Question

Rework Example 3.5; using ${\mathbf{S}}_{\mathbf{base}\mathbf{3}\mathbf{\varphi }}\mathbf{=}\mathbf{100}\mathbf{}\mathbf{kVA}$and${\mathbf{V}}_{\mathbf{base}\mathbf{,}\mathbf{LL}}\mathbf{=}\mathbf{600}\mathbf{}\mathbf{V}$.

Short answer

(a) The per-unit current is $0.268\angle -73.78°\mathrm{pu}$.

(b) The actual current is $25.78\angle -73.78°\mathrm{A}$.

Step-by-step solution

Given data

The base power is $S_{base}=100\mathrm{kVA}$, and the base voltage is $V_{base}=600\mathrm{V}$.

Determine the formulas of impedance, current and voltages.

Write the formula of transformation of delta load into star load.

${\mathbf{Z}}_{\mathbf{Y}}\mathbf{=}\frac{{\mathbf{Z}}_{\mathbf{∆}}}{\mathbf{3}}$ ……. (1)

Here,$Z_{∆}$is delta load, and$Z_Y$ is star load.

Write the formula of the base impedance.

${\mathbf{Z}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{base}}^{\mathbf{2}}}{{\mathbf{S}}_{\mathbf{base}}}$ ……. (2)

Here,$Z_{base}$is base impedance; $S_{base}$is base power;$V_{base}$ is base voltage.

Write the formula of per-unit load impedance.

${\mathbf{Z}}_{\mathbf{pu}\left(\mathrm{load}\right)}\mathbf{=}\frac{{\mathbf{Z}}_{\mathbf{Y}}}{{\mathbf{Z}}_{\mathbf{base}}}$ ……. (3)

Write the formula of per-unit line impedance.

${\mathbf{Z}}_{\mathbf{pu}\left(\mathrm{line}\right)}\mathbf{=}\frac{{\mathbf{Z}}_{\mathbf{line}}}{{\mathbf{Z}}_{\mathbf{base}}}$ ……. (4)

Here,$Z_{line}$is line impedance.

Write the formula of base voltage.

${\mathbf{V}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{base}\left(\mathrm{line}\right)}}{\sqrt{\mathbf{3}}}$ ……. (5)

Here,$V_{base}$is the base voltage.

Write the formula of phase voltage.

${\mathbf{E}}_{\mathbf{an}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{ab}}}{\sqrt{\mathbf{3}}}$ ……. (6)

Here,$E_{an}$is phase voltage.

Write the formula of per-unit voltage.

${\mathbf{E}}_{\mathbf{pu}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{an}}}{{\mathbf{V}}_{\mathbf{base}}}$ ……. (7)

Write the formula of per-unit current.

role="math" localid="1655284260611" ${\mathbf{I}}_{\mathbf{pu}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{pu}}}{{\mathbf{Z}}_{\mathbf{pu}\left(\mathrm{line}\right)}\mathbf{+}{\mathbf{Z}}_{\mathbf{pu}\left(\mathrm{load}\right)}}$ ……. (8)

Write the formula of the base current.

${\mathbf{I}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{base}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{base}}}$ ……. (9)

Write the formula of the actual current.

${\mathbf{I}}_{\mathbf{actual}}\mathbf{=}{\mathbf{I}}_{\mathbf{pu}}{\mathbf{I}}_{\mathbf{base}}$ ……. (10)

Determine the per-unit current

(a)

Transform the delta load impedance into star impedance.

Substitute $30\angle 40°\mathrm{\Omega }\mathrm{for}{Z}_{∆}$in equation (1).

$$\begin{gathered} Z_Y=\frac{30\angle 40°\mathrm{\Omega }}{3} \\ =10\angle 40°\mathrm{\Omega } \end{gathered}$$

Determine the base impedance.

Substitute $600\hspace{0.17em}\mathrm{V}\mathrm{for}{V}_{base}\mathrm{and}100\mathrm{kVA}\mathrm{for}{S}_{base}$in equation (2).

$$\begin{gathered} Z_{base}=\frac{{\left(600\mathrm{V}\right)}^2}{100\mathrm{kVA}} \\ =3.6\mathrm{\Omega } \end{gathered}$$

Determine the per-unit load impedance.

Substitute $3.6\mathrm{\Omega }\mathrm{for}{Z}_{base}\mathrm{and}10\angle 40°\mathrm{\Omega }\mathrm{for}{Z}_Y$in equation (3).

$$\begin{gathered} Z_{pu\left(load\right)}=\frac{10\angle 40°\mathrm{\Omega }}{3.6\mathrm{\Omega }} \\ =2.78\angle 40°\mathrm{pu} \end{gathered}$$

Determine the per-unit line impedance.

Substitute $3.6\mathrm{\Omega }\mathrm{for}{Z}_{base}\mathrm{and}1\angle 85°\mathrm{\Omega }\mathrm{for}{Z}_{line}$in equation (4).

$$\begin{gathered} Z_{pu\left(line\right)}=\frac{1\angle 85°\mathrm{\Omega }}{3.6\mathrm{\Omega }} \\ =0.278\angle 85°\mathrm{pu} \end{gathered}$$

Determine the base voltage.

Substitute $600\mathrm{V}\mathrm{for}{V}_{base\left(line\right)}$in equation (5).

$$\begin{gathered} V_{base}=\frac{600\mathrm{V}}{\sqrt{3}} \\ =346.41\mathrm{V} \end{gathered}$$

Determine the phase voltage of the supply.

Substitute $480\angle -30°\mathrm{V}\mathrm{for}{E}_{ab}$in equation (6).

$$\begin{gathered} E_{an}=\frac{480\angle -30°\mathrm{V}}{\sqrt{3}} \\ =277.13\angle -30°\mathrm{V} \end{gathered}$$

Determine the per-unit voltage.

Substitute $277.13\angle -30°\mathrm{V}\mathrm{for}{E}_{an}\mathrm{and}346.41\mathrm{V}\mathrm{for}{V}_{base}$in equation (7).

$$\begin{gathered} E_{pu}=\frac{277.13\angle -30°\mathrm{V}}{346.41\mathrm{V}} \\ =0.8\angle -30°\mathrm{pu} \end{gathered}$$

Determine the per-unit current.

Substitute $0.8\angle -30°\mathrm{pu}\mathrm{for}{E}_{pu},0.278\angle 85°\mathrm{pu}\mathrm{for}{Z}_{pu\left(line\right)}\mathrm{and}2.78\angle 40°\mathrm{for}{Z}_{pu\left(load\right)}$in equation (8).

$$\begin{gathered} I_{pu}=\frac{0.8\angle -30°\mathrm{pu}}{0.278\angle 85°\mathrm{pu}+2.78\angle 40°\mathrm{pu}} \\ =0.268\angle -73.78°\mathrm{pu} \end{gathered}$$

Therefore, the per-unit current is$0.268\angle -73.78°\mathrm{pu}$ .

Determine the actual load current

(b)

Determine the base current.

Substitute $100\mathrm{kVA}\mathrm{for}{S}_{base}\mathrm{and}600\mathrm{V}\mathrm{for}{V}_{base}$in equation (9).

$$\begin{gathered} I_{base}=\frac{100\mathrm{kVA}}{600\mathrm{V}\times \sqrt{3}} \\ =96.22\mathrm{A} \end{gathered}$$

Determine the actual current.

Substitute$96.22\mathrm{A}\mathrm{for}{I}_{base}\mathrm{and}0.268\angle -73.78°\mathrm{pu}\mathrm{for}{I}_{pu}$ in equation (10).

$$\begin{gathered} I_{actual}=\left(96.22\mathrm{A}\right)\left(0.268\angle -73.78°\mathrm{pu}\right) \\ =25.78\angle -73.78°\mathrm{A} \end{gathered}$$

Therefore, the actual current is $25.78\angle -73.78°\mathrm{A}$.