Amps of 240 volt heat elements at 208

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Does someone have the formula to determine the amps of 240 volt heat elements connected to a 208 volt system?

240 volt 10 kw HS = 42 amps

What are the amps at 208 volts?


Thanks!!!
 
Sparky5150 said:
Does someone have the formula to determine the amps of 240 volt heat elements connected to a 208 volt system?
Ohms Law for constant R
V1 = I1 X R which can be changed to V1/I1 = R
V2 = I2 X R and V2/I2 = R

This means V1/I1 = V2/I2

Further manipulation gives

I2 = I1 x V2/V1
 
You need to determine the resistance of the heater.
E/I=R
240/42 = 5.7 ohms

Now we can figure amperage at 208 volts since E/R=I.

208/5.7 = 36.5 amps.

Now instead of 10KW the heater will provide
208 X 36.5 amps = 7592 watts (7.6 KW).

Hope that helps.
 
I would check with the heater mfr- most heating elements (actually most metals) have positive CTR so they are non-linear. The current may actually be higher than the simple analysis suggests.
 
What about P = V^2 / R
P = 240V^2 / 5.76 ohms = 10,000W
P = 208V^2 / 5.76 ohms = 7,511W

Then
I = 7,511W / 208V = 36.1A

This would mean the heater only has about 75% of the heating capacity at 208V.
 
When the resistance is fixed such as is the case here with heating elements, you don't have to calculate its value. All you need are the 2 different voltages involved, and the known number of rated Watts.

We can determine the Amps at 240 Volts by taking 10,000/240 = 41.67

Next, set up a proportion equation thus:

240 Volts = 41.67 Amps
208 Volts = (x) Amps

Solve for (x)

208 times 41.67 = 240(x)

8667.36/240 = (x)

36.11 = (x)


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A loose rule of thumb is the watts will be 75% value at 208 Volts, VS 240 Volts, as shown in other calculations in this thread.

The reduction of the Amps will be proportional to the change in Volts, in this case 86.7%
 
Another rule of thumb, if you double the voltage you will get 4X the output in watts.
 
Another rule of thumb, if you double the voltage you will get 4X the output in watts.
Yes, the watts change with the square of the change in voltage. 208 is 86.6% of 240 and 86.6%^2 is 75.1%.
Don
 
I dont want to seem difficult but; None of the formuli shown here to date have taken account of the reduced element temperature.
The original post didn't say what type of element, a water heater element may not have a very great temperature difference being held somewhat constant by (near) water contact but there will be a larger temperature difference if it is a duct heater.

The result of the lower voltage will be a decreased temperature and a decreased resistance ; thus increased wattage, depending on the element type used.
I would feel safer in saying the new calculated amp draw is a MINIMUM.
 
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