Mathcad - 02-11

Mathcad - 02-11 - kVAR kW Problem 2-11: Solution Repeat...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
When connected in WYE, the impedance use one-third of the power they use when connected in Δ . The reason of course is they only have the line to neutral voltage imposed on them in WYE versus the line voltage when connected in Δ . Q 3.84 kVAR = P y 7.68 kW = Q3 V I L sin θ () := P y 3 V I L F p := Calculating the real and reactive power, yields: θ 26.565 deg = θ arg I L := F p 0.894 = F p cos atan X R := The power factor can be found from the impedance: arg I L 26.565 deg = I L 10.328amp = I L 9.238 4.619j amp = I L V br Z := Thus the line current is: V br 277.128volt = The branch voltage lags the line voltage by 30 degrees
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

Mathcad - 02-11 - kVAR kW Problem 2-11: Solution Repeat...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online