## Calculate No of Lighting Fixtures / Lumen for Indoor Lighting

• An office area is 20meter (Length) x 10meter (width) x 3 Meter (height). The ceiling to desk height is 2 meters. The area is to be illuminated to a general level of 250 lux using twin lamp 32 watt CFL luminaires with a SHR of 1.25. Each lamp has an initial output (Efficiency) of 85 lumen per watt. The lamps Maintenance Factor (MF) is 0.63 ,Utilization Factor is 0.69 and space height ratio (SHR) is 1.25

## Calculation:

### Calculate Total Wattage of Fixtures:

• Total Wattage of Fixtures= No of Lamps X each Lamp’s Watt.
• Total Wattage of Fixtures=2×32=64Watt.

### Calculate Lumen per Fixtures:

•  Lumen per Fixtures = Lumen Efficiency(Lumen per Watt) x each Fixture’s Watt
• Lumen per Fixtures= 85 x 64 = 5440Lumen

### Calculate No’s of Fixtures:

•  Required No of Fixtures = Required Lux x Room Area / MFxUFx Lumen per Fixture
• Required No of Fixtures =(250x20x10) / (0.63×0.69×5440)

### Calculate Minimum Spacing Between each Fixture:

•  The ceiling to desk height is 2 meters and Space height Ratio is 1.25 so
• Maximum spacing between Fixtures =2×1.25=2.25meter.

### Calculate No of Row Fixture’s Row Required along with width of Room:

•  Number of Row required = width of Room / Max. Spacing= 10/2.25
• Number of Row required=4.

### Calculate No of Fixture’s required in each Row:

•  Number of Fixture Required in each Row = Total Fixtures / No of Row = 21/4
• Number of Fixture Required in each Row = 5 No’s:

### Calculate Axial Spacing between each Fixture:

•  Axial Spacing between Fixtures = Length of Room / Number of Fixture in each Row
• Axial Spacing between Fixtures =20 / 5 = 4 Meter

### Calculate Transverse Spacing between each Fixture:

•  Transverse Spacing between Fixtures = width of Room / Number of Fixture’s row
• Transverse Spacing between Fixtures = 10 / 4 = 2.5 Meter.

## Conclusion:

•  No of Row for Lighting Fixture’s= 4 No
• No of Lighting Fixtures in each Row= 5 No
• Axial Spacing between Fixtures= 4.0 Meter
• Transverse Spacing between Fixtures= 2.5 Meter
• Required No of Fixtures =21 No’s

## Calculate Size of Capacitor Bank / Annual Saving & Payback Period

• Calculate Size of Capacitor Bank Annual Saving in Bills and Payback Period for Capacitor Bank.
• Electrical Load of (1) 2 No’s of 18.5KW,415V motor ,90% efficiency,0.82 Power Factor ,(2) 2 No’s of 7.5KW,415V motor ,90% efficiency,0.82 Power Factor,(3) 10KW ,415V Lighting Load. The Targeted Power Factor for System is 0.98.
• Electrical Load is connected 24 Hours, Electricity Charge is 100Rs/KVA and 10Rs/KW.
• Calculate size of Discharge Resistor for discharging of capacitor Bank. Discharge rate of Capacitor is 50v in less than 1 minute.
• Also Calculate reduction in KVAR rating of Capacitor if Capacitor Bank is operated at frequency of 40Hz instead of 50Hz and If Operating Voltage 400V instead of 415V.
• Capacitor is connected in star Connection, Capacitor voltage 415V, Capacitor Cost is 60Rs/Kvar. Annual Deprecation Cost of Capacitor is 12%.

## Calculation:

• For Connection (1):
• Total Load KW for Connection(1) =Kw / Efficiency=(18.5×2) / 90%=41.1KW
• Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 41.1 /0.82=50.1 KVA
• Total Load KVA (new) for Connection(1)= KW /New Power Factor= 41.1 /0.98= 41.9KVA
• Total Load KVAR= KWX([(√1-(old p.f)2) / old p.f]- [(√1-(New p.f)2) / New p.f])
• Total Load KVAR1=41.1x([(√1-(0.82)2) / 0.82]- [(√1-(0.98)2) / 0.98])
• OR
• tanǾ1=Arcos(0.82)=0.69
• tanǾ2=Arcos(0.98)=0.20
• Total Load KVAR1= KWX (tanǾ1- tanǾ2) =41.1(0.69-0.20)=20.35KVAR
• For Connection (2):
• Total Load KW for Connection(2) =Kw / Efficiency=(7.5×2) / 90%=16.66KW
• Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 16.66 /0.83=20.08 KVA
• Total Load KVA (new) for Connection(1)= KW /New Power Factor= 16.66 /0.98= 17.01KVA
• Total Load KVAR2= KWX([(√1-(old p.f)2) / old p.f]- [(√1-(New p.f)2) / New p.f])
• Total Load KVAR2=20.35x([(√1-(0.83)2) / 0.83]- [(√1-(0.98)2) / 0.98])
• For Connection (3):
• Total Load KW for Connection(3) =Kw =10KW
• Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 10/0.85=11.76 KVA
• Total Load KVA (new) for Connection(1)= KW /New Power Factor= 10 /0.98= 10.20KVA
• Total Load KVAR3= KWX([(√1-(old p.f)2) / old p.f]- [(√1-(New p.f)2) / New p.f])
• Total Load KVAR3=20.35x([(√1-(0.85)2) / 0.85]- [(√1-(0.98)2) / 0.98])
• Total KVAR=KVAR1+ KVAR2+KVAR3
• Total KVAR=20.35+7.82+4.17
• Total KVAR=32 Kvar

## Size of Capacitor Bank:

•  Site of Capacitor Bank=32 Kvar.
• Leading KVAR supplied by each Phase= Kvar/No of Phase
• Leading KVAR supplied by each Phase =32/3=10.8Kvar/Phase
• Capacitor Charging Current (Ic)= (Kvar/Phase x1000)/Volt
• Capacitor Charging Current (Ic)= (10.8×1000)/(415/√3)
• Capacitor Charging Current (Ic)=44.9Amp
• Capacitance of Capacitor = Capacitor Charging Current (Ic)/ Xc
• Xc=2 x 3.14 x f x v=2×3.14x50x(415/√3)=75362
• Capacitance of Capacitor=44.9/75362= 5.96µF
• Required 3 No’s of 10.8 Kvar Capacitors and

## Protection of Capacitor Bank

### Size of HRC Fuse for Capacitor Bank Protection:

•  Size of the fuse =165% to 200% of Capacitor Charging current.
• Size of the fuse=2×44.9Amp
• Size of the fuse=90Amp

### Size of Circuit Breaker for Capacitor Protection:

•  Size of the Circuit Breaker =135% to 150% of Capacitor Charging current.
• Size of the Circuit Breaker=1.5×44.9Amp
• Size of the Circuit Breaker=67Amp
• Thermal relay setting between 1.3 and 1.5of Capacitor Charging current.
• Thermal relay setting of C.B=1.5×44.9 Amp
• Thermal relay setting of C.B=67 Amp
• Magnetic relay setting between 5 and 10 of Capacitor Charging current.
• Magnetic relay setting of C.B=10×44.9Amp
• Magnetic relay setting of C.B=449Amp

### Sizing of cables for capacitor Connection:

•  Capacitors can withstand a permanent over current of 30% +tolerance of 10% on capacitor Current.
• Cables size for Capacitor Connection= 1.3 x1.1 x nominal capacitor Current
• Cables size for Capacitor Connection = 1.43 x nominal capacitor Current
• Cables size for Capacitor Connection=1.43×44.9Amp
• Cables size for Capacitor Connection=64 Amp

### Maximum size of discharge Resistor for Capacitor:

•  Capacitors will be discharge by discharging resistors.
• After the capacitor is disconnected from the source of supply, discharge resistors are required for discharging each unit within 3 min to 75 V or less from initial nominal peak voltage (according IEC-standard 60831).
• Discharge resistors have to be connected directly to the capacitors. There shall be no switch, fuse cut-out or any other isolating device between the capacitor unit and the discharge resistors.
• Max. Discharge resistance Value (Star Connection) = Ct / Cn x Log (Un x√2/ Dv).
• Max. Discharge resistance Value (Delta Connection)= Ct / 1/3xCn x Log (Un x√2/ Dv)
• Where Ct =Capacitor Discharge Time (sec)
• Un = Line Voltage
• Dv=Capacitor Discharge voltage.
• Maximum Discharge resistance =60 / ((5.96/1000000)x log ( 415x√2 /50)
• Maximum Discharge resistance=4087 KΩ

### Effect of Decreasing Voltage & Frequency on Rating of Capacitor:

•  The kvar of capacitor will not be same if voltage applied to the capacitor and frequency changes
• Reduced in Kvar size of Capacitor when operating 50 Hz unit at 40 Hz
• Actual KVAR = Rated KVAR x(Operating Frequency / Rated Frequency)
• Actual KVAR = Rated KVAR x(40/50)
• Actual KVAR = 80% of Rated KVAR
• Hence 32 Kvar Capacitor works as 80%x32Kvar= 26.6Kvar
• Reduced in Kvar size of Capacitor when operating 415V unit at 400V
• Actual KVAR = Rated KVAR x(Operating voltage / Rated voltage)^2
• Actual KVAR = Rated KVAR x(400/415)^2
• Actual KVAR=93% of Rated KVAR
• Hence 32 Kvar Capacitor works as 93%x32Kvar= 23.0Kvar

## Annual Saving and Pay Back Period

### Before Power Factor Correction:

•  Total electrical load KVA (old)= KVA1+KVA2+KVA3
• Total electrical Load KW =62kw
• KVA Demand Charge=KVA X Charge
• KVA Demand Charge=82x60Rs
• KVA Demand Charge=8198 Rs
• Annual Unit Consumption=KWx Daily usesx365
• Annual Unit Consumption=62x24x365 =543120 Kwh
• Annual charges =543120×10=5431200 Rs
• Total Annual Cost= 8198+5431200

### After Power Factor Correction:

•  Total electrical load KVA (new)= KVA1+KVA2+KVA3
• Total electrical Load KW =62kw
• KVA Demand Charge=KVA X Charge
• KVA Demand Charge=69x60Rs =6916 Rs————-(1)
• Annual Unit Consumption=KWx Daily usesx365
• Annual Unit Consumption=62x24x365 =543120 Kwh
• Annual charges =543120×10=5431200 Rs—————–(2)
• Capital Cost of capacitor= Kvar x Capacitor cost/Kvar = 82 x 60= 4919 Rs—(3)
• Annual Interest and Deprecation Cost =4919 x 12%=590 Rs—–(4)
• Total Annual Cost= 6916+5431200+4919+590

## Pay Back Period:

•  Total Annual Cost before Power Factor Correction= 5439398 Rs
• Total Annual Cost After Power Factor Correction =5438706 Rs
• Annual Saving= 5439398-5438706 Rs
• ### Annual Saving= 692 Rs

• Payback Period= Capital Cost of Capacitor / Annual Saving
• Payback Period= 4912 / 692