Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

a) The following loads are being fed from a 3-phase, 4.16kVsystem. Determine tot

ID: 1830356 • Letter: A

Question

a) The following loads are being fed from a 3-phase, 4.16kVsystem. Determine total line current, power factor, real, power,reactive power, and apparent power: 2 MVA @ 0.80 power factor lagging. 1 MW @ 0.90 power factor lagging. 700 kW + j700 kVAR. b) Using the answers determined above, determine the reactivepower compensation required to improve the combined load's powerfactor to 0.95, and draw the P-Q diagram. (Answers are: I = 565A, P = 3,300 kW, Q = 2,384 kVAR, S =4,071 KVA, PF = 0.81 lagging, Q (compensation) = 1,299 kVAR ,but please show all steps.) Thank you. a) The following loads are being fed from a 3-phase, 4.16kVsystem. Determine total line current, power factor, real, power,reactive power, and apparent power: 2 MVA @ 0.80 power factor lagging. 1 MW @ 0.90 power factor lagging. 700 kW + j700 kVAR. b) Using the answers determined above, determine the reactivepower compensation required to improve the combined load's powerfactor to 0.95, and draw the P-Q diagram. (Answers are: I = 565A, P = 3,300 kW, Q = 2,384 kVAR, S =4,071 KVA, PF = 0.81 lagging, Q (compensation) = 1,299 kVAR ,but please show all steps.) Thank you.

Explanation / Answer

S1=2MVA at an angle 36.87(lag) S2=(1/.9)=1.11MVA at an angle 25.84(lag) S3=(7002+7002)=7002=.99MVAat an angle 45(lag) total apparant power S=4.07MVA at an angle 35.88(lag) line current=Iline=4.07*10^6/(3*4.16*10^3)=564.8A power factor= cos35.88=.81(lag) real power P= 4.07*cos35.88=3.3MVA=3300KVA reactive power Q=4.07*sin35.88=2.38MVA let x be the reactive compensation then, (2.38-x)/3.3=tan(cos-1.95) x=1.28MVA