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)
  • Required No of Fixtures =21 No’s

 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.

   Untitled

 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])
  • Total Load KVAR1=20.35 KVAR
  • 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])
  • Total Load KVAR2=7.82 KVAR
  • For Connection (3):
  • Total Load KW for Connection(2) =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 Load KVAR1=4.17 KVAR
  • 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=2×3.14xfxv=2×3.14x50x(415/√3)=75362
  • Capacitance of Capacitor=44.9/75362= 5.96µF
  • Required 3 No’s of 10.8 Kvar Capacitors and
  • Total Size of Capacitor Bank is 32Kvar

 

 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)
  • Cn=Capacitance  Farad.
  • 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= 50.1+20.08+11.76
  • Total electrical load=82 KVA
  • Total electrical Load KW=kW1+KW2+KW3
  • Total electrical Load KW=37+15+10
  • Total electrical Load KW =62kw
  • Load Current=KVA/V=80×1000/(415/1.732)
  • Load Current=114.1 Amp
  • 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
  • Total Annual Cost before Power Factor Correction= 5439398 Rs

 After Power Factor Correction:

  •  Total electrical load KVA (new)= KVA1+KVA2+KVA3
  • Total electrical load= 41.95+17.01+10.20
  • Total electrical load=69 KVA
  • Total electrical Load KW=kW1+KW2+KW3
  • Total electrical Load KW=37+15+10
  • Total electrical Load KW =62kw
  • Load Current=KVA/V=69×1000/(415/1.732)
  • Load Current=96.2 Amp
  • 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
  • Total Annual Cost After Power Factor Correction =5438706 Rs

 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
  • Payback Period = 7.1 Years

Electrical Thumbs Rules (Part-10)


Economical Voltage for Power Transmission:

  •  Economic generation voltage is generally limited to following values (CBIP Manual).

Economic generation voltage (CBIP Manual)

Total Load

Economical Voltage

Up to 750 KVA

415 V

750 KVA to 2500 KVA

3.3 KV

2500 KVA to 5000 KVA

6.6 KV

Above 5000 KVA

11 KV or Higher

  •  Generally terminal voltage of large generators is 11 kV in India. Step up voltage depends upon Length of transmission line for interconnection with the power system and Power to be transmitted.
  • High voltage increases cost of insulation and support structures for increased clearance for air insulation but decreases size and hence Cost of conductors and line losses.
  • Many empirical relations have been evolved to approximately determine economic voltages for power evacuation. An important component in transmission lines is labor costs which are country specific.
  • An empirical relation is given below.
  • Voltage in kV (line to line) = 5.5x√0.62L + kVA/150

  • where kVA is total power to be transmitted;
  • L is length of transmission line in km.
  • American practice for economic line to line voltage kV (based on empirical formulation) is
  • Voltage in kV line to line = 5.5x√0.62L + 3P/100

  • For the purpose of standardization in India transmission lines may be classified for operating at 66 kV and above. 33 kV is sub transmission, 11 kV and below may be classified as distribution.
  • Higher voltage system is used for transmitting higher amounts of power and longer lengths and its protection is important for power system security and requires complex relay systems.

Required Power Transfer (MW)

Distance (KM)

Economical Voltage Level (KM)

3500

500

765

500

400

400

120

150

220

80

50

132

 Factor affected on Voltage Level of system:

  • Power carrying capability of transmission lines increases roughly as the square of the voltage. Accordingly disconnection of higher voltage class equipment from bus bars get increasingly less desirable with increase in voltage levels.
  • High structures are not desirable in earthquake prone areas. Therefore in order to obtain lower structures and facilitate maintenance it is important to design such sub-stations preferably with not more than two levels of bus bars.

 Size of Cable according to Short circuit (for 11kV,3.3kV only)

  • Short circuit verification is performed by using following formula:
  • Cross Section area of Cable (mm2)S = I x√t / K

  • Where:
  • t = fault duration (S)
  • I = effective short circuit current (kA)
  • K = 0.094 for aluminum conductor insulated with XLPE
  • Example: Fault duration(t)= 0.25sec,Fault Current (I) = 26.24 kA
  • Cross Section area of Cable = 26.24 x √ (0.25) / 0.094= 139.6 sq. mm
  • The selected cross sectional area is 185 sq. mm.

Ground Clearance:

  • Ground Clearance in Meter = 5.812 + 0.305 X K

  • Where K= (Volt-33) / 33
Voltage Level Ground Clearance
<=33KV 5.2  Meter
66KV 5.49 Meter
132KV 6.10 Meter
220KV 7.0   Meter
400KV 8.84  Meter

 Voltage Rise in Transformers due to Capacitor Bank:

  • The voltage drop and rise on the power line and drop in the transformers. Every transformer will also experience a voltage rise from generating source to the capacitors. This rise is independent of load or power factor and may be determined as follows:
  • % Voltage Rise in Transformer=(Kvar / Kva)x Z

  • Kvar =Applied Kvar
  • Kva = Kva of the transformer
  • z = Transformer Reactance in %
  • Example: 300 Kvar bank given to 1200 KVA transformer with 5.75% reactance.
  • % Voltage Rise in Transformer=(300/1200)x 5.75 =1.43%

Electrical Thumbs Rules (Part-9)


 

Load in Multi-storied Building (Madhyanchal Vidyut Vitran Nigam)

Type of Load

Calculation

Diversity

Domestic  (Without Common Area)

50 watt / sq. meters

0.5

Commercial  (Without Common Area) 150 watt / sq. meters

0.75

Lift, Water Pump, Streetlight ,Campus Lighting ,Common Facilities,

Actual load shall be calculated

0.75

 

Load in Multi-storied Building (Noida Power Company Limited)

Type of Load

Calculation

Diversity

Domestic (Constructed area) 15 watt / sq. Foot

0.4

Commercial(Constructed area) 30 watt / sq. Foot

0.8

Industrial (Constructed area) 100 watt/ 1 sq. Foot

0.5

Lift, Water Pump, Streetlight ,Campus Lighting ,Common Facilities, 0.5Kw / Flat

 

Voltage Drop: 2% Voltage drop from Transformer to Consumer end.

T&D Losses: 2% T&D Losses from Transformer to Consumer end.

 

Approximate % Cost or Sq.Foot Cost

Project Item

% of Total Project Cost

Rs per Sq.Foot

Articheture (Consultancy)

0.7%

13.1 Rs / Sq.Foot

Structural (Consultancy)

1.2%

21.8 Rs / Sq.Foot

Service Design (Consultancy)

0.4%

7.2 Rs / Sq.Foot

Fire Fighting Work

1.3%

23 Rs / Sq.Foot

Electrical Work (Internal)

4.1%

76 Rs / Sq.Foot

Lift Work

4.4%

82 Rs / Sq.Foot

 

Street Light Costing (CPWD-2012)

Fluorescent Lamp

95 Rs/Sq.Meter

With HPMV Lamp

130 Rs/Sq.Meter

With HPSV Lamp

165 Rs/Sq.Meter

Electrical Sinage

85 Rs/Sq.Meter

 

Other Electrical Cost

Area Required for Solar Light

10 Watt/Sq.Foot

Solar Power Installation

1.5 Lacs Rs/1Kw

HVAC Cost

18 Watt/Sq.Foot

 

Distribution Losses (Gujarat Electricity Board)

Voltage (Point of Injection)

At 11 KV

Point of Energy Delivered

11KV / 22KV / 33KV

10%

1082%

400 Volt

-

16.77%

 

Rate Analysis (CPWD-2012)

Description

Amount
Sub Station Equipment 7000 Rs/ KVA
D.G Set with installation 1000 Rs / KVA
UPS with 30min Breakup 20000 Rs / KVA add 8000 Rs / KVA additional each 30 min
Solar Power Generation 1.25 Lacs / KW
Solar Water System (200Liter/Day) 46000 Rs
Solar Water System (300Liter/Day) 64000 Rs
Solar Water System (1000Liter/Day) 210000 Rs
Central AC Plant 75000 RS / Ton
VRF / VRV System 55000 Rs / HP
Air condition System 11000 Rs / Ton
CCTV System 300 Rs / Sq Meter
Access Control system 200 Rs / Sq Meter
Hydropenumatic Water system 2000 Rs / LPM
Building Management System 300 Rs /  Sq Meter add 100 Rs / Sq Meter additional area beyond 10000 Sq Meter

 

Rate Analysis (Rs per Sq. Meter) (CPWD-2012)

Work

Office/College/Hospital

School

Hostel

Residence

Fire Fighting (with Wet Riser)

500

500

500

500

Fire Fighting (with Sprinkler)

750

750

750

750

Fire Alarm (Manually)

-

-

-

300

Fire Alarm (Automatic)

500

500

500

500

Pressurized  Mechanical Ventilation

650

650

650

650

 

Rate Analysis (% of Total Project Cost) (CPWD-2012)

Work

Office/College/Hospital

School

Hostel

Residence

Internal Water Supply & Sanitary

4%

10%

5%

12%

Internal Electrical Installation

12.5%

12.5% 12.5% 12.5%

 

Lift Speed (Indian Army Manual)

No of Floor

Lift Speed

4 to 5

0.5 to 0.7 meter/Sec

6 to 12

0.75 to 1.5 meter/Sec

3 to 20

1.5 to 2.5 meter/Sec

Above 20

 Above 2.5 meter/Sec

 

Lift  Details (CPWD-2012)

Type of Lift 

Persons

Weight

Speed  M/Sec

Travel

Price

Add Rs /Floor

Passenger Lift 

8 Person 544 Kg 1.0 G+4 18 Lacs 1.25 Lacs

Passenger Lift 

13 Person 844 Kg 1.5 G+4 22 Lacs 1.25 Lacs

Passenger Lift 

16 Person 1088 Kg 1.0 G+4 28 Lacs 1.50 Lacs

Passenger Lift 

120 Person 1360 Kg 1.5 G+4 24 Lacs 1.50 Lacs

 

MCB Class according to Appliances

Appliance

Capacity / watt

MCB Rating

MCB Class

Air Conditioner

1.0 Tone

10A

C Class

1.5 Tone

16A

C Class

2.0 Tone

20A

C Class

Freeze

165 Liter

3 A

C Class

350 Liter

4 A

C Class

Oven /Grill

4500 Watt

32 A

B Class

1750 Watt

10 A

B Class

Oven / Hotplate

750 Watt

6 A

B Class

2000 Watt

10 A

B Class

Room Heater

1000 Watt

6 A

B Class

2000 Watt

10 A

B Class

Washing Machine

300 Watt

2 A

C Class

1300 Watt

8 A

C Class

Water Heater

1000 Watt

6 A

B Class

2000 Watt

10 A

B Class

3000 Watt

16 A

B Class

6000 Watt

32 A

B Class

Iron

750 Watt

6 A

B Class

1250 Watt

8 A

B Class

Toaster

1200 Watt

8 A

B Class

1500 Watt

10 A

B Class

Calculate Technical Losses of Transmission / Distribution Line:


Introduction:

  • There are two types of Losses in transmission and distribution Line.
  • (1) Technical Losses and
  • (2) Commercial Losses.
  • It is necessary to calculate technical and commercial losses.Normally Technical Losses and Commercial Losses are calculated separately .Transmission (Technical) Losses are directly effected on electrical  tariff but  Commercial losses are not implemented to all consumers.
  • Technical Losses of the Distribution line mostly depend upon Electrical Load, type and size of conductor, length of line etc.
  • Let’s try to calculate Technical Losses of one of following 11 KV Distribution Line

Example:

  • 11 KV Distribution Line have following parameter.
  • Main length of 11 KV Line is 6.18 Kms.
  • Total nos. of Distribution Transformer on Feeder 25 KVA= 3 No, 63 KVA =3 No,100KVA=1No.
  • 25KVA Transformer Iron Losses = 100 W, Copper Losses= 720 W, Average LT Line Loss= 63W.
  • 63KVA Transformer Iron Losses = 200 W, Copper Losses= 1300 W, Average LT Line Loss= 260W.
  • 100KVA Transformer Iron Losses = 290 W, Copper Losses= 1850 W, LT Line Loss= 1380W.
  • Maximum Amp is12 Amps.
  • Unit sent out during to feeder is  490335 Kwh
  • Unit sold out during from Feeder is 353592 Kwh
  • Normative Load diversity Factor for Urban feeder is 1.5 and for Rural Feeder is 2.0

Calculation:

Total Connected Load=No’s of Connected Transformer.

  • Total Connected Load= (25×3) + (63×3) + (100×1).
  • Total Connected Load=364 KVA.

 Peak Load = 1.732 x Line Voltage x Max Amp

  • Peak Load = 1.732x11x12
  • Peak Load =228 KVA.

 Diversity Factor (DF) = Connected Load (In KVA) / Peak Load.

  • Diversity Factor (DF) = 364 /228
  • Diversity Factor (DF) =1.15

 Load Factor (LF)= Unit Sent Out (In Kwh) / 1.732 x Line Voltage x Max Amp. x P.F. x 8760

  • Load Factor (LF)=490335 / 1.732x11x12x0.8×8760
  • Load Factor (LF)=0.3060

 Loss Load Factor (LLF)= (0.8 x LFx LF)+ (0.2 x LF)

  • Loss Load Factor (LLF)= ( 0.8 x 0.3060 x 0.3060 ) + (0.2 x 0.306)
  • Loss Load Factor (LLF)= 0.1361

 Calculation of Iron losses:

  • Total Annual Iron loss in Kwh =Iron Loss in Watts X Nos of TC on the feeder X8760 / 1000
  • Total Annual Iron loss (25KVA TC)=100x3x8760 /1000 =2628 Kwh
  • Total Annual Iron loss (63KVA TC)=200x3x8760 /1000 =5256 Kwh
  • Total Annual Iron loss (100KVA TC)=290x3x8760 /1000 =2540 Kwh
  • Total Annual Iron loss =2628+5256+2540 =10424Kwh

 Calculation of Copper losses:

  • Total Annual Copper loss in Kwh =Cu Loss in Watts XNos of TC on the feeder LFX LF X8760 / 1000
  • Total Annual Copper loss (25KVA TC)=720x3x0.3×0.3×8760 /1000 =1771 Kwh
  • Total Annual Copper loss (63KVA TC)=1300x3x0.3×0.3×8760 /1000 =3199 Kwh
  • Total Annual Copper loss (100KVA TC)=1850x1x0.3×0.3×8760 /1000 =1458 Kwh
  • Total Annual Copper loss =1771+3199+1458=6490Kwh

 HT Line Losses (Kwh)=0.105 x (Conn. Load x 2) x Length x Resistance x LLF /( LDF x DF x DF x 2 )

  • HT Line Losses= 1.05 x(265×2) x 6.18 x 0.54 x 0.1361 /1.5 x 1.15 x1.15 x 2
  • HT Line Losses = 831 Kwh

 Peak Power Losses= (3 x Total LT Line Losses) / (PPLxDFxDFx 1000)

  • Peak Power Losses= 3 x (3×63+3×260+1×1380) /1.15 x 1.15 x 1000
  • Peak Power Losses= 3.0

 LT Line Losses (Kwh)= (PPL.) x (LLF) x 8760

  • LT Line Losses = 3 x 0.1361 x 8760
  • LT Line Losses = 3315 Kwh

 Total Technical Losses= (HT Line Losses + LT Line Losses + Annual Cu Losses + Annual Iron Losses)

  • Total Technical Losses = ( 831+ 3315 + 10424 + 6490)
  • Total Technical Losses = 21061 Kwh

% Technical Loss= (Total Losses) / (Unit Sent Out Annually) x 100

  • % Technical Loss= (21061/490335) x100= 4.30%

% Technical Loss=4.30%

Calculate Transformer Losses (As per Test Results)


111

  • Calculate Form Factor at Full Load & Actual Load.
  • Calculate Core Losses at Full Load & Actual Load.
  • Calculate Copper Losses at Full Load & Actual Load.
  • Calculate Stray  Losses at Full Load & Actual Load.
  • Calculate Total Losses at Full Load & Actual Load.
  • Calculate Transformer Efficiency  at Full Load & Actual Load.

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Pirating of Technical Works.


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Lots of time has been spent to read Books,Manuals,Handbooks and combined it with  practical experience to serve Handy Electrical tools,Notes to serve the Electrical Community.This Blog is a fusion of Theoretical and Practically knowledge to make all technical things easier to understand.

Please look at following totally copy past material of  this  Blog. 

After publishing Post on this Blog. Other Blogger republished the same post on their Blog .the blogger did not even change THE COLOR OF THE FONT while copy past and not even Chang the Title of the Post ….. All reader keeps their eyes on Date of original Post and date of copy past blog. Following blogger neither take any permission of the original blogger to republished the post nor give any credit by putting Reference of original post.

Blogger No-1

(1) Original post :http://electricalnotes.wordpress.com/2013/07/27/electrical-thumb-rules-part-1-2/

Copy past post of other blogger: http://myelectricalnote.blogspot.in/2013/08/electrical-thumb-rules.html

(2)Original post :http://electricalnotes.wordpress.com/2013/03/08/calculate-numbers-of-platepipestrip-earthing-for-system/

Copy past post of other blogger:http://myelectricalnote.blogspot.in/2013/04/calculate-numbers-of-platepipestrip.html

(3) Original post : http://electricalnotes.wordpress.com/2013/04/01/calculate-numbers-of-platepipestrip-earthings-part-3/

Copy past post of other blogger: http://myelectricalnote.blogspot.in/2013/04/calculate-numbers-of-platepipestrip_10.html

(4) Original post :http://electricalnotes.wordpress.com/2011/09/05/type-of-earthing-systems-in-electrical-distribution/

Copy past post of other blogger:http://myelectricalnote.blogspot.in/2013/05/type-of-electrical-power-distribution.html

(5) Original post :http://electricalnotes.wordpress.com/2012/07/17/parallel-operation-of-transformers/

Copy past post of other blogger: http://myelectricalnote.blogspot.in/2013/03/parallel-operation-of-transformers_7.html

Blogger No-2

(1) Original post :http://electricalnotes.wordpress.com/2013/10/01/electrical-thumb-rules-part-8/

Copy past post of other blogger http://www.electricalinfo.co.in/electrical-thumb-rules-part-8/

(2)Original post :http://electricalnotes.wordpress.com/2013/11/01/type-of-mcb-distribution-board/

Copy past post of other blogger:http://www.electricalinfo.co.in/difference-tpn-4p-spn-dp/#comment-1119

(3) Original post :http://electricalnotes.wordpress.com/2013/07/27/electrical-thumb-rules-part-1-2/

Copy past post of other blogger:http://www.electricalinfo.co.in/electrical-thumb-rules-part-1/

It is wast of time to put all lists of copy past items because this copy past list is long…………………..

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