MIZUNO REPLICATION AND MATERIALS ONLY

Quote from JedRothwell

Take an example from before heat engines were developed. In France, they wanted to move water uphill. They used gigantic water wheels to run belts, that ran a series of smaller wheels, that carried some of the water uphill. This was tremendously inefficient. The water that reached the higher level was a tiny fraction of the water going down the river, pushing the wheels.

Here is a modern version of that! Much more efficient. This is a water driven water pump. As you see, it takes far more water going downstream to push this than the amount of water that goes a mile up the hill to the fields. Overall, it produces no net energy. You cannot use it as a perpetual motion machine. It is useful because these people have a lot of moving water in a large stream, and they only need to pump a small amount of that water uphill.

These Water Wheels Can Pump Water Over A Mile Without Electricity

Here is my different take on the external temperature debate...

Changing the internal thermal resistance should not change the external thermal resistance. So the thermal resistance between the outside of the vessel and infinity/ambient is the same for control and experiment.

All the power leaving the vessel must flow through this external resistance.

If the power flowing through this external resistance is the same for control and experiment then the temperature gradient between the outside of the vessel and infinity must also be the same. Therefore the outside temperature must be the same. I agree with Jed.

Let's look at the limits of the problem, a thermos in a vacuum. the thermos can only transfer energy through radiation (modeled as blackbody). The temperature inside the thermos will increase until it becomes hotter than any electrical coil can get. Heat energy transfer from the heater to the thermos must slow to the rate of energy loss. I assume that rate of tranfer is less than the nominal power of the heater. No heater can be ideal and supply a constant energy input in this case. So using this assumption of a perfect heater is a flawed and must be accounted for in these scenarios. I think that is what Mark is getting at. Please correct me if I am wrong.

Quote from PhysicsForDummies

Heat energy transfer from the heater to the thermos must slow to the rate of energy loss. I assume that rate of tranfer is less than the nominal power of the heater.

Sorry that's not right.

If you put 100W into a heating element then 100w must go somewhere. It must either escape the element or start raising its temperature or both.

Typically what happens is the element temperature increases and that increases losses from the element. Eventually a steady state constant temperature condition is reached. Under steady state 100w must escape the element.

The more you insulate the element the hotter it has to get to dissipate the 100w. But it must dissipate 100w under steady state conditions.

If it was ideally insulated so no heat can escape it's temperature would go on rising indefinitely (until melted or otherwise failed).

Replacing air with hydrogen would increase conducted losses meaning the element would loose the 100w at a lower temperature. There are two effects that cancel out exactly..

1) the increased conductivity would tend to increase the external temperature but

2) the lower element temperature would tend to reduce the external temperature.

I can see a potential source of error...

If the element runs cooler when hydrogen is used then less power would be conducted out of reactor down the wires that power the heater. This would mean there is more power escaping the walls of the reactor raising the external temperature so it looks like excess heat is generated.

However the scale of the effect should be small or you can measure it and account for it.

Quote from CWatters

If the element runs cooler when hydrogen is used then less power would be conducted out of reactor down the wires that power the heater.

That is a practical problem with a real-world device. There are all kinds of similar issues with actual calorimeters. There are always heat losses you cannot pin down. The surface temperature is never uniform. Etc., etc. My comments above are about ideal devices. Imaginary devices.

Quote from CWatters

Sorry that's not right.

If you put 100W into a heating element then 100w must go somewhere.

My point is you no longer would be able to "put in" 100W, in the real world. Besides the fact that heating elements like Nichrome melt at 1400C, resistance of materials rises with temperature. The voltage needed to drive this current increases. Not ignoring these real life limitations is important. It is physically impossible to continually raise the temperature using an electric heater to any arbitrarily high temperature. Since no known substance can be used for the filament, you would have to make up some ideal substance "unobtainium". But then what is its resistance vs temperature? Pretty soon you need a power supply with infinite voltage to maintain 100w. So Mark general idea that it takes more "work" to apply the same current is essentially correct, if I paraphrase his thoughts correctly.

If one needs more evidence of the difficulty of real life power supplies in generating high temperatures by powering heating elements, simply look at the cooling issues of Ross's power supply during the ecat SK "industrialization" demo a couple years ago.

Quote

So using this assumption of a perfect heater is a flawed and must be accounted for in these scenarios.

I believe so.

In a nutshell, here is what I think will be observed when power is fed into the core of a cylinder vs the core of an insulated cylinder, with a power supply set to a constant according to the power it draws from the electrical mains itself.

The insulated cylinder will (of course) require much longer to come to thermal equilibrium, where its surface temperature no longer increases but stays the same.

Surprisingly, that equilibrium temperature of the insulated cylinder will be a little lower than the equilibrium temperature of the uninsulated cylinder.

This is because the insulated cylinder will draw slightly less power from the power supply. To make up for this in energy balance, the power supply (to the insulated cylinder) will itself emit slightly more heat.

But that power supply may emit even a bit more heat than this excess heat ; this is because there is a temperature differential in the insulated cylinder, even at equilibrium (when the outer wall temperature is constant). This temperature differential between the core and outer wall of the cylinder amounts to potential energy stored in the insulated cylinder. But I'm not sure if this aspect is simply part and parcel of the previous aspect.

It should be noted that the above scenario assumes that the entire apparatus is 'outside' in the 'cold', where the temperature is constant.

If the entire apparatus was in, say, a small perfectly insulated room, things would be different!

In real life the scenario is somewhere between those two extremes, making things that much more tricky.

Disclaimer : What I've said above is just my relatively naive opinion and does not reflect the opinion of any sponsor, corporation or higher power.

Quote from Mark U

I believe so.

In a nutshell, here is what I think will be observed when power is fed into the core of a cylinder vs the core of an insulated cylinder, with a power supply set to a constant according to the power it draws from the electrical mains itself.

The insulated cylinder will (of course) require much longer to come to thermal equilibrium, where its surface temperature no longer increases but stays the same.

Surprisingly, that equilibrium temperature of the insulated cylinder will be a little lower than the equilibrium temperature of the uninsulated cylinder.

This is because the insulated cylinder will draw slightly less power from the power supply. To make up for this in energy balance, the power supply (to the insulated cylinder) will itself emit slightly more heat.

But that power supply may emit even a bit more heat than this excess heat ; this is because there is a temperature differential in the insulated cylinder, even at equilibrium (when the outer wall temperature is constant). This temperature differential between the core and outer wall of the cylinder amounts to potential energy stored in the insulated cylinder. But I'm not sure if this aspect is simply part and parcel of the previous aspect.

It should be noted that the above scenario assumes that the entire apparatus is 'outside' in the 'cold', where the temperature is constant.

If the entire apparatus was in, say, a small perfectly insulated room, things would be different!

In real life the scenario is somewhere between those two extremes, making things that much more tricky.

Disclaimer : What I've said above is just my relatively naive opinion and does not reflect the opinion of any sponsor, corporation or higher power.

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If the power supply is set to a constant output, then it cannot also reduce the power going out. The heater coil will not refuse to accept any more input, even if the heat cannot transfer away, although it’s resistance could climb until less power is made. A constant power input arrangement, however, will simply increase the voltage and/or current until the power setting is reached in order to compensate for increased resistance.

If power is applied to a heater coil (with a special resistance that does not appreciably increase) and it cannot exit a sort of thermal vault, and the vault cannot dissipate the heat, then the heater will just keep heat up hotter and hotter until it melts, or the thermal vault melts, or both.

I am not sure if this is related, but...

Back in the days when arguing about the Lugano device was fashionable, we once worked out that the internal, central core temperature of a solid ceramic cylinder wrapped in a heater coil (but is below the cylinder surface), reaches at steady state, the same temperature as the (adjacent) heater coil wire. The outer surface will be cooler than both, fairly obviously, and the core cannot be hotter than the heat source at equilibrium.

Quote from Paradigmnoia

If the power supply is set to a constant output, then it cannot also reduce the power going out

Sorry I believe that is a wrong assumption. It can and will reduce the power going out if it us impossible by the laws of physics and it's electronic design, to do so. Increasing resistance of the filament will insure this at some point. There is not an even nearly perfect power supply. Physics is responsible for how materials behave, leading to the impossibility of such a power supply. The thought experiment of the thermos in a vacuum shows such a power supply is impossible since the increase in resistance would require ever increasing voltage.

Quote from Mark U

The insulated cylinder will (of course) require much longer to come to thermal equilibrium, where its surface temperature no longer increases but stays the same.

Surprisingly, that equilibrium temperature of the insulated cylinder will be a little lower than the equilibrium temperature of the uninsulated cylinder.

That would be surprising. It would mean you have a perpetual motion machine. I suggest you do that. You will win a Nobel prize and a lot of money for revolutionizing physics.

I find this discussion intellectually lazy to a disturbing degree. There's a general tendency to confuse temperature with power that leads to incorrect application of thermodynamics. For example in the most recent "gedanken experiment" the insulated cylinder will have a larger surface area than the plain one, and the surface emissivity will also be different. So to be precise, we should say that at equilibrium, the thermal power leaving each of the cylinders as heat will be equal. This is clearly required by thermodynamics, and can be satisfied even with different surface temperatures.

Quote from magicsound

I find this discussion intellectually lazy to a disturbing degree. . . .

For example in the most recent "gedanken experiment" the insulated cylinder will have a larger surface area than the plain one, and the surface emissivity will also be different.

No. I specified that the two cylinders are exactly the same size and shape, with the same surface area. Obviously they would be at different temperatures if one was bigger.

You are a little intellectually lazy for not seeing that is what I said. I repeated it several times.

So then you are proposing insulation of zero thickness and emissivity identical to bare metal?

Quote from magicsound

So then you are proposing insulation of zero thickness and emissivity identical to bare metal?

The insulation is on the inside of the cylinders. The discussion started when someone mentioned that some gasses insulate better than others. In the thought experiment, I described replacing the gasses with insulation. The effect is larger, and easier to measure. A bare heater is placed in a cylinder, and another heater wrapped in insulation and placed in an identical cylinder. The heater wrapped in insulation would reach a higher temperature with the same input power, and it would take longer for the outside surface of the cylinder to heat up to the terminal temperature and then plateau. But, after a while, the surfaces of both cylinders would reach the same temperature.

After you turn off both heaters, the insulated cylinder surface would take longer to cool down.

OK, I missed that detail in your original proposal, and my comment applied mainly to those responding to it. Each of the cylinders should be considered as a "black box" containing a source of heat, and I agree with your conclusion that given identical exterior qualities, the equilibrium external temperature will be the same for each.

Quote from magicsound

I agree with your conclusion that given identical exterior qualities, the equilibrium external temperature will be the same for each.

But in real life, with actual cylinders, there are bound to be small but measurable differences, as I said before. Even if you use the same cylinder for the two different tests, I expect it would give slightly slightly different results for various reasons. Mainly because the surface temperatures are not uniform.

Quote from PhysicsForDummies

Sorry I believe that is a wrong assumption. It can and will reduce the power going out if it us impossible by the laws of physics and it's electronic design, to do so. Increasing resistance of the filament will insure this at some point. There is not an even nearly perfect power supply. Physics is responsible for how materials behave, leading to the impossibility of such a power supply. The thought experiment of the thermos in a vacuum shows such a power supply is impossible since the increase in resistance would require ever increasing voltage.

if you cannot put anything more in, when before you were, then it is not at steady state.
At a moment of change, almost anything goes.