Electric Machines snc Generators, Lecture notes of Electric Machines

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EEE 316

ELECTRICAL MACHINERY II

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

QUESTION:

• A shunt dc generator has an armature resistance of 𝑅! = 0. 4 Ω , a field resistance of 𝑅" = 125 Ω, a rated armature current of 𝐼!,$ = 30 A at a speed of 800 rpm, an induced emf of 𝐸 = 211 V at no-load.
• (a) While terminal voltage is 200 V, what will be the output power to keep the induced emf 𝐸 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. (𝐸 is same as in part–a.)

• @ full load : 𝐼! = 30 A then 𝑉 = 𝐸 − 𝐼!𝑅! = 211 − 30 × 0. 4 = 199 V
• 𝐼& = 30 − ( ('/

= 28. 4 A

• 𝑃% = 𝑉 𝐼& = 199 × 28. 4 = 5651. 6 W

' 𝑅! + 𝐼" ' 𝑅9: = 30 ' × 0. 4 + 1. 6 '

× 125 = 676. 8 W

(, (**

= 1. 14 × 5651. 6 + 676. 8 = 7123 W

• 𝑃9:!"; > 7123 W

QUESTION

• A series dc motor has an armature resistance of 𝑅! = 0. 4 Ω , a field resistance of 𝑅" = 0. 1 Ω, a rated armature current of 𝐼!,$ = 20 A at a speed of 1100 rpm, an applied terminal voltage of 𝑉 = 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 − 10 0. 4 + 0. 1 = 195 V @𝐼! = 10 A

• 195 = 0. 0824 × 10 ×𝜔

• 𝜔 = 236 rad/s ⇒ 𝑛 = C* 'D 𝜔 = 2260 rpm @𝐼! = 10 A
• b) 𝐼! = 10 A ⇒ 𝑃 1 + 𝑃 234 = 8% 𝑃 0
• 𝑃 0 = 𝑉 𝐼! = 200 × 10 = 2000 W
• ⇒ 𝑃 1 + 𝑃 234 = 160 W
• 𝑃 17 = 𝑃 17 ,! + 𝑃 17 ," = 𝐼! ' 𝑅! + 𝑅9> = 10 ' 0. 4 + 0. 1 = 50 W

• 𝑃% = 𝑃9:!"; = 1790 W

(^66) (^67)

(E8* '***

𝑃^ 𝑃$%^8 ' 𝑃'( 𝑃) = 𝑃-./0* 𝑃+ 𝑃 1

• SOLUTION:

• (a)
• 𝑉 = 120 𝑉, 𝑅9: = 15 Ω, 𝑛 = 600𝑟𝑝𝑚 → 𝐼 =? 𝐼" = - .!"

('* (/

• From the table; 𝐼" = 8 𝐴 → 𝐸 = 126 𝑉
• 𝐼! = ?)- . 2

('C)('* .'

• (b)
• 𝑉 = 144 𝑉, 𝑅!" = 18 Ω, 𝑛 = 700 𝑟𝑝𝑚 → 𝐼 =? 𝐼# = $ %!" = &'' &( = 8 𝐴
• From the table 𝐼# = 8 𝐴 → 𝐸 = 126 𝑉 @𝑛 = 600𝑟𝑝𝑚
• 𝐸 = 𝑘)𝜙𝑛 →
  • = 𝑐𝑜𝑛𝑠𝑡 *# +# = *$ +$ → 𝐸, = 𝐸& +$ +# = 126 -.. /.. = 147 𝑉
• 𝐼 0 =
  • 1 $ %% = &'- 1 &'' ..., = 150 𝐴 𝐼 = 𝐼 0 − 𝐼# = 150 − 8 = 142 𝐴

• SOLUTION:

• J

→ 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 𝑛 𝑟𝑝𝑚 𝐼! 𝐴 1400 1200 𝐼!" 𝐼!# 𝑉" 𝑉#

• (b)
• 𝑃 1 =? 𝜂 = (^2)! (^2) " ⇒ 0. 82 = 34 × 678 (^2) " → 𝑃 1 = 8975. 6 𝑊
• 𝑃 1 = 𝑃 9 + 𝑃: = 𝑃 9 + 𝑉 𝐼:
• Operating condition at point 1:
• 𝑃 1 = 𝑃 9 + 𝑉 3 𝐼:, 3
  8975. 6 = 100 + 200 𝐼:, 3 ⇒ 𝐼:, 3 = 44. 38 𝐴 𝑃 0 ⇒ ⇐^ 𝑃/

• (c)
• 𝑃9:!"; = 𝑃= − 𝑃" 3 F + 𝑃 1 = 𝐸 𝐼! − 𝑃" 3 F + 𝑃 1 1%G9;
• 𝐸(𝐼!,( − 𝑃9:!";,( = 𝐸'𝐼!,' − 𝑃9:!";,' 200 − 44. 38 × 0. 5 44. 38 − 7360 = 152. 4 × 50. 03 − 𝑃9:!";,' ⇒ 𝑃9:!";,' = 7093. 4 W
• 𝑇9:!";,' = (^6) !"2:;, 3 A 3

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

• ii)
• (a)
• Operating condition at point 1: , 𝑘>𝜙 = 0. 127 𝑉/𝑟𝑝𝑚
• Operating condition at point 2 :
• 𝑃 0 = 𝑐𝑜𝑛𝑠𝑡 , 𝑉 = 𝑐𝑜𝑛𝑠𝑡 ⇒ 𝑃 0 = 𝑃" + 𝑉𝐼! ⇒ 𝐼! = 𝑐𝑜𝑛𝑠𝑡 1 2 𝑛 𝑟𝑝𝑚 𝐼! 𝐴 1400 1200 𝐼!" = 𝐼!# 𝑅!," = 0. 5 Ω 𝑅!,#

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, 𝑋 9 = 2. 5 Ω, 𝑅! = 0. 02 Ω,
• 𝑃 1 = 1023 𝑊 , 𝑃 234 = 834 W (These losses are constant in all load condition)
• Dc motor;
• 55 kW, 400 V, 155.7 A, 𝑅! = 0. 24 Ω, 𝑉" = 400 V, 2 Δ𝑉I = 0 ,
• 𝑃 1 + 𝑃 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 𝜙 = 0. 003125 𝐼". 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.