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I was given in class a problem about power transform but I don't understand which current is used in a stage of the resolution. It is asked to find the efficiency of a transformer in a period of a day (24 hours) given that;

  • \$S_{N}=110kVA\$
  • \$2200V/110V\$
  • \$ 60~ Hz\$
  • \$R_{1}=.22\Omega\$
  • \$R_{2}=.5m\Omega\$
  • \$X_{1}=2\Omega\$
  • \$X_{2}=5m\Omega\$
  • \$R_{c}=5.5k\Omega\$
  • \$X_{m}=1.1k\Omega\$ with this power cycle
  • 4 hrs with no load
  • 8 hrs at 1/4 full load and pf .8
  • 8 hrs at 1/2 at pf 1
  • 4 hrs at full load and pf 1
  • Constant core loss of 1.346 kW Both in the resolution given and my idea is to make this fraction

\$\eta=\frac{output@24hrs}{input@24hrs}\$

first
\$output=(4\times0+8\times.25\times.8+8\times.5\times1+4\times1\times1)=1056kWh\$
then the core loss is
\$1346\times24=32.3kWh\$
and finally the ohmic losses in the windings

  • 8 hrs at .25 full load
    (\$250^2\times0.005+14.1^2\times.22)8=2.85\$
  • 4hrs at .5 full load
    (\$500^2\times0.005+26.6^2\times.22)4=5.62\$
  • 8hrs at full load
    (\$1000^2\times0.005+51.7^2\times.22)8=44.70\$

but I don't understand why 14.1, 26.6 and 51.7, isn't supposed to be the nominal current on the high voltage side \$I_{1N}=\frac{110000}{2200}=50A\$?

and that would be

  • 8 hrs at .25 full load
    (\$250^2\times0.005+12.5^2\times.22)8=2.85\$
  • 4hrs at .5 full load
    (\$500^2\times0.005+25^2\times.22)4=5.62\$
  • 8hrs at full load
    (\$1000^2\times0.005+50^2\times.22)8=44.70\$

If not then which current is this and how to calculate?

UPDATE
the total ohmic loss will be 53.17kWh and then

\$\eta=\frac{1056}{1056+32.3+53.17}=92.5\%\$
But I still don't get how the 14.1...and respective currents are calculated

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    \$\begingroup\$ What exactly are you asking? Why you should include the magnetizing current in the I^2*R loss calculations? Or why it has that specific value? (The details you gave are fine, but pretty much meaningless without the equivalent model they refer to) \$\endgroup\$ – Brian Drummond Mar 13 '18 at 18:05
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    \$\begingroup\$ Rated voltage is done at full load, you are assuming Vout at no load So kVA rating= Ploss + Pload for R loads and effic = Ploss/kVA \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 13 '18 at 18:05
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    \$\begingroup\$ How many circuits could be made from 3 resistors and 3 inductors? \$\endgroup\$ – Andy aka Mar 13 '18 at 18:59
  • \$\begingroup\$ @BrianDrummond, what are the 14.1, 26.6 and 51.7 values? what current does stand for? \$\endgroup\$ – riccs_0x Mar 13 '18 at 22:19
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    \$\begingroup\$ Primary current. Your second set of figures only accounts for the current delivered in the secondary, omitting the magnetizing current and any due to power factor considerations. \$\endgroup\$ – Brian Drummond Mar 13 '18 at 22:21

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