Capacitance power factor question

[oxon88]

Homework Statement

A 50 KW load operates from a 60 Hz 10KV rms line with a power factor of 60% lagging. Determine the capacitance that must be placed in parallel with the load to achieve a 90% lagging power factor.

The Attempt at a Solution

for 60% p.f:-

θ₁ = Cos^-1(0.6) = 53.13 degrees

Q₁ = 50,000*tan(53.13) = 66.667 kVAR

for 90% p.f:-

θ₂ = Cos^-1(0.9) = 25.84 degrees

Q₂ = 50,000*tan(25.84) = 24.22 kVAR
Q_capacitor = Q₂ - Q₁ = -42.45 kVAR
X_c = V_RMS /Q_c = (10⁴)² / -42450 = -2356 Ωω = 2.∏.f = 2.∏.60 = 377

C = 1 / ω.X_c = 1 / 377*2356 = 1.126 μF
can anyone check if I'm on the right path here please?

 

[oxon88]

anyone?

 

[NascentOxygen]

I posted a response to this about 6 hours ago, saying that's the answer I got.

Not sure where my reply went. https://www.physicsforums.com/images/icons/icon9.gif

 

[oxon88]

Ok that's good. thank you

 

[Luk-e-2012]

Was this answer correct as i am getting a different value..?

 

[oxon88]

yes it is correct.

 

[agreaves]

message deleted

 

[grinder76]

Homework Statement

A 50 KW load operates from a 60 Hz 10KV rms line with a power factor of 60% lagging. Determine the capacitance that must be placed in parallel with the load to achieve a 90% lagging power factor.

The Attempt at a Solution

for 60% p.f:-

θ₁ = Cos^-1(0.6) = 53.13 degrees

Q₁ = 50,000*tan(53.13) = 66.667 kVAR

for 90% p.f:-

θ₂ = Cos^-1(0.9) = 25.84 degrees

Q₂ = 50,000*tan(25.84) = 24.22 kVAR
Q_capacitor = Q₂ - Q₁ = -42.45 kVAR
X_c = V_RMS /Q_c = (10⁴)² / -42450 = -2356 Ωω = 2.∏.f = 2.∏.60 = 377

C = 1 / ω.X_c = 1 / 377*2356 = 1.126 μF
can anyone check if I'm on the right path here please?

why isn't it - 2356 in the final formula ?

 

[gneill]

why isn't it - 2356 in the final formula ?

oxon88 calculated the signed reactance of the capacitor to be -2356 Ω. The minus sign reflects the fact that it is capacitive reactance. He then used the formula for the magnitude of the reactance, which is unsigned, to find the value of the capacitor.

An alternative approach might have been to use complex impedance which retains the sign throughout, but that would achieve the same end result.

 

[grinder76]

Thanks for the reply i now understand