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The ratings of a three-phase three-winding transformer are

Primary (1) : Y connected, 66kV, 15MVA

Secondary (2) : Y connected, 13.2kV, 10MVA

Tertiary (3): A connected, 2.3kV, 5MVA

Neglecting winding resistances and exciting current, the per-unit leakage reactances are

X12 = 0.08on a 15MVA, 66kV base

X13 = 0.10on a 10 MVA, 13.2kV base

X23= 0.09on a 5MVA, 2.3kV base

(a) Determine the reactances X1X2X3 of the equivalent circuit on a 15MVA, 66kV base at the primary terminals. (b) Purely resistive loads of 7.5MW at 13.2kV and 5MW at 2.3kV are connected to the secondary and tertiary sides of the transformer, respectively. Draw the per-unit diagram, showing the per-unit impedances on a 15MVA, 66kV base at the primary terminals.

Short Answer

Expert verified

(a) The reactances X1,X2 and X3 are 0.0225 pu, 0.0575 pu, and 0.775 pu respectively on the base value 66kV, 15MVA.

(b) The per unit impedance diagram on base 66kV, 15MVA is.

Step by step solution

01

Write the given data by the question:

The primary base value is 66kV, 15MVA.

The secondary base value is 13.2kV, 10MVA.

The tertiary base values are 2.3kV, 5MVA.

The reactance X12 = 0.08 on a 15MVA, 66kV base.

The reactance X13 = 0.10on a 10 MVA, 13.2kV base.

The reactance X23= 0.09on a 5MVA, 2.3kV base.

The power rating of load connected to secondary is 7.5MW on 13.2kV .

The power rating of load connected to tertiary is 5MW on 2.3kV .

The base voltage Vbase = 66kV.

The base is Sbase= 15MVA .

02

Determine the formula to calculate the values of X1X2X3 and to calculate the per unit values of the per unit diagram. 

Since winding resistance and excitation current is neglected, hence the transformer is considered as ideal.

The expression to convert the reactance from old base to new base is given by,

Xnew=Xold(basevaluetakengivenbasevalue)…… (1)

The expression of per unit reactance for localid="1656745837287" X1is given by,

localid="1656745842898" X1=12(X12+X12-X23)…… (2)

The expression of per unit reactance forlocalid="1656745846518" X2 is given by,

localid="1656745850208" X2=12(X12+X23-X13)…… (3)

The expression of per unit reactance for localid="1656745854395" X3 is given by,

localid="1656745870452" X3=12(X13+X23-X12)…… (4)

The expression of three phase transformer power is given by,

localid="1656745882602" P3ϕ=3VLN2R…… (5)

Here,localid="1656745893735" P3ϕis the three-phase transformer power, localid="1656745905522" VLN is the line-to-line voltage of load and localid="1656745917334" Ris the resistance.

The expression to calculate the base impedance is given by,

localid="1656745948251" Zbase=Vbase2Sbase…… (6)

The expression to calculate the per unit value is given by,

localid="1656745961222" Perunit=ActualvalueBasevalue…… (7)

All the value should belong to the same parameter.

03

Calculate the value of reactances .X1X2X3

(a)

The reactance X12&X13 already given on the base66 kV, 15 MVA.

The reactance X23 can be calculate on the base values 66 kV, 15 MVA by using the equation using the equation (1).

X23=0.091510X23=0.135 pu

Calculate the reactance X1by using the equation (2).

X1=120.08+0.100.135X1=0.0225 pu

Calculate the reactance X2 by using the equation (3).

X1=120.08+0.1350.10X1=0.0575 pu

Calculate the reactance X3by using the equation (4).

X3=120.10+0.1350.08X3=0.0775 pu

Hence, the reactances X1,X2 and X3are 0.0225 pu, 0.0575 puand respectively on the base value66 kV, 15 MVA.

04

 Step 4: Draw the per unit diagram of the system on the base value66 kV, 15 MVA .

(b)

Calculate the secondary resistive load by using the equation (5).

R2=13.2×10327.5×106R2=174.24×1067.5×106R2=23.23Ω

Calculate the tertiary resistive load by using the equation (5).

R3=2.3×10325×106R3=5.29×1065×106R3=1.06Ω

Calculate the base impedance for the secondary winding by using the equation (6).

Zbase2=13.2×103215×106Zbase2=174.24×10615×106Zbase2=11.61Ω

Calculate the base impedance for the tertiary winding by using the equation (6).

Zbase3=2.3×103215×106Zbase3=5.29×10615×106Zbase3=0.35Ω

Calculate the per unit value of the secondary resistance by using the equation (7).

R2 pu=23.2311.61R2 pu=2 pu

Calculate the per unit value of the tertiary resistance by using the equation (7).

R3 pu=1.060.35R3 pu=3.03 pu

Hence, the per unit impedance diagram on base 66 kV, 15 MVA is shown below

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Most popular questions from this chapter

A single-phase, 50kVA,2400/240V,60Hz distribution transformer has the following parameters:

Resistance of the 2400 - V winding:R1=0.75Ω ,

Resistance of the 240 - V winding:R2=0.0075Ω

Leakage reactance of the 2400 - V winding:X1=1.0Ω ,

Leakage reactance of the 240 - V winding:X2=0.01Ω

Exciting admittance on the 240 - V side=0.003j0.02S

(a) Draw the equivalent circuit referred to the high-voltage side of the transformer.

(b) Draw the equivalent circuit referred to the low-voltage side of the transformer. Show the numerical values of impedances on the equivalent circuits.

Three single-phase two-winding transformers, each rated 25MVA,54.25.42kV , are connected to form a three-phaseY-Δ bank with a balanced Y-connected resistive load of 0.6Ωper phase on the low-voltage side. By choosing a base of 75MVA (three phase) and 94kV (line-to-line) for the high-voltage side of the transformer bank, specify the base quantities for the low-voltage side. Determine the per-unit resistance of the load on the base for the low-voltage side. Then determine the load resistance in ohms referred to the high-voltage side and the per-unit value of this load resistance on the chosen base

Consider a single-phase three-winding transformer with the primary excited winding of N1turns carrying a current I1and two secondary windings of N2andN3turns, delivering currents of I2and I3respectively. For an ideal case, how are the ampere-turns balanced?

(a) N1I1=N2I2-N3I3

(b) N1I1=N2I2+N3I3

(c) N1I1=-(N2I2-N3I3)

The two parallel lines in Example 3.13 supply a balanced load with a load current of 1-30°per unit. Determine the real and reactive power supplied to the load bus from each parallel line with (a) no regulating transformer, (b) the transformer in Example 3.13(a), and (c) the phase-angle-regulating transformer in Example Assume that the voltage at bus abcis adjusted so that the voltage at bus a'b'c'remains constant at 10°per unit. Also assume positive sequence. Comment on the effects of the regulating transformer.

In order to avoid difficulties with third-harmonic exciting current, which three-phase transformer connection is seldom used for step-up transformers between a generator and a transmission line in power systems.

(a)Y-(b)-Y(c)Y-Y

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