Question

Good day all.

I am facing the following problem and would like to seek some kind help, please.

1. An 10 MVA, 33 kV, 34 pole, 60 Hz, three-phase wye-connected synchronous generator has a synchronous reactance of 10 and a negligible armature resistance. a. Find the synchronous speed of this generator. b. At a certain excitation, the generator delivers 50% of rated load at 0.7 leading power factor at '33' kV. Find the internal excitation voltage of the generator. Is the generator overexcited or underexcited? c. Draw the phasor diagram for the generator. d. Draw the Power Triangle for the generator e. For the same excitation as (b), find the armature current and power factor when the input torque is reduced such that the generator is delivering 140% of the real power (P) load in (b). 2. The power consumption of two three phase loads L1 and L2 connected in parallel is respectively 16 KVAR at 0.8 p.f. lagging and '5' kW at '0.9' p.f leading at 420 V, 50Hz as shown in the following Figure 2. 4+6 420 V 3-0,4 Connected, Voltage Source 4+6j L1: Y- Connected 16 KVAR, 0.8 power factor lagging 4+6j L2: A-Connected, 5 kW, 0.9 pf leading 420 V a. Determine the rms voltage of the 3-phase source. b. Determine the total complex power delivered by the source. C. The consumer wishes to have the power factor of the combined load to 0.5 leading power factor. What will be the value of 3-phase delta connected capacitor bank that needs to be connected in parallel to the load?

Sincerely appreciate all help and input.

- Ms Tan

Answer 1

I. at Synchronus speed (s) = 120f Ns & 120 x 60 Ns = 21-76 rpm by at 50% wad I = 5x100 87.47 amp (line-line for leading Giren eta oit leading from. Es cemf) eqn of sun.mic &x=(rosa) + (rsino - I a Xs) - = 33 kr (X Connected (Ra? neglected] Ip = 87.67 = 50.50 top E = (33000 x0.7) + ( 33000x 0.7141 - 50.50 x 10) G = 32.64 kv) a internal excitation vollage.

t x 10 ū + jtats 33 000 Lo + 50.50 | Cos 0,7 +90 33000 10 + 505.0 190+ 45.57 € = 32.641 16.206 JlaXs LjIaxs 145.54) since, ti coss : 32.641 Cos 6. 206 = 32.44 KN 1 Ef Cossar under excited clearly Ecosse

@ Power Trangle of Gen @ 50% of rated load Q: 3.57MVAN, S= 5 MVA 45.572 P= 3.5 (50% load , MW l 3.5 P = 1x1080.7 1xcox 0.714 = 3.57 MVAYix.

Load- Ip (current per phase) - Ie (leare current) - 16X10² 16600 * 420 x * sings 3 x 420 x0.6 420 I 36.65 |-36. to ad - 2 Ile/53 420 Ip current per phase) I = 5x10² 3 * 420*0.9 In - 7. 636 | 75.841 From regure I e I = I total I 40,42 | - 2.72 Now, applying per phase kul t; 420 LO + (4+63) (10.72 | -2.72 ) 154 639.55 12.168

- 250*50i, & 100,- 1080 +80 (1,-0.2x10ci-io) +5) 20 -250 & 60 i, - 101 & 8oti, -21, +28+5) = 0 - 250 & 600, -10%. & 160i -soi, + 40020 isoin - doi, + 150 co 15i - zi,+ 15 20 24 - 151 =15 (11) torson eq ④ and eam - 2 (lobio) - 1510=15 3.2 i-1510 =15 5 - 1.2711 Amp] also i, a 1.6 io = - 2.0338 Amp .: Vo = 101,- io) Ivo : -7.627 volls) - l. 2711 Amp NOTE: at voltage across current source is not known then KvL containing current source element is an ionaid equation.