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jolli1 [7]
3 years ago
13

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.
Engineering
1 answer:
Softa [21]3 years ago
5 0

Answer: 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

<em>I</em>₁ = ((400 × 10⁶) / 3(138.56 × 10³)) ∠ -cos⁻¹ (0.8)

<em>I</em>₁  = 962.28∠ -36.87° A

To find the voltage V₁, we say

V₁ = ( 1.2 + j6) <em>I</em>₁ + 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 )<em>I</em>₁ + 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

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Given that:

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cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]

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replacing their values;

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Also, to find the angle \phi between the stress [001] & normal slip plane [111]

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cos \  \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]

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i.e d_1 = 0 ,e_1 = 0 f_1 =  1 & d_3 = 1 , e_3 = 1 , f_3 = 1

cos \  \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]

cos \phi= \dfrac{1} {\sqrt{3} }

However, the critical resolved SS(shear stress) \mathbf{\tau_c} can be computed using the formula:

\tau_c = (\sigma )(cos  \phi )(cos \lambda)

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\tau_c =13.9\times (  \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})

\mathbf{\tau_c =5.675 \ MPa}

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