what controls the incoming voltage to the autotransformer?

Autotransformer

An autotransformer demonstrates several advantages over an isolation transformer designed to perform the same job.

From: AC Ability Conditioners , 1990

Fundamental Concepts: Transformers

George Patrick Shultz , in Transformers and Motors, 1989

Variable Autotransformers

Autotransformers are also made to be adjustable. They are rated in kVA and are available in sizes from 0.25 to 200 kVA. Voltage ratings are 120, 240, and 480 volts.

These devices are made as individual units or ganged together from 2 up to 27 units. Ganged units may be fabricated to adjust manually, or they may exist motor driven. A schematic diagram of this device is shown in Figure 1-19.

FIGURE 1-19. Variable autotransformer.

Although solid-country silicon-controlled rectifiers are replacing these transformers for such applications as light dimmers and in other loftier-power control circuits, many of these installations remain in operation today.

Variable autotransformers are particularly useful for test situations where precise values are desired. They are often used in conjunction with DC power supplies to adjust the Air-conditioning input.

Variable autotransformers are designed to stride the voltage up besides as down. The footstep-up ratio is usually limited to 20%. They are also manufactured with fixed taps rather than a movable arm.

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Modelling of transformers, phase shifters, static power plant and static load

Nasser Tleis BSc (Hons), MSc, PhD, CEng, FIET, Thou-CIGRE , in Power Systems Modelling and Mistake Analysis (Second Edition), 2019

Positive-sequence and negative-sequence equivalent circuits in physical units

Autotransformers that interconnect extra high-voltage transmission systems are not mostly equipped with tap-changers due to high costs. However, those that interconnect the manual and subtransmission or distribution networks are usually equipped with on-load tap-changers in social club to control or improve the quality of their LV output voltage under heavy or low-cal load system conditions. In autotransformers, tap-changers are connected on either the HV winding or the LV winding. In the latter case, the connection may be either at the LV winding line-end or the LV winding neutral-terminate.

A single-phase representation of the full general example of an autotransformer with a third winding is shown in Fig. 4.25B. The tertiary winding is assumed, as is by and large the practice, the innermost winding closest to the core. Using S, C and T to announce the series, common and tertiary windings, we can write in bodily physical units

(4.34a) $V_{\mathrm{H}}-E_{\mathrm{H}}=Z_{\mathrm{S}}{I}_{\mathrm{H}}+Z_{\mathrm{C}}(I_{\mathrm{H}}+I_{\mathrm{L}})$

(4.34b) $5_{\mathrm{Fifty}}-E_{\mathrm{50}}=Z_{\mathrm{C}}(I_{\mathrm{H}}+I_{\mathrm{50}})$

(4.34c) $5_{\mathrm{T}}-E_{\mathrm{T}}=Z_{\mathrm{TT}}{I}_{\mathrm{T}}$

The MMF rest equation is given past

(4.35a) $N_{\mathrm{S}}{I}_{\mathrm{H}}+{\mathrm{Due\; north}}_{\mathrm{C}}(I_{\mathrm{H}}+I_{\mathrm{L}})+N_{\mathrm{T}}{I}_{\mathrm{T}}={\mathrm{North}}_{\mathrm{T}}{I}_{\mathrm{Thou}}$

(four.35b) ${\mathrm{Due\; north}}_{\mathrm{H}}=N_{\mathrm{Due\; south}}+N_{\mathrm{C}}\mathrm{and}{N}_{\mathrm{50}}=N_{\mathrm{C}}$

(four.35c) $\frac{E_{\mathrm{H}}}{{\mathrm{Due\; east}}_{\mathrm{Fifty}}}=\frac{{\mathrm{Due\; north}}_{\mathrm{H}}}{N_{\mathrm{L}}}\frac{E_{\mathrm{L}}}{E_{\mathrm{T}}}=\frac{{\mathrm{North}}_{\mathrm{L}}}{N_{\mathrm{T}}}\frac{{\mathrm{Due\; east}}_{\mathrm{H}}}{E_{\mathrm{T}}}=\frac{N_{\mathrm{H}}}{N_{\mathrm{T}}}$

Neglecting initially the no-load exciting electric current I _M and using Eqs. (iv.35b), (4.35c) and (4.34a), Eq. (4.34b) and (4.34c) can be written equally

(four.36a) $\frac{V_{\mathrm{L}}}{N_{\mathrm{50}}}-N_{\mathrm{50}}{I}_{\mathrm{L}}\frac{1}{N_{\mathrm{L}}^2}\left(\frac{{\mathrm{Due\; north}}_{\mathrm{H}}-N_{\mathrm{L}}}{N_{\mathrm{H}}}\right)Z_{\mathrm{C}}=\frac{V_{\mathrm{H}}}{N_{\mathrm{H}}}-{\mathrm{North}}_{\mathrm{H}}{I}_{\mathrm{H}}\frac{\mathrm{one}}{N_{\mathrm{H}}^2}\left[Z_{\mathrm{S}}-\left(\frac{{\mathrm{Northward}}_{\mathrm{H}}-N_{\mathrm{50}}}{N_{\mathrm{L}}}\right)Z_{\mathrm{C}}\right]$

(4.36b) $\frac{{\mathrm{Five}}_{\mathrm{T}}}{N_{\mathrm{T}}}-N_{\mathrm{T}}{I}_{\mathrm{T}}\frac{\mathrm{one}}{{\mathrm{Due\; north}}_{\mathrm{T}}^2}\left(Z_{\mathrm{TT}}+\frac{{\mathrm{North}}_{\mathrm{T}}^{\mathrm{ii}}}{N_{\mathrm{L}}{N}_{\mathrm{H}}}{Z}_{\mathrm{C}}\right)=\frac{V_{\mathrm{H}}}{{\mathrm{Northward}}_{\mathrm{H}}}-N_{\mathrm{H}}{I}_{\mathrm{H}}\frac{\mathrm{one}}{{\mathrm{Due\; north}}_{\mathrm{H}}^{\mathrm{ii}}}\left[Z_{\mathrm{Due\; south}}-\left(\frac{{\mathrm{Northward}}_{\mathrm{H}}-N_{\mathrm{Fifty}}}{N_{\mathrm{50}}}\right)Z_{\mathrm{C}}\right]$

Eq. (4.36a) and (iv.36b) tin can be represented by the star or T equivalent circuit shown in Fig. four.26A, which includes three ideal transformers placed at the H, 50 and T terminals of the autotransformer.

Figure 4.26. Positive-sequence/negative-sequence equivalent circuit of an autotransformer with a tertiary winding: (A) equivalent circuit in bodily physical units in terms of series, common and tertiary impedances; (B) every bit (A) in a higher place but leakage impedances are referred to H side; (C) as (B) in a higher place but for an autotransformer without a tertiary winding; (D) as (C) in a higher place but with impedances referred to H side and one ideal transformer (E) every bit (C) above but with impedances referred to L side and one ideal transformer.

The measurement of the positive-sequence and zero-sequence impedances of an autotransformer with a 3rd winding using short-circuit tests between ii winding terminals is dealt with after in this section. However, it is instructive to use Eq. (4.36a) and (4.36b) to explain the results that are obtained from such tests in the construction of the autotransformer equivalent circuit. Using Eq. (4.36a) and (4.36b), the positive-sequence impedance measured from the H side with the L side brusk-circuited and T side open-circuited is $Z_{\mathrm{HL}}=\frac{{\mathrm{Five}}_{\mathrm{H}}}{I_{\mathrm{H}}}{|}_{V_{\mathrm{L}}=0,I_{\mathrm{T}}=0}$ giving

(4.37a) $Z_{\mathrm{HL}}=Z_{\mathrm{S}}+{\left(\frac{N_{\mathrm{H}}-N_{\mathrm{L}}}{{\mathrm{North}}_{\mathrm{L}}}\right)}^2{Z}_{\mathrm{C}}$

Also, the impedance measured from the H terminals with the T terminals short-circuited and 50 terminals open up-circuited is $Z_{\mathrm{HT}}=\frac{5_{\mathrm{H}}}{I_{\mathrm{H}}}{|}_{{\mathrm{Five}}_{\mathrm{T}}=0,I_{\mathrm{L}}=0}$ giving

(4.37b) $Z_{\mathrm{HT}}=Z_{\mathrm{S}}+Z_{\mathrm{C}}+\frac{N_{\mathrm{H}}^2}{N_{\mathrm{T}}^2}{Z}_{\mathrm{TT}}$

Similarly, the impedance measured from the L terminals with the T terminals brusk-circuited and H terminals open-circuited is $Z_{\mathrm{LT}}=\frac{V_{\mathrm{50}}}{I_{\mathrm{50}}}{|}_{V_{\mathrm{T}}=0,I_{\mathrm{H}}=0}$ giving

(4.37c) $Z_{\mathrm{LT}}=Z_{\mathrm{C}}+\frac{{\mathrm{Due\; north}}_{\mathrm{L}}^2}{{\mathrm{Northward}}_{\mathrm{T}}^2}{Z}_{\mathrm{TT}}$

At present, given the three measured impedances Z _HL, Z _HT and Z _LT in ohms, the impedance of each co-operative of the T equivalent circuit with all impedances referred to the H side voltage base of operations tin be calculated. Let

(4.38a) $Z_{\mathrm{H}}+Z_{\mathrm{L}}^{\prime }=Z_{\mathrm{HL}}$

(iv.38b) $Z_{\mathrm{H}}+Z_{\mathrm{T}}^{\prime }=Z_{\mathrm{HT}}$

(4.38c) $Z_{\mathrm{L}}^{\prime }+Z_{\mathrm{T}}^{\prime }=\frac{N_{\mathrm{H}}^2}{N_{\mathrm{L}}^{\mathrm{two}}}{Z}_{\mathrm{LT}}$

Or in matrix grade

(4.38d) $\left[\begin{array}{c}{Z}_{\mathrm{HL}}\\ Z_{\mathrm{HT}}\\ Z_{\mathrm{LT}}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 0\\ \mathrm{i}& 0& 1\\ 0& \frac{1}{\left(\frac{{\mathrm{Northward}}_{\mathrm{H}}^{\mathrm{two}}}{{\mathrm{Northward}}_{\mathrm{L}}^2}\right)}& \frac{1}{\left(\frac{{\mathrm{Due\; north}}_{\mathrm{H}}^{\mathrm{two}}}{N_{\mathrm{L}}^{\mathrm{two}}}\right)}\end{array}\right]\left[\begin{array}{c}{Z}_{\mathrm{H}}\\ Z_{\mathrm{L}}^{\prime }\\ Z_{\mathrm{T}}^{\prime }\end{array}\right]$

where the prime indicates impedances referred to the H side voltage base of operations.

Solving Eq. (iv.38a)–(4.38d) for each co-operative impedance, we obtain

(4.39a) $Z_{\mathrm{H}}=\frac{1}{2}\left(Z_{\mathrm{HL}}+Z_{\mathrm{HT}}-\frac{N_{\mathrm{H}}^{\mathrm{two}}}{{\mathrm{Northward}}_{\mathrm{L}}^2}{Z}_{\mathrm{LT}}\right)$

(4.39b) $Z_{\mathrm{L}}^{\prime }=\frac{1}{\mathrm{ii}}\left(Z_{\mathrm{HL}}-Z_{\mathrm{HT}}+\frac{N_{\mathrm{H}}^2}{N_{\mathrm{L}}^2}{Z}_{\mathrm{LT}}\right)$

(4.39c) $Z_{\mathrm{T}}^{\prime }=\frac{1}{\mathrm{two}}\left(-Z_{\mathrm{HL}}+Z_{\mathrm{HT}}+\frac{{\mathrm{North}}_{\mathrm{H}}^2}{N_{\mathrm{L}}^2}{Z}_{\mathrm{LT}}\right)$

Or in matrix grade

(4.39d) $\left[\begin{array}{c}{Z}_{\mathrm{H}}\\ \\ Z_{\mathrm{L}}^{\prime }\\ \\ Z_{\mathrm{T}}^{\prime }\end{array}\right]=\frac{1}{\mathrm{two}}\left[\begin{array}{ccc}1& \mathrm{one}& -\left(\frac{N_{\mathrm{H}}^{\mathrm{ii}}}{{\mathrm{North}}_{\mathrm{L}}^2}\right)\\ 1& -1& \left(\frac{N_{\mathrm{H}}^{\mathrm{two}}}{{\mathrm{Due\; north}}_{\mathrm{L}}^{\mathrm{ii}}}\right)\\ -\mathrm{ane}& \mathrm{i}& \left(\frac{{\mathrm{Northward}}_{\mathrm{H}}^2}{{\mathrm{North}}_{\mathrm{Fifty}}^{\mathrm{ii}}}\right)\end{array}\right]\left[\begin{array}{c}{Z}_{\mathrm{HL}}\\ \\ Z_{\mathrm{HT}}\\ \\ Z_{\mathrm{LT}}\end{array}\right]$

Now, substituting Eq. (four.37a) and (4.37b) into Eq. (iv.39a)–(iv.39d), nosotros obtain

(4.40a) $Z_{\mathrm{H}}=Z_{\mathrm{South}}-(\frac{N_{\mathrm{H}}-N_{\mathrm{L}}}{N_{\mathrm{50}}})Z_{\mathrm{C}}$

(4.40b) $Z_{\mathrm{L}}^{\prime }=\frac{N_{\mathrm{H}}}{N_{\mathrm{L}}}\frac{(N_{\mathrm{H}}-N_{\mathrm{Fifty}})}{N_{\mathrm{Fifty}}}{Z}_{\mathrm{C}}$

(4.40c) $Z_{\mathrm{T}}^{\prime }=\frac{{\mathrm{North}}_{\mathrm{H}}}{N_{\mathrm{L}}}{Z}_{\mathrm{C}}+\frac{N_{\mathrm{H}}^{\mathrm{ii}}}{N_{\mathrm{T}}^2}{Z}_{\mathrm{TT}}$

Or in matrix form

(iv.40d) $\left[\begin{array}{c}{Z}_{\mathrm{H}}\\ \\ Z_{\mathrm{L}}^{\prime }\\ \\ Z_{\mathrm{T}}^{\prime }\end{array}\right]=\left[\begin{array}{ccc}1& -\left(\frac{{\mathrm{Due\; north}}_{\mathrm{H}}-N_{\mathrm{L}}}{N_{\mathrm{50}}}\right)& 0\\ 0& \frac{{\mathrm{Northward}}_{\mathrm{H}}}{N_{\mathrm{50}}}\left(\frac{N_{\mathrm{H}}-N_{\mathrm{L}}}{N_{\mathrm{L}}}\right)& 0\\ 0& \frac{{\mathrm{Northward}}_{\mathrm{H}}}{N_{\mathrm{L}}}& \frac{N_{\mathrm{H}}^2}{N_{\mathrm{T}}^2}\end{array}\right]\left[\begin{array}{c}{Z}_{\mathrm{S}}\\ \\ Z_{\mathrm{C}}\\ \\ Z_{\mathrm{TT}}\end{array}\right]$

Substituting Eq. (four.40a)–(4.40c) in Eq. (iv.36a) and (4.36b), we obtain

(4.41a) $\frac{{\mathrm{Five}}_{\mathrm{L}}}{{\mathrm{Due\; north}}_{\mathrm{L}}}-{\mathrm{Due\; north}}_{\mathrm{L}}{I}_{\mathrm{Fifty}}\frac{Z_{\mathrm{L}}^{\prime }}{{\mathrm{Northward}}_{\mathrm{H}}^{\mathrm{ii}}}=\frac{V_{\mathrm{H}}}{N_{\mathrm{H}}}-N_{\mathrm{H}}{I}_{\mathrm{H}}\frac{Z_{\mathrm{H}}}{N_{\mathrm{H}}^2}$

(4.41b) $\frac{V_{\mathrm{T}}}{{\mathrm{North}}_{\mathrm{T}}}-{\mathrm{North}}_{\mathrm{T}}{I}_{\mathrm{T}}\frac{Z_{\mathrm{T}}^{\prime }}{{\mathrm{Due\; north}}_{\mathrm{H}}^{\mathrm{ii}}}=\frac{V_{\mathrm{H}}}{{\mathrm{Northward}}_{\mathrm{H}}}-N_{\mathrm{H}}{I}_{\mathrm{H}}\frac{Z_H}{N_{\mathrm{H}}^2}$

Fig. 4.26B shows the equivalent circuit of Eq. (4.41a) and (iv.41b).

Where the autotransformer does not have a tertiary winding or where the tertiary winding is unloaded, the T terminal in Fig. 4.26B is unconnected to the power system network and the T branch impedance has no effect on the network currents and voltages. Thus, this co-operative can exist overlooked and the effective autotransformer impedance would then be the sum of the H and L branch impedances. Fig. four.26B can be simplified to the equivalent circuit shown in Fig. 4.26C which, in turn, can be farther simplified to the equivalent circuits shown in Fig. 4.26D and Eastward, where the equivalent leakage impedance is referred to the H side and L side, respectively.

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Modelling of transformers, static power institute and static load

Nasser D. Tleis BSc, MSc, PhD, CEng, FIEE , in Power Systems Modelling and Fault Analysis, 2008

PPS and NPS equivalent circuits

Autotransformers that interconnect extra high voltage transmission systems are not more often than not equipped with tap-changers due to high costs. However, those that interconnect the transmission and subtransmission or distribution networks are usually equipped with on-load tap-changers in social club to command or better the quality of their LV output voltage under heavy or light load system conditions. Although some tap-changers are connected on the HV winding, most tend to be connected on the LV winding. Most of these are continued at the LV winding line-cease and only a few are connected at the winding neutral-end.

A single-phase representation of the general case of an autotransformer with a tertiary winding is shown in Figure 4.20(b). Using Due south, C and T to denote the series, mutual and tertiary windings, we can write in actual physical units

(4.33a) $5_{\text{H}}-{\mathrm{East}}_{\text{H}}=Z_{\text{S}}{I}_{\text{H}}+Z_{\text{C}}\left(I_{\text{H}}+I_{\text{50}}\right)$

(4.33b) $V_{\text{L}}-{\mathrm{Eastward}}_{\text{L}}=Z_{\text{C}}\left(I_{\text{H}}+I_{\text{50}}\right)$

(4.33c) $5_{\text{T}}-{\mathrm{East}}_{\text{T}}=Z_{\text{TT}}{I}_{\text{T}}$

Neglecting the no-load current, the MMF balance is expressed as

$N_{\text{S}}{I}_{\text{H}}+{\mathrm{Northward}}_{\text{C}}\left(I_{\text{H}}+I_L\right)+N_{\text{T}}{I}_{\text{T}}=0$

or

(4.34a) $I_{\text{H}}+\frac{I_{\text{Fifty}}}{{\text{N}}_{\text{HL}}}+\frac{I_{\text{T}}}{{\text{N}}_{\text{HT}}}=0$

where

(four.34b) $\begin{array}{ccc}{\mathrm{Due\; north}}_{\text{HL}}=\frac{N_{\text{H}}}{N_{\text{50}}}=\frac{N_{\text{S}}+N_{\text{C}}}{{\mathrm{North}}_{\text{C}}}& \text{and}& N_{\text{HT}}=\frac{{\mathrm{Northward}}_{\text{H}}}{{\mathrm{Due\; north}}_{\text{T}}}=\frac{N_{\text{Due south}}+N_{\text{C}}}{N_{\text{T}}}\end{array}$

Also

(4.34c) $\begin{array}{ccc}\frac{E_{\text{H}}}{E_{\text{L}}}={\mathrm{Due\; north}}_{\text{HL}}=\frac{\mathrm{ane}}{N_{\text{LH}}}& \text{and}& \frac{{\mathrm{Due\; east}}_{\text{H}}}{E_{\text{T}}}=N_{\text{HT}}=\frac{1}{N_{\text{TH}}}\end{array}$

Using Equations (4.34b), (four.34c) and (4.33a), Equations (4.33b) and (4.33c) can be written every bit

(four.35a) $\frac{\mathrm{i}}{N_{\text{LH}}}\left[{\mathrm{Five}}_{\text{50}}-I_{\text{L}}\left(\frac{N_{\text{HL}}-\mathrm{ane}}{N_{\text{HL}}}\right)Z_{\text{C}}\right]={\mathrm{Five}}_{\text{H}}-I_{\text{H}}\left[Z_{\text{S}}-\left(N_{\text{HL}}-1\right)Z_{\text{C}}\right]$

(four.35b) $\frac{1}{{\mathrm{Due\; north}}_{\text{TH}}}\left[5_{\text{T}}-I_{\text{T}}\left(Z_{\text{TT}}+\frac{N_{\text{HL}}}{N_{\text{HL}}^{\text{ii}}}{Z}_{\text{C}}\right)\right]=5_{\text{H}}-I_{\text{H}}\left[Z_{\text{S}}-\left(N_{\text{HL}}-\mathrm{i}\right)Z_{\text{C}}\right]$

Equation (4.35) tin can exist represented by the star equivalent circuit shown in Figure 4.21(a) containing two ideal transformers equally for a three-winding transformer.

Figure 4.21. PPS/NPS equivalent circuit of an autotransformer with a tertiary winding: (a) equivalent circuit in actual physical units; (b) equally (a) above but with L and T branch impedances referred to H voltage base; (c) as (b) above but an autotransformer without a tertiary winding and (d) equally (b) higher up but all quantities are in pu

The measurement of the autotransformer PPS and ZPS impedances using brusk-circuit tests betwixt 2 winding terminals is dealt with after in this section. However, information technology is instructive to utilize Equation (4.35) to demonstrate the results that can be obtained from such tests. Using Equations (4.34a) and (iv.35), the PPS impedance measured from the H terminals with the 50 terminals brusk-circuited and T terminals open up-circuited is

${Z_{\text{HL}}=\frac{5_{\text{H}}}{I_{\text{H}}}|}_{5_{\text{50}}=0,I_{\text{T}}=0}$

hence

(4.36a) $Z_{\text{HL}}=Z_{\text{S}}+{\left(N_{\text{HL}}-1\right)}^2{Z}_C$

Also, the impedance measured from the H terminals with the T terminals short-circuited and L terminals open-circuited is

${Z_{\text{HT}}=\frac{V_{\text{H}}}{I_{\text{H}}}|}_{V_{\text{T}}=0,I_{\text{L}}=0}$

hence

(4.36b) $Z_{\text{HT}}=Z_{\text{South}}+Z_C+{\mathrm{Due\; north}}_{\text{HT}}^2{Z}_{\text{TT}}$

Similarly, the impedance measured from the Fifty terminals with the T terminals brusk-circuited and H terminals open-circuited is

${Z_{\text{LT}}=\frac{V_{\text{L}}}{I_{\text{L}}}|}_{5_{\text{T}}=0,I_{\text{H}}=0}$

hence

(four.36c) $Z_{\text{LT}}=Z_{\text{C}}+\frac{N_{\text{HT}}^{\mathrm{two}}}{N_{\text{HL}}^2}{Z}_{\text{TT}}$

To calculate the impedance of each co-operative of the T equivalent circuit in ohms with all impedances referred to the H side voltage base of operations, let usa define the measured impedances as follows:

(four.36d) $Z_{\text{HL}}=Z_{\text{H}}+{Z^{\prime }}_{\text{L}}$

(four.36e) $Z_{\text{HT}}=Z_{\text{H}}+{Z^{\prime }}_{\text{T}}$

(4.36f) $Z_{\text{LT}}=\frac{\mathrm{one}}{{\mathrm{Northward}}_{\text{HL}}^{\mathrm{ii}}}\left({Z^{\prime }}_{\text{L}}+{Z^{\prime }}_{\text{T}}\right)$

where the prime indicates quantities referred to the H side.

Solving Equations (4.36d), (4.36e) and (4.36f) for each branch impedance, we obtain

(4.37a) $Z_{\text{H}}=\frac{\mathrm{ane}}{2}\left(Z_{\text{HL}}+Z_{\text{HT}}-N_{\text{HL}}^2{Z}_{\text{LT}}\right)$

(iv.37b) ${Z^{\prime }}_{\text{50}}=\frac{\mathrm{i}}{2}\left(Z_{\text{HL}}+N_{\text{HL}}^2{Z}_{\text{LT}}-Z_{\text{HT}}\right)$

(iv.37c) ${Z^{\prime }}_{\text{T}}=\frac{1}{2}\left({\mathrm{Due\; north}}_{\text{HL}}^2{Z}_{\text{LT}}+Z_{\text{HT}}-Z_{\text{HL}}\right)$

At present, substituting Equations (4.36a), (iv.36b) and (iv.36c) into Equations (4.37a), (iv.37b) and (four.37c), we obtain

(4.38a) $Z_{\text{H}}=Z_{\text{S}}-\left(N_{\text{HL}}-1\right)Z_{\text{C}}$

(4.38b) ${Z^{\prime }}_{\text{L}}={\mathrm{Due\; north}}_{\text{HL}}\left(N_{\text{HL}}-1\right)Z_{\text{C}}$

(iv.38c) ${Z^{\prime }}_{\text{T}}=Z_{\text{HL}}{Z}_{\text{C}}+N_{\text{HT}}^{\mathrm{two}}{Z}_{\text{TT}}$

Effigy iv.21(b) shows the autotransformer PPS T equivalent circuit with all impedances in ohms referred to the H terminals voltage base. In the absence of a 3rd winding, Figure four.21c shows the equivalent circuit of the autotransformer. Using Equations (four.38) in Equations (four.35), we obtain

(four.39a) $\frac{1}{N_{\text{LH}}}\left[V_{\text{50}}-I_{\text{Fifty}}\frac{{Z^{\prime }}_{\text{L}}}{{\text{Due north}}_{\text{HL}}^{\text{2}}}\right]=V_{\text{H}}-I_{\text{H}}{Z}_{\text{H}}$

(4.39b) $\frac{1}{N_{\text{Th}}}\left[{\mathrm{Five}}_{\text{T}}-I_{\text{T}}\frac{{Z^{\prime }}_{\text{T}}}{{\text{N}}_{\text{HT}}^{\text{2}}}\right]=5_{\text{H}}-I_{\text{H}}{Z}_{\text{H}}$

Now, nosotros will convert Equations (iv.39) from actual units to pu values. To do so, we define the following pu quantities

(iv.40a) $\begin{array}{ccccc}{V}_{\text{pu}}=\frac{\mathrm{Five}}{V_{\left(\text{B}\right)}}& I_{\text{pu}}=\frac{I}{I_{\left(\text{B}\right)}}& Z_{\text{H}\left(\text{pu}\right)}=\frac{Z_{\text{H}}}{Z_{\text{H}\left(\text{B}\right)}}& Z_{\text{L}\left(\text{pu}\right)}=\frac{{Z^{\prime }}_{\text{Fifty}}}{Z_{\text{H}\left(\text{B}\right)}}& Z_{\text{T}\left(\text{pu}\right)}=\frac{{Z^{\prime }}_{\text{T}}}{V_{\text{H}\left(\text{B}\right)}}\end{array}$

(four.40b) $\begin{array}{cc}{V}_{\text{H}\left(\text{B}\right)}=Z_{\text{H}\left(\text{B}\right)}{I}_{\text{H}\left(\text{B}\right)}& 5_{\text{L}\left(\text{B}\right)}=Z_{\text{50}\left(\text{B}\right)}{I}_{\text{L}\left(\text{B}\right)}\end{array}$

(4.40c) ${\mathrm{South}}_{\text{H}\left(\text{B}\right)}=S_{\text{L}\left(\text{B}\right)}=S_{\text{T}\left(\text{B}\right)}=V_{\text{H}\left(\text{B}\right)}{I}_{\text{H}\left(\text{B}\right)}=V_{\text{50}\left(\text{B}\right)}{I}_{\text{Fifty}\left(\text{B}\right)}=5_{\text{T}\left(\text{B}\right)}{I}_{\text{T}\left(\text{B}\right)}$

(4.40d) $\begin{array}{cc}\frac{V_{\text{H}\left(\text{B}\right)}}{V_{\text{L}\left(\text{B}\right)}}=\frac{N_{\text{H}\left(\text{nomianal}\right)}}{N_{\text{H}\left(\text{nomianal}\right)}}& \frac{{\mathrm{Five}}_{\text{H}\left(\text{B}\right)}}{V_{\text{T}\left(\text{B}\right)}}=\frac{N_{\text{H}\left(\text{nomianal}\right)}}{{\mathrm{North}}_{\text{T}\left(\text{nomianal}\right)}}\end{array}$

Using Equations (iv.40) in Equations (four.39a) and (4.39b), we obtain

(4.41a) $\frac{V_{\text{50}\left(\text{pu}\right)}}{\frac{{\mathrm{Due\; north}}_{\text{LH}}{V}_{\text{H}\left(\text{B}\right)}}{V_{\text{Fifty}\left(\text{B}\right)}}}-\frac{N_{\text{LH}}{\mathrm{Five}}_{\text{H}\left(\text{B}\right)}}{{\mathrm{Five}}_{\text{L}\left(\text{B}\right)}}{I}_{\text{L}\left(\text{pu}\right)}{Z}_{\text{L}\left(\text{pu}\right)}={\mathrm{Five}}_{\text{H}\left(\text{pu}\right)}-Z_{\text{Fifty}\left(\text{pu}\right)}{I}_{\text{L}\left(\text{pu}\right)}$

(4.41b) $\frac{5_{\text{T}\left(\text{pu}\right)}}{\frac{{\mathrm{Northward}}_{\text{TH}}{V}_{\text{H}\left(\text{B}\right)}}{V_{\text{T}\left(\text{B}\right)}}}-\frac{N_{\text{Thursday}}{V}_{\text{H}\left(\text{B}\right)}}{5_{\text{T}\left(\text{B}\right)}}{I}_{\text{T}\left(\text{pu}\right)}{Z}_{\text{T}\left(\text{pu}\right)}=V_{\text{H}\left(\text{pu}\right)}-Z_{\text{H}\left(\text{pu}\right)}{I}_{\text{H}\left(\text{pu}\right)}$

Equations (iv.41a) and (4.41b) can be rewritten as

(four.42a) $\frac{V_{\text{L}\left(\text{pu}\right)}}{t_{\text{LH}\left(\text{pu}\right)}}-t_{\text{LH}\left(\text{pu}\right)}{I}_{\text{L}\left(\text{pu}\right)}{Z}_{\text{L}\left(\text{pu}\right)}=V_{\text{H}\left(\text{pu}\right)}-Z_{\text{H}\left(\text{pu}\right)}{I}_{\text{H}\left(\text{pu}\right)}$

(4.42b) $\frac{V_{\text{T}\left(\text{pu}\right)}}{t_{\text{Thursday}\left(\text{pu}\right)}}-t_{\text{TH}\left(\text{pu}\right)}{I}_{\text{T}\left(\text{pu}\right)}{Z}_{\text{T}\left(\text{pu}\right)}=V_{\text{H}\left(\text{pu}\right)}-Z_{\text{H}\left(\text{pu}\right)}{I}_{\text{H}\left(\text{pu}\right)}$

where the post-obit pu tap ratios are defined

(four.43a) $t_{\text{LH}\left(\text{pu}\right)}=\frac{N_{\text{LH}}{V}_{\text{H}\left(\text{B}\right)}}{V_{\text{L}\left(\text{B}\right)}}=\frac{N_{\text{L}}{V}_{\text{H}\left(\text{B}\right)}}{N_{\text{H}}{V}_{\text{50}\left(\text{B}\right)}}=\frac{{\mathrm{Due\; north}}_{\text{L}}{V}_{\text{H}\left(\text{nominal}\right)}}{N_{\text{H}}{N}_{\text{L}\left(\text{nominal}\right)}}=\frac{\frac{{\mathrm{Due\; north}}_{\text{L}\left(\text{at\hspace{0.17em}a\hspace{0.17em}given\hspace{0.17em}tap\hspace{0.17em}position}\right)}}{N_{\text{L}\left(\text{nominal\hspace{0.17em}tap\hspace{0.17em}position}\right)}}}{\frac{N_{\text{H}\left(\text{at\hspace{0.17em}a\hspace{0.17em}given\hspace{0.17em}tap\hspace{0.17em}position}\right)}}{N_{\text{H}\left(\text{nominal\hspace{0.17em}tap\hspace{0.17em}position}\right)}}}=\frac{t_{\text{L}\left(\text{pu}\right)}}{t_{\text{H}\left(\text{pu}\right)}}$

(4.43b) $t_{\text{TH}\left(\text{pu}\right)}=\frac{N_{\text{TH}}{V}_{\text{H}\left(\text{B}\right)}}{5_{\text{T}\left(\text{B}\right)}}=\frac{N_{\text{T}}{V}_{\text{H}\left(\text{B}\right)}}{{\mathrm{North}}_{\text{H}}{V}_{\text{T}\left(\text{B}\right)}}=\frac{N_{\text{T}}{5}_{\text{H}\left(\text{nominal}\right)}}{{\mathrm{North}}_{\text{H}}{N}_{\text{T}\left(\text{nominal}\right)}}=\frac{\frac{N_{\text{T}\left(\text{at\hspace{0.17em}a\hspace{0.17em}given\hspace{0.17em}tap\hspace{0.17em}position}\right)}}{{\mathrm{North}}_{\text{T}\left(\text{nominal\hspace{0.17em}tap\hspace{0.17em}position}\right)}}}{\frac{{\mathrm{Northward}}_{\text{H}\left(\text{at\hspace{0.17em}a\hspace{0.17em}given\hspace{0.17em}tap\hspace{0.17em}position}\right)}}{{\mathrm{Due\; north}}_{\text{H}\left(\text{nominal\hspace{0.17em}tap\hspace{0.17em}position}\right)}}}=\frac{t_{\text{T}\left(\text{pu}\right)}}{t_{\text{H}\left(\text{pu}\right)}}$

Equations (4.42) are represented by the pu equivalent circuit shown in Figure 4.21(d) which represents the autotransformer PPS/NPS equivalent circuit ignoring the delta 3rd stage shift. The autotransformer is clearly represented as three 2-winding transformers that are star or T connected. Two of these transformers take off-nominal tap ratios that tin represent whatever off-nominal tap ratios on any winding or a combination of tap-ratios. In some cases, the two variable ratios must be consistent and coordinated where an agile tap-changer on simply one winding can in effect change the effective turns ratio on another. For example, for a 400 kV/132 kV/thirteen kV autotransformer having a tap-changer acting on the neutral terminate of the common winding, the variation of t _LH(pu) caused by HV to LV turns ratio changes will also cause corresponding changes in the HV to Boob tube turns ratio and hence in t _TH(pu). Therefore, t _Thursday(pu) is a function of t _LH(pu) which varies as a event of controlling the LV (132 kV) last voltage to a specified target value around a deadband.

Where the autotransformer does not have a tertiary winding or where the tertiary winding is unloaded, the T terminal in Figure 4.21(d) would be unconnected to the power system network and its branch impedance has no effect on the network currents and voltages. Thus, this branch can be disregarded and the effective autotransformer impedance would then be the sum of the H and Fifty branch impedances given by Z _HL(pu) = Z _H(pu) + Z _L(pu). In this case, the PPS/NPS equivalent circuit of an autotransformer is similar to that already derived for a 2-winding transformer and shown in Figures iv.8(c) or 4.nine(c). These can be used to correspond an autotransformer with 'series' winding tap-changer or 'mutual' winding tap-changer, respectively. The latter represents British practice irrespective of whether the tap-changer is connected to the line-end or neutral-end of the 'common' winding.

The impedances of the autotransformer required in the equivalent circuit of Figure four.21(d) are calculated from short-excursion test information supplied by the manufacturer. This is covered in detail in Section 4.two.9.

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Magnetic circuits and power transformers

Easwaran Chandira Sekaran , in Electric Renewable Free energy Systems, 2016

11.thirteen Autotransformers

In an autotransformer, the main and secondary windings are linked together both electrically and magnetically. Therefore information technology is economical for the same VA rating equally windings are reduced, but the disadvantage is that it does not have isolation between main and secondary windings. The winding tin can be designed with multiple tapping points, to provide unlike voltage points along its secondary winding. The winding diagram and the number of windings in chief and secondary ( Due north _p and Northward _s, respectively), current, and voltage across primary and secondary are shown in Figure 11.xiii.

Effigy 11.13. Winding diagram of an autotransformer.

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Generator configuration

David Stephen , in Contained Generation of Electric Power, 1994

Voltage regulator

A simple autotransformer with on-load tap-changing can be used to inject a voltage difference between the two busbar voltages and and so recoup for function of the impedance drop betwixt the systems. This would require tapping steps of no greater than 2½ per cent, and with normal AVRs should requite a reasonable control of VAR circulation.

When the distance between the ii systems is considerable or the load to be transmitted is high, it volition exist necessary to use a distribution voltage greater than the busbar voltage. This will require transformers at both ends of the line, and by selecting suitable tapping steps on these a convenient control function can be obtained. With this arrangement, of class, in that location is no need for the two busbar voltages to exist equal or even of the same social club.

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The Transformer

Thousand.One thousand. SMITH CENG., Grand.I.E.E. , in Electrical Engineering Principles for Technicians, 1970

six.9 The autotransformer

Consider the primary winding AB of a single-phase transformer (Fig. 6.13). There is an equal voltage beyond each plough and if a tapping were fabricated at C, there would be a p.d. existing between C and B . This arrangement is called an autotransformer and when a load impedance Z is continued across terminals C and B, a current flows in it.

FIG. 6.thirteen.

In the part of the winding CB, the primary and secondary currents may exist regarded equally being superimposed. As with the two-winding transformer, the ampere-turns due to the secondary current must exist in opposition to those ready by the primary electric current. The current in this part of the winding is the phasor difference betwixt the primary and secondary currents. Neglecting all losses, the phasor difference is equal to the numerical divergence. This part of the winding can have conductors of smaller cross-sectional area than the remainder of the winding and and then there is a saving in the corporeality of copper required as compared with the equivalent ratio double-wound transformer. The nearer the secondary voltage is to the primary voltage, the greater is the saving in copper and the cheaper the transformer relative to a double-wound transformer of the same voltage ratio.

Some autotransformers take several tappings to enable alternative values of secondary voltage to exist obtained. Such three-stage autotransformers may be used for starting induction and synchronous motors. Autotransformers are also used for interconnecting the 275-kV grid to the 132-kV system. Their rating can exist in backlog of 200 MVA.

There are a number of applications where, for reasons of safety, autotransformers may not be used. If a pause should occur in the common part of the winding (Fig. 6.14), the secondary voltage could become equal to the primary voltage. Autotransformers must not, for example, exist used with toys and portable tools and reference should be made to the 14th edition of the I.E.Due east. Regulations for the Electrical Equipment of Buildings.

FIG. vi.fourteen.

There are continuously variable autotransformers which are often known as Variacs. They are frequently used in laboratories and can exist made to have an output up to virtually 20 kVA at 240 V.

They have a toroidal winding wound on a round core. The enamelled wire conductors have the enamel removed over a modest width so equally to enable a narrow carbon castor to brand electrical contact with the turns. This constitutes a variable contact as the castor is moved either manually or by an electric drive.

The voltage between adjacent turns is not allowed to exceed about 1 volt in lodge to limit the circulating electric current when the narrow high resistance castor short-circuits next turns. If the brush were wide, large circulating currents would be fix up through the castor. The brush tends to run hot and the heavy copper castor holder assists in the dissipation of this heat.

Instance half-dozen.3

An autotransformer has a 200-V input and a 150-V output. Calculate the current in each part of the winding if the load is 12 kW at 0·8 power gene. Neglect all losses.

$\begin{array}{ll}\text{Secondary electric current}{I}_{\mathrm{ii}}& =\frac{12,000}{150\times 0\cdot \mathrm{viii}}\text{amperes}\\ & =100\text{undefined}A.\\ \text{Primary current}{I}_1& =\frac{12,000}{200\times 0\cdot 8}\text{amperes}\\ & =75\text{undefined}A.\\ \text{Common winding current}& =I_2-I_{\mathrm{ane}}\\ & =(100-75)\text{ amperes}\\ & =\mathrm{two}\mathrm{five}\text{undefined}A.\end{array}$

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Induction motor—Operation from 50/60Hz supply

Austin Hughes , Beak Drury , in Electrical Motors and Drives (Fifth Edition), 2019

vi.2.3 Autotransformer starter

A three-phase autotransformer is usually used where star-delta starting provides insufficient starting torque. Each stage of an autotransformer consists of a unmarried winding on a laminated cadre. The incoming supply is connected across the ends of the coils, and one or more tapping points (or a sliding contact) provide a reduced voltage output, as shown in Fig. 6.3.

Fig. half-dozen.3. Autotransformer starter for cage induction motor.

The motor is first connected to the reduced voltage output, and when the current has fallen to the running value, the motor leads are switched over to the full voltage.

If the reduced voltage is chosen then that a fraction α of the line voltage is used to start the motor, the starting torque is reduced to approximately α^two times its directly-on-line value, and the electric current drawn from the supply is likewise reduced to α² times its direct value. Every bit with the star/delta starter, the torque per ampere of supply current is the same as for a direct commencement.

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Network Analysis

MG Say PhD, MSc, CEng, FRSE, FIEE, FIERE, ACGI, DIC , MA Laughton BASc, PhD, DSc(Eng), FREng, FIEE , in Electrical Engineer's Reference Book (Sixteenth Edition), 2003

(b) Star-star transformer

For a ii-circuit three-phase transformer or autotransformer connected in a star-star arrangement, the equivalent circuit is as shown in linear-graph grade in Effigy three.52 . Parallel transformer windings are taken to represent equivalent single-phase transformers. The circuit is synthetic from the simple connection of three of the full general circuits shown in Figure iii.51 with taps on both windings. In practise, of course, either α or β, or both, would be 1.0 p.u.

Figure iii.52. A three-phase equivalent circuit of a star–star connected transformer

In a more than concise grade, the equivalent excursion of Effigy three.52 may exist described by the connection tabular array given in Table iii.6 where, for case, an admittance of value y/α² is connected betwixt North and A, also N and B, etc. If the neutrals are earthed or connected together either solidly or through an impedance, the appropriate additions or deletions can be made to the circuit and corresponding terms inverse in the connection table. From inspection of the circuit, the corresponding admittance matrix tin exist assembled with or without rows and columns for the neutral nodes, depending on the earthing arrangements.

Table three.six. Connection table for the star—star transformer equivalent circuit shown in Effigy three.52 (α = i + t _α p.u.; β = 1 + t _β p.u.)

Comprisal	Between nodes
y/αii	North—A, N—B, Northward—C
y/βtwo	n—a, north—b, northward—c
y/αβ	A—a, B—b, C—c
−y/αβ	north—A, n—B, northward—C; North—a, Due north—b, Northward—c
3y/αβ	Northward—n

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Lines and Loads

P.S.R. Murty , in Power Systems Assay (Second Edition), 2017

9.2 Transformers

Transformers operate with taps on lines. The tap setting volition change the line flows and the voltage. Modeling of different types of transformers with off nominal turns ratio is required in load flow solution.

Further, a phase shifting transformer may be present on the lines for control purpose. The exact modeling of these devices will exist presented now.

9.ii.i Transformer with Nominal Turns Ratio

Consider a transformer with turns ratio a :one. This can be represented every bit an ideal autotransformer in serial with an admittance. Let p–q represents the input and output buses of the transformer. The ideal autotransformer is shown between p and t buses, while the series admittance is shown betwixt t and q (run into Fig. 9.5).

Effigy 9.5. Transformer with taps.

(9.3) $\begin{array}{ll}{I}_{\mathrm{tq}}\hfill & =\mathrm{current}\mathrm{flowing}\mathrm{from}t\mathrm{to}q\hfill \\ I_{\mathrm{tq}}\hfill & =({\mathrm{five}}_{\mathrm{t}}-{\mathrm{five}}_{\mathrm{q}})y_{\mathrm{pq}}\hfill \end{array}$

The terminal current at p,

(ix.4) ${\mathrm{I}}_{\mathrm{p}}=({\mathrm{V}}_{\mathrm{t}}-{\mathrm{V}}_{\mathrm{q}})\frac{{\mathrm{Y}}_{\mathrm{pq}}}{\mathrm{a}}$

The terminal current at q is similarly

(ix.6) $I_{\mathrm{q}}=(V_{\mathrm{q}}-V_{\mathrm{t}})y_{\mathrm{pq}}$

Substituting for

(9.seven) $v_{\mathrm{t}}{I}_{\mathrm{q}}=\left(V_{\mathrm{q}}-\frac{V_{\mathrm{p}}}{a}\right)Y_{\mathrm{pq}}=(a{\mathrm{five}}_{\mathrm{q}}-v_{\mathrm{p}})·\frac{y_{\mathrm{pq}}}{a}$

Now, let u.s. consider an equivalent π-network mode "fifty" for the transformer as shown in Fig. 9.6.

Figure 9.6. Equivalent π network for variable plow ratio transformer.

For the π-network

(9.8) $I_{\mathrm{p}}=(V_{\mathrm{p}}-5_{\mathrm{q}})A+{\mathrm{Five}}_{\mathrm{p}}B$

(ix.9) $I_{\mathrm{q}}=(V_{\mathrm{q}}-V_{\mathrm{p}})A+V_{\mathrm{q}}C$

Let V _p=0 and V _q=1 in Eqs. (9.four) and (9.8).

Letting E _p=0 and E _q=1 in Eqs. (9.vii) and (9.9)

$\begin{array}{ll}{I}_{\mathrm{q}}\hfill & =Y_{\mathrm{pq}}\mathrm{and}{I}_{\mathrm{q}}=A+C\hfill \\ \hfill & =\frac{Y_{\mathrm{pq}}}{a}+C\hfill \end{array}$

hence

(9.11) $C=\left(\mathrm{ane}-\frac{1}{a}\right)Y_{\mathrm{pq}}$

Equating the currents in Eqs. (9.3) and (9.8) and substituting for A from equation

(9.12) $B=\frac{\mathrm{i}}{a}\left(\frac{1}{a}-\mathrm{ane}\right)Y_{pq}$

Thus we obtain the equivalent π-model in terms of admittance and off-nominal turns ratio as shown in Fig. 9.seven.

Figure nine.7. Equivalent π-network model.

nine.ii.2 Phase Shifting Transformers

A phase shifting transformer can be represented by its impedance or comprisal in series with an ideal autotransformer having a circuitous turns ratio as shown in Fig. ix.8.

Figure ix.8. Phase shifting transformer.

(9.thirteen) $\frac{V_{\mathrm{p}}}{V_{\mathrm{s}}}=a+jb$

Since there is no power loss in an ideal autotransformer

(9.14) $5_{\mathrm{p}}^{\ast }{i}_{\mathrm{pr}}={\mathrm{Five}}_{\mathrm{s}}^{\ast }{i}_{\mathrm{sq}}$

i.e.,

(9.fifteen) $\frac{i_{\mathrm{pr}}}{i_{\mathrm{sq}}}=\frac{V_{\mathrm{due\; south}}^{\ast }}{V_{\mathrm{p}}^{\ast }}=\frac{1}{a-jb}$

Likewise, i _sq=(5 _s–V _q) Y _pq and hence

(9.xvi) ${\mathrm{i}}_{\mathrm{pr}}=({\mathrm{V}}_{\mathrm{s}}-{\mathrm{V}}_{\mathrm{q}})\frac{{\mathrm{V}}_{\mathrm{pq}}}{\mathrm{a}-\mathrm{jb}}$

Substituting for 5 _south from Eq. (9.13)

(ix.17) $i_{\mathrm{pr}}=\left(\frac{V_{\mathrm{p}}}{a+jb}-{\mathrm{Five}}_{\mathrm{q}}\right)\frac{Y_{\mathrm{pq}}}{a-jb}$

(9.18) $[V_{\mathrm{p}}-V_{\mathrm{q}}(a+jb)]\frac{Y_{\mathrm{pq}}}{a^{\mathrm{two}}+b^2}$

similarly, we can prove that

(nine.nineteen) $i_{\mathrm{qs}}=(V_{\mathrm{q}}-V_{\mathrm{due\; south}})Y_{\mathrm{pq}}\mathrm{and}\mathrm{substituting}$

for Five _s again from Eq. (ix.13)

(nine.20) $i_{\mathrm{qs}}=[(a+jb)V_{\mathrm{q}}-V_{\mathrm{p}}]·\frac{Y_{\mathrm{pq}}}{a+jb}$

To evaluate the constants, nosotros shall substitute known purlieus conditions into relevant equations.

Let 5 _p=0; permit all other buses be brusk circuited. The phase shifting transformer lies between buses p and q. The total bus admittance

(9.21) $Y_{\mathrm{PP}}=i_{\mathrm{P1}}+i_{\mathrm{P2}}+\cdots +i_{\mathrm{Pr}}+\cdots +i_{\mathrm{Pn}}$

where n is the number of buses continued to bus p.

Note: $I_{\mathrm{p}}={\sum }_{k=1}^n{i}_{\mathrm{p}k}=V_{\mathrm{p}}{Y}_{\mathrm{pp}}$ and 5 _p=ane.0   p.u.

Hence

(nine.22) $\begin{array}{l}{i}_{\text{p1}}=Y_{\text{p1}}\hfill \\ i_{\text{p2}}=Y_{\text{p2}}\hfill \\ \overline{i_{\text{p}n}=Y_{\text{p}\mathrm{due\; north}}}\hfill \end{array}\}$

and

(ix.23) $i_{\mathrm{pr}}=\frac{Y_{\mathrm{pq}}}{a^{\mathrm{ii}}+b^2}$

from Eq. (9.17) with 5 _p=1.0 and since all other buses are short circuited V _q=0.

The current flowing out of bus p is –i _sq, the common admittance

(ix.24) $y_{\mathrm{qp}}=-i_{\mathrm{sq}}$

And so

(9.25) $y_{\mathrm{qp}}=-i_{\mathrm{sq}}=-(V_{\mathrm{s}}-5_{\mathrm{q}})Y_{\mathrm{pq}}$

Since

$V_{\mathrm{q}}=0$

we obtain

(9.26) $y_{\mathrm{qp}}=-Y_{\mathrm{pq}}{V}_{\mathrm{s}}$

Similarly letting V _q=i.0   p.u. and curt circuiting all other buses, the cocky-admittance at bus q is

(9.27) $Y_{\mathrm{qq}}=i_{\mathrm{q1}}+i_{\mathrm{q2}}+\cdots +i_{\mathrm{qs}}+\cdots +i_{\mathrm{qn}}$

i.e.,

(9.28) $Y_{\mathrm{qq}}=i_{\mathrm{q1}}+i_{\mathrm{q2}}+\cdots +i_{\mathrm{qp}}+\cdots +i_{\mathrm{qn}}$

The current flowing out of bus p to jitney q is given by

(9.29) $i_{\mathrm{pq}}=+i_{\mathrm{pr}}$

Therefore, the mutual comprisal

(9.30) $Y_{\mathrm{pq}}=V_{\mathrm{q}}{i}_{\mathrm{pr}}=i_{\mathrm{pr}}$

And then

(9.31) $Y_{\mathrm{pq}}=i_{\mathrm{pr}}=({\mathrm{Five}}_{\mathrm{southward}}-5_{\mathrm{q}})\frac{Y_{\mathrm{pq}}}{a-jb}=\frac{-Y_{\mathrm{pq}}}{a-jb}$

The circuitous terms ratio a+jb can be completed for a specified angular displacements and tap setting from

(9.32) $A+jb=A(\cos \theta +j\sin \theta )$

where

(nine.33) $\left|V_{\mathrm{p}}\right|=A\left|V_{\mathrm{s}}\right|$

Thus all the required parameters are determined.

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Power menstruum in electrical systems

Arturo S. Bretas , ... Breno E.B. Carvalho , in Cyber-Concrete Power Systems State Estimation, 2021

In-phase transformer

In general, the in-phase transformer modeling comprises a series impedance or admittance and an ideal autotransformer (no losses in the core) whose transformation ratio is given past one  : a _kl . Fig. three.4 illustrates this type of transformer connecting buses one thousand and l.

Fig. 3.four. Representation of an in-phase transformer.

Every bit can exist seen, p denotes a reference point for the transformation ratio. Thus the relationship between the magnitude of voltage at this point and autobus grand is given by a _kl , that is, 5 _p   = a _kl V _grand . Since in this case in that location is no angular lag betwixt k and p (θ _k   = θ _p ), the human relationship betwixt the circuitous voltages is given by

(3.12) $\frac{{\mathrm{East}}_p}{E_{\mathrm{yard}}}=\frac{V_p{\mathrm{east}}^{j{\mathrm{\theta }}_p}}{V_k{\mathrm{east}}^{j{\mathrm{\theta }}_k}}=a_{\mathit{kl}}$

From the platonic model, that is, without considering the losses in the transformer, the following relationship is valid:

(3.13) $\begin{array}{c}{\mathrm{Due\; east}}_k{I}_{\mathit{kl}}^{⁎}+E_p{I}_{\mathit{lk}}^{⁎}=0\\ {\mathrm{Eastward}}_k{I}_{\mathit{kl}}^{⁎}=-a_{\mathit{kl}}{E}_{\mathrm{one\; thousand}}{I}_{\mathit{lk}}^{⁎}=0\end{array}$

therefore $\frac{I_{\mathit{kl}}}{I_{\mathit{lk}}}=-\frac{\left|I_{\mathit{kl}}\right|}{\left|I_{\mathit{lk}}\right|}=-a_{\mathit{kl}}$ .

By performing the nodal analysis of the transformer model, shown in Fig. 3.4, in terms of the complex currents I _kl and I _lk , the following equations are obtained:

(3.14) $\begin{array}{c}{I}_{\mathit{kl}}=a_{\mathit{kl}}{I}_{\mathit{pl}}=a_{\mathit{kl}}\left(-I_{\mathit{lk}}\right)\\ I_{\mathit{kl}}=-a_{\mathit{kl}}{y}_{\mathit{kl}}\left(E_l-{\mathrm{Due\; east}}_p\right)=-a_{\mathit{kl}}{y}_{\mathit{kl}}{E}_l+a_{\mathit{kl}}{y}_{\mathit{kl}}{E}_p\end{array}$

as Due east _p   = E _chiliad a _kl , we have that:

(3.xv) $I_{\mathit{kl}}=-a_{\mathrm{k}\mathrm{fifty}}{y}_{\mathit{kl}}{E}_l+a_{\mathit{kl}}^2{y}_{\mathit{kl}}{\mathrm{Due\; east}}_k$

and

(3.sixteen) $\begin{array}{c}{I}_{\mathit{lk}}=y_{\mathit{kl}}\left({\mathrm{Eastward}}_l-{\mathrm{East}}_p\right)=y_{\mathit{kl}}{\mathrm{Eastward}}_l-y_{\mathit{kl}}{E}_p\\ I_{\mathit{lk}}=y_{\mathit{kl}}{\mathrm{Eastward}}_{\mathrm{fifty}}-a_{\mathit{kl}}{y}_{\mathit{kl}}{E}_k\end{array}$

Based on the relationships of voltages and currents, we must follow the equation of the circuitous power menstruum from double-decker g to motorbus l:

(iii.17) $\begin{array}{c}{S}_{\mathit{kl}}^{⁎}=P_{\mathit{kl}}-j{Q}_{\mathit{kl}}=E_{\mathrm{1000}}^{⁎}{I}_{\mathit{kl}}\\ S_{\mathit{kl}}^{⁎}={\mathrm{Due\; east}}_{\mathrm{one\; thousand}}^{⁎}\left[-a_{\mathit{kl}}{y}_{\mathit{kl}}{\mathrm{Eastward}}_l+a_{\mathit{kl}}^2{y}_{\mathit{kl}}{\mathrm{Eastward}}_k\right]\\ S_{\mathit{kl}}^{⁎}=V_{\mathrm{chiliad}}{\mathrm{east}}^{-j{\theta }_k}\left[-a_{\mathit{kl}}{y}_{\mathit{kl}}{V}_l{\mathrm{east}}^{j{\theta }_l}+a_{\mathit{kl}}^2{y}_{\mathit{kl}}{V}_k{\mathrm{eastward}}^{j{\theta }_m}\right]\\ S_{\mathit{kl}}^{⁎}=y_{\mathit{kl}}{a}_{\mathit{kl}}{\mathrm{Five}}_k{e}^{-j{\theta }_k}\left[-V_{\mathrm{50}}{e}^{j{\theta }_l}+a_{\mathit{kl}}{5}_k{\mathrm{east}}^{j{\theta }_g}\right]\\ S_{\mathit{kl}}^{⁎}=-y_{\mathit{kl}}{a}_{\mathit{kl}}{V}_k{V}_l{e}^{-j\left({\theta }_{\mathrm{grand}}-{\theta }_l\right)}+y_{\mathit{kl}}{a}_{\mathit{kl}}^{\mathrm{ii}}{V}_k^2\end{array}$

Considering θ _kl   = θ _k   − θ _fifty , e ^{−  jθ _kl}   =   cos θ _kl   − j  sin θ _kl and y _kl   = g _kl   + jb _kl , we take

(3.18) $S_{\mathit{kl}}^{⁎}=-\left(g_{\mathit{kl}}+j{b}_{\mathit{kl}}\right)a_{\mathit{kl}}{V}_k{V}_l\left(cos{\mathrm{\theta }}_{\mathit{kl}}-\mathit{jsin}{\mathrm{\theta }}_{\mathit{kl}}\right)+\left(g_{\mathit{kl}}+j{b}_{\mathit{kl}}\right)a_{\mathit{kl}}^{\mathrm{two}}{5}_{\mathrm{1000}}^2$

Separating the existent and imaginary part of Eq. (iii.18), we obtain the equations for the active and reactive power flows:

(iii.19) $\begin{array}{c}{P}_{\mathit{kl}}=g_{\mathit{kl}}{a}_{\mathit{kl}}^2{V}_k^2-a_{\mathit{kl}}{V}_k{V}_l{k}_{\mathit{kl}}cos{\theta }_{\mathit{kl}}-a_{\mathit{kl}}{V}_{\mathrm{thousand}}{V}_{\mathrm{50}}{b}_{\mathit{kl}}\mathit{sin}{\theta }_{\mathit{kl}}\\ Q_{\mathit{kl}}=-b_{\mathit{kl}}{a}_{\mathit{kl}}^{\mathrm{two}}{V}_k^{\mathrm{ii}}+a_{\mathit{kl}}{V}_{\mathrm{1000}}{V}_l{b}_{\mathit{kl}}cos{\theta }_{\mathit{kl}}-a_{\mathit{kl}}{V}_k{V}_{\mathrm{50}}{\mathrm{thou}}_{\mathit{kl}}\mathit{sin}{\theta }_{\mathit{kl}}\end{array}$

Following the aforementioned procedure, we take the equation of the ability catamenia from bus l to passenger vehicle m:

(3.twenty) $\begin{array}{c}{\mathrm{South}}_{\mathit{lk}}^{⁎}=P_{\mathit{lk}}-j{Q}_{\mathit{lk}}=E_{\mathrm{fifty}}^{⁎}{I}_{\mathit{lk}}\\ {\mathrm{Due\; south}}_{\mathit{lk}}^{⁎}=E_l^{⁎}\left[y_{\mathit{kl}}{\mathrm{Eastward}}_{\mathrm{50}}-a_{\mathit{kl}}{y}_{\mathit{kl}}{E}_{\mathrm{thousand}}\right]\\ {\mathrm{Due\; south}}_{\mathit{lk}}^{⁎}=V_l{\mathrm{east}}^{-j{\theta }_{\mathrm{50}}}\left[y_{\mathit{kl}}{V}_l{e}^{j{\theta }_{\mathrm{fifty}}}-a_{\mathit{kl}}{y}_{\mathit{kl}}{5}_k{e}^{j{\theta }_{\mathrm{chiliad}}}\right]\\ S_{\mathit{lk}}^{⁎}=y_{\mathit{kl}}{V}_l^{\mathrm{two}}-a_{\mathit{kl}}{y}_{\mathit{kl}}{V}_{\mathrm{yard}}{\mathrm{Five}}_l{e}^{-j\left({\theta }_l-{\theta }_{\mathrm{chiliad}}\right)}\end{array}$

If θ _lk   = θ _l   − θ _grand , e ^{−  jθ _lk}   =   cos θ _lk   − j  sin θ _lk and y _kl   = one thousand _kl   + jb _kl , we accept:

(three.21) ${\mathrm{Due\; south}}_{\mathit{lk}}^{⁎}=\left(k_{\mathit{kl}}+j{b}_{\mathit{kl}}\right)V_l^2-a_{\mathit{kl}}\left(g_{\mathit{kl}}+j{b}_{\mathit{kl}}\right){\mathrm{Five}}_k{V}_{\mathrm{fifty}}\left(cos{\mathrm{\theta }}_{\mathit{lk}}-\mathit{jsin}{\mathrm{\theta }}_{\mathit{lk}}\right)$

Separating the real and imaginary part of Eq. (3.21), we obtain the equations for the agile and reactive flows:

(iii.22) $\begin{array}{c}{P}_{\mathit{lk}}={\mathrm{grand}}_{\mathit{kl}}{\mathrm{Five}}_{\mathrm{50}}^2-a_{\mathit{kl}}{V}_k{\mathrm{Five}}_{\mathrm{fifty}}{g}_{\mathit{kl}}cos{\theta }_{\mathit{lk}}-a_{\mathit{kl}}{5}_k{V}_{\mathrm{50}}{b}_{\mathit{kl}}\mathit{sin}{\theta }_{\mathit{lk}}\\ Q_{\mathit{lk}}=-b_{\mathit{kl}}{5}_{\mathrm{50}}^2+a_{\mathit{kl}}{V}_{\mathrm{thou}}{V}_l{b}_{k\mathrm{l}}cos{\theta }_{\mathit{lk}}-a_{\mathit{kl}}{V}_k{V}_l{g}_{\mathit{kl}}\mathit{sin}{\theta }_{\mathit{lk}}\end{array}$

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