Question

3.1 When two four pole, 50 Hz synchronous generators are paralleled their phase displacement is 2 mechanical. The synchronous reactance of each machine is 10 n/phase and the common busbar voltage is 66kV. Calculs the s cho nisimlengud AN 2 A synchronous generator has a synchronous impedance of 2p and a resist ance of 0.01 p.u. It is connected to an infinite busbar of voltage 1 pu. through a transformer of reactance 01 pou. If the generated (no-load) emfis 1.1 p. calculate the current and power factor for maximum output Answer: 0.708 p.u.;0.74 leading) A 66kV synchronous generator has negligible resistance and synchronous reactance of 49/phase. It is connected to an infinite busbar and gives 2000 A at unity power factor. If the excitation is increased by 255 find the maximum power output and the corresponding power factor. State any assumption made. (Answer: 31.6 MW;0.95 leading) A synchronous generator whose characteristic curves are given in Figure 3.4 eves T o wing power factors 0.8 lagging, 10, and 0.9 lead- ing. Determine the percentage regulation at these loads (Answer: 167, 119,76%) 35 A salient-pole, 75 MVA, 11 kV synchronous generator is connected to an infinite busbar through a link of reactance 0.3 p.u. and has Xd=15p.u. and Xq1p.u., and negligible resistance. Determine the power output for a load Components of a Power System angle 30 if the excitation e.m.f. is 1.4 times the rated terminal voltage Calcu late the synchronizing coefficient in this operating condition. All pu values are on a 75 MVA base (Answer: P0.48 pudP/=0.78 p.u.) 3.6 A synchronous generator of open-circuit terminal voltage Ipuis on no-load and then suddenly short-circuited the trace of current against time is shown in Figure 3.6(b). In Figure 3.6(b) the current Oc1.8 p.u., a 57 pu., and Obsp.uCalculate the values of Xs, X and X. Resistance may be neglected If the machine is delivering 1 pu current at 0.8 power factor Lagring at the rated terminal voltage before the short circuit occurs, sketch the new envelope of the SO Hz current waveform. (Answer: Xs -08p.u.;X-0.25 p.u.;X-0.18p.u.) 3.7 Construct a performance chart for a 22KV, 500 MVA, 0.9p.. generator having a short-circuit ratio of 0.55. 3.8 A 275 kV three-phase transmission line of length 96 km is rated at 800 A The values of resistance, inductance and capacitance per phase per kilometre are 0.078 , 1.056 mH and 0.029 F, respectively. The receiving-end voltage is 275 kV when full load is transmitted at 0.9 power factor lagging Calculate the sending-end voltage and current, and the transmission efficiency, and com- pare with the answer obtained by the short-line representation. Use the mom nal and methods of solution. The frequency is 60 Hz. (Answer: Vs=179 KV per phase) 3.9 A 220 kV, 60 Hz three-phase transmission line is 320 km long and has the following constants per phase per km: Inductance 0.81 mH Capacitance 12.8F

Answer 1

Summary- The part which is specified contains two numericals,so they are solved using various formulae of Synchronous Generator. Solution is provided below.

32. Syncasonous impedence = 2 = 2pu. synchronous resistance = Rao 0.0lpu - Syncisonous Reactomce o Xs » V22 6.01)2 o;.99 pe : Transformer Reactionce - jotpu. Infinite bus bas voltage, Vto Apu os 4 Loopu _ Generaled Emy o los pu or I'd Z6 puak step-1 so power angle / Angle b/w Ef e re phosos. Equation of all power. sa J J * . Ia a Armature current VELO | 9cm v * O Ozir Angle between 02 L 25 ZOZ Angle between total Impedence & Resistonice. 1 - Active power = Pogo fue as 82-5) - ve as oz. 25 step-2 Condition for masumum is the By [- schi (02-) x-4)-0 =0 ett sm (02-8)=0. » Oz-soo » Oz?S

Step-3) nding Ozo tom Toful Reactance - Total Resistance = tn | 1.99 +011.89.72. S = Oz a 89.72 por mas pour i. calculating the value of cussent € 28-ve co* 2520z 76.(2.99 + 0.1)2 + (0-01)2 = 2:09 pu Oz 8 2 89.72° i 11 289.72 Izo' 2:09 289.72. 209289572 leted , - 05263 LO- 0.4784 2-89.72 i 09709_42: 39. Armature aussent at morsumum power - 0709 pu & power factors ars (12.39) 0738 (Gading) 1 4074 leading)

83. Nameplate Name plate data of synchronous generator, Terminal voltage = V (line) 8 6 kV. ve (phase) - 66 - 3.81 kV V3 unity. = cosa. Resistance - 0. ; Synchronous reactomce = 402 Iphone. Current > 2000 AD QKA. power factor unity. = cose... Step- By using the magnitude formula, the generated voltage per phase will be ff (phase) - V (Vecarse + Ia Ra)? -+ (ve sin q + Ia xs) ² we have I cosa al sind.o, Raz 0 I - QkA, re = 3.81 kv) 8 25= 4040h + (phase). (381)2+ (2x 472 886 kV. Step-2 Now, the generated voltage in increased asr. no Ę pnew) / phase - 8-86*4:25 = 11:075 kv. - 2 step-3 We know off power & Vt as (02-8) - Ve ² ars Oz. ncusonous resistance As we have no O2 = 95. ou & zgo Xs pog vt sm (8) 20. 013 (90-5) , Sing For more power of sins=do, sago Xs