Delta (∆)-Star(Y) Conversion And Star-Delta Conversion

DELTA (∆)-STAR(Y) CONVERSION AND STAR-DELTA CONVERSION

Before taking up the examples, the formula for Delta (∆)-Star(Y) conversion and also Star-Delta conversion, using impedances as needed, instead of resistance as elements. The formulas for delta-star conversion, using resistances (Figure 1.70) are:

If three equal resistance (R₁ = R₂ = R₃ = R) connected in delta, are converted into its equivalent star, the resistances obtained are equal, its value being, R_a = R_b = R_c = (R/ 3)= R', which is derived using formulas given earlier. Similarly, if three equal resistances connected in star, are converted into its equivalent delta, the resultant resistances, using formulas, are equal (R₁ = R₂ = R₃ = 3 • R' = 3 ⋅ (R/3) = R).

The formula for the above conversions using impedances, instead of resistances, are same, replacing resistances by impedances, as the formula for series and parallel combination using impedances, instead of resistances, remain same as shown in the previous module on ac single phase circuits.

Please note that all the impedances in the formula given here are cofmplex quantities, like Z1 201... Za Za..., having both magnitude and angle as given.

An example is taken up, when three equal impedances connected in delta are to be converted into its equivalent star. The impedances are equal, both in magnitude and angle, such that |Z_c|=|Z₂|=|Z₃|=|Z, and  The impedances connected in delta are of the form . Using the formula given here, the impedances of the star Labai doidw equivalent are also equal, having the magnitude as:

The angles of the equivalent impedance connected in star are equal to the angles of the impedances connected in delta. The impedances connected in delta are also equal, both in magnitude and angle, and are of the form

Similarly, if three equal impedances connected in star are converted into its equivalent delta, the magnitude and angle of the impedances using the formulas given here, are  and 1=2=03=0 respectively. This shows that three impedances are equal, both in magnitude and angle, with its value being

which can also be obtained simply from the result given earlier.

Now, let us use the above formula for the circuits (Figure 1.72), using inductances only. The symbols used for the inductances are same (L1, ..., La...). The impedances of the inductances connected in delta, are computed as Z11 =0.0+ jœL1 =X1 ≤90°, the angles in three cases are 90°. The magnitudes of the impedances are proportional to the respective inductances as |Z1 = X, ∞ L1. Converting the combination into its equivalent star, the inductances using the formulas given here, are:

These relations can also be derived. Further, these are of the same form, as has been earlier obtained for resistances. It may be observed here that the formulas for series and parallel combination using inductances, instead of resistances, remain same, as shown in the previous module on ac single phase circuits, and also can be derived from first principles, such as relationship of induced emf in terms of inductance, as compared with Ohm's law for resistance. The inductances are all ideal, i.e., lossless, having no resistive component. The formulas for star-delta conversion using inductances (conversion of star-connected inductances into its equivalent delta) are:

These are of the same form as derived for circuits with resistances. If three equal inductances (L1 =L2 =L3 =L) connected in delta, are converted into its equivalent star, the inductances obtained are equal, its value being La=Lb Le=(L/3)=L', which is derived using formulas given earlier. Similarly, if three equal inductances connected in star, are converted into its equivalent delta, the resultant inductances, using formulas, are equal (L=L2=L3=3.L'=3.(L/3)=L).

The formulas for the circuits (Figure 1.73) using capacitances are derived here. The symbols used for the capacitances are same (C1, ..., Ca, ...). The impedances of the inductances connected in delta, are computed as Z1 ≤1 =0.0-jX1 = X1 ≤-90°, the angles in three cases are (- 90°). The magnitudes of the impedances are inversely proportional to the respective capacitances as, Z1 = X = X1 = (1/∞ C) (1/C1). Converting the combination into its equivalent star, the resultant capacitances using the formulas given here, are:

The capacitances in this case are all ideal, without any loss, specially at power frequency, which is true in nearly all cases, except otherwise stated. The formulas for star- delta conversion using capacitances (conversion of star-connected capacitances into its equivalent delta) are:

If three equal capacitances (C1 =C2=C3=C) connected in delta, are converted into its equivalent star, the capacitances obtained are equal, its value being Ca=Cb =Cc =(3⋅C)=C', which is derived using formulas given earlier. Similarly, if three equal capacitances connected in star, are converted into its equivalent delta, the resultant capacitances, using formula, are equal (C1 =C2=C3=C'/3=(3.C)73=C).

The formulas for conversion of three equal inductances/capacitances connected in delta into its equivalent star and vice versa (star-delta conversion) can also be obtained from the formulas using impedances as shown earlier, only by replacing inductance with with impedance, and for capacitance by replacing it reciprocal of impedance (in both cases using magnitude of impedance only, as the angles are equal (90° for inductance and - 90° for capacitance). Another point to note is left for observation by the reader. Please have a close look at the formula needed for delta-star conversion and vice versa (star-delta conversion) for capacitances, including those with equal values of capacitances, and then compare them with the formulas needed for such conversion using resistances/inductances (may be impedances also). The rules for conversion of capacitances in series/parallel into its equivalent one can be compared to the rules for conversion of resistances/inductances in series/parallel into its equivalent one. lios sonstoubai od 107

Exmple

The star-connected load consists of a resistance of 15 Q, in series with a coil having resistance of 5 Q, and inductance of 0.2 H, per phase. It is connected in parallel with the delta-connected load having capacitance of 90 μF per phase (Figure (a)). Both the loads being balanced, and fed from a three-phase, 400 V, 50 Hz, balanced supply, with the phase sequence as R-Y-B. Find the line current, power factor, total power & reactive VA, and also total volt-amperes (VA).

This example can be solved by converting the star-connected part into its equivalent

After starting with the generation of three-phase balanced voltage system, the phase and line voltages, both being balanced, first for star-connection, and then or delta-connection (both on source side), are discussed. The currents (both phase and line) for balanced star- connected load, along with total power consumed.