Demand factor (in IEC, Max.Utilization factor (Ku)):

  1. The word “demand” itself says that means of Demand issue. The magnitude relation of the utmost coincident demand of a system, or a part of a system, to the full connected load of the system.
  2. Demand Factor = Maximum demand / Total connected load
  3. For example, an oversized motor 20 kW drives a constant 15 kW load whenever it is ON. The motor demand factor is then 15/20 =0.75= 75 %.
  4. Demand Factor is expressed as a percentage (%) or in a ratio (less than 1).
  5. Demand factor is always < =1.
  6. The lower the demand factor, the less the system capacity required to serve the connected load.

Calculation:

  • A Residence shopper has ten No’s Lamp of four hundred W however at the identical time, it’s doable that solely nine No’s of Bulbs are used at an identical time. Here Total Connected load is 10×40=400 W. shopper most demand is 9×40=360 W. Demand Facto of this Load = 360/400 =0.9 or 90%.
  •  One shopper has ten lights at sixty power unit every in room, the load is sixty power unit x ten = 600 power unit. this can be true as long as All lights are activated identical time (Demand factor=100% or 1)
  • For this shopper, it’s ascertained that solely half the lights being turned ON at a time thus we will say that the demand issue is zero.5 (50%). The calculable load = 600 power unit X zero.5 = 300 kW.

Use of  demand factors:

  • Feeder conductors ought to have sufficient Ampere capability to hold the load. The Ampere capability doesn’t invariably be up to the overall of all masses on connected branch-circuits.

  • This issue should be applied to every individual load, with specific attention to electrical motors, that square measure terribly seldom operated at full load.

  • As per the National Electrical Code, (NEC), the demand issue is also applied to the overall load. The demand issue permits a feeder ampacity to be but one hundred pc of all the branch-circuit masses connected thereto.

  • Demand issue is often applied to calculate the scale of the sub-main that is feeding a Subpanel or a set load sort of a motor etc. If the panel incorporates a total load of 250 kVA, considering a requirement issue of zero.8, we will size the feeder cable for 250 x zero.8= 200 kVA.

  • Demand factors for buildings usually vary between fifty and eighty you look after the connected load.

  • In Associate in Nursing industrial installation, this issue is also calculable on a mean at zero.75 for motors.

  • For incandescent-lighting masses, the issue invariably equals one.

Demand Factor

Text Book of Principal of Power System-V.K.Mehta

Utility

Demand Factor

Residence Load (<0.25 KW)

1

Residence Load (<0.5 KW)

0.6

Residence Load (>0.1 KW)

0.5

Restaurant

0.7

Theatre

0.6

Hotel

0.5

School

0.55

Small Industry

0.6

Store

0.7

Motor Load (up to 10HP)

0.75

Motor Load (10HP to 20HP)

0.65

Motor Load (20HP to 100HP)

0.55

Motor Load (Above 100HP)

0.50

Demand Factor For Industrial Load

Text Book of Design of Elect. Installation- Jain

Electrical Load

Demand Factor

1 No of Motor

1

Up to 10 No’s of Motor

0.75

Up to 20 No’s of Motor

0.65

Up to 30 No’s of Motor

0.6

Up to 40 No’s of Motor

0.5

Up to 50 No’s of Motor

0.4

Demand Factor

Saudi Electricity Company Distribution Standard

Utility

Demand Factor

Residential

0.6

Commercial

0.7

Flats

0.7

Hotel

0.75

Mall

0.7

Restaurant

0.7

Office

0.7

School

0.8

Common Area in building

0.8

Public Facility

0.75

Street Light

0.9

Indoor Parking

0.8

Outdoor Parking

0.9

Park / Garden

0.8

Hospital

0.8

Workshops

0.6

Ware House

0.7

Farms

0.9

Fuel Station

0.7

Factories

0.9

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