EM-Problems First Semester 2019-2020 Problem 1 An 8-pole, wave-connected armature has 600 conductors and is driven at 625rev/min. If the flux per pole is 20mWb. - Determine the generated e.m.f. Solution: ๐ = ๐๐๐, ๐ = ๐ (๐๐๐ ๐ ๐๐๐๐ ๐๐๐๐ ๐๐๐), ๐ = ๐ ๐๐๐๐๐ ๐= ๐๐๐ ๐๐๐ [ ], ๐๐ ๐ ๐ = ๐๐ ∗ ๐๐−๐ ๐พ๐ ๐ฎ๐๐๐๐๐๐๐๐ ๐. ๐. ๐. , ๐ฌ = ๐๐ท๐๐๐ ๐ ๐๐๐ ๐ (๐) (๐๐ ∗ ๐๐−๐ ) ( ) (๐๐๐) ๐๐ ๐ฌ= = ๐๐๐ ๐ฝ๐๐๐๐ ๐ Problem 2 A 4-pole generator has a lap-wound armature with 50 slots with 16 conductors per slot. The useful flux per pole is 30mWb. - Determine the speed at which the machine must be driven to generate an e.m.f. of 240V. Solution: ๐ธ = 240 ๐, ๐ = 2๐ ( ๐๐๐ ๐ ๐๐๐ ๐ค๐๐๐๐๐๐), ๐ = 50 ∗ 16 = 800, ๐ = 30 ∗ 10−3 ๐๐ ๐ฎ๐๐๐๐๐๐๐๐ ๐. ๐. ๐. , ๐ฌ = ๐น๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐, ๐๐๐๐๐ , ๐ = ๐๐๐๐๐ ๐๐๐๐๐ = = ๐๐๐ ๐ ๐๐ ๐ฌ ๐๐๐ ๐๐๐ ๐๐๐ = = ๐๐ ๐๐ ๐๐๐ (๐๐ ∗ ๐๐−๐ ) (๐๐๐) ๐๐ ๐ ๐๐๐ Problem 3 An 8-pole, lap-wound armature has 1200 conductors and a flux per pole of 0.03Wb. - Determine the e.m.f. generated when running at 500rev/min. Solution: ๐ฎ๐๐๐๐๐๐๐๐ ๐. ๐. ๐. , ๐ฌ = ๐๐ท๐๐๐ ๐๐ท๐๐๐ = = ๐๐๐ ๐ ๐๐ท ๐ช = ๐ ๐๐๐ ๐๐๐ − ๐๐๐๐๐ ๐๐๐๐๐๐๐ ๐ฌ = ๐ ๐ ๐ = (๐. ๐๐) ( ๐๐๐ ) (๐๐๐๐) = ๐๐๐ ๐๐๐๐๐ ๐๐ 1 Problem 4 - Determine the generated e.m.f. in Problem 3 if the armature is wave-wound. Solution: ๐บ๐๐๐๐๐๐ก๐๐ ๐. ๐. ๐. , ๐ธ = 2๐๐๐๐ 2๐๐๐๐ = ๐ 2 "๐ = ๐ ๐๐๐ ๐๐๐๐ − ๐๐๐๐๐ ๐๐๐๐๐๐๐" ๐ฌ = ๐ท ๐ ๐ ๐ = (๐) (๐ ๐ ๐) = ๐๐๐ ๐๐๐๐๐ ๐ฌ = (๐) (๐๐๐) = ๐๐๐๐ ๐๐๐๐๐ Problem 5 A d.c. shunt-wound generator running at constant speed generates a voltage of 150 V at a certain value of field current. - Determine the change in the generated voltage when the field current is reduced by 20%, assuming the flux is proportional to the field current. Solution: The generated e.m.f. E of a generator is proportional to ะค ω, i.e. is proportional to ะค n, where ะค is the flux and n is the speed of rotation. It follows that E = k ะค n, where k is a constant. At speed n1 and flux ะค1, E1 = k ะค1 n1 At speed n2 and flux ะค2, E2 = k ะค2 n2 Thus, by division: ๐ธ1 ๐ ๐1 ๐1 ๐1 ๐1 = = ๐ธ2 ๐ ๐2 ๐2 ๐2 ๐2 The initial conditions are E1 =150V, ะค = ะค1 and n = n1. When the flux is reduced by 20%, the new value of flux is 80/100 or 0.8 of the initial value, i.e. ะค2 = 0.8 ะค1. Since the generator is running at constant speed, n2 = n1. ๐โ๐ข๐ ๐กโ๐๐ก ๐๐ , ๐ธ1 ๐1 ๐1 ๐1 ๐1 1 = = = ๐ธ2 ๐2 ๐2 0.8 ๐1 ๐1 0.8 ๐ธ2 = 150 ∗ 0.8 = 120 ๐ Thus, a reduction of 20% in the value of the flux reduces the generated voltage to 120 V at constant speed. 2 Problem 6 A d.c. generator running at 30 rev/s generates an e.m.f. of 200V. - Determine the percentage increase in the flux per pole required to generate 250 V at 20 rev/s. Solution: E ∝ ะค ω and since ω =2πn, E ∝ n. Let E1 = 200 V, n1 = 30 rev/s and flux per pole at this speed be ะค1 Let E2 = 250 V, n2 = 20 rev/s and flux per pole at this speed be ะค2 ๐๐๐๐๐ ๐ธ ∝ ๐ ๐ ๐กโ๐๐ ๐ป๐๐๐๐ ๐๐๐๐ ๐คโ๐๐โ, ๐2 = ๐ธ1 ๐1 ๐1 = ๐ธ2 ๐2 ๐2 200 ๐1 (30) = 250 ๐2 (20) ๐1 (30) (250) = ๐. ๐๐๐ ๐๐ (20) (200) Hence the increase in flux per pole needs to be 87.5 % Problem 7 - Determine the terminal voltage of a generator which develops an e.m.f. of 200 V and has an armature current of 30 A on load. Assume the armature resistance is 0.30 โฆ Solution: With reference to the Figure, terminal voltage, VT = E – (Ia * Ra) = 200 – (30) (0.30) = 200 – 9 = 191 Volts Problem 8 A generator is connected to a 60 โฆ load and a current of 8 A flows. If the armature resistance is 1โฆ - Determine (a) the terminal voltage, and (b) the generated e.m.f. Solution: a- Terminal voltage, V = Ia * RL = (8) (60) = 480 Volts b- Generated e.m.f., E = V + (Ia* Ra) = 480 + (8 *1) = 480 + 8 = 488 Volts 3 Problem 9 A separately-excited generator develops a no-load e.m.f. of 150V at an armature speed of 20 rev/s and a flux per pole of 0.10Wb. Determine the generated e.m.f. when: (a) the speed increases to 25 rev/s and the pole flux remains unchanged (b) the speed remains at 20 rev/s and the pole flux is decreased to 0.08Wb. (c) the speed increases to 24 rev/s and the pole flux is decreased to 0.07Wb. Solution: ๐๐๐๐ ๐คโ๐๐โ, Hence, ๐๐๐๐ ๐คโ๐๐โ, ๐ธ2 = ๐ธ1 ∅1 ๐1 = ๐ธ2 ∅2 ๐2 (0.10)(20) 150 = (0.10)(25) ๐ธ2 (150)(0.10)(25) = 187.5 ๐๐๐๐ก๐ (0.10)(20) (0.10)(20) 150 = (0.08)(20) ๐ธ3 ๐๐๐๐ ๐คโ๐๐โ, ๐. ๐. ๐. , ๐ธ3 = (150)(0.08)(20) = 120 ๐๐๐๐ก๐ (0.10)(20) (0.10) (20) 150 = (0.07) (24) ๐ธ4 ๐๐๐๐ ๐คโ๐๐โ, ๐. ๐. ๐. , ๐ธ4 = 4 (150)(0.07)(24) = 126 ๐๐๐๐ก๐ (0.10)(20) Problem 10 A shunt generator supplies a 20 kW load at 200 V through cables of resistance, R= 100 mโฆ. If the field winding resistance, Rf =50 โฆ and the armature resistance, Ra =40 mโฆ, Determine: (a) the terminal voltage (b) the e.m.f. generated in the armature. Solution: (a) The circuit is as shown in the Figure. ๐ฟ๐๐๐ ๐๐ข๐๐๐๐๐ก, ๐ผ = 20000 ๐๐๐ก๐ก๐ = 100 ๐ด 200 ๐๐๐๐ก๐ ๐๐๐๐ก ๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐๐๐๐ ๐ก๐ ๐กโ๐ ๐๐๐๐ = ๐ผ ∗ ๐ = (100 ๐ด) ∗ (100 ∗ 10−3 ๐บ) = 10 ๐๐๐๐ก๐ ๐ป๐๐๐๐ ๐ก๐๐๐๐๐๐๐ ๐ฃ๐๐๐ก๐๐๐, ๐ = 200 ๐๐๐๐ก๐ + 10 ๐๐๐๐ก๐ = 210 ๐๐๐๐ก๐ ๐ด๐๐๐๐ก๐ข๐๐ ๐๐ข๐๐๐๐๐ก ๐ผ๐ = ๐ผ๐ + ๐ผ ๐น๐๐๐๐ ๐๐ข๐๐๐๐๐ก, ๐ผ๐ = ๐ 210 ๐๐๐๐ก๐ = = 4.2 ๐ด ๐ ๐ 50 Ω ๐ป๐๐๐๐ ๐ผ๐ = ๐ผ๐ + ๐ผ = 4.2 ๐ด + 100 ๐ด = 104.2 ๐ด ๐บ๐๐๐๐๐๐ก๐๐ ๐. ๐. ๐. ๐ธ = ๐ + ๐ผ๐ ๐ ๐ ๐ธ = 210 ๐๐๐๐ก๐ + (104.2 ๐ด) (40 ∗ 10−3 Ω) = 210 ๐๐๐๐ก๐ + 4.168 ๐๐๐๐ก๐ = 214.17 ๐๐๐๐ก๐ 5 Problem 11 A short-shunt compound generator supplies 80 A at 200 V. If the field resistance, Rf = 40 โฆ, the series resistance, Rse = 0.02 โฆ and the armature resistance, Ra = 0.04 โฆ, determine the e.m.f. generated. Solution: The circuit is shown in the Figure. ๐๐๐๐ก ๐๐๐๐ ๐๐ ๐ ๐๐๐๐๐ ๐ค๐๐๐๐๐๐ = ๐ผ ๐ ๐๐ = (80 ๐ด) (0.02 Ω) = 1.6 ๐ ๐๐๐ก๐๐๐ก๐๐๐ ๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐ ๐ ๐กโ๐ ๐๐๐๐๐ ๐ค๐๐๐๐๐๐ = ๐๐๐ก๐๐๐ก๐๐๐ ๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐ ๐ ๐๐๐๐๐ก๐ข๐๐ ๐1 = 200 ๐๐๐๐ก๐ + 1.6 ๐๐๐๐ก๐ = 201.6 ๐๐๐๐ก๐ ๐น๐๐๐๐ ๐๐ข๐๐๐๐๐ก ๐ผ๐ = ๐1 ๐ ๐ = 201.6 ๐๐๐๐ก๐ = 5.04 ๐ด 40 Ω ๐ด๐๐๐๐ก๐ข๐๐ ๐๐ข๐๐๐๐๐ก, ๐ผ๐ = ๐ผ + ๐ผ๐ = 80 ๐ด + 5.04 ๐ด = 85.04 ๐ด ๐บ๐๐๐๐๐๐ก๐๐ ๐. ๐. ๐. , ๐ธ = ๐1 + (๐ผ๐ ∗ ๐ ๐ ) ๐ธ = 201.6 ๐๐๐๐ก๐ + (85.04 ๐ด ∗ 0.04 Ω) = 201.6 ๐๐๐๐ก๐ + 3.4 ๐ฃ๐๐๐ก๐ = 205 ๐๐๐๐ก๐ Problem 12 A 10kW shunt generator having an armature circuit resistance of 0.75 โฆ and a field resistance of 125 โฆ, generates a terminal voltage of 250V at full load. - Determine the efficiency of the generator at full load, assuming the iron, friction and windage losses amount to 600W. Solution: The circuit is shown in the Figure. ๐๐ข๐ก๐๐ข๐ก ๐๐๐ค๐๐ = 10000 ๐ = ๐ ∗ ๐ผ ๐๐๐๐ ๐คโ๐๐โ, ๐๐๐๐ ๐๐ข๐๐๐๐๐ก ๐น๐๐๐๐ ๐๐ข๐๐๐๐๐ก, ๐ผ๐ = ๐ ๐ ๐ = ๐ผ= 250 ๐๐๐๐ก๐ 125 Ω 10000 ๐ = 10000 ๐๐๐ก๐ก 250 ๐๐๐๐ก๐ = 40 ๐ด =2๐ด ๐ด๐๐๐๐ก๐ข๐๐ ๐๐ข๐๐๐๐๐ก, ๐ผ๐ = ๐ผ๐ + ๐ผ = 2 ๐ด + 40 ๐ด = 42 ๐ด Efficiency, ๐∗๐ผ ๐= ( ) ∗ 100 % 2 ๐ ∗ ๐ผ + ๐ผ๐ ๐ ๐ + ๐ผ๐ ∗ ๐ + ๐ถ ๐= ( 10000 10000 ∗ 100 % = ∗ 100 % = 80.5 % ) 10000 + (422 )(0.75) + (2)(250) + 600 12423 6 Problem 13 A d.c. motor operates from a 240 V supply. The armature resistance is 0.2 โฆ. - Determine the back e.m.f. when the armature current is 50 A. Solution: ๐น๐๐ ๐ ๐๐๐ก๐๐, ๐ = ๐ธ + ๐ผ๐ ∗ ๐ ๐ โ๐๐๐๐ ๐๐๐๐ ๐. ๐. ๐. , ๐ธ = ๐ − (๐ผ๐ ∗ ๐ ๐ ) ๐ธ = 240 ๐๐๐๐ก๐ − (50 ๐ด ∗ 0.2 Ω) = 240 ๐๐๐๐ก๐ − 10 ๐๐๐๐ก๐ = 230 ๐๐๐๐ก๐ Problem 14 The armature of a d.c. machine has a resistance of 0.25 โฆ and is connected to a 300 V supply. - Calculate the e.m.f. generated when it is running: (a) as a generator giving 100 A. (b) as a motor taking 80 A. Solution: (a) As a generator, generated e.m.f., ๐ธ = ๐ + (๐ผ๐ ∗ ๐ ๐ ) ๐ธ = 300 ๐๐๐๐ก๐ + (100๐ด ∗ 0.25 Ω) = 300 ๐๐๐๐ก๐ + 25 ๐ฃ๐๐๐ก๐ = 325 ๐๐๐๐ก๐ (b) As a motor, generated e.m.f. (or back e.m.f.), ๐ธ = ๐ − (๐ผ๐ ∗ ๐ ๐ ) ๐ธ = 300 ๐๐๐๐ก๐ − (80 ๐ด ∗ 0.25 Ω) = 280 ๐๐๐๐ก๐ Problem 15. An 8-pole d.c. motor has a wave-wound armature with 900 conductors. The useful flux per pole is 25 mWb. Determine the torque exerted when a current of 30A flows in each armature conductor. Solution: P = 4, c = 2 for wave winding, ะค = 25 * 10-3 Wb, Z= 900, Ia = 30 A (4)(25 ∗ 10−3 )(900)(30) ๐ ๐ ๐ ๐ผ๐ ๐ก๐๐๐๐ข๐ ๐ = = = 429.7 ๐๐ ๐๐ ๐(2) 7 Problem 16. - Determine the torque developed by a 350V d.c. motor having an armature resistance of 0.5 โฆ and running at 15 rev/s. The armature current is 60 A. Solution: ๐ = 350 ๐, ๐ ๐ = 0.5 Ω, ๐ = 15 ๐๐๐ฃ , ๐ ๐ผ๐ = 60 ๐ด ๐ต๐๐๐ ๐. ๐. ๐. ๐ธ = ๐ − (๐ผ๐ ∗ ๐ ๐ ) = 350 ๐๐๐๐ก๐ − (60 ๐ด ∗ 0.5 Ω) = 320 ๐๐๐๐ก๐ ๐ก๐๐๐๐ข๐ ๐ = (320 ๐๐๐๐ก๐ ) (60 ๐ด) ๐ธ ๐ผ๐ = = 203.7 ๐๐ ๐๐๐ฃ 2๐๐ 2 ๐ (15 ๐ ) Problem 17 A six-pole lap-wound motor is connected to a 250V d.c. supply. The armature has 500 conductors and a resistance of 1โฆ. The flux per pole is 20mWb. 2p = one pair of poles Calculate: - This means (a) the speed C = number of poles (b) the torque developed when the armature current is 40A. Solution: ๐ = 250 ๐๐๐๐ก๐ , ๐ = 500, ๐ ๐ = 1 Ω, Φ = 20 ∗ 10−3 Wb, ๐ผ๐ = 40 A, c = 2P for lap winding ๐ต๐๐๐ ๐. ๐. ๐. ๐ธ = ๐ − (๐ผ๐ ∗ ๐ ๐ ) = 250 ๐๐๐๐ก๐ − (40 ๐ด ∗ 1 Ω) = 210 ๐๐๐๐ก๐ ๐ธ. ๐. ๐. 2๐๐๐๐ 2๐ (20 ∗ 10−3 ๐๐) ๐ (500) ๐ธ= = ๐. ๐. 210 ๐๐๐๐ก๐ = ๐ 2๐ ๐ป๐๐๐๐ ๐ ๐๐๐๐ ๐ = 210 ๐๐๐๐ก๐ ๐๐๐ฃ ๐๐๐ฃ (21 = 21 ๐๐ ∗ 60) = 1260 (20 ∗ 10−3 ๐๐)(500) ๐ ๐๐๐ ๐ก๐๐๐๐ข๐ ๐ = (210 ๐๐๐๐ก๐ )(40 ๐ด) ๐ธ ๐ผ๐ = = 63.66 ๐๐ ๐๐๐ฃ 2๐๐ 2 ๐ (21 ๐ ) 8 Problem 18 The shaft torque of a diesel motor driving a 100V d.c. shunt-wound generator is 25Nm. The armature current of the generator is 16 A at this value of torque. If the shunt field regulator is adjusted so that the flux is reduced by 15%, the torque increases to 35Nm. - Determine the armature current at this new value of torque. Solution: The shaft torque T of a generator is proportional to Ia, where is the flux and Ia is the armature current. Thus, T = k ะค Ia, where k is a constant. The torque at flux ะค1 and armature current Ia1 is T1 = k ะค1 Ia1. Similarly, T2 = k ะค2 Ia2 ๐1 ๐ ∅1 ๐ผ๐1 ∅1 ๐ผ๐1 = = ๐2 ๐ ∅2 ๐ผ๐2 ∅2 ๐ผ๐2 ๐ต๐ฆ ๐๐๐ฃ๐๐ ๐๐๐ ๐ป๐๐๐๐ ๐. ๐. 25 ∅1 ∗ 16 = 35 0.85 ∅1 ∗ ๐ผ๐2 ๐ผ๐2 = 16 ∗ 35 = 26.35 ๐ด 0.85 ∗ 25 That is, the armature current at the new value of torque is 26.35 A. 9 Problem 19 A 100V d.c. generator supplies a current of 15A when running at 1500 rev/min. If the torque on the shaft driving the generator is 12 Nm. Determine: (a) the efficiency of the generator. (b) the power loss in the generator. Solution: The efficiency of the generator = ๐๐ข๐ก๐๐ข๐ก ๐๐๐ค๐๐ ∗ 100 % ๐๐๐๐ข๐ก ๐๐๐ค๐๐ The output power is the electrical output, i.e. V I [Watts]. The input power to a generator is the mechanical power in the shaft driving the generator, i.e. T ω or T (2πn) [Watts], where T is the torque in Nm and n is speed of rotation in rev/s. Hence, for a generator ๐๐๐๐๐๐๐๐๐๐ฆ, ๐. ๐. ๐= ๐= ๐∗๐ผ ∗ 100 % ๐ (2๐๐) (100) (15) ∗ 100 % = 79.6 % 1500 (12) (2๐) ( 60 ) ๐โ๐ ๐๐๐๐ข๐ก ๐๐๐ค๐๐ = ๐๐ข๐ก๐๐ข๐ก ๐๐๐ค๐๐ + ๐๐๐ ๐ ๐๐ ๐ป๐๐๐๐, ๐ (2๐๐) = (๐ ∗ ๐ผ) + ๐๐๐ ๐ ๐๐ ๐. ๐. ๐๐๐ ๐ ๐๐ = ๐ (2๐๐) − (๐ ∗ ๐ผ) ๐ฟ๐๐ ๐ ๐๐ = 12 (2 ๐ 1500 ) − (100 ∗ 15) 60 ๐. ๐. ๐๐๐ค๐๐ ๐๐๐ ๐ ๐๐ = 1885 ๐ − 1500 ๐ = 385 ๐ 10 Problem 20 A 240V shunt motor takes a total current of 30 A. If the field winding resistance Rf =150 โฆ and the armature resistance Ra = 0.4 โฆ. Determine: (a) the current in the armature. (b) the back e.m.f. Solution: ๐น๐๐๐๐ ๐๐ข๐๐๐๐๐ก, ๐ผ๐ = ๐ ๐ ๐ = 240 ๐๐๐๐ก๐ 150 Ω = 1.6 ๐ด ๐๐ข๐๐๐๐ฆ ๐๐ข๐๐๐๐๐ก ๐ผ = ๐ผ๐ + ๐ผ๐ ๐ป๐๐๐๐ ๐๐๐๐๐ก๐ข๐๐ ๐๐ข๐๐๐๐๐ก, ๐ผ๐ = ๐ผ − ๐ผ๐ = 30 ๐ด − 1.6 ๐ด = 28.4 ๐ด ๐ต๐๐๐ ๐. ๐. ๐. ๐ธ = ๐ − ๐ผ๐ ๐ ๐ ๐ธ = 240 ๐๐๐๐ก๐ − (28.4 ๐ด) (0.4 Ω) = 228.64 ๐๐๐๐ก๐ Problem 21 A 200V, d.c. shunt-wound motor has an armature resistance of 0.4 โฆ and at a certain load has an armature current of 30A and runs at 1350 rev/min. If the load on the shaft of the motor is increased so that the armature current increases to 45A, determine the speed of the motor, assuming the flux remains constant. Solution: ๐โ๐ ๐๐๐๐๐ก๐๐๐๐ โ๐๐ ๐ธ ∝ ๐ ๐ ๐๐๐๐๐๐๐ ๐ก๐ ๐๐๐กโ ๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐ ๐๐๐ก๐๐๐ . ๐น๐๐ ๐ ๐๐๐ก๐๐๐ , ๐ธ = ๐ − ๐ผ๐ ๐ ๐ ๐ธ1 = 200 ๐๐๐๐ก๐ − (30 ๐ด) (0.4 Ω) = 188 ๐๐๐๐ก๐ ๐ธ2 = 200 ๐๐๐๐ก๐ − (45 ๐ด) (0.4 Ω) = 182 ๐๐๐๐ก๐ ๐โ๐ ๐๐๐๐๐ก๐๐๐๐ โ๐๐, ๐ธ1 ๐1 ๐1 = ๐ธ2 ๐ 2 ๐2 ๐๐๐๐๐๐๐ ๐ก๐ ๐๐๐กโ ๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐ ๐๐๐ก๐๐๐ . ๐๐๐๐๐ ๐กโ๐ ๐๐๐ข๐ฅ ๐๐ ๐๐๐๐ ๐ก๐๐๐ก, ๐1 = ๐2 1350 ๐1 ∗ ( 188 60 ) , ๐ป๐๐๐๐ = 182 ๐1 ∗ ๐2 ๐. ๐. ๐2 = 22.5 ∗ 182 ๐๐๐ฃ = 21.78 188 ๐ ๐โ๐ข๐ ๐กโ๐ ๐ ๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐ก๐๐ ๐คโ๐๐ ๐กโ๐ ๐๐๐๐๐ก๐ข๐๐ ๐๐ข๐๐๐๐๐ก ๐๐ 45๐ด ๐๐ 21.78 ∗ 60 11 ๐๐๐ฃ ๐๐๐ , ๐. ๐. 1307 ๐๐๐ฃ ๐๐๐ Problem 22 A 220V, d.c. shunt-wound motor runs at 800 rev/min and the armature current is 30 A. The armature circuit resistance is 0.4 โฆ. Determine: (a) the maximum value of armature current if the flux is suddenly reduced by 10 %. (b) the steady state value of the armature current at the new value of flux, assuming the shaft torque of the motor remains constant. Solution: (a) For a d.c. shunt-wound motor, E = V – Ia Ra. Hence initial generated e.m.f., E1 = 220 – 30 × 0.4 = 208V. The generated e.m.f. is also such that E ∝ n, so at the instant the flux is reduced, the speed has not had time to change, and E = 208 × 90/100 = 187.2V. Hence, the voltage drop due to the armature resistance is 220 −187.2, i.e. 32.8V. The instantaneous value of the current is 32.8/0.4, i.e. 82A. This increase in current is about three times the initial value and causes an increase in torque, (T ∝ Ia). The motor accelerates because of the larger torque value until steady state conditions are reached. (b) T ∝ Ia and since the torque is constant, ะค1 Ia1 = ะค2 Ia2. The flux is reduced by 10 %, hence ะค2 = 0.9 ะค1 Thus, ะค1 × 30 = 0.9 ะค1 × Ia2 i.e. the steady state value of armature current, ๐ผ๐2 = 30 1 = 33 ๐ด 0.9 3 12 Problem 23 A series motor has an armature resistance of 0.2 โฆ and a series field resistance of 0.3 โฆ. It is connected to a 240V supply and at a particular load runs at 24 rev/s when drawing 15 A from the supply. (a) Determine the generated e.m.f. at this load. (b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A. Assume that this causes a doubling of the flux. Solution: With reference to the Figure, generated e.m.f., E, at initial load, is given by ๐ธ1 = ๐ − ๐ผ๐ (๐ ๐ + ๐ ๐ ) = 240 ๐๐๐๐ก๐ − [(15๐ด) ∗ (0.2 Ω + 0.3Ω)] = 232.5 ๐๐๐๐ก๐ ๐โ๐๐ ๐กโ๐ ๐๐ข๐๐๐๐๐ก ๐๐ ๐๐๐๐๐๐๐ ๐๐ ๐ก๐ 30 ๐ด, ๐กโ๐ ๐๐๐๐๐๐๐ก๐๐ ๐. ๐. ๐. ๐๐ ๐๐๐ฃ๐๐ ๐๐ฆ: ๐ธ2 = ๐ − ๐ผ๐ (๐ ๐ + ๐ ๐ ) = 240 ๐๐๐๐ก๐ − [(30๐ด) ∗ (0.2 Ω + 0.3Ω)] = 225 ๐๐๐๐ก๐ ๐๐๐ค ๐. ๐. ๐. ๐ธ ∝ Φ๐ ๐ธ1 ๐1 ๐1 = ๐ธ2 ๐ 2 ๐2 ๐โ๐ข๐ ๐. ๐. 232.5 ๐1 ∗ (24) = (2๐1 ) ∗ (๐2 ) 225 ๐ป๐๐๐๐ ๐ ๐๐๐๐ ๐๐ ๐๐๐ก๐๐, ๐2 = ๐ ๐๐๐๐ ๐2 = 2๐1 (24) (225) ๐๐๐ฃ = 11.6 (232.5) (2) ๐ As the current has been increased from 15A to 30A, the speed has decreased from 24 rev/s to 11.6rev/s. Its speed/current characteristic is similar to the Figure. 13 Problem 24 A 320V shunt motor takes a total current of 80 A and runs at 1000 rev/min. If the iron, friction and windage losses amount to 1.5 kW, the shunt field resistance is 40 โฆ and the armature resistance is 0.2 โฆ. - Determine the overall efficiency of the motor. Solution: The circuit is shown in the Figure. ๐น๐๐๐๐ ๐๐ข๐๐๐๐๐ก, ๐ผ๐ = ๐ ๐ ๐ = 320 ๐๐๐๐ก๐ 40 Ω =8๐ด ๐ด๐๐๐๐ก๐ข๐๐ ๐๐ข๐๐๐๐๐ก ๐ผ๐ = ๐ผ − ๐ผ๐ = 80๐ด − 8๐ด = 72๐ด C = iron, friction and windage losses = 1500 W Efficiency, ๐ ∗ ๐ผ − ๐ผ๐2 ๐ ๐ − ๐ผ๐ ∗ ๐ − ๐ถ ๐= ( ) ∗ 100 % ๐∗๐ผ (320)(80)− (722 )(0.2)− (8)(320)−1500 ๐= ( (320)(80) ) ∗ 100 % = 14 20503 25600 ∗ 100 % = 80.1 % Problem 25 A 250V series motor draws a current of 40 A. The armature resistance is 0.15 โฆ and the field resistance is 0.05โฆ. - Determine the maximum efficiency of the motor. Solution: The circuit is as shown in the Figure. The efficiency, ๐ ๐ผ − ๐ผ๐2 ๐ ๐ − ๐ผ๐ ๐ − ๐ถ ๐= ( ) ∗ 100 % ๐๐ผ ๐ป๐๐ค๐๐ฃ๐๐ ๐๐๐ ๐ ๐ ๐๐๐๐๐ ๐๐๐ก๐๐, ๐ผ๐ = 0 ๐๐๐ ๐กโ๐ ๐ผ๐2 ๐ ๐ ๐๐๐ ๐ ๐๐๐๐๐ ๐ก๐ ๐๐ ๐ผ 2 (๐ ๐ + ๐ ๐ ) ๐ ๐ผ − ๐ผ 2 (๐ ๐ + ๐ ๐ ) − ๐ถ ๐ป๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ฆ, ๐ = ( ) ∗ 100 % ๐๐ผ ๐น๐๐ ๐๐๐ฅ๐๐๐ข๐ ๐๐๐๐๐๐๐๐๐๐ฆ ๐ผ 2 (๐ ๐ + ๐ ๐ ) = ๐ถ ๐ ๐ผ − 2๐ผ 2 (๐ ๐ + ๐ ๐ ) ๐ป๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ฆ, ๐ = ( ) ∗ 100 % ๐๐ผ (250)(40) − 2(40)2 (0.15 + 0.05) ๐= ( ) ∗ 100 % (250)(40) ๐= ( ๐= ( 10000 − 640 ) ∗ 100 % 10000 9360 ) ∗ 100 % = 93.6 % 10000 15 Problem 26 A 200V d.c. motor develops a shaft torque of 15 Nm at 1200 rev/min. If the efficiency is 80%. - determine the current supplied to the motor. Solution: The efficiency of a motor = ๐๐ข๐ก๐๐ข๐ก ๐๐๐ค๐๐ ∗ 100 % ๐๐๐๐ข๐ก ๐๐๐ค๐๐ The output power of a motor is the power available to do work at its shaft and is given by T ω or T(2πn) [watts], where T is the torque in Nm and n is the speed of rotation in [rev/s]. The input power is the electrical power in watts supplied to the motor, i.e. VI [watts]. Thus for a motor, efficiency, ๐= i.e. ๐ (2 ๐ ๐) ๐๐ผ ∗ 100 % 80 = [ (15) (2 ๐) ( 1200 ) 60 (200) (๐ผ) ] (100) ๐โ๐ข๐ ๐กโ๐ ๐๐ข๐๐๐๐๐ก ๐ ๐ข๐๐๐๐๐๐, ๐ผ = (15) (2๐) (20) (100) (200) (80) = 11.8 ๐ด Problem 27 A d.c. series motor drives a load at 30 rev/s and takes a current of 10 A when the supply voltage is 400V. If the total resistance of the motor is 2 โฆ and the iron, friction and windage losses amount to 300W, determine the efficiency of the motor. Solution: ๐ ๐ผ − ๐ผ 2 ๐ −๐ถ ๐ธ๐๐๐๐๐๐๐๐๐ฆ, ๐ = ( ๐๐ผ ) ∗ 100 % (400)(10)− (10)2 (2)−300 ๐ธ๐๐๐๐๐๐๐๐๐ฆ, ๐ = ( (400) (10) 4000−200−300 ๐ธ๐๐๐๐๐๐๐๐๐ฆ, ๐ = ( 4000 ) ∗ 100 % ) ∗ 100 % = 87.5 % 16 Problem 28 A 500V shunt motor runs at its normal speed of 10 rev/s when the armature current is 120A. The armature resistance is 0.2 โฆ. (a) Determine the speed when the current is 60A and a resistance of 0.5 โฆ is connected in series with the armature, the shunt field remaining constant. (b) Determine the speed when the current is 60A and the shunt field is reduced to 80 % of its normal value by increasing resistance in the field circuit. Solution: With reference to the Figure, back e.m.f. at 120A, ๐ธ1 = ๐ − ๐ผ๐ ๐ ๐ = 500 ๐ − [(120๐ด) ∗ (0.2 Ω)] = 476 ๐ ๐โ๐๐ ๐ผ๐ = 60 ๐ด, โซโซโซ ๐ธ2 = 500 ๐ − [(60๐ด) ∗ (0.2 Ω + 0.5Ω)] = 458 ๐ ๐๐๐ค ๐ธ1 ๐1 ๐1 = ๐ธ2 ๐ 2 ๐2 476 ๐1 (10) = 458 ๐ 2 ( ๐2 ) ๐. ๐. ๐๐๐๐ ๐๐๐๐๐, ๐๐๐๐๐ ๐๐ = ๐ ๐๐๐๐ ∅2 = ∅1 (๐๐) (๐๐๐) ๐๐๐ = ๐. ๐๐ [ ] (๐๐๐) ๐ ๐ฉ๐๐๐ ๐. ๐. ๐. ๐๐๐๐ ๐ฐ๐ = ๐๐ ๐จ, ๐ธ2 = 500 ๐ − [(60๐ด) ∗ (0.2 Ω)] = 488 ๐ ๐๐๐ค ๐ธ1 ๐1 ๐1 = ๐ธ2 ๐2 ๐2 ๐. ๐. 476 ๐1 (10) = 488 0.8 ๐1 (๐3 ) ๐๐๐๐ ๐๐๐๐๐, ๐๐๐๐๐ ๐๐ = ๐ ๐๐๐๐ ∅2 = 0.8 ∅1 (๐๐) (๐๐๐) ๐๐๐ = ๐๐. ๐๐ [ ] (๐. ๐)(๐๐๐) ๐ 17 Problem 29 On full-load a 300 V series motor takes 90 A and runs at 15 rev/s. The armature resistance is 0.1โฆ and the series winding resistance is 50 mโฆ. - Determine the speed when developing full load torque but with a 0.2 โฆ diverter in parallel with the field winding. (Assume that the flux is proportional to the field current.) Solution: ๐ด๐ก 300 ๐, ๐. ๐. ๐. ๐ธ1 = ๐ − ๐ผ๐ = ๐ − ๐ผ(๐ ๐ + ๐ ๐ ๐ ) = 300 − (90)(0.1 + 0.05) = 300 − 13.5 = 286.5 ๐ฃ๐๐๐ก๐ ๐๐๐กโ ๐กโ๐ 0.2 Ω ๐๐๐๐ฃ๐๐ ๐๐ ๐๐๐๐๐๐๐๐ ๐ค๐๐กโ ๐ ๐ ๐ (0.2)(0.05) (0.2)(0.05) = = 0.04 Ω (0.2) + (0.05) (0.25) 0.2 ๐ต๐ฆ ๐๐ข๐๐๐๐๐ก ๐๐๐ฃ๐๐ ๐๐๐, ๐๐ข๐๐๐๐๐ก ๐ผ1 = ( ) ๐ผ = 0.8 ๐ผ 0.2 + 0.05 ๐กโ๐ ๐๐๐ข๐๐ฃ๐๐๐๐๐ก ๐๐๐๐ ๐ก๐๐๐๐, ๐ = ๐๐๐๐๐ข๐, ๐ ๐ผ ๐ผ๐ ∅ ๐๐๐ ๐๐๐ ๐๐ข๐๐ ๐๐๐๐ ๐ก๐๐๐๐ข๐, ๐ผ๐1 ∅1 = ๐ผ๐2 ∅2 ๐ ๐๐๐๐ ๐๐๐ข๐ฅ ๐๐ ๐๐๐๐๐๐๐ก๐๐๐๐๐ ๐ก๐ ๐๐๐๐๐ ๐๐ข๐๐๐๐๐ก ∅1 ∝ ๐ผ๐1 ๐๐๐ ∅2 ∝ 0.8 ๐ผ๐2 ๐โ๐๐ (90)(90) = (๐ผ๐2 )(0.8 ๐ผ๐2 ) ๐๐๐๐ 2 ๐คโ๐๐โ, ๐ผ๐2 (90)2 90 = ๐๐๐ ๐ผ๐2 = = 100.62 ๐ด 0.8 √(0.8) ๐ป๐๐๐๐ ๐. ๐. ๐. ๐ธ2 = ๐ − ๐ผ๐2 (๐ ๐ + ๐ ) = 300 − (100.62)(0.1 + 0.04) = 300 − (100.62)(0.14) = 300 − 14.087 = 285.9 ๐ฃ๐๐๐ก๐ ๐๐๐ค ๐. ๐. ๐. ๐ธ ∝ ∅๐ ๐๐๐๐ ๐คโ๐๐โ, ๐ป๐๐๐๐ ๐ธ1 ∅1 ๐1 ๐ผ๐1 ๐1 = = ๐ธ2 ∅2 ๐2 0.8 ๐ผ๐2 ๐2 (286.5) (90)(15) = 285.9 (0.8)(100.62) ๐2 (285.9)(90)(15) = 16.74 ๐๐๐ฃ/๐ (286.5)(0.8)(100.62) ๐๐๐ฃ ๐โ๐ข๐ ๐กโ๐ ๐ ๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐ก๐๐ โ๐๐ ๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐ 15 ๐ ๐๐ ๐๐๐ ๐๐๐ค ๐ ๐๐๐๐, ๐2 = ๐. ๐. 900 ๐๐๐ฃ/๐๐๐ ๐ก๐ 16.74 ๐๐๐ฃ/๐ ๐๐ i.e., 1004 rev/min by inserting a 0.2 โฆ diverter resistance in parallel with the series winding. 18 Problem 30 A series motor runs at 800 rev/min when the voltage is 400 V and the current is 25 A. The armature resistance is 0.4 โฆ and the series field resistance is 0.2 โฆ. - Determine the resistance to be connected in series to reduce the speed to 600 rev/min with the same current. Solution: With reference to the Figure, at 800 rev/min, ๐. ๐. ๐. , ๐ฌ๐ = ๐ฝ − ๐ฐ (๐น๐ + ๐น๐๐ ) = ๐๐๐ ๐ฝ − (๐๐๐จ) ∗ (๐. ๐๐ + ๐. ๐๐) = ๐๐๐ ๐ฝ At 600 rev/min, since the current is unchanged, the flux is unchanged. ๐โ๐ข๐ ๐ธ ∝ ∅ ๐, ๐๐ ๐ธ ∝ ๐, ๐๐๐ ๐ธ1 ๐1 = ๐ธ2 ๐2 ๐๐๐๐ ๐๐๐๐๐, ๐ฌ๐ = ๐๐๐ ๐ฏ๐๐๐๐ ๐๐๐ ๐๐๐ = ๐ฌ๐ ๐๐๐ (๐๐๐) (๐๐๐) = ๐๐๐. ๐๐ ๐ฝ (๐๐๐) ๐ธ2 = ๐ − ๐ผ (๐ ๐ + ๐ ๐ ๐ + ๐ ) ๐ป๐๐๐๐ 288.75 = 400 ๐ − [(25๐ด) ∗ (0.4 Ω + 0.2Ω + R)] ๐ ๐๐๐๐๐๐๐๐๐๐ ๐๐๐ฃ๐๐ : 0.6 + ๐ = 400 − 288.75 = 4.45 25 ๐๐๐๐ ๐๐๐๐๐, ๐๐๐๐๐ ๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐, ๐น = ๐. ๐๐ − ๐. ๐ = ๐. ๐๐ ๐ Thus, the addition of a series resistance of 3.85 โฆ has reduced the speed from 800 rev/min to 600 rev/min. 19 Appendix 20 [๐ ] ๐ธ = ๐๐ ∅ ๐๐ ๐= ๐ ๐ธ ๐ผ๐ = = ๐๐ ∅ ๐ผ๐ ๐๐ ๐๐ [๐. ๐] The principal losses of machines are: (i) Copper loss, due to I 2R heat losses in the armature and field windings. (ii) Iron (or core) loss, due to hysteresis and eddy current losses in the armature. This loss can be reduced by constructing the armature of silicon steel laminations having a high resistivity and low hysteresis loss. At constant speed, the iron loss is assumed constant. (iii) Friction and windage losses, due to bearing and brush contact friction and losses due to air resistance against moving parts (called windage). At constant speed, these losses are assumed to be constant. (iv) Brush contact loss between the brushes and commutator. This loss is approximately proportional to the load current. The total losses of a machine can be quite significant and operating efficiencies of between 80% and 90% are common. Efficiency of a d.c. generator The efficiency of an electrical machine is the ratio of the output power to the input power and is usually expressed as a percentage. The Greek letter ‘η’ (eta) is used to signify efficiency and since the units are power/power, then efficiency has no units. Thus 21 If the total resistance of the armature circuit (including brush contact resistance) is Ra, then the total loss in the armature circuit is (๐ฐ๐๐ ∗ ๐น๐ ) If the terminal voltage is V and the current in the shunt circuit is If, then the loss in the shunt circuit is (If *V). If the sum of the iron, friction and windage losses is C then the total losses is given by: If the output current is I, then the output power is VI. Total input power = ๐ ๐ผ + ๐ผ๐2 ๐ ๐ + ๐ผ๐ ๐ + ๐ถ. Hence The efficiency of a generator is a maximum when the load is such that: i.e. when the variable loss = the constant loss 22 The efficiency of a d.c. motor the efficiency of a d.c. machine is given by: Also, the total losses = ๐ผ๐2 ๐ ๐ + ๐ผ๐ ๐ + ๐ถ (for a shunt motor) where C is the sum of the iron, friction and windage losses. For a motor, the input power = VI and the output power = VI – losses Hence The efficiency of a motor is a maximum when the load is such that: Also, the total losses = ๐ผ๐2 (๐ ๐ + ๐ ๐ ) + ๐ถ (for a series motor) where C is the sum of the iron, friction and windage losses. 23