High Voltage to bring electricity to the cities

[DaTario]

Hi All,

Is there a simple explanation for the fact that in power stations (where electric energy is created) the voltage applied to the conducting wires, which will carry the energy to the city, is much higher than 110 V ?

Thank you all

Best Wishes

DaTario

 

[russ_watters]

Yes, resistance losses are a function of amperage, so if you have a certain power requirement, using a higher voltage enables lower amperage and lower resistance losses.

 

[b.shahvir]

Hi, for the same power requirement, increasing the applied voltage will cause current to drop significantly. This is economical for the power utility companies as they will be saving on money invested in electrical equipment as reduction in amps will, for eg., reduce the size of transmission line conductors. Also, reduced amps results in reduced line losses as rightly pointed out by Russ.

You can play with Ohm's law by putting higher values of voltage in the eqn. P = V x I keeping power 'P' as constant. You will find that as voltage increases, amps decreases for same power requirement!

Regards,
Shahvir

 

[Bob S]

Also, there is actually power loss in long interstate lines due to radiation (think horizontal antennas) at 60 Hz. The radiation is proportional to current, so raising the voltage reduces the radiation.

 

[DaTario]

Ok,

First of all, thank you for the response.

Now, let me just touch one point in this discussion.

Joule effect gives us an expression to the power loss due to the passage of current through a conductor (P = R i^2).

But there seems to exist the power consumed by city users of electricity, which should be given by a different expression. The expression P = U i is an equivalent form of P = R i^2 (at least it seems to be).

I am a bit confused in articulating these two formulas (namely: P = R i^2 and P = Ui) as in both cases I guess they address the Power loss by heating (Joule effect).


Thank you once more,

Best wishes,

DaTario

 

[Bob S]

The two formulas for power consumed by city users are equivalent. The joule heating is the same. 1 watt = 1 joule per second. 1 kiowatt-hour is 1000 watts for 1 hour.

 

[DaTario]

Ok, let me check If I have correctly understood this:

By raising the voltage, for a given power requirement, the current assume a low value and this is the key concept, as low value for current represent low value for energy losses due to joule effect.

Is it correct?

What about capacitive and inductive effects?

Thank you

DaTario

 

[b.shahvir]

Ok, let me check If I have correctly understood this:

By raising the voltage, for a given power requirement, the current assume a low value and this is the key concept, as low value for current represent low value for energy losses due to joule effect.

Is it correct?

What about capacitive and inductive effects?

Thank you

DaTario

Your assumptions are correct! Low current means lower voltage drops across the capacitive and inductive components of the transmission line, as a result, lower losses in the lines itself...hence a more efficient power transfer to load.

 

[Bob S]

Your assumptions are correct! Low current means lower voltage drops across the capacitive and inductive components of the transmission line, as a result, lower losses in the lines itself...hence a more efficient power transfer to load.

There is one inductive effect that the utilities have to deal with that does not come from their own equipment. Nearly every electric motor appears to be slightly inductive to the utility. This includes electric motors in refrigerators and washing machines in homes. Unless corrected at the local level, the utility has to generate and transmit extra current to compensate for this. This is the basis for terms like "power factor" and "VA" (vs. watt) load. Our power meters measure only kilowatt-hours, and not KVA-hours, but some industrial companies have to pay a surcharge if they have too large (actually too small) a power factor.

 

[b.shahvir]

How does the 'self-inductance' of the transmission line itself contribute to power loss? ...as self-inductance is a reactive (wattless) component and power loss is due to active (wattful) components of a transmission line! please elaborate on this concept.
Thanx.

 

[DaTario]

How does the 'self-inductance' of the transmission line itself contribute to power loss?

Thanx.

Nice question!

Thank you for the debate.

DaTario

 

[b.shahvir]

The following are my thoughts on the topic;

In my opinion, power loss in transmission line means when the receiving end voltage Vr is less than the sending end voltage Vs. This happens as Vs is dropped across line components resulting in reduced voltage Vr. Reduction in Vr accounts for power loss.

The power loss is due to series line resistance, skin effect, series line loop inductance, shunt leakage capacitances (in long lines) shunt inductances (load side), power loss due to radiation and induction, etc.

Presently my topic of interest is power loss due to series line inductance which causes reduced Vr and hence in my opinion is equally responsible for power loss akin to series line resistance and skin effect.

Suitable guidance for the same will be greatly appreciated. Thanx.