Question

Q.2) A transformer has its maximum efficiency of 0.98 at 15 kVA at unity power factor. Compare its all-day efficiencies for the following load cycles: (a) Full load of 20 kVA 12 hours/day and no-load rest of the day. (b) Full load 4 hours/day and 0.4 full-load rest of the day. Assume the load to operate on unity power factor all day. (14 marks)

Answer 1

a) we have the definition of au day efficiency as . in = Energy, output over 24 hours Energy input over at hours to find the energy input we need to find the iron and copper losser. for that Masimum y is given. Since Maximum n occurs at which iron & cu losses are equal (ie Pi=Pc) we can write.

= Power) Mase Po (output Potapi :.98 = 156.98 (154.98)+2 Pi ap. = 14.7 -14.7 .98 Ri= 15-14.7 – 0.153 kw 2 Now; K= 15 (since 20lava is the full load Condition) +0.25 20 K2 = 0.562

Now; k²= Pi (iron loss) Pe c copper lose) :Pe=0.153 = 0.atako 0.562 a) Po=2014uA for 12 hor at upf (Given) . Wo=20x12=240 lawler. (where "w is the koher Energy ) Also Wo=0 (for the rest 12 her)

(imput Power) Pin = Pot Pit le²Pe. = 20+0.153+0.272 = 20.425 kw win = Pin x 12 = 245. 1 kwhere l for the first lather) For the Next 12 her Pin = 0+0.153 = 0.153 Win = 0.153x12 = 1.8lwher All day n = {wo Ewin = 240 X 100 245.1+1.8 = 97.2%.

b) Po = 20kwa for 4 har at upf. 40:4(20) live for do here at upf Wo = (20x4)+(8x20) = 80+160 = 240 lawler. Pin = Pot Pi + le? Pe (for the first 4 hores) - 20+0.153+0.272 = 20.495 (for first 4 heur) win = 20.425 x 4 = 81.7 Kwher

Now for the Next Qohas. Pin = 8+0.153+ ()x0.272 = 8.196 kw . Win=8.1964 20=163.9 kwher : All day n= Ewo Ewin - 240 Χιοο 81.7+163.9 =97.2.l.
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