laporan-percobaan motor induksi 3 fasa rotor sangkar 2

I. Tujuan Setelah percobaan ini mahasiswa/praktikan diharapkan dapat:

1. Menghitung rugi-rugi daya motor induksi 3 fasa rotor sangkar

2. Menentukan efesiensi dari perhitungan rugi-rugi

3. Membandingkan efisiensi motor langsung dan efisiensi tak langsung (mencari rugi-rugi

motor)

4. Mencari harga torsi maksimum motor induksi 3 fasa rotor sangkat

II. Data Hasil PercobaanA. Name Plate Motor Siemens

Tegangan = 380/660 Volt D/Y

Daya = 1,1 KW

Kecepatan = 1410 Rpm

Arus = 2,8/1,62 Ampere

Cosφ = 0,81

Frekuensi = 50 Hz

IP = 54

B. Pengukuran Tahanan Stator Pada Suhu 200 C

U – X = 20 Ω

V – Y = 20Ω

W – Z = 20 Ω

Rata-rata tahanan/fasa = 20 Ω

C. Pengukuran Motor Tanpa Kopel, Motor Terhubung “DELTA”.

VL = 380 Volt

IL = 2,5 A

Pin3ϕ = 200 Watt

D. Pengukuran Motor Disambung Dengan Kopel, Motor Terhubung “Delta”.

VL = 380 Volt

IL = 2,5 A

Pin3ϕ = 240 Watt

E. Tabel Percobaan Motor Induksi 3 Fasa Rotor Sangkar 2 (Pembebanan)

No VLL

(Volt)Pin3ϕ

(Watt)Nr

(Rpm)T

(Nm)Pout

(Watt)Σrugi2

(Watt)η% Cosφ S% Ket

1. 380 440 1450 1,8 265,77 174,23 60,4 0,26 3,332. 380 740 1445 3,6 531,55 208,45 71,83 0,45 3,663. 380 1000 1420 5,4 797,3 202,7 79,73 0,6 5,334. 380 1300 1410 7,2 1063,11 236,89 81,77 0,79 65. 380 1400 1400 7,5 1107 293 79,07 0,85 6,66

F. Table Pengukuran Torsi Maksimum

No VLL

(Volt)IL

(A)Pin3ϕ

(Watt)Nr

(Rpm)Torsi(Nm)

Ket

1 228 8,1 11 1255 6

III. Jawaban Pertanyaan1.

Pout

(Watt)Σrugi2

(Watt)η% Cosφ S%

265,77 174,23 60,4 0,26 3,33531,55 208,45 71,83 0,45 3,66797,3 202,7 79,73 0,6 5,331063,1

1236,89 81,77 0,79 6

1107 293 79,07 0,85 6,66

2. R200 C =

20+20+203

= 20 Ω

R750 C = R20

0 C 235+75 °C235+20 °C

R750 C = 20

235+75 °C235+20 °C

R750 C = 24,31 Ω

3. Motor Tanpa Kopel

IL = 2,5 A

Pin3ϕ= 200 W

R rata-rata = 20 Ω

Pcu stator = I2 R

Pcu stator = 2,5 2 x 20

Pcu stator = 125 W

Rugi bocor= 0,01 x Pin 3Ø

Rugi bocor= 0,01 x 200

Rugi bocor= 2 W

Pinti = Pin3ϕ - Pcu stator - Rugi bocor

Pinti = 200 – 125 – 2

Pinti= 73 W

4. ∆P mekanik = Pin kopel - Pin tanpa kopel

∆Pmekanik = 240 – 200

∆ Pmekanik = 40 Watt

5. Pmekanik Setiap Perubahan Torsi

T 25 % T 50 % T 75 % T 100 %Pcu str

(W)Pcu str = IL

2 RPcustr = (2,2)2 x 20Pcustr = 96,8

Pcustr = IL2 R

Pcustr = (2,3)2 x 20Pcustr = 105,8

Pcustr = IL2 R

Pcustr = (2,4)2 x 20Pcustr = 115,2

Pcustr = IL2 R

Pcustr = (2,7)2 20Pcustr = 145,8

Pbocor

(W)Pbocor= 0,01 x Pinstr

Pbocor= 0,01x 440Pbocor= 4,4

Pbocor= 0,01 x Pinstr

Pbocor= 0,01x 740Pbocor= 7,4


Pbocor= 0,01x 1000Pbocor= 10


Pbocor= 0,01x 1400Pbocor= 14

Pin rtr

(W)Pinrtr = Pin - Pcu - Pinti

Pinrtr = 440–96,8–73Pinrtr = 270,2

Pinrtr = Pin - Pcu - Pinti

Pinrtr = 740-105,8-73Pinrtr = 561,2


Pinrtr = 1000-115,2-73Pinrtr = 811,8


Pinrtr = 1400-145,8-73Pinrtr = 1181,2

Pcu rtr Pcurtr = P x S Pcurtr = P x S Pcurtr = P x S Pcurtr = P x S

(W)Pcurtr = P

ns−nrns

Pcurtr=270,21500−1440

1500Pcurtr =10,8

Pcurtr = P ns−nrns

Pcurtr = 561,21500−1445

1500Pcurtr = 20,5

Pcurtr = P ns−nrns

Pcurtr =811,81500−1420

1500Pcurtr = 43,3

Pcurtr= P ns−nrns

Pcurtr = 1181,21500−1400

1500Pcurtr = 63,02

Pmek

(W)Pmekanik = Pinrtr - Pcurtr

Pmekanik = 270,2- 10,8Pmekanik = 259,4

Pmekanik = Pinrtr - Pcurtr

Pmekanik = 561,2-20,5Pmekanik = 540,7

Pmekanik = Pinrtr - Pcurtr

Pmekanik =811,8-43,3Pmekanik =768,5

Pmekanik= Pinrtr - Pcurtr

Pmekanik = 1181,2-63,02Pmekanik =1118,18

6. Σrugi-rugi Setiap Perubahan Torsi Beban

T 25 % T 50 % T 75 % T100 %Σrugi (W)

Σrugi = Pin3ϕ - Pout3ϕ

Σrugi=440-265,77Σrugi= 174,23


Σrugi=740-531,55Σrugi= 208,45


Σrugi=1000-797,3Σrugi= 202,7


Σrugi=1400-1107Σrugi= 293

7. Pout Setiap Perubahan Torsi Beban

T 25 % T 50 % T 75 % T100 %Pout 3ϕ

(W)Pout3ϕ = Pin – ΣrugiPout3ϕ =440-174,23Pout3ϕ = 265,77

Pout3ϕ = Pin – ΣrugiPout3ϕ =740-208,45Pout3ϕ = 531,55

Pout3ϕ = Pin – ΣrugiPout3ϕ =1000-202,3Pout3ϕ = 797,3

Pout3ϕ = Pin – ΣrugiPout3ϕ =1400-293Pout3ϕ = 1107

8. η(%) Motor Setiap Perubahan Torsi Beban

T (Nm) η (%)1,8

η (%) = P∈−Σrugi−rugi

P∈¿¿ 100%

η (%) = 440−174,23

440 100%

η (%) = 60,4

3,6η (%) =

P∈−Σrugi−rugiP∈¿¿ 100%

η (%) = 740−208,45

740 100%

η (%) = 71,83

5,4η (%) =


η (%) = 1000−202,7

1000 100%

η (%) = 79,73

7,5η (%) =


η (%) = 1400−293

1400 100%

η (%) =79,07

9. Perbandingan η nomor 1 dan nomor 8

a. ∑Rugi1 = Pcu stator + Pinti + Pbocor + Pcu rotor + Pmekanik

= 96,8 + 73 + 4,4 + 10,8 + 40

= 225 Watt

Pout1 = Pin - ∑rugi

= 440 Watt – 225 Watt

= 215 Watt

ɳ1 = ( Pout : Pin ) x 100%

= ( 215: 440 ) x 100%

= 48,86%

b. ∑Rugi2 = Pcu stator + Pinti + Pbocor + Pcu rotor + Pmekanik

= 105,8+ 73 + 7,4 + 20,5+ 40

= 246,7 Watt


= 740 Watt – 246,7 Watt

= 493,3 Watt

ɳ2 = ( Pout : Pin ) x 100%

= ( 493,3 : 740 ) x 100%

= 66,66%

c. ∑Rugi3 = Pcu stator + Pinti + Pbocor + Pcu rotor + Pmekanik

= 115,2+ 73 + 10+ 43,3+ 40

= 281,5 Watt


= 1000 Watt – 281,5 Watt

= 718,5 Watt

ɳ3 = ( Pout : Pin ) x 100%

= ( 718,5 : 1000 ) x 100%

= 71,85%

d. ∑Rugi4 = Pcu stator + Pinti + Pbocor + Pcu rotor + Pmekanik

= 145,8+ 73 + 14 + 63,02 + 40

= 335,82 Watt


= 1400 Watt – 335,82 Watt

= 1064,18 Watt

ɳ4 = ( Pout : Pin ) x 100%

= ( 1064,18: 1400 ) x 100%

= 76,01%

10. Grafik η (%) = f (Pout)

265.77 531.55 797.3 11070

10

20

30

40

50

60

70

80

90Grafik η% = f (Pout)

secara langsung

secara tidak langsung

Pout (Watt)

η (

%)

11. Analisis Grafik di atas adalah

12. Torsi Maksimum

T max=T percobaan xV ¿nominal

V ¿ percobaan

¿6 x380228

=10Nm

13. Kesimpulan

laporan-percobaan motor induksi 3 fasa rotor sangkar 2

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