Week-11 Challenge: Braking

Q1. FOR A DEFINED DRIVING CYCLE, CALCULATE THE ENERGY REQUIRED FOR BRAKING.

SOLUTION:-

To calculate braking energy, we consider 9 points i.e. up to 300 sec. of Drive cycle & calculate braking energy during acceleration & deceleration.

A Driving Cycle is a series of data points representing the speed of a vehicle vs. time. Driving cycles are produced by different countries and organisations to access the performance of the vehicles in various ways, as for example fuel consumption and polluting emissions.

Formulas used in Calculations –

Where,

m = Mass of the vehicle (Payload + Kerb Weight) = 1000 Kg.

v = Velocity in Km/h

In case of Braking the high speed is reduced to low speed,

V_f = Final Velocity in m/s.

V_i = Initial Velocity in m/s.

By substituting the values in the above formulas we can get the output in the form of Joules or Kilojoules.

Total braking energy = Kinetic Energy of the Vehicle

= K.E. (Due to linear motion) + K.E. (Due to rotational motion)

But here we consider only Kinetic energy due to linear motion for simplicity

Let the mass of the vehicle (m) = 1000kg.

Now take K.E. between point 1 & point 2

Braking Energy = (K.E.)₂ – (K.E.)₁ = 0 ……… (As velocity is same = 0 m/s)

Between Point 2 & Point 3

Braking Energy = (K.E.)₃ – (K.E.)₂

= (0.5*M*V²) * 3 - (0.5*M*V²) * 2

= (0.5*1000*20²) - (0.5*1000*0²)

= 2,00,000 J

= 200 KJ

Between Point 3 & Point 4

Braking Energy = (K.E.)₄ – (K.E.)₃ = 0 ……(As Velocity is same)

Between Point 4 & Point 5

Braking Energy = (K.E.)₅ – (K.E.)₄

= (0.5*M*V²) - (0.5*M*V²)

= (0.5*1000*14²) - (0.5*1000*20²)

= 1,02,000 J

= - 102 KJ …. (Negative sign represents Deceleration)

Between Point 5 & Point 6

Braking Energy = (K.E.)₆ – (K.E.)₅ = 0 ……(As velocity is same)

Between Point 6 & Point 7

Braking Energy = (K.E.)₇ – (K.E.)₆

= (0.5*M*V²) - (0.5*M*V²)

= (0.5*1000*20²) - (0.5*1000*14²)

= 1,02,000 J

= 102 KJ

Between Point 7 & Point 8

Braking Energy = (K.E.)₈ – (K.E.)₇ = 0 ……(As velocity is same)

Between Point 8 & Point 9

Braking Energy = (K.E.)₉ – (K.E.)₈

= (0.5*M*V²) - (0.5*M*V²)

= (0.5*1000*28²) - (0.5*1000*20²)

= 1,92,000 J

= 192 KJ

Between Point 9 & Point 10

Braking Energy = (K.E.)₁₀ – (K.E.)₉ = 0 ……(As velocity is same)

Thus Net Braking Energy during Drive Cycle 0 + 200 + 0 + (-102) + 0 + 192 + 0

= 290 KJ

So, we required 290 KJ of Braking Energy up to 300 Sec. of Drive cycle.

Q2. WHY ELECTRIC MOTOR CAN'T DEVELOP BRAKING TORQUE AT HIGH SPEED SIMILAR TO STARTING? HOW ELECTRIC AND MECHANICAL BRAKES ARE COORDINATED?

SOLUTION:-

ELECTRIC MOTOR CAN'T DEVELOP BRAKING TORQUE AT HIGH SPEED SIMILAR TO STARTING

The torque developed by an electric motor during deceleration is very low compared to acceleration phase because during acceleration the motor operates at the constant torque region where the starting torque is very high.Whereas during deceleration phase the motor operates in the constant power region where the torque developed is very low.During deceleration ie when the motor is at high speeds ie at speeds greater than base speed, due to the field weakening effect the torque developed will be low.Hence it is not possible to obtain maximum regenerative braking efficiency due to the above explained motor limitation.Another limitation with respect to regenerative braking is due to the battery.The discharge rate of a battery is twice that of its charging rate.Hence during regenerative braking the energy saved cannot be used to charge the battery at the desired rate.The final limitation concerning regenerative braking is associated with the power converter.To supply the energy saved during regeneration from the motor back to the battery, we need a bi-directional power converter, which in turn increases the complexity of the system and also losses during conversion should be considered.

COORDINATION BETWEEN ELECTRIC AND MECHANICAL BRAKES

Coordination between electric and mechanical brakes can be basically obtained using two methods : Series strategy and Parallel strategy.

Under series strategy, both the electric and mechanical brakes are applied one after another.With serial type a blending between friction brake and regenerative brake can be obtained, leading to the vehicle deceleration corresponding to the driver's braking demand. This type offers a high regeneration efficiency and good braking feel, but requests several modifications in the hydraulic brake line to mechanically decouple the connection between the brake pedal and the brake pressure.

Under parallel strategy, both the electric and mechanical brakes are applied simultaneously.In parallel systems, the brake pressure is coupled to the brake pedal, and the regenerative braking is added directly to the friction brake operated by brake pedal, which results in a low regeneration efficiency and poor brake comfort.

Electric Driven Intelligent Brake (EDIB) developed by NISSAN is one of the most efficient type of coordinated braking established between electric and mechanical brakes.After stepping on the brake pedal, the system produces natural and adequate braking force that corresponds to the operation. The amount of energy regeneration also needs to be increased as much as possible.EDIB (Electric Driven Intelligent Brake) controls the regenerative brake and friction brake to support both of these requirements. Further, it also controls the reactive force from the pedal in order to unify the feeling when stepping down on the pedal and the sense of deceleration.EDIB is programmed such that during the initial phase of braking ie at high speeds both the regenerative and friction brakes are applied in a pre-determined ratio.Whereas at lower speeds the regenerative brakes are applied pre-dominantly.This is because at lower speeds, large torque is present which increases the efficiency of regenerative braking.

Q3. 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:-

While, motor works there are various types of losses –

1. Copper losses.
2. Iron losses.
3. Windage losses.
4. Constant losses.

1. Copper losses –

Copper losses are given by I².R

Where, R = Armature resistance.

And if we say Torque is proportional to current then, the copper loss is considered as K_c. T²

Where, K_c is to take care of resistance of brushes & also to take care of flux and its effect.

Hence Copper Losses = K_c. T²

2. Iron losses –

Iron losses are magnetic losses denoted by K_i. W, where K_i is the factor based on magnetic field effect.

3. Windage losses –

These are the mechanical or friction Windage losses. This is based on size and shape of the motor & the Windage losses that the motor is going to experience hence the Windage losses = K_w. W³

4. Constant losses –

This is denoted by C.

5. Total losses –

The total losses are given by,

Total Losses = K_c. T² + K_i. W + K_w. W³ + C.

% Plotting Efficiency Contours for Electric Motors

clear all
close all
clc

% Determining the Speed Range (rad/s)
x = linspace(1,1000);

% Determing the Torque Range (Nm)
y = linspace(1,250);

% Allocating Motor Loss Constants 

% For Copper Losses
kc = 0.2;

% For Iron Losses
ki = 0.008;

% For Windage Losses
kw = 0.00001;

% For Constant Motor Losses
ConL = 20;

% Meshing Speed and Torque ranges to create a 2D Array
[X,Y] = meshgrid(x,y);

% Output Power 
Po = (X.*Y);

% Copper Losses
B = kc.*(Y.^2);

% Iron Losses 
C = ki.*(X);

% Windage Losses
D = kw.*(X.^3);

% Input Power 
Pi = Po + B + C + D + ConL;

% Efficiency 
Z = Po./Pi;

% Setting the Efficiencies for which a contour will be plotted
V = [0.70,0.75,0.80,0.85,0.90,0.91,0.92,0.93,0.94,0.95];

% Plotting the Efficiency Contour 
box off 
grid off
contour(X,Y,Z,V)
xlabel('Speed (rad/s)')
ylabel('Torque (Nm)')
hold on 
% Plotting contours of Output Power at 8kW and 10kW
V = [8000,10000];
contour(X,Y,Po,V)

OUTPUT:-

OBSERVATION:-

* The above figure shows the contour plot and the X axis is speed in RPM, Y axis is Torque in Nm and the height (i.e) Z axis is Efficiency.

* The contour plot represents the operational condition of the Permanent Magnet DC Motor.

* The efficiency of the Motor is High at Lower values of speed & Torque at 0.93 . As Speed and Torque Increases the efficiency of the motor goes down upto 0.75 value of efficiency.

* Thus the contour plot is made using a Matlab program for an electric motor with given speed, Torque and efficiency values.

CONCLUSION:-

While completing this challenge we learnt the deacceleration calculation to find eneregy required while braking of any vehicle when it is at certain speed. We also learnt Why electric motor can’t develop braking torque at high speed similar to starting in detail, also How electric and mechanical brakes are coordinated with deatailed diagram and explaination.

We also did the programming in matlab to show the resultant contour graph of electric motor. We saw the different losses occurs while machine is on working. And learnt how to calculate the total losses and thus find the efficiency of the machine.