This section will calculate the kVAR rating of the capacitor required to increase the power factor to a more efficient value, while decreasing utility energy consumption.
Advantages of power factor improvement.
- Reduce energy bill.
- Increase of system capacity.
- Reduce energy loss in cable.
- Reduce cable size, transformer and switcher gear rating.
Methods of pf improvement.
- Provide shunt capacitors - for small and medium industries.
- Synchronous motors for large industries.
Capacitor bank design
Example:
Let us assume connected load 1328.25kW
Take diversity factor = 0.85
We know that,
Maximum demand = diversity factor x connected load
=> 0.85 x 1328.25 = 1129.01 kW
Consider the actual value of p.f. is 0.8 lagging
There fore, cosØ1 = 0.8
Ø1 = cos-1(0.8) = 36.870
Assume the required value of p.f. is 0.95 lagging
There fore, cosØ2 = 0.8
Ø2 = cos-1(0.95) = 18.200
kVAR Req = kVAR1 - kVAR2
= kW x tan Ø1 – kW x tan Ø2
= 1328.25 x tan (36.87) – 1325.25 x tan (18.2)
= 559.4 kVAR
There fore capacitor bank required is 600kVAR
tan (36.87) – tan (18.2) = 0.4211557
≈0.4212
So the multiplying factor is 0.4212
That is connected load x 0.4212 gives the required kVAR rating of capacitor bank.
The value of capacitor is depends on the power factor pf
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Common electrical units used in formulas and equations are:
ReplyDeleteVolt - unit of electrical potential or motive force - potential is required to send one ampere of current through one ohm of resistance
Ohm - unit of resistance - one ohm is the resistance offered to the passage of one ampere when impelled by one volt
Ampere - units of current - one ampere is the current which one volt can send through a resistance of one ohm
Watt - unit of electrical energy or power - one watt is the product of one ampere and one volt - one ampere of current flowing under the force of one volt gives one watt of energy
Volt Ampere - product of volts and amperes as shown by a voltmeter and ammeter - in direct current systems the volt ampere is the same as watts or the energy delivered - in alternating current systems - the volts and amperes may or may not be 100% synchronous - when synchronous the volt amperes equals the watts on a wattmeter - when not synchronous volt amperes exceed watts - reactive power
Kilovolt Ampere - one kilovolt ampere - KVA - is equal to 1,000 volt amperes
Power Factor - ratio of watts to volt amperes
Electric Power Formulas
W = E I (1a)
W = R I2 (1b)
W = E2/ R (1c)
where
W = power (Watts)
E = voltage (Volts)
I = current (Amperes)
R = resistance (Ohms)
Electric Current Formulas
I = E / R (2a)
I = W / E (2b)
I = (W / R)1/2 (2c)
Electric Resistance Formulas
R = E / I (3a)
R = E2/ W (3b)
R = W / I2 (3c)
Electrical Potential Formulas - Ohms Law
Ohms law can be expressed as:
E = R I (4a)
E = W / I (4b)
E = (W R)1/2 (4c)
Example - Ohm's law
A 12 volt battery supplies power to a resistance of 18 ohms.
I = (12 Volts) / (18 ohms)
= 0.67 Ampere
Electrical Motor Formulas
Electrical Motor Efficiency
μ = 746 Php / Winput (5)
where
μ = efficiency
Php = output horsepower (hp)
Winput = input electrical power (Watts)
or alternatively
μ = 746 Php / (1.732 E I PF) (5b)
Electrical Motor - Power
W3-phase = (E I PF 1.732) / 1,000 (6)
where
W3-phase = electrical power 3-phase motor (kW)
PF = power factor electrical motor
Electrical Motor - Amps
I3-phase = (746 Php) / (1.732 E μ PF) (7)
where
I3-phase = electrical current 3-phase motor (Amps)
PF = power factor electrical motor
ARUN MARTIN
The power factor of an AC electric power system is defined as the ratio of the active (true or real) power to the apparent power
ReplyDeletewhere
Active (Real or True) Power is measured in watts (W) and is the power drawn by the electrical resistance of a system doing useful work.
Apparent Power is measured in volt-amperes (VA) and is the voltage on an AC system multiplied by all the current that flows in it. It is the vector sum of the active and the reactive power.
Reactive Power is measured in volt-amperes reactive (VAR). Reactive Power is power stored in and discharged by inductive motors, transformers and solenoids
Reactive power is required for the magnetization of a motor but doesn't perform any action. The reactive power required by inductive loads increases the amounts of apparent power - measured in kilovolt amps (kVA) - in the distribution system. Increasing of the reactive and apparent power will cause the power factor - PF - to decrease.
Power Factor
It is common to define the Power Factor - PF - as the cosine of the phase angle between voltage and current - or the "cosφ".
The power factor defined by IEEE and IEC is the ratio between the applied active (true) power - and the apparent power, and can in general be expressed as:
PF = P / S (1)
where
PF = power factor
P = active (true or real) power (Watts)
S = apparent power (VA, volts amps)
A low power factor is the result of inductive loads such as transformers and electric motors. Unlike resistive loads creating heat by consuming kilowatts, inductive loads require a current flow to create magnetic fields to produce the desired work.
Power factor is an important measurement in electrical AC systems because
an overall power factor less than 1 indicates that the electricity supplier need to provide more generating capacity than actually required
the current waveform distortion that contributes to reduced power factor is caused by voltage waveform distortion and overheating in the neutral cables of three-phase systems
International standards such as IEC 61000-3-2 have been established to control current waveform distortion by introducing limits for the amplitude of current harmonics.
Example - Power Factor
A industrial plant draws 200 A at 400 V and the supply transformer and backup UPS is rated 200 A × 400 V = 80 kVA.
If the power factor - PF - of the loads is only 0.7 - only
80 kVA × 0.7
= 56 kVA
of real power is consumed by the system. If the power factor is close to 1 the supply system with transformers, cables, switchgear and UPS could be made considerably smaller.
A low power factor is expensive and inefficient and some utility companies may charge additional fees when the power factor is less than 0.95. A low power factor will reduce the electrical system's distribution capacity by increasing the current flow and causing voltage drops.
Power Factor for a Three-Phase Motor
The total power required by an inductive device as a motor or similar consists of
Active (true or real) power (measured in kilowatts, kW)
Reactive power - the nonworking power caused by the magnetizing current, required to operate the device (measured in kilovars, kVAR)
The power factor for a three-phase electric motor can be expressed as:
PF = P / [(3)1/2 U I] (2)
where
PF = power factor
P = power applied (W, watts)
U = voltage (V)
I = current (A, amps)
ARUN MARTIN
Light Level or Illuminance, is the total luminous flux incident on a surface, per unit area. The work plane is where the most important tasks in the room or space are performed.
ReplyDeleteMeasuring Units Light Level - Illuminance
Illumenance is measured in foot candles (ftcd, fc, fcd) (or lux in the metric SI system). A foot candle is actually one lumen of light density per square foot, one lux is one lumen per square meter.
lux = 10.752 fc
ftcd = lux / 10.752
Common Light Levels Outdoor
Common light levels outdoor at day and night can be found in the table below:Condition Illumination
(ftcd) (lux)
Sunlight 10,000 107,527
Full Daylight 1,000 10,752
Overcast Day 100 1,075
Very Dark Day 10 107
Twilight 1 10.8
Deep Twilight .1 1.08
Full Moon .01 .108
Quarter Moon .001 .0108
Starlight .0001 .0011
Overcast Night .00001 .0001
Common and Recommended Light Levels Indoor
The outdoor light level is approximately 10,000 lux on a clear day. In the building, in the area closest to windows, the light level may be reduced to approximately 1,000 lux. In the middle area its may be as low as 25 - 50 lux. Additional lighting equipment is often necessary to compensate the low levels.
Earlier it was common with light levels in the range 100 - 300 lux for normal activities. Today the light level is more common in the range 500 - 1000 lux - depending on activity. For precision and detailed works, the light level may even approach 1500 - 2000 lux.
The table below is a guidance for recommended light level in different work spaces:Activity Illumination
(lux, lumen/m2)
Public areas with dark surroundings 20 - 50
Simple orientation for short visits 50 - 100
Working areas where visual tasks are only occasionally performed 100 - 150
Warehouses, Homes, Theaters, Archives 150
Easy Office Work, Classes 250
Normal Office Work, PC Work, Study Library, Groceries, Show Rooms, Laboratories 500
Supermarkets, Mechanical Workshops, Office Landscapes 750
Normal Drawing Work, Detailed Mechanical Workshops, Operation Theatres 1,000
Detailed Drawing Work, Very Detailed Mechanical Works 1500 - 2000
Performance of visual tasks of low contrast and very small size for prolonged periods of time 2000 - 5000
Performance of very prolonged and exacting visual tasks 5000 - 10000
Performance of very special visual tasks of extremely low contrast and small size 10000 - 20000
ARUN MARTIN
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ReplyDeleteThe table below can be used as a guide to installed light power (Watts/Sq.Ft.) in some common types of buildings and rooms.Type of Building/Room Light Power
ReplyDelete(Watts/Sq.Ft.)
Apartments 1.0 - 4.0
Banks 2.0 - 5.0
Bars 1.5 - 2.0
Cafeterias 1.5 - 2.5
Churches 1.0 - 3.0
Clubhouses 1.5 - 2.0
Cocktail Lounges 1.5 - 2.0
Computer Rooms 1.5 - 5.0
Court Houses 2.0 - 5.0
Dental Centers 1.5 - 2.5
Department Stores 1.0 - 3.0
Dining Halls 1.5 - 2.5
Drug Store 1.0 - 3.0
Fire Stations 2.0 - 3.0
Hospitals, General Areas 1.5 - 2.5
Hotels Public Spaces 1.5 - 3.0
Kindergarten 1.0 - 3.0
Kitchens 1.5 - 3.0
Jail 1.0 - 2.5
Libraries 1.0 - 3.0
Luncheonettes 1.5 - 2.5
Lunch Rooms 1.5 - 2.5
Malls 1.0 - 3.0
Medical Centers 1.5 - 2.5
Motels Public Areas 1.5 - 3.0
Motels Guest Rooms 1.0 - 3.0
Municipal Buildings 2.0 - 5.0
Museums 1.0 - 3.0
Nightclubs 1.5 - 2.0
Nursing Home Patient Rooms 2.0 - 3.0
Police Stations 2.0 - 3.0
Post Offices 2.0 - 3.0
Precision Manufacturing 2.0 - 3.0
Residential 1.0 - 4.0
Retail Stores 2.0 - 6.0
Restaurants 1.5 - 3.0
School Classrooms 2.0 - 6.0
Shops 1.0 - 3.0
Supermarkets 1.0 - 3.0
Taverns 1.5 - 2.0
Town Halls 2.0 - 5.0
1 ft2 = 0.0929 m2 = 144 in2 = 0,1111 yd2
1 Watts/ft2 = 10.8 Watts/m2
ARUN MARTIN
thanks dear arun martin for posting
ReplyDelete