208v to 240v Buck Booster

[kwired]

Assuming the dryer is currently using 208 volts, and your boost transformer will boost this to 240 volts, then if your dryer is rated for 240 volts then the amount of current such as 30 ampere before will be proportionally less. In which case as explained by someone else there may be less voltage drop to the dryer and a better running dryer

Current on the 208 supply portion of the circuit will go up particularly on a resistance load and in reality would cause some additional voltage drop on the 208 supply circuit. If you have a short branch circuit length it might be a rather negligible amount though.

Better running dryer mostly only means more heat because of how Ohm's law would apply to the heating element and the higher input voltage to said element.

 

[ptonsparky]

Current on the 208 supply portion of the circuit will go up particularly on a resistance load and in reality would cause some additional voltage drop on the 208 supply circuit. If you have a short branch circuit length it might be a rather negligible amount though.

Better running dryer mostly only means more heat because of how Ohm's law would apply to the heating element and the higher input voltage to said element.

Shouldn't that be inductive?
The resistive load will go down on lower voltage.

 

[Hv&Lv]

120/208 incoming voltage
The dryer in question is 208/240 capable and has a 30amp rating.
I planned on installing a step up transformer, primary being 208v and secondary being 240v which should make the dryer heating element function better, right?
My question is about the amperage range, is 30amp the max the dryer would pull or would I be ok going 25 or 26 amp booster?
I had also read the conversion would change amperage or throw it off, anyone familiar with that, also, the booster will be very close to source and load so I shouldn’t have much loss in voltage due to distance. Thanks in advance for assistance!

That’s a lot of money wasted and aggravation for an extra 10-15 minutes of drying times.
There are thousands of apartment communities that run 240 dryers and stoves on 208 volts. It’s extremely common.

 

[kwired]

Shouldn't that be inductive?
The resistive load will go down on lower voltage.

Boosting a 208 supply volts to 240 will increase the volts across the resistance, watts will increase, input current will increase, resistance will still be same.

An inductive motor is going to try to maintain same RPM as long as the source can deliver what is needed to do so. So if supply volts is lowered then current will go up trying to maintain same work output.

Topic was involving a clothes dryer and the motor could be effected if it is 120 volts but ends up on the boosted portion of the circuit, but proper method to boost 208 to 240 with such a dryer would be to make sure the 120 volt portions of the load remain on the non boosted portion of the supply circuit and only connect the heating element to the boost transformer. That will increase current but also increase voltage drop in the supply conductors.

 

[ptonsparky]

Boosting a 208 supply volts to 240 will increase the volts across the resistance, watts will increase, input current will increase, resistance will still be same.

An inductive motor is going to try to maintain same RPM as long as the source can deliver what is needed to do so. So if supply volts is lowered then current will go up trying to maintain same work output.

Topic was involving a clothes dryer and the motor could be effected if it is 120 volts but ends up on the boosted portion of the circuit, but proper method to boost 208 to 240 with such a dryer would be to make sure the 120 volt portions of the load remain on the non boosted portion of the supply circuit and only connect the heating element to the boost transformer. That will increase current but also increase voltage drop in the supply conductors.

I need to keep up. I missed a post or two and read yours out of context. O bother.

 

[Jpflex]

Shouldn't that be inductive?
The resistive load will go down on lower voltage.

Current on the 208 supply portion of the circuit will go up particularly on a resistance load and in reality would cause some additional voltage drop on the 208 supply circuit. If you have a short branch circuit length it might be a rather negligible amount though.

Better running dryer mostly only means more heat because of how Ohm's law would apply to the heating element and the higher input voltage to said element.

So if most dryers use a resistive heating element then the power should be closer to 1 or unity. Are we then agreeing that the load current will drop with an increased voltage (minus negligible voltage drop such as wire resistance)

Also on a wiring diagram a resistive load is indicated by linear zigzag “teeth” while an inductive load sometimes represented as circular “spring”

Is all voltage dropped across the final inductive load for per ac cycle as all voltage is dropped across the final resistive load in a dc circuit?

What is the main construction difference in the wire of a resistive load and inductive load? they both seem to be wire bending or turns within itself?

 

[wwhitney]

So if most dryers use a resistive heating element then the power should be closer to 1 or unity. Are we then agreeing that the load current will drop with an increased voltage (minus negligible voltage drop such as wire resistance)

For a dumb heating resistive heating element, the current will increase with increasing voltage, as V = I*R. The instantaneous power will go up with the square of the voltage, as P = I * V = V²/R.

For a thermostatically controlled resistive heating element working against a fixed rate of heat loss, the total energy demanded over a given time period will be (basically) the same, regardless of applied voltage, as long as the voltage is high enough that the instantaneous power the heating element gives exceeds the average power demand. With a higher voltage, any demand for heat will simply be satisfied more quickly, so the thermostat will spend less of its time in the "on" position. The average square current I² will be independent of applied voltage, but the average current will be less with a higher voltage, even though the peak current rises.

Cheers, Wayne

 

[LarryFine]

If one is going to boost the voltage, it should be done at the source end of the source, not the load end.

 

[Besoeker3]

Is there really need to have 208V and/or 240V ? Surely just fix one of them - say 230V for Eueopean voltage.....

 

[ptonsparky]

Is there really need to have 208V and/or 240V ? Surely just fix one of them - say 230V for Eueopean voltage.....

I think it would be easier if we just moved to Europe.

 

[Besoeker3]

I think it would be easier if we just moved to Europe.

Now there's an idea !

 

[LarryFine]

I think it would be easier if we just moved to Europe.

But if we all move, we'll tilt the Earth.

 

[retirede]

Is there really need to have 208V and/or 240V ? Surely just fix one of them - say 230V for Eueopean voltage.....

The issue is being able to have 120V single phase naturally from either a single or 3 phase system.

 

[roger]

Since the OP does not meet the forum requirement of being a trade person and the discussion has covered most of the important points we will closed the thread.