How to properly size over current protection

[john2mcc]

if i have a 240vac 240w heater :ashamed1:and i run it at 240v my ampacity would be 1a which means i would need to upsize this by 20% for overcurrent protection. if i run this same heater at 120vac then i would multiply 240w X .5 (Half the efficiency) then divide the wattage by 120v. is this correct?

240*.5
120vac *1.2=overcurrent protection

 

[infinity]

If you half the voltage to the heater you will half the current as well since for a resistive load they're directly proportional. To prove this you would use Ohm's law with the resistance and the voltage to find the current.

 

[templdl]

if i have a 240vac 240w heater :ashamed1:and i run it at 240v my ampacity would be 1a which means i would need to upsize this by 20% for overcurrent protection. if i run this same heater at 120vac then i would multiply 240w X .5 (Half the efficiency) then divide the wattage by 120v. is this correct?

240*.5
120vac *1.2=overcurrent protection

What over current protective device does the heater manufacturer recommend? I guess probably none.
You may be making a mountain out of a mole hill.
This is my thought:
If you know what your load is take that load and if it is considered to be a continuous load multiply that times 125% and select the wire size base upon the 75deg column on table 310-16 of the NEC. The select the breaker base upon the wire rating and not the load. Breakers for the most part protect the wire.

 

[ToolHound]

If you half the voltage to the heater you will half the current as well since for a resistive load they're directly proportional. To prove this you would use Ohm's law with the resistance and the voltage to find the current.

Example problem with math can be seen here.

 

[ToolHound]

If you half the voltage to the heater you will half the current as well since for a resistive load they're directly proportional. To prove this you would use Ohm's law with the resistance and the voltage to find the current.

Now you have caused me to think. Your posts reminds me of the voltage being proportional to current in resistive loads. In light of that reminder then I will lose sleep till I re-learn/refresh as to why (I think) that smaller size transmission lines are required (presumably less current flows) by virtue of the higher voltage on the long distance utility power transmission lines. Is that an opposite situation (of the OP heater situation) where current and voltage are inversely proportional (or at least of a way different relationship than in a purely resistive situation), i.e., raising the transmission line voltage delivers the same power at the destination--but with less current flow (per watt delivered) than otherwise would be the case on the transmission lines. ???? In any case, I am wondering...everything else being the same, or constant, what generally would be the change in current magnitude on transmission lines if voltage is increased ? If transformers at the destination present a largely inductive load, on long distance transmission lines, what can be said about the relationship of voltage to current on such lines ?
Current is less proportional (but still nevertheless proportional?) to voltage than is the case in resistive loads?

But this is off topic. I gotta find a book on this.

 

[BAHTAH]

Over current Protectdion

Over current Protectdion

Use 100% of the non-continuous load plus 125% of the continuous load to determine your total load. A branch circuit for your heater would be #14cu with a 15amp 2-pole
breaker. The minimum size standard over current device is 15amps. There are smaller breakers but they are supplemental in their application. #14 is the smallest size building
wiring conductor. Some areas specify #12 as the smallest. Using Ohm's Law 240w/240v=1 amp. Heater is a long continuous load so 1 x 1.25= 1.25 amperes.

 

[GoldDigger]

Is that an opposite situation (of the OP heater situation) where current and voltage are inversely proportional (or at least of a way different relationship than in a purely resistive situation), i.e., raising the transmission line voltage delivers the same power at the destination--but with less current flow (per watt delivered) than otherwise would be the case on the transmission lines. ???? In any case, I am wondering...everything else being the same, or constant, what generally would be the change in current magnitude on transmission lines if voltage is increased ? If transformers at the destination present a largely inductive load, on long distance transmission lines, what can be said about the relationship of voltage to current on such lines ?
Current is less proportional (but still nevertheless proportional?) to voltage than is the case in resistive loads?

For a given size of transmission line, the resistance of the LINE is the same regardless of the voltage it carries. But the impedance of the LOAD at the end will have to change to continue to draw the same power.
So if you double the voltage at the load, you will need to cut the current in half. Half the current through the constant resistance of the line means half the voltage drop.
But wait..! Since you have doubled the transmission voltage, the half size voltage drop is now 1/4 of the original when expressed as a percentage of the line voltage. You will find books that talk about it more, but that is the basics in only a few lines.
The reactive current (into inductive or capacitive loads) is just a red herring in this calculation. In the ideal situation the power factor of the load the transmission line serves will be 1. If it is not, the difference in voltage drop will not change the above argument in any way.

 

[Dennis Alwon]

To follow Golddiggers point

Given a 4800 watt heater at 240V then we can calculate the resistance-

V^2 / P = R Thus 240*240/4800 = 12 ohms

P = V*V/ R
P= 120*120/12 = 1200 watts

so yes the wattage is 1/4 the 4800 watt element

 

[ToolHound]

For a given size of transmission line, the resistance of the LINE is ...

GoldDigger. Thanks. Transmission lines. Voltage up, less current, in transmission lines. ok. Inductive loads and resistive loads...different animals that act differently. --ToolHound

 

[kwired]

GoldDigger. Thanks. Transmission lines. Voltage up, less current, in transmission lines. ok. Inductive loads and resistive loads...different animals that act differently. --ToolHound

Yes but keep in mind what Dennis said. 1/2 voltage x half amps = 1/4 VA, but this only applies when impedance of the load remains the same at either voltage and current.

A transmission line is transformed to a specific voltage and loads are intended to be supplied at their rated voltage, they don't just change the voltage and expect whatever load is there to take that change in voltage.

IOW they don't just decide to run 345KV lines from a power plant to, wherever, they run them to specific substations with specific transformers designed for 345KV input and a specific output voltage. They then run to additional substations at that new voltage to transformers with specific input and output voltages for the design of the application. Should they decide they want to change that second transmission voltage, they also have to change transformers or at least taps on the transformers to make the equipment match the new conditions.