Chapter 29, Problem 4PP

To determine

The missing values in the given table.

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

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.

$E_{P\left(PRIMARY\right)}=E_{L(PRIMARY)}=23000\text{ V}$

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

$E_{P\left(SECONDARY\right)}=\frac{E_{L\left(SECONDARY\right)}}{1.732}=\frac{208}{1.732}=120\text{ V}$

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

$\begin{array}{l}Ratio=\frac{\text{Primary Voltage}}{\text{Secondary Voltage}}=\frac{23000}{120.09}=\frac{191.52}{1}\\ \\ Ratio=191.52:1\end{array}$

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:

$E_{L\left(LOAD\right)}=208\text{ V}$

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

$E_{P\left(LOAD\right)}=E_{L\left(LOAD\right)}=208\text{ V}$

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

$I_{P\left(LOAD\right)}=\frac{E_{P\left(LOAD\right)}}{Z}=\frac{208}{3}=\text{ 69}\text{.33 A}$

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

$\begin{array}{l}{I}_{L\left(LOAD\right)}=1.732\times I_{P\left(LOAD\right)}\\ =1.732\times 69.33\\ =120.08\text{ A}\end{array}$

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:

$I_{L\left(SECONDARY\right)}=120.08\text{ A}$

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

$I_{P\left(SECONDARY\right)}=I_{L\left(SECONDARY\right)}=120.08\text{ A}$

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.)

$I_{P\left(PRIMARY\right)}=\frac{I_{P\left(SECONDARY\right)}}{TurnsRatio}=\frac{120.08}{191.52}\text{= 0}\text{.626 A}$

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:

$\begin{array}{l}{I}_{L\left(PRIMARY\right)}=1.732\times I_{P\left(PRIMARY\right)}\\ =1.732\times 0.626\\ =1.084\text{ A}\end{array}$