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Synchronous Machines (AC Generators). Muhammad Abdul Majid Slides part 2. Induced voltage: Example.
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Synchronous Machines(AC Generators) Muhammad Abdul Majid Slides part 2
Induced voltage: Example • Example : The peak flux density of the rotor magnetic field in a simple 2-pole 3-phase generator is 0.2 T; the mechanical speed of rotation is 3600 rpm; the stator diameter is 0.5 m; the length of its coil is 0.3 m and each coil consists of 15 turns of wire. The machine is Y-connected. • What are the 3-phase voltages of the generator as a function of time? • What is the rms phase voltage of the generator? • What is the rmsline/terminal voltage of the generator?
Phasor diagram of a synchronous generator A phasor diagram of a synchronous generator with a unity power factor (resistive load) Lagging power factor (inductive load): a larger than for leading PF internal generated voltage EA is needed to form the same phase voltage. Leading power factor (capacitive load). For a given field current and magnitude of load current, the terminal voltage is lower for lagging loads and higher for leading loads.
Example • A 2300-V 1000-kVA 0.8-PF-lagging 60-Hz two pole Y-connected synchronous generator has a synchronous reactance of 1.1 Ω and an armature resistance of 0.15 Ω. • At 60 Hz, its friction and windage losses are 24 kW, and its core losses are 18 kW. The field circuit has a dc voltage of 200 V, and the maximum IFis 10 A. The resistance of the field circuit is adjustable over the range from 20 to 200 Ω
Example • The OCC is shown in Figure on next slide • (a) How much field current is required to make VT equal to 2300 V when the generator is running at no • load? • (b) What is the internal generated voltage of this machine at rated conditions? • (c) How much field current is required to make VT equal to 2300 V when the generator is running at rated • conditions? • (d) How much power and torque must the generator’s prime mover be capable of supplying?
Related Example • Assume that the field current of the generator in above Problem has been adjusted to a value of 4.5 A. • (a) What will the terminal voltage of this generator be if it is connected to a -connected load with an • impedance of 20 ∠ 30 °Ω? • (b) Sketch the phasor diagram of this generator. • (c) What is the efficiency of the generator at these conditions? • (d) Now assume that another identical ∆-connected load is to be paralleled with the first one. What • happens to the phasor diagram for the generator? • (e) What is the new terminal voltage after the load has been added? • (f) What must be done to restore the terminal voltage to its original value?
Measuring parameters of synchronous generator model: Ex • Example : A 200 kVA, 480 V, 50 Hz, Y-connected synchronous generator with a rated field current of 5 A was tested and the following data were obtained: • VT,OC = 540 V at the rated IF. • IL,SC = 300 A at the rated IF. • When a DC voltage of 10 V was applied to two of the terminals, a current of 25 A was measured. • Find the generator’s model at the rated conditions (i.e., the armature resistance and the approximate synchronous reactance).
Measuring parameters of synchronous generator model The approximate value of synchronous reactance varies with the degree of saturation of the OCC. The value is more accurate in unsaturated portion only.
The Synchronous generator operating alone The behavior of a synchronous generator varies greatly under load depending on the power factor of the load and on whether the generator is working alone or in parallel with other synchronous generators. Although most of the synchronous generators in the world operate as parts of large power systems, we start our discussion assuming that the synchronous generator works alone. Unless otherwise stated, the speed of the generator is assumed constant.
The Synchronous generator operating alone Effects of load changes (7.34.1)
The Synchronous generator operating alone The behavior of a synchronous generator varies greatly under load depending on the power factor of the load and on whether the generator is working alone or in parallel with other synchronous generators. Although most of the synchronous generators in the world operate as parts of large power systems, we start our discussion assuming that the synchronous generator works alone. Unless otherwise stated, the speed of the generator is assumed constant.
The Synchronous generator operating alone Effect of increasing inductive load
How to compensate the increase in Load Increase the field current, that increases flux and in turn EA causing the VT to remain same.
The Synchronous generator operating alone load increase effect on generators with Leading PF Unity PF
The Synchronous generator operating alone: Example Example : A 500 V, 60 Hz, Y-connected six-pole synchronous generator has a per-phase synchronous reactance of 1.0 . Its full-load armature current is 60 A at 0.8 PF lagging. Its friction and windage losses are 1.5 kW and core losses are 1.0 kW at 60 Hz at full load. Assume that the armature resistance can be ignored. The field current has been adjusted such that the no-load terminal voltage is 500 V. • What is the speed of rotation of this generator? • What is the terminal voltage of the generator if • It is loaded with the rated current at 0.8 PF lagging; • It is loaded with the rated current at 1.0 PF; • It is loaded with the rated current at 0.8 PF leading. • c. What is the efficiency of this generator (ignoring the unknown electrical losses) when it is operating at the rated current and 0.8 PF lagging? • d. How much shaft torque must be applied by the prime mover at the full load? • how large is the induced countertorque? • e. What is the voltage regulation of this generator at 0.8 PF lagging? at 1.0 PF? at 0.8 PF leading?
The Synchronous generator operating alone: Example Since the generator is Y-connected, its phase voltage is VT(no load) = 500V = EA (line, no load) Vφ = VT/(3)1/2 =288.6 = EA (phase, no load) At no load, IA= 0 and EA= 288.6 V and it is constant since the field current was initially adjusted that way. a. The speed of rotation of the synchronous generator is nm = 120fe/P = 120(60)/6 =1200 rpm 1200 rpm =125.7 rad/s
Example continued0.8 lagging PF jXSIA = (j 1) .( 60 ∠ -36.87) = 60 ∠ 53.13 from phasor diagram E2A = (Vφ + XSIASinθ)2 +(XSIA Cosθ)2
Example continued0.8 lagging PF (288.6)2 = (Vφ + (1.0)(60.0)(Sin 36.87))2 +(60( 1.0) Cos 36.87)2 Vφ = 248V Y connected therefore VT = (3)1/2 Vφ = 430 V
Example continuedUNITY PF E2A = (Vφ)2 +(XSIA)2 (288.6)2 = (Vφ)2 +((1.0)60)2 Vφ =282.V Y connected therefore VT = (3)1/2 Vφ = 488 V
Example continued0.8 leading PF E2A = (Vφ - XSIASinθ)2 +(XSIA Cosθ)2 V φ =320 v
Terminal characteristics of synchronous generators A typical speed vs. power plot A typical frequency vs. power plot shaft speed is linked to the electrical frequency as fe = nmP/120 the power output from the generator is related to its frequency: P = SP (fnl – fsys) Operating frequency of the system Slope of curve, W/Hz
Terminal characteristics of synchronous generators A similar relationship can be derived for the reactive power Q and terminal voltage VT. When adding a lagging load to a synchronous generator, its terminal voltage decreases. When adding a leading load to a synchronous generator, its terminal voltage increases.
Terminal characteristics of synchronous generators • When a generator is operating alone supplying the load: • The real and reactive powers are the amounts demanded by the load. • The governor of the prime mover controls the operating frequency of the system. • The field current controls the terminal voltage of the power system.