Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The High Voltage Rating for Auto - Transformer is 86kV
The Low Voltage Rating for Auto - Transformer is 78kV
The MVA rating is 268.75[tex]MVA[/tex]
b
The efficiency is 99.4%
Explanation:
From the question we are given are given that
The transformer has Mega Volt Amp rating of 25MVA
The frequency is 60-Hz
Voltage rating 8.0kV : 78kV
The short circuit test gives : 453kV,321A,77.5kW
The open circuit test gives : 8.0kV, 39.6A, 86.2kW
This can be represented on a diagram shown on the second uploaded image
From this diagram we can deduce that the The High Voltage Rating for Auto - Transformer is 86kV and the Low Voltage Rating for Auto - Transformer is 78kV
Now to obtain the current flowing through the 8kV coil in the Auto-transformer we have
[tex]\frac{25 \ Mega \ Volt\ Ampere }{8\ Kilo Volt}[/tex]
The volt will cancel each other
[tex]\frac{25*10^6}{8*10^3} = 3125\ A[/tex]
Now to obtain the required MVA rating we would multiply the value of Power obtained during the open circuit test by the value of the current calculated.we are making use of the power obtain during open circuit testing because the transformer at this point is not under any load.
[tex]MVA \ rating = (86*10^3)(3125) =268.75[/tex]
We need to understand that Iron losses is due to open circuit test which has power = 86.2kW
While copper loss is due to short circuit test which has power = 77.5kW
The the current flowing through the secondary coil [tex]I_2[/tex] as shown in the circuit diagram can be obtained as
[tex]I_2 = \frac{25*10^6}{78*10^3} =320.52 A \approx 321[/tex]
Now the efficiency can be obtained as thus
[tex]\frac{(operational \ MVA )*(Power factor \pf))}{(operational\ MVA (power factor pf) + copper loss + Iron loss)}*\frac{100}{1}[/tex]
=99.941%
