Fakultas Teknik Mesin / Industri Tarumanagara

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1 Fakultas Teknik Mesin / Industri Tarumanagara
MOTOR INDUKSI Fakultas Teknik Mesin / Industri Tarumanagara Edit : By. Ir. Harlianto Tanudjaja, Mkom,MM

2 Type of Electric Motors Classification of Motors
Alternating Current (AC) Motors Direct Current (DC) Motors Synchronous Induction Three-Phase Single-Phase Self Excited Separately Excited Series Shunt Compound Motors are categorized in a number of types based on the input supply, construction and principle of operation. We will start at looking at various forms of the DC motor such as shunt and series, followed by the AC motors including synchronous and induction motors. Motors are categorized on th ebasis of input supply, construction and operation principoles TEXT

3 Type of Electric Motors
AC Motors Electrical current reverses direction Two parts: stator and rotor Stator: stationary electrical component Rotor: rotates the motor shaft Speed difficult to control Two types Synchronous motor Induction motor Alternating current (AC) motors use an electrical current, which reverses its direction at regular intervals. An AC motor has two basic electrical parts: a "stator" and a "rotor". The stator is in the stationary electrical component. The rotor is the rotating electrical component, which in turn rotates the motor shaft. The main advantage of DC motors over AC motors is that speed is more difficult to control for AC motors. To compensate for this, AC motors can be equipped with variable frequency drives but the improved speed control comes together with a reduced power quality. There are two types of AC motors: synchronous (see figure) and induction. The main difference between the synchronous motor and the induction motor is that the rotor of the synchronous motor travels at the same speed as the rotating magnetic field.

4 Induction Motors

5 Type of Electric Motors AC Motors – Induction motor
Components Rotor Squirrel cage: conducting bars in parallel slots Wound rotor: 3-phase, double-layer, distributed winding (Automated Buildings) An induction motor has two main electrical components as shown in the figure Rotor. Induction motors use two types of rotors: A squirrel-cage rotor consists of thick conducting bars embedded in parallel slots. These bars are short-circuited at both ends by means of short-circuiting rings. A wound rotor has a three-phase, double-layer, distributed winding. It is wound for as many poles as the stator. The three phases are wired internally and the other ends are connected to slip-rings mounted on a shaft with brushes resting on them. Stator. The stator is made up of a number of stampings with slots to carry three-phase windings. It is wound for a definite number of poles. The windings are geometrically spaced 120 degrees apart Stator Stampings with slots to carry 3-phase windings Wound for definite number of poles

6 MOTOR INDUKSI TIGA PHASA
-. Motor induksi adalah suatu mesin listrik yang merubah energi listrik menjadi energi gerak dengan menggunakan gandengan medan listrik dan mempunyai slip antara medan stator dan medan rotor. -. Motor induksi merupakan motor yang paling banyak kita jumpai dalam industri.

7 KELEBIHAN MI Lebih murah dibanding DC Motor Secara mekanis lebih kuat
Lebih murah biaya pemeliharaan

8 Konstruksi Stator Dibuat dari pelat-pelat tipis dengan slot.
Belitan ditempatkan pada slot Gulungan tiga fasa dilingkarkan untuk sejumlah kutub tertentu Gulungan diberi spasi geometri sebesar 120° antar phasa

9 Bagian-Bagian Motor

10 Introduction Three-phase induction motors are the most common and frequently encountered machines in industry simple design, rugged, low-price, easy maintenance wide range of power ratings: fractional horsepower to 10 MW run essentially as constant speed from no-load to full load Its speed depends on the frequency of the power source not easy to have variable speed control requires a variable-frequency power-electronic drive for optimal speed control

11 Construction An induction motor has two main parts a stationary stator
consisting of a steel frame that supports a hollow, cylindrical core core, constructed from stacked laminations (why?), having a number of evenly spaced slots, providing the space for the stator winding

12 Construction Two basic design types depending on the rotor design
a revolving rotor composed of punched laminations, stacked to create a series of rotor slots, providing space for the rotor winding one of two types of rotor windings conventional 3-phase windings made of insulated wire (wound-rotor) » similar to the winding on the stator aluminum bus bars shorted together at the ends by two aluminum rings, forming a squirrel-cage shaped circuit (squirrel-cage) Two basic design types depending on the rotor design squirrel-cage: conducting bars laid into slots and shorted at both ends by shorting rings. wound-rotor: complete set of three-phase windings exactly as the stator. Usually Y-connected, the ends of the three rotor wires are connected to 3 slip rings on the rotor shaft. In this way, the rotor circuit is accessible.

13 Construction Slip rings
Cutaway in a typical wound-rotor IM. Notice the brushes and the slip rings Brushes

14 Komponen Stator Rangka. Inti stator Kumparan/gulungan Pelat penutup

15 Inti Stator Terbuat dari lempeng-lempeng baja silikon berlaminasi.
Untuk memperkecil rugi-rugi besi akibat arus pusar

16 Konstruksi Rotor Fungsi :mengubah daya dari stator menjadi tenaga mekanik. Terdapat dua tipe, yaitu : Rotor sangkar Rotor belitan Komponen-komponenRotor: Inti besi rotor, Kumparan/batang penghantar, Cincin Poros (shaft).

17 Construction Squirrel cage rotor Wound rotor Notice the slip rings

18 Kedua ujungnya dihubungsingkat dengan cincin
Rotor Sangkar Terdiri dari batang penghantar tebal yang diletakkan pada petak-petak slot paralel Kedua ujungnya dihubungsingkat dengan cincin

19 Rotor Belitan Konduktor yang digunakan adalah belitan
Belitan terhubung ke cincin geser yang dipasang pada shaft Belitan terhubung ke resistor melalui sikat karbon

20 Prinsip Kerja Prinsip kerja motor induksi mirip trafo
Rangkaian primer (stator) dan sekunder (rotor) tidak satu inti. Rangkain sekunder berputar

21 Prinsip Kerja Listrik dipasok ke sator sehingga menghasilkan medan magnet yang berputar dengan kecepatan sinkron Pada rangkaian rotor timbul arus segingga timbul kopel Rotor berputar searah putaran medan stator

22 Prinsip Kerja Bila Kumparan Stator diberi tegangan sumber maka timbul medan putar Medan putar tersebut akan memotong batang konduktor pada rotor Kumparan rotor timbul ggl ( E ), sehingga timbul arus ( I ) Adanya arus dalam medan magnet maka timbul F (ggl) pada rotor Bila F rotor cukup kuat memikul torka beban, maka rotor akan berputar Perbedaan kecepatan antara nr dan ns disebut slip yang dinyatakan: Bila nr = ns maka tidak dihasilkan torka, torka motor akan timbul apabila nr lebih kecil dari ns

23 Slip Dalam praktek rotor tidak pernah berputar pada kecepatan sinkron
Perbedaan kecepatan antara putaran medan stator dan kecepatan rotor disebut slip Ns = kecapatan sinkron (rpm) Nr = kecepatan putaran rotor (rpm)

24 Mode Operasi +T Bila ns > nr mesin berfungsi Sebagai motor.
Plugging Motor 1.0 -1.0 S Generator Bila ns < nr mesin berfungsi Sebagai generator. -T

25 Rotating Magnetic Field
Balanced three phase windings, i.e. mechanically displaced 120 degrees form each other, fed by balanced three phase source A rotating magnetic field with constant magnitude is produced, rotating with a speed Where fe is the supply frequency and P is the no. of poles and nsync is called the synchronous speed in rpm (revolutions per minute)

26 Synchronous speed P 50 Hz 60 Hz 2 3000 3600 4 1500 1800 6 1000 1200 8
750 900 10 600 720 12 500

27 Rotating Magnetic Field

28 Rotating Magnetic Field

29 Rotating Magnetic Field

30 Rotating Magnetic Field

31 Rotating Magnetic Field

32 Prinsip Kerja ia ic ib ia ic ib FT Fa Fb Fc FT Fa Fb Fc FT Fc Fa Fb FT
x a -b c -c -a b FT Fa Fb Fc FT Fa Fb Fc FT Fc Fa Fb FT Fb Fc Fa FT Fa Fb Fc FT Fc Fa Fb

33 Prinsip Kerja t0 t1 t2 t3 t4 t5 ia ib FT Fa Fb Fc FT Fa FT Fc Fa Fb FT
ic t0 ia t6 ib ia ic t5 t4 t3 t2 t1 x a -b c -c -a b FT Fa Fb Fc t0 t1 t2 t3 t4 t5 FT Fc Fa Fb FT Fc Fa Fb FT Fa Fb Fc FT Fb Fc Fa x a -b c -c -a b x a -b c -c -a b x a -b c -c -a b x a -b c -c -a b x a -b c -c -a b x a -b c -c -a b FT Fa Fb Fc FT Fa Fb Fc

34 Principle of operation
This rotating magnetic field cuts the rotor windings and produces an induced voltage in the rotor windings Due to the fact that the rotor windings are short circuited, for both squirrel cage and wound-rotor, and induced current flows in the rotor windings The rotor current produces another magnetic field A torque is produced as a result of the interaction of those two magnetic fields Where ind is the induced torque and BR and BS are the magnetic flux densities of the rotor and the stator respectively

35 Induction motor speed At what speed will the IM run?
Can the IM run at the synchronous speed, why? If rotor runs at the synchronous speed, which is the same speed of the rotating magnetic field, then the rotor will appear stationary to the rotating magnetic field and the rotating magnetic field will not cut the rotor. So, no induced current will flow in the rotor and no rotor magnetic flux will be produced so no torque is generated and the rotor speed will fall below the synchronous speed When the speed falls, the rotating magnetic field will cut the rotor windings and a torque is produced

36 Induction motor speed So, the IM will always run at a speed lower than the synchronous speed The difference between the motor speed and the synchronous speed is called the Slip Where nslip= slip speed nsync= speed of the magnetic field nm = mechanical shaft speed of the motor

37 The Slip Where s is the slip
Notice that : if the rotor runs at synchronous speed s = 0 if the rotor is stationary s = 1 Slip may be expressed as a percentage by multiplying the above eq. by 100, notice that the slip is a ratio and doesn’t have units

38 Induction Motors and Transformers
Both IM and transformer works on the principle of induced voltage Transformer: voltage applied to the primary windings produce an induced voltage in the secondary windings Induction motor: voltage applied to the stator windings produce an induced voltage in the rotor windings The difference is that, in the case of the induction motor, the secondary windings can move Due to the rotation of the rotor (the secondary winding of the IM), the induced voltage in it does not have the same frequency of the stator (the primary) voltage

39 Frequency The frequency of the voltage induced in the rotor is given by Where fr = the rotor frequency (Hz) P = number of stator poles n = slip speed (rpm)

40 Frequency What would be the frequency of the rotor’s induced voltage at any speed nm? When the rotor is blocked (s=1) , the frequency of the induced voltage is equal to the supply frequency On the other hand, if the rotor runs at synchronous speed (s = 0), the frequency will be zero

41 Sumber tegangan bolak-balik yang sinusoid menghasilkan fluks yang sinusoid pula
Fluks yang sinusoid ini hanya menghasilkan fluks (medan) pulsasi saja dan beban fluks yang berputar terhadap ruang.

42 E = tegangan induksi ggl
Tegangan GGL 2. Medan putar stator akan memotong konduktor yang terdapat pada sisi rotor, akibatnya pada kumparan rotor akan timbul tegangan induksi ( ggl ) sebesar E = tegangan induksi ggl f = frekkuensi N = banyak lilitan Q = fluks

43 Torque While the input to the induction motor is electrical power, its output is mechanical power and for that we should know some terms and quantities related to mechanical power Any mechanical load applied to the motor shaft will introduce a Torque on the motor shaft. This torque is related to the motor output power and the rotor speed and

44 Horse power Another unit used to measure mechanical power is the horse power It is used to refer to the mechanical output power of the motor Since we, as an electrical engineers, deal with watts as a unit to measure electrical power, there is a relation between horse power and watts

45 Torka Motor Induksi E1 Bila Z1 = R1 +jX1 dianggap kecil
a2R2/S a2X2 I0 I’2 V1 RC XM E1 Bila Z1 = R1 +jX1 dianggap kecil maka E1 = V1 , dan T adalah : Harga S untuk mendapatkan T maks maka didapat Tmaks pada dan

46 Kopel Motor Induksi

47 Kurva Torsi dan Slip

48 Daya motor Induksi Sehingga P2 : Pm : Pr = 1 : ( 1 - S ) : S
Daya masuk Stator : Daya masuk rotor : Daya keluar rotor ( P mekanis ) Rugi-rugi daya : Sehingga P2 : Pm : Pr = 1 : ( 1 - S ) : S

49 Pengaturan Motor Induksi
Mengubah frekuensi jala-jala dan jumlah kutup : Bila p ( jumlah kutup ) semakin besar maka semakin lambat kecepatan putaran dan se baliknya. Jumlah kutup dapat diubah2 dengan meren canakan kumparan stator sedemikian shg dapat menerima tegangan masuk pada posisi yang berbeda-beda . Dari persamaan diatas diketahui bahwa dengan mengubah f semakin besar maka Menyebabkan kecepatan motor akan semakin besar juga dan sebaliknya.

50 Pengaturan Motor Induksi
Mengatur tegangan jala – jala : Besarnya kopel motor induksi sebanding dengan pangkat dua tegangan yang di berikan ( V1) T = k V2. Karakteristik beban dapat dilihat seperti gambar disamping, kecepatan akan be rubah dari n1 ke n2 untuk tegangan masuk setengah dari tegangan semula. T V1 0.5V1 beban n2 n1 n Harmonic tinggi dan power factor ren dah , pengaturan ini biasanya dipakai untuk peralatan starting torque rendah

51 Pengaturan motor induksi
Pengaturan tahanan luar Penambahan tahanan luar R2 pada rotor belitan sampai harga tertentu dapat torka Start maksimum. T n R2 naik n4 n3 n2 n1 Penambahan tahanan luar juga diperlukan untuk memba tasi arus awal yg besa saat Start. Dengan mengubah2 tahanan luar juga diperlukan untuk me ngatur kecepatan motor. Cara ini mengakibatkan rugi daya yang cukup besar pada rotor Kurva T terhadap speed ( n ) dengan mengubah-ubah R2

52 Example A 208-V, 10hp, four pole, 60 Hz, Y-connected induction motor has a full-load slip of 5 percent What is the synchronous speed of this motor? What is the rotor speed of this motor at rated load? What is the rotor frequency of this motor at rated load? What is the shaft torque of this motor at rated load?

53 Solution

54 Rangkaian Ekivalen V1 = tegangan stator E1 = tegangan rotor
R1 = tahanan stator X1 = reaktansi bocor stator RC = reaktansi inti besi E1 = tegangan rotor R2 = tahanan stator X2 = reaktansi bocor stator Xm = reaktansi magnetisasi

55 Equivalent Circuit The induction motor is similar to the transformer with the exception that its secondary windings are free to rotate As we noticed in the transformer, it is easier if we can combine these two circuits in one circuit but there are some difficulties

56 Equivalent Circuit When the rotor is locked (or blocked), i.e. s =1, the largest voltage and rotor frequency are induced in the rotor, Why? On the other side, if the rotor rotates at synchronous speed, i.e. s = 0, the induced voltage and frequency in the rotor will be equal to zero, Why? Where ER0 is the largest value of the rotor’s induced voltage obtained at s = 1(loacked rotor)

57 Equivalent Circuit The same is true for the frequency, i.e.
It is known that So, as the frequency of the induced voltage in the rotor changes, the reactance of the rotor circuit also changes Where Xr0 is the rotor reactance at the supply frequency (at blocked rotor)

58 Rangkaian Pengganti Motor Induksi
Harga primer dipindah ke sekunder Harga sekunder dipindah ke primer

59 Equivalent Circuit Then, we can draw the rotor equivalent circuit as follows Where ER is the induced voltage in the rotor and RR is the rotor resistance

60 Equivalent Circuit Now we can calculate the rotor current as
Dividing both the numerator and denominator by s so nothing changes we get Where ER0 is the induced voltage and XR0 is the rotor reactance at blocked rotor condition (s = 1)

61 Equivalent Circuit Now we can have the rotor equivalent circuit

62 Equivalent Circuit Now as we managed to solve the induced voltage and different frequency problems, we can combine the stator and rotor circuits in one equivalent circuit Where

63 Power losses in Induction machines
Copper losses Copper loss in the stator (PSCL) = I12R1 Copper loss in the rotor (PRCL) = I22R2 Core loss (Pcore) Mechanical power loss due to friction and windage How this power flow in the motor?

64 Vektor Diagram rangkaian Motor Induksi

65 Power flow in induction motor

66 Power relations

67 Equivalent Circuit We can rearrange the equivalent circuit as follows
Resistance equivalent to mechanical load Actual rotor resistance

68 Power relations

69 Power relations 1 s 1-s

70 Example A 480-V, 60 Hz, 50-hp, three phase induction motor is drawing 60A at 0.85 PF lagging. The stator copper losses are 2 kW, and the rotor copper losses are 700 W. The friction and windage losses are 600 W, the core losses are 1800 W, and the stray losses are negligible. Find the following quantities: The air-gap power PAG. The power converted Pconv. The output power Pout. The efficiency of the motor.

71 Solution

72 Solution

73 Example A 460-V, 25-hp, 60 Hz, four-pole, Y-connected induction motor has the following impedances in ohms per phase referred to the stator circuit: R1= 0.641 R2= 0.332 X1=  X2=  XM= 26.3  The total rotational losses are 1100 W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 2.2 percent at the rated voltage and rated frequency, find the motor’s Speed Stator current Power factor Pconv and Pout ind and load Efficiency

74 Solution

75 Solution

76 Solution

77 Torque, power and Thevenin’s Theorem
Thevenin’s theorem can be used to transform the network to the left of points ‘a’ and ‘b’ into an equivalent voltage source VTH in series with equivalent impedance RTH+jXTH

78 Torque, power and Thevenin’s Theorem

79 Torque, power and Thevenin’s Theorem
Since XM>>X1 and XM>>R1 Because XM>>X1 and XM+X1>>R1

80 Torque, power and Thevenin’s Theorem
Then the power converted to mechanical (Pconv) And the internal mechanical torque (Tconv)

81 Torque, power and Thevenin’s Theorem

82 Torque-speed characteristics
Typical torque-speed characteristics of induction motor

83 Comments The induced torque is zero at synchronous speed. Discussed earlier. The curve is nearly linear between no-load and full load. In this range, the rotor resistance is much greater than the reactance, so the rotor current, torque increase linearly with the slip. There is a maximum possible torque that can’t be exceeded. This torque is called pullout torque and is 2 to 3 times the rated full-load torque.

84 Comments The starting torque of the motor is slightly higher than its full-load torque, so the motor will start carrying any load it can supply at full load. The torque of the motor for a given slip varies as the square of the applied voltage. If the rotor is driven faster than synchronous speed it will run as a generator, converting mechanical power to electric power.

85 Complete Speed-torque c/c

86 Maximum torque Maximum torque occurs when the power transferred to R2/s is maximum. This condition occurs when R2/s equals the magnitude of the impedance RTH + j (XTH + X2)

87 The maximum torque is independent of R2
The corresponding maximum torque of an induction motor equals The slip at maximum torque is directly proportional to the rotor resistance R2 The maximum torque is independent of R2

88 Maximum torque Rotor resistance can be increased by inserting external resistance in the rotor of a wound-rotor induction motor. The value of the maximum torque remains unaffected but the speed at which it occurs can be controlled.

89 Effect of rotor resistance on torque-speed characteristic
Maximum torque Effect of rotor resistance on torque-speed characteristic

90 Motor Induksi Tiga Fasa
Hubungan Delta Hubungan Bintang

91 Nameplate Motor Induksi

92 Informasi Pada Nameplate
Horse Power =: Kemampuan putaran rotor menggerakkan beban makimum. 1HP = 746 W Volt : biasanya mempunyai toleransi 10 % AMPS : Kemampuan motor dengan beban maksimum HERZT : Frekuensi jaringan listrik RPM : Kecepatan putaran rotor saat tersambung beban maksimum Service Factor : Faktor perkalian kemampuan daya mekanik dimana motor bisa dioperasikan

93 Insulation Class Class A, kemampuan isolasi hanya 105°C
Pembagian Kelas Isolasi : Class A, kemampuan isolasi hanya 105°C Class B, kemampuan isolasi hanya 130°C Class C, kemampuan isolasi hanya 155°C Class D, kemampuan isolasi hanya 180°C

94 Nema Design Menerangkan Karakteristik kemampuan torsi ouput rotor:
Nema A, motor mempunyai arus start tinggi dan torsi awal normal Nema B, motor mempunyai arus start rendah dan torsi awal normal Nema C, motor mempunyai arus start rendah dan torsi awal tinggi Nema D, motor mempunyai arus start rendah dan torsi awal sangat tinggi

95 Arus Start Mereferensikan terjadinya lock rotor,
Rotor terkunci sehingga akan menarik sumber sangat besar sekali Biasanya untuk motor Nema Design B sebesar 600 – 650 % arus beban penuh

96 Konversi Energi pada Motor

97 Efisiensi Motor Induksi
Ditentukan oleh kehilangan dasar yang hanya dapat dikurangi oleh perubahan pada rancangan motor dan kondisi operasi Jenis kehilangan Persentase kehilangan total (100%) Kehilangan tetap atau kehilangan inti 25 Kehilangan variabel: kehilangan stator I2R 34 Kehilangan variabel: kehilangan rotor I2R 21 Kehilangan gesekan & penggulungan ulang 15 Kehilangan beban yang menyimpang 5

98 Faktor-faktor yang Mempengaruhi Efisiensi
Usia Kapasitas Kecepatan Jenis Suhu Penggulungan ulang Beban

99 Perbandingan Motor Efisiensi Tinggi dengan Motor Standar

100 Hal yang harus Dipertimbangkan Saat Rewinding
Gunakan perusahaan yang bersertifikasi ISO 9000 atau anggota dari Assosasi Layanan Peralatan Listrik. Jika biaya pegulungan ulang melebihi 50% hingga 65% dari harga motor baru yang efisien energinya, lebih baik membeli motor yang baru Ukuran motor kurang dari 40 HP dan usianya lebih dari 15 tahun (terutama motor yang sebelumnya sudah digulung ulang) sebaiknya diganti. Untuk motor dibawah 15 HP sebaiknya mengganti motor baru, agar lebih ekonomis

101 Koreksi Faktor Daya Dengan Memasang Kapasitor
Faktor daya motor induksi < 1 Efisiensi seluruh sistem pabrik akan rendah Kapasitor yang dihubung paralel dapat digunakan untuk memperbaiki faktor daya. Kapasitas kapasitor ditentukan kVA R tanpa beban yang diserap motor Kapasitas kapasitor tidak boleh lebig dari 90% kVAR motor tanpa beban. Kapasitas terlalu besar dapat menyebabkan motor terbakar

102 Perawatan Motor Induksi
Perawatan yang buruk dapat memperburuk efisiensi Pelumasan yang tidak benar dapat menyebabkan meningkatkan gesekan motor dan penggerak transmisi peralatan Kondisi ambien juga akan mempengaruhi kinerja motor suhu ekstrim, kadar debu yang tinggi, atmosfir yang korosif, dan kelembaban dapat merusak sifat-sifat bahan isolasi

103 Periksa Perawatan Motor Induksi
Pemeriksaan motor secara teratur untuk pemakaian bearings dan Pemeriksaan kondisi beban untuk meyakinkan bahwa motor tidak kelebihan atau kekurangan beban Pemeriksaan secara berkala untuk sambungan motor yang benar dan peralatan yang digerakkan Dipastikan bahwa kawat pemasok dan ukuran kotak terminal dan pemasangannya benar Penyediaan ventilasi yang cukup dan menjaga agar saluran pendingin motor bersih

104 Pengendalian Kecepatan Motor
Motor dengan beberapa kecepatan Variable Speed Drives (VSDs) Penggerak Arus Seaarah (DC) Penggerak motor AC dengan gulungan rotor (motor induksi cincin geser)

105 Penurunan Pembebanan Beban yang kurang akan menurunkan efisiensi motor. Ukuran motor harus dipilih berdasarkan pada evaluasi beban dengan hati-hati Penyebab ketidak efisienan : Pembuat peralatan cenderung menggunakan faktor keamanan yang besar bila memilih motor Peralatan kadangkala digunakan dibawah kemampuan yang semestinya. Dipilih motor yang besar agar mampu mencapai keluaran pada tingkat yang dikehendaki, bahkan jika tegangan masuk rendah dalam keadaan tidak

106 Ukuran Motor untuk Beban yang Bervariasi
Motor industru sering beroperasi pada beban bervariasi Biasanya dipilih motor dengan antisipasi paling tinggi Alternatifnya: memilih motor sedikit lebi rendah dari beban antisipasi tertinggi Hal ini memungkinkan karena motor biasanya dirancang 15 % diatas nilai beban Kriteria pemilihan motor : Kenaikan suhu rata-rata diatas siklus operasi aktual harus tidak lebih besar dari kenaikan suhu pada operasi beban penuh yang berkesinambungan (100%)

107 Memperbaiki Kualitas Daya
Fluktuasi tegangan dan frekuensi dapat merigikan kinerja motor Ketidakseimbangan tegangan akan lebih merugikan . Dapat terjadi akibat penggunaan kabel dengan ukuran yang berbeda Keseimbangan fasa maksimum 1% Minimisasi ketidakseimbangan dapat dilakukan dengan Menyeimbangkan setiap beban phasa tunggal diantara seluruh tiga phasa Memisahkan setiap beban phasa tunggal yang mengganggu keseimbangan beban dan umpankan dari jalur/trafo terpisah

108 Penggulungan Ulang (Rewinding)
Biasanya dilakukan pada motor yang terbakar Faktor yang dapat mempengaruhi efisisensi motor: Desain slot dan gulungan Bahan gulungan Kinerja pengisolasi Suhu operasi Indikator keberhasilan penggulungan ulang adalah perbandingan arus dan tahanan sator tanpa sesudanh digulung ulang dan kondisi orisinil

109 Beban Motor Eff. = Efisiensi operasi motor dalam %
HP = Nameplate untuk HP Beban = Daya yang keluar sebagai % laju daya Pi = Daya tiga phasa dalam kW

110 Metode untuk Menentukan Beban Motor
Pengukuran daya masuk. Metode ini menghitung beban sebagai perbandingan antara daya masuk (diukur dengan alat analisis daya) dan nilai daya pada pembebanan 100%. Pengukuran jalur arus beban ditentukan dengan membandingkan amper terukur (diukur dengan alat analisis daya) dengan laju amper. Metode Slip. Beban ditentukan dengan membandingkan slip yang terukur bila motor beroperasi dengan slip untuk motor dengan beban penuh.

111 Faktor-faktor yang Memoengaruhi Kinerja Motor Listrik
Mengganti motor Standar dengan motor efisiensi tinggi Penurunan Pembebanan (menghindari motor yang ukurannya berlebih). Ukuran Motor untuk Beban Yang Bervariasi Memperbaiki Kualitas Daya Penggulungan Ulang (Rewinding) Koreksi Faktor Daya Dengan Memasang Kapasitor

112 Motor Efisiensi Tinggi
Efisiensinya sekitar 3% - 7% lebih besar dari motor standar Desain motor disesuaikan untuk menurungkan kehilangan dasar motor Karakteristik motor efisiensi tinggi : Menggunakan baja silikon Inti lebih panjang Kawat lebih tebal Laminasi lebi tipis Celah udara lebih tipis Bearing lebih bagus, dll

113 Example A two-pole, 50-Hz induction motor supplies 15kW to a load at a speed of 2950 rpm. What is the motor’s slip? What is the induced torque in the motor in N.m under these conditions? What will be the operating speed of the motor if its torque is doubled? How much power will be supplied by the motor when the torque is doubled?

114 Solution

115 Solution In the low-slip region, the torque-speed curve is linear and the induced torque is direct proportional to slip. So, if the torque is doubled the new slip will be 3.33% and the motor speed will be

116 Example A 460-V, 25-hp, 60-Hz, four-pole, Y-connected wound-rotor induction motor has the following impedances in ohms per phase referred to the stator circuit R1= 0.641 R2= 0.332 X1=  X2=  XM= 26.3  What is the maximum torque of this motor? At what speed and slip does it occur? What is the starting torque of this motor? If the rotor resistance is doubled, what is the speed at which the maximum torque now occur? What is the new starting torque of the motor? Calculate and plot the T-s c/c for both cases.

117 Solution

118 Solution The corresponding speed is

119 Solution The torque at this speed is

120 Solution The starting torque can be found from the torque eqn. by substituting s = 1

121 Solution If the rotor resistance is doubled, then the slip at maximum torque doubles too The corresponding speed is The maximum torque is still max = 229 N.m

122 Solution The starting torque is now

123 Determination of motor parameters
Due to the similarity between the induction motor equivalent circuit and the transformer equivalent circuit, same tests are used to determine the values of the motor parameters. DC test: determine the stator resistance R1 No-load test: determine the rotational losses and magnetization current (similar to no-load test in Transformers). Locked-rotor test: determine the rotor and stator impedances (similar to short-circuit test in Transformers).

124 DC test The purpose of the DC test is to determine R1. A variable DC voltage source is connected between two stator terminals. The DC source is adjusted to provide approximately rated stator current, and the resistance between the two stator leads is determined from the voltmeter and ammeter readings.

125 DC test then If the stator is Y-connected, the per phase stator resistance is If the stator is delta-connected, the per phase stator resistance is

126 No-load test The motor is allowed to spin freely
The only load on the motor is the friction and windage losses, so all Pconv is consumed by mechanical losses The slip is very small

127 No-load test At this small slip The equivalent circuit reduces to…

128 No-load test Combining Rc & RF+W we get……

129 No-load test At the no-load conditions, the input power measured by meters must equal the losses in the motor. The PRCL is negligible because I2 is extremely small because R2(1-s)/s is very large. The input power equals Where

130 No-load test The equivalent input impedance is thus approximately
If X1 can be found, in some other fashion, the magnetizing impedance XM will be known

131 Blocked-rotor test In this test, the rotor is locked or blocked so that it cannot move, a voltage is applied to the motor, and the resulting voltage, current and power are measured.

132 Blocked-rotor test The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value. The locked-rotor power factor can be found as The magnitude of the total impedance

133 Blocked-rotor test Where X’1 and X’2 are the stator and rotor reactances at the test frequency respectively

134 X1 and X2 as function of XLR
Blocked-rotor test X1 and X2 as function of XLR Rotor Design X1 X2 Wound rotor 0.5 XLR Design A Design B 0.4 XLR 0.6 XLR Design C 0.3 XLR 0.7 XLR Design D

135 Example The following test data were taken on a 7.5-hp, four-pole, 208-V, 60-Hz, design A, Y-connected IM having a rated current of 28 A. DC Test: VDC = 13.6 V IDC = 28.0 A No-load Test: Vl = 208 V f = 60 Hz I = 8.17 A Pin = 420 W Locked-rotor Test: Vl = 25 V f = 15 Hz I = 27.9 A Pin = 920 W Sketch the per-phase equivalent circuit of this motor. Find the slip at pull-out torque, and find the value of the pull-out torque.