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"source": [
"# CHAPTER 46 : ELECTRONIC CONTROL OF A.C MOTORS"
]
},
{
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"metadata": {},
"source": [
"## EXAMPLE 46.1 , PAGE NO :- 1827"
]
},
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"text": [
"Ton = 1.89 ms.\n",
"magnitude of the current pulse I0 = 102.31 A.\n"
]
}
],
"source": [
"'''The wound-rotor induction motor of Fig.43.5 is rated at 30-kW,975 rpm,440-V,50 Hz.The open-circuit line voltage is 400V and the load\n",
"resistance is 0.5 ohm.If chopper frequency is 200 Hz,calculate Ton so that the motor develops a gross torque of 200 N-m at 750 rpm.Also,\n",
"calculate the magnitude of the current pulses drawn from the capacitor.'''\n",
"\n",
"import math as m\n",
"Ns = 1000.0 #rpm (Synchronus speed)\n",
"N1 = 750.0 #rpm (rotating speed)\n",
"E2 = 400.0 #V (OC line voltage)\n",
"Tg = 200.0 #N-m (gross torque)\n",
"R0 = 0.5 #ohm (load resistance)\n",
"f = 200.0 #Hz (chopper frequency)\n",
"s = (Ns-N1)/Ns # slip\n",
"Vrl = s*E2 #V (rotor line voltage)\n",
"Vdc = 135.0 #V (DC voltage of 3-phase bridge rectifier)\n",
"#Now, Tg = P^2/(2*3.14*Ns).Therefore,\n",
"P2 = Tg*2*3.14*Ns/60 #W\n",
"#Power dissipated as heat\n",
"sP2 = s*P2 #W\n",
"#Power is actually dissipated in R and is equal to rectifier output Vdc*Idc.Therefore,\n",
"Idc = sP2/Vdc #A\n",
"#The apparent resistance at the input of chopper is\n",
"Ra = Vdc/Idc #ohm\n",
"#Now, Ra = Ro/(f*Ton)^2 .Therefore,\n",
"Ton = m.sqrt(R0/(f*f*Ra))*1000 #ms\n",
"#Current in R0 can be found from relation,I0^2*R0 = sP2\n",
"I0 = m.sqrt(sP2/R0) #A\n",
"print \"Ton = \",round(Ton,2),\"ms.\"\n",
"print \"magnitude of the current pulse I0 =\",round(I0,2),\"A.\""
]
},
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