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## Calculation of Cogging Torque and Frequency

This section involves finding the maximum magnitude of the cogging torque and its frequency (pulsations per mechanical revolution). In this problem, the cogging torque is the torque ripple,

which is produced from the preferred rotor alignment at certain angular positions at zero load. This means that if the stator is not excited.

The required calculation is done as shown below.

Cogging Torque = the Distance per second * sin (90) * g

Where g is the gravitational acceleration

Cogging Torque = 1500 / 60 * 100 / 1000 * sin (90) * 9.8

Cogging Torque = 21.90292N

Frequency

Frequency = 1500rpm / 60 * sin (90) = 22.34992

Frequency 22 pulses per revolution

rotorposition=0:0.01:10;

torque=sawtooth(2*pi*rotorposition);

plot(rotorposition,torque,'r')

axis([0 10 -5 5]);

grid on

SNR=10; % 10dB

sigma=std(torque)*10^(-SNR/20);

x=torque+sigma*randn(size(rotorposition));

figure,

plot(rotorposition,x,'r');

xlabel('Rotor Position');

ylabel('Torque');

title('Torque Vs. Rotor Position');

grid on

Figure 1:

This problem involves the plotting of the phase back-EMF of the machine at rated speed and finding the constant relating EMF to speed and the magnitude of the fundamental component.

The phase back-emf of the machine = Current * Slot opening factor * sin(θ)

The phase back-emf = 14.142 * 0.4 * sin(θ), where θ is the load angle from 0 to 360.

Code

clear;

theta = 0: 30: 360;

current = 14.142;

OpenFactor = 0.4;

EMF=current * OpenFactor * sin(theta);

figure

plot (theta, EMF);

title (“phase back-emf”);

ylabel(“EMF (V)”);

hold

The expected waveform of the torque pulsations is shown below, against the rotor position.

The result is shown below.

Figure 2: The Phase Back EMF Vs. Load Angle.

The phase back EMF is a factor of the sine function of the load angle.

The constant relating to speed is the rotation of 1500 rpm

The EMF magnitude ranges between -6V and 6V.

This problem involves calculation of the rated torque developed by the machine and is the frequency and magnitude of the torque ripple.

Rated torque developed by the machine

The required calculation is shown below.

The rated Torque = 1500 / 60 * 100 / 1000 * sin (360) * 9.8

The rated Torque = 23.49N

The frequency = 1500 / 60 = 25Hz

Magnitude of the torque ripple = 6

This problem involves the comparison of the flux density in various important parts of the machine such as the magnets, the centre of the stator teeth and the stator yoke for values of current from 0 to 10 times the rated current in steps of one time the rated current. In this case, no load condition is when the phase current is at 0A.

clear;

cur = 0: 10;

FluxMagnet = 0.5 * cur;

FluxStatorTeeth = 0.4 * cur;

FluxStatorYoke = 0.8 * cur;

disp(“Current  magnets  centre of the stator teeth stator yoke”);

disp(cur);

disp(FluxMagnet);

disp(FluxStatorTeeth);

disp(FluxStatorYoke); hold

The result is shown below.

Component flux density

Current            magnets           centre of the stator teeth         stator yoke

0          0          0          0

1          0.5       0.4       0.8

2          1          0.8       1.6

3          1.5       1.2       2.4

## Phase Back-EMF and Constant Relating EMF to Speed

4          2          1.6       3.2

5          2.5       2          4

6          3          2.4       4.8

7          3.5       2.8       5.6

8          4          3.2       6.4

9          4.5       3.6       7.2

10        5          4          8

From the calculation of the three variables, the flux density of the stator yoke is the highest followed by the current magnets and finally the stator teeth.

This task involves the plotting of the mean torque developed by the machine as a function of load current for values of current from 0 to 10 times the rated current in steps of one and the rated current. The linearity of the torque- current characteristic of this machine is shown in figure 2 below.

The code for the solution is shown below.

clear;

cur = 0: 10;

FluxMagnet = 0.5 * cur;

FluxStatorTeeth = 0.4 * cur;

FluxStatorYoke = 0.8 * cur;

Meanflux = (FluxMagnet + FluxStatorTeeth + FluxStatorYoke) / 3;

meanTorque = 1500 / 60 * 100 / 1000 * meanflux;

figure

plot(cur, meanTorque);

xlabel("Current");

ylabel("Mean Torque");

title("Mean Torque vs Current");

The result is shown below.

Figure 2: Mean Torque Vs. Current

This is related to task 4 in the sense that both are linear, having been generated as factors of the current.

This task involves the creation of a nameplate for the electrical machine, and a list of key parameters valid to the end user.

The key Parameters of the electrical Machine are:

Current = 14.142A

Mean Magnetic Flux = 4

This task involves finding the equivalent circuit parameters of the synchronous machine and using the equivalent circuit to calculate machine performance at rated conditions, to analyse the presence of any significant variation from the FEA results.

For Synchronous machine, the load angle alternates between a random value of either 0 or 90.

The performance measured by the is shown below:

Phase back-emf of the machine = Current * Slot opening factor * sin(θ)

The phase back-emf = 14.142 * 0.4 * sin(θ), where θ is the load angle from 0 to 360.

Code

clear;

theta = [0 90 0 0 90 0 90 90 0 0];

current = 14.142;

OpenFactor = 0.4;

EMF=current * OpenFactor * sin(theta);

figure

plot (theta, EMF);

title ("phase back-emf");

ylabel("EMF(V)");

The result is shown below

Figure 3: EMF (V) Vs. Load Angle

There is a variation from the FEA results in that the performance of the new rated machine ranges from 0 V to 5 V while that of the FEA results are from -6V to 6V. The synchronous machine has higher performance than the initial machine.

This task involves plotting of the motor output torque as a function of the load angle (both from the equivalent circuit and FEA) and analysis of any differences between the two calculations.

The Plotted motor output torque is shown below as a function of the load angle (both from the equivalent circuit and FEA).

theta = 0: 10: 360;

current = 14.142;

OpenFactor = 0.4;

Sync = current * sin(theta);

figure

plot (theta, Sync, 'g');

title ("phase back - emf");

ylabel("EMF(V)");

legend ("Sync");

The result is shown below.

Figure 4: Phase Back EMF Vs. Load Angle- equivalent circui.

Figure 5: Phase Back EMF Vs. Load Angle - FEA

There is a significant difference between the FEA and the equivalent circuit (Sync). The values of FEA are less than the values of the equivalent circuit.

This task involves the production of various values of the of the magnet pitch factor (the amount of each pole covered by a magnet) and plotting the effects of the calculated values on both the overall performance and airgap flux density.

To solve this, the magnet pitch factor is varied between 0 to 10, the amount of each pole covered by a magnet. The plotted result is shown below.

clear;

pitchFactor = 0:10;

FluxDensity = 0.5 * pitchFactor. ^ 0.5;

OveralPerformance = 1500 / 60 * 100 / 1000 * pitchFactor. ^ 0.5;

figure

plot (pitchFactor, FluxDensity, 'g', pitchFactor, OveralPerformance, 'r');

xlabel ("Pitch Factor");

ylabel ("Mean Torque");

legend ("Airgap Flux Density", "Overall Performance ");

title ("Flux Density and Overall Performance");

The result is shown below.

Figure 5: Airgap Flux Density and Overall Performance Vs. slot opening factor

The magnet pitch factor ranges from -4 and 10. The effects of the magnet pitch creates an incrementally wider gap between the overall performance and the airgap flux density as seen in the results above. The initial value for the Airgap Flux Density and Overall Performance begins from zero and both increases, with a widening distance between them as the pitch factor increases.

This task involves varying the slot opening factor (the variation between slot width and slot opening width) and plotting and analysing the effects on the magnitude of overall torque, cogging torque and torque ripple.

The slot opening factor is varied between 0 and 10, (the variation between slot width and slot opening width). The plotted result is shown below.

Code

clear;

slotOPFactor=0:180;

TorqueRipple =0.5 * sin(slotOPFactor).^0.5;

CoggingTorque = 1500 / 60 * 100 / 1000 * sin(slotOPFactor).^0.5;

OverallTorque = TorqueRipple + CoggingTorque;

figure

plot (slotOPFactor, TorqueRipple,'g', slotOPFactor, CoggingTorque,'r', slotOPFactor, OverallTorque, 'b');

xlabel ("Slot Factor");

ylabel("Torque");

legend ("Torque Ripple", "Cogging Torque","Overall Torque");

title ("Torque and Slot Factor");

hold

The result is shown below.

Figure 6: Torque Vs Slot Factor

From the analysis, the highest Torque value is the overall Torque, ranging between 0 and 3N. this torque is followed by the Cogging Torque, ranging from 0 to 2.5. There is minimal Torque ripple, ranging between 0 and 0.5, as shown in the plotted figure above. The effects on the magnitude of overall torque, cogging torque and torque ripple. The effect of varying the magnet pitch factor ranges from -4 and 10 reflects on the increase in the number of points of data generation, and a significant difference between the three calculated factors.

Cite This Work

My Assignment Help (2022) Cogging Torque And Back-EMF Analysis [Online]. Available from: https://myassignmenthelp.com/free-samples/eeee3114-electrical-machines-drive-systems-and-applications/telecommunication-electronics-file-A1D3B6F.html
[Accessed 24 February 2024].

My Assignment Help. 'Cogging Torque And Back-EMF Analysis' (My Assignment Help, 2022) <https://myassignmenthelp.com/free-samples/eeee3114-electrical-machines-drive-systems-and-applications/telecommunication-electronics-file-A1D3B6F.html> accessed 24 February 2024.

My Assignment Help. Cogging Torque And Back-EMF Analysis [Internet]. My Assignment Help. 2022 [cited 24 February 2024]. Available from: https://myassignmenthelp.com/free-samples/eeee3114-electrical-machines-drive-systems-and-applications/telecommunication-electronics-file-A1D3B6F.html.

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