Example 1: A 230V dc shunt motor drives a load at 900RPM and drawing a current of 30A. The resistance of armature circuit is 0.4Ω. The torque of the load is proportional to the speed. Calculate the resistance to be connected in series with the armature to reduce the speed to 600RPM. Ignore armature reaction. (ex.6.7p.158) SOLUTION Case I: Supply voltage: Armature current: Speed: Back emf: Case II: Supply voltage: Armature current: Speed: Back emf: V T = 230 V I a1 = 30 A (since no information is provided about the field circuit, otherwise this current should be considered as the total current withdrawn from the source, I T ) n1 = 900 RPM E b 1 =V T − I a1R a = 230 − 30 × 0.4 = 218 V E b 1 = k b φ n1 = k b φ 900 V T = 230 V I a2 = ? n 2 = 600 RPM E b 2 =V T − I a 2 ( R a + R ) = 230 − I a 2 ( 0.4 + R ) E b 2 = k b φ n 2 = k b φ 600 Since the load is proportional to the speed ( τ ∝ n ) and ( τ ∝ I a ) then ( n ∝ I a ). I φ n n 600 Hence, a 2 = 2 or I a 2 = I a1 × 2 = 30 × = 20A n1 900 I a1φ n1 E b 2 =V T − I a 2 (0.4 + R ) = 230 − 20(0.4 + R ) = 222 − 20R ⇒ E n 222 − 20R 600 = Since 2 = b 2 (magnetic flux is constant) ⇒ n1 E b 1 198 900 or R = 3.833Ω Example 2: A 250V dc shunt motor has an armature current of 20A when running at 1000RPM against full load torque. The armature resistance is 0.5Ω. What resistance must be inserted in series with the armature to reduce the speed to 500RPM at the same load torque, and what will be the speed if the load torque is halved with this inserted resistance? (ex.6.8p.159) SOLUTION Case I: Supply voltage: Armature current: Speed: Back emf: Case II: Supply voltage: Armature current: Speed: Back emf: V T = 250 V I a1 = 20 A n1 = 1000 RPM E b 1 =V T − I a1R a = 250 − 20 × 0.5 = 240 V E b 1 = k b φ n1 = k b φ 1000 V T = 250 V I a 2 = 20 A n 2 = 500 RPM (since the load torque remains constant at full load) E b 2 =V T − I a 2 ( R a + R ) = 250 − 20 ( 0.5 + R ) = 240 − 20R E b 2 = k b φ n 2 = k b φ 500 Since n2 Eb 2 (magnetic flux is constant) = n1 E b 1 Case III Supply voltage: Armature current: V T = 250 V I a 3 = 10 A Speed: n3 = ? Back emf: ⇒ 240 − 20R 500 = 240 1000 or R = 6Ω (since the load torque is halved: τ Load = k m I a1φ 1 1 I a3 τ Load = k m I a 3φ ⇒ = ) 2 2 I a1 E b 3 =V T − I a 3 ( R a + R ) = 250 − 10 ( 0.5 + 6 ) = 185 V E b 3 = k bφ n3 Since n3 E b 3 (magnetic flux is constant) = n1 E b 1 ⇒ n 185 = 3 240 1000 or n 3 = 771 RPM and Example 3: A 250V dc shunt motor has an armature resistance of 0.5Ω and a field resistance of 250Ω. When driving a constant load at 600RPM, the armature takes 20A. If it is desired to raise the speed from 600RPM to 800RPM, what resistance must be inserted in the shunt field circuit so that the change of speed occurs? Assume the magnetizing curve is a straight line. (ex.6.1p.151) SOLUTION Case I: Supply voltage: Armature current: Speed: Back emf: V T = 250 V I a1 = 20 A n1 = 600 RPM E b 1 =V T − I a1R a = 250 − 20 × 0.5 = 240 V E b 1 = k b φ1 n1 = k b φ1 600 Case II: Now let the magnetic flux and armature current be φ 2 and I a 2 , respectively. Supply voltage: V T = 250 V Armature current: I a2 = ? Speed: n 2 = 800 RPM Back emf: E b 2 =V T − I a 2 R a = 250 − 0.5I a 2 E b 2 = k b φ2 n 2 = k b φ2 800 Since the load is constant ∴ τ Load = k m I a 2φ 2 = k m I a1φ 1 and n 2 E b 2 φ1 = × n1 E b 1 φ2 since I a 2 ⇒ φ = 20 1 φ2 I a 2 = I a1 or φ1 φ = 20 1 φ2 φ2 800 250 − 0.5I a 2 φ1 = × φ2 600 240 or or ⎛ φ ⎞ φ 320 = ⎜ 250 − 10 1 ⎟ × 1 φ2 ⎠ φ2 ⎝ 2 ⇒ ⎛ φ1 ⎞ φ1 ⎜ ⎟ − 25 + 32 = 0 φ2 ⎝φ 2 ⎠ φ 1 25 ± 252 − 4 × 32 = = 23.65 φ2 2 or ⇒ we have a second order equation 1.35 The value 23.65 is rejected as it will not give the required increase in speed. Since magnetizing curve is a straight line φ1 I sh 1 V T / R sh 1 = = φ2 I sh 2 V T / R sh 2 R sh 2 φ1 = = 1.352 ⇒ R sh 1 φ2 ∴ since in general or V T = I sh R sh and φ = c I sh where c = const . R sh 2 = 1.353 × R sh 1 = 1.353 × 250 = 338Ω Hence, the resistance required to be inserted in the shunt field is R = R sh 2 − R sh 1 = 338 − 250 = 88Ω .