Electrical Machines Test 2

 \(X\left( {pu} \right) = \frac{{X\left( {\rm{\Omega }} \right) \times MVA}}{{{{\left( {kV} \right)}^2}}}\)

HV side: \(0.12 = \frac{{X\left( {\rm{\Omega }} \right) \times 75 \times {{10}^{ - 3}}}}{{{{\left( 5 \right)}^2}}} \Rightarrow X\left( {\rm{\Omega }} \right) = 40\;{\rm{\Omega }}\)

LV side: \(0.12 = \frac{{X\left( {\rm{\Omega }} \right) \times 75 \times {{10}^{ - 3}}}}{{{{\left( {0.4} \right)}^2}}} \Rightarrow X\left( {\rm{\Omega }} \right) = 0.256\;{\rm{\Omega }}\)

Percentage rated voltage required to produce short circuited current = % impedance of transformer

% resistance (%R) = 3 %

% reactance (%X) = 4 %

% impedance, \(\% Z = \sqrt {{{\left( {\% R} \right)}^2} + {{\left( {\% X} \right)}^2}} = 5\% \)

Percentage rated voltage required to produce short circuited current = 5%

Concept:

Voltage regulation is the change in secondary terminal voltage from no load to full load at a specific power factor of load and the change is expressed in percentage.

E2 = no-load secondary voltage

V2 = full load secondary voltage

Voltage regulation for the transformer is given by the ratio of change in secondary terminal voltage from no load to full load to no load secondary voltage.

Voltage regulation \(= \frac{{{E_2} - {V_2}}}{{{E_2}}}\)

To calculate the regulation of the transformer at load current I₂ and load power factor cos ϕ, we use the following formula in terms of HV side parameters.

Per unit regulation, VR \( = \frac{{{I_2}{R_{e2}}\cos \phi + {I_2}{X_{e2}}\sin \phi }}{{{V_{20}}}}\)

The difference between the reflected primary supply voltage magnitude \(V_1^{'}\) and the secondary load terminal voltage magnitude V₂ is the numerator of the voltage regulation formula.

\(V_1^{'} - {V_2} = {I_2}{R_{e2}}\cos \phi + {I_2}{X_{e2}}\sin \phi \)

\( \Rightarrow V_1^{'} = {V_2} + {I_2}{R_{e2}}\cos \phi + {I_2}{X_{e2}}\sin \phi \)

Calculation:

\({I_2} = \frac{{5 \times {{10}^3}}}{{400}} = 12.5\;A\)

\(V_1^{'}\) = 400 + 12.5 × 1.12 × 0.7 + 12.5 × 1.14 × 0.71

= 419.92 V

Turns ratio (a) = 200/400 = 0.5

\({V_1} = aV_1^{'} = 0.5 \times 419.92 \approx 210\;V\)

Core loss \(= V{I_0}\cos {\phi _0} = \frac{{400}}{{\sqrt 2 }} \times \frac{5}{{\sqrt 2 }}\cos 30^\circ = 866\;W\)

Concept:

Hysteresis losses: These are due to the reversal of magnetization in the transformer core whenever it is subjected to alternating nature of magnetizing force.

\({W_h} = \eta B_{max}^xfv\)

\({B_{max}} \propto \frac{V}{f}\)

Where

x is the Steinmetz constant

Bm = maximum flux density

f = frequency of magnetization or supply frequency

v = volume of the core

At a constant V/f ratio, hysteresis losses are directly proportional to the frequency.

Wh ∝ f

Eddy current losses: Eddy current loss in the transformer is I2R loss present in the core due to the production of eddy current.

\({W_e} = K{f^2}B_m^2{t^2}V\)

\({B_{max}} \propto \frac{V}{f}\)

Where,

K - coefficient of eddy current. Its value depends upon the nature of magnetic material

Bm - Maximum value of flux density in Wb/m2

t - Thickness of lamination in meters

f - Frequency of reversal of the magnetic field in Hz

V - Volume of magnetic material in m3

At a constant V/f ratio, eddy current losses are directly proportional to the square of the frequency.

We ∝ f2

Iron losses or core losses or constant losses are the sum of both hysteresis and eddy current losses.

Wi = Wh­ + We

At constant V/f ratio, Wi = Af + Bf2

Calculation:

The table below shows the given data.

The V/f ratio is constant in all the cases as shown in the above table.

Now, the equations for Case 1 and Case 2 are given below

Case 1: 39.2 = A (40) + B (40)²

Case 2: 73.2 = A (60) + B (60)²

By solving the above two equations,

A = 0.5, B = 0.012

Case 3: Hysteresis losses (W_h) = af = 0.5 × 50 = 25 W

Eddy current losses (W_e) = bf² = 0.012 × (50)² = 30 W

Concept:

The efficiency of the Transformer \((\eta ) = \frac{{XS\cos \phi }}{{XS\cos \phi + {P_i} + {X^2}{P_e}}}\)

Where, X = Fraction of load

S = Apparent power in kVA

Pi = Iron losses

Pcu = Copper losses

Maximum efficiency of transformer occurred at a fraction of load, \(X = \sqrt {\frac{{{P_i}}}{{{P_{cu}}}}}\)

In a transformer, a short circuit test is used to find copper losses and the open circuit is used to find core losses.

Calculation:

From the given table, iron losses (W_i) = 120 W

Copper losses (W_cu) = 600 W

kVA rating of the transformer at maximum efficiency \( = \sqrt {\frac{{{W_i}}}{{{W_{cu}}}}} \times kVA = \sqrt {\frac{1}{5}} \times 48 = 21.467\;kVA\)

Maximum efficiency, \(\eta = \frac{{21.467 \times {{10}^3} \times 0.8}}{{21.467 \times {{10}^3} \times 0.8 + 120 + 600{{\left( {\frac{1}{{\sqrt 5 }}} \right)}^2}}} \times 100 = 98.621\;\% \)

Primary voltage, \({V_1} = 4.44{\phi _m}{N_1}\)

\(\begin{array}{l} \Rightarrow {N_1} = \frac{{{V_1}}}{{4.44{\phi _m}{N_1}}}\\ \Rightarrow {N_1} = \frac{{5000}}{{4.44 \times 50 \times 8.21 \times {{10}^{ - 3}}}} = 2744\\\frac{{{V_2}}}{{{V_1}}} = \frac{{{N_2}}}{{{N_1}}}\\ \Rightarrow {N_2} = {N_1} \times \frac{{{V_2}}}{{{V_1}}} = 2744 \times \frac{{500}}{{5000}} = 274.4\end{array}\)

\(\eta = \frac{{600 \times 1}}{{600 \times 1+{P_i} + {P_c}}} = 0.95\)

P_i + P_c = 31.58     ----(1)

\(\frac{{600 \times 0.6}}{{600 \times 0.6 + {P_i} + 0.36{P_c}}} = 0.95\)

P_i + 0.36 P_C = 18.95     ----(2)

Salving equs (1) and (2) we get

P_i = 11.846

P_c = 19.734

at ¾ th full load.

\(\eta = \frac{{600 \times 0.75}}{{600 \times 0.75 + 11.846 + {{\left( { 0. 75} \right)}^2} \times 19.734}}\)

η = 95.15%