Aplikasi Perhitungan Model Motor Induksi

sekarang waktunya untuk mengaplikasikannya. Untuk latihan kita saat ini kita memakai data sheet motor ABB.

Pada datasheet diatas kita dapatkan data berikut ini :

P output (PN) 110 kW
Rated Voltage (UN) 400 VD (belitan delta)
Rated Frequency (fN) 50 Hz
Rated speed (nN) 2972 rpm
Rated Current (IN) 187 amps (\eta=94.8%, pf=0.89)
No Load current (INL) 33 Amps

Mari kita mulai:

1 P input (PIN) =P_{out}/\eta 116.034 x 103 W
2 Arus magnetisasi (Ima) =33 \angle -90^0  =-j33
3 Arus beban (Ia) =P_{in}/(\sqrt{3}xU_{N}xpf=116034/(\sqrt{3}x400x0.89 167.49-j 85.81 or 188.18 \angle -27.127^0 A
4 Arus ke rotor =step(3)-step(2)=Ia-Ima =167.49 –j 85.81 +j 33 167.49-j 52.81 A or 175.618 \angle -17.5^0 A.
5 \omega_{s}L_{ms} =400/(\sqrt{3}x33) 6.998\Omega
6 Pout (PN-cal) =110 x 103 W 110 x 103 W
7 Torsi output (TN) =Poutl/\omega_{m}=110000/(2972×2\pi/60) 353.44 N.m
8 Losses =(1-\eta)x PIN
=(1-0.948) x 116034
6.03376x 103 W
9 Losses Windage and Friction (10% losses)(PFW) = 0.1 x 6.03376 kW 0.60338x 103 W
10 Power to air gap =PFW+Pout(step 6 + step 9)= 109.31 +0.5996 109.91x 103 W
11 Torsi di air gap = step 10/\omega_{r}=109910/(2972x2x\pi/60) 353.149 N.m
12 Power which create Torque at point 11(Pma) = T.\omega_{sync}=353.149 x 4/2 x50 xpi 111.645×103 W
13 Slip (s) =(3000-2972)/3000 0.00933
14 R'_{r} =\frac{s.P_{ma}}{3.(I'_{a})^2} 0.01126 \Omega
15 Stator losses (34% total losses) per phase =34% X6033.76 /3 683.83 W
16 Stator resistance (RS) Refer to equivalent circuit di posting ini =step(15)/(step(4)^2)=683.83/175.618 0.02217\Omega
17 \bar{Z}_{r} =R_{s}+R'_{r}/s+j\omega_{s}L_{L}=0.02217+\frac{0.01126}{0.00933}+j\omega_{s}L_{L} 1.2288+j\omega_{s}L_{L} \Omega
18 \omega_{s}L_{L} =1.2378 x tan \angle (step 4)=1.2378 x tan (17.50) 0.3874\Omega

Bagaimana model motor induksi dari data sheet ini :

gambar 2

next >> how modelling the slip-torque curve?

 

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