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WIND ENERGY SYSTEMS - Cd3wd

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Chapter 5—Electrical Network 5–25<br />

The actual voltage is given in the problem as the rated or base voltage, so<br />

The per unit terminal voltage is then<br />

V = 2400 V<br />

V pu =<br />

V = 2400<br />

V base 2400 =1<br />

We now have to solve for the per unit current.<br />

I pu = V pu<br />

Z pu<br />

= 1/0o<br />

1.2 − j0.8 = 1/0 o<br />

1.44/ − 33.69 o =0.693/33.69o<br />

The actual current is<br />

= 0.577 + j0.384<br />

I = I pu I base =(0.693/33.69 o )(434) = 300/33.69 o<br />

A<br />

The per unit apparent power is<br />

S pu = |V pu ||I pu | = (1)(0.693) = 0.693<br />

The per unit real power is<br />

The per unit reactive power is<br />

P pu = |V pu ||I pu | cos θ = (1)(0.693) cos(−33.69 o )=0.577<br />

The actual powers per phase are<br />

Q pu = |V pu ||I pu | sin θ = (1)(0.693) sin(−33.69 o )=−0.384<br />

S = (0.693)(1042) = 722 kVA/phase<br />

P = (0.577)(1042) = 600 kW/phase<br />

Q = (−0.384)(1042) = −400 kvar/phase<br />

The total power delivered to the three-phase load would then be 2166 kVA, 1800 kW, and -1200 kvar.<br />

The generated voltage E in per unit is<br />

Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001

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