Problems solved about Autotransformer

Problem 94-5 Source: Problem prepared by the author of the site.

A single-phase autotransformer of 50 kVA, 60 Hz and voltage 13,200 / 9,000 V has a ratio of 6 V / turns. Consider that the maximum magnetic flux density is 1.2 T. Find:

a) the core area.

b) the number of turns of the autotransformer.

c) in which turn should the 9,000 V tap be removed.

d) Calculate the copper savings compared to a conventional transformer.

Solution of the Problem 94-5

Item a

To solve this item, we will use eq. 91-12 presented below.

eq. 91-12

Remembering the eq. 75-11a, where we have Φ_m = B_m A and, when we substitute into eq. 91-12, we obtain:

V₁ = 4.44 f N1 B_m A ⇒ V₁ / N₁ = 4.44 f B_m A

Where V₁ / N₁ is the volt / turns ratio. This value is provided by the problem statement, being V₁ / N₁ = 6 V/turn. Therefore, replacing the variables by their respective numerical values, we have:

6 = 4.44 x 60 x 1.2 x A

Performing the calculation, we find:

A = 0.01877 m² = 187.7 cm²

Item b

To calculate the number of turns we will use the relationship volt / turn = 6 V/esp.

N₁ = 13,200 / 6 = 2,200 turns

Item c

9,000 V

N₉₀₀₀ = 2,200 (9,000 / 13,200) = 1,500 esp.

Item d

eq. 94-10

k_w

a = 13,200 / 9,000 = 1.467

k_w = ( 1.467 - 1 ) / 1.467 = 0.3181

31.81%

Copper Savings = 100% - 31,81% = 68.19%

We can also use the equation below. In this case, the copper savings value is straightforward.

K_cu = 1 / 0.3181 = 0.6819 = 68.19%

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