1. Lecture 32Electro Mechanical System1 Synchronous Reactance  The value of X S can be determined by measurements of the open-circuit and short-circuit tests  Test are conducted under an unsaturated core condition  Open-circuit test is conducted at rated speed with the exciting current I xn adjusted until the generator terminals are at rated voltage, E n  Short-circuit test is conducted at rated speed with the exciting current I xn gradually raised from 0 amps up to the original value used in the open-circuit test  The resulting short-circuit current Isc is measured, allowing the calculation of X S X S = E n /I SC Where: X S = Synchronous reactance per phase[Ω] E n = Rated open circuit voltage line to neutral [V] I SC = Short-circuit current, per phase, using same exciting current Ixn that was required to produce En [A]

2. Lecture 32Electro Mechanical System2 Example A 3-phase synchronous generator produces an open-circuit line voltage of 6928 V when the dc exciting current is 50 A. The ac terminals are then short-circuited under the same excitation, the three line currents are 800 A 1)Calculate the per-phase synchronous reactance 2)Calculate the resulting terminal voltage if 12-ohm resistors are connected in wye across the terminals Solution: 1)Line to neutral induced voltage is E O = E L /√3 = 6928 / √3 = 4000V Neglecting resistance, synchronous reactance per phase is X S = E n /I SC = 4000/800 = 5Ω

3. Lecture 32Electro Mechanical System3 Example 2)Equivalent circuit per phase is shown The impedance can be calculated as: The current is : I = E O / Z = 4000 / 13 = 308 A Voltage across the resistor is: E = I R = 308 x 12 = 3696 V The line voltage under load is: E L = √3 E = √3 x 3696 = 6402 V

4. Lecture 32Electro Mechanical System4 Synchronous generator under load 1)Current I lags behind terminal voltage E by an angle . 2)Cosine  = power factor of the load. 3)Voltage E x across the synchronous reactance leads current I by 90°. Given by the expression E x = jIX s. 4)Voltage E o generated by the flux  is equal to the phasor sum of E + E s. 5)Both E o and E x are voltages that exist inside the synchronous generator windings and cannot be measured directly. 6)Flux  is that produced by the dc exciting current I x  Phasor diagram shows E o leads E by δ degrees & E o is greater than E  Behavior of a synchronous generator depends upon the type of load it has to supply. Loads, can be of two basic categories: 1)Isolated loads, supplied by a single generator 2) The infinite bus  For isolated loads; Consider a generator that supplies power to a load with a lagging pf. For phasor diagram, we list the following facts:

5. Lecture 32Electro Mechanical System5 Synchronous generator under load  If the load is capacitive, current I leads the terminal voltage by .  In the new phasor diagram, the voltage E x across the synchronous reactance is still 90° ahead of the current.  E o is again equal to the phasor sum of E and E x.  The terminal voltage is now greater than the induced voltage E o, which is a very surprising result.  The inductive reactance X s enters into partial resonance with the capacitive reactance of the load.  It may appear we are getting something for nothing, the higher terminal voltage does not yield any more power.  If the load is entirely capacitive, a very high terminal voltage can be produced with a small exciting current.

6. Lecture 32Electro Mechanical System6 Example A 36 MVA, 20.8 kV, 3-phase alternator has a synchronous reactance of 9Ω and a nominal current of 1 kA. The no- load saturation curve giving the relationship between E O and I x is given. If the excitation is adjusted so that the terminal voltage remains fixed at 21 kV, calculate the exciting current required and draw the phasor diagram for the following conditions: a)No-load b)Resistive load of 36 MW a)Capacitive load of 12 Mvar We shall immediately simplify the circuit to show only one phase. The line-to-neutral terminal voltage for all cases is fixed at E = 20.8/√3 = 12 kV a) At no-load there is no voltage drop in the synchronous reactance; consequently, E o = E= 12 kV Excitation current : I x = 100 A

7. Lecture 32Electro Mechanical System7 Example b) With a resistive load of 36 MW: The power per phase is: P = 36/3 = 12 MW The full-load line current is: I = P/E = 12 x 10 6 /12 000 = 1000 A = 1kA Current is in phase with the terminal voltage. The voltage across X s is E x = jIX s =j1000 x 9 = 9 kV  90° This voltage is 90° ahead of I. The voltage E 0 generated by I x is equal to the phasor sum of E and E x. Referring to the phasor diagram, its value is The required exciting current is: I x = 200 A

8. Lecture 32Electro Mechanical System8 Example c) With a capacitive load of 12 Mvar: The reactive power per phase is: Q = 12/3 = 4 M VAR I =Q/E = 4 X 10 6 /12 000. = 333 A The voltage across X s is : E x = jIX s = j333 x 9 = 3 kV  90 o As before E x leads I by 90° The voltage E O generated by I x is equal to the phasor sum of E and E x. E o = E + E x = 12 + (-3) = 9kV The corresponding exciting current is : I x = 70 A Note that E O is again less than the terminal voltage E.