This topic contains 10 replies, has 0 voices, and was last updated by  vaithilingam 8 years, 5 months ago.

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• #10247

vaithilingam
Member

If i have 3 phase input power KW is 35 what will be the load end power distance 35 km whether it will increase or decrease

#11555

Keymaster

its increase brcause of resistance of wires that is called  VOLTAGE DROP

#11577

Member

considering your answer engr. saad… kindly elaborate further by showing your calculation using the voltage drop formula..

#11581

Keymaster

considering distance and size of cable,it is likely to decrease.

#11587

Calculate your current from

 kW=√3VIcosф

Work out your impedance Z from point A to B where A is at the sending end and B is 35 m away. Your voltage drop will be approximately I x Z. Your  voltage will drop by this amount.

#11589

Sorry, the drop is not I x Z as indicated, but should be

 I²xZ
#11578

Leynin
Member

There will naturally be a decrease in the value of voltage at your load terminals due to I²R loss (heat) plus reactive drops if electric motors are connected. Determine the supply current from

I = 35 X 10³ ⁄ (I x V x cosΦ√3

and the total impedance of the circuit from the source to the load, ie, the R of a 35m length cable (what is the csa? and is it made of CU?) + the impedance of the connected appliances. There is a standard allowable voltage drop(mV/A/m) for your particular cable, not more than 4% of your supply voltage (IEEE). Check this and use the formula,

Total voltage drop = (mV/A/m) x I x 35 / 1000

That will be the value at the load terminals.

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#11648

Keymaster

vaithilingam said:

If i have 3 phase input power KW is 35 what will be the load end power distance 35 km whether it will increase or decrease

If I get your question right, you are asking for the available power at the end of 35km distance? the “load end power” will depend on the size of your conductor in this 35Km distance and this can be calculated using the voltage drop equation. In addition, the type of load (motors, resistive load, etc) should also be considered.

#11649

Keymaster

If I get your question right, you are asking for the available power at the end of 35km distance? the “load end power” will depend on the size of your conductor in this 35Km distance and this can be calculated using the voltage drop equation. In addition, the type of load (motors, resistive load, etc) should also be considered.

Moreover, available power will never increase at the end of the line.

#11657

Keymaster

Assume

35m cable length

10amps / phase

Cable 10mm² = 0.00343Volts drop per amp per meter from cable manufactures tables) copper

0.00343 x 35m x10 amps = 1.20v drop (at the end of the 35m)

#11663