Tugas Rangkaian Listrik(Perbaikan faktor daya)

Nama : Norman Eko PutroKelas: 2ENIM: 1211020023

Sumber 1: http://maruzar.blogspot.com/2012/01/perbaikan-power-factor-dengan-kapasitor.html

Perbaikan Power Factor dengan Kapasitor

Untuk memperbaiki power factor (pf, Cos Phi) pada alternating current (AC) biasanya digunakan kapasitor. Cara ini cukup populer diterapkan, misalnya pada lampu neon, motor, transformator, dll. Skema rangkaiannya dapat dilihat pada gambar dibawah. Instrument yang dibutuhkan adalah Power meter (P) dan Ammeter (A). Kapasitor (C) mempunyai suatu harga tertentu, tidak boleh terlalu besar, dan tidak boleh terlalu kecil, agar power factor mendekati 1 (satu, ideal)sehingga konsumsi daya listrik jadi irit. Posisi kapasitor harus sedekat mungkin dengan beban.

Untuk melihat grafiksimulasi gelombang tegangan dan aruspada power factor tertentu dibahas di artikel terpisah. Demikian pula tentang definisi:effisiensi, power factor, watt, voltampere; dibahas di artikel tersendiri.

Contoh perhitungan disini terkandung padalembar Excel terlampiryang mempunyai rumus untuk menghitung koreksi power factor dengan kapasitor. Sebelum kapasitor dipasang untuk memperbaiki power factor, maka perlu diketahui dulu seberapa besar power factor yang terjadi pada suatu beban daya (real power) tertentu.

Misalkan didapat data sebagai berikut:Frequensi : 60 HzTegangan : 220 voltArus : 1 ampereDaya aktual (real power): 160 watt

Daya yang diberikan oleh sumber arus:apparent power = 220 volt x 1 ampere = 220 voltampere

Power factor = pf = real power / apparent power = 160 / 220 = 0,73

Lalu reactive power dihitung dengan menggunakan formula Phytagoras:

Reactive power = ((apparent power x apparent power) (real power x real power))1/2Reactive power = (2202 1602)1 / 2= 151 var (voltampere reactive)

Dari data reactive power, maka akan didapat data reactive load:Reactive load = Voltage2/ reactive power = 2202/ 151 = 320,5W

Untuk menghitung besarnya kapasitas maka digunakan rumus Faraday:Capacity = 1 / ( 2px frequencyx reactive load ) = 1/( 2px 60 x 320,5 ) = 8,27 f

Misal dipilih dan dipasang paralel kapasitor sebesar 8 f untuk mengkoreksi power factor, maka nilai untuk capasitive reactance yang ditimbulkan capasitor tersebut adalah:xc= 1 / (2px 60 x 8 x 10-6) = 331.44W

Arus yang mengalir untuk capasitive reactance ini adalah:ac= 220 / 331.44 = 0.66 amperes

Daya untuk capasitive reactance, perlu diketahui bahwa vektor reactive power ini arahnya berlawanan dengan vektor reactive power sebelumnya, sehingga diberi tanda negatif ( - ),adalah:pc= 220 x 0,66 = - 146.03 var

Reactive power yang baru atau total adalah:Q1= 151 146.03 = 4,97 var

Apparent power yang terkoreksi adalah :S1= ((real power)2+ (new reactive power)2)1/2= (1602+ 4,972)1/2= 160.08 va

Sehingga power faktor yang terkoreksi atau diperbaiki adalah:pf1= 160 / 160,08 =0,9995Posted9th JanuarybyHeru Maruza

Sumber 2: http://www.allaboutcircuits.com/vol_2/chpt_11/4.html

Practicalpowerfactor correctionWhen the need arises to correct for poorpowerfactorin an ACpowersystem, you probably won't have the luxury of knowing the load's exact inductance in henrys to use for your calculations. You may be fortunate enough to have an instrument called apowerfactormeterto tell you what thepowerfactoris (a number between 0 and 1), and the apparentpower(which can be figured by taking a voltmeter reading in volts and multiplying by an ammeter reading in amps). In less favorable circumstances you may have to use an oscilloscope to compare voltage and current waveforms, measuring phase shift indegreesand calculatingpowerfactorby the cosine of that phase shift.Most likely, you will have access to a wattmeter for measuring truepower, whose reading you can compare against a calculation of apparentpower(from multiplying total voltage and total current measurements). From the values of true and apparentpower, you can determine reactivepowerandpowerfactor. Let's do an example problem to see how this works: (Figurebelow)

Wattmeter reads truepower; product of voltmeter and ammeter readings yields appearantpower.First, we need to calculate the apparentpowerin kVA. We can do this by multiplying load voltage by load current:

As we can see, 2.308 kVA is a much larger figure than 1.5 kW, which tells us that thepowerfactorin this circuit is rather poor (substantially less than 1). Now, we figure thepowerfactorof this load by dividing the truepowerby the apparentpower:

Using this value forpowerfactor, we can draw apowertriangle, and from that determine the reactivepowerof this load: (Figurebelow)

Reactivepowermay be calculated from truepowerand appearantpower.To determine the unknown (reactivepower) triangle quantity, we use the Pythagorean Theorem backwards, given the length of the hypotenuse (apparentpower) and the length of the adjacent side (truepower):

If this load is an electric motor, or most any other industrial AC load, it will have a lagging (inductive)powerfactor, which means that we'll have to correct for it with acapacitorof appropriate size, wired in parallel. Now that we know the amount of reactivepower(1.754 kVAR), we can calculate the size of capacitor needed to counteract its effects:

Rounding this answer off to 80 F, we can place that size of capacitor in the circuit and calculate the results: (Figurebelow)

Parallel capacitor corrects lagging (inductive) load.An 80 F capacitor will have a capacitive reactance of 33.157 , giving a current of 7.238 amps, and a corresponding reactivepowerof 1.737 kVAR (for the capacitoronly). Since the capacitor's current is 180oout of phase from the the load's inductive contribution to current draw, the capacitor's reactivepowerwill directly subtract from the load's reactivepower, resulting in:

This correction, of course, will not change the amount of truepowerconsumed by the load, but it will result in a substantial reduction of apparentpower, and of the total current drawn from the 240 Volt source: (Figurebelow)

Powertriangle before and after capacitor correction.The new apparentpowercan be found from the true and new reactivepowervalues, using the standard form of the Pythagorean Theorem:

This gives a correctedpowerfactorof (1.5kW / 1.5009 kVA), or 0.99994, and a new total current of (1.50009 kVA / 240 Volts), or 6.25 amps, a substantial improvementover the uncorrected value of 9.615 amps! This lower total current will translate to less heat losses in the circuit wiring, meaning greater system efficiency (lesspowerwasted).