5kW water heater - 30.04A - can I rounding down to a 30A breaker?

[kwired]

Hm, my understanding of voltage/power/current is that a 5 kw heater is going to draw the amps required for 5 kw regardless of the voltage. That is, it would draw 20.83A if wired at 240V and 24.04A if wired at 208.

If you have a 240V rated water heater and power it with a 208V system, it's going to run at 208V. I don't see how getting a 240V heater is going to help anything in this situation.

If the load is inductive like a motor the power will remain fairly constant, if voltage goes up, current goes down, if voltage goes down current goes up. There will be limitations but to go from 208 to 240 will result in similar net power.

For resistive loads (water heater most likely is unless it is a heat pump) the resistance remains constant no matter what voltage is applied. If it is rated @ 208 volts and you apply 240 instead the resistance still remains the same and an increase in voltage will result in an increase in current.

 

[JoeStillman]

Hm, my understanding of voltage/power/current is that a 5 kw heater is going to draw the amps required for 5 kw regardless of the voltage. That is, it would draw 20.83A if wired at 240V and 24.04A if wired at 208.

If you have a 240V rated water heater and power it with a 208V system, it's going to run at 208V. I don't see how getting a 240V heater is going to help anything in this situation.

The rated kW of a heater depends on the voltage. The 5kW, 240V heater only puts out 5 kW at 240V. If you apply a different voltage, you change the kW. Some heaters list their rating as 208/240V. When they do, two currents will be listed the same way, such as 18.1/20.8A for the 5 kW heater. You could look it up in the Qmark catalog.

Electric heaters are purely resistive. Changing the voltage does not change the resistance. For pure resistance, V=IR and P=IV. Solve the first equation for I and substitute into the second; I=V/R & P=(V/R)xV or P=V²/R. You can use the rated kW and voltage to figure out the resistance; R=V²/P or 240²/5000 = 11.52 ohms. Put 208 Volts onto 11.52 ohms and you get I=V/R=208/11.52=18.06 amps. At 240, I=V/R=240/11.52=20.83 amps. The power output of the heater at 208V is only 208x18.06=3,756 watts.

Just remember that, for pure resistance loads, power is proportional to the square of the voltage. P₁/P₂=V₁²/V₂².

To solve for the current at a new voltage, the equation is I₂=P₁V₂/V₁². If you plug in the given values, I₂₀₈=(5000x208)/240².

Remember, this ONLY applies to purely resistive loads. You can't use this sleight-of-hand with anything but heaters. Motors and non-linear loads are partly reactive, not purely resistive and don't behave the same way.

 

[apelk]

Thanks for the answers, especially with the formulas. I will have to write this one down

 

[kwired]

Thanks for the answers, especially with the formulas. I will have to write this one down

All you need is Ohm's Law and other basic electricity formulas. Any other equations written here are nothing more than re-written equations to find the missing variable.

 

[JoeStillman]

All you need is Ohm's Law and other basic electricity formulas. Any other equations written here are nothing more than re-written equations to find the missing variable.

He's right. The only equations above that I know by heart are V=IR and P=IV. The rest were worked out on a scratch pad for the post.

 

[T.M.Haja Sahib]

I would say you need a 35 amp breaker and #8 wire.

I agree, because resistance at 'cold' start will take a much higher current than its rated current before it reaches its operating temperature and so the 30A breaker is liable to trip on start up.

 

[Dennis Alwon]

Here is a pie chart I got off the internet. I printed it and have it on the wall above my desk. The internet version was incorrect so I fixed it. It had V= √I x R, It also has V- I x R we know they can't be correct. Anyway here is the corrected version

 

[david luchini]

I agree, because resistance at 'cold' start will take a much higher current than its rated current before it reaches its operating temperature and so the 30A breaker is liable to trip on start up.

The resistance change from cold start to operating temperature would be negligible. But even if it was a large as say, 10% lower at startup, the heater would draw 26.7A at startup. This wouldn't trip a 30A breaker. If it was as large as 20% lower, the heater would draw 30A at startup. This wouldn't trip a 30A breaker.

 

[Open Neutral]

I agree, because resistance at 'cold' start will take a much higher current than its rated current before it reaches its operating temperature and so the 30A breaker is liable to trip on start up.

CalRod type heaters are very stable, R-wise. Don't expect much starting surge.
(Not so incandescent lamps; they and their cousins have seen service as current regulators.)