ee301 lesson 36 three phase power

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Lesson 36 AC Three Phase Power

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Page 1: EE301 Lesson 36 Three Phase Power

Lesson 36 AC Three Phase Power

Page 2: EE301 Lesson 36 Three Phase Power

Learning Objectives Compute the real, reactive and apparent power in

three phase systems

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

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

Page 3: EE301 Lesson 36 Three Phase Power

AC Power SummaryReview

Real Power P = VI (W)

P = I2R =V2/R

P = 0 (W) P = 0 (W)

Reactive Power

Q = 0 (VAR) Q = I2XL =V2/XL = I2XC =V2/XC

Resistance Reactance

R XL = L XC = 1/C

Page 4: EE301 Lesson 36 Three Phase Power

Power Triangle

The power triangle graphically shows the relationship between real (P), reactive (Q) and apparent power (S).

P

QL

S

S

P

QC

Review

cos cos

sin sin

P VI S

Q VI S

(W)

(VAR)

Page 5: EE301 Lesson 36 Three Phase Power

Y-load

Active Power to Wye (Y) Load

Single phase of Y-load

22cos R

Z R X j

VP V I I R

R

Z phase impedance

= = phase power

Page 6: EE301 Lesson 36 Three Phase Power

Because we are considering a balanced system, the power per phase (P) is identical and the total active power (PT) is simply PT = 3 P.

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

3T an bn cnP P P P P

Active Power (P) to Wye (Y) Load

3 3 cos 3 cos3

3 cos

LT L

L L

VP P V I I

V I

(W)

3LV V

Page 7: EE301 Lesson 36 Three Phase Power

Example Problem 1a

EAN = 277-30 V . Compute PΦ, PT.

Page 8: EE301 Lesson 36 Three Phase Power

The reactive power per phase (Q) is given

Reactive Power (Q) to Wye (Y) Load

22

sin

X

Q V I

VI X

X

(VAR)

(VAR)

Q = V I sin

P

S

Page 9: EE301 Lesson 36 Three Phase Power

Because we are considering a balanced system, the power per phase (Q) is identical and the total reactive power (QT) is simply QT = 3 Q.

Using line voltage (VL ) and line current (IL):

3T an bn cnQ Q Q Q Q

Reactive Power (Q) to Wye (Y) Load

3 sinT L LQ V I (VAR)

Page 10: EE301 Lesson 36 Three Phase Power

Example Problem 1b

EAN = 277-30 V . Compute QΦ, QT.

Page 11: EE301 Lesson 36 Three Phase Power

The apparent power per phase (S) is given

Apparent Power (S) to Wye (Y) Load

22

3T L L

S V I

VI Z

Z

S V I

(VA)

(VA)

(VA)

Q

P

S = V I

Page 12: EE301 Lesson 36 Three Phase Power

The power factor (FP) is given

Power Factor (FP)

cosTP

T

PPF

S S

Q

P

S

Page 13: EE301 Lesson 36 Three Phase Power

Example Problem 1c

EAN = 277-30 V . Compute SΦ, ST, and FP.

Page 14: EE301 Lesson 36 Three Phase Power

3T ab bc caP P P P P

-load Single phase of -load

cos

phase impedance

phase power

Z

P V I

Z

Power to a Delta () Load

Page 15: EE301 Lesson 36 Three Phase Power

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

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

Which was the EXACT same equation as for Y loads

3T ab bc caP P P P P

Active Power (P) to Delta () Load

3 3 cos 3 cos3

3 cos

LT L

L L

IP P V I V

V I

(W)

3LI I

Page 16: EE301 Lesson 36 Three Phase Power

The equations for calculating total reactive and apparent power are also identical to the Wye load versions:

Reactive and apparent power to Delta (Δ) Load

3 sinT L LQ V I (VAR)

3T L LS V I (VA)

Page 17: EE301 Lesson 36 Three Phase Power

Example Problem 2aEAN=120-30 V.

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

Page 18: EE301 Lesson 36 Three Phase Power

Example Problem 2bEAN=120-30 V.

Determine total powers (active, reactive, and apparent) by using these formulas: 3

cos

sin

T L L

T T

T T

S V I

P S

Q S

Page 19: EE301 Lesson 36 Three Phase Power

Power in Advanced 3 phase You must pay attention to the problem statement! Does it ask for total or per-phase power? What kind of power? S, P, or Q? Where is the power?

Generator Line Impedances Load

Pline=?Qline =?

Sgen =?Pgen =?Qgen =?

Sload =? Pload =?Qload =?

Page 20: EE301 Lesson 36 Three Phase Power

Power Factor Power factor (FP) tells us what portion of the

apparent power (S) is actually real power (P).

FP = P / S = cos

Power factor angle

= cos-1(P / S)=cos-1(FP)

For a pure resistance, = 0º For a pure inductance, = 90º For a pure capacitance, = -90º

P

QS

NOTE: is the phase angle of ZT, not the current or voltage.

Review

Page 21: EE301 Lesson 36 Three Phase Power

Power Factor Correction In order to cancel the reactive component of

power, we must add reactance of the opposite type. This is called power factor correction.

Review

Page 22: EE301 Lesson 36 Three Phase Power

Three Phase Power Correction Capacitors will be connected in parallel with

each load phase

Page 23: EE301 Lesson 36 Three Phase Power

Power Factor Correction 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 QΦ=512 VAR, insert a capacitor with QΦ=-512 VAR.

3. Calculate the reactance (X) that will give this value of Q Normally the Q=V2/X formula will work

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

Page 24: EE301 Lesson 36 Three Phase Power

Example Problem 3EAB=4800 V. Frequency 60 Hz.

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