nota_pengubah.doc

Bab 6 Tajuk:Pengubah

PENGUBAH / TRANSFORMER

Objektif Pelajaran : Pada akhir bab ini pelajar akan dapat :

Menyatakan fungsi pengubah Mengenal dan melakar simbol pengubah Mengenal jenis teras udara, teras besi dan serbuk besi Mengenal teras jenis kelompang ( E-I ) dan bentuk teras

tingkap ( U-I ) Mengenal jenis belitan utama dan sekunder Mengenal kaedah belitan sepusat dan belitan apit Menerangkan kendalian asas pengubah Menghitung nisbah belitan, nisbah voltan dan nisbah arus.

(Formula : V1/V2 = N1/N2 = I2/I1 ) Menyatakan kuasa masukan, kuasa keluaran dan kehilangan

pengubah Menerangkan dan menghitung kecekapan pengubah Menyatakan penggunaan pengubah kuasa

Zahari Zain PKE SMT Langkawi 1

UNIT 6


PengenalanPengubah adalah satu alat elektrik yang statik yang digunakan untuk menaikkan ataumenurunkan voltan.Pengubah terdiri dari dua lilitan yang disambungkan melalui terasmagnet. Satu belitan disambung kepada bekalan manakala satu lagi belitan disambungkepada beban.

Binaan asas:1.Teras

diperbuat daripada kepingan keluli silikon yang berlapis dengan penebat tebal keluli antara 1mm-5mm keluli silikon digunakan kerana kehilangan histerisis rendah teras berlapis bertujuan untuk mengurangkan kehilangan kuasa akibat arus pusar

2.Belitan gegelung dibuat daripada dawai kuprum berpenebat terdiri daripada belitan primer dan belitan sekunder


N1 N2 Bekalan AU Beban

Fluks magnet

Belitan sekunderBelitan Primer

Teras

Binaan asas pengubah

Histerisis : kehilangan kuasa elekto magnet yang disebabkan oleh perubahan kutub pada teras akibat berlaku perubahan kitar voltan AU dalam masa yang singkat

Arus pusar (eddy current) : arus yang terhasil pada teras pengubah dan mengakibatkan kepanasan


Jenis teras

a.bentuk kelompang ( E-I ) b.bentuk tingkap ( U- I )

Tujuan:kurangkan kelesapan kuasa dalam teras

Jenis Belitan

a.belitan sepusat b.belitan apit

Simbol Pengubah


Belitan primer

Belitan sekunder

Belitan sekunder

Belitan primer


Jenis Teras Simbol

1.Teras Udara

2.Teras Besi

3.Teras besi serbuk

4.Pengubah Auto

5.Pengubah Sadap Tengah (centre tap)

6.Pengubah berbilang sadap (multi tap)


Simbol am pengubah


Prinsip Kendalian Pengubah

E1 : d.g.e. belitan utamaE2 : d.g.e. belitan sekunderV1 : voltan primer (bekalan/masukan)V2 : voltan sekunder (keluaran ke beban)I1 : arus beliatan primerI2 : arus belitan sekunderN1 : bilangan belitan primerN2 : bilangan belitan sekunder

Kendalian asas pengubahi. belitan primer yang disambungkan ke bekalan AU (V1) akan menghasilkan fluksii. fluks magnet yang mengembang dan menguncup (mengikut frekuensi arus bekalan) akan memotong belitan primer (Hukum Faraday)iii. d.g.e. akan terhasil dalam belitan primer (E1) kesan daripada aruhan diri.iv. fluks yang terhasil akan memotong belitan sekunder melalui teras (E2) kesan dari aruhan saling. v. d.g.e teraruh ini akan disalurkan kepada punca keluaran untuk disambungkan kepada beban.Voltan ini dikenali sebagai voltan sekunder (V2)

Konsep pengubah unggul


V1 V2E1E2

N1 N2

I1I2


1.Secara praktikal akan berlaku kehilangan kuasa di dalam pengubah.Bagi tujuan pengiraan, dianggap pengubah tidak mengalami sebarang kehilangan kuasa atau dianggap unggul.2.Pengubah unggul adalah pengubah di mana kehilangan yang berlaku di dalamnya diabaikan dan ianya berkendali dengan kecekapan 100%.

Bagi pengubah unggul, kadar perubahan fluks adalah sama pada kedua-dua belitan primer dan sekunder,

Oleh itu : = .

D.g.e. teraruh per lilit adalah malar (tetap).Diandaikan tiada kehilangan, E1 = V1 dan E2 = V2

= atau =

Sekiranya satu beban disambungkan kepada belitan sekunder, arus I2 akan mengalir.Bagi pengubah unggul, kehilangan kuasa adalah diabaikan dan kecekapan pada 100%.Oleh itu, kuasa masukan = kuasa keluaran atau V1 I1 = V2 I2 ,

Jadi: =

Dengan menggabungkan persamaan menjadikan:

= =

Nisbah PengubahNisbah antara nilai voltan primer dengan voltan sekunder atau nisbah antara lilitan primer dengan lilitan sekunder.



n = atau

n = , di mana n ialah nisbah pengubah

Jika nilai n adalah lebih besar dari 1, voltan sekunder adalah lebih besar daripada voltan primer.Oleh itu pengubah ini adalah jenis pengubah langkah turun (step down transformer)

Jika n < 1 i.e. N2 < N1 : pengubah jenis langkah naik(step-up transformer)Jika n > 1 i.e. N2 > N1 : pengubah langkah turun (step-down transformer)Jika n = 1 i.e. Ns = Np : pengubah penjodoh atau pengasing (coupling transformer)

Contoh Berikut adalah bilangan belitan bagi tiga pengubah:

(a) N1 = 100, N2 = 2000(b) N1 = 3000, N2= 2000(c) N1 = 100, N2 =100

Berapakah nisbah pengubah n dan nyatakan jenis pengubah berdasarkan nisbah pengubah.

(a)

pengubah langkah naik

(b)

pengubah langkah turun

(c)

pengubah penjodoh

Contoh latihan 1


N1

(100)N2

(2000)

N1

(3000)N2

(2000)

N2

(100)N1

(100)


Satu pengubah mempunyai 600 lilitan utama dan 400 lilitan sekunder. Jika bekalan voltan sebanyak 240V,50 Hz dikenakan ke bahagian utama, hitungkan :

(a) nisbah belitan(b) voltan sekunder

Jawapan :

(a) Nisbah belitan

n = 3 : 2

(b)voltan sekunder, V2

=

V2 = 160 Volt

Contoh latihan 2


V1=240V V2 = ?

N1=600 N2=400


A transformer is to be used to provide a 60 V output from a 240 V A.C supply. Calculate (a) the turns of ratio required and (b) the number of primary turns, if the secondary is wound with 500 turns.Satu pengubah direka untuk mengeluarkan voltan sebanyak 60 V AU daripada voltan bekalan 240 V AU. Kirakan :

(a) nisbah belitan yang diperlukan(b) bilangan belitan lilitan primer jika belitan sekunder dililit sebanyak 500 lilit.

Jawapan Latihan 2

V2 = 60 VV1 = 240 VN2 = 500

a) = =

jadi nisbah belitan ialah n = = atau 4 : 1

b) =

belitan primer, N1 = 2000 lilit

Latihan 3Satu pengubah 2000/200V, 20kVA mempunyai 66 lilitan belitan sekunder.Hitung:



(i) bilangan belitan primer(ii) arus primer dan arus sekunder semasa beban penuh

Abaikan kehilangan kuasa.

Jawapan Latihan 3

Rajah 3

Rajah 3 mewakili keadaan pengubah di atas.

(i)

(ii) (20 kVA=20,000 VA)

arus sekunder,

arus primer, I1 = 10 A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE TO THE NEXT INPUT…!



8.1 A transformer has 500 primary turns and 3000 secondary turns. If the primary voltage is 240 V, determine the secondary voltage, assuming an ideal transformer.

8.2 An ideal transformer with a turns ratio of 2 : 7 is fed from 240 V supply. Determine its output voltage.

8.3 An ideal transformer has a turns ratio of 8 : 1 and the primary current is 3 A when it is supplied at 240 V. Calculate the secondary voltage and current.

8.4 An ideal transformer connected to a 240 V mains, supplies a 12 V, 150 W lamp. Calculate the transformer turns ratio and the current taken from the supply.

8.5 How many turns are required for a 36 V secondary if a 240 V primary has 600 N.8.6 True or false ? If the primary winding of a 1:2 step-up transformer has

3 N/ V, the secondary winding will have 6 N/ V.8.7 True or false ? The secondary flux of a transformer cancels part of the

primary flux.Jawapan

8.1 1440 V8.2 840 V8.3 V2 = 30 volts, I2 = 24 A8.4 turns ratio = 20, I1 = 0.625 A8.5 908.6 False8.7 True

EMF equation of a transformerSuppose the maximum value of flux to be m webers and the frequency to be f hertz. From Fig. 8.6 it is seen that the flux has to change from +m to -m in the half cycle, namely in

seconds.

Average rate of change of flux = 2 m

=4 f m webers per second

and average e.m.f induced per turn is = 4 f m volts



Source: Electrical Electronic Technology by Edward Hughes

Figure 8.6 Waveform of flux variation.

But for a sinusoidal wave the r.m.s or effective value is 1.11 times the average value,RMS value of e.m.f induced per turn = 1.11 4 f m

Hence r.m.s value of e.m.f induced in primary is E1 = 4.44 N1 f m volts

And r.m.s value of e.m.f induced in secondary isE2 = 4.44 N2 f m volts

Example 8.4

A 250 kVA, 1100 V / 400 V, 50 Hz single-phase transformer has 80 turns on a secondary. Calculate :a) the approximate values of the primary and secondary currents.b) the approximate number of primary turns.c) the maximum values of flux.

Solution to Example 8.4a) Full – load primary current

= 250 1000 = 22.7 A 1100and full – loaded secondary current

= 250 1000 = 625 A400

b) Number of primary turns



= 80 11000 = 2200 400

c) From expression ,E2 = 4.44 N2 f m volts

400 = 4.44 80 50 m

m = 22.5 mWb

Example 8.5

An ideal 25 kVA transformer has 500 turns on the primary winding and 40 turns on the secondary winding. The primary is connected to 3000 V, 50 Hz supply. Calculate

(i) primary and secondary currents on full-load(ii) secondary e.m.f. and(iii) the maximum core flux

Solution to Example 8.5

(i)

(ii)

(iii)

Activity 8B




8.8 A 100 kVA, 4000 V/200 V, 50 Hz single phase transformer has 100 secondary turns. Determine :(a) the primary and secondary current(b) the number of primary turns(c) the maximum value of the flux

8.9 A 4500 V/225 V, 50 Hz single phase transformer is to have an approximate e.m.f per turns of 15 V and operate with a maximum flux of 1.4 T. Calculate :(a) the number of primary and secondary turns(b) the cross sectional area of the core

8.10 A single phase 2200/250V, 50 Hz transformer has a net core area of 36cm 2 and a maximum flux density of 6 wb/m2. Calculate the number of primary and secondary turns.

Feedback to the Activity 8B

8.8 (a) I1 = 25 A, I2 = 500 A(b) N1 = 2000 turns(c) m = 9.01 mWb

8.9 (a) N1=300, N2= 15

(b) A = 0.0483 m2

8.10 N1 = 459, N2 = 52

Kehilangan dan kecekapan pengubah



a.arus pusar ( eddy current ) -arus yang terhasil di dalam teras apabila berlakunya perubahan fluks magnet dalam

teras -arus ini akan memanaskan teras dan menyebabkan kuasa kemagnetan berkurangan -cara atasi : teras dibina berlapis-lapis (Rujuk m.s. 134)

b.histerisis -kehilangan kekuatan magnet disebabkan oleh perubahan arah arus bekalan AU -perubahan arus yang kerap akan menyebabkan domain magnet dalam teras berubah -kehilangan kuasa berlaku kerana sebahagian tenaga diperlukan untuk menukar arah

domain tersebut. -keadaan ini dinamakan histerisis -cara atasi : teras yang baik seperti besi lembut/ferrite

c.kebocoran fluks -tidak semua fluks magnet memotong pengalir sekunder, sebahagiannya akan hilang

dalam litar kemagnetan -kehilangan amat kecil dan diabaikan dalam pengiraan -cara atasi : lilitan bertindih atau sepusat antara primer dan sekunder

Kecekapan Pengubah


Kehilangan Kuasa

Kehilangan Besi, Pi

Kehilangan Kuprum, Pc

Berlaku dalam belitan Bergantung kepada nilai

arus pada belitan utama & belitan sekunder

Pc = I 2 R

Berlaku dalam teras besi Nilai kehilangan adalah tetap

samada beban penuh atau separuh beban.

3 bentuk kehilangan: arus pusar histerisis kebocoran fluks

magnet


Kecekapan pengubah adalah nisbah antara kuasa keluaran terhadap kuasa masukan.Kecekapan dinyatakan dalam peratus atau perpuluhan.Kecekapan ditandakan dengan simbol (eta).

Kecekapan = x 100 %

= x 100 %

= x 100 %

Kecekapan = x 100 %

dengan V2 = voltan keluaran I2 = arus sekunder Kos = f.k. pada beban Pi = kehilangan besi

Pc = kehilangan kuprum


Pengubahmasukan keluaran

PinPout

=95% -98%

kehilangan

kuprum,Pc besi, Pi


Hasil darab I2V2 adalah kadaran kuasa pengubah atau kuasa keluaran ketara.Unitnya adalah dalam Volt-Ampere (VA) Kecekapan pengubah bergantung kepada :

Faktor kuasa Beban Kehilangan kuasa

Contoh kadaran pengubah

Contoh Pengiraan Kecekapan Pengubah:


Kadaran kuasa maksimum pengubah

Frekuensi bekalan

Voltan Masukan

Voltan Keluaran


Contoh 1Satu pengubah 4 kVA fasa tunggal mengalami kehilangan besi 400 W dan kehilangan kuprum 600 W. Jika faktor kuasa pengubah adalah 0.8, hitungkan kecekapan pengubah tersebut.

Kecekapan = x 100 %

= x 100 %

= x 100 %

= 76.19 %

Contoh 2Satu pengubah 110 kVA, 2 000/100 V fasa tunggal mengalami kehilangan besi 750 W dan kehilangan kuprum 850 W.Faktor kuasa pengubah adalah 0.85.



Hitungkan kecekapan pengubah itu.

Kecekapan = x 100 %

= x 100 %

= x 100 %

= 98.32 %

Contoh 3In a 50 kVA transformer, the iron loss is 500 W and full-load copper loss is 800W. Find the efficiency at full-load and half-load at 0.8 p.f. lagging.



Sebuah pengubah 50 kVA mempunyai kehilangan besi sebanyak 500W dan kehilangan kuprum sebanyak 800W pada faktor kuasa 0.8.Hitung kecekapan pengubah tersebut:

(a) beban penuh(b) beban separuh

(a) Kecekapan pada beban penuh :

Kecekapan = x 100 %

= x 100 %

= 96.85 %

(b) Kecekapan pada separuh beban

Kecekapan 1/2 = x 100 %

= x 100 %

1/2 = 96.6 %



Contoh soalan SPM 2000

Rajah di atas menunjukkan satu pengubah jenis langkah naik 6 kVA, 240/400V 50 Hz.

(a) Hitungkan:(i) Arus I2

(ii) Rintangan RL semasa beban penuh.(i) Jumlah belitan primer, N1 sekiranya N2 bersamaan 125 belitan.

[ 9 markah](b) Sekiranya pengubah mempunyai kehilangan besi 35 W dan kehilangan kuprum

48 W pada faktor kuasa 0.9 , hitungkan:(i) Kecekapan pengubah ketika beban penuh, η(ii) Kecekapan pengubah ketika beban separuh, η1/2

(iii) Kehilangan kuprum apabila faktor kuasa pengubah menurun ke 0.7 pada kecekapan 97%.

[ 11 markah ]



Jawapan: (a)i Arus I2 :

( ii)Rintangan RL

= 26.6

(iii) belitan primer, N1

=

(b) i. Kecekapan beban penuh,

Kecekapan = x 100 %

= x 100 %

= x 100

= x 100

= 0.9845 x 100 %

= 98.45 %

(b) ii. Kecekapan beban separuh, 1/2



Kecekapan 1/2 = x 100 %

= x 100 %

= x 100 %

1/2 = 98.29 %

(ii) Kehilangan kuprum

= x 100 %

masukan =

=

= 4330 Watts

Masukan = Keluaran + kehilangan Kehilangan = 4330W – 6 kVA x 0.7 = 130 W Kehilangan kuprum Pc: = 130W – 35 W = 95 W




8.11 The ideal transformer provides 20 V at 9.5 A to a pure resistance load. Determine its efficiency.

8.12 The open circuit test measures the _______________ losses of a transformer.

8.13 Why are the core losses negligible during the short-circuit test?

8.14 A transformer has a 120 V primary and 20 V secondary rate at 4A. The measured winding resistances are 0.3 and 5 for secondary and primary, respectively. Estimate the copper loss.

Feedback to Activity 8C

8.10 0.968.11 core8.12 Because the primary voltage is very low.8.13 7 W

SELF-ASSESSMENT 8



You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback on Self-Assessment 8 given on the next page. If you face any problems, discuss it with your lecturer. Good luck.

Question 8-1

(a) What is a transformer ?(b) Draw the circuit diagram symbol for a transformer.(c) How is a transformer rated. (d) Name two types of transformer construction.(e) A transformer has 600 primary turns connected to a 1.5 kV supply. Determine the

number of secondary turns for a 240 V output voltage, assuming no losses.(f) An ideal transformer with a turns ratio of 2 : 9 is fed from 220 V supply. Determine

its output voltage.(g) A step-down transformer having a turns ratio of 20 : 1 has a primary voltage of 4 kV

and a load of 10kW. Neglecting losses, calculate the value of the secondary current.(h) An ideal transformer has a turns ratio of 12 : 1 and is supplied at 192 V. Calculate the

secondary voltage.(i) A transformer primary winding connected across a 415 V supply has 750 turns.

Determine how many turns must be wound on the secondary side if an output of 1.66 kV is required.

(j) An ideal transformer has a turns ratio of 12 : 1 and is supplied at 180V when the primary current is 4A. Calculate the secondary voltage and current.

Question 8-2

(a) A 50kVA, 6600/250V transformer has 52 secondary turns. Find (i) the number of primary turns(ii) full-load primary and secondary currents.

(Neglect losses)

(b) The net cross-sectional area of the core of 400/3000V, 50Hz transformer is 600cm2. If the maximum flux density in the core is 1.3 wb/m2, find the number of primary and secondary turns.



(c) A 60 kVA, 1600 V/100 V, 50 Hz, single phase transformer has 50 secondary windings. Calculate :(i) the primary and secondary current(ii) the number of primary turns(iii) the maximum value of the flux

(d) A single phase, 50 Hz transformer has 40 primary turns and 520 secondary turns. The cross sectional area of the core is 270 cm2. When the primary winding is connected to a 300 V supply, determine :(i) the maximum value of flux density in the core(ii) the voltage induced in the secondary winding

(e) A 3.3 kV/110 V, 50 Hz, single phase transformer is to have an approximate e.m.f per turn of 22 V and operate with a maximum flux of 1.25 T. Calculate (i) the number of primary and secondary turns(ii) the cross sectional area of the core

Question 8-3

(a) A 200 kVA rated transformer has a full loaded copper loss of 1.5 kW and an iron loss of 1 kW. Detemine the transformer efficiency at full load and 0.85 power factor.

(b) A 400 kVA transformer has a primary winding resistance of 0.5 and a secondary winding resistance of 0.001. The iron loss is 2.5 kW and the primary and secondary voltage are 5 kV and 320 V respectively. If the power factor of the load is 0.85, determine the efficiency of the transformer :i. on a full loadii. on half load

(c) A single phase transformer has a voltage ratio of 6 : 1 and the h.v winding is supplied at 540 V. The secondary winding provides at full load current of 30 at a power factor of 0.8 lagging. Neglecting losses, find :i. a rating of the transformerii. the power supplied to loadiii. the primary current.

FEEDBACK TO SELF-ASSESSMENT 8

Answer of Question 8-1



(a) A transformer is a device which uses the phenomenon of mutual induction to change the values of alternating voltages and currents.In fact, one of the main advantages of AC transmission and distribution is the ease with which an alternating voltage can be increased or decreased by transformers.

(b)

(c) K =

(d) i. core type ii. shell type

(e) 96

(f) 990 V

(g) 50 A

(h) 16 V

(i) 3000 turns

(j) 15 V, 48 A Answer of Question 8-2

(a) (i) 1373 (ii) 7.58 A; 200 A

(b) 24; 198

(c) (i) 37.5 A, 600 A (ii) 800 (iii) 9.0 mWb

(d) (i) 1.25 T (ii) 3.90 kV

(e) (i) 150, 5 (ii) 792.8 cm2

Answer of Question 8-3

(a) 0.9855 or 98.55 %(b) i. 97.91 % ii. 97.87 %


Np Ns


(c) i. 2,7 kVA ii. 2.16 kW iii. 5 A


nota_pengubah.doc

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