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- 2010/10/16 at 11:34 am #10247
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

2010/10/18 at 3:42 am #11555its increase brcause of resistance of wires that is called VOLTAGE DROP

2010/10/26 at 3:47 am #11577considering your answer engr. saad… kindly elaborate further by showing your calculation using the voltage drop formula..

2010/10/26 at 8:10 pm #11581considering distance and size of cable,it is likely to decrease.

2010/10/27 at 10:25 am #11587Calculate 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.

2010/10/28 at 9:11 am #11589Sorry, the drop is not I x Z as indicated, but should be

I²xZ 2010/11/01 at 9:04 am #11578There 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.

In 2004 in the United States, there were 104 (69 pressurized water reactors and 35 boiling water reactors)

commercial nuclear generating units licensed to operate, producing a

total of 97,400 megawatts (electric), which is approximately 20% of the

nation’s total electric energy consumption2010/11/30 at 8:09 am #11648**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.2010/11/30 at 8:11 am #11649If 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.

2010/12/02 at 8:33 am #11657Assume

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)

2010/12/05 at 2:55 pm #11663The input is 35 KW ,3 phase voltage is specified.We ned to calculate resistance of cable conductor /Km & apply the fourmula of

Receiving end Power = sending end power ( 35 KW) – ( power Losses in conductor /cable+ shut capacitence impedence along the O.H.line for 35 Km),Gives you the answer.It will decrese if you use 440 Volts ,if transmit same through 3.3 KV it will increases,as per my little knowledge goes.

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