Demonstrative Video
Fundamentals & EMF Equation
Problem-1
A \(3000/200\) V, 50 Hz, single-phase transformer has a cross-sectional area of 150 cm2 for the core. If the number of turns on the low-voltage winding is 80, determine the number of turns on the high-voltage winding and maximum value of flux density in the core.
Given data: \[\begin{aligned} E_1 & = 3000~\mathrm{V} \qquad E_2 = 200~\mathrm{V} \qquad f = 50~\mathrm{Hz}\\ A & = 150~\mathrm{cm}^2=150 \times 10^{-4}~\mathrm{m}^2\qquad N_2 = 80 \end{aligned}\]
Number of turns on the high-voltage winding \[\begin{aligned} \dfrac{E_2}{E_2} & = \dfrac{N_2}{N_1} \\ \dfrac{200}{3000} & = \dfrac{80}{N_1} \\ N_1 & = 1200 \end{aligned}\]
Maximum value of flux density \[\begin{aligned} E_1 & = 4.44f\phi_mN_1 = 4.44fB_mAN_1\\ \Rightarrow~3000&=4.44\times 50\times B_m\times150\times10^{-4}\times1200\\ \Rightarrow~B_m & = 0.75~\mathrm{Wb/m^2} \end{aligned}\]
Problem-2
A single-phase \(50 \mathrm{~Hz}\) transformer has 80 turns on the primary winding and 280 turns in the secondary winding. The voltage applied across the primary winding is 240 V at 50 Hz. Calculate (i) maximum flux density in the core, and (ii) induced emf in the secondary. The net cross-sectional area of the core is \(200 \mathrm{~cm}^2\).
Given data: \[\begin{aligned} f & =50 \mathrm{~Hz} \qquad N_1 =80 \qquad N_2 =280 \\ V_1 & =240 \mathrm{~V} \qquad A =200 \mathrm{~cm}^2=200 \times 10^{-4} \mathrm{~m}^2 \end{aligned}\] (i) Maximum flux density in the core \[\begin{aligned} & V_1 \approx E_1=240 \mathrm{~V} \\ & E_1=4.44 f \phi_m N_1=4.44 f B_m A N_1 \\ \Rightarrow & 240=4.44 \times 50 \times B_m \times 200 \times 10^{-4} \times 80 \\ \Rightarrow & B_m=0.68 \mathrm{~Wb} / \mathrm{m}^2 \end{aligned}\]
(ii) Induced emf in the secondary \[\begin{aligned} \frac{E_2}{E_1} & =\frac{N_2}{N_1} \\ \Rightarrow \frac{E_2}{240} & =\frac{280}{80} \\ \Rightarrow E_2 & =840 \mathrm{~V} \end{aligned}\]
Problem-3
An \(80 ~\mathrm{kVA},~ 3200 / 400 ~\mathrm{~V},~ 50 \mathrm{~Hz}\) single-phase transformer has 111 turns on the secondary winding. Calculate (i) number of turns on primary winding, (ii) secondary current, and (iii) cross-sectional area of the core, if the maximum flux density is 1.2 T.
Given data: \[ \begin{aligned} \mathrm{kVA} \text { rating } & =80~ \mathrm{kVA} \\ E_1 & =3200 \mathrm{~V} \\ E_2 &=400 \mathrm{~V} \\ f &=50 \mathrm{~Hz} \\ N_2 &=111 \\ B_m &=1.2 \mathrm{~T} \end{aligned} \] (i) Number of turns of primary winding \[\begin{aligned} \frac{E_2}{E_1} & =\frac{N_2}{N_1} \\ \Rightarrow~\frac{400}{3200} & =\frac{111}{N_1} \\ \Rightarrow~N_1 & =888 \end{aligned}\]
(ii) Secondary current \[\begin{aligned} & V_2 \approx E_2=400 \mathrm{~V} \\ & I_2=\frac{\text { kVA rating } \times 1000}{V_2}=\frac{80 \times 1000}{400}=200 \mathrm{~A} \end{aligned}\] (iii) Cross-sectional area of the core \[\begin{aligned} E_2 & =4.44 f \phi_m N_2=4.44 f B_m A N_2 \\ \Rightarrow~400 & =4.44 \times 50 \times 1.2 \times A \times 111 \\ \Rightarrow~A & =0.0135 \mathrm{~m}^2=135 \mathrm{~cm}^2 \end{aligned}\]
Core Losses & No-load operation
Problem-4
A \(50 ~\mathrm{kVA}, 2300 / 230~ \mathrm{V}, 50 \mathrm{~Hz}\) transformer takes \(200 ~\mathrm{W}\) and \(0.3~ \mathrm{A}\) at no load, when \(2300 ~\mathrm{V}\) are applied to the high-voltage side. The primary resistance is \(3.5~ \Omega\). Determine: core loss, and no-load pf.
\[ \begin{aligned} W_i & =200 \mathrm{~W} \\ I_0 & =0.3 \mathrm{~A} \\ V_1 & =2300 \mathrm{~V} \\ R_1 & =3.5 \Omega \end{aligned} \] (i) Core loss \[\begin{aligned} \text{Copper loss in primary}~ &=I_0^2 R_1=(0.3)^2 \times 3.5=0.315 \mathrm{~W}\\ \text{Core loss}~&=\text { Input power }- \text { Copper loss }\\ &=200-0.315=199.685 \mathrm{~W} \end{aligned}\] (ii) No load pf \[\begin{aligned} W_i & =V_1 I_0 \cos \phi_0 \\ \Rightarrow~200 & =2300 \times 0.3 \times \cos \phi_0 \\ \Rightarrow~\cos \phi_0 & =0.29(\text { lagging) } \end{aligned}\]
Problem-5
A 230/110 V, single-phase transformer takes an input of 350 VA at no load and at rated voltage. The core loss is \(110 \mathrm{~W}\). Find (i) no-load power factor; (ii) the iron loss component of no-load current, and (iii) magnetizing component of no-load current.
Given data : \[\begin{aligned} S & =350 \mathrm{VA} \qquad W_i =110 \mathrm{~W} \qquad E_1 =230 \mathrm{~V} \end{aligned}\] (i) No-load power factor \[\begin{aligned} S & =V_1 I_0=E_1 I_0 \\ \Rightarrow~350 & =230 \times I_0 \\ \Rightarrow~I_0 & =1.52 \mathrm{~A} \\ W_i & =V_1 I_0 \cos \phi_0 \\ \Rightarrow~110 & =350 \cos \phi_0 \\ \Rightarrow~\cos \phi_0 & =0.314 \end{aligned}\]
(ii) Iron loss component of no-load current \[I_\omega=I_0 \cos \phi_0=1.52 \times 0.314=0.48 \mathrm{~A}\] (iii) Magnetizing component of no-load current \[I_\mu=I_0 \sin \phi_0=1.52 \times \sin \left(\cos ^{-1} 0.314\right)=1.44 \mathrm{~A}\]
Transformer on Load, Copper Loss & Equivalent Parameters
Problem-6
A \(6600 / 400 \mathrm{~V}\) transformer has a primary resistance of \(2.5 \Omega\) and a reactance of \(3.9 \Omega\). The secondary resistance is \(0.01 \Omega\) and the reactance is \(0.025 \Omega\). Determine the equivalent circuit parameters referred to primary and secondary.
Given data: \[ \begin{aligned} E_1 & =6600 \mathrm{~V} \\ E_2 & =400 \mathrm{~V} \\ R_1 & =2.5 \Omega \\ X_1 & =3.9 \Omega \\ R_2 & =0.01 \Omega \\ X_2 & =0.025 \Omega \end{aligned} \] Transformation ratio: \[K =\dfrac{E_2}{E_1}=\frac{400}{6600}=0.06\]
(i) Equivalent resistance referred to primary \[R_{01}=R_1+\frac{R_2}{K^2}=2.5+\frac{0.01}{(0.06)^2}=5.28 \Omega\]
(ii) Equivalent reactance referred to primary \[X_{01}=X_1+\frac{X_2}{K^2}=3.9+\frac{0.025}{(0.06)^2}=10.84 \Omega\] (iii) Equivalent resistance referred to secondary \[R_{02}=K^2 R_{01}=(0.06)^2 \times 5.28=0.02 \Omega\] (iv) Equivalent reactance referred to secondary \[X_{02}=K^2 X_{01}=(0.06)^2 \times 10.84=0.04 \Omega\]
Problem-7
A \(30~ \mathrm{kVA}, 2400 / 120 \mathrm{~V}, 50 \mathrm{~Hz}\) transformer has high-voltage winding resistance of \(0.1 \Omega\) and leakage reactance of \(0.22 \Omega\). The low voltage winding resistance is \(0.035 \Omega\) and leakage reactance is \(0.012 \Omega\). Calculate equivalent resistance as referred to primary and secondary, equivalent reactance as referred to primary and secondary, equivalent impedance as referred to primary and secondary, copper loss at full load and at \(75 \%\) of full load.
Given data: \[ \begin{aligned} \mathrm{kVA} \text { rating } &=30 \mathrm{kVA} \\ E_1 &=2400 \mathrm{~V} \\ E_2 &=120 \mathrm{~V} \\ R_1 &=0.1 \Omega \\ X_1 &=0.22 \Omega \\ R_2 &=0.035 \Omega\\ X_1 &= 0.012 \Omega \end{aligned} \] Transformation ratio: \[K =\dfrac{E_2}{E_1}=\frac{120}{2400}=0.05\]
(i) Equivalent resistance as referred to primary and secondary \[\begin{aligned} & R_{01}=R_1+\frac{R_2}{K^2}=0.1+\frac{0.035}{(0.05)^2}=14.1 \Omega \\ & R_{02}=K^2 R_{01}=(0.05)^2 \times 14.1=0.035 \Omega \end{aligned}\] (ii) Equivalent reactance as referred to primary and secondary \[\begin{aligned} & X_{01}=X_1+\frac{X_2}{K^2}=0.22+\frac{0.012}{(0.05)^2}=5.02 \Omega \\ & X_{02}=K^2 X_{01}=(0.05)^2 \times 5.02=0.013 \Omega \end{aligned}\] (iii) Equivalent impedance as referred to primary and secondary \[\begin{aligned} & Z_{01}=\sqrt{R_{01}^2+X_{01}^2}=\sqrt{(14.1)^2+(5.02)^2}=14.97 \Omega \\ & Z_{02}=K^2 Z_{01}=(0.05)^2 \times 14.97=0.037 \Omega \end{aligned}\]
(iv) Copper loss at full load and at \(75 \%\) of full load \[E_2 \simeq V_2=120 \mathrm{~V}\] Full load secondary current
\[\begin{aligned} I_2 & =\dfrac{\mathrm{kVA} \text { rating } \times 1000}{V_2}\\ &=\dfrac{30 \times 1000}{120}=250 \mathrm{~A} \end{aligned}\]
Copper loss
\[\begin{aligned} W_{\mathrm{Cu}} & =I_2^2 R_{02}=(250)^2 \times 0.035=2.18 \mathrm{~kW} \\ \text{At 75\% of F.L.} & =x^2 W_{\mathrm{Cu}}=(0.75)^2 \times 2.18=1.23 \mathrm{~kW} \end{aligned}\]
Voltage Regulation
Problem-8
A single-phase, 440/220 V, \(10 ~\mathrm{kVA}, 50 \mathrm{~Hz}\) transformer has a resistance of \(0.2 \Omega\) and reactance of \(0.6 \Omega\) on H.V. side. The corresponding values of L.V. side are \(0.04 ~\Omega\) and \(0.14~ \Omega\). Calculate the percentage regulation on full load for (i) 0.8 lagging pf (ii) 0.8 leading pf, (iii) unity pf.
Given data: \[\begin{aligned} \mathrm{kVA} \text { rating } & =10 \mathrm{~kVA} \quad E_2 =220 \mathrm{~V} \quad E_1=440 \mathrm{~V} \\ R_2 & =0.04 \Omega \quad R_1=0.2 \Omega \quad X_2=0.14 \Omega \quad X_1=0.6 \Omega \\ \end{aligned}\]
\[\begin{aligned} V_2 & \approx E_2=220 \mathrm{~V}\\ I_2 & =\frac{\mathrm{kVA} \text { rating } \times 1000}{V_2}=\frac{10 \times 1000}{220}=45.45 \mathrm{~A} \\ K & =\frac{E_2}{E_1}=\frac{220}{440}=0.5 \end{aligned}\]
\[\begin{aligned} & R_{02}=R_2+K^2 R_1=0.04+(0.05)^2 \times 0.2=0.09 \Omega \\ & X_{02}=X_2+K^2 X_1=0.14+(0.5)^2 \times 0.6=0.29 \Omega \end{aligned}\] (i) Percentage regulation on full load for 0.8 lagging \(\mathrm{pf}\) \[\begin{aligned} \cos \phi & =0.8 \qquad \sin \phi =0.6 \\ \% \text { regulation } & =\frac{I_2\left(R_{02} \cos \phi+X_{02} \sin \phi\right)}{E_2} \times 100 \\ & =\frac{45.45(0.09 \times 0.8+0.29 \times 0.6)}{220} \times 100 \\ & =5.08 \% \end{aligned}\]
(ii) Percentage regulation on full load for 0.8 leading pf \[\begin{aligned} \% \text { regulation } & =\frac{I_2\left(R_{02} \cos \phi-X_{02} \sin \phi\right)}{E_2} \times 100 \\ & =\frac{45.45(0.09 \times 0.8-0.29 \times 0.6)}{220} \times 100 \\ & =-2.11 \% \end{aligned}\]
(iii) Percentage regulation on full load for unity pf \[\begin{aligned} \cos \phi & =1 \qquad \sin \phi =0 \\ \% \text { regulation } & =\frac{I_2\left(R_{02} \cos \phi \pm X_{02} \sin \phi\right)}{E_2} \times 100 \\ & =\frac{45.45(0.09 \times 1-0.29 \times 0)}{220} \times 100 \\ & =1.86 \% \end{aligned}\]
Efficiency
Problem-9
Iron loss of \(80 \mathrm{kVA}, 1000 / 250 \mathrm{~V}\), single-phase, \(50 \mathrm{~Hz}\) transformer is \(500 \mathrm{~W}\). The full-load copper loss is \(400 \mathrm{~W}\). Find (i) area of cross section of limb if working flux density is \(1 \mathrm{~T}\) and there are 1000 turns on the primary, (ii) efficiency at full load and pf 0.8 lagging, and (iii) efficiency at \(75 \%\) of full load and unity pf.
Given data: \[\begin{aligned} \text { Full load kVA } & =80 \mathrm{~kVA} \quad E_1 =1000 \mathrm{~V} \quad E_2 =250 \mathrm{~V} \quad f =50 \mathrm{~Hz} \\ W_i & =500 \mathrm{~W}=0.5 \mathrm{~kW} \quad W_{C u} =400 \mathrm{~W}=0.4 \mathrm{~kW} \\ B_m & =1 \mathrm{~T} \quad N_1 =1000 \end{aligned}\] (i) Area of cross section of limb \[\begin{aligned} E_1 & =4.44 f \phi_m N_1=4.44 f B_m A N_1 \\ \Rightarrow~1000 & =4.44 \times 50 \times 1 \times \mathrm{A} \times 1000 \\ \Rightarrow~A & =4.5 \times 10^{-3} \mathrm{~m}^2 \end{aligned}\]
(ii) Efficiency at full load and 0.8 lagging pf \[\begin{aligned} x & =1 \quad \mathrm{pf} =0.8 \\ \% \eta & =\frac{x \times \text { full-load } \mathrm{kVA} \times \mathrm{pf}}{x \times \text { full-load } \mathrm{kVA} \times \mathrm{pf}+W_i+x^2 W_{C u}} \times 100 \\ & =\frac{1 \times 80 \times 0.8}{1 \times 80 \times 0.8+0.5+(1)^2 \times 0.4} \times 100 \\ & =98.61 \% \end{aligned}\] (iii) Efficiency at \(75 \%\) of full load and unity pf \[\begin{aligned} x & =0.75 \quad \mathrm{pf} =1 \\ \% \eta & =\frac{x \times \text { full-load } \mathrm{kVA} \times \mathrm{pf}}{x \times \text { full-load } \mathrm{kVA} \times \mathrm{pf}+W_i+x^2 W_{C u}} \times 100 \\ & =\frac{0.75 \times 80 \times 1}{0.75 \times 80 \times 1+0.5+(0.75)^2 \times 0.4} \times 100 \\ & =98.81 \% \end{aligned}\]
Problem-10
A \(150 \mathrm{kVA}\), single-phase transformer has iron loss of \(1.4 \mathrm{~kW}\) and full-load copper loss of \(1.6 \mathrm{~kW}\). Determine
the KVA load for maximum efficiency and the maximum efficiency at 0.8 lagging pf
the efficiency at half full load and 0.8 lagging pf
Given Data: \[\begin{aligned} \text{Full load kVA} & =150 \mathrm{~kVA}\quad W_i =1.4 \mathrm{~kW} \quad W_{cu} =1.6 \mathrm{~kW} \end{aligned}\]
(i) Load kVA for maximum efficiency and the maximum efficiency \[\begin{aligned} \text { Load kVA } & =\text { Full load kVA } \times \sqrt{\frac{W_i}{W_{C u}}}\\ &=150 \times \sqrt{\frac{1.4}{1.6}}=140.31~ \mathrm{kVA} \end{aligned}\]
For maximum efficiency, \[\begin{aligned} W_i & =W_{C u}=1.4 \mathrm{~kW} \\ \mathrm{pf} & =0.8 \\ \% \eta_{\max } & =\frac{\text { load } \mathrm{kVA} \times \mathrm{pf}}{\text { load } \mathrm{kVA} \times \mathrm{pf}+W_i+W_i} \times 100 \\ & =\frac{140.31 \times 0.8}{140.31 \times 0.8+1.4+1.4} \times 100 \\ & =97.57 \% \end{aligned}\] (ii) Efficiency at half full load and \(0.8 \mathrm{pf}\) \[\begin{aligned} x & =0.5 \qquad \mathrm{pf} =0.8 \\ \% \eta & =\frac{x \times \text { full-load kVA } \times \mathrm{pf}}{x \times \text { full-load kVA } \times \mathrm{pf}+W_i+x^2 W_{C u}} \times 100 \\ & =\frac{0.5 \times 150 \times 0.8}{0.5 \times 150 \times 0.8+1.4+(0.5)^2 \times 1.6} \times 100 \\ & =97.08 \% . \end{aligned}\]
OC and SC Testing
Problem-11
A \(5 ~\mathrm{kVA}\), \(200 / 400 \mathrm{~V}\), \(50 \mathrm{~Hz}\), single-phase transformer gives the following test results:
\[\begin{aligned} \text{OC test (LV side)} & ~~200 \mathrm{~V} ~~ 0.7 \mathrm{~A} ~~ 60 \mathrm{~W} \\ \text{SC test (HV side)} & ~~22 \mathrm{~V} ~~ 16 \mathrm{~A} ~~ 120 \mathrm{~W} \end{aligned} \]
Draw the equivalent circuit of the transformer and insert all parameter values.
Find efficiency and regulation at \(0.9 \mathrm{pf}\) (lead) if operating at rated load.
Find current at which efficiency is maximum.
(i) Equivalent circuit of the transformer
From OC test (meters are connected on LV side, i.e., primary), \[\begin{aligned}
& W_i=60 \mathrm{~W} \quad V_1=200 \mathrm{~V} \quad I_0=0.7
\mathrm{~A} \\
& \cos \phi_0=\frac{W_i}{V_1 I_0}=\frac{60}{200 \times 0.7}=0.43
\\
& \sin \phi_0=0.9 \\
& I_w=I_0 \cos \phi_0=0.7 \times 0.43=0.3 \mathrm{~A} \\
& R_0=\frac{V_1}{I_w}=\frac{200}{0.3}=666.67 \Omega \\
& I_\mu=I_0 \sin \phi_0=0.7 \times 0.9=0.63 \mathrm{~A} \\
& X_0=\frac{V_1}{I_\mu}=\frac{200}{0.63}=317.46 \Omega
\end{aligned}\]
From \(\mathrm{SC}\) test (meters are connected on HV side, i.e., secondary), \[\begin{aligned} & W_{\mathrm{sc}}=120 \mathrm{~W} \quad V_{\mathrm{sc}}=22 \mathrm{~V} \quad I_{\mathrm{sc}}=16 \mathrm{~A} \\ & Z_{02}=\frac{V_{\mathrm{sc}}}{I_{\mathrm{sc}}}=\frac{22}{16}=1.375 \Omega \\ & R_{02}=\frac{W_{\mathrm{sc}}}{I_{\mathrm{sc}}^2}=\frac{120}{(16)^2}=0.47 \Omega \\ & X_{02}=\sqrt{\left(Z_{02}\right)^2-\left(R_{02}\right)^2}=\sqrt{(1.375)^2-(0.47)^2}=1.29 \Omega \end{aligned}\]
\[\begin{aligned}
K & = \dfrac{400}{200}=2\\
R_{01} & = \dfrac{R_{02}}{K^2} = \dfrac{0.47}{2^2} =
0.12~\Omega \\
X_{01} & = \dfrac{X_{02}}{K^2} = \dfrac{1.29}{2^2} =
0.32~\Omega \\
\end{aligned}\]
(ii) Efficiency at rated load and \(0.9~ \mathrm{pf}\) leading \[W_i=60 \mathrm{~W}=0.06 \mathrm{~kW}\] Since meters are connected on secondary in SC test, \[\begin{aligned} I_2 & =\frac{5 \times 1000}{400}=12.5 \mathrm{~A} \\ W_{C u} & =I_2^2 R_{02}\\ &=(12.5)^2 \times 0.47=73.43 \mathrm{~W}=0.073 \mathrm{~kW} \\ x & =1 \qquad \mathrm{pf} =0.9 \\ \% \eta & =\frac{x \times \text { full load kVA } \times \mathrm{pf}}{x \times \text { full-load kVA } \times \mathrm{pf}+W_i+x^2 W_{C u}} \times 100 \\ & =\frac{1 \times 5 \times 0.9}{1 \times 5 \times 0.9+0.06+(1)^2 \times 0.073} \times 100 \\ & =97.13 \% \end{aligned}\]
Regulation at rated load and \(0.9 ~\mathrm{pf}\) lead \[\begin{aligned} \cos \phi & =0.9 \qquad \sin \phi =0.44 \\ \% \text { regulation } & =\frac{I_2\left(R_{02} \cos \phi-X_{02} \sin \phi\right)}{E_2} \times 100 \\ & =\frac{12.5(0.47 \times 0.9-1.29 \times 0.44)}{400} \times 100 \\ & =-0.45 \% \end{aligned}\] (iii) Current at maximum efficiency \[\begin{aligned} W_i & =I_2^2 R_{02} \\ I_2 & =\sqrt{\frac{W_i}{R_{02}}}=\sqrt{\frac{60}{0.47}}=11.3 \mathrm{~A} \end{aligned}\]