Electric power distribution from powerplant to homes

[EN1986]

I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage.

The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me.

I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable consumption. And according to this a higher voltage (while assuming a constand resistance that comprises of the line resistance and the load resistance) leads to a higher current that results in a higher power loss.

What am I missing?

 

[.Scott]

Power lines are resistors. The more current that flows through them, the greater the voltage drop from source to user, and thus the more power is dissipated.
The total power delivered is volts times amps. So, given a certain amount of required power, higher voltage means less current and thus less transmission losses.

 

[Baluncore]

W = I ⋅ V ;
If you double the line voltage, then for the same power, the load current will be halved.
Then consider power lost in the line resistance, R.
Ohm's Law; V = I ⋅ R ;
W = I² ⋅ R ;
By doubling the voltage, power loss in the line is reduced to one quarter.

 

[russ_watters]

I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable consumption. And according to this a higher voltage (while assuming a constand resistance that comprises of the line resistance and the load resistance) leads to a higher current that results in a higher power loss.

What am I missing?

What you are saying is true for a simple circuit with one source and one load/resistor, but the power line isn't the only thing in in the circuit. The things the power line is feeding needs specific amperage, voltage and therefore power.

 

[SredniVashtar]

What you are missing is that, since the wire resistance and the load resistance (let's keep it all resistive for simplicity) are in series, they share the same current, no matter what is the power required by the load.

Voltage is more or less the same at the transmitting station and at the receiving station - the difference being the (hopefully small) voltage drop in the wires. So the loss in the wires is given by Rwire x Iload^2 and for the same power at the load this loss is lower the smaller Iload is.

 

[Fisherman199]

there's a distinction in voltage and voltage drop. voltage is determined by generation and by operating requirements of the system.

you seem to be confusing the two (system voltage and voltage drop). V_drop = i*r

voltage drop is calculated of irrespective of the nominal voltage of the system. it makes no difference if it's a 12kV, 25kV, 69kV, 230kV, or 500kV system, voltage loss is the same calculation.

power losses are similar. P_loss = i^2*r

 

[Rive]

What am I missing?

For one: the transformers are not just adjusting the voltage, they are also modifying the apparent resistance (well, more like impedance) accordingly. And, of course, the other transformer at the other end of the line which will set the lower voltage will also 'reset' the resistance.

Thus: the whole 'transmission through elevated voltages' business is kind of 'transparent'.

 

[Guineafowl]

a higher voltage […] while assuming a constant resistance […] leads to a higher current that results in a higher power loss.

I think your problem is in red. Although true, it doesn’t apply to the power grid.

What you need to make things happen is a certain amount of power, so, for a given (constant) amount of power:
$$P=VI$$
You can either have high $V$ and low $I$, or vice versa.

Because of $P={I^2}R$ you’re better off with low $I$.

 

[renormalize]

You can either have high $V$ and low $I$, or vice versa.

Because of $P={I^2}R$ you’re better off with low $I$.

I think a more cogent argument is the following.
Define $V_0$ to be the voltage at the generator, $R_W$ the resistance of the delivery wiring and $I$ the current flowing in that wiring. Then the power that can be delivered to a load is:$$P_L=V_L\,I=\left(V_0-IR_W\right)I=V_0\,I-I^2 R_W\tag{1}$$This power vanishes at the two limiting currents $0$ and $V_0/R_W$ and so it must be maximized for some $I$ between those values. Thus:$$0=\frac{dP_{L}}{dI}=V_{0}-2IR_{W}\:\Rightarrow\:I_{max\,pow}=\frac{V_{0}}{2R_{W}}\tag{2}$$Using this in eq.(1) gives then:$$P_{L\:max}=\frac{V_{0}^{2}}{4R_{W}}\tag{3}$$Thus, maximizing the generator voltage $V_0$ always maximizes the delivered power $P_L$.

 

[DaveE]

I think a more cogent argument is the following.
Define $V_0$ to be the voltage at the generator, $R_W$ the resistance of the delivery wiring and $I$ the current flowing in that wiring. Then the power that can be delivered to a load is:$$P_L=V_L\,I=\left(V_0-IR_W\right)I=V_0\,I-I^2 R_W\tag{1}$$This power vanishes at the two limiting currents $0$ and $V_0/R_W$ and so it must be maximized for some $I$ between those values. Thus:$$0=\frac{dP_{L}}{dI}=V_{0}-2IR_{W}\:\Rightarrow\:I_{max\,pow}=\frac{V_{0}}{2R_{W}}\tag{2}$$Using this in eq.(1) gives then:$$P_{L\:max}=\frac{V_{0}^{2}}{4R_{W}}\tag{3}$$Thus, maximizing the generator voltage $V_0$ always maximizes the delivered power $P_L$.

But as others have said $R_W$ isn't necessarily a constant.

I don't think maximum power delivery is the only metric to optimize, this leads to very inefficient solutions. Other things a utility may value:
1) Infrastructure costs.
2) Operating costs.
3) Power quality.
4) Reliability.

Several of these strongly favor a low loss system, if they can afford to build it.

 

[renormalize]

Several of these strongly favor a low loss system, if they can afford to build it.

Sure, but the OP specifically asked:

I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage.

That's the basic question I aimed to answer in my post #9.

 

[DaveE]

Sure, but the OP specifically asked:

I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage.

He asked about "more effective", not just more. Your example is good, but maximum power for a given line isn't really the point. Utilities simply don't want to operate in this very lossy region with poor load regulation.

I interpret this more like minimum losses (cost, really) for a given user load. I think as @Rive said, it's more about transformers (and equivalents) in distribution. You could also approach this solution as higher voltage requires less copper for the same efficiency. In this, overly simplistic, model there is no optimum voltage. Higher voltage always results in lower resistive losses without limit.

 

[Averagesupernova]

The poor OP is likely thoroughly confused by now. The end goal always is to have a steady unchanging voltage where the end user utilizes it. Whatever the user loads on this voltage source will draw the current as needed while the supply voltage stays steady. So with steady voltage, current and watts delivered are proportional. The idea is to have as little loss as possible in the non perfect wires throughout the distribution system. Using transformers as previously described accomplishes this. I'd say this is simply a case of the OP mis-applying ohms law.

 

[Baluncore]

The poor OP is likely thoroughly confused by now.

The OP has not been back since posting the question.

 

[Averagesupernova]

The OP has not been back since posting the question.

No surprise there...

 

[Kenwstr]

I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage.

The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me.

I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable consumption. And according to this a higher voltage (while assuming a constand resistance that comprises of the line resistance and the load resistance) leads to a higher current that results in a higher power loss.

What am I missing?

I believe, your missing the load to convert transmission voltage down to local substation voltage & then consumer load on the substation.

 

[Rive]

The poor OP is likely thoroughly confused by now.

Likely right. So - ...
...there are multiple facets of this 'higher voltage' thing.

One is, the transmission. Somewhere around the generators there will be a transformer, which will elevate the voltage to transmission level to be fed into the grid with lower current (= > lower losses) for transmission. Somewhere around the consumers there will be another transformer which brings it back to consumer level voltage, and will be able to supply high currents. The voltage, current and apparent impedance are all different at different sections.

One other is on the the consumption side. At higher voltage less current (thus: less copper) is needed for a required power. This part is about elevating the 'nominal' voltage from the outlet, thus: consumer equipment needs to be ready for the change. In Europe, this was the 220V to 230V change, with equipment already required to be ready for up to 240V. Higher voltage is more efficient, but change requires time and old equipment needs to be phased out. For this part, equipment is designed to take what's needed at nominal voltage, so at higher voltage you need less current for the same 1kW motor, for example. It's not the same for old 220V and for new 230V, but a new equipment => they will indeed eat up less, despite Ohm's laws: because they were designed so.

Third is the short term stability of the grid. The actual voltage is not 230V, but fluctuating around it. At 235V, resistive loads will take more power than nominal, while at 225V they take less => if there is momentary overgeneration on the grid, then the voltage will rise and through the higher current consumption will just take it: if there is undergeneration, the consumption will graciously eat up less current without any intervention.
Though there are equipment (anything with electronics power supplies) which will consume at constant power, thus requiring less current at higher voltage and more at lower voltage - kind of opposite to regular resistive loads. They work against this shelf-regulation of the grid, so too many of them is a problem.

 

[Guineafowl]

I think a more cogent argument is the following.
Define $V_0$ to be the voltage at the generator, $R_W$ the resistance of the delivery wiring and $I$ the current flowing in that wiring. Then the power that can be delivered to a load is:$$P_L=V_L\,I=\left(V_0-IR_W\right)I=V_0\,I-I^2 R_W\tag{1}$$This power vanishes at the two limiting currents $0$ and $V_0/R_W$ and so it must be maximized for some $I$ between those values. Thus:$$0=\frac{dP_{L}}{dI}=V_{0}-2IR_{W}\:\Rightarrow\:I_{max\,pow}=\frac{V_{0}}{2R_{W}}\tag{2}$$Using this in eq.(1) gives then:$$P_{L\:max}=\frac{V_{0}^{2}}{4R_{W}}\tag{3}$$Thus, maximizing the generator voltage $V_0$ always maximizes the delivered power $P_L$.

Cogent, yes, but overcomplicated for a beginner. Beware the curse of knowledge.

 

[renormalize]

Beware the curse of knowledge.

Agreed, but also beware of answers that are too "hand-waving" or "dumbed-down".
Again, quoting you:

What you need to make things happen is a certain amount of power, so, for a given (constant) amount of power:$$P=VI$$You can either have high $V$ and low $I$, or vice versa.
Because of $P={I^2}R$ you’re better off with low $I$.

But by that logic one could just as well say:
"Because of $P=V^2/R$ you're better off with low $V$."
How is a beginner to know which is the correct reasoning? That's the ambiguity in your answer that I try to address in my "overcomplicated" post #9.

 

[Guineafowl]

Agreed, but also beware of answers that are too "hand-waving" or "dumbed-down".
Again, quoting you:

But by that logic one could just as well say:
"Because of $P=V^2/R$ you're better off with low $V$."
How is a beginner to know which is the correct reasoning? That's the ambiguity in your answer that I try to address in my "overcomplicated" post #9.

Ah, I did miss something out there. How about, for the last line:

Because the power loss from heating in the lines $P=I^2{R}$ , you’re much better off with low $I$.

We did this quite early on in physics, perhaps age 12, hence the assumption of the OP’s prior knowledge level. I don’t think someone throwing calculus at us would have got very far. But the OP appears to have disappeared, so we might never know!

 

[Averagesupernova]

But by that logic one could just as well say:
"Because of P=V2/R you're better off with low V."

The OP isn't the only one mis-applying ohms law. The resistance R in the above formula is the resistance of the wires where the loss we are concerned with is occuring. So a mile of wire with a finite resistance will have a voltage from one end to the other. This is definitely not a voltage we want at all to say nothing of 'the bigger the better'.

 

[sophiecentaur]

how transferring electric from the powerplant to my house is more effective using high voltage.

You need to consider the whole system if you want to discuss efficiency. The appliances in your home are specified to work at a certain Voltage and to use a certain amount of Power. That requires a certain value of Resistance. (R = Vsquared)/P).The supply cables always have some resistance and that's due to the thickness of the 'copper' conductors. It's an engineering compromise between the thickness of wire you can afford and the power loss due to the consumer's current going through the cable resistance The lost power is often referred to as "Isquared r" loss (r is the cable resistance). There is also a voltage drop across the cables, which means the consumer gets less power delivered to existing appliances.

In any practical power distribution system, transformers are used to match the operating voltage of a link in the chain to the current capability of the cables and the Power to be transferred. Higher voltages need better insulation so the operating voltage is limited by cable insulation and height of transmission poles etc.. The overall system behaves just like a simple generator-cable-load.

 

[Babadag]

First, you are limited by what it is called “ampacity”-in USA-or current carrying capacity in IEC.
The current running through conductor produces losses as was stated above. The losses turn into heat, and the heat arise the temperature of conductor. Each conductor is covered by insulation. If the temperature is very high the insulation is damaged.

In order to limit the temperature, rise for the same current you have to reduce the resistance of the conductor by increasing the cross-section area of the conductor.

Second, you have to reduce the voltage drop.
The equipment-lighting bulbs, induction or other electric motors and other-as heaters for instance- need supply of a required level of voltage. The current passing through the conductor produces a voltage drop. The voltage drop reduces the voltage supplied at equipment then you have to reduce the voltage drop in order to keep the required level of voltage at end of supply point. If the supply point is so far that you need so big conductor cross-section area the possibilities are limited, at least, by price.

You have then to reduce the current. Then you have to rise the voltage in order to convey the same power, since the power is the product of current by voltage.
However, if you are the sole consumer at a very long distance from generators, you are in trouble. Usually the utillity-the electricity supplier- has a facility in vicinity so the equipment- mainly the cables- usually will be cheaper.

But the utility has to convey large power from big distances. So, they have to use more and more high level of voltage. As a simple calculation you need 1000 V per each km.
The losses are more in high-voltage-not only through resistance of the conductor but through discharge in air-corona phenomenon- and other effects and the voltage drop is bigger due to what is called reactance produced by magnetic field in alternative current. However, this is very low compared with the use of low voltage at high power conveying and large distances.

 

[berkeman]

Each conductor is covered by insulation. If the temperature is very high the insulation is damaged.

I don't think high voltage distribution wires (in air on poles or towers) are coated with insulation. That would just add weight and cost, and not be needed for safety insulation purposes. Only the final 120V/240V wires that come down from the pole transformers to the homes would have insulation on them.

 

[Averagesupernova]

I don't think high voltage distribution wires (in air on poles or towers) are coated with insulation. That would just add weight and cost, and not be needed for safety insulation purposes. Only the final 120V/240V wires that come down from the pole transformers to the homes would have insulation on them.

This is basically correct. Overhead wiring ahead of the transformer (primary wires) is never insulated as far as I know. There is overhead wiring that is insulated such as triplex but those are 600 volts or less at least here in the USA.

 

[Rive]

I don't think high voltage distribution wires (in air on poles or towers) are coated with insulation.

Insulation may reduce current capacity, actually.
Though there are special (wildlife-protection, for example) cases when it may be required. At least, in the EU.

 

[Averagesupernova]

Insulation may reduce current capacity, actually.
Though there are special (wildlife-protection, for example) cases when it may be required. At least, in the EU.

Also, thicker wire will hold more ice. Ice on overhead wire is bad news. Heavy ice on wires alone is not necessarily why they come down. Wind causes the ice to form an irregular shape of ice surrounding the wire. This causes it to act like a wing and move in normal winds that would not otherwise matter. Also, when the ice falls off, the wires will snap/jerk. This causes a cascade of adjacent wires with ice to do the same thing. This can often mean wires breaking. With severe ice load this can lead to poles breaking off. So the insulation making for a larger area to collect ice is not something we want.

 

[Baluncore]

There are several reasons why extra high voltage lines are not insulated. If insulation was present on EHV lines, corona discharge could pyrolyse, or ignite, the insulation.

The drag forces due to wind are typically greater than the force due to gravity. The total force on the catenary is the vector sum of the two. For that reason, to reduce windage, the conductor must be of the smallest possible section. Insulation would increase the windage and necessitate stronger, and so more expensive, support structures.

The formation of ice on conductors can be relieved by circulating a reactive current in the line, that heats the wire. Electrical insulation on the line, would increase the thermal insulation, and so increase the reactive current needed to melt the external ice. Thicker wires, with greater ampacity, collect more ice, and so require more reactive current to melt that ice.

Thicker conductors may be more I²R efficient in operation, but they cost more to support, have higher windage, and are harder to maintain free of ice. Indeed, in cold climates, thinner wires can reduce both installation and maintenance costs.

 

[jrmichler]

The 7200 volt distribution line past my house has steel conductors without insulation. We had a big storm last year that closed the road in nine places around my house. I had a good look at the downed lines while helping the town road crew chainsaw trees. The wires did not break in one place, so were under a lot of tension. When I made the last cut, it made like a slingshot and flung the log a good distance. It's a good thing I was ready for that.

The steel conductors look exactly like the guy wires used to brace the power poles. Same diameter and number of wires in the cable.

As a simple calculation you need 1000 V per each km.

Sounds about right. I'm 5 miles from the substation.

 

[Babadag]

If you speak about conductors and not about cables you are right. However, in order to connect equipment with air transmission line you need cables, usually. The access to equipment has to be insulated. Of course, there are high voltage equipment connected directly to transmission line but there are not close to residential installations as open post requested:
“I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage.”
At least low voltage cables are involved in this case.