Question

1. (10 pts) Consider the diagram showing the governor characteristics of two Benerators. Unit 1 is rated 1200 MW and Unit 2 is rated 600 MW. They are the only generators in the control area. The base frequency is 60 Hz (a) When the system is operating at 60.00 Hz, what are the respective outputs from Units 1 and 2, and what is the total load in MW? (b) What is the regulation constant R for each generator, in Hz/MW? (c) Within the operating range shown, what is the area frequency response character istic B? (d) If 300 MW of load is suddenly lost while the system is operating at 60.00 Hz, what is the new system frequency? (e) By how many MW does each unit reduce its output? Indicate the new operating points on the graph, approximately. (f) Suppose that the control area is synchronously interconnected with another, larger area with many more generators. Qualitatively, how would the frequency response change? frequency (P.1). 1.04 - t Unitz Unit power (p.u.) 1.00 ez 04 0.6 0.98

Answer 1

7. Ans- unit-I rated 1200mw. unit-2 rated s boomw. Base frequency, 60Hz. frecuencap.) । unita 1.02 units 1.00 0.2 0.4 0.6 0& ip 0.98 + power (a). system operating at 60Hz. facta Gore Đ

from graph at 1. pu frequency unit-1 op isI p.o. Unit - Actual olpiss ox unit-o raketolp 1200 MW. unit-2 olp is so spu. at Ipu. frequency unit-2 Actual olp = 0.8x unit-2 ratetolp -0.8 x 600mw 480 Mw. loadis s 1200+ 480 : Total 1680mw b) = ? Regulation constant R

Regulation constant (R). Tento Af= charge in frequeny lin42) ap = change in power of (mw) Ris ab There is in unit-1 I. p.u for Af = 0.02pu power charge lic Af> Api -1.2 0.02x60 HZ = 1.2 H2 I pu x 1200 , 1200 mW -103 -0.00l. H2/mw 1200 AP2 for unita2 0.06 pu frequeey chary for power olp chann for 'l pu

Дfa дүр , ook х 60 - 3.6к2 три = 6овото » Ко» спод Goo (hu. ДР : хботи - р, - 4 к се ооос хэнэ. foo characterstic (B) | (c) frequency response р» тр, . 1000 + 146 47 р , 1166 67 present total load (Pt=1680mw P, 2 2оо то Pa 48о то

ДРТ , дра 2.9 for дү, cha cy in pnt -1 Apr chary in unit-2 А, В, дү, Afa> R2 AP2. Atp Ata As both core in synch. R,API = R2 AP2 DP2 |АР, АР, - Зооно. ( САР, t ДР. - Зоогоо 7 ДР. » Зь» то | гр.» за 2 ч 2. 8ѕ nо ДР, 6 4 ДР. де , asa I, ст.

New frequency ftat 60p R,x AP, , 60+ 0.00lx 257.15 mw f new , 60.25 HZ 1. As solved in part (2) Api= 257.15mW AP2 - 42.85 mw H frequency respoke (B) ches As the No. of Generators increases P valen increases. and frequency response of system improves le freency chenye reduces with load changes.