 DETERMINATION OF INDUCTION MOTOR PARAMETERS
The motor is a three phase 158-W, 240-V induction motor (Model 295 Bodine Electric Co.)
The motor is Y-connected with no access to the neutral point.
DC Resistance Test:
To determine R1;
Connect any two stator leads to a variable voltage DC power supply.
Adjust the power supply to provide rated stator current.
Determine the resistance from the voltmeter and ammeter readings.
As shown in figure 3.7, a DC voltage VDC is applied so that the current IDC is close to the motor rating.
Because the machine is Y-connected: RS = Rdc/2 = (VDC/IDC)/2.
From measurement, VDC = 30.6V, IDC = 1.05A.
Hence,
RS = RDC = (31.5/1.04) = 15.14?/phase.
2 2

Figure 3.7 Circuit for DC resistance test.
BLOCKED – ROTOR TEST
To determine X1 and X2
Determine R2 when combined with data from the DC test.
Block the rotor so that it will not turn.
Connect to a variable voltage supply and adjust until the blocked – rotor current is equal to the rated current.
To determine the magnetizing reactance, Xm and combined core, friction, and wind age losses.
Connect as in block rotor test below.
The rotor is unblocked and allowed to run unloaded at rated voltage and frequency.
The set up for no load test and blocked rotor test is shown in the figure below:

We Will Write a Custom Essay Specifically
For You For Only \$13.90/page!

order now

We Will Write a Custom Essay Specifically
For You For Only \$13.90/page!

order now

Figure 3.8 Circuit for no load and locked rotor test.
With the motor running at no load, measure V, I and P to find the machine reactance Xn =Xis+Xm
Table 4.3 Measured value
Frequency (Hz) 50
Voltage (V) 230
Current (A) 1.32
Real power (W) 158

At no load the per-unit slip is approximately zero, hence the equivalent circuit is as shown in figure 3.9 below.

Figure 3.9 Equivalent circuit of three phase induction motor under no load test.
The real power P represents,
Hysteresis and Eddy current losses (core losses)
Friction and wind age losses (rotational losses)
Copper losses in stator and rotor (usually small as no load)
Phase voltage:
Va =V = 220 = 132V
?3?3
Phase current:
la = 1.32A
Phase real power:
Pa = Pa/3 = 138.2 ÷3 = 46.1W
Phase reactive power:
Q_a = ??(VaIa)2-P2a= ?(((137 x 1.32)2)-(46.1)2)=174.86VAr?_
Xn = Qa=174.86 =100.36?
I2a 1.322
Since S ~ 0,
Xn~ Xls +Xm
3. Locked rotor test
With the rotor locked, the rotor speed is zero and per- unit slip is equal to unity. The equivalent circuit is as shown in Figure 3.10 or Figure 3.11.

Figure 3.10 Equivalent circuit of three phase induction motor under locked rotor test.

Figure 3.11 Simplified equivalent circuit of three phase induction motor under locked rotor rest.

Table 4.4 the tested value
Frequency (Hz) 50
Voltage (V) 68.52
Current (A) 1.3
Real power (W) 105.33

Phase voltage:
Va =V = 68.52 = 39.56V
?3 ?3
Phase current:
la = 1.3A
Active power per phase
Pa = P = 105.33=35.1W
3 3
Reactive power phase
Q_a = ??(VaIa)2-P^2 a= ?(((35.56 x1.3)2)-(35.1)2)=30.08VAr?_

For a class C motor.
Xls = 0.3 x Qa= 0.3 x 30.08 = 5.34?
I2a 1.32
Xlr = 0.7 xQa = 0.7 x 30.08 = 12.46?
12a 1.32
From the no – load test, Xn = 100.36?, so
Xm = Xn – Xls = 100.36 – 5.34 = 95.02?
R = Pa = 35.1 = 20.77?
12a 1.32
From figures 3.11,
R2 = R – Ris= 20.77 – 5.34 = 1 5.43?
Comparing figures 3.10 and 3.11,
R2 + jX2 = (Rr + jXir) x jXm
(Rr + jXir) + jXm
R2 =Rr X2m
Rr + (Xlr + Xm)2
Rr = R2 x (Xir + Xm)2 = 15.43 x (12.46 + 95.02)2 = 19.74?
Xm95.02
Summarizing,
Stator winding resistance Rs = 15.14?/phase
Rotor winding resistance Rr = 19.74?/phase
Magnetizing reactance Xm = 95.02?/phase
The magnetizing inductance per phase is
Lm = Xm = 95.02 = 0.3024H
2?f 2? x 50
Stator leakage reactance Xls= 5.34?/phase
The stator inductance per phase is
Lls = Xls= 5.34 = 0.0169H.
2?f 2nx50
Rotor leakage reactance Xlr = 12.46?/phase,
The rotor leakage inductance per phase is
Llr = Xlr =12.46 = 0.0396H.
2?f 2?x50
Table 4.5: Motor parameters
Rated voltage 240V
Maximum torque 1.5N-m
Poles 4
Rated speed 1440rpm
Stator resistance 15.14?
Rotor resistance 19.74?
Stator leakage inductance 0.0169H
Rotor leakage inductance 0.0396H
Mutual inductance 0.3024H

3.3 IMPLEMENTATION
MATLAB fuzzy logic tool box was used in the implementation of the duty ratio fuzzy controller. The graphic user interface included in the tool box was used to edit the membership functions for the inputs (the torque error and the flux position), the output (the duty ratio). A Mamdani type fuzzy inference engine was used in the simulation. The membership functions and the fuzzy rules were adjusted using the simulation until a particular torque ripple reduction was achieved.
To know the performance of the duty ratio controller, the simulation was run at switching frequency of 5KHz. The difference between the conventional DTC and DTC with duty ratio fuzzy control was clearly realized by monitoring the switching behavior of the stator voltage and the electric torque. The selected voltage vector is applied for the complete sampling period and the torque keeps increasing for the complete period, then a zero voltage is applied and the torque keeps decreasing for the complete sampling period and these results in high torque ripple.
The selected voltage vector is applied for part of the sampling period and removed for the rest of the period. As a result, the electric torque increases for part of the sampling period and then starts to decrease. By adjustment of the duty ratio, the desired average torque may be continuously maintained. The duty ratio controller smoothly adjusts the average stator voltage.

x Hi!
I'm Chloe

Would you like to get a custom essay? How about receiving a customized one?

Check it out
x Hi!
I'm Ted!

Would you like to get a custom essay? How about receiving a customized one?

Check it out