Phasor diagram of transformer for leading power factor

Example 2-5:

A 15-kVA, 2300/230-V transformer is to be tested (short circuit & open circuit tests as shown

below) to determine its excitation branch components, its series impedances, and its voltage

regulation (See figures below):

+ +

!" !# $# $"

2300V 230V

+ +

!%& !# $# $"

2300V 230V

'%&

()*

+

!+* $# $"

2300V 230V

'"&

()*

The following test data have been taken from the transformer:

Open-circuit test

(High voltage side open)

Short-circuit test

(Low voltage side shorted)

,!-. = 230,! !/. = 47,!

'-. = 215,6 '/. = 810,6

(-. = 90,: (/. = 580,:

(a) --- Find the equivalent circuit for this transformer referred to the high-voltage side.

(b) --- Find the equivalent circuit for this transformer referred to the low-voltage side.

(c) --- Calculate the full-load voltage regulation at 0.8 lagging power factor, unity power factor & at

0.8 leading power factor using the exact equation for,!; --- Do a phasor diagram for each case

(d) --- Plot the voltage regulation as load is increased from no-load to-full load at power factors of

0.8 lagging, unity, and 0.8 leading.

(e) --- Find the efficiency of the transformer at full load with a power factor of 0.8 lagging

45

Solution --- The figure at left & the eq’s are repeated from previous page(s):

+ +

!%& !# $# $"

2300V 230V

'%&

()*

<" >"

>?

@2

'%&

!%&

+

(%&

<*

@2

<#

@2

>#

@2

'%&

!%&

+

(%&

>?

@2

<*

@2

15 KVA, 2300/230-V (Transformer Rating) --- [ !-. = 230,!A '-. = 215,6A (-. = 90,:]

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(a) --- Find the equivalent circuit of this transformer referred to the high-voltage side:

From the figures at left, we have:

@ = 2300B230 = 50 (The turn-ratio or, the voltage-ratio)

Use the OC-test to find the shunt parameters:

· The figure above (middle) shows the circuit for the OC-test case

· The figure above (right) shows the approximate circuit for the OC-test case --- Using this circuit:

C-. = cos DE ;FG

HFG,I ! = cos"#

$%&'

()*%&+,().#&-, = 84°

/0 = |1 !&|

|+ !| &&2 3 567 --- The excitation admittance or, the shunt admittance

= ).#&-

)*%&+ 2 3 84°&9 = :.::;<>&2 3 84°&9 = :.:::;?4 3 @:.::;:8&9 = &ABCD 3 @EFCD

G

HBCD = #

IJCK =

#

%.%%%L$M = <:?:&N --- The value of&HB referred to the secondary

OFCD = #

PQCK =

#

%.%%L%R = <<:&N --- The value of&OF referred to the secondary

46

Solution [Continued] --- The first two figures and the eq’s are repeated from previous page(s):

+

3 S9T UV UW

2300V 230V

XWY

Z[T

HV @OV \2HW @\2OW

@O] HY

XV

SV

3

+

XW

\

H^_V @O^_V XV

SV

3

+

XW

\

\ = `>::a`>: = <:

15 KVA, 2300/230-V --- Transformer Rating

[Sb7 = 4d&S , Xb7 = e.:&f , Zb7 =

----------------------------------------------------------------------------------------------------------------------------------

(a) --- Find the equivalent circuit of this transformer referred to the high-voltage side [Continued]:

Use the SC-test to find the series parameters:

· The figure above (middle) shows the circuit for the SC-test case

· The figure above (right) shows the approximate circuit for the SC-test case --- Using this circuit:

5b7 = cos "# hi!

+i!&1i! = cos"#

#j%&'

(Mk&+,(j&-, = ??.4° --- The series impedance angle

lb0 = &|+i!&|

|1i!| &25b7 --- The equivalent series impedance

= Mk&+2$$.M°&

j&- = d.8>>2??.4° = 4.4? m @e.4?&N = HnpCh m @OnpCh

G

HnpCh = 4.4?&N & OnpCh = e.4?&N --- The series elements referred to the primary side

47

Solution [Continued] --- The first figure & the eq’s are repeated from previous page(s):

+ +

3 3

SW SV UV UW

2300V 230V

\ = `>::a`>: = <:

H7Cb = <:?:&N & OFCb = <<:&N (The values are of referred to the secondary)

HnpCh = 4.4?&N & OnpCh = e.4?&N (The values are of referred to the primary)

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(a) --- Find the equivalent circuit of this transformer referred to the high-voltage side [Continued]:

· Next, we need to bring the shunt elements to the primary (or, high-voltage) side:

H7Ch = \ )&H7Cb = (<:,

)(<:?:&N, = <:?&qN & OFCh = \ )&OFCb = (<:,

)(<<:&N, = <<&qN

· This equivalent circuit is shown in figure above, middle

(b) --- Find the equivalent circuit of this transformer referred to the low-voltage side (secondary):

· We have previously found the shunt components referred to secondary and the series

components referred to primary --- See above

· So, in order to find the model referred to secondary, we need to refer the series components

to the secondary as follows:

HnpCb = rtuCv

&wx =

. !"#

"($%)& = 0.0445"' & *+,-/ =

123-6 "7&

= 8. !"# "($%)&

= 0.0945"'

The resulting equivalent circuit is shown in figure above, right.

48

Solution [Continued] --- Repeated from previous page(s):

+ +

: :

;< ;> ?> ?<

2300V 230V

15 KVA, 2300/230-V Referred to primary Referred to secondary

;@ = A6 7B CAD-EF

AD-EF × G00

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(c) --- Calculate the full-load voltage regulation at 0.8 lagging power factor, unity power factor, and

at 0.8 leading power factor:

H/-I7J+K = /LMN2O AD-LMN2O

= $!-%%%"AP QR%"A

= 95.S"T (The magnitude of the full-load current on secondary side)

UV = "0.W (Lagging): H/ = |H/|"X : YZ

A6 7 = ;/ ^ @+,H/ ^ _*+,H/ --- ;/ = S\0X0°"; (Full-load secondary voltage)

= S\0X0°"; ^ (0.0445"')`95.SX : \9.]°"Ta ^ _(0.0945"')`95.SX : \9.]°"Ta = S\0X0°"; ^ S.]0X : \9.]°"; ^ 4.SGX5\.G°"; = S\0 ^ S.\S : _G.b4 ^ S.5S ^ _\.\9 = S\4.W4 ^ _G.9S = S\4.W5X0.40° (Full-load primary voltage, referred to secondary) Now,

;@ = A6 7B CAD-EF

AD-EF × G00 = QR .c!"ACQR%"A

QR%"A × G00d = S.Gd (The voltage regulation)

49

Solution [Continued] --- Repeated from previous page(s): 15 KVA, 2300/230-V (Transformer Rating)

+ +

: :

;< ;> ?> ?<

2300V 230V

----------------------------------------------------------------------------------------------------------------------------------

(c) --- Calculate the full-load voltage regulation at 0.8 lagging power factor, unity power factor, and

at 0.8 leading power factor [Continued]:

UV = "G (Unity): H/ = |H/|"X : YZ

= S\0X0°"; ^ S.]0X0°"; ^ 4.SGX]0° = S\0 ^ S.]0 ^ _4.SG = S\S.] ^ _4.SG = S\S.]4XG.04° [ ;@ =

A6 7B CAD-EF AD-EF

× G00"" = QRQ.g "ACQR%"A QR%"A

× G00d = G.SWd (The voltage regulation)

UV = "0.W (Leading): H/ = |H/|"X ^ YZ

A6 7 = S\0X0°"; ^ (0.0445"')`95.SX\9.]°"Ta ^ _(0.0945"')`95.SX\9.]°a

= S\0X0°"; ^ S.]0X\9.]°"; ^ 4.SGXGS9.]°"; = S\0 ^ S.\S ^ _G.S4 : S.5S ^ _\.\9 = SS].W0 ^ _5.G0 = SS].W5XG.Sb°"; [ ;@ =

A6 7B CAD-EF AD-EF

× G00""" = QQg.c!"ACQR%"A QR%"A

× G00d = :0.09Sd (The voltage regulation)

50

Solution [Continued] --- Repeated from previous page(s): 15 KVA, 2300/230-V (Transformer Rating)

+ +

: :

;< ;> ?> ?<

2300V 230V

----------------------------------------------------------------------------------------------------------------------------------

(c) --- Calculate the full-load voltage regulation at 0.8 lagging power factor, unity power factor, and

at 0.8 leading power factor --- Do a phasor diagram for each case:

A6 7 = ;/ ^ @+,H/ ^ _*+,H/ --- Using the figure above at right

· We have all components of the above equation (as shown in the above figure, right)

· We also, found the currents for the 3 cases of 0.8 lagging power factor, unity power factor,

and 0.8 leading power factor

· Using these values, we simply show the above equation using phasors --- Each of these three

phasor diagrams is shown in figures below:

51

Solution [Continued] --- Repeated from previous page(s):

+ +

: :

;< ;> ?> ?<

2300V 230V

----------------------------------------------------------------------------------------------------------------------------------

(d) --- Plot the voltage regulation as load is increased from no-load to-full load at power factors of

0.8 lagging, unity, and 0.8 leading:

The best way to plot the voltage regulation as a function of load is to repeat the calculations

presented in part (c) for many different values of the loads by writing a MATLAB program (See

your text) --- The plot produced by this program is shown in figure below:

;@ = A6 7 B CAD AD

× G00 ---

· Can you interpret these curves?

· Do this part and submit a report (See Design 1)

52

Solution [Continued] --- Repeated from previous page(s): 15 KVA, 2300/230-V (Transformer Rating)

+ +

: :

;< ;> ?> ?<

2300V 230V

H/ = 95.S"X : \9.]°"T (Full-load current for the case of power factor of 0.8 lagging)

h = eijN eFikkl !"#

× 100% = !"#

( $"& '!)*+& !"# × 100% --- The efficiency at this condition

----------------------------------------------------------------------------------------------------------------------------------

(e) --- Find the efficiency of the transformer at full load with a power factor of 0.8 lagging:

To find the transformer efficiency of the transformer, we calculate its output power and

its losses --- Using the circuit at right:

,-. = |/2| 3456 = (789:;<+

3(090>>8;?+ = 1@A;B (The copper losses)

,-CD5 = ;|EF GH |

I

J$

KI

= (3LM9NO;E+I

PQOQ;R = 8:98;B (The core losses)

,S.T = |U2||/2|;,V (The output power of the transformer at this power factor)

= (:W0;U+(789:;<+09@ = 1:X000;B

Finally, the efficiency is:

Y = Z"#

( $";&; $!)*+;&; Z"# × 100% =

P3XQQQ;[

(PN\;[;&;O39O;[+;&;P3XQQQ;[ × 100% = A@90W%

53

Design 1 [Due next week today]:

This design is related to the previous example problem --- Figures below are repeated:

+ +

] ]

Û U_ _̀ ^̀

2300V 230V

----------------------------------------------------------------------------------------------------------------------------------

(1) --- Plot the voltage regulation of the transformer as its loading is increased from no-load to full-

load at power factors of 0.8 lagging, unity, and 0.8 leading --- This is part (d) of the previous

example problem (You should plot the 3 curves in one graph for easy comparison)

(2) --- Plot the efficiency of the transformer as its load is increased from no-load to-full load at

power factors of 0.8 lagging, unity, and 0.8 leading --- This is part (e) of the previous example

problem, but extended to include the various loading as well as loads with the unity, and

leading power factors (You should plot the 3 curves in one graph for easy comparison)

(3) --- Prepare a report including: a summary of the problem, your MATLAB code, the plots and an

engineering conclusion derived from each of the plots

(4) --- The report should be submitted to me via email (Gradman@TNTech.edu) in one PDF-file and

the file should be named as follows:

Your class no -- Your last name -- Design 1 -- ECE 3610 -- Summer 2017

For example, assume my number is 99 --- Then, I will name my PDF-file as follows:

99 -- Radman -- Design 1 -- ECE 3610 -- Summer 2017