Transformer Voltage Drop Mastery: Solved Problems

Demonstrative Video

Problem-1

A 30 KVA, 2400/120 V, 50-Hz transformer has a high voltage winding resistance of 0.1 \(\Omega\) and a leakage reactance of 0.22 \(\Omega\). The low voltage winding resistance is 0.035 \(\Omega\) and the leakage reactance is 0.012 \(\Omega\). Find the equivalent winding resistance, reactance and impedance referred to the

  1. high voltage side

  2. the low voltage side.

Solution-1

\[\begin{aligned} K &=120 / 2400=1 / 20 \\ R_{1}&=0.1 \Omega,~~ X_{1}=0.22 \Omega \\ R_{2} &=0.035 \Omega \text { and } X_{2}=0.012 \Omega \end{aligned}\] (i) High-voltage side : \[\begin{aligned} R_{01}=R_{1}+R_{2}^{\prime}=R_{1}+R_{2} / K^{2}=0.1+0.035 /(1 / 20)^{2}=14.1 \Omega \\ X_{01}=X_{1}+X_{2}^{\prime}=X_{1}+X_{2} / K^{2}=0.22+0.12 /(1 / 20)^{2}=5.02 \Omega \\ Z_{01}=\sqrt{R_{01}^{2}+X_{01}^{2}}=\sqrt{14.1^{2}+5.02^{2}}=15 \Omega \end{aligned}\]

(ii) Low Voltage Side: \[\begin{aligned} R_{02} &=R_{2}+R_{1}^{\prime}=R_{2}+K^{2} R_{1}=0.035+(1 / 20)^{2} \times 0.1=0.03525 \Omega \\ X_{02} &=X_{2}+X_{1}^{\prime}=X_{2}+K^{2} X_{1}=0.012+(1 / 20)^{2} \times 0.22=0.01255 \Omega \\ Z_{02} &=\sqrt{R_{02}^{2}+X_{02}^{2}}=\sqrt{0.0325^{2}+0.01255^{2}}=0.0374 \Omega \\ \text {or} ~Z_{02} &=K^{2} Z_{01}=(1 / 20)^{2} \times 15=0.0375 \Omega \end{aligned}\]

Problem-2

The following data refer to a 1-phase transformer:

Turn ratio \(19.5: 1,~ R_{1}=25 \Omega,~ X_{1}=100 \Omega,~ R_{2}=0.06 \Omega,~ X_{2}=0.25 \Omega\) No-load Current \(=1.25\) A leading the flux by \(30^{\circ}\). The secondary delivers \(200 \mathrm{~A}\) at a terminal voltage of \(500 \mathrm{~V}\) and p.f. of 0.8 lagging.

Determine by the aid if a vector diagram, the primary applied voltage, the primary pf and the efficiency.

Solution-2

\[\begin{aligned} \mathrm{V}_{2} &=500 \angle 0^{\circ}=500+j 0 \\ \mathrm{I}_{2} &=200(0.8-j 0.6)=160-j 120 \\ \mathrm{Z}_{2} &=(0.06+j 0.25) \\ \mathrm{E}_{2} &=V_{2}+\mathrm{I}_{2} \mathrm{Z}_{2} \\ &=539.6+j 32.8\\ &=541 \angle 3.5^{\circ} \end{aligned}\] Obviously, \(\beta=3.5^{\circ}\) image

\[\begin{aligned} \mathrm{E}_{1} &=\mathrm{E}_{2} / \mathrm{K}=19.5\times \mathrm{E}_{2}\\ &=19.5(539.6+j 32.8) =10,520+j 640 \\ \therefore- \mathrm{E}_{1}&=-10,520-j 640=10.540 \angle 183.5^{\circ} \\ \end{aligned}\]

\[\begin{aligned} \mathrm{I}_{2}^{\prime} &=-I_{2} K=(-160+j 120) / 19.5 \\ &=-8.21+j 6.16 \end{aligned}\] Now, \(I_{0}\) leads \(V_{2}\) by an angle \(=3.5^{\circ}+90^{\circ}+30^{\circ}=123.5^{\circ}\) \[\begin{aligned} I_{0} &=1.25 \angle 123.5^{\circ} \\ &=-0.69+j 1.04 \\ I_{1} &=I_{2}^{\prime}+I_{0}\\ &=-8.9+j 7.2=11.45 \angle 141^{\circ} \\ V_{2} &=-E_{1}+I_{1} Z_{1} \\ &=-11,462-j 1350 \\ &=11,540 \angle 186.7^{\circ} \end{aligned}\] Phase angle between \(V_{1}\) and \(I_{1}\) is \(=186.7^{\circ}-141^{\circ}=45.7^{\circ}\)

\(\therefore\) primary p.f. \(=\cos 45.7^{\circ}=0.698\) (lag)

\[\begin{aligned} \text{NL primary input power} &=V_{1} I_{0} \sin \phi_{0} \\ &=11,540 \times 1.25 \times \cos 60^{\circ}\\ &=7.210 \mathrm{~W} \\ R_{02} &=R_{2}+K^{2} R_{1}\\ &=0.06+25 / 19.5^{2}=0.1257 \Omega \\ \text{Total Cu loss (secondary)} &=I_{2}^{2} R_{02}\\ &=200^{2} \times 0.1257=5,030 \mathrm{~W} \\ \text { Output } & =V_{2} I_{2} \cos \phi_{2}\\ &=500 \times 200 \times 0.8=80,000 \mathrm{~W} \\ \text { Total losses } & =5030+7210=12,240 \mathrm{~W} \\ \text{Input} & =80,000+12,240=92,240 \mathrm{~W} \\ \eta&=80,000 / 92,240=0.8674 \\ & 86.74 \% \end{aligned}\]

Problem-3

A \(50 \mathrm{KVA}, 4400 / 220 \mathrm{~V}\) transformer has \(\mathrm{R}_{1}=3.45\) \(\Omega, \mathrm{R}_{2}=0.009 \Omega\). The values of reactances are \(\mathrm{X}_{1}\) \(=5.2 \Omega\) and \(X_{2}=0.015 \Omega\). Calculate for the transformer

  1. equivalent resistance as referred to primary

  2. equivalent resistance as referred to secondary

  3. equivalent reactance as referred to both primary and secondary

  4. equivalent impedance as referred to both primary and secondary

  5. total Cu loss, first using individual resistances of the two windings and secondly using equivalent resistances as referred to each side

Solution-3

\[\begin{aligned} \text{Full-load}~I_{1}&=50,000 / 4,400=11.36 \mathrm{~A} \\ \text{Full-load}~I_{2}&=50,000 / 2220=227 \mathrm{~A} \\ K & =220 / 4,400=1 / 20 \\ R_{01}&=R_{\mathrm{1}}+\dfrac{R_{2}}{K^{2}}=3.45+\frac{0.009}{(1 / 20)^{2}}=3.45+3.6=7.05 \Omega \\ R_{02}&=R_{2}+K^{2} R_{1}\\ &=0.009+(1 / 20)^{2} \times 3.45=0.0176 \Omega \\ \text{OR} &=K^{2} R_{01}=(1 / 20)^{2} \times 7.05=0.0176 \Omega \\ X_{01}&=X_{1}+X_{2}^{\prime}=X_{1}+X_{2} / K^{2}=5.2+0.015 /(1 / 20)^{2}=11.2 \Omega \\ X_{02}&=X_{2}+X_{1}^{\prime}=X_{2}+K^{2} X_{1}=0.015+5.2 / 20^{2}=0.028 \Omega \\ \text{OR} &=K^{2} X_{01}=11.2 / 400=0.028 \Omega \end{aligned}\]

\[\begin{aligned} Z_{01}&=\sqrt{\left(R_{01}^{2}+X_{01}^{2}\right)}=\sqrt{\left(7.05^{2}+11.2\right)^{2}}=13.23 \Omega \\ Z_{02}&=\sqrt{\left(R_{02}^{2}+X_{02}^{2}\right)}=\sqrt{\left(0.0176^{2}+0.028\right)^{2}}=0.03311 \Omega \\ \text{OR}&=K^{2} Z_{01}=13.23 / 400=0.0331 \Omega \\ \text{Culoss} &=I_{1}^{2} R_{2}+I_{2}^{2} R_{2}=11.36^{2} \times 3.45+227^{2} \times 0.009=910 \mathrm{~W} \\ \text { OR } &=I_{1}^{2} R_{01}=11.36^{2} \times 7.05=910 \mathrm{~W} \\ \text{OR} &=I_{2}^{2} R_{02}=227^{2} \times 0.0176=910 \mathrm{~W} \end{aligned}\]

Problem-4

A \(230 / 460 \mathrm{~V}\) transformer has a primary resistance of \(0.2 \Omega\) and reactance of \(0.5 \Omega\) and the corresponding values for the secondary are \(0.75 \Omega\) and \(1.8 \Omega\) respectively. Find the secondary terminal voltage when supplying \(10 \mathrm{~A}\) at \(0.8 \mathrm{pf}\) lagging.

Solution-4

\[\begin{aligned} K&=460 / 230=2 \\ R_{02}&=R_{2}+K^{2} R_{1}\\ &=0.75+2^{2} \times 0.2=1.55 \Omega \\ X_{02} &=X_{2}+K^{2} X_{1}\\ &=1.8+2^{2} \times 0.5=3.8 \Omega \\ \text { Voltage drop } &=I_{2}\left(R_{02} \cos \phi+X_{02} \sin \phi\right)\\ &=10(1.55 \times 0.8+3.8 \times 0.6)=35.2 \mathrm{~V} \\ V_2 &=460-35.2=424.8 \mathrm{~V} \end{aligned}\]

Problem-5

Calculate the percentage voltage drop for a transformer with a percentage resistance of 2.5 % and a percentage reactance of 5% of rating 500 KVA when it is delivering 400 KVA at 0.8 pf lagging.

Solution-5

\[\% \text { drop }=\frac{(\% R) I \cos \phi}{l_{f}}+\frac{(\% X) I \sin \phi}{I_{f}}\] where \(I_{f}\) is the full-load current and \(I\) the actual current.

\[\% \text { drop }=\frac{(\% R) k W}{k V A \text { rating }}+\frac{(\% X) k V A R}{k V A \text { rating }}\] In the present case, \[\mathrm{kW}=400 \times 0.8=320 \text { and } \mathrm{kVAR}=400 \times 0.6=240\] \[\text { drop }=\frac{2.5 \times 320}{500}+\frac{5 \times 240}{500}=4 \%\]