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Analysis and Calculations of Resistances, Wire Sizes, and Magnetomotive Force in DC Machine Field Windings

Analysis and calculations related to DC machine field windings.

Emma Thompson
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Analysis and Calculations of Resistances, Wire Sizes, and Magnetomotive Force in DC Machine Field Windings - Page 1 preview image

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Analysis and Calculations of Resistances, Wire Sizes, and Magnetomotive Forcein DC Machine Field WindingsMeasurements taken on a DC Machine during a lab: Rated at 125V, 2.4A1.Shunt Field = .57A, 125.3V, 219.82 Ohms.2.Series Field = 2.4 A,19.008V, 7.92 Ohms.3.Armature = 2.97 A, 21.48 V, 8.95 Ohms.Explain the difference in the resistances found for the shunt field, series field, and armature circuits.In particular, why is the resistance of the shunt field so different from the resistances of thearmature and series field? What would I expect therelative wire size to be (heavy or light gauge)and why, and the number of turns (many or few) on each of the three windings?Answer:-series field and armature resistances need to be as minimum as possible becausetominimize the Ia^2 R loss due to armature current Iawhereas in the case of shunt field, the value ofresistance must be largest so that we canreduce V^2/Rf losses.Relative wire size should be very light gauge for the shunt field and heavy gauge for the other two i.e.series field and armature.Number of turns is more in shunt field due to high resistance and less in other two.2. If the values obtained were taken at a temperature of 20 C, what would the resistance valuesabove be at 55 C? could you please show the calculation, and how to do it.( I researched andfound that copper wire increases in resistance by .0039% for each degree rise in temperature, inCelsius.Answer:-R = R1[1+⍺(T2-T1)], R1is initial resistance as 20 CHere T1 = 20 Celsius, T2 = 55Celsius and⍺= 0.0039% = 0.000039On setting these values, we getR = R1[1+0.000039(55-20)] = R1(1.001365)So the new resistance at 55C will be 1.001365 times of the initial resistance R1 which is measuredat 20C4.If the shunt field windings are composed of #34 AWG copper wire, and each turn isapproximately 12 inches long, how many turns of wire are there on the shunt field of the DCmachine? (I looked this up and found the resistance coefficient of #34 copper wire to be 260.9Ohms.Area = 0.0201mm^20.129mm

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