The per unit impedance of a generator on a 60 MVA, 20 kV base is j0.09 p.u. The per unit impedance to a 100 MVA, 20 kV base will be (a.) j0.054 p.u. (c.) j 1.5 p.u. (b.) į0.54 p.u. (d.) j0.15 p.u.
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- The reactance of a generator is given as 0.25 per-unit based on the generator’s of 18 kV, 500 MVA. Find its per-unit reactance on a base of 20 kV, 100 MVA. a. 0.0505 b. 0.0605 c. 0.0405 d. 0.0606The p.u. impedance value of an alternator corresponding to base values of 13.2 kV and 30 MVA is 0.2 p.u. Then the p.u. impedance value of an alternator for the new base values of 13.8 kV and 50 MVA is......The reactance of a generator is given as 0.1 pu based on the generator of 17 KV, 300 MVA. Determine the pu reactance on a base of 17 KV, 250 MVA. New pu reactance is.......
- In the system shown in Figure 1, the transformers are connected star-star with both star points grounded and the generators are connected in star with thier star points grounded. The system base is 15 MVA. The transformers all have reactances of 0.04 p.u. on this 15 MVA base. The reactances of all other elements are given in Table 1 (in 2) and the voltage levels are given in Table 2. p.u. G1 p.u. T1 jö Per-Unit Convert all values to p.u. on a 15 MVA base. Xa= p.u. Xc₂= XL = V BABE G1 2 X 9 T3 Figure 1: A section of the distribution system T1 L Table 1: Sequence reactances (2) 3 G1 L G2 0.3 0.59 0.01 4 L 9/10 10 Fault Voltage What is the voltage at bus 3 (in Volts) after the fault has occurred? Vp= V T2 5 T2 34 10/4 Table 2: Voltage bases (kV) G2 4 T3 10/9 | G2 Fault Current A three-phase fault with a fault reactance of 0.01 p.u. occurs at bus 3. Calculate the fault current flowing at the fault point in KA. Ip=-j KA SoA 30 MVA, 11 kV generator has a reactance of 0.10p.u.on its own base. Determine the per-unit reactance when referred to base kVA of 50,000 kVA and base kV of 33 kV.An alternator with internal vallage of 125, p.u. and synchronous reactance of 0.4 p.u. is connected by a transmission line of reactance 0.1 p.u. c a synchronous meto having synchronous reactance 2.35 p u. and internal voltage of 0.85/6, p.u. If the real power supplied by the alternator is D 806 p... than (8, & is. degrees. (Round off to 2 decimal places) (Machines are of non-salient typa. Naglact resistances)
- A synchronous generator has been synchronized to an infinite grid of 13.8 kV and 60 Hz. The generator prime mover C/C is such that the no-load frequency is 62.5 Hz and the power slope is 1 MW/Hz. The AVR C/C (The reactive power against the terminal voltage C/C) is such that the zero reactive power voltage is 14.6 kV, while the slope is 0.4 MVAR/kV. Calculate: (a) the kVA loading of the generator and its power factor. (b) the settings of the prime mover and AVR such that the generator delivers 3 MW at 0.85 lagging power factor.two generators supplying a load. Generator I has a no-load frequency of 62.5 Hz and a slope Sp1 of I MW/Hz. Generator 2 has a no-load frequency of 62.0 Hz and a slope sp2 of I MW/Hz. The two generators are supplying a real load totaling 2.5 MW 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? (b) Suppose an additional I-MW load were attached to this power system. What would the new system frequency be, and how much power would Gl and G2 supply now? Generator 1 VT V2 Generator 2 VTí KVAR KVARTwo alternators rated 400 MW and 600 MW are operating in parallel. The droopcharacteristics of their governors are 3% and 4% respectively from no loadto full load. Assuming that the alternators are operating at 50 Hz at noload, how would a load of 1000 MW be shared between them? What will be thesystem frequency at this load?
- Why rating of an Alternator is expressed in terms of VA,KVA or MVA ? Option 1: As losses are square times of the power factor Option 2: As losses depends on power factor Option 3: As losses are independent of power factor Option 4: As losses are inversely proportional to the power factor.In the system shown in Figure 1, the transformers are connected star-star with both star points grounded and the generator is connected in star with its star points grounded. The per unit impedances of each element on a 40 MVA base are given in Table 1 and the voltage levels are given in Table 2. Z [p.u.] 0 2 tööt Generator 0.01 +j0.08 p.u. L Figure 1: A section of the distribution system Transformer T1 Line V BASE [KV] 0.04 + j 0.03 + j 0.15000000000000002 0.06000000000000000 Generator Table 1: Sequence impedances (p.u. on 40 MVA base) 3 T2 181. Line 3 10 Table 2: Voltage bases (kV) Load 1 Transformer T2 Load Current A load current of 11.316 kA is flowing with a lagging power factor of 90 %. Convert this to a current vector in per-unit. IL = 0.04 + j 0.06000000000000000What is the main direct cause of reactive power in AC system?A. Resistance of transmission linesB. Inductance and capacitance in the loadsC. Ideal transformer connected in the systemD. Power produced by generator