Question

A three-phase, 415 V load takes a line current of 800 A from a 3300/415V delta – star transformer. The 3300V system is supplied from an 11000/3300V star – star transformer. Draw the circuit diagram and, assuming no losses, find both line and phase values of voltages and currents in each part of the circuit. What will be the turns ratios of both transformers?

Answer 1

Delta star transformer

Secondary side

`V_L` = 415 V

`V_P` = 239.6 V

`I_P` = Line current,
`I_L` = 800 A

Primary side

`V_L` = Phase voltage,
`V_P` = 3300 V

Primary phase current = 58.08 A

`I_L` = 100.6 A

Turns ratio = 83/(660·√3)

Star-star transformer

Secondary side

`V_L` = 3300 V

`V_P` = 1905.26 V

`I_P` = Line current,
`I_L` = 100.6 A

Primary side

`V_L` = 11000 V

`V_P` = 6350.9 V

Line current,
`I_L` = Phase current,
`I_P` = 3.018 A

Turns ratio = 3/100.

Step-by-step explanation:

Delta star transformer

For the star Connection to the 3-Phase load,

Secondary line voltage,
`V_L` = 415 V

Secondary phase voltage,
`V_P` =
`V_L` ÷ √3 = 415 ÷ √3 = 239.6 V

Secondary phase current,
`I_P` = Line current,
`I_L` = 800 A

For the primary side of the delta star transformer

Line voltage,
`V_L` = Phase voltage,
`V_P` = 3300 V

K = 415/(3300×√3) = 83/(660·√3)

Primary phase current = 83/(660·√3)× 800 A = 58.08 A

Line current,
`I_L` = Phase current,
`I_P` × √3 = 100.6 A

The turns ratio = 83/(660·√3)

For the star-star transformer, we have;

Secondary line voltage,
`V_L` = 3300 V

Secondary phase voltage,
`V_P` =
`V_L` ÷ √3 = 3300 ÷ √3 = 1905.26 V

Secondary phase current,
`I_P` = Line current,
`I_L` = 100.6 A

For the primary side of the star-star transformer

Line voltage,
`V_L` = 11000 V

Phase voltage,
`V_P` = 11000 V ÷ √3 = 6350.9 V

K = 3300/11000 = 3/100

Primary phase current = 3/100× 100.6 = 3.018 A

Line current,
`I_L` = Phase current,
`I_P` = 3.018 A

The turns ratio = k = 3/100.