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AIM To understand Braking in automobiles. INTRODUCTION 1. BRAKE: [1] Brake is a mechanical device that inhibits motion by absorbing energy from a moving system. It is used for slowing or stopping a moving vehicle, wheel, axle, or to prevent its motion, most often accomplished by means…
Laasya Priya Nidamarty
updated on 06 May 2021
To understand Braking in automobiles.
[1] Brake is a mechanical device that inhibits motion by absorbing energy from a moving system. It is used for slowing or stopping a moving vehicle, wheel, axle, or to prevent its motion, most often accomplished by means of friction.
Figure 1. Disc brake on a motorcycle.
Most brakes commonly use friction between two surfaces pressed together to convert the kinetic energy of the moving object into heat, though other methods of energy conversion may be employed. For example, regenerative braking converts much of the energy to electrical energy, which may be stored for later use. Other methods convert kinetic energy into potential energy in such stored forms as pressurized air or pressurized oil. Eddy current brakes use magnetic fields to convert kinetic energy into electric current in the brake disc, fin, or rail, which is converted into heat. Still other braking methods even transform kinetic energy into different forms, for example by transferring the energy to a rotating flywheel.
Brakes are generally applied to rotating axles or wheels but may also take other forms such as the surface of a moving fluid (flaps deployed into water or air). Some vehicles use a combination of braking mechanisms, such as drag racing cars with both wheel brakes and a parachute, or airplanes with both wheel brakes and drag flaps raised into the air during landing.
Since kinetic energy increases quadratically with velocity (K = 0.5mv2), an object moving at 10 m/s has 100 times as much energy as one of the same mass moving at 1 m/s, and consequently the theoretical braking distance, when braking at the traction limit, is up to 100 times as long. In practice, fast vehicles usually have significant air drag, and energy lost to air drag rises quickly with speed.
NEED FOR BRAKING SYSTEM: [2] In an automobile vehicle braking system is needed:
CHARACTERISTICS OF BRAKES: [1] Brakes are often described according to several characteristics including:
Characteristic |
Description |
Peak force |
The peak force is the maximum decelerating effect that can be obtained. The peak force is often greater than the traction limit of the tires, in which case the brake can cause a wheel skid. |
Continuous Power dissipation |
Brakes typically get hot in use and fail when the temperature gets too high. The greatest amount of power (energy per unit time) that can be dissipated through the brake without failure is the continuous power dissipation. Continuous power dissipation often depends on e.g., the temperature and speed of ambient cooling air. |
Fade |
As a brake heats, it may become less effective, called brake fade. Some designs are inherently prone to fade, while other designs are relatively immune. Further, use considerations, such as cooling, often have a big effect on fade. |
Pedal Feel |
Brake pedal feel encompasses subjective perception of brake power output as a function of pedal travel. Pedal travel is influenced by the fluid displacement of the brake and other factors. |
Smoothness |
A brake that is grabby, pulses, has chatter, or otherwise exerts varying brake force may lead to skids. For example, railroad wheels have little traction, and friction brakes without an anti-skid mechanism often lead to skids, which increases maintenance costs and leads to a "thump" feeling for riders inside. |
Power |
Brakes are often described as "powerful" when a small human application force leads to a braking force that is higher than typical for other brakes in the same class. This notion of "powerful" does not relate to continuous power dissipation and may be confusing in that a brake may be "powerful" and brake strongly with a gentle brake application yet have lower (worse) peak force than a less "powerful" brake. |
Drag |
Brakes have varied amount of drag in the off-brake condition depending on design of the system to accommodate total system compliance and deformation that exists under braking with ability to retract friction material from the rubbing surface in the off-brake condition. |
Durability |
Friction brakes have wear surfaces that must be renewed periodically. Wear surfaces include the brake shoes or pads, and also the brake disc or drum. There may be tradeoffs, for example, a wear surface that generates high peak force may also wear quickly. |
Weight |
Brakes are often "added weight" in that they serve no other function. Further, brakes are often mounted on wheels, and un-sprung weight can significantly hurt traction in some circumstances. "Weight" may mean the brake itself or may include additional support structure. |
Noise |
Brakes usually create some minor noise when applied, but often create squeal or grinding noises that are quite loud. |
Table 1. Characteristics of Brake employed in an automobile.
[2] In an automobile vehicle, a braking system is an arrangement of various linkages and components (brake lines or mechanical linkages, brake drum or brake disc , master cylinder or fulcrums etc.) that are arranged in such a fashion that it converts the vehicle’s kinetic energy into the heat energy which in turn stops or decelerates the vehicle. The conversion of kinetic energy into heat energy is a function of frictional force generated by the frictional contact between brake shoes and moving drum or disc of a braking system.
Figure 2. Classification of Braking Systems.
1. BASED ON POWER SOURCE: The power source which carries the pedal force applied by the driver on brake pedal to the final brake drum or brake disc in order to de accelerate or stop the vehicle the braking systems are of six types:
Figure 3. Depiction of Mechanical Brake.
Figure 4. Depiction of Hydraulic Braking System.
Figure 5. Depiction of Pneumatic Braking System.
Figure 6. Depiction of Vacuum Braking System.
Figure 7. Depiction of Magnetic Braking System.
Figure 8. Depiction of Electric Braking System.
2. BASED ON FRICTIONAL BRAKING CONTACT: On the basis of the final friction contact made between the rotating brake components i.e., brake drum or disc rotor and the brake shoe the braking systems are of two types:
Figure 9. Depiction of Disc and Drum brakes.
3. BASED ON APPLICATION: On the basis of method of applying brakes, braking systems are of two types:
4. BASED ON BRAKE FORCE DISTRIBUTION: On the basis of the type of force application of the brake, there are two types:
[3] Regenerative braking in EVs and HEVs adds some complexity to the braking system design. Two basic questions arise:
Usually, regenerative braking is effective only for the driven axle. The traction motor must be controlled to produce the proper amount of braking force for recovering the kinetic energy as much as possible and, at the same time, the mechanical brake must be controlled to meet the braking force command from the driver. Basically, there are three different brake control strategies: series braking with optimal braking feel; series braking with optimal energy recovery; and parallel braking.
a) SERIES BRAKING WITH OPTIMAL BRAKING FEEL: The series braking system with optimal feel consists of a braking controller that controls the braking forces on the front and rear wheels. The control objective is to minimize the stopping distance and optimize the driver’s feel. The shortest braking distance and good braking feel require the braking forces on the front and rear wheels to follow the ideal braking force distribution curve I. When the commanded deceleration (represented by the braking pedal position) is less than 0.2 g, only the regenerative braking on the front wheels is applied, which emulates the engine retarding function in conventional vehicles. When the commanded deceleration is greater than 0.2 g, the braking forces on the front and rear wheels follow the ideal braking forces distribution curve I, as shown in Figure 10. by the thick solid line.
The braking force on the front wheels (driven axle) is divided into two parts: regenerative braking force and mechanically frictional braking force. When the braking force demanded is less than the maximum braking force that the electric motor can produce, only electrically regenerative braking will apply. When the commanded braking force is greater than the available regenerative braking force, the electric motor will operate to produce its maximum braking torque, and the remaining braking force is met by the mechanical brake system.
It should be noted that the maximum regenerative braking force produced by an electric motor is closely related to the electric motor’s speed. At low speed (lower than its base speed), the maximum torque is constant. However, at high speed (higher than its base speed), the maximum torque decreases hyperbolically with its speed. Therefore, the mechanical brake torque at a given vehicle deceleration varies with vehicle speed.
Figure 10. Illustration of braking forces on the front and rear axle for series braking — optimal feel.
b) SERIES BRAKING WITH OPTIMAL ENERGY RECOVERY: The principle of the series braking system with optimal energy recovery is to recover the braking energy as much as possible in the condition of meeting the total braking force demanded for the given deceleration. This principle is illustrated in Figure 11. When the vehicle is braked with an acceleration rate of j/g < μ, the braking forces on the front and rear wheels can be varied in a certain range, as long as the Fbf + Fbr = Mvj is satisfied. This variation range of the front and rear axles is shown in Figure 11 by the thick solid line ab, where μ = 0.9 and j/g = 0.7.
In this case, regenerative braking should be used in priority. If the available regenerative braking force (maximum braking force produced by the electric motor) is in this range (point c in Figure 11, for example), the braking force on the front wheels should be developed only by regenerative braking without mechanical braking. The braking force on the rear wheels, represented by point e, should be developed in order to meet the total braking force requirement. If, on the same road, the available regenerative braking force is less than the value corresponding to point a (e.g., point i in Figure 11), the electric motor should be controlled to produce its maximum regenerative braking force. The front and rear braking forces should be controlled at point f so as to optimize the driver feel and reduce braking distance. In this case, additional braking force on the front wheels must be developed by mechanical braking by the amount represented by Fbf-mech, and the braking force on the rear axle is represented by point h.
Figure 11. Demonstration of series braking — optimal energy recovery.
When the commanded deceleration rate, j/g, is much smaller than the road adhesive coefficient (j/g = 0.3 in Figure 11 for example), and the regenerative braking force can meet the total braking force demand, only regenerative braking is used without mechanical braking on the front and rear wheels (point j in Figure 11). When the commanded deceleration rate, j/g, is equal to the road adhesive coefficient μ, the operating point of the front and rear braking forces must be on the curve I. On a road with a high adhesive coefficient (μ = 0.7, operating point f, in Figure 11, for example), the maximum regenerative braking force is applied, and the remaining is supplied by the mechanical brake. On a road with a low adhesive coefficient (μ = 0.4, operating point k, in Figure 11, for example), regenerative braking alone is used to develop the front braking force. When the commanded deceleration rate, j/g, is greater than the road adhesive coefficient μ, this commanded deceleration rate will never be reached due to the limitation of the road adhesion. The maximum deceleration that the vehicle can obtain is (a/g)max = μ. The operating point of the front and rear braking forces is on the curve I, corresponding to μ (μ = 0.4 and j/g > 0.4 in Figure 11, for example); the operating point is point k and the maximum deceleration rate is j/g = 0.4.
It should be noted that the series brake with both optimal feel and energy recovery needed active control of both electric regenerative braking and mechanical braking forces on the front and rear wheels. At present, such a braking system is under research and development.
[4] There are two types of losses occur in three phase induction motor. These losses are:
Figure 12. Demonstration of losses in an induction motor.
Figure 13. Distribution of the power in an induction motor.
A part of this power input is used to supply stator losses which are stator iron loss and stator copper loss. The remaining power i.e., (input electrical power – stator losses) are supplied to rotor as rotor input. Now, the rotor has to convert this rotor input into mechanical energy, but this complete input cannot be converted into mechanical output as it has to supply rotor losses. As explained earlier the rotor losses are of two types of rotor iron loss and rotor copper loss. Since the iron loss depends upon the rotor frequency, which is very small when the rotor rotates, so it is usually neglected. So, the rotor has only rotor copper loss. Therefore the rotor input has to supply these rotor copper losses. After supplying the rotor copper losses, the remaining part of Rotor input, is converted into mechanical power. Therefore, this mechanical power developed is given to the load by the shaft but there occur some mechanical losses like friction and windage losses. So, the gross mechanical power developed has to be supplied to these losses. Therefore the net output power developed at the shaft is finally given to the load.
The three-phase induction motor efficiency is given as:
For a defined driving cycle, calculate the energy required for braking.
EXPLANATION AND OBSERVATION:
Time(sec) |
Velocity(m/s) |
Braking Energy |
0 |
0 |
… |
1 |
2 |
… |
2 |
3 |
… |
3 |
5 |
… |
4 |
7 |
… |
5 |
10 |
… |
6 |
14 |
… |
7 |
16 |
… |
8 |
22 |
… |
9 |
20 |
42000 |
10 |
17 |
55500 |
11 |
12 |
72500 |
12 |
6 |
54000 |
13 |
3 |
13500 |
14 |
0 |
4500 |
15 |
7 |
… |
16 |
15 |
… |
17 |
25 |
… |
18 |
19 |
132000 |
19 |
17 |
36000 |
20 |
20 |
… |
21 |
13 |
115500 |
22 |
8 |
52500 |
23 |
11 |
… |
24 |
2 |
58500 |
25 |
5 |
… |
26 |
19 |
… |
27 |
29 |
… |
28 |
18 |
258500 |
29 |
7 |
137500 |
30 |
0 |
24500 |
Table 2. The drive cycle depicting the values of time and velocity along with braking energy column.
Why electric motor cannot develop braking torque at high speed similar to starting? How electric and mechanical brakes are coordinated?
EXPLANATION AND OBSERVATION:
[5] In the case of steady state operation (the angular acceleration α is zero) the motor torque has to make friction torque correspond proportionally to the angular speed and load torque at that specific angular speed. The braking torque and power need in respect to time varies greatly in these two different load types.
The motor torque is given as:
For, constant load torque conditions, i.e., TLOAD = Constant then the power is defined as:
EVALUATING BRAKE TORQUE AND POWER:
In the case of steady state operation (the angular acceleration α = 0) the motor torque has to make friction torque correspond proportionally to the angular speed and load torque at that specific angular speed. The braking torque and power need in respect to time varies greatly in these two different load types.
Let us first consider the case where the load is constant torque type and the drive system is not able to generate braking torque, i.e., the drive itself is single quadrant type. In order to calculate the braking time needed one can apply the following equation. Please note that formula (1) underlines that the torque needed for inertia accelerating (or decelerating), friction and load torque is in the opposite direction to the motor torque.
In practice, it is difficult to define the effect of friction exactly. By assuming friction to be zero the time calculated is on the safe side. (β = 0)
On rewriting the motor torque equation:
This applies for those applications where the load torque remains constant when the braking starts. In the case where load torque disappears the kinetic energy of the mechanics remains unchanged but the load torque that would decelerate the mechanics is now not in effect. In that case if the motor is not braking the speed will only decrease as a result of mechanical friction. Therefore, in the case of constant torque condition, the braking torque cannot be developed as high as the starting torque. The natural braking effect is at its maximum at the beginning of the braking. This clearly indicates that it is not necessary to start braking the motor till it reaches a certain threshold value.
COORDINATION OF ELECTRIC AND MECHANCIAL BRAKES: Basically, there are three different brake control strategies: series braking with optimal braking feel; series braking with optimal energy recovery; and parallel braking.
Figure 14. Illustration of parallel braking strategy.
The parallel brake system has a conventional mechanical brake which has a fixed ratio of braking force distribution on the front and rear wheels. Regenerative braking adds additional braking force to the front wheels, resulting in the total braking force distribution curve. The mechanical braking forces on the front and rear axles are proportional to the hydraulic pressure in the master cylinder. The regenerative braking force developed by the electric motor is a function of the hydraulic pressure of the master cylinder, and therefore a function of vehicle deceleration. Because the regenerative braking force available is a function of motor speed and because almost no kinetic energy can be recovered at low motor speed, the regenerative braking force at high vehicle deceleration (e.g., a/g = 0.9) is designed to be zero so as to maintain braking balance. When the demanded deceleration is less than this deceleration, regenerative braking is effective.
Figure 15. Braking forces varying with deceleration rate.
When the braking deceleration commanded is less than a given value, say 0.15 g, only regenerative braking is applied. This emulates the engine retarding in a conventional vehicle. In Figure 14, the regenerative braking force, Fbf-regen, and mechanical braking forces on the front and rear wheels, Fbf-mech and Fbr, are illustrated. Figure 15 shows the total braking force, regenerative braking force, and mechanical braking force on the front wheels as well as the braking force on the rear wheels in the parallel braking system of a passenger car.
The parallel braking system does not need an electronically controlled mechanical brake system. A pressure sensor senses the hydraulic pressure in the master cylinder, which represents the deceleration demand. The pressure signal is regulated and sent to the electric motor controller to control the electric motor to produce the demanded braking torque. Compared with the series braking of both optimal feel and energy recovery, the parallel braking system has a much simpler construction and control system. However, the driver’s feeling, and amount of energy recovered are compromised.
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.
EXPLANATION AND OBSERVATION:
Constant |
Value |
Copper loss coefficient (k_c) |
0.3 |
Iron loss coefficient (k_i) |
0.008 |
Winding loss coefficient (k_w) |
0.00001 |
Coefficient accounting for constant motor loss (c ) |
25 |
Table 3. Description of the values of constants.
Constant |
Value |
Copper loss |
(Torque)2 (k_c) |
Iron loss |
(Angular velocity * k_i) |
Winding loss |
(angular velocity)3 (k_w) |
Table 4. Calculation of various losses incurred by an induction motor.
%% PLOTTING CONTOURS OF MOTOR SPEED, TORQUE AND EFFICIENCY
clear
close all
clc
w = linspace(0,1800,100); %angular velocity in rad/sec
tor = linspace(0,100,100); %torque in Nm
%Defining the motor constants
k_c = 0.3; %Copper loss coefficient
k_i = 0.008; %Iron loss coefficient
k_w = 0.00001; %Windage loss coefficient
c = 25; %constant motor loss coefficient
%Creating a mesh i.e. 2D array of angular velocity and torque
[W,T] = meshgrid(w,tor);
Power_output = [W.*T]; % product of torque and angular velocity
Copper_Loss = (T.^2) * k_c;
Iron_Loss = (W * k_i);
Windage_Loss = (W.^3)*k_w;
Power_input = Copper_Loss + Iron_Loss + Windage_Loss + Power_output + c;
efficiency = Power_output./Power_input;
%PLOTTING
figure(1);
surf(W,T,efficiency);
colorbar
xlabel('Angular velocity (rad/sec)');
ylabel('Torque (Nm)');
zlabel('Efficiency');
title('Efficiency Curve for an Induction Motor');
grid off;
E = [0.70,0.75,0.80,0.85,0.90,0.91,0.92,0.93,0.94,0.95]; %selecting the efficiencies to appear on the plot
figure(2);
box off;
grid off;
[C,h] = contourf(W,T,efficiency,E);
clabel(C,h);
colormap(winter);
colorbar;
xlabel('Angular speed (rad/sec)');
ylabel('Torque (Nm)');
zlabel('Efficiency');
title('Torque-Speed characteristic curve for an induction motor');
Figure 16. Layout of the drive cycle.
It can be observed from the above figure of the drive cycle that the braking energy is needed from the time period as shown below:
Time (sec) |
Velocity (m/s) |
Braking Energy (J) |
9 |
20 |
42000 |
10 |
17 |
55500 |
11 |
12 |
72500 |
12 |
6 |
54000 |
13 |
3 |
13500 |
14 |
0 |
4500 |
18 |
19 |
132000 |
19 |
17 |
36000 |
21 |
13 |
115500 |
22 |
8 |
52500 |
24 |
2 |
58500 |
28 |
18 |
258500 |
29 |
7 |
137500 |
30 |
0 |
24500 |
Table 5. Depiction of the braking energy produced in the drive cycle.
The maximum value of the braking energy occurs when the difference in the velocities is the maximum and that occurs when the drive cycle moves from 27th to the 28th second. The velocity changes from 29m/s to 18 m/s. The braking energy produced during this time is the maximum braking energy and it amounts to 2,58,500 J. Whereas the least braking energy occurs when the velocity decreases from 3 m/s to 0 m/s as the drive cycle moves from 13th second to 14th second. The least braking energy obtained is 4,500 J.
Figure 17. Efficiency curve of an induction motor.
Figure 18. Torque-Speed characteristics of an induction motor..
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