Pushing 240V through a 208V heating element

[Jraef]

Things made in China are somewhat dangerous when used according to their directions (if you can find directions). Using them another way is just asking for trouble.

Case in point:

Chinese Safety- Can't make this stuff up

Bought one of these $5-8 Chinese made hair dryers from Amazon just to take to bits. They have no UL or NRTL listing, heck they don't even list the voltage or wattage on the casing. First thing first: When I plugged it did not turn on. Instead it started to smell like burning rubber and...

forums.mikeholt.com

 

[Jraef]

So how can we use NEMA standard motors? The nameplates are required to be marked 230 volts for use on a nominal 240 volt system, and 460 for use on a nominal 480 volt system.

"Nominal"...

 

[don_resqcapt19]

"Nominal"...

And the rule requires that the nameplate voltage not be less than the nominal system voltage. 460 is less than 480. (this is just broken code)

 

[A/A Fuel GTX]

I have had absolutely no response from the Chinese manufacturer regarding this question. Surprise! I'm figuring that as long as the circuit ampacity and OCPD is adequate, the only consequence may be shortened element life.

 

[Jraef]

I have had absolutely no response from the Chinese manufacturer regarding this question. Surprise! I'm figuring that as long as the circuit ampacity and OCPD is adequate, the only consequence may will be shortened element life.

FIFY
...in which case the cheap Chinese oven was not the bargain they thought it was.

 

[hillbilly1]

You could always use a buck/boost transformer if it doesn’t require a neutral.

 

[drcampbell]

One possible problem with applying a voltage higher than rated is what's called the "watt density" of the heating elements themselves, basically how hot the SURFACE temperature of the element will get. ...

Not just the surface temperature, but the temperature of every material & component part of the heating element assembly -- the shell, insulator, nichrome wire, mounting base, electrical terminals, the crimps holding it together, ... everything.

 

[jaggedben]

I'm inclined to believe that with any kind of appropriate safety factors, a 15% increase in voltage should not set the dryer on fire. (Surface temperature increase should be pretty close to linear.) But I don't design dryers for a living.

 

[drcampbell]

The temperature rise will be roughly proportional to the power, not the voltage.

 

[kwired]

I'm inclined to believe that with any kind of appropriate safety factors, a 15% increase in voltage should not set the dryer on fire. (Surface temperature increase should be pretty close to linear.) But I don't design dryers for a living.

The temperature rise will be roughly proportional to the power, not the voltage.

and ~15% increase in voltage results in about 33% increase in power.

add: which was brought up in first couple replies of this thread.

 

[jaggedben]

The temperature rise will be roughly proportional to the power, not the voltage.

Why is there a square in the increase in power? Isn't this just Ohms law?

 

[wwhitney]

Why is there a square in the increase in power? Isn't this just Ohms law?

Yes, in the form P = V^2/R.

Wayne

 

[powerpete69]

Yes, in the form P = V^2/R.

Wayne

Remember my post a month ago how when voltage doubles, power to the same resistance quadruples!!!!
And when the voltage goes up 4 times, the power 16 "tiples" to the same resistance load.

P=(V^2)/R is the real deal and is quite SHOCKING!!

Remember this too. Take V=IR and P=IV and you can derive many a formula. For example, P=(V^2)/R is definitely derived from those two!
P=I^2*R is as well. See if you can algebraically derive them. When you do this for yourself, you tend to go "ahhhhh" and believe the formula a bit more!!!

For example, since V=IR....you can definitely take IR and replace it were V was in P=IV....you can do this little trick with the other letters too.....

 

[jaggedben]

Thanks for walking me through it. I should have known better, I'm just too used to thinking about current instead of power.

 

[sameguy]

Just tell them not to turn it to high, you know #11 on the dial.

 

[drcampbell]

Why is there a square in the increase in power? Isn't this just Ohms law?

Think of each electron as a wee little dump truck.
The voltage is the amount of energy each one is carrying.
The current is how many trucks arrive during each shift.

Suppose the resistance is constant and you triple the voltage. The current will triple in response.
Now you have triple the number of trucks arriving, and each one is carrying triple the load it was carrying before. 9x the total delivery.

 

[Artical430]

The only problem with appliance 240 volts on an appliance rated for 208 volts ...is the appliance may not last or work for a very long period of time...

 

[Another C10]

I'm probably way off here but the required 208v for this oven may very well be able to handle another 32v, good chance too the wiring is rated for 300-600v, also Increasing the voltage decreases the amperage ....right, or did that change in some code along the way. Although ! .. If the spec of the oven says only rated at 208v then applying another 2 phase potential could void a warranty or liability.
208v =48.07A 240=41.66A hope I did that right.

 

[Another C10]

Think of each electron as a wee little dump truck.
The voltage is the amount of energy each one is carrying.
The current is how many trucks arrive during each shift.

Suppose the resistance is constant and you triple the voltage. The current will triple in response.
Now you have triple the number of trucks arriving, and each one is carrying triple the load it was carrying before. 9x the total delivery.

I don't know if that's right but I like the analogy, how about the little dumps trucks stay the same number but can handle a larger load because of their beefier engines. Sorry couldn't resist.

 

[LarryFine]

I'm probably way off here but the required 208v for this oven may very well be able to handle another 32v, good chance too the wiring is rated for 300-600v, also Increasing the voltage decreases the amperage ....right, or did that change in some code along the way.

That is very incorrect. Except for motors, which attempt to be constant-power loads, the only time increasing voltage means less current is when the equipment is designed and built to consume a given power with a different applied voltage.

In other words, for a constant impedance load, current increases proportionately with increased voltage, not inversely.