Boosting 208V to 240V

Ok, so my friend got the B&B transformer. It is a 240/24. The problem is the "H" leads are not tagged. The "X" leads are. How would you determine which "H" is which? What would the voltage be with this transformer?
I came up with 236V.
 
Little Bill said:
Ok, so my friend got the B&B transformer. It is a 240/24. The problem is the "H" leads are not tagged. The "X" leads are. How would you determine which "H" is which? What would the voltage be with this transformer?
I came up with 236V.
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He called back and said the "H" leads were printed and not tagged, so we're good there. Still wondering about the voltage, given the transformer used here.
 
Little Bill said:
He called back and said the "H" leads were printed and not tagged, so we're good there. Still wondering about the voltage given the transformer used here.
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The ratio is 10:1, so in a boost configuration you'll get 208 + 208/10 = 229V. Should be easy to tell if the two X leads end up reversed, as then you'd get 208 - 208/10 = 187V.

Cheers, Wayne
 
wwhitney said:
The ratio is 10:1, so in a boost configuration you'll get 208 + 208/10 = 229V. Should be easy to tell if the two X leads end up reversed, as then you'd get 208 - 208/10 = 187V.

Cheers, Wayne
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Thanks, so that would mean one of the legs would be 109V to neutral, correct?
 
Little Bill said:
Thanks, so that would mean one of the legs would be 109V to neutral, correct?
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No. In cartesian coordinates, and for one choice of layout, your starting voltages are N = (0,0), L1 = (-104,-60), L2 = (+104, -60). That gives you the expected voltages: L1-N = sqrt(104^2+60^2) = 120V, likewise L2-N = 120V. And L2-L1 = (104,-60) - (-104,-60) = (208,0), or 208V.

Now you boost L2 relative to L1 to get L2' = L2 + (20.8,0) = (124.8,-60). So the L2'-N voltage is sqrt(124.8^2+60^2) = 138V.

A diagram would make this all clearer, but I don't have the means to quickly produce one. I could draw it out and photograph it if you want to better understand the above calculation.

Cheers, Wayne
 
wwhitney said:
No. In cartesian coordinates, and for one choice of layout, your starting voltages are N = (0,0), L1 = (-104,-60), L2 = (+104, -60). That gives you the expected voltages: L1-N = sqrt(104^2+60^2) = 120V, likewise L2-N = 120V. And L2-L1 = (104,-60) - (-104,-60) = (208,0), or 208V.

Now you boost L2 relative to L1 to get L2' = L2 + (20.8,0) = (124.8,-60). So the L2'-N voltage is sqrt(124.8^2+60^2) = 138V.

A diagram would make this all clearer, but I don't have the means to quickly produce one. I could draw it out and photograph it if you want to better understand the above calculation.

Cheers, Wayne
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Your previous post said 229V. So 120V on one leg and 109V on the other.
 
P.S. The above means that if you supply the pump with only L1, N, and L2', you must ensure that all the 120V loads are connected to L1 and N (or else ensure that they are all happy with 138V). Or if the 120V loads are spread across both legs, you could isolate the 240V load and supply it with L1-L2', while supplying the rest of the loads with L1-N-L2.
 
Little Bill said:
Your previous post said 229V. So 120V on one leg and 109V on the other.
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That logic only applies when L1, N, and L2' are all on the same straight line in voltage space, i.e. that N-L1 and N-L2' are 180 degrees out of phase. But L1, N, and L2' are not on a straight line, the phase angle between them is a bit over 120 degrees. That's why the answer is determined by the more complicated calculation in my previous post.

Cheers, Wayne
 
wwhitney said:
That logic only applies when L1, N, and L2' are all on the same straight line in voltage space, i.e. that N-L1 and N-L2' are 180 degrees out of phase. But L1, N, and L2' are not on a straight line, the phase angle between them is a bit over 120 degrees. That's why the answer is determined by the more complicated calculation in my previous post.

Cheers, Wayne
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Ok, he shows the following
L-L = 233
L1-N = 120
L2-N = 140
 
Little Bill said:
Ok, he shows the following
L-L = 233
L1-N = 120
L2-N = 140
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I think it's clearer to call the boosted leg L2' and keep L2 for the original leg.

So assuming you mean L2' instead of L2 in the above, the numbers are consistent with the L1-L2' and N-L2' calculations I made, they are both just a bit higher. Which, if the L2-N voltage is also 120V, suggests that the 240V:24V transformer puts out a no-load voltage on the 24V side a bit higher than spec. Which is perhaps intentional so that the voltage under load is closer to spec.

Cheers, Wayne
 
wwhitney said:
I think it's clearer to call the boosted leg L2' and keep L2 for the original leg.

So assuming you mean L2' instead of L2 in the above, the numbers are consistent with the L1-L2' and N-L2' calculations I made, they are both just a bit higher. Which, if the L2-N voltage is also 120V, suggests that the 240V:24V transformer puts out a no-load voltage on the 24V side a bit higher than spec. Which is perhaps intentional so that the voltage under load is closer to spec.

Cheers, Wayne
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I don't actually know which is L1 or L2, I just listed it that way to show there was a difference between the two. He found a diagram for the pump unit and is looking to see if he can see what requires the 120V and will put the common leg there.
 
LarryFine said:
Exception No. 1: An autotransformer shall be permitted without the connection to a grounded conductor where transforming from a nominal 208 volts to a nominal 240-volt supply or similarly from 240 volts to 208 volts.
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But your intention is to derive a multiwire output and not a simple two wire output. I don't necessarily see the exception applying to that situation.
 
fastline said:
I'm lost in the conversation but what in the world does 120* have to do with anything here? Load is single phase. Any single phase, whether in a 1P or 3P service is only a single sine wave. There is no phase offset. but if you split the phase, it would be 180*.

If someone actually had even a pencil to show how a load was configured, it might be solvable. Otherwise, this is like a "how much does the moon weigh?" sort of question. 120V is already present between neutral and any selected phase currently. So there are ways to work out the potential issue.
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There is 180 degrees between two points yes. Three points with one being centered between the other two is also 180 degrees between any two points. Two phase conductors and a neutral conductor from a wye supply is only 180 degrees between any two points , that 120 degree angle is what makes it 208 and not 240 between the two outer points unlike a true single phase three wire system is.

That said I maybe did not fully understand how this may been desired to be connected, if you only connect the autotransformer to two points of the supply and the output is limited to no more than one conductor of the supply then all the output is at a 180 degree angle. The thing is your "output neutral" in the example given is not at same potential of the supply neutral which at very least may make matters confusing when someone that doesn't understand the arrangement comes along to troubleshoot or modify something let alone whether it is code compliant or not.
 
The transformer is connected and the neutral is confirmed to only be for the alarm. Also confirmed L1 is the common and goes to the alarm.
 
LarryFine said:
I never claimed that the source would include the neutral; it would be only two lines, 208v 1ph.
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Sorry for thinking you did.

But, in that case there is no angle involved at all. You only have two points, there is no way then can be 180° apart from each other.
 
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