KW Calculation

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

Calculate your current from

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.

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.

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.

The 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.