DC transmission is common on long feeders and at grid interconnections. DC transmission is more efficient when the circuits are very long (500km+) and at ultra high voltages (above 500kV) because of the high inductance and capacitance that AC transmission lines have and because of the skin effect. You also only need 2 cables instead of 3, saving on conductor material. The up front cost for the converter stations is the biggest hurdle, which is why they're more economical above a certain circuit length.
greatly explained the energy loss part and why we need transformers ( high voltage on long distances and low voltage where energy consumption takes place)
I calculate the low voltage load resistance to be 53 ohm and 4.3 amps for 230v and 1kW. How are you calculating for this example? Resistance =Voltage^2 / Power is what I used.
Hello sir, I have a question. Since the energy that goes through the wire is the same, regardless of the voltage, what is the fundamental change that causes the energy loss. What I can imagine is that with the low voltage we have current through the wire and with the high voltage we have current to the perimeter and around the wire. Can you please give your explanation? Thank you
In both AC and DC transmission lines, Joule heating is the main power loss mechanism. From a pure loss-reduction standpoint, DC transmission is superior due to the lack of skin effect and radiative losses. The difficulty with DC is the efficient conversion of voltages since only high voltage transmission can minimize Joule heating along the line.
I'm still getting Leyden jar deliveries from the milk man, er I mean elecrticity man. (That's a joke, I'm a licensed and certified electrician... I say I'm an electrician not to imply that I'm too knowledgeable about electricity to believe in the Leyden jar delivery man... just to imply that I can't pass up an electricity joke.)
I was thinking about the exact same thing! Nikola deserves at least 15 sec of introduction here - he's the father (and mother) of AC. But more seriously - fantastic videos Professor Danner! Very interesting and very well explained
Tesla unfortunately worked on 2 phase not the 3 phase that became the world standard and although he did contribute a lot in the world of electricity he was not the only engineer working on this there were hundreds around the world
Using V^2/R you have just verified that putting 200V across a 2 ohm resistor consumes more power than putting 100V across it. We always want as small voltage drop as possible for this reason. But voltage drop is different than voltage with respect to ground. Assume we have 100V across a line resistance but each end is 100100V and 100000V. This delivers far more power to the load than say 1100V and 1000V.
@@adanner i was confused why 200v saves more electric energy than 100v. do the power companies limit the house's power consumption? they fix the voltage but i can use the Ampere in the limit they provide. if 1000watt is all that power company allows me 1000watt = P(100v)= IV = 10A *100v 1000watt = P(200v)= IV = 5A *200v if 2 ohm is from copper wire resistance that causes a power loss 1. P(100v) = P =I^2 * R = (10A)^2 * 2 ohm = 200w 2. P(200v) = P =I^2 * R = (5A)^2 * 2 ohm = 50w 3. P(100v) = P =V^2 / R = (100v)^2 / 2 ohm = 10000w/2 4. P(200v) = P =V^2 / R = (200v)^2 / 2 ohm = 40000w/2 3 and 4. cant have 100v and 200v, i understood 3. 4. were wrong thanks for reply, asking you fixed my brain,
@@abcdef2069 Yes the power company strictly limits the power households can consume: It is limited to the maximum output power of the generators connected to the grid. If the power is exceeded, they shut off electricity to some customers for awhile (rolling blackouts).
in the DC Transmission, you made calculations to show that low voltage has a bigger power loss than high voltage. The problem with what you just showed is that you put in the RLoad yourself. I could make a video, wit the same calculations and say oh we're using RLoad to be 1k Ohms in low voltage and RLoad to be 1k ohms as well on high voltage. It just makes no sense.
It wouldn't be a fair comparison if you used the same RLoad in both the high and low voltage cases. I deliberately picked values for RLoad that would make the power delivered to the load the same. For the same power delivered, we can then safely compare losses in each system.
concise and useful in clear and beautiful English. what a pleasure to watch your lectures.
DC transmission is common on long feeders and at grid interconnections. DC transmission is more efficient when the circuits are very long (500km+) and at ultra high voltages (above 500kV) because of the high inductance and capacitance that AC transmission lines have and because of the skin effect. You also only need 2 cables instead of 3, saving on conductor material. The up front cost for the converter stations is the biggest hurdle, which is why they're more economical above a certain circuit length.
greatly explained the energy loss part and why we need transformers ( high voltage on long distances and low voltage where energy consumption takes place)
beautifully explained. what a gifted lecturer!
Very nice, concise and clear course on Edx, Thank you
You're most welcome. Keep watching and learning. Please do spread the word so others can learn too.
Nice explanation 😊
Thank you Aaron...
How do you get the Rload value?
nice video , Thanks
Thank you sir
I calculate the low voltage load resistance to be 53 ohm and 4.3 amps for 230v and 1kW. How are you calculating for this example? Resistance =Voltage^2 / Power is what I used.
Exactly, that's where I am stuck too. 53-13 ohm =40 ohm should be the load resistance.
Hello sir, I have a question.
Since the energy that goes through the wire is the same, regardless of the voltage, what is the fundamental change that causes the energy loss. What I can imagine is that with the low voltage we have current through the wire and with the high voltage we have current to the perimeter and around the wire.
Can you please give your explanation?
Thank you
In both AC and DC transmission lines, Joule heating is the main power loss mechanism. From a pure loss-reduction standpoint, DC transmission is superior due to the lack of skin effect and radiative losses. The difficulty with DC is the efficient conversion of voltages since only high voltage transmission can minimize Joule heating along the line.
What if the load for the both voltage sources is the same i.e 10 ohms. Surely the load is not fixed
Why are long submarine cable DC and not AC?
If we have U=230V and P(load)=1000W then R(load)=Upower2/P=53Ohm, not 10Ohm. Then P(line)=157W, not 1300W. Am I right?
Title AC vs DC, content HV vs LV😂
I'm still getting Leyden jar deliveries from the milk man, er I mean elecrticity man.
(That's a joke, I'm a licensed and certified electrician... I say I'm an electrician not to imply that I'm too knowledgeable about electricity to believe in the Leyden jar delivery man... just to imply that I can't pass up an electricity joke.)
incredible that it does not mention Nikola Tesla.
I was thinking about the exact same thing! Nikola deserves at least 15 sec of introduction here - he's the father (and mother) of AC. But more seriously - fantastic videos Professor Danner! Very interesting and very well explained
Tesla unfortunately worked on 2 phase not the 3 phase that became the world standard and although he did contribute a lot in the world of electricity he was not the only engineer working on this there were hundreds around the world
@@drmindbender8616 Yes of course. 😆You do not have the absolute truth. Edison and Westinghouse only contributed to the monopolization of electricity.
at 4:29 P =I^2 * R (V=I * R) = V^2/ R
P100 = (100v)^2 /2 ohm
P200 = (200v)^2 /2 ohm
P200 >> P100 whats wrong in this way
Using V^2/R you have just verified that putting 200V across a 2 ohm resistor consumes more power than putting 100V across it. We always want as small voltage drop as possible for this reason. But voltage drop is different than voltage with respect to ground. Assume we have 100V across a line resistance but each end is 100100V and 100000V. This delivers far more power to the load than say 1100V and 1000V.
@@adanner i was confused why 200v saves more electric energy than 100v.
do the power companies limit the house's power consumption?
they fix the voltage but i can use the Ampere in the limit they provide.
if 1000watt is all that power company allows me
1000watt = P(100v)= IV = 10A *100v
1000watt = P(200v)= IV = 5A *200v
if 2 ohm is from copper wire resistance that causes a power loss
1. P(100v) = P =I^2 * R = (10A)^2 * 2 ohm = 200w
2. P(200v) = P =I^2 * R = (5A)^2 * 2 ohm = 50w
3. P(100v) = P =V^2 / R = (100v)^2 / 2 ohm = 10000w/2
4. P(200v) = P =V^2 / R = (200v)^2 / 2 ohm = 40000w/2
3 and 4. cant have 100v and 200v, i understood 3. 4. were wrong
thanks for reply, asking you fixed my brain,
@@abcdef2069 Yes the power company strictly limits the power households can consume: It is limited to the maximum output power of the generators connected to the grid. If the power is exceeded, they shut off electricity to some customers for awhile (rolling blackouts).
in the DC Transmission, you made calculations to show that low voltage has a bigger power loss than high voltage. The problem with what you just showed is that you put in the RLoad yourself. I could make a video, wit the same calculations and say oh we're using RLoad to be 1k Ohms in low voltage and RLoad to be 1k ohms as well on high voltage. It just makes no sense.
It wouldn't be a fair comparison if you used the same RLoad in both the high and low voltage cases. I deliberately picked values for RLoad that would make the power delivered to the load the same. For the same power delivered, we can then safely compare losses in each system.
And once again Tesla is ignored as the other inventor.
he wasn't "the other inventor" he was THE inventor