Example 1 A 4-pole d.c. shunt generator has an armature resistance of 0.018 Ω. The armature is lap-wound with 520 conductors. When driven at 750 rev/min the machine produces a total armature current of 400 A at a terminal voltage of 200 V. Calculate the useful flux/pole . Solution P = 4; Ia = 400 A; Ra = 0.018 Ω; V = 200 V Z = 520; n = 750 rev/min; E = V + Ia Ra = 200 + (400 x 0.018) = 207.2 V Emf equation for generator; E = ΦPNZ/(60 A) For lap winding A = P So, E = ΦNZ / 60 Φ = (E x 60) / (N x Z) = (207.2 x 60) / (750 x 520) = 31.9 mWb Example 2 A short-shunt compound generator has armature, shunt field and series field resistances of 0.5 Ω, 100 Ω, and 0.3 Ω respectively. When supplying a load of 8 kW at a terminal voltage of 250 V the input power supplied by the driving motor is 10.4 kW. Calculate (a) the generated emf, (b) the efficiency, (c) the iron, friction and windage loss, (d) the total fixed losses. (e) total I2R losses at full load (f) total losses at full load Solution Ia Rse = 0.3 Ω IL If Ra = 0.5 Ω Rf = 100 Ω E Pi = 10.4 kW (a) IL = Po / V = 8000 / 250 = 32 A Vf = V + IL Rse = 250 + 32 x 0.3 = 259.6 V If = Vf / Rf = 259.6 / 100 = 2.596 A V = 250 V Ω Po = 8 kW Ia = IL + If = 32 + 2.596 = 34.596 A E = Vf + Ia Ra = 259.6 + 34.596 x 0.5 = 276.898 V (b) η = Po / Pi = 8 / 10.4 = 0.769 = 76.9% (c) From the power flow diagram PFe = Pi – E x Ia = 10.4 x 103 – 276.898 x 34.596 = 820.437 W (d) Total fixed losses = PFe + If2 x Rf = 820.437 + 2.5962 x 100 = 1494.36 W (e) Total I2R losses at full load = Ia2 x Ra + If2 x Rf + IL2 x Rse = 34.5962 x 0.5 + 2.5962 x 100 + 322 x 0.3 = 1579.5 (f) Total losses at full load = PFe + I2R losses = 820.437 + 1579.563 = 2400 W Pi – Total losses = Po 10400 – 2400 = 8000 Motor Example A 4-pole series motor has 944 wave-connected armature conductors. At a certain load, the flux per pole is 34.6 mWb and the total mechanical torque developed is 209 N-m. Calculate the line current taken by the motor and the speed at which it will run with an applied voltage of 500 V. Total motor resistance is 3 ohm. Solution Τa = PZϕIa / (2ᴨA) = 0.159 ϕZIa(P/A) 209 = 0.159 x 34.6 x 944 x Ia (4/2) Ia = 209 / (0.159 x 34.6 x 10-3 x 944 x 2) = 20.1 A Eb = V - IaRa = 500 – 20.1 x 3 = 439.7 V Speed may be found either by using the relation for Eb or Τa. Eb = ΦPNZ/(60 A) N = Eb x 60 x A / ΦPZ = (439.7 x 60 x 2) / (34.6 x 10-3 x 4 x 944) = 403.86 rpm Example A d.c. series motor operates at 800 r.p.m. with a line current of 100 A from 230-V mains. Its armature circuit resistance is 0.15 W and its field resistance 0.1 W. Find the speed at which the motor runs at a line current of 25 A, assuming that the flux at this current is 45 per cent of the flux at 100 A. Solution N2 N1 = Eb2 Eb1 Φ1 x Φ2 Φ2 = 0.45 Φ1 Eb1 = V - IaRa = 230 – (0.15 + 0.1) x 100 = 205 V Eb2 = V - IaRa = 230 - (0.15 + 0.1) x 25= 223.75 V N2 800 = 223.75 205 N2 = 1940 rpm x 1 0.45