Complex Power Nov09

8/2/2019 Complex Power Nov09

1/18


2/18

2

Voltage! Current! Power!Nov, 2001, 03, 06 Denard Lynch! EE201.3 (03)

Resistive Case!

( )t()(VIP MM 02cos0cos2

+!= "

With which we are familiar! !

( t)(VIP MM !2cos12

"=

0=!

Z-Diagram!

"!


Note: the following waveforms were generated in MATLAB!

Using:!

ivp

ttv

tti

=

!=+=

!==

"#$#

"##

40where);sin(3

40where);sin(2


3/18

3


0 5 10-6

-4

-2

0

2

4

6


Resistive Inductive (R-L) Case!!!

"

#

$$

%

&+'= )t()(VIP MM4

2cos4

cos2

()

(

Which is shifted in both value and time! !

!!

"

#

$$

%

&+'= )t(VIP MM4

2cos2

1

2

()

4

!"=

Z-Diagram!

"!


4/18

4


0 5 10-6

-4

-2

0

2

4

6


Purely Inductive (L) Case!!!

"

#

$$

%

&+'= )t()(VIP MM2

2cos2

cos2

()

(

Which has an average value of 0! !

!!

"

#

$$

%

&+'= )t(VIP MM2

2cos02

()

2

!

"=

( t)(VIP MM !2sin2

=Z-Diagram!

"!


5/18

5


0 5 10-6

-4

-2

0

2

4

6


Resistive Capacitive (R-C) Case!!!

"

#

$$

%

& '+''= )t()(VIP MM

42cos

4cos

2

()

(

Note the average value is the same as the R-L case! !

!!!!

"

#

$$$$

%

&

''+= )t(cosVIP MM4

2

2

1

2

()

4

!

"

#

=

Z-Diagram!

"!


6/18

6


0 5 10-6

-4

-2

0

2

4

6


Purely Capacitive (C) Case!!!

"

#

$$

%

&'''= )t()(VIP MM2

2cos2

cos2

()

(

Note: just the inverse of the L case! !

!!

"

#

$$

%

&''= )t(VIP MM2

2cos02

()

2

!

"

#

=

( t)(VIP MM !2sin2

"=

Z-Diagram!

"!


7/18

7


0 5 10-6

-4

-2

0

2

4

6


! Nownote the offset and phaseshift as the load changes from

inductive through resistive!


8/18

8


0 5 10-6

-4

-2

0

2

4

6

L!

j

I

V


0 5 10-6

-4

-2

0

2

4

6

L-R!

j

I

V


9/18

9


0 5 10-6

-4

-2

0

2

4

6

R!

j

I

V


0 5 10-6

-4

-2

0

2

4

6

R-C!

j

I

V


10/18

10


0 5 10-6

-4

-2

0

2

4

6

C!j

I

V


R-L!

R-C!

R!

C!

L!

+"!

-"!

+j!

-j!

S+Q

-Q

P

S

View Power in a complex domain (Complex Power)!

*

22

22

22

or,

,

,

VIS

jQPjQPS

SS

QQPP

QPS

XIX

VQRI

R

VP

CAPIND

iiT

i

iT

i

iT

=

!+=

=

==

+=

+=+=

"

""

P = real power, Watts (W)!

Q = reactive power, Volt-Amp-

!Reactive (VAR)!

S = apparent power, Volt Amp (VA)!


11/18

EE201.3 Complex Power Examples Nov. 2006

Denard Lynch Page 1 of 2

Example #1

The loading of a factory on a 1000-V, 60-Hz system includes:

20-kW heating and incandescent lighting (unity power factor) 10-kW* of induction motors (0.7 lagging power factor, ! = .9) 5-kW fluorescent lighting (0.85 lagging power factor)

a) Establish the power triangle for the total loading on the supply.

b) Determine the power-factor capacitor required to raise the power factor to unity.

c) Determine the change in supply current and power savings from the uncompensated to thecompensated system. !

a)

Load P(W) Q(VAR) S(VA)

Heating

Motors*

Fluorescents

total

*the default assumption is that this is the input power, i.e. before any efficiency losses.


12/18

EE201.3 Complex Power Examples Nov. 2006

Denard Lynch Page 2 of 2

Example #2

(2004 Final)

A single-phase 6hp motor is connected to a 240VAC, 60Hz supply. On average, the motor isloaded to 75% of its rated power. Its efficiency is 81%, and the name plate rating states 6.9kVA

at full load.

a) What is the Power Factor. FP, for the motorwithoutany correction?

b) What element (C or L, in and value in F or H) would be required to improve the PowerFactor to 0.9Lagging?

c) What is the minimum VA rating for this correction element?


13/18

EE201 Complex Power Notes

Denard Lynch Page 1 of 2 Nov. 2007

Power Factor Correction

The goal is to improve (i.e. get closer to 1) the power factor of a device (like a motor)or combined load (like a shop or factory) by adding a reactor (a corrective reactive

element: an inductor or capacitor) to reduce the net quantity of VARs (reactive power)

being consumed, reducing the apparent power and thus the line losses resulting from theextra current required to provide those VARs.

Procedure:

(Note: the key to this process is remembering that the real power, P, must remain the

same.)

Step 1) Determine the power triangle (i.e. P, Q, and S) for the device orcombined load. (Drawing the power triangle usually helps too!)

Step 2) Determine the new apparent power, SNew: using the improved powerfactor, FP_New:

SNew

=

P

FP_ New

Step 3) Determine the new reactive power, QNew, the net Q that will be left aftercorrection:

QNew

= SNew

2

!P2

Step 4) Calculate the reactive power, !Q, that must be added: New Old Q Q Q! = "

A negative resultindicated that negative VARS must be added (i.e. a capacitor); a

positive resultindicates positive VARs must be added (i.e. an inductor).

For a load fed from a voltage source, VS, the impedance of the reactor is:

XR=

VS

2

!Q=

1

"C="L

#

$%&

'(

The reactive element must be placed across the source in parallel with the load (so the

power delivered to the load will not be affected by the addition of the element).Since, for a purely reactive element, the apparent power, S, is equal in magnitude to the

reactive power, Q, the rating of the element is |!Q|VA @ VS Volts (e.g. 1200VA,

600VAC)

For a load fed from a current source, IS, the admittance of the reactor is:2

1SR

IY C

Q L!

!

" #= = =$ %& ' (

The reactive element must be placed in series, between the source and load (so, again, thepower delivered to the load will not be affected by the addition of the element).


14/18

EE201 Complex Power Notes

Denard Lynch Page 2 of 2 Nov. 2007

Since, for a purely reactive element, the apparent power, S, is equal to the reactive power,

Q, the rating of the element is again |!Q|VA @ VS Volts (e.g. 1200VA, 600VAC)

An Example:

Given: a combined load fed by a 240VAC, 60Hz source and represented by the power

triangle shown in the figure:j

=4144.9W

!Q=1100VARSorig

ower!

Qnew=2007VARSnew

where the original Q was 1100VAR + 2007VAR = 3307VAR. SOrig = 5561VA"38.6

0,

and the original power factor, FP = cos(38.60) = .78Lagging .

We wish to correct the power factor to 0.9Lagging. Following the steps above, we first find

the new S:

SNew

=

P

FP_ New

=

4145

.9= 4606VA

Next we calculate the Q that will be left when the power factor is improved to 0.9:

QNew

= SNew

2

!P2= 4606

2+ 4145

2= 2007VAR

IND

Then we figure out the change in Q that is needed:

!Q = QNew

"QOld

= 2007VAR " 3107VAR = "1100VAR

or 1100VARCAP.

Finally, we can determine the value and rating for the capacitive element:

XR=

VS

2

!Q=

240V2

1100VAR= 52.36"

C=1

!XR

=

1

377 52.36"( )= 50.7F

This element must be placed across the source in parallel with the load, and must thenhandle an apparent power of 1100VA, and will be subject to a voltage os 240VAC. WE

would then state the specified rating as a minimum of1100VA at 240VAC.


15/18

EE201.3 Notes on Complex Power 2001/2002 T1

Denard Lynch Page 1 of 4 Nov. 22, 2001

( )

( )

.capacitiveifve-'andinductiveifve'isXwhere

:whenMaximumRelative

when

:TransferPowerMax

cos,

:..),(,

tan,

or,

)(

,

,

22

*

__

***

122

22

22

22

+

++=

=

==

===

!"

#$%

&=+=

+=

=

==

+=

====

'''

LoadSourceSourceL

SourceLoadMax

LeadingORLaggingP

CAPIND

i

iT

i iTi iT

XXRR

ZZP

S

P

FrPowerFacto

AAgeconjugatecomplexIVIS

P

QQPS

jQPjQPS

additionvectorcomplexSS

QQPP

QPS

XIX

VQRI

R

VP

The Power Triangle

P Q S

+Q

-Q

S (R-L)

S (R-C)

Rea

ctivePower,Q

Active Power, P

App

arentP

ower

,SInductive Load

Capacitive Load

FP


16/18



+

-

0.14

Transmission lineV

S 117V

Isupply

j0.08700W

1000VARIND

200VARIND

40W460W

800VARCAP

1000W

700VARCAP

60Hz

a) Find P, Q, S for the combined load. Draw the power triangle.

b) with VL = 117V, find the total IL.

c) What type of element should be added to correct the power factor of the load to Unity?Where should it be placed?

d) What voltage would be present at the load after FP correction?e) What is the Power Factor seen by the source?

Determine the total load:

a) Total the real and reactive powers in the load:P = 700W + 40W + 460W + 1000W = 2200W

Q = 1000VAR + 200VAR 800VAR 700VAR = -300VAR = 300VARCAPS =

Add the complex power vectors:S = (700+j1000)VA + (0+j200)VA + (40+j0)VA + (460-j800)VA + (1000-j700)VA

S = (2200-j300)VA = 2220.4-7.770VA

2220VA

2200W

-300VAR

b) Find the total Current:

S = VI*

I* = S/V = = 18.987.770A

c) To correct to unity power factor, need to add +300VAR or 300VAR Inductive:

Inductive element, QL = V2/XL

XL =

L =

It should be placed across the mains in parallel with the customer load.(Note: it should also have a rating of 300VA!)

d) First, we need to find the voltage at source. To do this we use the current drawn before

correction and work back:

VS = 11700V + ISZTx


17/18



VS =

Total impedance of the load before correction was:

Which is now in parallel with the Power Factor Correction element:

All of which is in series with the transmission line impedance:

New ZS =

And the new source current IS =

The new voltage across the load, VL =

And the power factor seen by the supply, FP, is the cosine of the angle between the source

voltage and current (draw a phasor diagram to help visualize where voltages and currents arewith respect to each other):

New FP-source =


18/18


Denard Lynch Page 4 of 4 Nov 22 2001

Lets check total power both before and after the correction was added to the system:

Before:

IS = 18.987.770A SS = VSIS* = (119.44.894

0)(18.98-7.77

0) = 2250+j271.4VA

Tx Line:P =

Q =

Load:

P =

Q =

Total:

After:

New SS = VSIS* = (119.44.8940)(18.774-.179

0) = 2242.4.715

0VA = 2242.2+j27.98VA

Tx Line:

P =

Q =

Load: P =

Q =

Total:

Q: You are a Customer Services Engineer with a manufacturer that makes stand-bygenerators. You have been called to a site because an irate customer thinks your 120VAC

1.8kVA model is defective as it seems to have burned out some of his equipment. At the site

you find an electro plating machine, along with a fan used to ventilate fumes, set up in atemporary building and both powered by the portable generator. You check the specs on

these two appliances and note that the Plater (which is a capacitive load) is rated at 120VAC

60Hz, 95W @ 3.08A and the fan is 0.1 HP, 120VAC 60Hz @ 3A, with an efficiency, ,of 61.3%. (Remember: 1HP = 746W). You investigate and report. What do you conclude?

Complex Power Nov09

Documents

Transcript of Complex Power Nov09