I have a question on resistive loads (first post)

[LouisianaApprentice]

I'm new to the forum sorry if this question is in the wrong section. I tried searching but couldn't find an answer for my question. My question is with resistive loads when you add voltage how can you calculate amperage drop if there even is any along with heat output. What made me think of this is when you have a 480v heating element (my employer recently changed out 2 cooling towers) the load should still be the same right whether 277 or 480 the output would just be different? Or is it just that these types of elements are really just 3 with a phase tapped into each one which I cannot see as there was no required neutral so I have no idea how voltage would return.

So let's say it's a 1500w resistive load would this load be divided over the three legs (which I couldn't see) or would u divide by the voltage to find wire sizing. Also would it be negligible the amount of resistance decreased from 277 to 480 to reduce amperage. Sorry Im kinda just trying to wrap my head around resistance loads.

 

[augie47]

Not sure I fully understand your question.
On your heating elements, the resistance is a constant. If you change the voltage supplying the heater, the current would also change as would the power.
The neutral is not necessary on three phase loads as the load will be distributed between the phases.
Calculating the current per phase on three phase loads is a bit complicated as you can see in this thread:
https://forums.mikeholt.com/showthread.php?t=195632&page=6

 

[LouisianaApprentice]

I did some reading

I did some reading

Ok I researched a little deeper and someone correct me if my new understanding is incorrect. So from the material I have read resistive loads use all of the current supplied so there is no need for a return so then you can calculate wire sizing by leg for the load of the element? There were three taps for straight 480 so if I'm right you would have to calculate the load (watts) divide by 3 then find 270v amp ratings for a single conductor using 1/3 the load?

 

[LouisianaApprentice]

Ok Augie sorry I think I overcomplicated things in my mind. Could you provide me with a reference. This is just for informational purposes. But could you show how to calculate wire size with a 480v element with 3 taps. Such as the screw in ones on cooling towers using the information on the element.

 

[augie47]

I hope I get this right.. if not, one of the sharper tools on the Forum will give you the right answer.

If your heater was 1500 total watts and designed for the 3 phase supply, it would be made so there would be (3) 500 watt loads, one on phases A-B, one on B-c, and one one A-C.
Using ohms law I= P/E I = 1500/480 X 1.732 = 1.8 amps per phase.

 

[LouisianaApprentice]

Sorry just saw reply

Sorry just saw reply

Thank you for the formula now I'm more confused as to what situation applies to which lol after watching a video on a guy calculating a balanced load on a Delta config. So I guess you would have to know 2 things to figure out each individual wire ampacity... The resistance between each leg and input voltage for each leg. Is the formula u gave a universal one?

 

[jaggedben]

It seems like you are conflating several different questions or issues.

Start with Ohms law. Voltage = current * resistance

Mathematically then current = voltage/resistance. Which means that if you increase voltage and resistance doesn't change then current will increase. This is why things 'fry' when you apply too much voltage to them.

 

[jaggedben]

... Is the formula u gave a universal one?

(Typing at the same time as augie...)

Yes, Ohm's Law is fundamental physics of electricity.

Precise calculations get more complicated with alternating current and impedance. But understanding Ohm's Law will give you a solid basis of understanding of 'why stuff happens', if not how to do engineering calculations.

Another thing is that the resistance of a heating element typically changes (increases) as it heats up. But typically not so much as to change 'why bad stuff happens'.

Important point that follows from Ohm's law: a wattage rating on a device or appliance is typically only for the voltage listed on that appliance. Change the voltage and it will no longer be accurate.

 

[jaggedben]

I hope I get this right.. if not, one of the sharper tools on the Forum will give you the right answer.

If your heater was 1500 total watts and designed for the 3 phase supply, it would be made so there would be (3) 500 watt loads, one on phases A-B, one on B-c, and one one A-C.
Using ohms law I= P/E I = 1500/480 X 1.732 = 1.8 amps per phase.

You got the right answer at the end. Using the square root of 3 was correct.

 

[LouisianaApprentice]

Alright I tried the formula. I checked it by finding the resistance of each load. E=I*R
270=500r
R=.54ohms
So each element has a .54 ohm resistance

 

[LouisianaApprentice]

So when doing it for total voltage of legs u multiplied by the sqrrt of 3

 

[Adamjamma]

Augie, isnt the basic the amps divided by the nominal voltage? So, if it is a 1500 amp single phase on 120 volts then 1500/120, or if it was 1500 amps in UK then 1500/240?

Or do we use the cube root because he has three phase? I am confused here as well.

 

[LouisianaApprentice]

Thank you guys so much you really helped my understanding sorry for jumping the gun the entire time lol. From rereading your replies I think I finally grasp it. So if a 480v heating element has a labeled wattage u just do an amp calculation for 3 phase to find the correct wire size? Such as this one I(A) = P(W) / (√3 × PF × VL-L(V) ) ..... So in pure resistive loads power factor is 1? Correct? Sorry if I keep trailing off just trying to figure out how u would calculate this in field.

 

[LouisianaApprentice]

I got 1.8 amps

 

[Little Bill]

Augie, isnt the basic the amps divided by the nominal voltage? So, if it is a 1500 amp single phase on 120 volts then 1500/120, or if it was 1500 amps in UK then 1500/240?

Or do we use the cube root because he has three phase? I am confused here as well.

It's not amps divided by anything. It's watts or power divided by voltage divided by sq rt of 3 Or P/(E x 1.732) You can say P divided by E divided by sq rt 3 or P divided by (E x sq rt 3) and get the same answer.

 

[Adamjamma]

It's not amps divided by anything. It's watts or power divided by voltage divided by sq rt of 3 Or P/(E x 1.732) You can say P divided by E divided by sq rt 3 or P divided by (E x sq rt 3) and get the same answer.

OK. Thanks.

 

[LouisianaApprentice]

So from the job I remember we ran #10s the run was only around 100ft in total as the MCC was close. When I plugged that into a wire size calculator it said 14Cu, 12 Al. The wattage was 15000 for these heaters (2 per tower@ 7500w) I got 18.04 amps @ 480. So learning these calculations would have saved a good bit of money. That is if I'm correct lol.

 

[augie47]

You will find a whole different topic when you deal with termination allowances (110.14) and termination restrictions in 240.4 plus the possibility your heaters might be a continuous load.. A #10 could have been the correct size but it would have been a #12 at minimum.

 

[jaggedben]

So from the job I remember we ran #10s the run was only around 100ft in total as the MCC was close. When I plugged that into a wire size calculator it said 14Cu, 12 Al. The wattage was 15000 for these heaters (2 per tower@ 7500w) I got 18.04 amps @ 480. So learning these calculations would have saved a good bit of money. That is if I'm correct lol.

Well, let's see. If the amps are 18.0 then you cannot put that on 14awg wire. Not sure I'd trust the app you're using. Also you may have a code requirement for a heater for continuous load. So that's 22.5amps, which does require 10awg. Maybe you could have done 12awg.

 

[LouisianaApprentice]

Ok thanks you guys. seems I have a lot to learn as far as necessary code and how it applies to everything I've only been in this field 2 years and just started looking into how everything works on this side of things instead of common practices my employers have used.