Week-11 Challenge: Braking
1) For a defined driving cycle, calculate the energy required for braking.
step 1: Insert drive cycle xlsx data into matlab
step 2: Add physical parameters of vehicle
step 3:Assign 1st column to T and column 2 data to V
step 4: for loop comparing if there is any declaration in values .
step 5: if declaration found calculate braking energy and add it into an array .
Step6: Converting row data to column data .
%parameters of vehicle
M =1000
%custom Drive cycle
T = finaluddscycle(:,1)
V = finaluddscycle(:,2)
for i = 1:(length(T)-1)
if V(i) > V(i+1)
brakingenergy(i) = (0.5*M*((V(i)-V(i+1)))^2)/3600;
end
end
brakingenergy(i+1)=0
B = brakingenergy.'
plot(T,V,'linewidth',2)
xlabel('Time')
ylabel('Velocity')
fprintf('Total Braking energy required is %.3f Kh',sum(B))Output:
Total Braking energy required is 47.350 Kh>>
2 ) i)Why electric motor can’t develop braking torque at high speed similar to starting?
Regenerative braking is an energy recovery mechanism that slows down a moving vehicle or object by converting its kinetic energy into a form that can be either used immediately or stored until needed. In this mechanism, the electric traction motor uses the vehicle's momentum to recover energy that would otherwise be lost to the brake discs as heat.
There are practically two limitations associated for reason behind low torque at high speed
1) There is a quantative limitation to which a battery can accept charging .
(Most LiPo batteries need to be charged rather slowly, compared to NiMH or NiCd batteries. While we would routinely charge a 3000mAh NiMH battery at four or five amps, a LiPo battery of the same capacity should be charged at no more than three amps. Just as the C Rating of a battery determines what the safe continuous discharge of the battery is, there is a C Rating for charging as well. For the vast majority of LiPos, the Charge Rate is 1C. The equation works the same way as the previous discharge rating, where 1000mAh = 1A. So, for a 3000mAh battery, we would want to charge at 3A, for a 5000mAh LiPo, we should set the charger at 5A, and for a 4500mAh pack, 4.5A is the correct charge rate.)
(NOTE:The safest charge rate for most LiPo batteries is 1C, or 1 x capacity of battery in Amps.)
2 ) For the 2nd reason we need to look at torque speed curve:
As we can see at higher speed the the motor starts operating in constant power region instead of constant torque region ,where available torque is less , so similarly motor is unable to provide maximum torque while braking at high speeds .As the motor speed is reduced the braking torque producing capacity will also increase .So as a common practice by companies we see mechanical brakes available .
ii)How electric and mechanical brakes are coordinated?
Figure 30 shows the simulation results. The motor accelerates until reach the cruise speed (1350 rpm), when the torque reduces to maintain the speed constant. The braking process starts at t=5 s. The torque becomes negative as well as the supply power. The motor follows the speed reference but, at low speed, that corresponds to a low inverter frequency, the torque becomes oscillatory, affecting the deceleration. Another important fact is that the supply power becomes positive. This means that the powertrain losses overcome the power recovery, making the regenerative procedure useless.
Such results indicate the regenerative braking must be interrupted before the effective vehicle stop. The final braking will be performed by the mechanical brake.
Strategies to the Scalar Control
A closed loop torque control is added. During the braking,This way the induction machine slip is negative, resulting a negative torque. The inverter turns-off at low speeds, avoiding the torque oscillations. Then, Mechanical Torque block applies a friction
torque to maintain the same deceleration
It is necessary to know if the required torque is greater than electric motor limit torque. The use of the block "Mechanical Torque" is divided into two parts: (i) Brake1" and (ii) Brake2" to apply friction torque according to the torque demand
The torque developed by the induction motor is directly proportional to the ratio of the applied voltage and the frequency of supply. By varying the voltage and the frequency, but keeping their ratio constant, the torque developed can be kept approximately constant.
In Brake1 mode the initial friction torque is 0, for a maximum speed reference ωre f .The electrical brake is performed with the desired torque while the current in the SC bank is greater than a minimum value. When the inverter turns-off, the desired torque is applied by the friction brake, maintaining the deceleration and the stability of the vehicle. In the Brake2 mode the friction torque is greater than zero for a maximum speed reference ωre f . The desired torque is divided between the electrical and the mechanical system. When the regenerative brake losses efficacy, the inverter turns-off and the friction torque assumes the full torque.
The Brake1 has a longer braking distance than Brake2, due to the fact that brake demand is lower. The reference speed is applied while the inverter is working.
Figure 35 shows the inverter DC bus voltage (where the SC Bank is connected).The recovered energy in Brake1 is lower than Brake2 since the initial kinetic energy is lower.
Strategies for DTC
Direct torque control (DTC) is based on two PI controllers for torque and flux, which allows better response to the driver’s command.
"Brake1" and "Brake2". The desired torque or torque demand is made by the brake pedal, according to the motor speed vs. torque curve. If the desired torque is greater than the motor limit torque, it is necessary to add the mechanical torque. Brake1 is performed when the electric brake is enough. The mechanical brake is applied only when the inverter turns-off. Brake2 uses both, the electric and the mechanical brakes concurrently.when the brake commands 60% of the maximum motor torque (Brake1). If the command corresponds to 100% or more the strategy is Brake2.
It can be observed that more energy is recovered by Brake1 than Brake2, since both start with the same kinetic energy.
Now when the braking power requirement is high two further strategies are used generally which are:
Two versions of regenerative braking[6] are currently available. The first type is serial regenerative braking that is based on a combination of a friction-based adjustable braking system with a regenerative braking system that transfers energy to the electric motors and batteries under an integrated control strategy. The second type is a parallel braking system [7,8] in which the
friction- based braking system and the regenerative braking system are operated in tandem, without integrated control which means that neither the friction braking force nor the regenerative braking force can be adjusted easily. The parallel regenerative braking system has the advantage of not requiring a complex control system and the existing friction based braking system can
be used without alteration. However, the energy recovered by the regenerative braking system will be small, and the drivability of the vehicle may be adversely affected. On the other hand, a series regenerative braking system is more complex to implement than the parallel regenerative braking system, but the energy recovered will be higher and the drivability of the vehicle can be preserved.
file:///D:/downloadd/wevj-02-00363.pdf )
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
A contour plot is a 2d plot representing certain 3 variable in a 2 d plot specifying the 3rd variable using colors.
For efficiency calculation certain losses are to be considered which are:
Iron loss:Ki*w (ki = constant , w = speed)
As the armature core is made of iron and it rotates in a magnetic field, a small current gets induced in the core itself too. Due to this current, eddy current loss and hysteresis loss occur in the armature iron core. Iron losses are also called as Core losses or magnetic losses.
Copper loss:= I^2R (I=current ,R = resistance)term often given to heat produced by electrical currents in the conductors of transformer windings
Copper loss means there is power loss due to both internal & external load resistances on Load condition.
Windage loss:= kw*(w)^3
windage losses occur due to the air inside the rotating coil of the machine. kw= constant ,w=speed
Constant motor loss:=c
This are losses which are unaccounted for in above 3 ,happening due to unknown reasons.
clear all
close all
clc
%define torque and speed
Torque = linspace(0,300,150);
Velocity = linspace(0,1000,1000);
%Ironloss
ki = 0.002
%copperloss
kc= 0.1
%windage
kw=0.00001
%constant motor loss
c=25
%creating contour
[w,t] = meshgrid(Velocity,Torque);
powerout = (w.*t)
%total loss
Loss = (((t.^2).*kc) + (ki.*w) + ((w.^3).*kw) + c)
Inputpower = Loss + powerout
Efficiency = powerout./Inputpower
%efficiency curve value
E = [0.64,0.67,0.68,0.70,0.73,0.75,0.78,0.80,0.83,0.85,0.87,0.90,0.91,0.915,0.92,0.925,0.93,0.94,0.95]
%creating plots:
contourf(w,t,Efficiency,E)
colorbar
xlabel('SPEED')
ylabel('torque')
zlabel('efficiency')
Step 1:Provide torque and speed values
Step 2:Calculate diffrent losses and add it to power output
Step 3:Specify scale of contour data and Calculate efficiency and plot it using contourf command
Output: