Power Factor (PF)

Learning Outcomes

At the end of the lecture, students are expected to be able

to calculate the size of a capacitor for PF correction, and

to calculate the total energy save per year, penalty per year and the payback period.

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Introduction

PF is used to measure the efficiency level of electricity usage.

=

=

cos

= cos

where = real power (W)

= apparent power (VA) = 2 + 2

= reactive power (Var)

& = rms current & rms voltage

cos = angle between & I P

Q

S

q

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Causes of Low PF

1. Transformer and induction motor

Magnetic current magnetic field

2. Arc lamps, electric discharge lamps and industrial heating system

They require control gear in a form of chokes, ballasts and etc.

3. Varying load condition

During low load period, supply voltage is increase which increase magnetizing current.

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Effects of low PF

1. Large kVA rating of equipment (more expensive).

2. Greater conductor size (size of cable depends on current).

3. Large copper losses (poor efficiency).

4. Poor Voltage Regulation (VR).

VR is a measure of change in the voltage magnitude between the sending and receiving end or from no load to full load conditions.

5. Reduced handling capacity of system.

6. Penalty from power utilities.

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PF Correction

1. Static capacitor (bank).

2. Synchronous condenser.

3. Phase advancer.

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Static Capacitor (1)

Install in shunt

Utility side pole mounted racks, enclosed bank (LV) & substation banks (HV EHV).

Load side individual or group loads.

http://www.tamintablo.com/english/product/Capacitorbanks.aspx

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Static Capacitor (2)

Advantages: cheapest, less maintenance and simplest method.

Disadvantages:

Resonance amplification of harmonic voltage overvoltage.

Oversize capacitor and switching operation transient voltage tripping.

repair is uneconomical.

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Synchronous Condenser (1)

Connected in parallel with power supply; running on no load condition.

It takes a leading current when over-exited, and therefore behaves as a capacitor.

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https://jawnsy.wordpress.com/page/3/

Synchronous Condenser (2)

Advantages

Stepless control varying the field excitation

Long useful life the windings have high thermal stability

No resonance, voltage spike and static components

Disadvantages losses in the motor, high maintenance cost, auxiliary circuit and noise

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Phase Advancer

Used to improve the PF of induction motors.

It is mounted on the same shaft as the main motor and is connected in the rotor circuit.

It provides required magnetomotive force (MTF) or ampere turns to the rotor at slip frequency.

Uneconomical to be applied on motors below 200hp.

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Example

A 5 MVA transformer is loaded to 4.5 MVA at a power factor of 0.82 lag. The transformer has a rated conductor loss equal to 1% of the transformer rating. The tariff given by power utility is RM0.17/kWhr and the penalty charge is 0.20sen/kVARhr (if PF falls below 0.95). Assume that the transformer operates for 24 hours per day.

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a) Calculate the leading kVar necessary to correct the PF to 0.95 lag.

Reactive power needed for PF correction, = 1 2

where

1 = reactive power before PF correction (Var)

2 = reactive power after PF correction (Var)

(Answer: 1,363kVar)

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b) Calculate the size of a PF-corrector capacitor. The capacitor will be installed in 3-phase system (400V, 50Hz, Y-configuration).

PF capacitor, =

2

where

= angular frequency = 2

= operating frequency

= rms voltage accross the capacitor terminal

(Answer: 81,348F)

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c) Calculate energy saved per year.

Energy saved per day(kWhr) = loss base operating hour

loss (p.u.) = lossold PF lossnew PF

loss p.u. = %loss real

2

(Answer: 90,490.8kWhr)

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d) Calculate cost saved per annum.

Cost saved per year (RM) = energy saved per year (kWhr) tariff

(Answer: RM15,383.44)

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e) Calculate the penalty charge per year

Penalty RM = 1 in kVar operating hour penalty 0.01

(Answer: RM23,879.76)

Other alternative calculation, please refer TNB regulation.

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f) If the maximum demand charge per kVA is RM14 (assume per month), calculate the reduction in maximum demand and its saving in a month.

Maximum demand, =

(Answer: 616kVar, RM8,624)

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g) Calculate the cost of the PF capacitor and its payback period. Assume that the capacitor cost per kVar is RM150, the rate of depreciation per year is 7% and the bank interest charge per year is 8%,

payback period =capacitor cost

Net saving per year

Net saving per year

= Saving demand charge per year total depreciation interest

total depreciation interest

= capacitor cost depreciation rate + bank interest

(Answer: RM204,450, 2.8 years)

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