TEXTBOOK OF ELECTRICITY W/MINDTAP >BI<
TEXTBOOK OF ELECTRICITY W/MINDTAP >BI<
6th Edition
ISBN: 9781305812505
Author: Herman
Publisher: Cengage Learning
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Chapter 29, Problem 4PP
To determine

The missing values in the given table.

Expert Solution & Answer
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Answer to Problem 4PP

Primary Secondary Load
EP = 23000 V EP = 120 V EP = 208 V
IP = 0.626A IP = 120.08 A IP = 69.33 A
EL = 23000 V EL = 208 V EL = 208 V
IL = 1.084 A IL = 120.08 A IL = 120.08 A
Ratio = 191.52:1 Z = 3 Ω

Explanation of Solution

TEXTBOOK OF ELECTRICITY W/MINDTAP >BI<, Chapter 29, Problem 4PP

In the figure, three single-phase transformers have been connected to form a delta–wye bank.

The primary is connected to a three-phase line of 23000 V.

The secondary voltage is 208 V.

A three-phase resistive load with an impedance of 3 Ω per phase is connected to the secondary of the transformer.

The primary windings of the three single-phase transformers are connected to form a delta connection. In a delta connection, the phase voltage is equal to line voltage.

EP(PRIMARY)=EL(PRIMARY)=23000 V

The secondary windings are connected as a wye. In a wye connection, the phase voltage is less than the line voltage by a factor of 1.732 (the square root of 3). Therefore, the phase value of the primary voltage can be calculated using the formula

EP(SECONDARY)=EL(SECONDARY)1.732=2081.732=120 V

The turns ratio can be calculated by comparing the phase voltage of the primary with the phase voltage of the secondary:

Ratio=Primary VoltageSecondary Voltage=23000120.09=191.521Ratio=191.52:1

The load is connected directly to the output of the secondary. The line voltage applied to the load must therefore be the same as the line voltage of the secondary:

EL(LOAD)=208 V

The load bank is connected in a delta connection. The voltage across the phase of the load bank will equal to the line voltage.

EP(LOAD)=EL(LOAD)=208 V

The phase current of the load can be calculated using Ohm’s law:

IP(LOAD)=EP(LOAD)Z=2083= 69.33 A

The amount of line current supplying a delta-connected load will be 1.732 times the phase current of the load:

IL(LOAD)=1.732×IP(LOAD)           =1.732×69.33           =120.08 A

Since the secondary of the transformer is supplying current to only one load, the line current of the secondary will be the same as the line current of the load:

IL(SECONDARY)=120.08 A

The phase current in a wye connection is equal to the line current.

IP(SECONDARY)=IL(SECONDARY)=120.08 A

The phase current of the transformer primary can now be calculated using the phase current of the secondary and the turns ratio. Because the primary has a higher voltage than the secondary, it will have a lower current. (Volts times amperes input must equal volts times amperes output.)

IP(PRIMARY)=IP(SECONDARY)Turns Ratio=120.08191.52= 0.626 A

All the transformed values of voltage and current take place across the phases, the primary has a phase current of 6.68 A. In a delta connection, the line current is 1.732 times the phase current:

IL(PRIMARY)=1.732×IP(PRIMARY)              =1.732×0.626              =1.084 A

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