HW4 Solutions

# HW4 Solutions - Homework 4 Solutions 1 Problem 12.1(Haadat...

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Homework # 4 Solutions 1) Problem 12.1 (Haadat) Solution: Speed Regulation, R = 5% = 0.05 59.7 – 60 = - 0.3 Hz. = -0.3/60 pu. = 0.3/(60*0.05) = 0.1 pu. = 0.1*250 = 25 MW. 2) Problem 12.2 (Haadat) Solution: Let, the load on Gen A = x Then, the load on Gen B = 500 – x Now, the change of frequency = Therefore, Also, Now, equating the frequency deviation, or, or, or, Therefore, Gen A takes up 200 MW Gen B takes up = 500 – 200 = 300 MW. 3) Problem 12.3 (Haadat) Solution:

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Given: Single area-Two Generators-rated at 400 and 800 MVA Speed Regulation for Gen 1, Speed Regulation for Gen 1, Let the common base be 1000 MVA. ( ) = / = . . Load increase pu 130 1000 0 13 pu ) = Case a when D 0 , = – . = . . New frequency f 60 0 3 Hz 59 7 Hz : Change in Generations = = . = 0 05 pu 50 MW = = . = . 0 08 pu 80 MW : New Generations ) = . Case a when D 0 804 = , = – . = . . New frequency f 60 0 291 Hz 59 709 Hz : Change in Generations
= = . pu 48 5 MW = = . . pu 77 6 MW : New Generations ) 4 : Solution Speed Regulation for Gen 1, Speed Regulation for Gen 1, Speed Regulation for Gen 1, ( ) On their respective bases , = Let Sbase 100 MVA ( a , Unit area frequency characteristics on 100 MVA base = . ( ) 165 pu on 100 MVA base ( b , Load increase , = Now frequency change

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