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3 years ago

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Suppose you have a Y-connected balanced three-phase load which consumes 200 kW with pf of 0.707 lagging. The line-to-line voltag

e of the load is 440 V. Use the line-to-line load voltage as your angle reference. (a) Draw the per-phase equivalent and label all the components. (b) Calculate the a-phase load current. (c) Calculate the capacitive reactive power per phase needed to correct the power factor to 0.8 lagging (d) Draw two power triangles - before and after the power factor correction in (c).

1 answer:

elena55 [62]3 years ago

5 0

Given Information:

three-phase Y-connected load = P = 200 kW

PF = 0.707 lagging

line-to-line load voltage = VL-L = 440 V

Required Information:

(a) Draw the per-phase equivalent circuit ?

(b) Calculate the a-phase load current = ?

(c) Calculate the capacitive reactive power = ?

(d) Power triangles - before and after the power factor correction = ?

Answer:

a) See attached drawing

b) I = 123.72<-45°

c) Qc = 16.66 kVAR

d) See attached drawing

Explanation:

b) Calculate the a-phase load current

VL-N = VL-L/ $\sqrt{3}$ = 440/ $\sqrt{3}$ = 254 V

Three phase load current can be found by

P = 3*VL-N*I*cos(θ)

θ = cos⁻¹(PF) = cos⁻¹(0.707) = 45°

I = P/3*VL-N*cos(45°)

I = 200,000/3*254*cos(45°)

I = 371.18 A

single phase current is

I = 371.18/3 = 123.72 A

In polar form,

I = 123.72<-45° A ( minus sign due to lagging PF)

Since the load is balanced, the current in other phases is same with 120° phase shift

(c) Calculate the capacitive reactive power

Three phase reactive compensation power is

Qc = P (tan(θold) - tan(θnew))

θnew = cos⁻¹(PF) = cos⁻¹(0.8) = 36.86°

Qc = 200 (tan(45°) - tan(36.86°))

Qc = 200 (0.250)

Qc = 50 kVAR

Per phase reactive compensation power is

Qc = 50/3 = 16.66 kVAR

(d) Power triangles - before and after the power factor correction

Before

P = 200 kW

Q = 3*VL-N*I*sin(45° ) = 3*254*123.72*sin(45° ) = 66.66 kVAR

S = P + jQ = 200 + j66.66 kVA = 210.81 kVA

After

I = P/3*VL-N*cos(36.86°) = 200,000/3*254*cos(36.86°) = 328 < -36.86 A

single phase current

I = 328/3 = 109.33 < -36.86 A

Q = 3*VL-N*I*sin(36.86° ) = 3*254*109.33*sin(36.86° ) = 50 kVAR

S = P + jQ = 200 + j50 kVA = 206 kVA

As you can see the current and reactive power are reduced after power factor correction.

The power triangle before and after power factor correction is attached.

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