Panel Design / Calculate Size of Bus bar


 

ScreenHunter_01 Jul. 21 20.06

  • Calculate Bus Bar Cross Section Area
  • Calculate Current Rating of Bus Bar
  • Calculate Peak elector-magnetic Force between Phase Conductors
  • Calculate Force on Insulator of Bus bar Support.
  • Calculate Mechanical Strength of Bus Bar.
  • Calculate Maximum Temperature Rise of Bus Bar.

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Calculate Voltage Drop and No’s of Street Light Pole


ScreenHunter_01 Jul. 14 20.14

  • Calculate Voltage drop in Street Light Pole
  • Calculate Distance between Each Street Light Pole
  • Calculate Wattage of each Street Light Luminaire
  • Calculate Required Street Light Watt as per Given Area.
  • Calculate Luminaire Payback Period

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Calculate Size of Neutral Earthing Transformer (NET)


  • Calculate Total Capacitive charging current (It).
  • Calculate Neutral Earthing Transformer  Secondary Current.
  • Calculate Required Grounding Resistor value at primary side (Rp).
  • Calculate Required Resistance value at secondary side (Rsec).
  • Calculate Neutral Earthing Transformer X/R Ratio
  • Calculate Fault current through Neutral (single line to ground fault) (If)
  • Calculate Rating of Neutral Earthing Transformer (P).

ScreenHunter_01 Jun. 06 22.20

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Calculate Street Light Pole’s Distance / Fixture Watt / Lighting Area


(1) Calculate Distance between each Street Light Pole:

Example: Calculate Distance between each streetlight pole having following Details,

  • Road Details: The width of road (w) is 11.5 Foot.
  • Pole Details: The height of Pole is 26.5 Foot.
  • Luminaire of each Pole: Wattage of Luminaries is 250 Watt, Lamp Out Put (LL) is 33200 Lumen, Required Lux Level (Eh) is 5 Lux, Coefficient of Utilization Factor (Cu) is 0.18, Lamp Lumen Depreciation Factor (LLD) is 0.8, Lamp Lumen Depreciation Factor (LLD) is 0.9.
  • Space Height Ratio should be less than 3.

Calculation:

  • Spacing between each Pole=(LL*CU*LLD*LDD) / Eh*W
  • Spacing between each Pole=(33200×0.18×0.8×0.9) / (5×11.5)
  • Spacing between each Pole= 75 Foot.
  • Space Height Ratio = Distance between Pole / Road width
  • Space Height Ratio = 3. Which is less than define value.
  • Spacing between each Pole is 75 Foot.

(2) Calculate Street Light Luminaire Watt:

Example: Calculate Streetlight Watt of each Luminaire of Street Light Pole having following Details,

  • Road Details: The width of road (w) is 7 Meter. Distance between each Pole (D) is 50 Meter.
  • Required Illumination Level for Street Light (L) is 6.46 Lux per Square Meter. Luminous efficacy is 24 Lumen/Watt.
  • Maintenance Factor (mf) 0.29, Coefficient of Utilization Factor (Cu) is 0.9.

Calculation:

  • Average Lumen of Lamp (Al) = 8663 Lumen.
  • Average Lumen of Lamp (Al) =(LxWxD) / (mfxcu)
  • Average Lumen of Lamp (Al)= (6.46x7x50) / (0.29×0.9)
  • Average Lumen of Lamp (Al)=8663 Lumen.
  • Watt of Each Street Light Luminar = Average Lumen of Lamp / Luminous efficacy
  • Watt of Each Street Light Laminar = 8663 / 24
  • Watt of Each Street Light Luminaire = 361 Watt

(3) Calculate Required Power for Street Light Area:

Example: Calculate Streetlight Watt of following Street Light Area,

  • Required Illumination Level for Street Light (L) is 6 Lux per Square Meter.
  • Luminous efficacy (En) is 20 Lumen per Watt.
  • Required Street Light Area to be illuminated (A) is 1 Square Meter.

Calculation:

  • Required Streetlight Watt = (Lux per Sq.Meter X Surface Area of Street Light) / Lumen per Watt.
  • Required Streetlight Watt = (6 X 1) / 20.
  • Required Streetlight Watt = 0.3 watt per Square Meter.

IP Rating for Electrical Enclosure


IP rating:

  •  IP letters stand for “International Protection” rating or “Ingress Protection” rating. IP ratings are defined in international standard (British BS EN 60529, IEC 60509).
  • It is used to explain levels of sealing effectiveness of electrical enclosures against foreign bodies (tools, dirt etc) and moisture.

 

Meaning of IP Rating:

  • The IP rating code is a two-digit (or optionally three-digit) designator to standardize the rating of protection level against intrusion of solids and liquids into mechanical and electrical enclosures. An enclosure can be a piece of equipment, an assembly unit, a cable or simply a connector.
  • The numbers of IP of each have a specific meaning.
  • First Number: The first Number indicates the degree of protection from moving parts, as well as the protection of enclosed equipment from foreign bodies.
  • Second Number: The second Number indicates the protection level that the enclosure enjoys from various forms of moisture (drips, sprays, submersion etc).
  • Third Number: The third digit in the designator is not part of the official IEC standard and is sometimes included (but more often omitted) to reference additional protections.

Abbreviation of IP Rating:

 

                                                                IP Rating Digits

IP Rating First Digit Second Digit Third Digit (Optional)
Solid Objects Protection Liquids Protection Mechanical impacts

0

No special protection No protection. No protection.
1 Protected against solid objects greater than 50mm in diameter ( such as large part of the body like hand) Protection against vertically falling drops of water e.g. condensation. Protects against impact of 0.225 joule (150 g weight falling from 15 cm height)
2 Protected against solid objects over 12 mm in diameter (person’s fingers) Protection against direct sprays of water up to 15° from the vertical. Protected against impact of 0.375 joule (250 g weight falling from 15 cm height)
3 Protected against solid objects not greater than 80mm in length and 12mm in diameter (tools and wires). Protected against direct sprays of water up to 60° from the vertical. Protected against impact of 0.500 joule (250 g weight falling from 20 cm height)
4 Protected against solid objects larger than 1 mm diameter (tools, wires, and small wires). Protection against water sprayed from all directions (limited ingress permitted). Protected against impact of 2.0 joule (500 g weight falling from 40 cm height)
5 Protected against dust limited ingress (no harmful deposit). Protected against low pressure jets of water from all directions (limited ingress). Protected against impact of 6.0 joule (1.5 kg weight falling from 40 cm height)
6 Totally dust tight. Protected against temporary flooding of water, e.g. for use on ship decks (limited ingress permitted). Protected against impact of 20.0 joule (5 kg weight falling from 40 cm height)
7 N/A Protected against the effect of immersion between 15 cm and 1 m. N/A
8 N/A Protects against long periods of immersion under pressure. N/A

 

Example:

  • IP65 Enclosure: IP rated as protection against dust (6) and protection from low water pressure (5). 
  • IP66 Enclosure – IP rated as protection against dust (6) and protected against heavy seas or powerful jets of water (6)

 

 

Calculate No of Street Light Poles


Typical Calculation of Road Lighting:

  •  Luminaries are properly selected and mounted on a location most feasible and effective with minimum cost. For a 230 volts system, a voltage drop of 5% is allowed although in extreme cases 15% voltage drop is sometimes tolerated.  3
  • Street illumination level in Lux (E)=(Al x (cu x mf)) / (w x d)
  • E = The illumination in Lux
  • w = Width of the roadway
  • d = Distance between luminaries
  • cu = Coefficient of utilization. Which is dependent on the type of fixture, mounting height, width of roadway and the length of mast arm of outreach?
  • Al = Average lumens, Al = (E x w x d) / Cu x mf
  • The typical value of Al is
  • 20500 lumens for 400 watts
  • 11500 lumens for 250 watts
  • 5400 lumens for 125 watts
  • The value of Al varies depending upon the type of lamp specified.
  • mf : It is the maintenance factor (Normally 0.8 to 0.9)

 (1) Calculate Lamp watt for street Light Pole:

  • Calculate Lamp Lumen for street Light Pole having Road width of 7 meter, distance between two Pole is 50 meter, Maintenance factor is 0.9, Coefficient of utilization factor is 0.29, light pedestrian traffic is medium and Vehicular traffic is very light and Road is concrete road.
    Solution:
    From Above table Recommended of illumination (E) in Lux is 6.46 per sq. meter.
    w = 7.00 meters , d = 50 meters , mf = 0.9, cu = 0.29
    To decide Lamp Watt It is necessary to calculate Average Lumens of Lamp (Al).
  • Average Lumen of Lamp (Al)=(E x w x d) / Cu x mf
  • Al=(6.46x7x50)/(0.29×0.9)= 8662.83 Average lumen
    Lamp lumen of a 250 watts lamp is 11,500 lm which is the nearest value to 8662.83 lumen. Therefore, a 250 watts lamp is acceptable.
    Let’s Computing for the actual illumination E for 250 Watt Lamp
  • Illumination (E)=(Al x (cu x mf)) / (w x d)
  • E= (11500×0.29×0.9) / (7×50) = 8.57 lumen per sq meter.
    Conclusion:
    Actual illumination (E) for 250 Watt is 8.57 lumen per sq meter which is higher than recommended illumination (E) 6.46.
  • Hence 250 watt gives adequately lighting.

 (2) Calculate Spacing between two Light Poles:

  • Calculate Space between Two Pole of Street Light having Fixture Watt is 250W , Lamp output of the Lamp (LL) is 33200 lumens , Required Lux Level (E) is 5 lux , Width of the road (W) = 11.48 feet (3.5 M),Height of the pole (H) = 26.24 feet (8 M) ,Coefficient of utilization (CU) = 0.18, Lamp Lumen Depreciation Factor (LLD) = 0.8 ,Luminaries dirt Depreciation Factor (LDD) = 0.9
    Solution:
  • Luminaries Spacing (S) = (LLxCUxLLDxLDD) / (ExW)
  • Luminaries Spacing (S) = (33200×0.18×0.9×0.8) / (5×11.48)
  • Luminaries Spacing (S) = 75 feet (23 Meters)

 (3) Calculation of the allowed illumination time:

  • The allowed illumination time in hours T = k.t.1000/E.
  • Where: k = extension factor
  • t = permissible time in hours at 1000 lux, unfiltered daylight
  • E = luminance (lx)

 

Extension Factor
Lamp Extension Factor
Incandescent lamps, 2.7 to 3.2
Halogen reflector lamps 2.5 to 3.5
Halogen capsules 2.5 to 3.5
High-pressure metal-halide 1.1 to 2.1
High-pressure sodium lamps 4
Fluorescent lamps 1.9 to 2.7

 

  • Example:
  • In sunlight (100000 lux) and extension factor 1: The permissible illumination time (T) =1 x 70 x 1000/100 000 = 0.7 hour.
  • In halogen light (200 lux) and extension factor 2.3: The permissible illumination time (T) = 2.3 x 70 x 1000/200 = 805 hours.
  • In UV-filtered halogen light (200 lux) and extension factor 3.5: The permissible illumination time (T) = 3.5 x 70 x 1000/200 = 1225 hours.

 (4) Calculate Uniformity Ratio:

  • Once luminaries spacing has been decided It is necessary to check the uniformity of light distribution and compare this value to the selected lighting
  • Uniformity Ratio ( UR) = Eav /Emin
  • Eav= average maintained horizontal luminance
  • Emin = maintained horizontal luminance at the point of minimum illumination on the pavement

 (5) Energy Saving Calculations:

  • At a simplistic level, the cost of running a light is directly related to the wattage of the globe plus any associated ballast or transformer. The higher the wattage, the higher the running cost and it is a straightforward calculation to work out the running cost of lamp over its lifetime:
  • Running cost = cost of electricity in $/kWh x wattage of lamp x lifetime in hours.

 Calculate Lux Level for Street Lighting

  • The Average Lux Level of Street Light is measured by 9 point method. Make two equal quadrants between two Street light poles. on the lane of light poles( one side pole to road).
  • We have 3 points P1,P2 and P3 under the light pole then P4 & P7 are points opposite pole 1 or Point P3 same is applicable for P6 and P9 for Pole 2.
  • The average lux = [(P1+P3+P7+P9)/16]+[(P2+P6+P8+P4)/8]+[P5/4]

 2

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 

20 Person 1360 Kg 1.5 G+4 30 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

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