How much heat is generated

[Ponchik]

I was wondering if there is a way to calculate how much heat current flow generate.

For example: How much heat is generated by a continuous current flow of 80Amps?

Is there a ratio of so many amps will generate so much heat?

I am not even sure if this is a valid question.:?

 

[K8MHZ]

I was wondering if there is a way to calculate how much heat current flow generate.

For example: How much heat is generated by a continuous current flow of 80Amps?

Is there a ratio of so many amps will generate so much heat?

I am not even sure if this is a valid question.:?

Is this an 80 heater or an 80 amp motor?

 

[jim dungar]

The formula is P(watts of heat) = I?R.
So square the current and then multiply it by the resistance. Oh wait, you didn't give us the resistance.

 

[Ponchik]

The formula is P(watts of heat) = I?R.
So square the current and then multiply it by the resistance. Oh wait, you didn't give us the resistance.

I should have been specific. I was asking for heat generated when a current flows on a conductor not the load.

So is it fair to assume the following: If we have a circuit that is 200' long with #2 copper conductors and there is 80 amp flowing, then the heat generated from the conductor is 197 watts?

(80x80) x .0308 (resistance of #2 copper for 200')
6400 x .0308 = 197 watts

 

[broadgage]

For relatively small conductors, heat production may calculated very easily indeed.

Multiply the current by the voltage drop.
For example 80 amp load, 4 volts voltage drop, 320 watts of heat produced.

No great accuracy can be expected in real world conditions since the actual supply voltage varies a bit, and depending on the type of load this may increase or decrease the current draw.

The voltage drop will also vary according to the conductor temperature which is far from fixed.

A large power cable loaded for only a few seconds each hour will be at virtually ambient temperature and have a lower voltage drop than a similar cable which is hot due to continuall loading.

For very large cables carrying AC part of the voltage drop is due to inductance, which unlike resistance does not produce heat. This is negligable on all but very large cables.

 

[mike_kilroy]

I should have been specific. I was asking for heat generated when a current flows on a conductor not the load.

So is it fair to assume the following: If we have a circuit that is 200' long with #2 copper conductors and there is 80 amp flowing, then the heat generated from the conductor is 197 watts?

(80x80) x .0308 (resistance of #2 copper for 200')
6400 x .0308 = 197 watts

remember too that what goes out must come back so you have TWO wires 200' long so if you are asking because you want to know total heat that will be in a conduit or something, double it if that is appropriate......

 

[Rick Christopherson]

I should have been specific. I was asking for heat generated when a current flows on a conductor not the load.

So is it fair to assume the following: If we have a circuit that is 200' long with #2 copper conductors and there is 80 amp flowing, then the heat generated from the conductor is 197 watts?

(80x80) x .0308 (resistance of #2 copper for 200')
6400 x .0308 = 197 watts

I didn't check your numbers, but assuming your resistance is correct, then this is correct for a single wire (double it for the return). However, generally knowing just the heat generated is not all that important without also examining the heat dissipation too. If the heat dissipation rate is lower than the heat generation, then you will get a temperature rise until the two reach equilibrium. The heat dissipation rate will vary as the temperature difference (ΔT) changes. The greater the ΔT, the more heat it will dissipate, and this will determine the maximum temperature rise of the conductor for a given amperage when they reach equilibrium.

What is interesting about this, is that this is the reason why the inner coils of a retractable cord reel will melt when the cord is not fully extended. The heat generation is the same for the entire reel, but the inner coils cannot dissipate the heat as fast as the outer coils because they effectively have more insulation around them (the outer coils). So the ΔT of the inner coils is higher.

 

[BPoindexter]

If you are using the DC resistance of uncoated #2 copper then you have a transcription error. R = .0388 ohms for 200 ft. As Rick stated above you would need to double that for out and return.

Brings up a question. Does it matter whether you use Z (impedance) or R (resistance)? I would assume that you would want to use Z as this is an AC circuit but am not sure if that affects heating of the cables or is it stricly resistance? There is a difference. Table 8 gives the DC resistance and Table 9 gives the impedance which is affected by the type of raceway, power factor, and the frequency.

 

[broadgage]

If calculating voltage drop in large conductors then take account of Z or impedance.

If calculating heat produced, then use R or resistance.

Inductance is insignificant for small conductors at power line frequencies, but has to be allowed for at high frequencies or in very large conductors.

 

[barclayd]

The formula is P(watts of heat) = I?R.
So square the current and then multiply it by the resistance. Oh wait, you didn't give us the resistance.

Watts is Watts - Watts is not heat.
A #18 wire carrying 20 amps will be a lot hotter than 20 amps on a 500MCM wire.
There is no direct conversion factor for 'heat'.
sorry
db

 

[mivey]

Watts is Watts - Watts is not heat.
A #18 wire carrying 20 amps will be a lot hotter than 20 amps on a 500MCM wire.
There is no direct conversion factor for 'heat'.
sorry
db

Jim's answer is correct since heat is the energy transfer. Heat does not mean temperature, except in loose terms. The resistance of the conductors is different of course but the energy lost as heat is indeed I²R.

 

[jim dungar]

Watts is Watts - Watts is not heat. A #18 wire carrying 20 amps will be a lot hotter than 20 amps on a 500MCM wire. There is no direct conversion factor for 'heat'. sorry db

Sorry, but you need to rethink your position.
There is a direct conversion between Watts, cal/sec, and BTU/HR which are all 'units' of heat.
In your example, the difference between the wire sizes is the resistance of the conductor.

 

[kingpb]

1 KW = 3412 BTU/Hr

 

[barclayd]

Watt, Cal/Sec, BTU/Hr, etc. are units of POWER.
Watt-Hour, Cal, BTU, etc. are units of ENERGY.
None of them are HEAT.
If you add energy to a substance, its temperature will rise (it will heat up).
For a specific amount of energy applied to a specific amount of a specific substance under specific conditions, the temperature rise can be calculated.
You can call watts 'heat' if you want to, but I don't think I will.
db

 

[barclayd]

Jim's answer is correct since heat is the energy transfer. Heat does not mean temperature, except in loose terms. The resistance of the conductors is different of course but the energy lost as heat is indeed I²R.

I2R is Power, not Energy.
db

 

[rcwilson]

Average W/FT heat from cables.

Average W/FT heat from cables.

I was asked to calculate the operating heat load for a large electrical room to verify HVAC sizing and to specifically include the heat load from cables in trays. We developed an Excel sheet that listed all cable sizes, cable resistance per foot, typical average amps based on 75% of the cables rated ampacity and heat loss per foot. When we graphed W/ft versus cable size, we found we could get close enough using an average W/ft for all cables. Larger cables had less resistance but more current so the W/ft were relatively constant in the cable sizes we were using.

Unfortunately, I can't find the file or remember the resulting W/ft number to share. I do recall that when the plant operated at light loads you could hang meat in the elelctrical room. (Maybe it wasn't such a good design method).

 

[Hv&Lv]

The power loss may produce "X" watts of heat energy, but it isn't the same as degrees of heat energy.
In order to know what temperature the wire will be at a specific current, other variables will have to come into play. Environmental conditions will need to be figured in.

 

[jim dungar]

You can call watts 'heat' if you want to, but I don't think I will.
db

http://www.spiraxsarco.com/resource...iples-and-heat-transfer/engineering-units.asp

The transfer of energy as a result of the difference in temperature alone is referred to as heat flow. The watt, which is the SI unit of power, can be defined as 1 J/s of heat flow.

Other units used to quantify heat energy are the British Thermal Unit (Btu: the amount of heat to raise 1 lb of water by 1?F) and the kilocalorie (the amount of heat to raise 1 kg of water by 1?C). Conversion factors are readily available from numerous sources.

http://www.engineeringtoolbox.com/heat-units-d_664.html

The unit of heat in the imperial system - the BTU - is

• the amount of heat required to raise the temperature of one pound of water through 1^oF (58.5^oF - 59.5^oF) at sea level (30 inches of mercury).
• 1 Btu (British thermal unit) = 1055.06 J = 107.6 kpm = 2.931 10^-4 kWh = 0.252 kcal = 778.16 ft.lbf = 1.05510¹⁰ ergs = 252 cal = 0.293 watt-hours

An item using one kilowatt-hour of electricity generates 3412 Btu.

 

[kingpb]

Heat = Energy transferred

Therefore, I believe barclayd is correct that KW is not heat.

The BTU, which is equal to approx. 1.055 KJoules, is defined as amount of heat required to raise the temperature of one 1 pound (0.454 kg) of liquid water by 1 ?F (0.556 ?C) at a constant pressure of one atmosphere.

To convert KW to BTU, you need to add the time component, typically Hrs.

Therefore, 1KW-Hr (KWH) = 3412 BTU would be the appropriate way to convert electrical energy to mechanical heat.

Certainly there is the SI unit of joules which could be used in place of British Thermal Units (BTU).

 

[barclayd]

Yes, yes, yes - you can convert kwh to btu. But some of that energy actually does WORK. Not all is converted into heat. That's why that room was "cold enough to hang meat". Calculate away.
Another way to look at it is a 2x4 fluorescent troffer. We can measure the current and voltage and, thereby, calculate watts. Over time there will be an amount of watt-hours. Some of the watt-hours produced light, some of them produced heat. But all the watt-hours look the same.
db