Question from my inbox:
I have a question regarding problem 11 of 4.2 Electric Power Devices. The problem is regarding a transformer bank in which one of the banks is removed. The reduced capacity is 57.7% or 288.5 MVA. How do we arrive at this answer?
The open delta transformer is identical to a normal delta transformer except one of the transformers is not installed.
See below.
Let us say that each of these single phase transformers are rated for 300 kVA and 4160V to 480V.
First, let us solve these problem as if it was a regular delta-delta transformer. In other words, it was not an open delta configuration.
The rating would be:
(3)×300 kVA)=900 kVA.
The current on the secondary side of the transformer would be:
I=S/(V×√3)=900,000/(480×√3)=1,083 amps
See below.
Let us say that each of these single phase transformers are rated for 300 kVA and 4160V to 480V.
First, let us solve these problem as if it was a regular delta-delta transformer. In other words, it was not an open delta configuration.
The rating would be:
(3)×300 kVA)=900 kVA.
The current on the secondary side of the transformer would be:
I=S/(V×√3)=900,000/(480×√3)=1,083 amps
Now with an open delta configuration, the transformer carries the current of the missing phase. The current is limited by the rating of the individual transformer.
I=S/V=300,000/480= 625 amps
The current in any one phase is limited by 625 amps.
Given this, the maximum capacity is:
S=V×I×√3=480×625×√3=519 kVA
I=S/V=300,000/480= 625 amps
The current in any one phase is limited by 625 amps.
Given this, the maximum capacity is:
S=V×I×√3=480×625×√3=519 kVA
You can see that 519/900=0.577 and 625/1083=0.577. This is 57.7%.