Write a short note on copper saving in auto- transformer. Also derive expression.
1 Answer
  • The primary and secondary windings are electrically connected so that a part of the winding is common to the two sides, the transformer is known as an autotransformer.

  • Common winding (BC) : its voltage appears on both sides (or common for both sides) of the transformer.

  • Series winding (AC) : it is connected in series with the common winding.

  • An auto transformer transfers power through induction (transformed power) and through conduction (conducted power).

  • The power of common winding or series winding represents the power transferred by induction.
  • The current entering in load from source known as conduction current. се
  • The apparent power of an auto transformer is sum of an transformed apparent power (2-winding transformer power) and conducted apparent power.

Saving in conductor material (Saving of Copper)

weight of Copper = volume of copper x copper density

As copper density is constant

weight of Copper ∝ volume of copper

weight of Copper ∝ conductor cross section area x length of conductor

Conductor cross section area depends on current rating (I) and length of conductor depends on number of turns (N) weight of Copper ∝ NI (mmf)

Saving of Copper in Auto Transformer

Two winding transformer

  • Weight of Cu required $\alpha\left(I_{1} N_{1}+I_{2} N_{2}\right)$ Autotransformer
  • Weight of Cu required in section 1-2 $\alpha I_{1}\left(N_{1}-N_{2}\right)$
  • Weight of Cu required in section 2-3 $\alpha\left(I_{2}-I_{1}\right) N 2$
  • Total weight of $\mathrm{Cu}$ required $\alpha \mathrm{I}_{1}\left(\mathrm{~N}_{1}-\mathrm{N}_{2}\right)+\left(\mathrm{I}_{2}-\mathrm{I}_{1}\right) \mathrm{N}_{2}$ $\frac{\text { Weight of } \mathrm{Cu} \text { in autotransformer }}{\text { Weight of } \mathrm{Cu} \text { in ordinary transformer }}=\frac{\mathrm{I}_{1}\left(\mathrm{~N}_{1}-\mathrm{N}_{2}\right)+\left(\mathrm{I}_{2}-\mathrm{I}_{1}\right) \mathrm{N}_{2}}{\mathrm{I}_{1} \mathrm{~N}_{1}+\mathrm{I}_{2} \mathrm{~N}_{2}}$

$\frac{\text { Weight of } \mathrm{Cu} \text { in autotransformer }}{\text { Weight of } \mathrm{Cu} \text { in ordinary transformer }}=\frac{\mathrm{I}_{1}\left(\mathrm{~N}_{1}-\mathrm{N}_{2}\right)+\left(\mathrm{I}_{2}-\mathrm{I}_{\mathrm{f}}\right) \mathrm{N}_{2}}{\mathrm{I}_{1} \mathrm{~N}_{1}+\mathrm{I}_{2} \mathrm{~N}_{2}}$ $$ \begin{aligned} &=\frac{N_{1} I_{1}-N_{2} I_{1}+\widetilde{N_{2} I_{2}-N_{2} I_{1}}}{N_{1} I_{1}+N_{2} I_{2}}=\frac{N_{1} I_{1}+N_{2} I_{2}-2 N_{2} I_{1}}{N_{1} I_{1}+N_{2} I_{2}} \\ &=1-\frac{2 N_{2} I_{1}}{N_{1} I_{1}+N_{2} I_{2}}=1-\frac{2 N_{2} I_{1}}{2 N_{1} I_{1}} \quad\left(\because N_{2} I_{2}=N_{1} I_{1}\right) \\ &=1-\frac{N_{2}}{N_{1}}=1-K \end{aligned} $$ Weight of Cu.in auto $t f r\left(W_{a}\right)=(1-K) \times$ weight of Cu.in ordinary $t f r\left(W_{o}\right)$

$$ \mathrm{W}_{\mathrm{a}}=(1-\mathrm{K}) \times \mathrm{W}_{\mathrm{o}} $$ $\therefore$ Saving in $\mathrm{Cu}=\mathrm{W}_{\mathrm{o}}-\mathrm{W}_{\mathrm{a}}=\mathrm{W}_{\mathrm{o}}-(1-\mathrm{K}) \mathrm{W}_{\mathrm{o}}=\mathrm{K} \mathrm{W}_{\mathrm{o}}$ Saving in $\mathrm{Cu}=\mathrm{K} \times \mathrm{Wt}$. ofCu in ordinary transformer - Thus if $\mathrm{K}=0.1$, the saving of $\mathrm{Cu}$ is only $10 \%$ but if $\mathrm{K}=0.9$, saving of $\mathrm{Cu}$ is $90 \%$. - Therefore, saving of $\mathrm{Cu}$ is more when $\mathrm{K}$ is nearer to $\mathrm{I}$.

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