How much heat is generated

[mivey]

I2R is Power, not Energy.
db

Yes, yes, yes, the area under the power vs time graph is the energy. We still call those I²R losses. For that rate of energy, pick whatever time interval you want as a multiple and you get the energy lost to heat for that time interval.

Jim did use "watts". That is the same as the rate of heat conversion. It seems you were confusing heat with temperature.

 

[Besoeker]

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

From that link the unit for energy or heat is the joule.
The unit for power is the watt which is also joules per second.
The watt is thus not a measure of heat.

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

And from that link:

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 1oF (58.5oF - 59.5oF) 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

Note that. Watt-hours so power times time. Not watts.

 

[mivey]

Heat = Energy transferred

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

Yes. Power is the rate of energy exchange/flow/etc.

But the point he was trying to make was that two different sized conductors would have a different temperature. That is not the same thing as heat energy. Jim's answer to calculate the amount of heat generated is still correct. The resistance varies with the wire size (and wire temperature, and signal frequency). The I²R formula is still what is used to find the heat generated (over whatever time period is relevant).

 

[Besoeker]

Jim's answer to calculate the amount of heat generated is still correct.

It calculates power dissipation. Not heat.
What you are concerned with, in practical terms for rating conductors, is that they do not exceed their temperature rating.
If you wanted to calculate that (rather than just accept what tables give you) you'd need thermal impedance in degC/W.

 

[mivey]

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".

But the I²R
losses are heat losses.

Some of the watt-hours produced light, some of them produced heat. But all the watt-hours look the same.
db

The light that gets absorbed becomes heat. Still does not change the fact that the I²R losses generate heat.

 

[Besoeker]

But the I²R
losses are heat losses.

Power losses.

 

[mivey]

It calculates power dissipation. Not heat.
What you are concerned with, in practical terms for rating conductors, is that they do not exceed their temperature rating.
If you wanted to calculate that (rather than just accept what tables give you) you'd need thermal impedance in degC/W.

Did the OP mention the conductor's thermal rating? The question was how to calculate the heat generated due to wire resistance. See Bob's #16 for an example of where this might be useful where HVAC loading is of interest. We can also calculate the current needed to melt ice off of power lines. Of course, we also calculate the thermal rating of power lines to make sure they do not have excess sag or temperatures under heavy loading.

 

[mivey]

Power losses.

Rate of energy transfer.

 

[Besoeker]

Did the OP mention the conductor's thermal rating? The question was how to calculate the heat generated due to wire resistance.

Heat is not measured in watts.

 

[Besoeker]

Rate of energy transfer.

For watts, rate of energy, yes. J/s
Energy, no. J.

 

[mivey]

Heat is not measured in watts.

Heat is energy. The rate at which energy is converted to heat is watts. The watts we use to calculate this conductor heat loss is given by I²R. Thus, we call the energy lost "I²R losses".

 

[mivey]

For watts, rate of energy, yes. J/s
Energy, no. J.

And the heat energy generated by current in a resistive conductor is calculated as I²R over time. That is what Jim said. He was talking about the heat generated, not the wire temperature.

 

[Besoeker]

Heat is energy. The rate at which energy is converted to heat is watts.

The conversion rate yes. That doesn't make it heat.
That would analogous to calculating distace from miles per hour alone. You can't do it.

 

[Besoeker]

And the heat energy generated by current in a resistive conductor is calculated as I²R over time.

The formula is P(watts of heat) = I?R.

is what he posted. No mention of time.
Energy is i²rt in joules. Not watts.

 

[jim dungar]

Do you consider energy (kwh) or power(kw) when choosing the following electric heating devices?
Water heater
Iron
Stove
Oven
Tea kettle
Soldering Iron
Space heater

for example:
http://www.21stcenturyheating.com/ch-finesse-radiator1

 

[mivey]

The conversion rate yes. That doesn't make it heat.

We are in agreement.

That would analogous to calculating distace from miles per hour alone. You can't do it.

But given the speed, we can calculate the distance traveled for the period of interest.

 

[mivey]

is what he posted. No mention of time.

The OP did not mention over what time period either. He just wanted to know if the heat loss could be calculated using the current and wire resistance. Jim gave him the formula for the rate of heat loss.

The exclusion of time was not the point of contention initially raised. It was that a smaller wire would get hotter. That is a temperature difference and does not negate the validity of an I²R calculation.

 

[Besoeker]

We are in agreement.

Thought we might be....

But given the speed, we can calculate the distance traveled for the period of interest.

Except that MPH, like watts doesn't have a period.

 

[Besoeker]

The OP did not mention over what time period either. He just wanted to know if the heat loss could be calculated using the current and wire resistance. Jim gave him the formula for the rate of heat loss.

Yes. But rate of heat loss isn't heat loss.

 

[Besoeker]

Do you consider energy (kwh) or power(kw) when choosing the following electric heating devices?
Water heater
Iron
Stove
Oven
Tea kettle
Soldering Iron
Space heater

You ought to. It's kWh you pay for.