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60 Hz; 4 poles Circuit parameters: RS = 15 s = 005
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; RR = 3
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Assumptions: The motor is operated at rated voltage and frequency Analysis: With reference to the equivalent circuit of Figure 1826, you can easily show
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that the impedance seen by the backward emf, Eb , is much smaller than that seen by the forward emf, Ef This corresponds to stating that the backward component of the magnetizing impedance (which is in parallel with the backward component of the rotor impedance) is much larger than the backward component of the rotor impedance, and can therefore be neglected This approximation is generally true for values of slip less than 015, and corresponds to stating that j Xm 05Zb = 05 RR + j XR 2 s RR + j XR 2 s
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RR + j (Xm + XR ) 2 s
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The approximate circuit based on this simpli cation is shown in Figure 1828
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Is +
XS + Ef
05XR
05Xm
05RR s
Vs + Eb 05XR 05RR 2 s
Figure 1828 Approximate circuit model for single-phase induction motor
Using the approximate circuit, we nd that the impedance seen by the backward emf is given by the expression 05Zb = 05 RR + j XR 2 s = 05(1538 + j 2) = 05(Rb + j Xb )
The impedance seen by the forward emf is, on the other hand, given by the exact expression: j Xm 05Zf = 05 RR + j XR s
RR + j (Xm + XR ) s j 50(60 + j 2) = 119 + j 1469 = 05Rf + j 05Xf = 05 60 + j (50 + 2)
18
Special-Purpose Electric Machines
If we let ZS = RS + j XS = 15 + j 2 impedance of the motor as follows:
, we can write an expression for the total
Z = ZS + 05Zf + 05Zb = 14169 + j 1769 = 2266 513 Knowing the total series impedance we can calculate the stator current: 110 V VS = = 485 513 A Z 2266 513 We can now calculate the power absorbed by the motor by separately computing the real power absorbed in the forward and backward elds: IS =
2 Pf = IS 05Rf = (485)2 119 = 2799 W 2 Pb = IS 05Rb = (485)2 0769 = 181 W
The net power is the difference between the two components, thus, P = Pf Pb = 2618 W, and the torque developed by the motor is equal to the ratio of the power to the motor speed The synchronous speed can be computed to be: s = 4 f = 1885 rad/s p 2618 W P P = = = 146 N-m (1 s) s 095 1885 rad/s
and, if we assume negligible rotational losses, we have: T =
Comments: Note that the power factor of the motor is pf = cos( 513 ) = 0625 Such low power factors are typical of single-phase motors
EXAMPLE 1811 Analysis of Single-Phase Induction Motor
Problem
Find the following quantities for the single-phase machine of Example 1810: 1 2 3 Output torque Output power Ef ciency
Solution
Known Quantities: Motor operating characteristics Find: Motor torque, T ; output power, Pout ; ef ciency, Schematics, Diagrams, Circuits, and Given Data: Motor operating data:
60 Hz; 4 poles; s = 005
hp; 110 V;
Assumptions: The motor is operated at rated voltage and frequency The combined rotational and core losses are Prot + Pcore = 30 W Analysis:
Output power calculation The motor generated power is the difference between the forward and backward components, as explaind in Example 1810 Thus, P = Pf Pb = 2618 W
Part III
Electromechanics
The motor power is the difference between the generated power and the losses: Pout = P Ploss = 2618 30 W = 2318 W 2 Shaft torque calculation The shaft speed is: = (1 s) s = (1 s) and the torque is: 2318 W Pout = = 129 N-m 179 rad/s Ef ciency calculation To calculate the overall ef ciency of the motor we must account for three loss mechanisms: mechanical losses, core losses, and electrical losses The rst two are given as a lumped number; the electrical losses can be computed by calculating the I 2 R losses in the stator and forward and backward circuits: T =
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