Solving 3 Phase 11kV Line w/ 4.1 MW & 2.6 MVAR Lag

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SUMMARY

The discussion focuses on calculating the sending end voltage and load angle for a 3-phase 11kV line delivering a load of 4.1 MW and 2.6 MVAR lag. The series impedance of the line is given as (7.2 + j12.1). To determine the power factor, participants suggest using the power triangle, where the apparent power S can be calculated using the formula S = √(P² + Q²), leading to a power factor of cosθ = P/S.

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  • Understanding of 3-phase power systems
  • Knowledge of complex impedance in electrical engineering
  • Familiarity with power factor calculations
  • Ability to use the power triangle for electrical calculations
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  • Calculate apparent power S using the formula S = √(P² + Q²)
  • Learn about power factor correction techniques in 3-phase systems
  • Study the impact of series impedance on voltage drop in transmission lines
  • Explore load angle calculations in AC circuits
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Electrical engineers, power system analysts, and students studying 3-phase power systems will benefit from this discussion.

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hi there.

can anybody advise with this one please?

a short 3 phase 11kv line delivers a load of 4.1 mw,2.6 mvar lag if the series impedance of the line is (7.2 +j12.1 ) calculate the sending end voltage and load angle.

the thing which is confusing me is the 2.6mvar lag bit,i know i need to get a power factor from this to work things out, how do i change this into a power factor value ie 0-1?

any pointers appreciated.
 
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You have P=4 MW and Q = 2.6MVAR, so you can get S using the power triangle and then you can easily get the power factor, cosθ = P/S.
 

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