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EE301 Lesson 35

Reading: 23.9

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P

QL

S

S

P

cos cos

sin sin

P VI S

Q VI S

= =

= =

(W)

(VAR)

2

2cos

RV

P V I I RR

= = = phase power

3T an bn cn

P P P P P= + + =

3L

V V=

2

2sin

XV

Q V I I X QX

= = = (VAR) =

3 sinT L LQ V I = (VAR)

cosT

P

T

PPF

S S

= = =

THREE PHASE POWER

Learning Objectives

(a) Compute the real, reactive and apparent power in three phase systems.

(b) Calculate currents and voltages in more challenging three phase circuit arrangements.

(c) Apply the principles of Power Factor Correction to a three phase load.

Recall that the power trianglegraphically shows the relationship between

real (P), reactive (Q) and apparent power (S).

We will first examine three-phase power in the context of a wye-load; then well examine a delta load.

Power to the Wye-Load

Active (Real) Power. Suppose that each phase has impedance .

Then the active (real)power per phase(P) is given

Because we are considering a balanced system, the power per phase (P) is identical in all three phases,

and thus the total active power (PT) is simply PT= 3 P.

Using line voltage ( ) and line current(IL=I), we have

Reactive Power The reactivepower per phase(Q) is given by

The total reactive power can be calculated similar to the total active power:

Apparent Power

The apparentpower per phase(S) is given

The power factor (FP) is given

3 3 cos 3 cos 3 cos3

LT L L LVP P V I I V I

= = = =

(W)

2

2

3T L L

VS V I I Z

Z

S V I

= = =

=

(VA)

(VA)

Z R X j = = +Z

7/27/2019 EE301Topic35.pdf

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EE301 Lesson 35

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3T ab bc ca

P P P P P= + + =

3LI I=

3 3 cos 3 cos 3 cos3

L

T L L L

IP P V I V V I

= = = =

(W)

Power to the Delta () LoadActive (Real) Power.

Total active power (PT) is simply PT= 3 P.

Using line voltage (VL=I) and line current( ):

Which was the EXACT same equation as for Y loads

Reactive and Apparent Power The equations for calculating total reactive and apparent power are also

identical to the Wye load versions:

The applicable portion of the equation sheet:

3 sinT L L

Q V I = (VAR)

3T L L

S V I= (VA)

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Example In the Y-Y circuit shown, EAN= 277-30V.

(a) Compute P

(b) ComputePT.

(c) Compute Q

(d) Compute QT.

(e) Compute S(f) Compute ST

(g) Compute FP.

Solution:

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3 cos sinT L L T T T T S V I P S Q S = = =

Example

In the circuit shown EAN= 120-30V(a) Determine per phase powers (active, reactive, and apparent).(b) Determine total powers (active, reactive, and apparent) by

multiplying the per-phase powers by 3.(c) Determine total powers (active, reactive, and apparent) by

using these formulas:

Solution:

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Example In the circuit shown, EAB= 2080V.

a. Determine the line currentsb. Determine total real power delivered by the generator

c. Total real power dissipated by the load.

d. Determine the load phase voltage Van

Solution:

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Example. In the circuit shown, EAB= 2080V

a. Find the load phase voltage Vabb. Find ST, QT, and PTdelivered by the generator.

c. Find S

T

, Q

T

, and P

T

of the load.

Solution:

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EE301 Lesson 35

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Power Factor Correction

Recall: In order to cancel the reactive component of power, we mustadd reactance of the opposite type. This is called power factor

correction.

In a three phase circuit, capacitors are connected in parallel with

each load phase (presuming the actual load is inductive, which is

usually the case)

Solution steps:1. Calculate the reactive power (Q) of ONE PHASE of the load

2. Insert a component in parallel of the load that will cancel out that reactive power

e.g. If the load has Q=512 VAR, insert a capacitor with Q= - 512 VAR.

3. Calculate the reactance (X) that will give this value of QNormally the Q=V

2/X formula will work

4. Calculate the component value (F or H) required to provide that reactance.

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Example

In the system shown we have EAB= 4800V.

The frequency is 60 Hz.

Determine value of capacitor which must be placed across eachphase of the motor to correct to a unity power factor.

Solution:

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Example (from Robbins and Miller, Circuit Analysis Theory and Practice, 4th

ed, Delmar Cengage,

2007)

In the circuit below, the 60 Hz motor is providing 100 hp at an efficiency of 80%. The power factor of

the motor is 0.85 (lag). The line voltage is 575 V.

The capacitors are connected in a Y-configuration, and each capacitor has the value C = 120.

F Compute the overall power factor of the system.

Solution.