Answer Posted / shubha_ee

If motor is fed from Delta connected supply.....
P=15hp = 11.19kW
so P per phase = 11.19/3 = 3.73kW..
pf = 11190/(1.732*380*31) = 0.55(lag)
then Vp=VL = 380V..
and
Ip=IL/1.732 = 31/1.732 = 17.9Amp..
So, R per phase = (380)^2/P = (380)^2/3730 = 38.71 ohm..
so, Heat dissipation for 1sec due to I^2R loss,
H1 = (17.9)^2*38.71*1 = 12403.07 Joules...

Now, if the VL=250V = Vp then IL= 11190/(1.732*250*0.55) = 46.98Amp...
so, Ip=46.98/1.732 = 27.12Amp...
So, Heat dissipation due to I^2R loss for 1sec ,
H2 = (27.12)^2*38.71*1 = 28470.98 Joules...

Now, increase in heat dissipation = (H2 - H1) = 16067.91 Joules
so, increase in temperature is 8.46 deg Celcius.....(as 1degC = 1899.1005 Joules)

hi frenz,its aswathy.i hve aplied for the hpcl test on feb,pls send the test patern & the placement papers as soon as possible. mail-id :kjaswathy@gmail.com

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