A technique for resolving complex repetitive waveforms into sine or cosine waves and a DC component is known as:
1 answer:
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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/ = 440/
= 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.
