Condition for the maximum efficiency

\[\eta = \dfrac{V_2I_2\cos\Phi_2}{V_2I_2\cos\Phi_2+P_i+I_2^2R_T} = \dfrac{V_2\cos\Phi_2}{V_2\cos\Phi_2+P_i/I_2+I_2R_T}\] Now, \(V_2\) is constant. Thus for a given \(\cos\Phi_2\), \(\eta\) depends upon \(I_2\). Hence, efficiency will be maximum when \[\begin{aligned} \dfrac{d}{dI_{2}} & =\left(V_{2}cos\Phi_{2}+\dfrac{P_{i}}{I_{2}}+I_{2}R_{T}\right)=0\\ & \Rightarrow0-\dfrac{P_{i}}{I_{2}^{2}}+R_{T}=0\\ & \Rightarrow P_{i}=I_{2}^{2}R_{T}\\ & \Rightarrow\mbox{Iron loss}=\mbox{copper loss} \end{aligned}\] Value of output current for maximum efficiency \[I_2 = \sqrt{\dfrac{P_i}{R_T}}\] If \(x\) is the fraction of full load KVA at which \(\eta_{max}\). Then \[\begin{aligned} \mbox{Iron loss} & =P_{i}\\ \mbox{Copper loss} & =x^{2}P_{c} \end{aligned}\] For maximum efficiency \(x^2P_c = P_i\) \[x = \sqrt{\dfrac{P_i}{P_c}}\] The output KVA corresponding to maximum efficiency \[\begin{aligned} \eta_{max} & =x\times KVA_{FL}\\ \Rightarrow\eta_{max} & =KVA_{FL}\times\sqrt{\dfrac{P_{i}}{P_{c}}} \end{aligned}\]

All-Day Efficiency

Power transformer (at generating station) and Distribution transformer (distribution substation)

Power transformers are switched in or out of the circuit depending upon the load to be handled

Distribution transformer is never switched off remain irrespective of the load

In such case, constant loss continues to be dissipated

Concept of energy based efficiency arise - "all day efficiency"

\[\begin{aligned} \%\mbox{all-day}\eta & =\dfrac{\mbox{Output energy in KWh during a day}}{\mbox{Input energy in KWh during a day}}\times100\\ & =\dfrac{\mbox{Output energy in KWh during a day}}{\mbox{Output energy +Energy spent for total losses}}\times100 \end{aligned}\]

Transformer with one winding only

Part of the winding is common to both primary and secondary

No electrical isolation as in 2-winding transformer

Theory and operation remains similar

Because of one winding, uses less copper and hence cheaper

It is used where transformation ratio varies little from unity

Saving of Cu

Volume (or weight) \(\propto\) length and area of cross-section

length of conductor \(\propto\) number of turns

cross-section depends on current

Weight \(\propto\) product of current and number of turnsWt. of Cu in section AC is \(\propto\) \((N_1-N_2)I_1\)

Wt. of Cu in section BC is \(\propto\) \(N_2(I_2-I_1)\)

\(\therefore\) Total Wt. of Cu \(\propto\) \((N_1-N_2)I_1+N_2(I_2-I_1)\)

If a 2-winding T/F perform the same task, then

Wt. of Cu in primary \(\propto\) \(N_1I_1\)

Wt. of Cu in secondary \(\propto\) \(N_2I_2\)

\(\therefore\) Total Wt. of Cu \(\propto\) \(N_1I_1+N_2I_2\) \[\therefore \dfrac{\mbox{Wt. of Cu in auto-transformer}}{\mbox{Wt. of Cu in ordinary transformer}} = \dfrac{(N_1-N_2)I_1+N_2(I_2-I_1)}{N_1I_1+N_2I_2} = (1-K)\]\(\Rightarrow\) \(W_a\) = \((1-K) \times W_0\)

equation* Saving=W_0-W_a=W_0-(1-K)W_0=KW_0

Hence, saving will increase as \(K\) approaches unity

\(\Rightarrow\) Power transferred inductively isequation* P_ind = Input (1-K)

\(\Rightarrow\) The rest of the power is conductively transferred

equation* P_cond = Input K

Advantages of Auto transformer

Less costly

Better regulation

Low losses as compared to 2-winding T/F of the same rating

Disadvantages of Auto transformer

The secondary winding is not insulated from primary so if a low supply voltage is used from a high voltage and if a break occurs in secondary winding then full primary voltage comes across the secondary terminal which is dangerous to the operator and the equipment.

used only in limited places where a slight variation of the output voltage from input voltage is required.

Applications of Auto transformer

Used as a starter to give 50-60% of full voltage to the stator of a squirrel cage induction motor during starting

used to give a small boost to a distribution cable, to correct the voltage drop

used as a voltage regulator

in power T & D system and also in audio system and railways