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CIRCUITO EQUIVALENTE (MODELO POR FASE)
ENSAIO EM VAZIO (sem carga)
Aplica-se ππππ e mede-se πΌ0 e π0.
Circuito equivalente do motor em vazio.
π β 0 β π2β² Β·(1 β π )
π β β
Β·(π β π¬)
π¬
ππ
ππ
Em vazio, πΌ β 0,3 pu
π0 =ππππ2
ππ β ππ =
ππππ2
π0
ππ
ππ§π¨π¦ ππ
ππ« ππ±
πΌ0 senπ =ππππππ
β ππ =πππππΌ0 senπ
cosπ =π0
ππππ Β· πΌ0 β senπ = β1 β (
π
ππππ Β· πΌ0)2
=βππππ
2 Β· πΌ02 β π2
ππππ Β· πΌ0
ππ =ππππ Β· ππππ Β· πΌ0
πΌ0βππππ2 Β· πΌ0
2 β π2=
ππππ2
βππππ2 Β· πΌ0
2 β π2
π
πΌ0
π
π
πΌ0 senπ
ENSAIO EM CURTO-CIRCUITO (CC)
Curto-circuito β π2β² Β·(1βπ )
π = 0 β π = 1 (rotor bloqueado)
Aplica-se πΌπππ e mede-se ππΆπΆ (menor que ππππ) e ππΆπΆ.
Circuito equivalente em curto-circuito:
(π1 + π2β²)β
este modelonΓ£o distingue
Β· πΌπππ2 = ππΆπΆ β π1 + π2
β² =ππΆπΆπΌπππ2
ππ§π¨π¦
πππ πππ
π1 β π2β² na maioria dos casos. TambΓ©m pode ser π1 = 1 a 3 Γ π2
β².
Para motores de alto escorregamento β ~0,4 π1 e ~0,6 π2β²
Como, analogamente, senπ =βππΆπΆ
2 Β·πΌπππ2 βππΆπΆ
2
ππΆπΆΒ·πΌπππ, temos:
(π1 + π2β²) =
ππΆπΆπΌπππ
Β·βππΆπΆ
2 Β· πΌπππ2 β ππΆπΆ
2
ππΆπΆ Β· πΌπππ=βππΆπΆ
2 Β· πΌπππ2 β ππΆπΆ
2
πΌπππ2
Distribuição entre π1 e π2β² :
De π2β² = π1 para motores com curva normal de conjugado
atΓ© π2β² = 1 a 2 Γ π1 para motores de alto escorregamento.
π
ππΆπΆ ππΆπΆ senπ = (π1 + π2
β²) Β· πΌπππ
EXERCΓCIO
Dado o seguinte circuito equivalente:
de um motor de 2.000 hp, 2.300 V, 3Γ, 4 polos, 60 Hz e ligação em Y
com π πππ = 0,0375.
Calcular a eficiΓͺncia do motor.
1 hp = 550 lbf Β· ft sβ = 745,7 W
1 cv = 75 kgf Β· m sβ = 735,5 W
ππ
ππ π, ππ Β·(π β π¬)
π¬ π
π, ππ π ππ, ππ π ππ, ππ π π, ππ π
πππ, π π πππ π
RESOLUΓΓO
Reduzindo o circuito, temos:
ΞοΏ½ΜοΏ½π β (0,02 + π0,32 Ξ©) Β· (408,16β -14,8Β° A) β 130,9 β 71,6Β° V
οΏ½ΜοΏ½π = οΏ½ΜοΏ½π β ΞοΏ½ΜοΏ½π = 1.292,6 β -5,51Β° V
π, ππ + ππ, ππ π π, ππ + ππ, ππ π
π, ππ + πππ, π π ππ ππβ²
π«ππ
π, ππ + ππ, ππ π
π, πππ + ππ, πππ π
οΏ½ΜοΏ½π β πππ, ππβ -ππ, πΒ° π
π, πππ + ππ, πππ π
οΏ½ΜοΏ½π =οΏ½ΜοΏ½π
οΏ½ΜοΏ½ππͺ; οΏ½ΜοΏ½π =
π. πππ
βπ π
Conferindo:
πΌ2β² =
ππ
ππππβ(1.292,6 β -5,51Β° V)
(3,20 + π0,32 Ξ©)β 401,9 β -11,22Β° V
ππππΓΊπ‘ππ = π2β² Β· πΌ2
β² β (3,08 Ξ©) Β· (401,9 A)2 β 497,5 kW
πππππππ =ππππππππ
3β(2.000 hp) Β· (745,7 W/hp)
3β 497,3 kW
A eficiΓͺncia do motor Γ© dada por:
π =ππππΓΊπ‘πππππΓ©πππ‘ππππ
=(497,5 kW)
(2.300
β3 V) Β· (408,16 cos(β14,8Β°) A)
β497,5 kW
524,02 kW
π β 94,9%
β
VARIAΓΓO DE VELOCIDADE EM MOTORES DE INDUΓΓO
MΓTODOS
(1) Motor de anΓ©is
β
ππ 0
ππ
βπππ₯ π2 > π1
βπ cte.
π1 > 0 rotor CC (r = 0)
π1
π2
π3
βπ β π2
πΆπ = cte.
π =ππ πΓππππππ‘π
=(0,7 pu)
(1 pu)= 70%
πππππππ = π Β· πππ = (0,3) Β· (1 pu) = 0,3 pu
βπ β π2
π =(0,343 pu)
(0,49 pu)= 70%
πππππππ = (0,3) Β· (0,49 pu) = 0,15 pu
(2) Variação de πππππ
πππ = ππππ‘π
ππ πΓππ = (1 β π ) Β· πππ
πππ ππππ = π Β· πππ
β
ππ 0 ππ
βπππ₯
βπ cte. βπ β π
2
β β π2