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A 50 KVA, 11000/400 V ΔY distribution transformer has a resistance of 1% and reactance of 7% per unit. What is the transformer phase impedance referred to the highvoltage side?
∴ Its base impedance
The PU impedance of the transformer is
Z_{eq} = [0.01 + j 0.07] PU
So the highside impedance in ohm is
Z_{eq} = Z_{eq pu} × Z_{base}
= (0.01 + j0.07) × 7260 Ω
= 72.6 + j 508.2
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 YY 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)
The transformer is connected yy 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
A 3phase bank of three single phase transformer are fed from 3 phase 33 kV (LL). It supplies a load of 5000 kVA at 11 kV (LL). Both supply and load three wire. Calculate the voltage rating of single phase transformer for startdelta connection.
For stardelta connection
Primary side phase voltage
Secondary side phase voltage = 11
∴ Transformer voltage rating 19.05/11 kV
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
Phase diagram of primary and secondary voltage
this connection is known as Dy_{11} connectiong
A 100,000KVA 230/115 – kV ΔΔ three phase power transformer has a perunit 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.
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%
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.
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
A1 ϕ 33 kVA, 3300 V/330 V, 50 Hz, distribution transformer is to be connected as an autotransformer to get an output voltage of 3630 V. It maximum kVA rating as an autotransformer is________kVA
Maximum kVA rating = V.A = V_{H}I_{H} = 3630 × 100 = 363 KVA
A 50 Hz, 3ϕ core type stardelta transformer has a line valtage ratio of 11,000/440 volts. The crosssection 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.
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
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 ?
line voltage at primary = V
Phase voltage at primary =V/√3
Line current at primary = I
Phase current at primary = I
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?
= 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
22 docs274 tests

22 docs274 tests
