Zack, I'm a Chartered Electrical Engineer and Technical Director for my company providing HV designs up to 132kV in the UK. I also provide training. Can I just say that your video is excellent. I was going to produce my own you tube presentation on this for internal training but I think I will just refer people to yours. No wasted words, no BS, no patronising explanations. 10/10. well done. I'll look at your other videos.
Ok... first of all... you are genius. The set up you arranged with "black" board in reverse is a brilliant idea. For many reasons. On top of this, your explanation is simple and clear! Thank you
This percentage impedance along with the transformer voltage determines if the maximum fault current could be handled properly by the breakers. The bigger the transformer, the lower the percentage impedance. Good video.
That second sentence is wrong. The larger the transformer, in general, the larger the impedance. If you drive down the road and see a 50kva pole mounted transformer it probably has a %IZ of 2.3-3% or close to that. If you see a 100MVA transformer it might be 6,7 or 8% IZ. In general the larger the transformer the larger the percent impedance.
It appears that the short circuit current can simply be calculated by realizing that if the current on the secondary is 37.5 A with 20 V on the primary then the current is (600/20)*37.5 = 1125 A when we have 600 V on the primary. (Ohms law). Your result was slightly different because of rounding error %IZ.
I'll just leave it there: the amount of voltage required to overcome the impedance of the transformer to reach max secondary current (i.e., secondary short-curcuit current, which is determined by the design)
The percent impedance is what is called a per unit value. The base values for the system are implied to be the rating of the transformer, so the base kva is 9kva and the base voltage would be 240v. This is just a way to model systems to make it much much easier for power system and primarily fault analysis. A transformer just becomes it's equivalent impedance in series. The voltage due to the selected base (pu = actual/base) is now just 1 pu. So in a short circuit you have a simple series circuit with a voltage of 1pu and a reactance of .0333. Using ohms law 1pu/.0333 gives you a per unit short circuit current of 30 pu. To convert from pu back to actual fault current, you would just need to multiply the 30 pu by the base current which is determined by the given power and voltage bases. This also happens to be your secondary current, so - 9kva/240v = 37.5 and 30*37.5 = 1125A. Or to put it all together (base voltage in per unit)/(Transformer Impedance in pu)*Base Current = (1pu/.0333pu)*37.5A The voltage is not mentioned because it is selected to be 1pu, but it is in there. Its still just ohms law. Your not actually dividing by a impedance you're multiplying by a unitless current value. Current = (per unit current * base current) = 37.5/.03333 The equation is just a quick and easy way of combining those steps to determine the maximum available short circuit current at the secondary of any transformer. You'll more likely use it in this form Isc=S/(Root(3)*Vline*%z) or S/%z for your available fault duty if you are looking to incorporate other equipment for a quick rough calculation.
I was explaining in the case where we know the impedance already, but testing for the impedance works the same. You adjust your voltage until you're at a known current to determine the impedance. So its the same simple series circuit with the same bases, but the tests adjust the voltage supply until the secondary is at rated current which is also its per unit base current so it equals 1pu, which eliminates an additional math step. This occurs at 20v or Vpu = 20/600=.03333 pu Using Ohms law Zpu = Vpu/Ipu = .0333pu/1pu.
Loved it.. Great explanation Could you please let me know the logic behind the formulae you've used .. For e.g: why should the short circuit current be equal to the rated secondary current divided by the percentage impedance? Many thanks .. subscribed!
So typically we say that short circuit has to do with available kVA and the voltage of the transformer, which using Rated current takes in to account both of those. The other thing to consider would be impedance of the short circuit path, that's why this video we are just talking about what's available in theory at the transformer, not what you would see actually at the short circuit point.
If you change the percent impedance into actual impedance, your calculation for short circuit is just Vs/Z = Isc or a circuit with only your voltage source and your transformer impedance in series. All primary impedances are transferred over to the secondary side as one z equivalent. To get to z from %z you do the following. (1) Isc = Vs/Z (2) Z= %Z*Zbase where (3) Zbase = V^2/S Plugging (3) into (2) and (2) into (1) Isc = V/(%Z*V^2/S) = S*V/(%Z*V^2) = S/(%Z*V) Where S= V*A so Isc = V*A/(%Z*V) = A/%Z It is just ohms law with a lot of convenient variables that cancel out. The calcs with the actual transformer impedance for this example are shown below which results in the same answer. Zbase = (240V)^2/9000VA = 6.4 ohms Z = Zbase*Z% = 6.4ohm*.03333 = .21333 ohms Isc = Vs/Z = 240 V/2.1333 ohms = 1125A For the 3 phase case - Say three of these transformers were hooked up in a Delta-Y, Vllpri=Vphasepri = 600, Vphasesec = 240V and Vllsec=416V, then the equivalent transformer would be a 27kVA 3 phase 600-416/240 transformer with %Z of 3.333%. Your results would still be 1125A for the max available fault current per phase Isc = 27000/(3*240)*30 = 1125A
He just assumed that as he kept on increasing the input voltage on the primary side, 20V was the number when the ammeter on the secondary side started reading 37.5A (37.5A being the rated secondary current). This could any number depending on what the actual impedance is. For e.g. if the impedance of that transformer was higher, the ammeter on the secondary side would have recorded 37.5A only when the applied voltage on the primary side was a higher number, let say 27V.
Why we put a short circuit on secondary side of transformer to calculate percentage impedance? Why don't we put a load on secondary side in order to calculate the impedance?
It probably is a silly question but since we short circuited the secondary does the voltage remanis at 240V? Is it a constant voltage source the secondaryy????
Sir , how we can calculate drop voltage at primary with calculation. As mentioned u 20V how it s came? Without meter any possible for find that drop voltage
The short circuit current of the secondary of 1136 A is in theory true. However, In reality in an example where the source is 7200 volts to ground primary on a pole mounted distribution transformer and the impedance is 3.3% (with a 240 volt secondary) , that current will ALWAYS be limited to less than the calculated value as in this video. Most distribution transformers at First Energy were 3%. The 7200 (7.2KV) distribution source side case, that source may not be able to supply the current to achieve the theoretical value on the low voltage secondary. In the power utility business to do this calculation for the secondary fault current you have to know the short circuit AVAILABLE fault current of the SOURCE. Once you know that you can solve the problem. You can use CYME (tm) or if you know the available high side fault current (the electric company can tell you) then u can use the FREE EATON fault current I Phone AP to get the answer. The AP is available in the Apple(tm) Store or www.eaton.com/us/en-us/products/electrical-circuit-protection/fuses/fault-current-calculator.html. Once again, you need to know the available fault current of the source to get the answer. Why do u need to know this? Suppose there's a 25 KVA pole transformer, 7.2 KV primary, 120/240 secondary,100 feet of 1/0 TPX to the weather-head and 100 feet of #2 CU to the main breaker in the load center. The electrical inspector needs to know what is the available fault current at the panel. If it's under 10 KA, then u can use off the shelf breakers from the home center. If its over 10 KA but less than 22 KA, then you need more expensive breakers that can handle the predicted fault current. The electrician needs to know this too. When the primary is 19.9 KV phase to neutral, the available fault currents are always high.
When I was in electrical engineering school I met a guy that was an electrician and proud that he had learned all about electricity and it sure didn't take six or seven years to learn it. I said you know about a transformer it's made of copper wire and insulation, and iron. He said "I know what it's made out of". I said good now that you know what you need to use how much of each one do you put in?" He said "how the hell would you know that". I said"first you'd fall off a turnip truck in front of the University math department and stick around there for about 4 years convincing you was willing to pay the money and worthy of sitting in that seat. Then a guy who gets paid to tell people that would take a break and teach her class and you would have already proven you was worthy to try and understand what he was explaining because you understood what the math department was explaining and then a couple of years you could answer that question just like he can? Electrician scratched his head and left the bar.
Zack, I'm a Chartered Electrical Engineer and Technical Director for my company providing HV designs up to 132kV in the UK.
I also provide training. Can I just say that your video is excellent. I was going to produce my own you tube presentation on this for internal training but I think I will just refer people to yours. No wasted words, no BS, no patronising explanations. 10/10. well done. I'll look at your other videos.
Thanks so much for watching!
I had so many lightbulb moments watching this, you explained it better than anyone has
I haven't had to do these calculation in over 25 years. WOW Brilliant breakdown. Everything just came flooding back. Kudos!
Ok... first of all... you are genius. The set up you arranged with "black" board in reverse is a brilliant idea. For many reasons. On top of this, your explanation is simple and clear! Thank you
Mr Hartle, this video is truly the best explanation of one of the foundational principle of transformer…thank you so much for sharing it
This percentage impedance along with the transformer voltage determines if the maximum fault current could be handled properly by the breakers. The bigger the transformer, the lower the percentage impedance.
Good video.
Absolutely. Thanks for watching!
That second sentence is wrong. The larger the transformer, in general, the larger the impedance. If you drive down the road and see a 50kva pole mounted transformer it probably has a %IZ of 2.3-3% or close to that. If you see a 100MVA transformer it might be 6,7 or 8% IZ. In general the larger the transformer the larger the percent impedance.
Zack is correct..... just checked the nameplate of a single-phase 15kVA 7620/13200V 60HZ transformer which had a 1.9% IZ
It appears that the short circuit current can simply be calculated by realizing that if the current on the secondary is 37.5 A with 20 V on the primary then the current is (600/20)*37.5 = 1125 A when we have 600 V on the primary. (Ohms law). Your result was slightly different because of rounding error %IZ.
Wow,
Thank you very much for this amazing explanation.
Understood crystal clear 🙏
Such a satisfying and helpful video, thank you so much!
You're a hell of a teacher love you maan
Great video! Love the explanations, drawings, and blackboard. Cheers.
Amazing … thanks for the excellent explanation
Thank you Zack.
Best explanation for me.
Thanks! Just got me an extra mark on my 3 phase exam!
Excellent!
Thank you sir
that was very, very interesting :)
I'll just leave it there:
the amount of voltage required to overcome the impedance of the transformer to reach max secondary current (i.e., secondary short-curcuit current, which is determined by the design)
Awesome video!
clear explanation thank you.
Great vid thank you
Extraordinary
Great explanation
Glad it was helpful!
Great video ! Thanks
You are welcome!
where does the I_sc = I / %z formula come from?
The percent impedance is what is called a per unit value. The base values for the system are implied to be the rating of the transformer, so the base kva is 9kva and the base voltage would be 240v. This is just a way to model systems to make it much much easier for power system and primarily fault analysis. A transformer just becomes it's equivalent impedance in series. The voltage due to the selected base (pu = actual/base) is now just 1 pu. So in a short circuit you have a simple series circuit with a voltage of 1pu and a reactance of .0333. Using ohms law 1pu/.0333 gives you a per unit short circuit current of 30 pu. To convert from pu back to actual fault current, you would just need to multiply the 30 pu by the base current which is determined by the given power and voltage bases. This also happens to be your secondary current, so - 9kva/240v = 37.5 and 30*37.5 = 1125A.
Or to put it all together
(base voltage in per unit)/(Transformer Impedance in pu)*Base Current = (1pu/.0333pu)*37.5A
The voltage is not mentioned because it is selected to be 1pu, but it is in there. Its still just ohms law. Your not actually dividing by a impedance you're multiplying by a unitless current value.
Current = (per unit current * base current) = 37.5/.03333
The equation is just a quick and easy way of combining those steps to determine the maximum available short circuit current at the secondary of any transformer. You'll more likely use it in this form Isc=S/(Root(3)*Vline*%z) or S/%z for your available fault duty if you are looking to incorporate other equipment for a quick rough calculation.
I was explaining in the case where we know the impedance already, but testing for the impedance works the same. You adjust your voltage until you're at a known current to determine the impedance. So its the same simple series circuit with the same bases, but the tests adjust the voltage supply until the secondary is at rated current which is also its per unit base current so it equals 1pu, which eliminates an additional math step.
This occurs at 20v or Vpu = 20/600=.03333 pu Using Ohms law Zpu = Vpu/Ipu = .0333pu/1pu.
Loved it.. Great explanation
Could you please let me know the logic behind the formulae you've used ..
For e.g: why should the short circuit current be equal to the rated secondary current divided by the percentage impedance?
Many thanks .. subscribed!
So typically we say that short circuit has to do with available kVA and the voltage of the transformer, which using Rated current takes in to account both of those.
The other thing to consider would be impedance of the short circuit path, that's why this video we are just talking about what's available in theory at the transformer, not what you would see actually at the short circuit point.
If you change the percent impedance into actual impedance, your calculation for short circuit is just Vs/Z = Isc or a circuit with only your voltage source and your transformer impedance in series. All primary impedances are transferred over to the secondary side as one z equivalent.
To get to z from %z you do the following.
(1) Isc = Vs/Z
(2) Z= %Z*Zbase where
(3) Zbase = V^2/S
Plugging (3) into (2) and (2) into (1)
Isc = V/(%Z*V^2/S) = S*V/(%Z*V^2) = S/(%Z*V)
Where S= V*A so
Isc = V*A/(%Z*V) = A/%Z
It is just ohms law with a lot of convenient variables that cancel out.
The calcs with the actual transformer impedance for this example are shown below which results in the same answer.
Zbase = (240V)^2/9000VA = 6.4 ohms
Z = Zbase*Z% = 6.4ohm*.03333 = .21333 ohms
Isc = Vs/Z = 240 V/2.1333 ohms = 1125A
For the 3 phase case - Say three of these transformers were hooked up in a Delta-Y, Vllpri=Vphasepri = 600, Vphasesec = 240V and Vllsec=416V, then the equivalent transformer would be a 27kVA 3 phase 600-416/240 transformer with %Z of 3.333%. Your results would still be 1125A for the max available fault current per phase
Isc = 27000/(3*240)*30 = 1125A
@@jaymoseley6216 this is what i was looking for when i came here. THANK YOU SIR!!!
How do you consider %Z of a bank of three single phase transformer for
3 phase use ?
Did you figure out the answer to that? I have my PE exam tomorrow fml
im sorry but does this means that at 1136.4 amps the transformer will trip?
Superb.
The whole concept of using 'ratings' to calculate actual values is what is throwing me off. I'm normally not this dumb. 😆
i thought you need to divide by sqrt(3) to get secondary rated current
If it was a 3-phase transformer you would. This example is a single phase
Could you explain where you got 20 volts. Seems like you just picked a random voltage? Did I miss something?
He just assumed that as he kept on increasing the input voltage on the primary side, 20V was the number when the ammeter on the secondary side started reading 37.5A (37.5A being the rated secondary current). This could any number depending on what the actual impedance is. For e.g. if the impedance of that transformer was higher, the ammeter on the secondary side would have recorded 37.5A only when the applied voltage on the primary side was a higher number, let say 27V.
love it
Thanks
Thank you!
Hello you mentioned in this video you can calculate losses of the transformer with the % impedance. How do you do that?
I have a video on it. Closed circuit test
Are any factors, such as sqrt(3), needed if you have a 3 phase transformer?
I am trying to find out that too!
can you give an example of a three phase tx
For percent impedance?
A three-phase transformer is just 3 single phase transformers wired together in Wye or Delta
Why we put a short circuit on secondary side of transformer to calculate percentage impedance? Why don't we put a load on secondary side in order to calculate the impedance?
How do I figure the 20 volts? How can it just be a guess?
It is a measured value that you would read with a meter
It probably is a silly question but since we short circuited the secondary does the voltage remanis at 240V? Is it a constant voltage source the secondaryy????
Yes
Sir , how we can calculate drop voltage at primary with calculation. As mentioned u 20V how it s came? Without meter any possible for find that drop voltage
You would need a meter to measure it
The short circuit current of the secondary of 1136 A is in theory true. However, In reality in an example where the source is 7200 volts to ground primary on a pole mounted distribution transformer and the impedance is 3.3% (with a 240 volt secondary) , that current will ALWAYS be limited to less than the calculated value as in this video. Most distribution transformers at First Energy were 3%. The 7200 (7.2KV) distribution source side case, that source may not be able to supply the current to achieve the theoretical value on the low voltage secondary. In the power utility business to do this calculation for the secondary fault current you have to know the short circuit AVAILABLE fault current of the SOURCE. Once you know that you can solve the problem. You can use CYME (tm) or if you know the available high side fault current (the electric company can tell you) then u can use the FREE EATON fault current I Phone AP to get the answer. The AP is available in the Apple(tm) Store or www.eaton.com/us/en-us/products/electrical-circuit-protection/fuses/fault-current-calculator.html. Once again, you need to know the available fault current of the source to get the answer. Why do u need to know this? Suppose there's a 25 KVA pole transformer, 7.2 KV primary, 120/240 secondary,100 feet of 1/0 TPX to the weather-head and 100 feet of #2 CU to the main breaker in the load center. The electrical inspector needs to know what is the available fault current at the panel. If it's under 10 KA, then u can use off the shelf breakers from the home center. If its over 10 KA but less than 22 KA, then you need more expensive breakers that can handle the predicted fault current. The electrician needs to know this too. When the primary is 19.9 KV phase to neutral, the available fault currents are always high.
Hi Zack, is this short circuit current only seen on the secondary? Could I reflect it to the primary?
It is the available fault current on the Secondary of the transformer. It's the maximum it can supply
@@ZackHartle Hello Zack, Could I reflect the fault current on the secondary to the primary?
I have a transformer with %IZ and %IX, what is the difference?
I'm not sure
It's the percentage reactance of the transformer.... %X = (IX / V ) * 100
Is this how you do it for a 3phase transformer ?
You would do each of the 3 single-phase transformers separately.
are you writing backwards?
Ha ha, I am not, but I can't reveal all my secrets! It's just fancy camera tricks
When I was in electrical engineering school I met a guy that was an electrician and proud that he had learned all about electricity and it sure didn't take six or seven years to learn it. I said you know about a transformer it's made of copper wire and insulation, and iron. He said "I know what it's made out of". I said good now that you know what you need to use how much of each one do you put in?" He said "how the hell would you know that". I said"first you'd fall off a turnip truck in front of the University math department and stick around there for about 4 years convincing you was willing to pay the money and worthy of sitting in that seat. Then a guy who gets paid to tell people that would take a break and teach her class and you would have already proven you was worthy to try and understand what he was explaining because you understood what the math department was explaining and then a couple of years you could answer that question just like he can? Electrician scratched his head and left the bar.