ECE 482 Electrical Drive Systems Topic: DC Drives – Rectifier controlled -Assignment_4-Questions And

downloaded from

Scholarfriends

https://scholarfriends.com Downloaded by: writerforacademic@gmail.com | https://scholarfriends.com/singlePaper/310923/ece-482-electrical-drive-systems-topic-dcdrives-rectifier-controlled-assignment-4-questions-a Distribution of this document is illegal.

ECE 482 Electrical Drive Systems Topic: DC Drives – Rectifier controlled

1

Q1 A 200 V, 875 rpm, 150 A separately excited dc motor has an armature resistance of 0.06 Ω. It is fed from a single phase fully controlled rectifier with an ac voltage source of 220 V, 50 Hz. Assuming continuous conduction, calculate:

(i) Firing angle for rated torque and 750 rpm (ii) Firing angle for rated torque and (-500) rpm (iii) Motor speed for α = 160˚ and rated torque 2

Q1 - Answer At rated operating condition

𝐸𝑏 = 𝑉 − 𝐼𝑎 𝑅𝑎 = 200 − 150 × 0.06 = 191 𝑉 (i) Assuming constant excitation for all operating conditions, when the motor is operating at rated torque & 750 rpm,

𝐼𝑎1 = 150 𝐴 750 𝐸𝑏1 = × 191 = 163.7 𝑉 875

𝑉𝑎 = 𝐸𝑏1 + 𝐼𝑎1 × 0.06 = 172.7 𝑉 3

Q1 - Answer Average voltage of a full bridge rectifier operating in continuous conduction mode is 2 𝑉𝑚 cos(𝛼) 𝜋

Therefore, 2 𝑉𝑚 𝑉𝑎 = 172.7 𝑉 = cos(𝛼) 𝜋

Thus, firing delay angle 𝛼= 29.3˚

4

Q1 - Answer (ii) When the motor is operating at rated torque & (-500) rpm, 𝐼𝑎 = 150 𝐴 −500 𝐸𝑏 = × 191 = −109 𝑉 875

𝑉𝑎 = 𝐸𝑏 + 𝐼𝑎 × 0.06 = −100 𝑉 2 𝑉𝑚 𝑉𝑎 = −100 = cos(𝛼) 𝜋

Thus, firing delay angle 𝛼= 120˚

5

Q1 - Answer (iii) When the motor is operating at rated torque & the firing delay angle is 160˚, 𝐼𝑎 = 150 𝐴 2 𝑉𝑚 𝑉𝑎 = cos 𝛼 𝜋

= −186 𝑉

𝑉𝑎 = −100 𝑉 = 𝐸𝑏 +𝐼𝑎 × 0.06 𝐸𝑏 = −195 𝑉 −195 𝑁= × 875 = −893.2 𝑟𝑝𝑚 191

6

Q2 A 220 V, 1500 rpm, 10 A separately excited dc motor is fed from a single phase fully controlled rectifier with an ac voltage source of 230 V, 50 Hz. Armature resistance = 2 Ω. Assuming continuous conduction, calculate:

(i) Firing angle for rated torque and 500 rpm (ii) Firing angle for rated torque and (-1000) rpm

7

Q2 - Answer At rated condition 𝐸𝑏 = 𝑉 − 𝐼𝑎 𝑅𝑎 = 220 − 10 × 2 = 200 𝑉 (i) Assuming constant excitation for all operating conditions, when the motor is operating at rated torque & 500 rpm, 500 𝐸𝑏1 = × 200 = 67 𝑉 1500

𝑉 = 𝐸𝑏1 + 𝐼𝑎 𝑅𝑎 = 87 V 8

Q2 - Answer Average voltage of a full bridge rectifier operating in continuous conduction mode is 2 𝑉𝑚 cos(𝛼) 𝜋

Therefore, 2 𝑉𝑚 𝑉𝑎 = 87 𝑉 = cos(𝛼) 𝜋

Thus, firing delay angle 𝛼= 65.1˚ 9

Q2 - Answer (ii) When the motor is operating at rated torque & (-1000) rpm, 𝐼𝑎 = 10 𝐴 −1000 𝐸𝑏 = × 200 = −133.3 𝑉 1500

𝑉𝑎 = 𝐸𝑏 + 𝐼𝑎 × 2 = −113.3 𝑉 2 𝑉𝑚 𝑉𝑎 = −113.3 = cos(𝛼) 𝜋

Thus, firing delay angle 𝛼= 123.2˚

10

Q3 A 220 V, 1200 rpm, 15 A separately excited dc motor has armature resistance and inductance of 1.8 Ω and 32 mH respectively. The motor is fed from a single phase fully controlled rectifier with an ac voltage source of 230 V, 50 Hz. Assuming continuous conduction, calculate:

Speed for α = 45˚ and torque = 40 Nm

11

Q3 - Answer At rated condition 𝐸𝑏 = 𝑉 − 𝐼𝑎 𝑅𝑎 = 220 − 15 × 1.8 = 193 𝑉 𝐸𝑏 𝐼𝑎 𝑇= = 23 𝑁𝑚 𝜔

12

Q3 - Answer When the motor is operating at torque of 40 Nm & the firing delay angle is 68.3˚,

𝐼𝑎 = 26 𝐴 2 𝑉𝑚 𝑉𝑎 = cos 𝛼 𝜋

= 146.4 𝑉

𝑉𝑎 = 146.4 𝑉 = 𝐸𝑏 +𝐼𝑎 × 1.8 𝐸𝑏 = 99.6 𝑉 99.6 𝑁= × 1200 = 619 𝑟𝑝𝑚 193

13

Q4 A 220 V, 600 rpm, 500 A separately excited dc motor has an armature resistance of 0.02 Ω and field resistance of 10 Ω. The load torque is given by the expression: TL = 2000-2N (Nm). Calculate

Flux as a percent of its rated value when the motor speed is 750 rpm

14

Q4 - Answer Rated back emf E1 = 220 – 500 x 0.02 = 210 V Rated speed ωm1 = 62.83 rad/s At rated condition, Keφ1 = 210/62.83 = 3.342 V-s/rad At 750 rpm, ωm2 = 78.54 rad/s At 750 rpm, let armature current be Ia2 & flux be φ2 Back emf E2 at 750 rpm = 78.54 Keφ2 Since armature voltage is fixed at rated value at 750 rpm 220 = 78.54 Keφ2 + 0.02 x Ia2 Torque at 750 rpm obtained using load torque equation = 500 Nm = Keφ2 Ia2 Solving the two equations, Ia2 = 181.5 A & Keφ2 = 2.755 V-s/rad Thus, φ2 as a percent of rated flux = 82.4%

15

Q5 A 220 V, 750 rpm, 200 A separately excited dc motor has an armature resistance of 0.05 Ω. The load torque is given by the expression: TL = 500-0.25N (Nm). Calculate

Flux as a percent of its rated value when the motor speed is 1500 rpm

16

Q5 - Answer Rated back emf E1 = 220 – 200 x 0.05 = 210 V Rated speed ωm1 = 78.54 rad/s At rated condition, Keφ1 = 210/78.54 = 2.674 V-s/rad At 1500 rpm, ωm2 = 157 rad/s At 1500 rpm, let armature current be Ia2 & flux be φ2 Back emf E2 at 1500 rpm = 157 Keφ2 Since armature voltage is fixed at rated value at 1500 rpm 220 = 157 Keφ2 + 0.05 x Ia2 Torque at 1500 rpm obtained using load torque equation = 125 Nm = Keφ2 Ia2 Solving the two equations, Ia2 = 91.1 A & Keφ2 = 1.372 V-s/rad Thus, φ2 as a percent of rated flux = 51.3%

17

Q6 A 220 V, 750 rpm, 200 A separately excited dc motor has an armature resistance of 0.05 Ω. The field winding is also rated for 220 V and has field resistance of 20 Ω.

The load torque is given by the expression: TL = 500-0.2N (Nm). Speeds below rated value are obtained by armature voltage control with full field and speeds above rated are obtained by field control at rated armature voltage.

Armature is supplied from a single phase Thyristor based full bridge rectifier from a supply of 250 V, 50 Hz. Field coil of the motor is also fed from a single phase Thyristor based full bridge rectifier from a 250 V, 50 Hz supply.

If the drive operates under continuous conduction mode, calculate the firing delay angle when drive operates at (i) 600 rpm and (ii) 1200 rpm 18

Q6 - Answer At rated operating condition 𝐸𝑏 = 𝑉 − 𝐼𝑎 𝑅𝑎 = 220 − 200 × 0.05 = 210 𝑉; 𝑇=

𝐸𝑏 𝐼𝑎 = 534.76 𝑁𝑚 𝜔

(i) when the drive is operating at 600 rpm, 𝑇𝐿 = 500 − 0.2 × 600 = 380 𝑁𝑚 At 600 rpm, excitation remains same as at rated conditions. Hence, 𝐼𝑎1 =

380 × 200 = 142.12 𝐴 534.76

𝐸𝑏1 =

600 × 210 = 168 𝑉 750

𝑉𝑎 = 𝐸𝑏1 + 𝐼𝑎1 × 0.05 = 175.1 𝑉

19

Q6 - Answer For the armature converter 2 𝑉𝑚 𝑉𝑎 = 175.1 𝑉 = cos 𝛼 𝜋

2×250√2 = cos 𝛼 𝜋

Thus, firing delay angle 𝛼= 38.927˚ For the field converter

2 𝑉𝑚 2 × 250√2 𝑉𝑓 = 220 𝑉 = cos 𝛼 = cos 𝛼 𝜋 𝜋 Thus, firing delay angle 𝛼= 12.195˚ 20

Q6 - Answer At rated operating condition 𝐸𝑏 = 𝑉 − 𝐼𝑎 𝑅𝑎 = 220 − 200 × 0.05 = 210 𝑉 𝐸𝑏 𝐾𝑒 ∅1 = = 2.6738 𝜔

𝑉 − 𝑠/𝑟𝑎𝑑

(i) when the drive is operating at 1200 rpm, 𝑇𝐿 = 500 − 0.2 × 600 = 260 𝑁𝑚 At 1200 rpm, excitation changes from its rated value. Hence,

𝐼𝑎2 =

260 𝐾𝑒 ∅2

𝐴

𝐸𝑏2 = 125.66𝐾𝑒 ∅2 𝑉

21

Q6 - Answer Above rated speed, the armature voltage is fixed at its rated value.

Hence, 220 = 125.66 𝐾𝑒 ∅2 + 0.05 ×

260 𝐾𝑒 ∅2

Solving the above equation, 𝐼𝑎2 = 153.89 𝐴 𝐾𝑒 ∅2 = 1.6895 𝑉 − 𝑠/𝑟𝑎𝑑 Flux reduction from its rated value is 63.19% 22

Q6 - Answer For the armature converter 2 𝑉𝑚 𝑉𝑎 = 220 𝑉 = cos 𝛼 𝜋

2×250√2 = cos 𝛼 𝜋

Thus, firing delay angle 𝛼= 12.195˚ For the field converter

2 𝑉𝑚 2 × 250√2 𝑉𝑓 = 0.6319 × 220 𝑉 = cos 𝛼 = cos 𝛼 𝜋 𝜋 Thus, firing delay angle 𝛼= 51.846˚ 23

Q7 A 220 V, 1500 rpm, 10 A separately excited DC motor has an armature resistance of 10 Ω. It is fed from a single phase fully controlled bridge rectifier with an AC source voltage of 230 V, 50 Hz. Assuming continuous conduction, compute (i) The motor speed at firing angle of 30˚ and torque of 5 Nm. (ii) Developed torque at firing delay angle of 45˚ and speed of 1000 rpm.

24

Q7 - Answer At rated operating condition 𝐸𝑏 = 𝑉 − 𝐼𝑎 𝑅𝑎 = 220 − 10 × 10 = 120 𝑉; 𝐸𝑏 𝐼𝑎 𝑇= = 7.64 𝑁𝑚 𝜔

(i) When the drive has firing angle of 30˚ and develops torque of 5 Nm. Assuming constant excitation, 𝐼𝑎1 = 6.545 𝐴 2 𝑉𝑚 𝑉𝑎1 = cos 𝛼 = 179.33 𝑉 𝜋

𝐸𝑏1 = 𝑉𝑎1 − 𝐼𝑎1 𝑅𝑎 = 113.88 𝑉

Speed 𝑁1 = 1423.5 rpm

25

Q7 - Answer Note: When the drive has firing angle of 30˚ and develops torque of 5 Nm, there is no guarantee the motor speed will be below rated speed.

Assuming the operating speed to be above rated speed, then excitation changes from its rated value. Also, armature current is unknown. Armature voltage will be rated voltage.

It is not possible to determine the speed without knowing either the armature current or the changed excitation value.

Hence, based on the information given, one can solve the problem only assuming the speed for the new operating condition is within rated speed.

26

Q7 - Answer (ii) When the firing delay angle of 45˚ and speed of 1000 rpm.

Assuming constant excitation, 2 𝑉𝑚 𝑉𝑎2 = cos 𝛼 𝜋

𝐸𝑏2 =

= 146.4 𝑉

1000 × 120 = 80 𝑉 1500

𝑉𝑎2 −𝐸𝑏2 𝐼𝑎2 = = 6.64 𝐴 𝑅𝑎

Torque under this operating condition would be 5.07 Nm 27

Q8 A 220 V, 1000 rpm, 10 A separately excited DC motor has an armature resistance of 1 Ω. Both the armature and the field windings are fed from single phase fully controlled bridge rectifiers with an AC source voltage of 250 V, 50 Hz. Assuming continuous conduction, Compute the motor speed when the armature current is 7 A and torque of 10 Nm.

28

Q8 - Answer At rated operating condition 𝐸𝑏 = 𝑉 − 𝐼𝑎 𝑅𝑎 = 220 − 10 × 1 = 210 𝑉; 𝐸𝑏 𝐼𝑎 𝑇= = 20 𝑁𝑚 𝜔

When the motor draws 7 A and develops torque of 10 Nm. As torque and current under new operating condition are not in same proportion as under rated conditions, the excitation of the system is changed.

For new torque and current, Keφ2 = 1.4286 V-s/rad Assuming armature voltage is fixed at rated value

220 = 1.4286 𝜔2 + 1 x 7 𝜔2 = 149.1 rad/s

𝑁2 = 1423.77 rpm

29