Microsoft Word - Synhronous Machines Units 5-8. Web view= 4.02 Ω. 0.8 pf lagging: E ......
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Synchronous Machine Ex.1. A 1200 kVA, 3300 volts, 50 Hz, three phase star connected alternator has an armature resistance of 0.25 Ω per phase. A field current of 40 Amps produces a short circuit current of 200 Amps and an open circuitemf of 1100 volts line to line. Find the % regulation at full load 0.8 pf lagging and leading by using emf method. Soln: Full load current = 1200 x 10 3 / (√3 x 3300) = 210 amps; Voltage per phase V ph = 3300/√3 = 1905 volts Synchronous impedance Z s = oc voltage per phase/ sc current per phase …….. for same excitation = (1100/√3) / 200 = 3.17 Ω Synchronous reactance = X s = √[( Z s ) 2 - (R a ) 2 ] = √ (3.17) 2 + (0.25) 2 = 3.16 Ω 0.8 pf lagging: referring to the phasor diagram E ph = √[ (V cosФ + IR a ) 2 + (V sinФ + IX S ) 2 ] =√[(1905 x 0.8 + 210 x 0.25) 2 + ( 1905 x 0.6 + 210 x 3.16) 2 = 2398 volts Voltage regulation = [(E ph – V ph )/ V ph ] x 100 = [(2398 – 1905) / 1905] x 100 = 25.9 % 0.8 pf leading: E ph = √[ (V cosФ + IR a ) 2 + (V sinФ - IX S ) 2 ] =√[(1905 x 0.8 + 210 x 0.25) 2 + ( 1905 x 0.6 - 210 x 3.16) 2 = 1647 volts Voltage regulation = [(E ph – V ph / V ph )] x 100 = [(1647 – 1905) / 1905] x 100 = - 13.54 % 1
Transcript of Microsoft Word - Synhronous Machines Units 5-8. Web view= 4.02 Ω. 0.8 pf lagging: E ......
Synchronous Machine
Ex.1. A 1200 kVA, 3300 volts, 50 Hz, three phase star connected alternator has an armature resistance of 0.25 Ω per phase. A field current of 40 Amps produces a short circuit current of 200 Amps and an open circuitemf of 1100 volts line to line. Find the % regulation at full load 0.8 pf lagging and leading by using emf method.
Soln: Full load current = 1200 x 103/ (√3 x 3300) = 210 amps;
Voltage per phase Vph= 3300/√3 = 1905 volts
Synchronous impedance Zs = oc voltage per phase/ sc current per phase …….. for same excitation
= (1100/√3) / 200 = 3.17 Ω
Synchronous reactance = Xs = √[( Zs)2 - (Ra)2] = √ (3.17)2 + (0.25)2 = 3.16 Ω
0.8 pf lagging: referring to the phasor diagram
Eph = √[ (V cosФ + IRa)2 + (V sinФ + IXS)2]
=√[(1905 x 0.8 + 210 x 0.25)2 + ( 1905 x 0.6 + 210 x 3.16)2
= 2398 volts
Voltage regulation = [(Eph – Vph)/ Vph] x 100
= [(2398 – 1905) / 1905] x 100
= 25.9 %
0.8 pf leading: Eph = √[ (V cosФ + IRa)2 + (V sinФ - IXS)2]
=√[(1905 x 0.8 + 210 x 0.25)2 + ( 1905 x 0.6 - 210 x 3.16)2
= 1647 volts
Voltage regulation = [(Eph – Vph/ Vph )] x 100
= [(1647 – 1905) / 1905] x 100 = - 13.54 %
Ex.2. A 3-phase star connected alternator is rated at 1600 kVA, 13500 volts. The armature resistance and
synchronous reactance are 1.5 Ω and30 Ω per phase respectively. Calculate the percentage
voltage regulation for a load of 1280 kW at a pf of 0.8 leading.
Soln: Full load current = 1600 x 103/ (√3 x 13500 x 0.8) = 68.4 amps;
Voltage per phase Vph= 13500/√3 = 7795volts
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Synchronous Machine
0.8 pf leading: Eph = √[ (V cosФ + IRa)2 + (V sinФ - IXS)2]
=√[(7795 x 0.8 + 68.4 x 1.5)2 + ( 7795 x 0.6 – 68.4 x 30)2
= 6861 volts
Voltage regulation = [(Eph – Vph/ Vph )] x 100
= [(6861 – 7795) / 7795] x 100
= - 12 %
Ex.3. A 3-phase star connected alternator is rated at 100 kVA. On short-circuit a field current of 50 amp gives the full load current. The e.m.f. generated on open circuit with the same field current is 1575 V/phase. Calculate the voltage regulation at (a) 0.8 power factor lagging, and (b) 0.8 power factor leading by synchronous impedance method. Assume armature resistance is 1.5 Ω.
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Soln:
Synchronous Machine
Ex. 4. A 10 MVA 6.6 kV, 3phase star connected alternator gave open circuit and short circuit data as follows. Field current in amps: 25 50 75 100 125 150 OC voltage in kV (L-L): 2.4 4.8 6.1 7.1 7.6 7.9 SC Current in Amps: 288 528 875 Find the voltage regulation at full load 0.8 pf lagging by emf method. Armature resistance per phase =
0.13 Ω. Soln: Full load current = 10 x 106/ (√3 x 6600) = 875 amps;
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Synchronous Machine
Voltage per phase Vph= 6600/√3 = 3810volts Corresponding to the full load current of 875 amps oc voltage from the oc and sc characteristics is 6100 volts
Hence synchronous impedance Zs = oc voltage per phase/ sc current per phase
= (6100/√3)/ 875
= 4.02 Ω0.8 pf lagging: Eph = √[ (V cosФ + IRa)2 + (V sinФ + IXS)2]
=√[(3810 x 0.8 + 875 x 0.13)2 + ( 3810 x 0.6 – 875x 4.01789)2
= 6607.26volts
Voltage regulation = [(Eph – Vph/ Vph )] x 100
= [(6607.26 – 3810) / 3810] x 100
= 73.42%
Ex. 5 The data obtained on 100 kVA, 1100 V, 3-phase alternator is : DC resistance test, E between line = 6 V dc, I in lines = 10 A dc. Open circuit test, field current = 12.5 A dc, line voltage = 420 V ac. Short circuit test, field current = 12.5 A, line current = rated value, calculate the voltage regulation of alternator at 0.8 pf lagging.
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Synchronous Machine
Soln:
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