulic brakes are applied and once the speed is reduced and the torque value is also high so the battery charging occurs efficiently.The ECU system with a feedback system to find the torque region and regenerative braking predominates in the high torque region.

TASK 3.

Make a MATLAB program which plots contour of given motor speed,torque ans efficiency values.Attach the code as a .m file attach a screenshot of all the plots.

AIM:
To plot the contour of given motor speed,torque ad efficiency values.

CODE :

close all
clear all
clc
omega=linspace(0,1000) %rpm
T=linspace(0,300) % torque (N.m)
kc=0.2
ki=0.008
kw=0.0001
[X,Y]=meshgrid(omega,T)
copper_losses=(Y.^2)*kc
Iron_losses=X.*ki
Windage_losses=(X.^3)*kw
con=20
o_p =X.*Y
Input_power = ( copper_losses)+(Iron_losses)+(Windage_losses)+(o_p)+(con)
Eff=(o_p)./(Input_power)
n=linspace(0.7,0.95,10)
grid off
box off
contour(X,Y,Eff,n)
xlabel('rotor rpm')
ylabel('torque')
zlabal('Efficiency')
title('torque speed curve of induction motor')
figure
surf(X,Y,Eff)
colorbar
xlabel('rotor speed (rpm)')
ylebal('torque induced (Nm')
zlabel('efficiecncy of a motor')

Plot result:

From the output graph,you could see that only at aparicular speed and torque the efficiency is at maximum i.e. 95% .

Where,speed x =151 rad/s @ 24 N-m torque.as the speed increases the efficiencies of the motor will be reduced.