Question

Three single-phase, two-winding transformers, each rated, $\mathbf{450}\mathbf{MVA}$,$\mathbf{20}\mathbf{kV}/\mathbf{288}\mathbf{.}\mathbf{7}\mathbf{kV}$ with leakage reactance ${\mathbf{X}}_{\mathrm{eq}}=\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{per}\mathbf{unit}$, are connected to form a three-phase bank. The high-voltage windings are connected in$\mathbf{Y}$ with a solidly grounded neutral. Draw the per-unit zero-, positive-, and negative-sequence networks if the low-voltage windings are connected (a) in$\mathbf{∆}$ with American standard phase shift or (b) inrole="math" localid="1656166741370" $\mathbf{Y}$ with an open neutral. Use the transformer ratings as base quantities. Winding resistances and exciting current are neglected.

Short answer

(a)

The value of per-unit zero sequence low voltage and high voltage winding are$V_{X0}$ and $V_{H0}$.

Draw the per-unit zero-sequence network for the three single phase, two-winding transformer with high-voltage windings connected in$Y$ and low-voltage windings connected in $∆$.

The value of per-unit positive sequence low voltage and high voltage winding is $1\angle -30°\text{\hspace{0.17em}per\hspace{0.17em}unit}$and $1\angle 0°\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

Draw the per-unit positive sequence network for the three single phase, two-winding transformer with high-voltage windings connected in $Y$and low-voltage windings connected in $∆$.

The value of per-unit negative sequence low voltage and high voltage winding is$1\angle -30°\text{\hspace{0.17em}per\hspace{0.17em}unit}$ and $1\angle 0°\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

Draw the per-unit negative sequence network for a normal transformer, the negative sequence network is same as the positive sequence network.

(b)

The value of per-unit zero sequence networks for three phase transformer of low voltage winding connected with an open neutral and high voltage star winding connected with solid neutral is$V_{X0}$ and$V_{H0}$ .

Draw the per-unit zero-sequence network for the three single phase, two-winding transformer with low-voltage windings are connected in$Y$ with an open neutral and low-voltage windings are connected in star with solid neutral.

The value of per-unit positive sequence voltage for normal transformer model and there is no phase shift between primary and secondary voltage is $1\angle 0°\text{\hspace{0.17em}V}$and $1\angle 0°\text{\hspace{0.17em}V}$.

Draw per-unit positive-sequence network for the three single phase, two-winding transformer with low-voltage windings are connected in$Y$ with an open neutral and low-voltage windings are connected in star with solid neutral.

The value of per-unit negative sequence voltage for star connected transformer model and there is no phase shift between primary and secondary voltage is$1\angle 0°\text{\hspace{0.17em}V}$ and $1\angle 0°\text{\hspace{0.17em}V}$.

Draw a per-unit negative sequence network for a normal transformer, the negative sequence network is same as the positive sequence network.

Step-by-step solution

Write the given data from the question.

Considerthree single-phase, two-winding transformer.

Each transformer rated at$\text{450MVA}$,$\text{20\hspace{0.17em}kV}/\text{288.7\hspace{0.17em}kV}$.

The leakage reactance is${\text{X}}_{\text{eq}}=\text{0.12\hspace{0.17em}per\hspace{0.17em}unit}$.

Base quantities of each transformer$S_{base1\varphi }=450\text{\hspace{0.17em}MVA}$ and $V_{baseHLN}=\text{288.7\hspace{0.17em}kV}$.

A base quantity of leakage reactance is $V_{baseXLN}=20\text{\hspace{0.17em}kV}$.

Now, Phase voltage can be given by.

$\left[\begin{array}{c}{V}_{ag}\\ V_{bg}\\ V_{cg}\end{array}\right]=\left[\begin{array}{c}288.7\angle 0°\\ 288.7\angle -120°\\ 288.7\angle 120°\end{array}\right]$

Formula of zero sequence voltage V0, positive sequence voltage V1  and negative sequence voltage V2.

Write the formula of positive sequence voltage.

${\mathbf{V}}_0=\frac{1}{3}({\mathbf{V}}_{\mathrm{ag}}\mathbf{+}{\mathbf{V}}_{\mathrm{bg}}\mathbf{+}{\mathbf{V}}_{\mathrm{cg}})$ …… (1)

Here, ${\mathbf{V}}_{\mathrm{ag}}$is phase voltage $a$with grounding, ${\mathbf{V}}_{\mathrm{bg}}$is phase voltage $b$with grounding and ${\mathbf{V}}_{\mathrm{cg}}$is phase voltage $c$with grounding.

${\mathbf{V}}_1=\frac{1}{3}({\mathbf{V}}_{\mathrm{ag}}\mathbf{+}{\mathbf{aV}}_{\mathrm{bg}}\mathbf{+}{\mathbf{a}}^2{\mathbf{V}}_{\mathrm{cg}})$ …… (2)

Here, ${\mathbf{V}}_{\mathrm{ag}}$is phase voltage role="math" localid="1656167848738" $a$with grounding, ${\mathbf{V}}_{\mathrm{bg}}$is phase voltage $b$with grounding and ${\mathbf{V}}_{\mathrm{cg}}$is phase voltage $c$with grounding $a$and $a^2$are conjugates.

${\mathbf{V}}_1=\frac{1}{3}({\mathbf{V}}_{\mathrm{ag}}\mathbf{+}{\mathbf{a}}^2{\mathbf{V}}_{\mathrm{bg}}\mathbf{+}{\mathbf{aV}}_{\mathrm{cg}})$ …… (3)

Here,${\mathbf{V}}_{\mathrm{ag}}$ is phase voltage$a$ with grounding,${\mathbf{V}}_{\mathrm{bg}}$ is phase voltage$b$ with grounding and${\mathbf{V}}_{\mathrm{cg}}$ is phase voltage$c$ with grounding,role="math" localid="1656168131229" $\mathit{a}$ and${\mathbf{a}}^{\mathbf{2}}$ are conjugates.

(a) Draw the per-unit zero-, positive-, and negative-sequence networks with high-voltage windings are connected in Y and low-voltage windings are connected in ∆ with American standard phase shift.

Determine the relation between zero sequences voltage $V_0$.

Substitute$288.7\angle 0°$ for $V_{ag}$,$288.7\angle -120°$ for$V_{bg}$ and$288.7\angle 120°$ for$V_{cg}$ into equation (1).

$\begin{array}{c}{V}_0=\frac{1}{3}(288.7\angle 0°+288.7\angle -120°+288.7\angle 120°)\\ =0\text{\hspace{0.17em}V}\end{array}$

Determine the relation between positive sequences voltage $V_1$.

Substitute$288.7\angle 0°$ for $V_{ag}$,$288.7\angle (-120°+120°)$ for$a{V}_{bg}$ and$288.7\angle (120°+240°)$ for$a^2{V}_{cg}$ into equation (2).

$\begin{array}{c}{V}_1=\frac{1}{3}(288.7\angle 0°+288.7\angle (-120°+120°)+288.7\angle (120°+240°))\\ =\frac{1}{3}(288.7\angle 0°+288.7\angle 0°+288.7\angle 0°)\\ =288.7\angle 0°\end{array}$

Determine the relation between negative sequences voltage $V_2$.

Substitute$288.7\angle 0°$ for $V_{ag}$, $288.7\angle (-120°+240°)$for$a^2{V}_{bg}$ and$288.7\angle (120°+120°)$ for$a{V}_{cg}$ into equation (3).

$\begin{array}{c}{V}_2=\frac{1}{3}(288.7\angle 0°+288.7\angle (-120°+240°)+288.7\angle (120°+120°))\\ =\frac{1}{3}(288.7\angle 0°+288.7\angle 120°+288.7\angle 240°)\\ =0\text{\hspace{0.17em}V}\end{array}$

Therefore, in secondary side there are no load connected, no transformer losses and current drawn in transformer will be zero.

Draw the per unit zero, positive and negative sequence network for the three single phase, two-winding transformer withhigh-voltage windings connected in$Y$ and low-voltage windings connected in $∆$.

Draw the per-unit zero-sequence network as shown in figure 1.

Figure 1

The HV side has the neutral point grounded to the Earth and it has the zero sequence current with the return path through the ground. The LV side is in delta connection and there is no ground connection for it, while the zero sequence current do not flow out of the LV winding and is open circuited.

Determine the positive sequence voltage network with primary side voltage will lead an angle of $30°$with secondary side voltage.

Also per unit low voltage side is

$V_{x1}\left(\text{per\hspace{0.17em}unit}\right)=\frac{V_{x1}}{V_{baseXLN}}$

Here, $V_{x1}$is low voltage reactance and $V_{baseXLN}$is line to neutral base reactance voltage.

$\begin{array}{c}{V}_{x1}\left(\text{per\hspace{0.17em}unit}\right)=\frac{20\angle -30°}{20\angle 0°}\\ =1\angle -30°\text{\hspace{0.17em}per\hspace{0.17em}unit}\end{array}$

Also per unit high voltage side is

$V_{H1}\left(\text{per\hspace{0.17em}unit}\right)=\frac{V_{H1}}{V_{baseHLN}}$

Here, $V_{H1}$is high voltage reactance and $V_{baseHLN}$is line to neutral base high voltage.

$\begin{array}{c}{V}_{H1}=\frac{288.7\angle 0°}{288.7\angle 0°}\\ =1\angle 0°\text{\hspace{0.17em}per\hspace{0.17em}unit}\end{array}$

Now, draw the per-unit positive sequence network as shown in figure 2.

Figure 2

For a normal transformer, the negative sequence network is same as the positive sequence network.

Draw the per-unit negative sequence network as shown in figure 3.

Figure 3

(b) Draw the per-unit zero-, positive-, and negative-sequence networks with low voltage side connected in Y with an open neutral.

Draw the per unit zero, positive and negative sequence network for the three single phase, two-winding transformer with low-voltage windings are connected in$Y$ with an open neutral and low-voltage windings are connected in star with solid neutral.

Draw the per-unit zero-sequence network as shown in figure 4.

Figure 4

Since, a zero sequence current can be flow in primary side of high voltage star winding with solid neutral.

The zero sequence current cannot be flow in secondary side of low voltage star winding with open neutral as shown in figure 4.

Now determine the positive sequence network, this network is same as normal transformer so there is no phase difference between primary and secondary current and voltage.

Draw per-unit positive-sequence network as shown in figure 5.

Figure 5

For all non-rotating machine positive sequence network is equal to negative sequence network. For a normal transformer, the negative sequence network is same as the positive sequence network.

Draw a per-unit negative sequence network as shown in figure 6.

Figure 6