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Q:- 1. For a defined driving cycle, calculate the energy required for braking. ANS:- ENERGY REQUIRED FOR BAKING BASED ON A DRIVING CYCLE:- The driving cycle taken into consideration is the UDDS cycle which stands for Urban Dynamometer Driving Schedule. Although the data is taken for a total driving time for the 1370s,…
Chandrakumar ADEPU
updated on 31 Jul 2022
Q:- 1. For a defined driving cycle, calculate the energy required for braking.
ANS:-
ENERGY REQUIRED FOR BAKING BASED ON A DRIVING CYCLE:-
The driving cycle taken into consideration is the UDDS cycle which stands for Urban Dynamometer Driving Schedule. Although the data is taken for a total driving time for the 1370s, for simplicity sake on 150s of driving time is considered. The speed which is in miles per hour has been converted to meters per second.
The drive cycle is given below.
The equation for kinetic energy from braking is given as,
Where,
m = Mass of the vehicle in Kg.
Vi = Initial speed of the vehicle in m/s.
Vf = Final speed of the vehicle in m/s.
From the drive cycle, 10 braking points were obtained. The mass is set at a constant 1500 Kg. The energy required for braking is given in the table below.
Therefore the total energy required for braking is 16.8 KW.
The excel sheet file of the drive cycle and braking data is provided attached below link
https://docs.google.com/spreadsheets/d/1oxMGUf9KOY1_7NyLyLJM2hex0yxGUa-yjgHXsgPBkGA/edit#gid=0
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Q:-2. Why electric motor can’t develop braking torque at high speed similar to starting? How electric and mechanical brakes are coordinated?
ANS:-
The torque-Speed characteristic of the motor is not exactly the same in the braking and motoring region.
This is torque-speed contour plot of motor with efficiency.
We can see that, at the time of accelerating vehicle at starting the maximum torque available motor is operating at constant torque region, in that motor will have maximum torque available, but as motor starts deaccelerating motor characteristics are shifted towards braking region, in this we don’t have max torque available for braking as motor is already running at higher speed.
Due to this operation, electric motor can’t develop braking torque at high speed similar to starting
Electrical and mechanical braking systems can are coordinated by using Brake strategy subsystem
1) Serial Braking Strategy
2) Parallel Braking Strategy
Implementation of serial braking strategy
Once the brake pedal is pressed and depending on the level of depression, a brake torque is sent. Now this brake torque passes into the brake control strategy block. Here, firstly it is checked if battery state-of-charge is below 80% which is set as the maximum permissible level of charge. If that condition is satisfied, then the battery is recharged. The brake torque requested is compared to the maximum regenerative torque limit of the motor-generator. If the overall brake torque requested is less than or equal to the maximum regenerative torque, then the vehicle brakes purely on regenerative torque and the battery is charged. The amount of charge ultimately put in depends also on the motor-generator operation efficiency and the battery charging efficiency. If the overall brake torque requested is greater than what the motor-generator can provide, then the friction brakes kick-in. Now since the vehicle is front driven, it is checked if the total brake torque requested from front is less than or equal to maximum regenerative torque capability of motor-generator. If yes, then front braking is provided purely by regeneration and rear (non-driven axle) by friction. Once the total brake torque request increases further, then front friction brakes (driven axle) comes into play in combination with rear friction brakes and front regenerative braking, to satisfy the overall brake torque demand. This has been represented in a chart manner.
Parallel Braking:-
Once the brake pedal is applied and a desired overall brake torque request is sent from the driver block. This brake torque on passing through the brake control strategy gets divided between regeneration and friction braking at all times. The strategy is so modeled with the comfort and brake pedal feel in mind. As a parallel regenerative braking is an add-on to the conventional systems, it is devoid of any integrated control. So to make the transition and co-performance of regenerative braking with friction smooth, it is assigned in such a way that, there is a clear division between the amount of brake torque that will be achieved from regeneration and friction braking. On application of brakes and in keeping with the design of the motor-generator, it is designed to operate generator in a constant generator torque region. Motor-generator can provide a maximum of 322 Nm of generator torque at or under the base speed. The generator is scaled such that the base speed of the generator corresponds to a velocity of 124 [kmph], which is the maximum speed limit on most highways. Hence regeneration works, as long as the vehicle is braking from a speed of under 124 [kmph] and the requested brake torque on the front driven wheels, after the front-rear brake split is greater than or equal to maximum generator torque. If that condition is satised, then regenerative and friction braking operates in tandem else it’s just the friction brakes at the driven and non-driven axle. This has been represented in a ow chart manner in the g.1.5.
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Q:-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.
ANS:
To plot contour line for a given Motor. A Motor have various types of losses happening inside. The various types of losses are:
Copper Losses - I^2 R, where R is the Armature resistance and we say that Torque is proportional to Current, this copper loss can be consider as .where is to take care of resistance of brushes and also take care of ux and its effect.
Copper losses = kc⋅T^2
Iron Losses - Iron Losses are nothing but Magnetic losses, which is denoted as , where factor is based on the Magnetic eld effect. For Permanent Magnets have constant Magnetic eld and this Iron losses is based on the value of and this are going to change according to the Speed. At Various Speed have different value of Back EMF and that is going to affect the Iron Losses.
Iron Losses = ki⋅ω
Windage Losses - This is a Mechanical or friction of Windage losses. This is based on size of shape of Motor and the amount of windage losses that the motor is going to experience.
Windage Losses = kw⋅ω^3
Constant Losses - This is denoted as C.
Total Losses - The Total Losses are considering the above four losses and denoted as,
Total Losses = kc⋅T^2+ki⋅ω+kw⋅ω^3+C
All Motors are model as per the above equation of losses. The above equation is not only applicable for Permanent Magnet DC Motor, but to other motors as well like an induction motor.
MATLAB program for calculating the losses of the EV motor:
clear all
close all
clc
w = linspace(0,1000);
T = linspace(0,250);
% motor constants
kc = 0.2; % for copper loss
ki = 0.008; % for iron loss
kw = 0.00001; % for windage loss
conL = 20; % for constant motor losses
% making mesh
[X,Y] = meshgrid(w,T);
OutputPower = (X.*Y); % Torque x Speed = Power
B = (Y.^2)*kc; % copper loss
C = (X.^3)*kw; % winding loss
D = X*ki; % Iron loss
InputPower = OutputPower + B + C + D + conL;
Z = OutputPower./InputPower;
% Set the efficiencies for which the contours will be plottted
V = [0.70 , 0.75 , 0.80 , 0.82 , 0.85 , 0.87 , 0.89 , 0.90 , 0.92];
box off
grid off
contour(X,Y,Z,V);
xlabel('Speed(rad/s)')
ylabel('Torque(N.m)')
hold on
V = [8000,10000];
contour(X,Y,OutputPower,V);
Result:-
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