The Equivalent Circuit Of A Synchronous Machine Operating On An Infinite Bus Is Given In Figure The Phasor Diagram Cor 1 (58.99 KiB) Viewed 36 times
The Equivalent Circuit Of A Synchronous Machine Operating On An Infinite Bus Is Given In Figure The Phasor Diagram Cor 2 (57.65 KiB) Viewed 36 times
The Equivalent Circuit Of A Synchronous Machine Operating On An Infinite Bus Is Given In Figure The Phasor Diagram Cor 3 (35.21 KiB) Viewed 36 times
The equivalent circuit of a synchronous machine operating on an infinite bus is given in Figure.. The phasor diagram corresponding to motoring operation is shown in Figure.z. E Z-4X Fig. 1 Equivalent Circuit of A Synchronous Motor on an Infinite Bus Fig 2 Phasor Diagram for Synchronous Motor Operation The mechanical output power, Pm, is given by: P₁ = 3 Re (E la} (1) and from Figure 2, la V-E 2-8 2,28 (2) Substitution of (2) into (1) gives, P. - (V, Cas(5-0)- E. Cos (0)) 3.E Z₂ Now, assume that a constant shaft load is applied to the motor and also assume that the synchronous Impedance, Z, is constant. Then, equations (2) and (3) can be properly combined to obtain a relation between the emf, E, and the armature current, I,. This relation can be further converted to a relation between the armature current, la, and field current, Irusing the open-circuit curve of the machine (The open-circuit curve gives the-relationship between the emf, E, and the field current, I. Although a major allowance for saturation is made by using the open circuit curve, some error is introduced due to the assumption of constant synchronous impedance. A typical set of curves of armature current vs. field current for various values of constant shaft loads is shown in Figure 3. These curves are known as V-curves because of their characteristic "V-shapes". For a constant load on the motor, the armature current is minimum at a particular field current which corresponds to unity power factor. If the field current is gradually decreased below this value, the armature current will increase and therefore the power factor will decrease (lagging) until it is m
practically limited by a point where the machine becomes unstable. A similar phenomenon is observed when the field current is gradually increased. In this case, the power factor decreases in the leading region. If the points corresponding to the same power factor for different loads are connected together, the constant power factor lines are obtained, as shown in Figure 3 with the dashed lines. P Armature Current. I, P>P>P Lagging power factor Power factor = 1.0 Field current, I Fig. 3 Typical V-Curves for A Synchronous Motor EXPERIMENT Fix Vf = 100 V and adjust If by rheostat. Record your readings tables below. No-Load (P3) Load Step I (P2) If [A] la [A] If [A] 0.5 1.53 0.5 1.0 1.0 1.26 1.5 1.03 1.5 2.0 2.0 082 2.5 2.5 0.64 3.0 3.0 0.53 3-5 0.62 4-0 4-5 5.0 5.5 6.0 3-5 4.0 4-5 5.0 5-5 6.0 10.80 1:02 11.30 1.52 B P Leading power factor la [A] 1.50 1.22 1.01 0.80 062 0.51 0.52 0.64 0.82 1'05 1:51 1.57 Load Step II (PI) If [A] la [A] 0.5 11.46 1.0 1.19 1.5 0.97 2.0 0.79 2.5 0.58 3.0 0.50 3-5 4.0 0-68 4-5 5.0 11.07 1.35 5-5 6.0 1:19
1) Does the excitation affect the synchronous motor speed? 2) Does the excitation affect the power factor ?How? 3) Define following with one sentence each: Under-excitation: Over-excitation: 4) What does V-curves tell us? 5) Plot V-curves of the synchronous motor using the obtained data at the experiment. 6) Calculate electrical power values drawn by the synchronous motor: P1 = P2 = P3=
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