How much heat is generated

[mivey]

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

But the period is known for the average values.

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

But you do use the rate to find the quantity.

There are times when there is a definite separation between quantities and rates. One example is when you have the kWh consumed (the quantity) and would like to know the maximum kW demand (the maximum rate). Because the demand varies during the consumption period, there is no way to directly connect the two with time. The best we can calculate is the average demand (the average rate).

With a steady load, the average demand and consumption can readily be related over the consumption period. Think about a meter with an hourly demand interval. The kWh (energy consumed) and the kW demand (rate of consumption) have the same value.

With the OP's question, a steady load can be assumed for the consumption period. The average I²R (average rate) can readily be used to get the heat generated (the quantity) during the period. The period would not be an unknown in this case. So once one knows how to calculate the average power for the period of interest, the heat generated during that period can be readily obtained.

This is not the same issue we normally face with questions about power and energy conversions. Usually the poster wants to use an average value (the average demand or the energy consumed) to calculate the peak power, which we know can't be done.

The minutia about time was really making a mountain out of a molehill for the given question since time would not be an unknown and we assume a steady load for the period of interest.

The other question raised was about how a smaller wire would get hotter. It was posed as though this fact would invalidate the use of the average power to find the heat generated. The fact remains that the heat generated is different than the wire temperature. Completely different.

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

Many would say the power would be the primary focus for most of those items.

 

[Ponchik]

What i meant with my questions is:

If i install a temperature probe in the conduit with 200' of #2 copper with 80Amp of current, how hot will the probe read?

I understand the ambient will have an effect. But what i want to know if there is a relationship between so many amps and so much heat that is generated.

 

[Besoeker]

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

But the period is known for the average values. MPH tells you nothing about period.

But you do use the rate to find the quantity.

But you can't determine quantity from rate alone. Watts give you just rate.

There are times when there is a definite separation between quantities and rates.

They are separate things.

With the OP's question, a steady load can be assumed for the consumption period. The average I²R (average rate) can readily be used to get the heat generated

It can be be used to get the rate at which heat is being generated.

The minutia about time was really making a mountain out of a molehill for the given question since time would not be an unknown and we assume a steady load for the period of interest.

Yes, you can assume steady state conditions. That then gives you a steady state value for the rate at which heat is being generated. Steady state temperature will be attained when the rate of heat generation and rate of heat dissipation are in equilibrium. Call it minutia if you like but the distinction between power and heat is real and matter of fact.

The fact remains that the heat generated is different than the wire temperature. Completely different.

I agree. And, like the difference between heat and power, it is a difference that seems to be misunderstood or overlooked quite often. That the OP asked about heat when he meant temperature illustrates that point.

 

[T.M.Haja Sahib]

Watt is joule per second i.e a measure of heat generated per second.

 

[T.M.Haja Sahib]

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

Please do not make 'rate' mysterious again as it was when the calculus was invented by Newton.

 

[Besoeker]

Watt is joule per second i.e a measure of heat generated per second.

Yes. Heat per second.

 

[Besoeker]

Please do not make 'rate' mysterious

I don't believe I made it mysterious at all.

 

[T.M.Haja Sahib]

Yes. Heat per second.

So watt is nothing but heat given off every second.

 

[T.M.Haja Sahib]

I don't believe I made it mysterious at all.

Did 'rate' become a clear concept then?

 

[Besoeker]

So watt is nothing but heat given off every second.

But heat is joules, not watts.

 

[T.M.Haja Sahib]

But heat is joules, not watts.

But watt is also joule (per second) and so also a measure of heat.

 

[Besoeker]

Did 'rate' become a clear concept then?

Given that the derivation is from Medieval Latin, I feel that the meaning of the word rate was understood a while before Newton used it.

Anyway, back on topic the watt is the rate of doing work. The measure of work is the joule. Different thing. They are not interchangeable terms.

 

[Besoeker]

But watt is also joule (per second) and so also a measure of heat.

It absolutely is not.
Say 5A through a 10ohm resistor is 250W. Allow that to pas for say 10 seconds and the heat generated would be 2.5kJ
For say, 1us, and the heat generated would be 250uJ. Nothing like the same heat but still 250W.
You cannot possible consider the 250W to be the heat generated.

 

[mivey]

MPH tells you nothing about period.

Of course not. But on a trip, one would know the period. If I average 75 MPH, I can use this information to determine my arrival time. I can also use it to determine how far I drove during my time at the wheel.

You are acting as if the period can't be known and thus the MPH can't be used to determine other trip values. All that has been said is that the average MPH can be used to determine distance, and it can.

But you can't determine quantity from rate alone.

Of course you can't. But why the assumption that the time is an unknown?

They are separate things.

Of course they are but as I explained, the manner of separation is not always the same (the difference between the average rate vs. energy and a varying rate vs. energy).

It can be be used to get the rate at which heat is being generated.

And that rate, in turn, can be used in the quantity calculation.

Yes, you can assume steady state conditions. That then gives you a steady state value for the rate at which heat is being generated. Steady state temperature will be attained when the rate of heat generation and rate of heat dissipation are in equilibrium.

See Bob's #16 for that very use.

Call it minutia if you like but the distinction between power and heat is real and matter of fact.

And it is a matter of fact that the rate can be used to determine the heat generated. Given that the user would know the time period, finding the energy is a simple calculation. With the period selected by the user, the only real hurdle is calculating the rate.

I you know you lose a constant $50/hour in a casino, calculating your losses is simple. Repeatedly stating that you can't calculate the losses without the # of hours played is focusing on the minutia.

That the OP asked about heat when he meant temperature illustrates that point.

Agreed

 

[mivey]

What i meant with my questions is:

If i install a temperature probe in the conduit with 200' of #2 copper with 80Amp of current, how hot will the probe read?

I understand the ambient will have an effect. But what i want to know if there is a relationship between so many amps and so much heat that is generated.

There is a relationship. The calculations are quite complex but they make software that will calculate the steady-state temperature.

The calculations for finding the steady-state temperature for aerial conductors on a power system are not extremely complicated and I make these from time-to-time. Bundled cables in buried conduit with other power cables nearby is a complex calculation and is not what I would consider fun (except maybe once or twice) and I would just let the canned software do the math.

If you want to have fun with cable calculations, I have found these books by Anders to be valuable:
"Rating of Electric Power Cables"
and
"Rating of Electric Power Cables In Unfavorable Thermal Environment"

 

[Besoeker]

Of course not. But on a trip, one would know the period. If I average 75 MPH, I can use this information to determine my arrival time.

But not from the MPH alone.

You are acting as if the period can't be known and thus the MPH can't be used to determine other trip values.

I'm not suggesting that it can't be known.

All that has been said is that the average MPH can be used to determine distance, and it can.

OK. My average speed is 58.6 MPH. What distance did I travel?

Of course you can't. But why the assumption that the time is an unknown?

Watts are a measure of power. That tells you nothing about time. That's not an assumption.

Of course they are but as I explained, the manner of separation is not always the same.]
It is. J=Ws. Always the same separation.

I you know you lose a constant $50/hour in a casino, calculating your losses is simple. Repeatedly stating that you can't calculate the losses without the # of hours played is focusing on the minutia.

You can't.
Play for 5 hours and lose $250. Play for 10 and lose $500. ($250 is minutia? )
Those calculations required the number of hours to be taken into account.

 

[mivey]

But not from the MPH alone.

But on a trip, one would know the period.

My average speed is 58.6 MPH. What distance did I travel?

You tell me. It is your trip. On a trip, you would know the period. You can use the average speed to calculate the distance.

The question is: Can the rate be used to find the quantity? It can if we know the average rate for the period of concern.

Watts are a measure of power. That tells you nothing about time.

Time does not have to be assumed to be an unknown.

Of course they are but as I explained, the manner of separation is not always the same.

It is. J=Ws. Always the same separation.

Not the same for every quantity.

Consider the monthly power bill where the period of interest is the month. The kWh used for the month and the average kW for the month are directly related by the hours for the month. Not so with the peak kW for the month and kWh used for the month.

Given the average kW, it is trivial to calculate the billed kWh. Given the peak kW, we can't calculate the billed kWh.

On a full day trip, your average speed was 58.6 MPH. You traveled 58.6 miles/hour * 24 hours = 1406.4 miles. Your maximum speed has no relationship to the trip period. The separation between the two speeds and the trip distance is different.

You can't.
Play for 5 hours and lose $250. Play for 10 and lose $500. ($250 is minutia? )
Those calculations required the number of hours to be taken into account.

And that is what the player can calculate once the rate is known.

 

[kwired]

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

But how fast you get the task done is what you are usually most interested in when selecting which appliance to use.

 

[Besoeker]

You tell me. It is your trip. On a trip, you would know the period. You can use the average speed to calculate the distance.

All that has been said is that the average MPH can be used to determine distance, and it can.

Is what you posted.
You can?t.
You also need to know the period as well as the rate.
And that, to cut to the chase, is the analogy with watts. You cannot calculate heat from watts alone.
Which makes this from post #3 just plain wrong:

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

Watts are not a measure of heat.
And no amount of discussion can make them so.

 

[T.M.Haja Sahib]

It absolutely is not.
Say 5A through a 10ohm resistor is 250W. Allow that to pas for say 10 seconds and the heat generated would be 2.5kJ
For say, 1us, and the heat generated would be 250uJ. Nothing like the same heat but still 250W.
You cannot possible consider the 250W to be the heat generated.

But you have to agree on units of measurement for energy and time before arriving at the rate of energy usage so that a meaning can be attached to it i.e power. If you take one second as unit of time, 250w means 250joules energy given off every second. If you take 1us as unit of time, 250 uJ of energy given off every microsecond.