Commercial kitchen stove loads -

[sundowner]

So the job requires moving several stoves that are 208/240 dual rated - 3800W/5100W. I need to move these from a 240 volt system to a 208 system. So my question is - how are these guys on a 20 A breaker now with no problems - my calcs show 5100Watts equals 21.25 Amps, but no breaker trips! So now I do the math at the new 208 Volt level and my calcs say even more Amps.

I realise I should really know how this works - but I'm really confused at this point...

Any help or further explanation would be great guys!

Thanks

Steve

 

[infinity]

Your calculation for the 208 volts is incorrect. When you lower the voltage you proportionally lower the current. Use the wattage (3800) and volts (208) to find the current.

 

[wwhitney]

To expand, the stove behaves (close to) a fixed resistance R. So the power use is given by P = V²/R.

3800W is the rating at 208V, and 5100W is the rating at 240V. The power usage goes down with lower voltage, as you can see from the numbers and from the formula.

(208/240)² = 0.75, so a fixed resistance at 208V dissipates only 75% of the power it would dissipate at 240V. And indeed, 5100W * 75% = 3800W.

Cheers, Wayne

 

[sundowner]

Its making more sence guys!

Thanks!

 

[LarryFine]

Steve, you need to remember which are the constants and which are the variables when doing the math.

To maintain a given power level when changing the applied voltage, you must alter the load impedance.

When altering the applied voltage to a given impedance, the resultant current varies proportionately.

Making a new load for a different voltage vs. using the same load on a different voltage is very different.