Taming Three-Phase Transformers: Solved Problems

Problem-1

A 3-phase, 500 kVA, 6000V/400V, 50Hz, delta-star connected transformer is delivering 300 kW, at 0.8 pf lagging to a balanced 3-phase load connected to the LV side with HV side supplied from 6000 V, 3- phase supply. Calculate the line and winding currents in both the sides. Assume the transformer to be ideal.

Solution-1

Note: it is not a bank of single phase transformers.
It is a single unit of \(3-\phi\) transformer with 500 kVA, 6000 V/400 V, 50Hz, \(\Delta/Y\) connection
500 kVA represents total kVA & voltages specified are always line to line.
Similarly unless otherwise specified, kW rating of a \(3-\phi\) load is the total kW absorbed by the load.

\[\begin{aligned} \text{Load KVA,}S & =\dfrac{P}{\cos\theta}=\dfrac{300}{0.8}=375\mathrm{kVA}=\text{input kVA}\\ \text{Line current drawn by load,}I_{2L} & =\dfrac{S}{\sqrt{3}V_{LL}}=\dfrac{375\times1000}{\sqrt{3}\times400}=541.3\mathrm{A}\\ \text{HV side line current,}I_{1L} & =\dfrac{375\times1000}{\sqrt{3}\times6000}=36.1\mathrm{A}\\ \text{LV Star-connection}I_{2P} & =I_{2L}=541.3\mathrm{A}\\ \text{HV delta-connected,}I_{1P} & =\dfrac{I_{1L}}{\sqrt{3}}=\dfrac{36.1}{\sqrt{3}}=20.84\mathrm{A} \end{aligned}\]

Problem-2

A 3-phase, 50Hz transformer has a delta-connected primary and star connected secondary, the line voltage being 22,000 V and 400 V respectively. The secondary has a star connected balanced load at 0.8 power factor lagging. The line current on the primary side is 5 A. Determine the current in each coil of the primary and in each secondary line. What is the output of the transformer in kW?

Solution-2

Note: In 3-phase T/F, the phase transformation ratio is equal to the turn ratio but the terminal or line voltages depends upon the connection employed

\[\begin{aligned} \text{Phase voltage on primary side} &=22,000 \mathrm{~V}\\ \text{ Phase voltage on secondary side} &=400 / \sqrt{3}\\ K&=400 / 22,000 \times \sqrt{3}\\ &=1 / 55 \sqrt{3}\\ \text{Primary phase current} &=5 / \sqrt{3} \mathrm{~A}\\ \text{Secondary phase current} &=\frac{5}{\sqrt{3}} \div \frac{1}{55 \sqrt{3}}=275 \mathrm{~A}\\ \text { Secondary line current } & =275 \mathrm{~A} \\ \text { Output } & =\sqrt{3} V_{L} I_{L} \cos \phi\\ &=\sqrt{3} \times 400 \times 275 \times 0.8\\ &=15.24 \mathrm{~kW} \end{aligned}\]

Problem-3

A 500-KVA, 3-phase, 50-Hz transformer has a voltage ratio (line voltages) of 33/11-KV and is delta/star connected. The resistance per phase are: high voltage 35 \(\Omega\), low voltage 0.876 \(\Omega\) and the iron loss is 3050 W. calculate the value of efficiency at full-load and one-half of full-load respectively

at unity p.f.
0.8 p.f.

Solution-3

\[\begin{aligned} K & =\dfrac{11/\sqrt{3}}{33}=0.192\\ R_{02} & =R_{1}^{\prime}+R_{2}\\ & =K^{2}R_{1}+R_{2}\\ & =0.192^{2}\times35+0.876\\ & =2.166\Omega\\ \text{Star-Side,}I_{2p} & =I_{2L}=\dfrac{S}{\sqrt{3}\times V_{2L}}\\ & =\dfrac{500\times10^{3}}{\sqrt{3}\times11\times10^{3}}\\ & =26.243\mathrm{A} \end{aligned}\]

Full-Load Condition:

\[\begin{aligned} \text{Cu-loss} & =3I_{2p}^{2}R_{02}\\ & =3\times26.243^{2}\times2.166\\ & =4475.14~\mathrm{W}\\ \text{Iron-loss} & =3050\mathrm{W}\\ \text{Total-loss} & =4475.14+3050=7525.14~\mathrm{W}\\ \text{At UPF}~\eta & =\dfrac{500\times10^{3}\times1}{500\times10^{3}\times1+7525.14}\times100\\ & =98.51\%\\ \text{At 0.8 PF,}~\eta & =\dfrac{500\times10^{3}\times0.8}{500\times10^{3}\times0.8+7525.14}\times100\\ & =98.15\% \end{aligned}\]

Half-Load Condition:

\[\begin{aligned} \text{Cu-loss} & =\left(\dfrac{1}{2}\right)^{2}\times4475.14=1118.78~\mathrm{W}\\ \text{Total-loss} & =1118.78+3050=4168.78~\mathrm{W}\\ \text{At UPF}\eta & =\dfrac{0.5\times500\times10^{3}\times1}{500\times10^{3}\times1+4168.78}\times100\\ & =98.35\%\\ \text{At 0.8 UPF}\eta & =\dfrac{0.5\times500\times10^{3}\times0.8}{500\times10^{3}\times0.8+4168.78}\times100\\ & =97.95\% \end{aligned}\]

Problem-4

A 3-phase transformer 33/6.6-KV, delta-star, 2 MVA has a primary resistance of 8 \(\Omega\) per phase and a secondary resistance of 0.08 ohm per phase. The percentage impedance is 7%. Calculate the secondary voltage with rated primary voltage and hence the regulation for full-load 0.75 p.f. lagging conditions.

Solution-4

\[\begin{aligned} \text{F.L.}~I_{2} & =\frac{2 \times 10^{6}}{\sqrt{3} \times 6.6 \times 10^{3}}=175 \mathrm{~A} \\ K&=6.6 / \sqrt{3} \times 33=1 / 8.65 \\ R_{02}&=0.08+8 / 8.65^{2}=0.1867 \Omega ~\text{per phase} \\ Z_{02}~ \text{drop per phase} &=\frac{7}{100} \times \frac{6,600}{\sqrt{3}}=266.7 \mathrm{~V}\\ Z_{02}&=266.7 / 175=1.523 \Omega~ \text{per phase}\\ X_{02}&=\sqrt{Z_{02}^{2}-R_{02}^{2}}\\ &=\sqrt{1.523^{2}-0.1867^{2}}=1.51 \Omega / \text{phase} \end{aligned}\]

\[\begin{aligned} \text{Drop per phase} & =I_{2}\left(R_{02} \cos \phi+X_{02} \sin \phi\right)\\ &=175(0.1867 \times 0.75+1.51 \times 0.66)=200 \mathrm{~V}\\ E_2~ \text{per-phase} &=6,600 / \sqrt{3}=3,810 \mathrm{~V} \\ V_{2}&=3,810-200=3,610 \mathrm{~V}\\ V_{2L}&=3,610 \times \sqrt{3}=6,250 \mathrm{~V}\\ \%\text{regn.} &=200 \times 100 / 3,810=5.23 \% \end{aligned}\]

Problem-5

A 5000 KVA, 3-phase transformer, 6.6/33-KV, delta-star has a no-load loss of 15 KW and a full-load loss of 50 KW. The impedance drop at full-load is 7%. Calculate the primary voltage when a load of 3200 KW at 0.8 pf is delivered at 33 KV.

Solution-5

\[\begin{aligned} \text{Full-load}~I_{2}&=5 \times 10^{6} / \sqrt{3} \times 33,000=87.5 \mathrm{~A} \\ \text{Impedance drop/phase}&=7 \% \text { of }(33 / \sqrt{3})=7 \% \text { of } 19 \mathrm{kV}=1,330 \mathrm{~V}\\ Z_{02}&=1.330 / 87.5=15.3~ \Omega \text { /phase} \\ \text{F.L. Cu loss }&=50-15=35 \mathrm{~kW}\\ 3 I_{2} R_{02}&=35,000 \\ R_{02}&=35,000 / 3 \times 8.75^{2}=1.53 \Omega ~\text { /phase }\\ X_{02}&=\sqrt{15.3^{2}-1.53^{2}}=15.23~ \Omega \end{aligned}\]

\[\begin{aligned} & \text{When load is}~3,200~ \mathrm{~kW}~\text{at 0.8 p.f}\\ I_{2}&=3,200 / \sqrt{3} \times 33 \times 0.8 = 70 \mathrm{~A}\\ \mathrm{drop}&=70(1.53 \times 0.8+15.23 \times 0.6)=725 \mathrm{~V} / \text { phase } \\ \%\text { regn. } &=\frac{725 \times 100}{19,000}=3.8 \% \\ &\text{Primary voltage will have to be increased by}~3.8 \%\\ \text{Primary voltage} & =6.6+3.8 \% \text { of } 6.6=6.85 \mathrm{kV}=6.850 \mathrm{~V} \end{aligned}\]