Transformer Voltage Drop Mastery: Solved Problems

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

high voltage side
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 :

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

(ii) Low Voltage Side:

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,

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

\[\begin{aligned} \mathrm{I}_{2}^{\prime} &=-I_{2} K=(-160+j 120) / 19.5 \\ &=-8.21+j 6.16 \end{aligned}\]

is and Phase angle between by an angle leads Now,

\(\therefore\) 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

equivalent resistance as referred to primary
equivalent resistance as referred to secondary
equivalent reactance as referred to both primary and secondary
equivalent impedance as referred to both primary and secondary
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

\[\% \text { drop }=\frac{(\% R) I \cos \phi}{l_{f}}+\frac{(\% X) I \sin \phi}{I_{f}}\]

the actual current. is the full-load current and where

\[\% \text { drop }=\frac{(\% R) k W}{k V A \text { rating }}+\frac{(\% X) k V A R}{k V A \text { rating }}\]

In the present case,