Current carrying capacity of conductor

Current Carrying Capacities

Active and reactive current both contribute to power loss in wiring. It is only at the load that reactive current does no actual work. The power loss of the wires comes from the actual current running through them. However, active and reactive current sum in quadrature not arithmetically. In other words, 4 amperes of active current and 3 amperes of reactive current does not make 7 amperes of current through your wires, but 5 amperes of current. 3²+4²=25=5²

But What are the losses where reactive current contributes?

If due to that loss heat is generated, then it should not be reactive current.

Does those losses due to reactive Current lead conductor to fail? Which type of failure? (Melting of conductor or some thing else?)

Reactive current is an expression of the frequency mismatch between supply and load. As it flows in the wiring connecting them, it contributes to the heat that has to be dissipated in the conductors, even though it takes no part in the power that is transmitted to the load.

Power factor correction is the technique of minimising conductor losses.

AP,

Each conductor in AC behaves like an impedance, it has a resistance (R), and an impedance X ( L inductive and C capacitive part), dependent on the type of cable, relative position, etc.

But under normal circumstances ( normal industrial frequency 50/60 Hz, cable lengths), their (L,C cable components) influence is can be neglected.

The resulting power factor ( active and reactive part) is given almost exclusive by the character of the load. If your load is purely resistive (RL), the resulting current (Ia) will be in phase with the voltage, contribution of the cable inductivity & capacity being negligible, losses in the cable are Ia2R.

If the load has both resistive (RL) and reactive component (XL), the current through the cable (Ib) has to carry the currents for both (sum of squares as stated in a previous posting), so it (the resulting current) is higher than for the case of a purely ohmic load.

Consequently the losses (Ib2R) in the cable will be higher (Ib>Ia).

Higher currents in the cable represent higher losses, they lead to an increased temperature of the copper, increased stress on the cable insulation, insulation damage if the copper temperature exceeds the maximum insulation material listed temperature, etc.

These losses are caused by a higher current, irrespective on what causes that higher current through the feeding cables ( additional reactive or ohmic load).

There seems to be a general misconception regarding the active and reactive power.

The Active power is the power that is converted to another form ie that takes part in energy exchange across the boundary of the system (or the circuit, if you want).

Any power is defined as

P = V.I (dot product)

The I is the current flowing through the element

and V is the voltage drop across the element.

If you have a resistive element the Voltage and currents are in phase (as if aligned vectors) and hence the

P = VI (simple multiplication)

V = IR

so P = I²R

This I is the value of I (being the arithmetic multiplication)

When you have an inductive circuit,

P = V.I

V = ZI = ωLI

However it is at 90^o to I

hence the dot product is zero.

Similar is the case with the capacitive, at 90 degrees the dot product is zero.

Note here these voltages are the drops across the elements and there Will be other elements where the drops will takes place so that in the complete circuit as per KVL ΣV=0

Also note the Voltages, being phasors (treat them almost like vectors) the sum will be the vector sums and not arithmetic sums.

Now try to get the meaning, though the V and I are different in phase (sorry PWS, not frequency ) across the pure elements either it will be in phase or at 90⁰.

So though over all reactive power exist due to the reactive component of the current, this whole current will be treated as the active current by the resistors, since this whole will create a voltage drop across it.

In other way think in mechanical terms,

Say there is a pump storage system of electricity generation. 12 hours per day it gives out power and other 12 hours it pumps back the water (assume 100% water is pumped back).

So the whole thing is like a pure reactive system is it? half cycle inductor stores energy, other half it gives out the energy.

Now look at the friction loss in the pipe line - is it equal to... the flow of water through it. Irrespective of what it is used for, even though we know that after 12 hours the water is going to be pushed back, but the pipe doesn't know it, for it the water quantity is what matters.

Now substitute the water by current, and if you want, let there be nett energy exchange across the boundary ie the water used for peak energy demand is not wholly pushed back in off peak time.

Can We say as under:

All theories & Formulas are saying resonsible current for losses in cable is Resultant current (i+ji or real + imaginary). Hence, we must consider resutant current for selecting conductor size.

But Still,

Assume 2 separate circuit. Practiclly these circuts are not possible, but near to this can be possible.

Circuit 1 : Purely inductive load. Currnet will be 90 degree lagging. Means 0 Power Factor. Say 10Amps Passing through conductor. Conducor resistance is say 1 Ohms. Means Loss in the conductor will be I^2R = 100 W. This loss will be in terms of Heat.

Circuit 2 : Purely Resistive load. Currnet will be in Phase. Means 1 Power Factor. Say 10Amps Passing through conductor. Conducor resistance is say 1 Ohms. Means Loss in the conductor will be I^2R = 100 W. This loss will be in terms of Heat.

Does it mean, In both the cases heat generated will be same?

Does it mean, Imaginary current also produces heat same heat as real current ?

If this is true then heat is not active Power, since it is work done by imaginery componenet of current !!!!!!!!!!

REDFRED,

GREAT.....

I got correct solution for my unsolved question.

Let me confirm once again what I understood.

Voltage Drop accross resistor (i.e. cable) will be in phase with current passing through the circuit, irrespective of Phase difference between Total Load & Total Voltage at source.

Same way, Voltage Drop accross inductor (i.e. assumed inductive load) will be in 90 degree behind current passing through the circuit, irrespective of Phase difference between Total Load & Total Voltage at source.

Fred

I like your answer. 1 GA

man the rective power flows in the circuit i.e reactive current in tro and fro motion from iductive load/capacitive load to the generator ,hence it is not achived as power out put but the current does flow in the cables soooooooooo consider apparent current for power transmission .....and if 100% reactive power there will be heat loss in conductors

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