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barcode reader code in c# net Hz; 4 poles Circuit parameters: RS = 15 s = 005 in Software
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XS + Ef
05XR
05Xm
05RR s
Vs + Eb 05XR 05RR 2 s
Figure 1828 Approximate circuit model for singlephase 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
SpecialPurpose 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 Nm (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 singlephase motors EXAMPLE 1811 Analysis of SinglePhase Induction Motor
Problem
Find the following quantities for the singlephase 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 Nm 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 =

