Ohm Law with High Voltage Transformers

[Voltz]

Okay here's something that I've been thinking about for a while, I know that in Power Plants they use step up transformers to reduce waste:

W = I x R
W = 500 x 100 = 50,000 vs W = 20 x 100 = 2000

But what about ohms law, I mean I know a lot of people get mixed up when they think of the transformer as load (with a resistance/impedance value) but if the wire stays the same and a higher voltage is pumped through it doesn't that mean the current flowing through it should increase?

Similarly stun guns, they step up the voltage to voltages like 20kV but however do not exceed 5-10ma when this voltage is pumped through the resistance of the human body

I'm sure there's something I'm missing? Enlighten me please

 

[Hero999]

What don't you understand?

At the power station, a step-up transformer increases the voltage so the same amount of power can be distributed with a lower current. Distributing large amounts of power at low voltages is impractical because the cables have to be really thick to carry the large currents. For safety reasons, the voltage is then stepped down to a more usable low voltage to power homes and offices.

Suppose you need to distribute 500MW of power.

At 230V:
[latex] I = \frac{P}{V}=\frac{500 \times 10^6}{230} = 2.17 \times 10^6A[/latex]

At 400kV:
[latex]I = \frac{500 \times 10^6}{400 \times 10^3} = 1.25 \times 10^3A[/latex]

Imagine how much thicker the cables would need to be to handle 2.17MA compared to 1.25kA.

Of course, in real life the power is generated in three phases but I kept the calculations for single phase because it simplifies things.

 

[Reloadron]

To add to Hero's post this is a good read on the subject. It pertains to a North American power grid distribution but pretty much only the voltages change elsewhere.

During the past few months I have been involved with our power company (First Energy) planning a new sub station for our facility. Our current sub station is exceeding its capacity now and we are limited as to what we can run and when. Currently we bring in 5 KV three phase in two separate lines feeding 4 transformers. Each transformer delivers 480 volts 3 phase 800 amp service. All indoor with underground feed.

The new sub we had a choice of 26 KV or 69 KV and opted for the 26 KV even though the 69 KV service would have a lower cost for actual power. The deal breaker was when we saw the cost of 69 KV switch gears. It would take us like 50 years to break even on the switch gears.

Once you get the power to the sub you need upstream and downstream switches that look like this. That part is all outside which is fine with me.

Oh yeah, the 26 KV switches carry a price tag of about $200,000 USD for a three phase switch. Things will be going in this summer.

Ron

 

[Voltz]

What I don't understand is that if I = V/R then what does this mean for the voltage and current across the resistance of the wire

 

[Hero999]

Sorry, I still don't get what you're saying.

If the same power can be transferred with a lower current, the I²R losses will be lower, meaning the amount of power dissipated, i.e. lost in the cable so the efficiency will be higher.

EDIT:
Did you read all of the above posts a couple of times?

Try plugging different numbers into the formulae posted above.

Consider a 2400W kettle being powered from a very long piece of cable with a resistance of 2Ω.

The power supply voltage to the heater is 240V

How much current does the kettle require?

What will the losses in the cable be at 240?

Now add a step up transformer to increase the voltage to 2.4kV at the start of the cable and a step down transformer to convert the 2400V to 240V at the other end of the cable.

For simplicity's sake, ignore the losses in the transformers, assume they're perfect.

Now calculate the current through the cable, remember the load at the other end now requires 2400V with the power remaining at 2400W.

What will the losses in the cable now be?

 

[ericgibbs]

hi voltz,

Similarly stun guns, they step up the voltage to voltages like 20kV but however do not exceed 5-10ma when this voltage is pumped through the resistance of the human body

Its whats called the source resistance of the voltage/current supply
If you have a 20kV voltage source that will deliver say 10mA, this means the source resistance is 20,000/0.01 = 2 meg Ohms, so the 10K resistance of the human body is small compared to 2megs source resistance.

 

[Diver300]

What I don't understand is that if I = V/R then what does this mean for the voltage and current across the resistance of the wire

If the transmission voltage is increased, then the load resistance is increased, so that the power is kept the same. The load resistance is either increased by putting a transformer in the way, or by redesigning the load. If a transformer is used, it's actually the impedance not the resistance that is increased, but the concept is the same.

Because the power is the same, the current is less if the voltage is bigger. If the same cable is used, the resistance of the cable is the same. Less current means less voltage drop in the cable.

For example, a vehicle headlight is 55 W. It can't be more, or the casing would melt.

55 W bulbs are available at 12 V for cars or 24 V for trucks. The 24 V ones are made with longer, thinner filaments. The resistance of the 12 V ones is 12 x 12 / 55 = 2.62 Ω. The resistance of the 24 V one is 24 x 24 / 55 = 10.47 Ω

If the cable has a resistance of 0.1 Ω, the voltage drop at 12 V is 55 / 12 x 0.1 = 0.45833 V and 2.1 W of power is lost in the cable.

At 24 V, the voltage drop is 0.2292 V and 0.525 W of power is lost in the cable.

 

[Voltz]

If the transmission voltage is increased, then the load resistance is increased, so that the power is kept the same. The load resistance is either increased by putting a transformer in the way, or by redesigning the load. If a transformer is used, it's actually the impedance not the resistance that is increased, but the concept is the same.

Because the power is the same, the current is less if the voltage is bigger. If the same cable is used, the resistance of the cable is the same. Less current means less voltage drop in the cable.

For example, a vehicle headlight is 55 W. It can't be more, or the casing would melt.

55 W bulbs are available at 12 V for cars or 24 V for trucks. The 24 V ones are made with longer, thinner filaments. The resistance of the 12 V ones is 12 x 12 / 55 = 2.62 Ω. The resistance of the 24 V one is 24 x 24 / 55 = 10.47 Ω

If the cable has a resistance of 0.1 Ω, the voltage drop at 12 V is 55 / 12 x 0.1 = 0.45833 V and 2.1 W of power is lost in the cable.



At 24 V, the voltage drop is 0.2292 V and 0.525 W of power is lost in the cable.

Thanks, thats the explanation I was looking for
EDIT: Okay I still don't fully understand, so the load resistance must change, I know that for the Power to remain the same if the voltage goes up the current has to go down (P = V x I) but for the voltage to go up and the current to go down the resistance has to go up (I = V/R) - we're talking totally about the resistance of the wire here though I'm not talking about load as that's on the other side of a transformer

Okay so if the voltage is 2kV and the wire's total resistance is 500Ω then the current is 2000/500 = 4A and P = I² x R = 16 x 500 = 8kW

If the power stays at 8kW and the voltage goes up to say 200kV then for the current to go down by the inverse factor (10²) so it should be 0.04A but if the wire's resistance remains at 500Ω then I=V/R is I = 200,000/500 = 4kA

So basically if the voltage goes up and resistance stays the same why does current go down?

 

[crutschow]

The line resistance does not determine the current in a power line, expect under short circuit conditions. Normally it's a very small percentage of the load resistance. If the voltage goes up and the resistance stays the same, then both current and power dissipated will, of course, increase. But for a given load power, then if the voltage goes up, the current will decrease.

 

[Hero999]

I think attempting the problems I posted previously will help you to learn why increasing the voltage reduces the losses.