gdub0000
Member
- Location
- Concord, Ca., USA
Am trying to arrive at 1ph amps needed. 10.1 kw @ 208 3ph, what amperage is needed at 240 1ph
converting 3ph to single ph
- Thread starter gdub0000
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- Chapel Hill, NC
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- Retired Electrical Contractor
Well, a 10.1 kW load is a 10.1 kW load.
How you end up with it still being 10 kW on single phase 240 V might need a bit of explaining.
Mr Alwon is right. There are needs to be clarification.
junkhound
Senior Member
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- Renton, WA
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- EE, power electronics specialty
Sounds like a beginning circuits class problem and you don't want to read the whole chapter to answer and the homework assignment is due tomorrow morning?*
Take phase current, multiply by 3, divide by 2. You do know how to get the phase current? You need to use sqrt 3 for that, eh?
If still uncertain, read the chapter on phase vectors.
Hmm, better check the above, maybe is not correct?
Were you told what the power factor or negative sequence part of the 3 phase is, that will change the answer, possibly drastically.
*heh, heh, been there, done that, only no internet back when we only had slide rules.
gdub0000
Member
- Location
- Concord, Ca., USA
This is a oven
gar
Senior Member
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- Ann Arbor, Michigan
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- EE
140331-0844 EDT
Very simple.
I = 10100/240 = 42.08 A.
If the load was non-resistive, then you would need to know much more about the 3 phase and single phase loads.
.
Very simple.
I = 10100/240 = 42.08 A.
If the load was non-resistive, then you would need to know much more about the 3 phase and single phase loads.
.
kwired
Electron manager
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- NE Nebraska
Good chance you have three elements in a delta configuration, if you only apply power to two input leads you are only powering two of them.
It will need some reconnecting done to get full 10.1 kW out of it.
Plus since you are applying 240 instead of 208 volts you will get higher kW out of it unless the connections are somehow multitap designed to make that kind of selection.
It will need some reconnecting done to get full 10.1 kW out of it.
Plus since you are applying 240 instead of 208 volts you will get higher kW out of it unless the connections are somehow multitap designed to make that kind of selection.
- Location
- Lockport, IL
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- Retired Electrical Engineer
The question does not make sense. If you have a single phase load, there is no way to connect it to a three phase power supply, and use all of the available wires. So what exactly are you trying to do with this oven? What do you mean by giving a power value at 208V 3 phase, if the load is single phase?
Let me pose a different question. This might actually be the question you are trying to pose, but you have worded it in a confusing manner. Is this what you meant to say:
- I have a three phase, 208 volt power supply. I have a single phase oven that is rated at 240 volts, and at 10.1 kW. What current will this oven draw, if I were to connect it to two legs of the power supply?
The answer to that question is found by taking the 10.1kw, multiplying it by (208/240) squared, to get a new power rating of 7.59 kW. Then you divide that number by 208 volts, to get a current of 36.5 amps.
- Location
- Lockport, IL
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- Retired Electrical Engineer
See, that is part of the confusion here. I interpreted it the other way, that the equipment was rated at 240 and the available power source is 208. That is why we need clarification of the question.
kwired
Electron manager
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- NE Nebraska
I am on board with that, for all we know we could both be wrong with what the application is.
gar
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- Ann Arbor, Michigan
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- EE
140331-1053 EDT
The question as written seems to imply that an existing oven rated at 10.1 kW and 3 phase is to be replaced with a single phase 240 V oven also rated at 10.1 kW.
Realistically you can not take an existing 3 phase Y or delta heater and connect it single phase and get the same power output unless each resistance element's leads are individually available. Then voltage rating of an individual element may be important, and require a transformer to get the correct voltage.
If the goal is to take an existing Y connected resistive load with 4 wires, and a 120-0-120 supply is available, then the Y center wire could connect to the 120-0-120 neutral, and two of the other legs connected together and to one 120 V leg, and the remaining heater leg connected to the other 120 leg. The result is an unbalanced load of the same output power as the original 3 phase oven.
.
The question as written seems to imply that an existing oven rated at 10.1 kW and 3 phase is to be replaced with a single phase 240 V oven also rated at 10.1 kW.
Realistically you can not take an existing 3 phase Y or delta heater and connect it single phase and get the same power output unless each resistance element's leads are individually available. Then voltage rating of an individual element may be important, and require a transformer to get the correct voltage.
If the goal is to take an existing Y connected resistive load with 4 wires, and a 120-0-120 supply is available, then the Y center wire could connect to the 120-0-120 neutral, and two of the other legs connected together and to one 120 V leg, and the remaining heater leg connected to the other 120 leg. The result is an unbalanced load of the same output power as the original 3 phase oven.
.
kwired
Electron manager
- Location
- NE Nebraska
And I envisioned a delta connected load and easy conversion to single phase by paralleling all three elements - but most equipment I have seen designed for that is rated 240 volts but can operate at reduced output at 208 volts - again more details from the OP would help a lot.
gdub0000
Member
- Location
- Concord, Ca., USA
My apology for not being specific.
This is a 3ph 208 oven that draws 10.1 KW according to its name plate. The power available is 1ph 240 volts. This will have to
have a buck and boost to bring the single phase to 208, and a phase inverter. What amperage will be needed
Thanks
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kwired
Electron manager
- Location
- NE Nebraska
Can it be reconfigured to run on single phase? Some but not all of this type of equipment has that capability. That would be much simpler and less costly than adding phase conversion equipment, you possibly will still want to buck the voltage. Depending on what you have it may even cost less to find an oven with the right voltage/phases then to use phase conversion and buck transformers.
- Location
- Placerville, CA, USA
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- Retired PV System Designer
Many heating appliances nameplated for three phase will have alternative internal connections for straight 240 or 120/240 single phase 3 wire supply.
If yours does not, then you may have use a phase converter (not inverter) to get what it needs.
One such converter is essentially a stripped down VFD and using that you might not need to transform 240 to 208 in a separate device.
But it may well be cheaper to just use an oven configured for 240 in the first place.
Among other things, if the equipment was designed for 208Y/120 four wire, the controls may not withstand 240V to ground.
BTW that would be either a buck or a boost, depending on the details, rather than a buck and boost.
Tapatalk!
If yours does not, then you may have use a phase converter (not inverter) to get what it needs.
One such converter is essentially a stripped down VFD and using that you might not need to transform 240 to 208 in a separate device.
But it may well be cheaper to just use an oven configured for 240 in the first place.
Among other things, if the equipment was designed for 208Y/120 four wire, the controls may not withstand 240V to ground.
BTW that would be either a buck or a boost, depending on the details, rather than a buck and boost.
Tapatalk!
gar
Senior Member
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- Ann Arbor, Michigan
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- EE
140331-2117 EDT
gdub0000:
What is the internal wiring of this oven?
If it is a Y configuration, then you have three 120 V heating elements.
.
gdub0000:
What is the internal wiring of this oven?
If it is a Y configuration, then you have three 120 V heating elements.
.
JoeStillman
Senior Member
- Location
- West Chester, PA
If the 10.1kW 3? heater is delta connected, and you connect 240V single phase to two of the terminals, the power is 6.7kW.
10.1 kW 3? = 3 @ 33.67kW/?. R/? = 208?/3,367 = 12.8ohms
The total 1? resistance is one phase leg in parallel with two phase legs in series.
R1? = 1/((1/12.8)+(1/25.6)) = 8.55ohms
P1? = 240?/8.55 = 6,738 W or 6.73 kW
If you go with a phase converter, you probably will not need a buck-boost transformer. You may be able to do the phase change with a VFD. It would need to be rated for 42 FLA (10.1 kW @ 240V, 1?).
10.1 kW 3? = 3 @ 33.67kW/?. R/? = 208?/3,367 = 12.8ohms
The total 1? resistance is one phase leg in parallel with two phase legs in series.
R1? = 1/((1/12.8)+(1/25.6)) = 8.55ohms
P1? = 240?/8.55 = 6,738 W or 6.73 kW
If you go with a phase converter, you probably will not need a buck-boost transformer. You may be able to do the phase change with a VFD. It would need to be rated for 42 FLA (10.1 kW @ 240V, 1?).
Last edited:
gdub0000
Member
- Location
- Concord, Ca., USA
gdub0000
gdub0000
to 208 single ph and phase converted. If the convertor pulls 7 amps how do i determine single ph ampere draw.
gdub0000
The equipment is and oven rated at 10.1 kw 208 three phase. The supply is 240 single ph. This will probably have to be transformed
to 208 single ph and phase converted. If the convertor pulls 7 amps how do i determine single ph ampere draw.
gar
Senior Member
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- Ann Arbor, Michigan
- Occupation
- EE
140401-2343 EDT
gdub0000:
You are possibly asking the wrong question as a starting point.
If you want to know how much current at 240 V is required to supply a resistive load of 10.1 kW, then the answer is 10100/240 = 42.08 A. But I do not think that is the question you want to ask.
Joe Stillman in post 17 gave you one answer on the assumption that the oven (load) is wired delta. If it is delta and you applied 240 to two terminals, then you would overheat one winding, and that might be a problem. There is no solution to the delta configuration that allows application of a single single phase voltage to two terminals and produce a total load of 10.1 kW without grossly overheating one element, and the internal heat distribution would not be uniform.
I gave you a partial answer if the load was wired as a Y and had 4 wires.
You need to provide specific details on the internal wiring of the oven, or buy a new oven designed for single phase 240 V.
If the load is a 4 wire Y with a neutral, then you could supply 120 V to the parallel combination of the three internal heaters. Then the current would be 84.16 A.
.
gdub0000:
You are possibly asking the wrong question as a starting point.
If you want to know how much current at 240 V is required to supply a resistive load of 10.1 kW, then the answer is 10100/240 = 42.08 A. But I do not think that is the question you want to ask.
Joe Stillman in post 17 gave you one answer on the assumption that the oven (load) is wired delta. If it is delta and you applied 240 to two terminals, then you would overheat one winding, and that might be a problem. There is no solution to the delta configuration that allows application of a single single phase voltage to two terminals and produce a total load of 10.1 kW without grossly overheating one element, and the internal heat distribution would not be uniform.
I gave you a partial answer if the load was wired as a Y and had 4 wires.
You need to provide specific details on the internal wiring of the oven, or buy a new oven designed for single phase 240 V.
If the load is a 4 wire Y with a neutral, then you could supply 120 V to the parallel combination of the three internal heaters. Then the current would be 84.16 A.
.
kwired
Electron manager
- Location
- NE Nebraska
The simple answer is V x A = W or rewritten to find A is A = W/V
So 10100W / 240V = 42.08333333...
This is assuming that value is supplying transformation and phase conversion so the elements still see their rated voltage and phases, though there will be some losses in transformation and conversion that will require extra input.
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