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

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 = ?

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.

the equation above shows how temperature f, measured in degrees fahrenheit, relates to a temperature c, measured in degrees celsius. based on the equation, which of the following must be true?

a temperature increase of 1 degree fahrenheit is equivalent to a temperature increase of

5

9

degree celsius.

a temperature increase of 1 degree celsius is equivalent to a temperature increase of 1.8 degrees fahrenheit.

a temperature increase of

5

9

degree fahrenheit is equivalent to a temperature increase of 1 degree celsius.

a) i only

b) ii only

c) iii only

d) i and ii only

answer:

sure does

explanation: