Question

Three single-phase, 10 kVA, 460/120 V, 60 Hz transformers are connected to form a three-phase 460/208 V transformer bank. The equivalent impedance of each transformer referred to HV side is 1.0 +j2.0 Ω. The transformer delivers 20 kW at 0.8 pf leading. Answer the following questions:
(a) Draw a schematic diagram showing the transformer connection.
(b) Determine the magnitude of transformer primary and secondary winding currents.
(c) Determine the primary voltage magnitude for this operating condition. Determine the voltage regulation

Answer 1

Answer:

A) attached below

B) I₁ = 18.1 A ,  I₂ = 69.39 A

C)  V( magnitude) = 454.5 ∠ 5.04° V ,  Voltage regulation = ≈  -1.2%

Explanation:

A) Schematic diagram attached below

attached below

B) magnitude of primary and secondary winding currents

I₂ ( secondary current ) = P / √3 * VL * cos∅ ---------- ( 1 )

VL = Line voltage = 208

cos∅  ( power factor ) = 0.8

P = 20 * 10^3 watts

insert values into equation 1

I₂ = 69.39 A

I₁ ( primary current ) = I₂V2 / V1

                               I₁ = ( 69.39 * 120 ) / 460  = 18.1 A

C ) Calculate the Primary voltage magnitude and the Voltage regulation

V(magnitude ) = Vp + ( I₁ ∠∅ ) Req                            ( 1 + j2 = 2.24 ∠63.43° )

                       = 460 + ( 18.1 * cos^-1 (0.8) ) ( 1 + j2 )

                       = 460 + 40.544 ∠ 100.3°

∴ V( magnitude) = 454.5 ∠ 5.04° V

Voltage regulation

= ((Vmag - V1) / V1 )) * 100

= (( 454.5 - 460 / 460 )) * 100

= -1.195 % ≈  -1.2%