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QUESTION: 1

A 50 KVA, 11000/400 V Δ-Y distribution transformer has a resistance of 1% and reactance of 7% per unit. What is the transformer phase impedance referred to the high-voltage side?

Solution:

∴ Its base impedance

The PU impedance of the transformer is

Z_{eq} = [0.01 + j 0.07] PU

So the high-side impedance in ohm is

Z_{eq} = Z_{eq pu} × Z_{base}

= (0.01 + j0.07) × 7260 Ω

= 72.6 + j 508.2

*Answer can only contain numeric values

QUESTION: 2

An auto transformer is used to connect a 12.6 kV distribution line to a 13.8 kV distribution line. It must be capable of handling 2000 kVA. There are three phases, connected Y-Y with their neutrals solidly grounded. What must the N_{C}/N_{SE} turn ratio be to accomplish this connection? (N_{C}→ common turn, N_{SE} → Series turn)

Solution:

The transformer is connected y-y so the primary and secondary phase voltage are the line voltage divided by 3–√3. The turn ratio of each auto transformer is given by

QUESTION: 3

A 3-phase bank of three single phase transformer are fed from 3 phase 33 kV (L-L). It supplies a load of 5000 kVA at 11 kV (L-L). Both supply and load three wire. Calculate the voltage rating of single phase transformer for start-delta connection.

Solution:

For star-delta connection

Primary side phase voltage

Secondary side phase voltage = 11

∴ Transformer voltage rating 19.05/11 kV

QUESTION: 4

Three single phase transformers are connected to form a3 - ϕ transformer bank the transformer are connected in the following manner.

the transformer connecting will be represented by

Solution:

Phase diagram of primary and secondary voltage

this connection is known as Dy_{11} connectiong

*Answer can only contain numeric values

QUESTION: 5

A 100,000-KVA 230/115 – kV Δ-Δ three phase power transformer has a per-unit resistance of 0.02 PU and a per unit reactance of 0.055. The excitation branch element are R_{C} = 110 PU and X_{M} = 20 PU. If this transformer supplies a load of 80 MVA at 0.85 P.F lagging. Find the voltage regulation (in %) of the transformer bank under these conditions.

Solution:

The transformer supplies a load 80 MVA at 0.85 P.F lagging.

∴ Secondary line current

= 402 A

The base value of the secondary line current is

So, the P.U secondary current

The per unit primary voltage of this transformer

V_{P} = V_{S} + IZ_{EQ} = 1∠0° +0.8∠ − 31.8° (0.02 + j0.55)

= 1.037 ∠1.6°

The voltage regulation 1.037−11×100 = 3.7%

*Answer can only contain numeric values

QUESTION: 6

The terminal voltage on the secondary side when supplying full load at 0.8 lagging P.F. is 440 V. the equivalent Resistance and reactance drops for the transformer are 1% and 5% respectively.

Find Induced voltage per phase (in volt) at the secondary side.

Solution:

Percentage voltage Regulation

= R cos ϕ + X sin ϕ

= 1 × 0.8 + 5 × 0.6 = 3.8%

We know that voltage regulation

I.V. – 440 = 0.038 I.V

I.V = 457.38

*Answer can only contain numeric values

QUESTION: 7

A1 ϕ 33 kVA, 3300 V/330 V, 50 Hz, distribution transformer is to be connected as an auto-transformer to get an output voltage of 3630 V. It maximum kVA rating as an auto-transformer is________kVA

Solution:

Maximum kVA rating = V.A = V_{H}I_{H} = 3630 × 100 = 363 KVA

*Answer can only contain numeric values

QUESTION: 8

A 50 Hz, 3ϕ core type star-delta transformer has a line valtage ratio of 11,000/440 volts. The cross-section of the core is square with a circumscribing circle of 0.5 m diameter. If maximum flux density of 1.30 ωb/m^{2} then find the number of turns per phase on high voltage windings. Assume insulation occupies 8% of the total core area.

Solution:

We known that diameter of the circumscribing is same as diagonal of the square.

Area of square core = l × b

∴ Net cross section area = 0.125 × 0.92 = 0.115

E.M.F/Turn = 4.44 fBA = 4.44 × 50 × 1.3 × 0.115 = 33.19

Phase turn ratio

∴ Number of turn per phase on high voltage side

QUESTION: 9

In a Star\Delta 3 phase transformer connection, the turns ratio is N_{1}/N_{2} with N_{1} being primary side turns and N_{2} being secondary side turns. What is the value of line voltage and current on secondary side, if the same for primary side are V & I respectively ?

Solution:

line voltage at primary = V

Phase voltage at primary =V/√3

Line current at primary = I

Phase current at primary = I

QUESTION: 10

A 3 – phase, Δ – Y, 33000 / 660 V, 150 kVA, 50Hz transformer is having resistance per phase of 4 Ω on primary side and 0.15 Ω on secondary side. If the efficiency of transformer and half load is 96.5% at 0.85 pf (lagging), then what are the iron losses?

Solution:

= 0.15053 Ω

Full load secondary current (phase)

= 3 × (131.216)^{2} × 0.15053

= 7,775.31 W

Cu loss at half load = (0.5)^{2} × 7775.31

= 1943.83 W

Losses = input - output

= 66062 – 63750

= 2312

Iron loss = total loss – cu loss

= 2312 – 1943.83

= 368.17 W

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