3 ms·

Ohms law: V = IR Power: P = VI Power loss due to resistive heating in the wires: P = I^2 * R Note that current (I) quickly dominates the loss calculation bec

Ohms law: V = IR Power: P = VI Power loss due to resistive heating in the wires: P = I^2 * R Note that current (I) quickly dominates the loss calculation because it is squared. So for transmission lines, efficiency is obtained by keeping the current as low as possible. Therefore, to transfer the same power you need to make the voltage as high as possible. The I^2*R heating loss applies equally to DC and AC transmission lines. So the DC lines still need to be high voltage for efficiency as well. By using AC we can 'transform' high-voltage-low-current to low-voltage-high-current power very easily. It's literally just two coils of wire around a chunk of iron. Step-down AC transformers are used both at your neighborhood substation and on the pole outside your house. But the article didn't say how they planned to transform the high-voltage-low-current DC into anything usable.

eigenvector 14y ago
A switch-mode DC/DC converter obtains the same result as an AC transformer with efficiency approaching that of a well-designed AC transformer. It is a solved problem.
- marshray 14y ago
  The century-old methods are still competitive at this high a voltage and power. http://en.wikipedia.org/wiki/High-voltage_direct_current#Disadvantages http://en.wikipedia.org/wiki/High-voltage_direct_current#Dis... HVDC is less reliable and has lower availability than alternating current (AC) systems, mainly due to the extra conversion equipment. Single-pole systems have availability of about 98.5%, with about a third of the downtime unscheduled due to faults. Fault-tolerant bipole systems provide high availability for 50% of the link capacity, but availability of the full capacity is about 97% to 98%.[19] The required converter stations are expensive and have limited overload capacity. At smaller transmission distances, the losses in the converter stations may be bigger than in an AC transmission line. The cost of the converters may not be offset by reductions in line construction cost and lower line loss.
breckinloggins 14y ago
I'm wondering if it would be cost-feasible (eventually) to incorporate pure sine wave inverters at the "last mile point" and do AC to the premises. Does anyone know?
ChuckMcM 14y ago
Historically the limit on voltage has been the dielectric constant for air. Once you exceed it, the air ionizes, and you get a short. Visible as 'lightning' going through the air from source to its ground counterpart. Transmission line effects limit the ability to send AC power long distances as the every time you switch the potential you generate magnetic fields and some of the energy is radiated off into the space around the wire. AC has been more desirable to date because of a relatively easy infrastructure for converting it, and for sharing it on the same line. Transforming DC into AC is easily done by driving a generator set but for most things you'd want the genset to convert to 'standard' three phase AC and then re-use existing city wide infrastructure. What bothered me about the article was that nowhere does it explain how you can make a DC breaker that can actually disconnect at those voltages. Since the dielectric constant is a function of distance, when the breaker first opens there is an arc because, the current really really wants to keep flowing. And the dielectric constant of ionized air is quite small. "Regular" breakers have a fan that kicks in once they get to a minimum distance apart to "blow out" the arc (it pushes it out to extend it and thus 'break' it, see the video here https://www.youtube.com/watch?v=hIkNY5xjy5k https://www.youtube.com/watch?v=hIkNY5xjy5k for how this works) Anyway, would love to hear actually what it was they invented.
- marshray 14y ago
  As ususal, looks like the best description made it into Wikipedia http://en.wikipedia.org/wiki/High-voltage_direct_current http://en.wikipedia.org/wiki/High-voltage_direct_current The ABB breaker contains four switching elements, two mechanical (one high-speed and one low-speed) and two semiconductor (one high-voltage and one low-voltage). Normally, power flows through the low-speed mechanical switch, the high-speed mechanical switch, and the low-voltage semiconductor switch. The last two switches are paralleled by the high-voltage semiconductor switch. Initially, all switches are closed (on). Because the high-voltage semiconductor switch has much greater resistance than the mechanical switch plus the low-voltage semiconductor switch, current flow through it is low. To disconnect, first the low-voltage semiconductor switch opens. This diverts the current through the high-voltage semiconductor switch. Because of its relatively high resistance, it begins heating very rapidly. Then the high-speed mechanical switch is opened. Unlike the low-voltage semiconductor switch, which is only capable of standing off the voltage drop of the closed high-voltage semiconductor switch, this is capable of standing off the full voltage. Because no current is flowing through this switch when it opens, it is not damaged by arcing. Then, the high-voltage semiconductor switch is opened. This actually cuts the power. However, it only cuts power to a very low level; it is not quite 100% off. A final low-speed mechanical switch disconnects the residual current.
lutorm 14y ago
Yes, you have to run both at high voltage, but I thought the efficiency difference was because DC has no inductive losses.

eigenvector 14y ago

A switch-mode DC/DC converter obtains the same result as an AC transformer with efficiency approaching that of a well-designed AC transformer. It is a solved problem.

marshray 14y ago
The century-old methods are still competitive at this high a voltage and power. http://en.wikipedia.org/wiki/High-voltage_direct_current#Disadvantages http://en.wikipedia.org/wiki/High-voltage_direct_current#Dis... HVDC is less reliable and has lower availability than alternating current (AC) systems, mainly due to the extra conversion equipment. Single-pole systems have availability of about 98.5%, with about a third of the downtime unscheduled due to faults. Fault-tolerant bipole systems provide high availability for 50% of the link capacity, but availability of the full capacity is about 97% to 98%.[19] The required converter stations are expensive and have limited overload capacity. At smaller transmission distances, the losses in the converter stations may be bigger than in an AC transmission line. The cost of the converters may not be offset by reductions in line construction cost and lower line loss.

breckinloggins 14y ago

I'm wondering if it would be cost-feasible (eventually) to incorporate pure sine wave inverters at the "last mile point" and do AC to the premises. Does anyone know?

ChuckMcM 14y ago

Historically the limit on voltage has been the dielectric constant for air. Once you exceed it, the air ionizes, and you get a short. Visible as 'lightning' going through the air from source to its ground counterpart. Transmission line effects limit the ability to send AC power long distances as the every time you switch the potential you generate magnetic fields and some of the energy is radiated off into the space around the wire. AC has been more desirable to date because of a relatively easy infrastructure for converting it, and for sharing it on the same line. Transforming DC into AC is easily done by driving a generator set but for most things you'd want the genset to convert to 'standard' three phase AC and then re-use existing city wide infrastructure. What bothered me about the article was that nowhere does it explain how you can make a DC breaker that can actually disconnect at those voltages. Since the dielectric constant is a function of distance, when the breaker first opens there is an arc because, the current really really wants to keep flowing. And the dielectric constant of ionized air is quite small. "Regular" breakers have a fan that kicks in once they get to a minimum distance apart to "blow out" the arc (it pushes it out to extend it and thus 'break' it, see the video here https://www.youtube.com/watch?v=hIkNY5xjy5k https://www.youtube.com/watch?v=hIkNY5xjy5k for how this works) Anyway, would love to hear actually what it was they invented.

marshray 14y ago
As ususal, looks like the best description made it into Wikipedia http://en.wikipedia.org/wiki/High-voltage_direct_current http://en.wikipedia.org/wiki/High-voltage_direct_current The ABB breaker contains four switching elements, two mechanical (one high-speed and one low-speed) and two semiconductor (one high-voltage and one low-voltage). Normally, power flows through the low-speed mechanical switch, the high-speed mechanical switch, and the low-voltage semiconductor switch. The last two switches are paralleled by the high-voltage semiconductor switch. Initially, all switches are closed (on). Because the high-voltage semiconductor switch has much greater resistance than the mechanical switch plus the low-voltage semiconductor switch, current flow through it is low. To disconnect, first the low-voltage semiconductor switch opens. This diverts the current through the high-voltage semiconductor switch. Because of its relatively high resistance, it begins heating very rapidly. Then the high-speed mechanical switch is opened. Unlike the low-voltage semiconductor switch, which is only capable of standing off the voltage drop of the closed high-voltage semiconductor switch, this is capable of standing off the full voltage. Because no current is flowing through this switch when it opens, it is not damaged by arcing. Then, the high-voltage semiconductor switch is opened. This actually cuts the power. However, it only cuts power to a very low level; it is not quite 100% off. A final low-speed mechanical switch disconnects the residual current.

lutorm 14y ago

Yes, you have to run both at high voltage, but I thought the efficiency difference was because DC has no inductive losses.