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

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To solve the problem of a 3-phase 11kV line delivering a load of 4.1 MW and 2.6 MVAR lagging, the first step is to calculate the apparent power (S) using the power triangle, where S = √(P² + Q²). The power factor can then be determined by the formula cosθ = P/S, converting the reactive power into a power factor value between 0 and 1. The series impedance of the line is given as (7.2 + j12.1), which will be used to calculate the sending end voltage and load angle. Understanding these calculations is crucial for accurately determining the electrical parameters of the system.
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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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