Transformer Efficiency

A \(11000 / 230 \mathrm{~V}, 150 \mathrm{KVA}, 1\) -phase \(, 50-\mathrm{Hz}\) transformer has core loss of \(1.4 \mathrm{KW}\) and F.L. Cu loss of \(1.6 \mathrm{KW}\). Determine

1. The KVA load for max efficiency and value of max efficiency at unity power factor

2. The efficiency at half F.L 0.8 of leading

Solution-1

• \[=\text { F.L. } k V A \times \sqrt{\frac{\text { Iron loss }}{\text { F.L. Cu loss }}}=250 \times \sqrt{\frac{1.6}{1.4}}=160 \mathrm{kVA}\]
  Load kVA corresponding to maximum efficiency is

• Since Cu loss equals iron loss at maximum efficiency,

  \[\begin{aligned} \text{total loss} &=1.4+1.4=2.8 \mathrm{~kW}\\ \text{output} & =160 \times 1=160 \mathrm{~kW} \\ \eta_{\max }&=160 / 162.8=0.982 \text { or } 98.2 \% \end{aligned}\]

• Cu loss at half full-load \(=1.6 \times(1 / 2)^{2}=0.4 \mathrm{~kW}\)

• Total loss \(=1.4+0.4=1.8 \mathrm{~kW}\)

• Half F.L. output at \(0.8 \mathrm{p.f}=(150 / 2) \times 0.8=60 \mathrm{~kW}\)

• \[60 /(60+1.8)=0.97~ \text{or}~ 97 \%\]
  Efficiency

A 20-KVA, \(440 / 220 \mathrm{~V}, 1\) -phase, \(50 \mathrm{~Hz}\) transformer has iron loss of \(324 \mathrm{~W}\). The Cu loss is found to be \(100 \mathrm{~W}\) when delivering half full load current. Determine

1. efficiency when delivering full-load current at 0.8 lagging pf and

2. the percent of full load when the efficiency will be maximum.

Solution-2

• F.L. Cu loss \(=2^{2} \times 100=400 \mathrm{~W} ;\)

• Iron loss \(=324 \mathrm{~W}\)

• F.L. efficiency at \(0.8 \mathrm{p} . \mathrm{f} .=\dfrac{20 \times 0.8}{(20 \times 0.8)+0.724} \times 100=95.67 \%\)

• \(\dfrac{\mathrm{kVA} \text { for maximum }}{\mathrm{F.L}, \mathrm{kVA}}=\sqrt{\dfrac{\text { Iron loss }}{\text { F.L. Cu loss }}}=\sqrt{\dfrac{324}{400}}=0.9\)

• Hence, efficiency would be maximum at \(90 \%\) of F.L.

A 600 KVA, 1-phase transformer has an efficiency of 92% both at full-load and half-load at UPF. Determine its efficiency at 60% of full load at 0.8 power factor lag.

• \[\begin{aligned} 92&=\frac{1 / 2 \times 600 \times 1}{(1 / 2) \times 600 \times 1+W_{i}+(1 / 2)^{2} W_{C u}} \times 100\\ W_{i}+0.25 W_{\mathrm{Cu}}&=26.087 \mathrm{~kW} \end{aligned}\]
  At half F.L. UPF. Here

• From \((i)\) and \((i i),\) we get, \(W_{i}=17.39 \mathrm{~kW}, W_{C_{u}}=34.78 \mathrm{~kW}\)

• \[\eta=\frac{0.6 \times 600 \times 0.8 \times 100}{(0.6 \times 600 \times 0.8)+17.39+(0.6)^{2} 34.78}=85.9 \%\]
  \(x=0.6\)\(60 \%\)

Solution-3

• \[\eta=\frac{x \times k V A \times \cos \phi}{(x \times k V A) \times \cos \phi+W_{i}+x^{2} W_{C u}} \times 100\]
  \(W_{C u}\)\(W_{i}\)\(x\)Efficiency:

• At FI. u.p.f. \(\quad\) Here \(x=1\)

  \[\begin{aligned} 92&=\frac{1 \times 600 \times 1}{1 \times 600 \times 1+W_{i}+1^{2} W_{C u}} \times 100 \\ W_{i}+W_{Cu}&=52.174 \mathrm{~kW} \end{aligned}\]

• The iron loss takes place throughout the day irrespective of the load on the transformer because its primary is energized all the 24 hours.

• Iron loss in 24 hours \(=24 \times 3.5=84 \mathrm{kWh}\)

• Total transformer loss \(=56.924+84=140.924 \mathrm{kWh}\)

• Transformer output in 24 hours \(=(6 \times 400)+(10 \times 300)+(4 \times 100)=5800 \mathrm{kWh}\)

• \[\begin{aligned} \eta_{\text {all-day }} &=\frac{\text { output }}{\text { output }+\operatorname{losses}}\\ &=\frac{5800}{5800+140.924}=0.976 \\ &=97.6 \% \end{aligned}\]
  All-day efficiency

Solution-4

• A load of \(400 \mathrm{~kW}\) at 0.8 p.f. is equal to \(400 / 0.8=500 \mathrm{kVA}\).

• \(300 \mathrm{~kW}\) at 0.75 p.f. means \(300 / 0.75=400 \mathrm{kVA}\)

• \[\begin{aligned} \text { Cu loss at } F . L \text { of } 500 \mathrm{kVA} &=4.5 \mathrm{~kW} \\ \text { Cu loss at } 400 \mathrm{kVA} &=4.5 \times(400 / 500)^{2}=2.88 \mathrm{~kW} \\ \text { Cu loss at } 125 \mathrm{kVA} &=4.5 \times(125 / 500)^{2}=0.281 \mathrm{~kW} \end{aligned}\]
  \(100 / 0.8=125 \mathrm{kVA}\)\(100 \mathrm{~kW}\)
• \[\begin{aligned} &=(6 \times 4.5)+(10 \times 2.88)+(4 \times 0.281)+(4 \times 0)\\ &=56.924 \mathrm{kWh} \end{aligned}\]
  Total Cu loss in 24 hours