A 400-MVA, 240-kV/24-kV, three-phase Y-A transformer has an equivalent series impedance of 1.2 + j6 N per phase referred to the high-voltage side. The transformer is supplying a three-phase load of 400-MVA, 0.8 power factor lagging at a terminal voltage of 24 kV (line to line) on its low-voltage side. The primary is supplied from a feeder with an impedance of 0.6 + ji.2 A per phase. Determine the line-to-line voltage at the high-voltage ter- minals of the transformer and the sending-end of the feeder.

the line-to-line voltage at the high-voltage terminals of the transformer and the sending-end of the feeder is 249.71∠1.8° kV

Explanation:

First we find the phase voltage per phase at the primary side connected in Y, so we say

V₂ = 240K/√3 = 138.56 kV

Now we find the primary current

I₁ = ((400 × 10⁶) / 3(138.56 × 10³)) ∠ -cos⁻¹ (0.8)

I₁  = 962.28∠ -36.87° A

To find the voltage V₁, we say

V₁ = ( 1.2 + j6) I₁ + V₂

we substitute

V₁ =  ( 1.2 + j6) 962.28∠ -36.87° + 138.56 × 10³

V₁  = 143∠1.57° kV

Now we find the phase voltage at the sending end

Vₓ = ( 0.6 + J1.2 )I₁ + V₁  

Vₓ = ( 0.6 + J1.2 ) 962.28∠ -36.87° + 143∠1.57° K

Vₓ = 144.17∠1.8° kV

So to Determine the line to line voltage at the sending end, we say:

Vₓ (line to line) = √3 × 144.17∠1.8° kV

Vₓ (line to line) = 249.71∠1.8° kV

hes mad if he says,

explanation:

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answer:

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