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Electric Machines snc Generators, Lecture notes of Electric Machines

Electric and electronic engineering course

Typology: Lecture notes

2022/2023

Uploaded on 10/26/2024

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EEE 316
ELECTRICAL MACHINERY II
Haliรง University
Electrical and Electronics Engineering
2023-2024 Spring Semester
WEEK 14
06.06.2024
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EEE 316

ELECTRICAL MACHINERY II

Haliรง University Electrical and Electronics Engineering 2023 - 2024 Spring Semester WEEK 14 06.06.

QUESTION:

  • A shunt dc generator has an armature resistance of ๐‘…! = 0. 4 ฮฉ , a field resistance of ๐‘…" = 125 ฮฉ, a rated armature current of ๐ผ!,$ = 30 A at a speed of 800 rpm, an induced emf of ๐ธ = 211 V at no-load.
  • (a) While terminal voltage is 200 V, what will be the output power to keep the induced emf ๐ธ constant.
  • (b) Since the sum of core losses and mechanical losses are 14% of the output power, what will be the power of the prime mover coupled to the generator operating at full-load to meet these losses and the losses in the field circuit. (๐ธ is same as in partโ€“a.)
  • @ full load : ๐ผ! = 30 A then ๐‘‰ = ๐ธ โˆ’ ๐ผ!๐‘…! = 211 โˆ’ 30 ร— 0. 4 = 199 V
  • ๐ผ& = 30 โˆ’ ( ('/

= 28. 4 A

โ€ข ๐‘ƒ% = ๐‘‰ ๐ผ& = 199 ร— 28. 4 = 5651. 6 W

' ๐‘…! + ๐ผ" ' ๐‘…9: = 30 ' ร— 0. 4 + 1. 6 '

ร— 125 = 676. 8 W

(, (**

= 1. 14 ร— 5651. 6 + 676. 8 = 7123 W

  • ๐‘ƒ9:!"; > 7123 W

QUESTION

  • A series dc motor has an armature resistance of ๐‘…! = 0. 4 ฮฉ , a field resistance of ๐‘…" = 0. 1 ฮฉ, a rated armature current of ๐ผ!,$ = 20 A at a speed of 1100 rpm, an applied terminal voltage of ๐‘‰ = 200 V. Magnetic saturation of the core is negligible. Calculate,
  • (a) the developed torque and speed of the motor when the current of the motor becomes 10 A.
  • (b) the motor efficiency, since the sum of core losses and mechanical losses are 8% of the input power, while the current is 10 A.

= 200 โˆ’ 10 0. 4 + 0. 1 = 195 V @๐ผ! = 10 A

โ€ข 195 = 0. 0824 ร— 10 ร—๐œ”

  • ๐œ” = 236 rad/s โ‡’ ๐‘› = C* 'D ๐œ” = 2260 rpm @๐ผ! = 10 A
  • b) ๐ผ! = 10 A โ‡’ ๐‘ƒ 1 + ๐‘ƒ 234 = 8% ๐‘ƒ 0
  • ๐‘ƒ 0 = ๐‘‰ ๐ผ! = 200 ร— 10 = 2000 W
  • โ‡’ ๐‘ƒ 1 + ๐‘ƒ 234 = 160 W
  • ๐‘ƒ 17 = ๐‘ƒ 17 ,! + ๐‘ƒ 17 ," = ๐ผ! ' ๐‘…! + ๐‘…9> = 10 ' 0. 4 + 0. 1 = 50 W

โ€ข ๐‘ƒ% = ๐‘ƒ9:!"; = 1790 W

(^66) (^67)

(E8* '***

๐‘ƒ^ ๐‘ƒ$%^8 ' ๐‘ƒ'( ๐‘ƒ) = ๐‘ƒ-./0* ๐‘ƒ+ ๐‘ƒ 1

โ€ข SOLUTION:

  • (a)
  • ๐‘‰ = 120 ๐‘‰, ๐‘…9: = 15 ฮฉ, ๐‘› = 600๐‘Ÿ๐‘๐‘š โ†’ ๐ผ =? ๐ผ" = - .!"

('* (/

  • From the table; ๐ผ" = 8 ๐ด โ†’ ๐ธ = 126 ๐‘‰
  • ๐ผ! = ?)- . 2

('C)('* .'

  • (b)
  • ๐‘‰ = 144 ๐‘‰, ๐‘…!" = 18 ฮฉ, ๐‘› = 700 ๐‘Ÿ๐‘๐‘š โ†’ ๐ผ =? ๐ผ# = $ %!" = &'' &( = 8 ๐ด
  • From the table ๐ผ# = 8 ๐ด โ†’ ๐ธ = 126 ๐‘‰ @๐‘› = 600๐‘Ÿ๐‘๐‘š
  • ๐ธ = ๐‘˜)๐œ™๐‘› โ†’
    • = ๐‘๐‘œ๐‘›๐‘ ๐‘ก *# +# = *$ +$ โ†’ ๐ธ, = ๐ธ& +$ +# = 126 -.. /.. = 147 ๐‘‰
  • ๐ผ 0 =
    • 1 $ %% = &'- 1 &'' ..., = 150 ๐ด ๐ผ = ๐ผ 0 โˆ’ ๐ผ# = 150 โˆ’ 8 = 142 ๐ด

โ€ข SOLUTION:

โ€ข J

โ†’ Thus speed cannot be controlled by field control methods.

  • Speed can be controlled by armature control methods;
    • i) adjusting the applied voltage V.
    • ii) adjusting the armature resistance Ra.
  • i)
  • (a) 1 2 ๐‘› ๐‘Ÿ๐‘๐‘š ๐ผ! ๐ด 1400 1200 ๐ผ!" ๐ผ!# ๐‘‰" ๐‘‰#
  • (b)
  • ๐‘ƒ 1 =? ๐œ‚ = (^2)! (^2) " โ‡’ 0. 82 = 34 ร— 678 (^2) " โ†’ ๐‘ƒ 1 = 8975. 6 ๐‘Š
  • ๐‘ƒ 1 = ๐‘ƒ 9 + ๐‘ƒ: = ๐‘ƒ 9 + ๐‘‰ ๐ผ:
  • Operating condition at point 1:
  • ๐‘ƒ 1 = ๐‘ƒ 9 + ๐‘‰ 3 ๐ผ:, 3
    1. 6 = 100 + 200 ๐ผ:, 3 โ‡’ ๐ผ:, 3 = 44. 38 ๐ด ๐‘ƒ 0 โ‡’ โ‡^ ๐‘ƒ/
  • (c)
  • ๐‘ƒ9:!"; = ๐‘ƒ= โˆ’ ๐‘ƒ" 3 F + ๐‘ƒ 1 = ๐ธ ๐ผ! โˆ’ ๐‘ƒ" 3 F + ๐‘ƒ 1 1%G9;
  • ๐ธ(๐ผ!,( โˆ’ ๐‘ƒ9:!";,( = ๐ธ'๐ผ!,' โˆ’ ๐‘ƒ9:!";,' 200 โˆ’ 44. 38 ร— 0. 5 44. 38 โˆ’ 7360 = 152. 4 ร— 50. 03 โˆ’ ๐‘ƒ9:!";,' โ‡’ ๐‘ƒ9:!";,' = 7093. 4 W
  • ๐‘‡9:!";,' = (^6) !"2:;, 3 A 3

E8H., 34 5# ('* = 56. 44 Nm ๐‘ƒ^ ๐‘ƒ$%^8 ' ๐‘ƒ'( ๐‘ƒ) = ๐‘ƒ-./0* ๐‘ƒ+ ๐‘ƒ 1

  • ii)
  • (a)
  • Operating condition at point 1: , ๐‘˜>๐œ™ = 0. 127 ๐‘‰/๐‘Ÿ๐‘๐‘š
  • Operating condition at point 2 :
  • ๐‘ƒ 0 = ๐‘๐‘œ๐‘›๐‘ ๐‘ก , ๐‘‰ = ๐‘๐‘œ๐‘›๐‘ ๐‘ก โ‡’ ๐‘ƒ 0 = ๐‘ƒ" + ๐‘‰๐ผ! โ‡’ ๐ผ! = ๐‘๐‘œ๐‘›๐‘ ๐‘ก 1 2 ๐‘› ๐‘Ÿ๐‘๐‘š ๐ผ! ๐ด 1400 1200 ๐ผ!" = ๐ผ!# ๐‘…!," = 0. 5 ฮฉ ๐‘…!,#

QUESTION

  • A separately excited dc motor is mechanically coupled to a cylindrical type synchronous generator.
  • Synchronous generator ;
  • 50 kVA, 190 V, 50 Hz, Y connected, 4 poles, ๐‘‹ 9 = 2. 5 ฮฉ, ๐‘…! = 0. 02 ฮฉ,
  • ๐‘ƒ 1 = 1023 ๐‘Š , ๐‘ƒ 234 = 834 W (These losses are constant in all load condition)
  • Dc motor;
  • 55 kW, 400 V, 155.7 A, ๐‘…! = 0. 24 ฮฉ, ๐‘‰" = 400 V, 2 ฮ”๐‘‰I = 0 ,
  • ๐‘ƒ 1 + ๐‘ƒ 234 = 1456 W, (These losses are constant in all load condition)
  • The armature and excitation windings of the DC motor are fed from two separate voltage supplies. The excitation current of the motor is adjusted by a reosta. It is assumed that there is no saturation in the magnetic core and the linear relationship is represented by a given function of ๐œ™ = 0. 003125 ๐ผ". The synchronous generator is required to be operated at constant voltage and frequency.
  • (a) While the sync. gen. operates at a full load and a lagging power factor of 0.8, and the dc motor has an excitation field of 4 A, calculate the armature current and the back emf constant, of the motor and the efficiencies of both motor and generator.
  • (b) While the sync. gen. operates at a half load and a lagging power factor of 0.75, calculate the field current of the motor and the efficiencies of both motor and generator.