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Electrical Power Systems

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336 <strong>Electrical</strong> <strong>Power</strong> <strong>Systems</strong><br />

for simulation studies must be done in accordance with Shannon's sampling theorem. Otherwise<br />

an error, proportional to the amount of aliasing, occurs.<br />

A mixed system, such as that shown in ig. 12.25, may be analyzed by modeling the entire<br />

system exactly as shown in ig. 12.25. The system output will be sampled at the normal<br />

sampling rate, i.e., if the sampling time T = 2 sec, the controller output will be updated every<br />

2 sec. Thus the control input to the power system in the continuous–time domain is held<br />

constant for 2 sec between consecutive samples.<br />

The control law in continuous mode is described as<br />

Ui(t) = – KIi z ACEi() t dt<br />

...(12.50)<br />

The discrete vevsion of eqn. (12.50) is computed directly as<br />

where K is sampling count.<br />

b g<br />

Ui (KT) = U K - 1 T<br />

in<br />

s– KIi * T * ACEi (KT) ...(12.51)<br />

EXERCISE<br />

12.1. A system consists of three identical 500 MVA generating units feeding a total load of 765 MW. The<br />

inertia constant H of unit is 5.0 on 500 MVA base. The load varies by 1% for 1% change in<br />

frequency. When there is a sudden increase in load by 15 MW, (Assume f0 = 50 Hz.)<br />

(a) Determine H and D expressed on 1500 MVA base.<br />

(b) ind the steady state value of frequency diviation and its mathematical expression. Assume<br />

there is no speed-governing action.<br />

Ans: Dfss = –1Hz.<br />

Df (t) = (e –t/20 –1)<br />

H = 5.0 sec.<br />

D = 0.01<br />

12.2. An isolated power system consists of three turbine-generating units rated 1000, 750 and 500 MVA,<br />

respectively. Regulation of each unit is 5 % based on its own rating. Initially each unit was<br />

operating at half of its own rating, when the system load suddenly increases by 200 MW. Determine<br />

area frequency response characteristic (ARC) b on a 1000 MVA base. Also steady-state frequency<br />

ss ss ss<br />

deviation and DPg1 , DPg2 and DPg3 . Assume U = 0 and D = 0, given system frequency<br />

f0 = 60 Hz.<br />

Ans: b = 45.0 pu<br />

Dfss = – 0.2667 Hz<br />

ss<br />

DPg1 = 88.88 MW<br />

ss<br />

DPg2 = 66.66 MW<br />

ss<br />

DPg3 = 44.44 MW.

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