Parallel Operation of ac generators Ee 107 AC Machinery Parallel Operation of AC Generators Parallel Operation of ac generators

Frequency-Power Characteristics Effect Of Additional Real Power To Generator Speed Prime mover speed decrease as generator load increase. Prime movers include a governor mechanism to make decrease linear with increasing load and provide a slight drooping. Pfl Power (kW) nnl nfl Mechanical speed (rpm) n

Frequency-Power Characteristics Effect Of Additional Real Power To Generator Speed Pfl Power (kW) nnl nfl Mechanical speed (rpm) n where: SD = speed droop of the prime mover nnl = no load speed of the prime mover nfl = full load speed of the prime mover Drooping can range from 2% - 4%.

Frequency-Power Characteristics Effect Of Additional Reactive Power To Generator Speed The speed should not droop for an increase in reactive power (kVAR), since this power is not actual power consumed by load. It does however affect the terminal voltage of the generator.

Frequency-Power Characteristics Adjusting Generator Speed Governor mechanism has set point to adjust the no load speed of the prime mover. When no load speed is adjusted, the whole speed-power curve will be adjusted. n n’nl n’’nl nnl nfl Mechanical speed (rpm) Power (kW) Pfl

Frequency-Power Characteristics Effect Of Additional Real Power To Generator Frequency Since prime mover speed is proportional to generator frequency, the latter also decreases with increasing load. Pfl Power (kW) fnl ffl Frequency(Hz) f where: P = power output of generator fnl = no load frequency ffl = full load frequency fsys = operating frequency of system at power P. sP = inverse of slope of curve, kW/Hz or MW/Hz. fsys P To adjust generator frequency, the governor mechanism can be adjusted accordingly.

Checkpoint § Express the power-frequency relationship of the generator in terms of the speed droop SD.

Illustrative Problem 1 § A 8-pole generator is operating at 60 Hz at no-load. At full-load of 100 kW, the operating frequency goes down to 58.5 Hz. The governor mechanism on the prime mover is set so that the frequency varies linearly with the power output. What is the prime mover rpm at no-load and full-load Express the power P delivered by the generator as a function of its operating frequency. Calculate for sP? If the operating frequency needs to be at 60 Hz at full load, at what frequency level should the generator be at no-load? What should be the prime mover rpm to attain this no-load frequency? If an additional load of 25 kW is connected to the generator without changing its frequency, what should be the prime-mover rpm at no-load?

Illustrative Problem 1 § Solution: Pfl Power (kW) fnl ffl Frequency(Hz) f P fsys Let fsys = operating frequency of generator for any given power output P.

Illustrative Problem 1 § Solution: f 60 Hz fnl ffl fsys f’nl f’fl fsys 58.5 Hz Since the slope of the curve is the same: Frequency(Hz) P Pfl Power (kW) Frequency – Power curve set at a higher no-load frequency

Illustrative Problem 1 § Solution: f fnl ffl fsys fsys fol 125 P 100 60 Hz fnl ffl fsys fsys 58.5 Hz fol Frequency(Hz) 125 P 100 Power (kW) Frequency – Power curve showing frequency at 125 kW

Illustrative Problem 1 § Solution: f f’nl fnl ffl fsys fsys 125 P 100 60 Hz fnl ffl fsys fsys 58.5 Hz Frequency(Hz) 125 P 100 Power (kW) Frequency – Power curve set at a higher no-load frequency to maintain 58.5 Hz at 125 kW

Terminal Voltage (VT)-Reactive Power (Q) Characteristics Effect Of Additional Reactive Power To Generator Speed Increase in reactive power (kVAR) supplied by generator, does not affect generator speed and thus its frequency, but it affects terminal voltage. Terminal voltage VT droops with additional unity or lagging load but increase with additional leading load

Terminal Voltage (VT)-Reactive Power (Q) Characteristics jXS IA δ Vφ IA θ Generator phasor diagram with lagging load

Terminal Voltage (VT)-Reactive Power (Q) Characteristics jXS IA δ Vφ θ IA Generator phasor diagram with additional reactive (inductive) load (Vφ and therefore VT decreases drastically!) (Note: Real power or load in this diagram is constant since IA cos θ remains constant, hence, frequency of generator (and speed) should NOT be affected)

Terminal Voltage (VT)-Reactive Power (Q) Characteristics jXS IA δ Vφ θ IA Generator phasor diagram with additional reactive (inductive) load (Vφ and therefore VT decreases drastically!) (Note: Real power or load in this diagram is constant since IA cos θ remains constant, hence, frequency of generator (and speed) should NOT be affected)

Terminal Voltage (VT)-Reactive Power (Q) Characteristics VT also decreases with increasing unity power factor load. EA jXS IA δ IA Vφ Generator phasor diagram with unity load (Note: Real power or load in this diagram increased since IA cos θ increased, hence, frequency of generator (and speed) will decrease)

Terminal Voltage (VT)-Reactive Power (Q) Characteristics jXS IA δ IA Vφ Generator phasor diagram with additional real power (Vφ and therefore VT decreases slowly)

Terminal Voltage (VT)-Reactive Power (Q) Characteristics Qfl Q (Reactive Power) kVAR supplied by generator VTnl VTfl Voltage(VT) Terminal -Q (Reactive Power) kVAR absorbed by generator

Terminal Voltage (VT)-Reactive Power (Q) Characteristics How do we maintain the terminal voltage for varying loads? We vary the field current, decreasing it for increasing leading power factors and increasing it for lagging and unity power factor loads.

Summary of Frequency and Terminal Voltage Characteristics To maintain the frequency of the generator for varying load, the governor set points of the prime mover is adjusted to control its rpm and thus the generator frequency. To maintain the terminal voltage of the generator, the field current of the generator field circuit is adjusted to control the internally generated voltage EA and thus the phase and terminal voltages, Vφ and VT.

Operation of Generators in Parallel with Large Power Systems or Other Generators Why Parallel Operation is Important? Several generators can supply bigger load than a single machine. Using several generators increases the reliability of the power system. Using several generators allows removal of one generator for maintenance without power interruption. Using a single large generator to serve a relatively smaller load can lead to the generator operating at less than full-load, which makes it less efficient, than operating several machines operating at near full load.

Operation of Generators in Parallel with Large Power Systems or Other Generators Conditions for Parallel Operation? The following conditions should be met when putting a generator in parallel with a running system: The RMS value of the oncoming generator must be equal to that of the running system. The phase angle of one phase of the oncoming generator must be the same as that of the same phase of the running system. For three-phase systems, the phase sequence of the oncoming generator must be the same as that of the existing system. The frequency of the oncoming generator must be slightly higher than that of the running system.

What happens if conditions for parallel operation are not met? Operation of Generators in Parallel with Large Power Systems or Other Generators What happens if conditions for parallel operation are not met?

Operation of Generators in Parallel with Large Power Systems The Infinite Bus Is a power system so large that its voltage and frequency do not vary regardless of how much real and reactive power is drawn or supplied to it. VTp f ffl ffl Voltage Frequency(Hz) Power (kW) Q, (kVAR) consumed supplied Infinite bus curves

Operation of Generators in Parallel with Large Power Systems Load Infinite Bus (Connected to an infinitely large generator)

Operation of Generators in Parallel with Large Power Systems Small cart (load) Large Freight Train (Infitely large generator) rigid cable (infinite bus)

Operation of Generators in Parallel with Large Power Systems What happens when a generator is connected on an infinite bus? G Generator Load Infinite Bus

Operation of Generators in Parallel with Large Power Systems Car (incoming generator) flexible cable Small cart (load) Large Freight Train (Infitely large generator) rigid cable (infinite bus)

Operation of Generators in Parallel with Large Power Systems After satisfying the conditions for parallel connection, when a generator is connected in parallel with an infinite bus, all generators (all generators comprising the infinite bus and the incoming generator) must have the same frequency and terminal voltage. At this point, the newly-connected generator is just “floating” on the line supplying little real and reactive power. Load G Generator Infinite Bus

Operation of Generators in Parallel with Large Power Systems Pib, kW Pib-load fnl Power supplied to load by infinite bus PG-load PG, kW Pload = Pib-load + PG-load Generator with exactly the same frequency with infinite bus when paralleled (PG-load ≈ 0) supplying a common load The freq. vs. power curve of the infinite bus curves and that of the incoming generator can be plotted “back-to-back”, having a common frequency axis.

Operation of Generators in Parallel with Large Power Systems Generator with Slightly Lower Frequency: f fnl Pload = Pib-load + PG-load Power supplied to load by infinite bus Pib, kW Pib-load PG-load PG, kW When the no-load frequency of generator is adjusted to become slightly lower than the infinite bus frequency (by decreasing its prime mover speed through the governor set points), EA and hence, the no-load VT will slightly decrease.

Operation of Generators in Parallel with Large Power Systems Generator with Slightly Lower Frequency: f fnl Pload = Pib-load + PG-load Power supplied to load by infinite bus Pib, kW Pib-load PG-load PG, kW When the no-load frequency of generator is adjusted to become slightly lower than the infinite bus frequency (by decreasing its prime mover speed through the governor set points), EA and hence, the no-load VT will slightly decrease. This action is the same as lowering the freq. vs. power curve of the incoming generator.

Operation of Generators in Parallel with Large Power Systems Generator with Slightly Lower Frequency: f fnl Pload = Pib-load + PG-load Power supplied to load by infinite bus Pib, kW Pib-load PG-load PG, kW If the generator is now connected to the bus, the latter will supply power to the generator and the generator will run as a motor. The bus is now supplying power to the load and the generator. Note: Generators are equipped with reverse-power trip that disconnects the generator from the line if it starts to run as a motor.

Operation of Generators in Parallel with Large Power Systems Generator with Slightly Lower Frequency: f fnl Pload = Pib-load + PG-load Power supplied to load by infinite bus Pib, kW Pib-load PG-load PG, kW This will cause the generator rpm to increase (and hence the frequency), raising the freq. vs. power curve and restoring EA and VT from its previous values. Note that the infinite bus is still supplying power to the generator.

Operation of Generators in Parallel with Large Power Systems Generator with Slightly Lower Frequency: f fnl Pload = Pib-load + PG-load Power supplied to load by infinite bus Pib, kW Pib-load PG-load PG, kW At this pt. the generator is still not supplying real and reactive power but in fact, getting power from the bus.

Operation of Generators in Parallel with Large Power Systems Generator with Slightly Higher Frequency: f A fnl Pload = Pib-load + PG-load Power supplied to load Power supplied by infinite bus Power supplied by generator G to load Pib, kW Pib-load PG-load PG, kW When the no-load frequency of generator is slightly higher than the infinite bus frequency, the generator no-load terminal voltage will slightly increase. When connected to the bus, the generator will supply power (instead of floating), up to a point where its frequency decreases to a level equal to the bus frequency (pt. A).

Operation of Generators in Parallel with Large Power Systems Generator with Slightly Higher Frequency: f A fnl Pload = Pib-load + PG-load Power supplied to load Power supplied by infinite bus Power supplied by generator G to load Pib, kW Pib-load PG-load PG, kW Note that the power supplied by the bus to the load will start to decrease as some of the load will be shouldered by the generator (assuming the generator do not supply power to the bus).

Operation of Generators in Parallel with Large Power Systems Generator at Higher Frequency: f fnl B Pload = Pib-load + PG-load Pib, kW Pib-load PG-load PG, kW When the frequency of generator is further increased (while supplying load), more power will be supplied by it. This is the same as raising the freq. vs. power curve of the generator.

Operation of Generators in Parallel with Large Power Systems Generator at Higher Frequency: f fnl B Pload = Pib-load + PG-load Pib, kW Pib-load PG-load PG, kW As more power is supplied, the generator frequency will start to decrease (since the prime mover rpm will start to decrease) up to a point where the generator frequency equalizes with the infinite bus frequency (pt. B). Note that as the governor mechanism is adjusted, VT changes in a transient manner but eventually settles to a constant value equal to the bus voltage, since it has to follow the bus voltage.

Operation of Generators in Parallel with Large Power Systems What happens to the generator phasor diagram as the real power of the generator increases?

Operation of Generators in Parallel with Large Power Systems Generator Phasor Diagram: The generator phasor diagram as the frequency is adjusted to higher value (by governor set points) for negligible resistance RA (leading power factor) IA Vφ EA δ EA sinδ IAXS θ When a generator is connected in parallel to an infinite bus, its operating frequency and terminal voltage cannot change and must conform with that of the bus. That means Vφ must be constant as well regardless of load even at no load conditions. Since EA = Vφ at no load, EA must be constant regardless of load provided field current is held constant. For this to happen, the generator has to operate at a leading power factor.

Operation of Generators in Parallel with Large Power Systems Generator Phasor Diagram: The generator phasor diagram as the frequency is adjusted to higher value (by governor set points) for negligible resistance RA (leading power factor) IA Vφ EA δ EA sinδ IAXS θ Hence for negligible RA (negligible copper loss) EA sinδ is proportional to the generator’s supplied real power since Vφ is constant. Since,

Operation of Generators in Parallel with Large Power Systems Generator Phasor Diagram: The generator phasor diagram as the frequency is adjusted to higher value (by governor set points) for negligible resistance RA (leading power factor) IA Vφ EA δ EA sinδ IAXS θ Also since EA is constant (field current remains untouched and generator freq. follows bus freq. quickly every time it’s incremented by governor set points), the torque angle δ must be changing. Thus, as real power is increased, the torque angle increases. Since,

Operation of Generators in Parallel with Large Power Systems IA I’A Vφ EA E’A δ δ’ EA sinδ E’A sinδ’ Operation of Generators in Parallel with Large Power Systems θ’ Generator Phasor Diagram: The generator phasor diagram as the frequency is adjusted to higher value (by governor set points) for negligible resistance RA (leading power factor) I’AXS IAXS Also since EA is constant (field current remains untouched and generator freq. follows bus freq. quickly every time it’s incremented by governor set points), the torque angle δ must be changing. Thus, as real power is increased, the torque angle increases. Since,

Operation of Generators in Parallel with Large Power Systems IA I’A Vφ EA E’A δ δ’ EA sinδ E’A sinδ’ Operation of Generators in Parallel with Large Power Systems θ’ Generator Phasor Diagram: The generator phasor diagram as the frequency is adjusted to higher value (by governor set points) for negligible resistance RA (leading power factor) I’AXS IAXS The generator is actually supplying negative reactive power which is the same as the generator is actually consuming reactive power.

Operation of Generators in Parallel with Large Power Systems Vφ EA IA δ θ jIAXS IA Cos θ EA Sin θ Constant Constant* Increasing increasing increasing proportionately with IA increasing proportionately with real power output Equal to IA Cos θ Behavior of variables as real power supplied by generator increases * Provided field current is held constant as real power drawn is increased. Note also that EA momentarily increases when rpm (and therefore frequency) of the generator is increased but eventually settles to the same value as it shares more real power to the load.

Operation of Generators in Parallel with Large Power Systems Thus to totally transfer the power source of a load from the infinite bus to the generator, we gradually increase the generator frequency until all power requirement of the load equals the power supplied by the generator. The generator frequency can be increased by adjusting the prime mover governor set points. Further increase in generator power generation, will result in the generator supplying power to the infinite bus. Note that while VT (and EA for that matter) change with change in frequency, this change is only momentary as the generator will have to adjust to the infinite bus frequency and terminal voltage. Hence, as VT and EA settles to a constant value, the current and associated power factor supplied by the generator changes.

Illustrative Problem 2 § A 200-kW, 0.8 lagging power factor load currently draws power from a 400-V, 60-Hz infinite bus. A 400-V, ∆-connected generator with synchronous reactance of 1.0 Ω is to be connected in parallel with an infinite bus to take over the load. Copper losses are negligible. What is the current drawn from the infinite bus (IB) before the generator is paralleled? What is the current drawn from the IB immediately after the generator is connected to the bus. What is the current drawn from the generator? The governor on the prime mover is gradually adjusted to transfer real power to the generator. At the point where the generator is delivering half of the real power requirements of the load, what is the current drawn from both the IB and the generator? What is the power factor of the generator? If the governor set point is adjusted such that the generator is now supplying 90% of the real power, find current supplied by the generator and the infinite bus.

Illustrative Problem 2 G2 IB IL Load 200-kW, 0.8 lagging power factor, 400-V, 60-Hz G2 Generator Infinite bus

Illustrative Problem 2 G2 At the time the generator is paralleled IB Load 200-kW, 0.8 lagging power factor, 400-V, 60-Hz G2 Generator IG = 0 Infinite bus At the time the generator is paralleled

Illustrative Problem 2 Solution: a) Before generator is put in parallel with infinite bus (put on line), the power and current drawn by the load will be supplied by infinite bus. Thus,

Illustrative Problem 2 b) At the instant the generator is paralleled to the infinite bus, its status will be floating supplying very little real power and current. Thus, The IB and generator curves and the phasor diagram of the generator at this time will look like this,

Illustrative Problem 2 VT f A A Pib, kW Qib, kVAR PG, kW QG, kVAR 60 Hz 400 V A A Pib, kW Qib, kVAR PG, kW QG, kVAR 200 150 Infinite bus and generator curves at the instant generator is put on line with bus VφG = EAG = 400 V Phasor diagram of generator at the instant generator is put on line with bus

Illustrative Problem 2 c) When the prime mover governor set points is adjusted for the generator to take on half of the load real power, the curves and phasor diagram will look like this, VT IAG VφG EAG δG EAG sinδG = XSIAG cos θi θG jXSGIAG f B 60 Hz 400 V A Pib, kW Qib, kVAR 100 100 PG, kW 160.5 -10.5 = QG QG, kVAR Infinite bus and generator curves when generator is supplying 100 kW Phasor diagram of generator supplying 100 kW

Illustrative Problem 2 Solution: c) Generator phasor diagram at 100 kW EAG jXSGIAG c) θG EAG sinδG = XSIAG cos θi IAG δG θG VφG Generator phasor diagram at 100 kW For the generator,

Illustrative Problem 2 Solution: c) Generator phasor diagram at 100 k VφG EAG δG EAG sinδG = XSIAG cos θG θG jXSGIAG Generator phasor diagram at 100 k Illustrative Problem 2 Solution: c) (for RA = 0) Since magnitude of EA does not change, |EAG| = |VφG|

Illustrative Problem 2 Solution: c) Generator phasor diagram at 100 kW VφG EAG δG EAG sinδG = XSIAG cos θi θG jXSGIAG Illustrative Problem 2 Solution: c) Generator phasor diagram at 100 kW (since generator is Δ-connected)

Illustrative Problem 2 Solution: c) Generator phasor diagram at 100 kW VφG EAG δG EAG sinδG = XSIAG cos θi θG jXSGIAG Illustrative Problem 2 Solution: c) Generator phasor diagram at 100 kW Alternatively, IAG can be found from the formula,

Illustrative Problem 2 Solution: c) Generator phasor diagram at 100 kW VφG EAG δG EAG sinδG = XSIAG cos θi θG jXSGIAG Illustrative Problem 2 Solution: c) Generator phasor diagram at 100 kW The reactive power on the generator would be, Since the reactive power is negative, the generator is actually consuming reactive power.

Illustrative Problem 2 Solution: (Using complex power) The above result shows that the generator is supplying real power but receiving reactive power (Q is negative).

Illustrative Problem 2 Solution: By energy conservation, This means, that the infinite bus is supplying real and reactive power. Note that the reactive power supplied by the infinite bus is the total of the reactive power consumed by the generator and the load.

Illustrative Problem 2 Solution: (not complex power)

Illustrative Problem 2 G2 IB IL Load 200-kW, 0.8 lagging power factor, 400-V, 60-Hz G2 Generator IG Infinite bus When generator is delivering ½ of the total real power.

Operation of Generators in Parallel with Large Power Systems Behavior of Reactive Power Supplied by Generator: While real power supplied by generator increased (thru small momentary increases in frequency thru the governor set points*), the reactive power did not. In fact, at this point, the generator is receiving reactive power from the infinite bus. How can the generator supply reactive power? Ans: by increasing the field current

Operation of Generators in Parallel with Large Power Systems Behavior of Reactive Power Supplied by Generator: When field current IF is increased, EA will proportionately increase assuming no rotor core saturation. While EA is increasing for increasing IF, the real power output of the generator (POUT) remains the same (ignoring losses) since this power supplied by the prime mover is constant (POUT=PIN=τAPP ωm ,ignoring losses).

Operation of Generators in Parallel with Large Power Systems EA IA cosθ = constant jIAXS EA sinδ = IAXscosθ = constant δ IA θ IA sinθ Vφ Phasor Diagram at IF = IF1 When POUT is constant, IAXScos θ must be constant. If IAXScosθ is constant, then EAsinδ must be constant as well since both expressions are equal (for negligible RA). If IAXScosθ is constant, then IAcosθ must be constant since XS does not change.

Operation of Generators in Parallel with Large Power Systems EA IA cosθ = constant jIAXS EA sinδ = IAXscosθ = constant δ θ IA sinθ Vφ IA Phasor Diagram at IF = IF2

Operation of Generators in Parallel with Large Power Systems EA IA cosθ = constant EA sinδ = IAXscosθ = constant jIAXS δ Vφ θ IA sinθ IA Phasor Diagram at IF = IF3 Based on the phasor movement for different values of IF, the magnitude of IAXS is increasing. Since XS is constant, IA must be increasing.

Operation of Generators in Parallel with Large Power Systems EA IA cosθ = constant EA sinδ = IAXscosθ = constant jIAXS δ Vφ θ IA sinθ IA Phasor Diagram at IF = IF3 In addition, IA sinθ is also increasing. Since Q = 3VφIAsinθ, this means Q, the supplied reactive power, must be increasing with increasing IF. Note: since Q is positive for this case, θ below the horizontal axis is considered positive. Above the horizontal, it’s negative.

Illustrative Problem 3 § After all real power requirements of the load has been transferred to the generator in IP No. 2, we now adjust the field current to do the same for the reactive power requirements. Assuming, 50% of the reactive power has been transferred, calculate The line current supplied by the generator The current supplied infinite bus as of this time. Solution: Before adjusting field current, the current supplied and power factor of the generator is: Therefore;

Illustrative Problem 3 Therefore: Although IAG and θG varies, together, this expression is proportional to the generator real output power Pout remains a constant while field current is changed. From the load, the reactive power drawn is: Therefore: From generator, the supplied reactive power is: (varies while IF is adjusted (see phasor diagram))

Illustrative Problem 3 (constant) Note: The generator is now operating at lagging power factor

Operation of Similar Generators in Parallel Load (Pload, Qload) f Pload PG1=Pload Power (kW) The frequency-power curve at the time G2 is connected

Operation of Similar Generators in Parallel At the time G2 is connected, the real and reactive power requirement of the load are still with G1. When frequency of G2 is gradually increased (through the governor set points of the prime mover) , G2 will slowly shoulder part of the real power of the load, while G1 will release part of the real power. The system operating frequency fsys of the two generators will increase. When field current of G2 is gradually increased (through the field current circuit in the G2), G2 will slowly shoulder part of the reactive power of the load, while G1 will release part of the reactive power. The system terminal voltage VT of the two generators will increase.

Operation of Similar Generators in Parallel At any given time, the sum of the real power supplied by G1 & G2 is equal to the real power demanded by the load. Likewise, the sum of the reactive power supplied by G1 & G2 is equal to the reactive power demanded by the load.

Operation of Similar Generators in Parallel Pload PG1=Pload Power (kW) The frequency-power curve at the time G2 is connected

Operation of Similar Generators in Parallel Pload PG1 PG2 Power (kW) The frequency-power curve when frequency of G2 is increased

Operation of Similar Generators in Parallel Pload PG1 PG2 Power (kW) The frequency-power curve when frequency of G2 is further increased

Operation of Similar Generators in Parallel Pload PG1 PG2 Power (kW) The frequency-power curve when frequency of G2 is further increased

Operation of Similar Generators in Parallel VT Qload QG1=Qload Reactive Power (kVAR) The terminal voltage-reactive power curve when G2 is connected

Operation of Similar Generators in Parallel VT Qload QG1 QG2 Reactive Power (kVAR) The terminal voltage-reactive power curve when G2 field current is further increased

Operation of Similar Generators in Parallel VT Qload QG1 QG2 Reactive Power (kVAR) The terminal voltage-reactive power curve when G2 field current is further increased

Operation of Similar Generators in Parallel VT Qload QG1 QG2 Reactive Power (kVAR) The terminal voltage-reactive power curve when G2 field current is further increased

Operation of Similar Generators in Parallel Is it possible to change the real power sharing of the two generators without changing the operating frequency? Increase the governor set point of one generator and simultaneously decrease the governor set point of the other

Operation of Similar Generators in Parallel fsys Pload PG1=Pload Power (kW) The frequency-power curve at the time G2 is connected

Operation of Similar Generators in Parallel fsys Pload PG1 PG2 Power (kW) The frequency-power curve when frequency of G2 is increased

Operation of Similar Generators in Parallel Is it possible to change the reactive power sharing of the two generators without changing the terminal voltage? Increase the field current of one generator and simultaneously decrease the field current of the other

Operation of Similar Generators in Parallel VT Qload QG1=Qload Reactive Power (kVAR) The terminal voltage-reactive power curve when G2 is connected

Operation of Similar Generators in Parallel VT Qload QG1 QG2 Reactive Power (kVAR) The terminal voltage-reactive power curve when G2 field current is further increased

Illustrative Problem 4 § Two generators are supplying a load. Generator 1 has a no-load frequency of 61.5 Hz and a slope of SP1 of 1 MW/Hz. Generator 2 has a no-load frequency of 60.0 Hz and a slope SP2 of 1 MW/Hz. The two generators are supplying a real load totaling 2.5 MW at 0.8 pf lagging. At what frequency is this system operating, and how much power is supplied by each of the two generators? Suppose an additional 1-MW load were attached to this power system. what would the new system frequency be, and how much power would G1 and G2 supply now? With the system in the configuration described in part b, what will the system frequency and generator powers be if the governor set points on G2 are increased by 0.5 Hz?

More Illustrative Problems There are two alternators, 100 kW, 3-ф in parallel are driven by shunt motors whose speed-load characteristics are as follows: Alternator A: NNL= 600 rpm and NFL=530 rpm; while alternator B: NNL= 590 rpm and NFL=550 rpm. What is the greatest load that can be delivered without overloading either alternator? (PT=171.428kW) Two alternators are operating in parallel, supplying a load drawing 1000 kVA at 0.80pf lagging. If alternator A contributes 500 kVA at 0.60pf lagging, determine the pf of alternator B. (0.928 lagging) Alternator A (100kVA, 3-ф, 240 v, 60hz, 1800 rpm) is operating in parallel with alternator B (125 kVA, 3-ф, 240 v, 60hz, 1800 rpm). The load of alternator A is 60kW at 90% pf lagging and the load of alternator B is 80kW at 70% pf lagging. Determine the pf of the load. (0.784 lagging) Two alternators are connected in parallel. The total load is 4250 kW at 0.85 pf lagging. Alternator A operates with a load of 2125kW at 0.707 pf lagging. Determine the kVA load of alternator B. (2185kVA)

More Illustrative Problems Two 3- φ 4160V, 60Hz alternators are operated in parallel. The total load of the system as 1050 W at 75% lagging pf. If alternator A is carrying 700kW at 80% pf lagging, determine kVAR of alternator B. (401kVAR) Two Y-connected alternators A and B running in parallel supply the following loads at 3.3kV: Load 1: 800 kW at unity pf Load 2: 600 kW at 0.8 pf lagging Load 3: 400 kW at 0.707 pf lagging If alternator A is adjusted to carry an armature current of 150A at 0.85pf lagging, what is the armature current of alternator B?

More Illustrative Problems Two generators connected in parallel is supplying a load. Generator 1 has a no-load frequency of 61.5Hz and a slope sP1 of 1MW/Hz. Generator 2 has a no-load frequency of 61Hz and a slope sP2 of 1MW/Hz. The two generators are supplying a real load totaling 2.5MW at 0.8 pf lagging. (a) At what frequency is this system operating, and how much power is supplied by each of the two generators? (60Hz , 1.5MW and 1MW) (b) Suppose an additional 1-MW load were attached to this power system. What would the new system frequency be,, and how much power would G1 and G2 supply now. (59.5Hz, 2MW and 1.5MW) (c) With the system in configuration described in part b, what will the system frequency and generator powers

References Electric Machinery Fundamentals, Stephen J. Chapman, 5th Edition, McGraw Hill

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