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= 0,01 x Pinstr
Pbocor= 0,01x 1000Pbocor= 10
Pbocor= 0,01 x Pinstr
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 = Pin - Pcu - Pinti
Pinrtr = 1000-115,2-73Pinrtr = 811,8
Pinrtr = Pin - Pcu - Pinti
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 = Pin3ϕ - Pout3ϕ
Σrugi=740-531,55Σrugi= 208,45
Σrugi = Pin3ϕ - Pout3ϕ
Σrugi=1000-797,3Σrugi= 202,7
Σrugi = Pin3ϕ - Pout3ϕ
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 (%) =
P∈−Σrugi−rugiP∈¿¿ 100%
η (%) = 1000−202,7
1000 100%
η (%) = 79,73
7,5η (%) =
P∈−Σrugi−rugiP∈¿¿ 100%
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
Pout2 = Pin - ∑rugi
= 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
Pout3 = Pin - ∑rugi
= 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
Pout4 = Pin - ∑rugi
= 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