Mathcad - 02-15

Mathcad - 02-15 - ⋅ = Solution Since this is a delta-load...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
I ph 22.019A = arg I ph ( 29 56.976deg = I L I ph 3 e j - 30 deg = I L 38.138A = arg I L ( 29 26.976deg = We can use the complex power formula on a per-phase basis S ph V ph I ph = S ph 2.88 4.431j - kVA = S 3ph 3 S ph = the apparent power is the magnitude of S 3ph : S 3ph 15.854 kVA = real power is the real part of S 3ph : P Re S 3ph ( 29 = P 8.64kW = reactive power is the imag part of S 3ph : Q Im S 3ph ( 29 = Q 13.292 - kVAR = Since the circuit is capacitive, the reactive power is negative P2-15. Repeat P2-14 with the three loads connected in delta. R 20 = X C 13 = V AB 240 V
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ⋅ = Solution: Since this is a delta-load with a series impedance, we need to find the total impedance and then solve for the phase current. Recall that the phase current leads the line current by 30 deg Z 1 1 R 1 j X C ⋅ ( 29- = define kVA kW ≡ kVAR kW ≡ Z 5.94 9.139j-Ω = Z 10.9 Ω = arg Z ( ) 56.976-deg = V ph V AB = V ph 240V = arg V ph ( 29 0deg = I ph V ph Z = I ph 12 18.462j + A =...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online