# barcode reader code in c# net rotational losses amount to 80 W in Software Drawer QR Code ISO/IEC18004 in Software rotational losses amount to 80 W

rotational losses amount to 80 W
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Analysis: The circuit model for the series machine is shown in Figure 1824 We shall
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Is Rf jX f
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use this model with the understanding that all currents and voltages are now phasors 1 Back emf computation To determine the back emf, we need to calculate the voltage across the armature coil and subtract it from the supply voltage: Eb = VS IS (Rf + j Xf + Ra + j Xa ) = VS (IS )(Rf + j Xf + Ra + j Xa ) The impedance angle is the only unknown quantity and it may be found from the power factor: pf = cos( ) = 0912 (lagging) Thus, Eb = VS (IS )(Rf + j Xf + Ra + j Xa ) = 120 (1785 2224 )(065 + j 12 + 136 + j 16) = 7356 248 V 2 Output power calculation The total power developed by the motor is equal to the product of the back emf times the series current: Ptotal = Eb IS = 7356 1785 = 1,31315 W The mechanical (output) power of the motor is the difference between the total power and the rotational losses: Pout = Ptotal Prot = 1,31315 80 = 1,23315 W 3 Output (shaft) torque calculation The output torque is equal to the ratio of output power to shaft speed: Tout = 1,23315 W Pout = 1472 N-m = 2 800 rad/s 60 = arccos(0912) = 2422
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Figure 1824 Equivalent circuit of a universal motor
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Ra + + V ~ s _ Eb
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Ef ciency calculation The ef ciency of the motor is de ned as the ratio of output power to input power: = Pout 1,23315 Pout = = = 6312% Pin VS IS cos 1,9535
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Comments: Note that the analysis of this machine is very similar to that of the series DC motor, except for the use of phasors It is very important to notice that in calculating the input power, one has to consider the power factor of the motor to obtain the real power
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Special-Purpose Electric Machines
EXAMPLE 188 Universal Motor Torque Expression
Problem
Compute an expression for the average torque generated by a universal motor, based on the circuit diagram of Figure 1824
Solution
Known Quantities: Circuit model of motor Find: Expression for average torque, Tave Assumptions: The motor operates in the linear region of the magnetization curve Analysis: With reference to 17, we know that the ux produced in a series motor by the series current iS (t) is = kS iS (t) The instantaneous torque produced by the machine is given by the expression
T (t) = kT (t)is (t) If the source waveform has period = 2 / , we can calculate the average power by integrating the instantaneous torque over one period:
2 / 2 /
Tave (t) = 2
2 kT kS iS (t
)dt = 2
2 kT kS IS (sin2 t )dt =
1 2 k T k S IS 2
where IS is the rms value of the (series) armature current
Comments: The series motor can produce a nonzero average torque when excited by an
AC current because of the quadratic nature of the instantaneous torque A PM DC machine, which has a linear torque-current relationship, would generate zero average torque if driven from an AC supply
Single-Phase Induction Motors A typical single-phase induction motor bears close resemblance to the polyphase squirrel-cage induction motor discussed in 17, the major difference being in the con guration of the stator winding A simpli ed schematic diagram of such a motor, with a single winding, is shown in Figure 1825; the winding is typically distributed around the stator so as to produce an approximately sinusoidal mmf Assume that the mmf for a practical motor can be generated so as to approximate the following function: F = Fmax cos( t) cos( m ) This function can be written as the sum of two components, as follows:
Figure 1825 Singlephase induction motor
1 F + = 2 Fmax cos( m t) 1 F = 2 Fmax cos( m + t)
Squirrelcage rotor
Stator winding
These two components may be interpreted as representing two mmf waves traveling in opposite directions around the stator Each of these mmf s produces torque