Electrical Transformers and Rotating Machines
Electrical Transformers and Rotating Machines
4th Edition
ISBN: 9781337264419
Author: Stephen L. Herman
Publisher: Cengage Learning US
Question
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Chapter 8, Problem 2P
To determine

The missing values.

Expert Solution & Answer
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Explanation of Solution

The given values are shown in below table:

Primary (P)Secondary (S)Load (L)
EP(P) EP(S) EP(L) 
IP(P) IP(S) IP(L) 
EL(P)7200EL(S)240EL(L) 
IL(P) IL(s) IL(L) 
Ratio Z4Ω  

Refer to the circuit shown in Figure 8-30.

The primary windings of three single phase transformer are connected in wye connection. In wye connection, the line voltage is equal to 3 times phase voltage.

  EL(P)=3EP(P)EP(P)=EL(P)3=72003=4156.92

The secondary windings of three single phase transformer are connected in delta connection. In delta connection, the line voltage is equal to phase voltage.

  EP(S)=EL(S)EP(S)=240

Calculate the turns ratio of primary and secondary winding.

  Turns ratio=EP(P)EP(S)=4156.92240=17.3205117.321

Thus, the turns ratio (EP(P):EP(S)) is 17.32:1.

Since, the secondary winding (S) is directly connected to the load (L). Thus, the line voltage of secondary winding and load are equal.

  EL(S)=EL(L)=240

Here, the load uses wye connection. In wye connection, the line voltage is equal to 3 times phase voltage.

  EL(L)=3EP(L)EP(L)=EL(L)3=2403=138.56

Calculate the phase current of the load (IP(L)).

  IP(L)=EP(L)Z=138.564Ω=34.64

In wye connection, the phase current (IP) and line current (IL) are equal.

  IL(L)=IP(L)IL(L)=34.64

Since, the secondary winding (S) is directly connected to the load (L). Thus, the line current of secondary winding and load are equal.

  IL(L)=IL(S)=34.64

Here, the secondary winding uses delta connection. In delta connection, the line current is equal to 3 times of phase current.

  IL(S)=3IP(S)IP(S)=IL(S)3=34.643=19.99920

Calculate the phase current in primary winding (IP(P)).

  IP(P)IP(S)=EP(S)EP(P)IP(P)=EP(S)EP(P)×IP(S)=117.32×20=1.15

Since, the primary winding uses wye connection. In wye connection, the line current is equal to phase current.

  IL(P)=IP(P)IL(P)=1.15

Thus, the all missing values are calculated and shown in below table:

Primary (P)Secondary (S)Load (L)
EP(P)4156.92EP(S)440EP(L)138.56
IP(P)1.15IP(S)20IP(L)34.64
EL(P)7200EL(S)240EL(L)240
IL(P)1.15IL(S)34.64IL(L)34.64
Ratio17.32:1Z4Ω  

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