Question

An infinite bus, which is a constant voltage source, is connected to the primary of the three-winding transformer Of Problem 3.53. A $\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{-}\mathit{M}\mathit{V}\mathit{A}\mathbf{,}\mathbf{}\mathbf{13}\mathbf{.}\mathbf{2}\mathbf{}\mathit{k}\mathit{V}$synchronous motor with a sub transient reactance of 0.5 per unit is connected to the transformer secondary. A localid="1656746464858" $\mathbf{5}\mathit{M}\mathit{W}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{3}\mathit{k}\mathit{V}$, three- phase resistive load is connected to the tertiary. Choosing a base of localid="1656746481214" $\mathbf{66}\mathit{k}\mathit{V}$and localid="1656746490341" $\mathbf{15}\mathit{M}\mathit{V}\mathit{A}\mathbf{}$in the primary, draw the impedance diagram of the system showing per-unit Neglect transformer exciting current, phase shifts, and all resistances except the resistive load.

Problem 3.53

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

Primary: Y connectedlocalid="1656746552467" $\mathbf{66}\mathit{k}\mathit{V}\mathbf{,}\mathbf{15}\mathit{M}\mathit{V}\mathit{A}$,

Secondary: localid="1656746559217" $Y$connectedlocalid="1656746566136" $\mathbf{13}\mathbf{.}\mathbf{2}\mathit{k}\mathit{V}\mathbf{,}\mathbf{10}\mathit{M}\mathit{V}\mathit{A}$,

Tertiary: localid="1656746571916" $\Delta$connectedlocalid="1656746577941" $\mathbf{2}\mathbf{.}\mathbf{3}\mathit{k}\mathit{V}\mathbf{,}\mathbf{5}\mathit{M}\mathit{V}\mathit{A}$,

Neglecting resistances and exciting current, the leakage reactance’s are:

localid="1656746584806" ${\mathit{X}}_{\mathbf{P}\mathbf{S}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{09}$per unit on alocalid="1656746592283" $\mathbf{15}\mathit{M}\mathit{V}\mathit{A}\mathbf{,}\mathbf{66}\mathit{k}\mathit{V}$ base

localid="1656746599787" ${\mathit{X}}_{\mathbf{P}\mathbf{T}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{08}$per unit on alocalid="1656746606719" $\mathbf{15}\mathit{M}\mathit{V}\mathit{A}\mathbf{,}\mathbf{66}\mathit{k}\mathit{V}$base

localid="1656746614327" ${\mathit{X}}_{\mathbf{S}\mathbf{T}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$per unit on alocalid="1656746620958" $\mathbf{10}\mathit{M}\mathit{V}\mathit{A}\mathbf{,}\mathbf{13}\mathbf{.}\mathbf{2}\mathit{k}\mathit{V}$base

Determine the reactance’s of the per-phase equivalent circuit using a base oflocalid="1656746628095" $\mathbf{15}\mathit{M}\mathit{V}\mathit{A}\mathbf{,}$and localid="1656746635563" $\mathbf{66}\mathit{k}\mathit{V}$ for the primary.

Short answer

Therefore, the per unit impedance diagram is shown below,

Step-by-step solution

Determine the formulas to calculate the quantities of the per unit impedance diagram

The reactance can be converted from one base to another,

${\mathit{X}}_{\mathbf{S}\mathbf{T}}\mathbf{=}{\mathit{X}}_{\mathbf{S}\mathbf{T}\left(old\right)}\left(\frac{baseMVA}{ActualMVA}\right)$…..(1)

The primary per unit reactance can be calculated by the expression,

${\mathit{X}}_{\mathbf{P}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(X_{PS}+X_{PT}-X_{ST}\right)$…..(2)

The secondary per unit reactance can be calculated by the expression,

${\mathit{X}}_{\mathbf{S}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(X_{SP}+X_{ST}-X_{PT}\right)$…..(3)

The tertiary per unit reactance can be calculated by the expression,

${\mathit{X}}_{\mathbf{T}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(X_{TP}+X_{TS}-X_{PS}\right)$…..(4)

The base impedance of the tertiary can be calculated by expression,

${\mathit{z}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}^{\mathbf{2}}}{{\mathbf{S}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}}$…..(5)

The per phase resistance can be calculated by the expression,

${\mathit{R}}_{\mathbf{p}\mathbf{h}}\mathbf{=}\frac{\mathbf{3}{\mathbf{V}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}^{\mathbf{2}}}{\mathbf{P}}$…..(6)

The per phase per unit impedance can be calculated by expression,

${\mathit{R}}_{\mathbf{p}\mathbf{.}\mathbf{u}}\mathbf{=}\frac{{\mathbf{R}}_{\mathbf{p}\mathbf{h}}}{{\mathbf{Z}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}}$

The per unit of reactance on the new base is given by,

${\mathit{X}}_{\mathbf{n}\mathbf{e}\mathbf{w}}\mathbf{=}{\mathit{X}}_{\mathbf{o}\mathbf{l}\mathbf{d}}\left(\frac{V_{motor}}{V_{base}}\right)\left(\frac{S_{base}}{S_{motor}}\right)$……(8)

Calculate the per unit values of all the parameter of the impedance circuit

Calculate the reactance value on new base by using the equation (1),

$\begin{array}{l}{X}_{ST}=0.08\left(\frac{15}{10}\right)\\ X_{ST}=0.12\text{\hspace{0.17em}p.u}\end{array}$

The per unit primary reactance of the impedance of per phase equivalent circuit by using equation (2),

$\begin{array}{l}{X}_P=\frac{1}{2}\left(0.07+0.09-0.12\right)\\ X_P=\frac{1}{2}\times 0.04\\ X_P=0.02\text{\hspace{0.17em}p.u}\end{array}$

The per unit secondary reactance of the impedance of per phase equivalent circuitby using equation (3),

$\begin{array}{l}{X}_S=\frac{1}{2}\left(0.07+0.12-0.09\right)\\ X_S=\frac{1}{2}\times 0.1\\ X_S=0.05\text{\hspace{0.17em}p.u}\end{array}$

The per unit tertiary reactance of the impedance of per phase equivalent circuit by using equation (4),

$\begin{array}{l}{X}_T=\frac{1}{2}\left(0.09+0.12-0.07\right)\\ X_T=\frac{1}{2}\times 0.14\\ X_T=0.07\text{\hspace{0.17em}p.u}\end{array}$

Calculate the base impedance of tertiary by using the equation (5),

$\begin{array}{l}{Z}_{base}=\frac{{\left(2.3\times {10}^3\right)}^2}{15\times {10}^6}\\ Z_{base}=\frac{5.29\times {10}^6}{15\times {10}^6}\\ Z_{base}=0.352\Omega \end{array}$

Calculate the per phase resistance by using the equation (6),

$\begin{array}{l}R=\frac{3\times {\left(2.3\times {10}^3\right)}^2}{5\times {10}^6}\\ R=\frac{3\times 5.29\times {10}^6}{5\times {10}^6}\\ R=3.174\Omega \end{array}$

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

$\begin{array}{l}{R}_{p.u}=\frac{3.174}{0.353}\\ R_{p.u}=8.99\text{\hspace{0.17em}p.u}\end{array}$

Calculate the per unit reactance of the synchronous motor by using the equation (8),

$\begin{array}{l}{X}_{new}=0.2\left(\frac{15}{7.5}\right){\left(\frac{13.2}{13.2}\right)}^2\\ X_{new}=0.2\times 2\times 1\\ X_{new}=0.4\text{\hspace{0.17em}p.u}\end{array}$

The per unit diagram of the system is shown below.