Demonstrative Video

Efficiency of a transformer

\[\begin{aligned} \eta & =\dfrac{\mbox{output power}}{\mbox{input power}}=\dfrac{\mbox{output power}}{\mbox{output power+losses}}\\ & =\dfrac{\mbox{output power}}{\mbox{output power+iron losses+copper losses}}\\ & =\dfrac{V_{2}I_{2}\cos\Phi_{2}}{V_{2}I_{2}\cos\Phi_{2}+P_{i}+P_{c}} \end{aligned}\] where \[\begin{array}{cc} V_{2} & \mbox{Secondary terminal voltage}\\ I_{2} & \mbox{Full-load secondary current}\\ \cos\Phi_{2} & \mbox{Power factor of the load}\\ P_{i} & \mbox{Iron losses =hysterises +eddy current losses}\\ P_{c} & \mbox{Full load copper losses = }I^{2}R_{T} \end{array}\] If \(x\) is the fraction of the full-load, then efficiency is given as \[\eta_{x}=\dfrac{x\times\mbox{Output}}{x\times\mbox{Output}+P_{i}+x^{2}P_{c}}=\dfrac{xV_{2}I_{2}\cos\Phi_{2}}{xV_{2}I_{2}\cos\Phi_{2}+P_{i}+x^{2}I_{2}^{2}R_{T}}\]

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

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