I saw the same peculiar behavior with Argon gas substituted for Hydrogen. Aside from the different thermal conductivity, Argon seems to exhibit anisotropic convection, and this was confirmed to me by two very experienced LENR researchers. They suggested Nitrogen is better for use in calibrations.
MIZUNO REPLICATION AND MATERIALS ONLY
-
-
Nope. If the two cylinders have the same surface area, the same internal power, and they are exposed to the same outside conditions (air or water), the surfaces have to come to the same temperature. Or you have a perpetual motion machine, as I said.
It may take a while for them to come to the same temperature.
Let's take your claim and exaggerate the conditions. We take two stainless steel thermoses of the same thermal mass and surface area. One of them is insulated on the inside (as it should be) and the other (strangely) not. We put water and a resistive heater into each, heating the water with the same power. We put a thermocouple on the outside of each thermos. Oh yes, we're outside in the winter and the temperature is about zero C. Will both thermocouples eventually read the same temperature? No. This is because the little heat that arrives at the surface of the insulated thermos is not allowed to build up on the surface over time ; it is conducted away by the cold air and its surface will remain cool. Meanwhile, with the uninsulated thermos, the cold air is not able to conduct heat away fast enough from the thermos' surface and the surface will be warm.
-
We put a thermocouple on the outside of each thermos. Oh yes, we're outside in the winter and the temperature is about zero C. Will both thermocouples eventually read the same temperature? No.
Yes, absolutely they will. Otherwise you can make a perpetual motion machine out of them, transferring heat through a heat engine from the hotter one to the cooler one. That is impossible if the power going into them is the same.
This is because the little heat that arrives at the surface of the insulated thermos is not allowed to build up on the surface over time
Nope. After a delay, the amount of heat reaching the surface of the well insulated container is the same as the poorly insulated one. The surface temperature is the same. Heat cannot be held back indefinitely. It has to come out eventually. The inside of the well insulated container will remain hotter than the poorly insulated one. For the same reason you stay warmer under a heavy blanket than a light one on a winter night. All of the heat from your body eventually reaches the surface of the heavy blanket and radiates into the room.
-
Yes, absolutely they will. Otherwise you can make a perpetual motion machine out of them, transferring heat through a heat engine from the hotter one to the cooler one. That is impossible if the power going into them is the same.
I think that is not correct. The heat is already coming out and an external heat engine can make use of it without violating conservation. There are also non-linear dynamics due to the 4th power of radiation, and that will act to balance the external temperature of the two surfaces as a negative feedback term.
That said, an ideal calorimeter will measure the heat by integration of all exit paths, including a "heat engine" within its detection volume. The Mizuno air flow calorimeter approximates that kind of heat engine. By your analogy, putting a blanket of insulation over an internally heated cell in the calorimeter will result in a higher temperature inside the cell, although the calorimeter measurement at equilibrium will be the same with or without the blanket. That is a potentially important factor in the behavior of the cell, and has not been considered in discussion of Mizuno replications.
-
That is a potentially important factor in the behavior of the cell, and has not been considered in discussion of Mizuno replications.
right ... you are basically saying that Mizuno's COP (coefficient of performance) is a function of the effectiveness of the thermal insulation of the vessel itself. A vessel that is thin and tall with a thin stainless steel wall is one version. A second version would be a short and fat vessel with ceramic fiber insulation around it.
These two examples would have vastly different internal temperatures and that internal temperature supposedly affects the COP.
-
right ... you are basically saying that Mizuno's COP (coefficient of performance) is a function of the effectiveness of the thermal insulation of the vessel itself. A vessel that is thin and tall with a thin stainless steel wall is one version. A second version would be a short and fat vessel with ceramic fiber insulation around it.
These two examples would have vastly different internal temperatures and that internal temperature supposedly affects the COP.
Yes, and even something as simple as changing the metal used for the cell has a profound effect. For example, the horizontal cylinder reactor of R20 would have very different thermal distribution if made of mild steel or aluminium instead of 30X alloy, even with identical dimensions.
-
Yes, absolutely they will. Otherwise you can make a perpetual motion machine out of them, transferring heat through a heat engine from the hotter one to the cooler one. That is impossible if the power going into them is the same.
I can't see how perpetual motion has to do with it. Generalizing your logic, why not feed power only into one thermos (thus requiring half the energy), and harvest the heat flow from the heated thermos to the cold one. Perpetual motion? Of course not.
Heat cannot be held back indefinitely. It has to come out eventually.
All things being equal, I would agree with this. But things are not equal. Your implicit assumption is that it's just as easy to pump a given amount of energy into the hot core of an insulated thermos than it is into the cooler core of an uninsulated thermos. Given that assumption is not correct, then to put in a given amount of energy into the hotter insulated thermos would require additional energy. But since we are considering total energy inventory we would have to reduce the given energy to the insulated thermos in order to maintain total energy expenditure. Thus, there is no contradiction that the insulated thermos in the long term cannot give off the same amount of heat as the uninsulated thermos, for the very fact that it is receiving less energy in the first place.
PS (Who carefully monitors the power draw or heat dissipation of their power supply? Low COPs may require this...)
-
There are also non-linear dynamics due to the 4th power of radiation, and that will act to balance the external temperature of the two surfaces as a negative feedback term.
Once they balance, the temperature is the same for as long as the power is the same.
-
Your implicit assumption is that it's just as easy to pump a given amount of energy into the hot core of an insulated thermos than it is into the cooler core of an uninsulated thermos.
The heat cannot go in. That would violate the Second Law. The heaters are inside the two vessels. The inside is hotter; the outside is cooler. The heat can only come out.
The heat engine would be something like a thermoelectric device with the cool side on the surface of what you suppose would be the cooler well-insulated vessel, and the hot side up against the hot vessel. The heat would go from the hot surface to the cooler surface. It would never go in.
But as I said, this is impossible. The surface of the two has to eventually become equal. It takes longer for the heat to emerge from the insulated vessel, but eventually it reaches the surface. After a while the surface temperature of the two vessels must be the same. Or you have a magic perpetual motion machine, as I said, with the thermoelectric device or any other type of heat engine.
Thus, there is no contradiction that the insulated thermos in the long term cannot give off the same amount of heat as the uninsulated thermos, for the very fact that it is receiving less energy in the first place.
The heaters are inside of both cylinders. It takes no "effort" to get the heat in, any more than it takes effort for you to warm a blanket when you are under it.
Thus, there is no contradiction that the insulated thermos in the long term cannot give off the same amount of heat as the uninsulated thermos, for the very fact that it is receiving less energy in the first place.
The resistance heaters are set at the same power level. They deliver the same amount of energy to both. They have to; energy is power over time. The insulated vessel gets hotter inside, with the same power level, for the same reason you get hotter under a heavier blanket. You are not producing any more heat. The outside of the cylinder (or the top of the blanket) radiates exactly the same amount of heat eventually, after plateauing.
-
PS (Who carefully monitors the power draw or heat dissipation of their power supply? Low COPs may require this...)
I do.
It shows that my 50V supply is only 85% efficient, besides being electrically noisy.
As far as insulation goes, it eventually heats up to whatever temperature it takes to dissipate the energy at the same rate as the input. It might take a very long time, but it will. The outside temperature it reaches will have the same effect inside a calorimeter as an uninsulated thing with the same input at steady state, no matter what the temperature is. This is why I am leery of temperature = power experiments. Everything has to be exactly the same in order to that method work well.
I have the reverse conversation every fall, when my spouse wants me to insulate the banana trees to protect them from freezing. I say that unless the bananas are making heat, they will still freeze inside the insulation, it will just take a little longer than uninsulated, and then they will just stay frozen longer when the freezing weather goes away. It is more important to keep the cold dry wind off, and keep them dry when they are dormant. (Wrapping in burlap, with a bucket or similar covering over the top works well.)
In my calorimeter, the 2 cm diameter x 20 cm long heater core heats the box (Delta T) the same at steady state as the same heater core installed in a 13 cm diameter 50 cm long stainless steel sealed(ish) tube. The heater core gets to around 493 C at 200W and the SS tube barely hits 100 C. The heater core does get hotter when inside the tube (maybe 550 C at 200W, I forget). And the steel tube just slows down the time to steady state. After several tests to confirm, I removed the SS tube to get the show on the road a bit quicker. The SS tube is just crappy insulation in this case. -
The heat engine would be something like a thermoelectric device with the cool side on the surface of what you suppose would be the cooler well-insulated vessel,
Just to clarify:
When you first turn on the two cylinders, the surface of the poorly-insulated one would get hot sooner. A thermoelectric device sandwiched between them would generate electricity. It would stop generating once the surfaces reached the same temperature. Later on, when you turn off the two, the well-insulated cylinder would remain hotter for a while. It would take longer to cool down. You could pull out the thermoelectric device, turn it around so the hot side is against the well-insulated cylinder surface, and it would generate power again. The energy generated in the first warm-up phase would equal the energy generated in the second, cool-down phase. So, no perpetual motion.
If you don't believe me, I suggest you try this with a heater wrapped in bubble insulation, placed in a glass container or something like that. Try it with the heater wrapped in insulation, and then without wrapping it. Measure the surface temperature and the internal temperature. You don't have to use two cylinders; one cylinder in two configurations will do.
-
I have the reverse conversation every fall, when my spouse wants me to insulate the banana trees to protect them from freezing. I say that unless the bananas are making heat, they will still freeze inside the insulation, it will just take a little longer than uninsulated, and then they will just stay frozen longer when the freezing weather goes away.
The trick is to make them very, very wet. Spray water all over them, all night. That's what they do with orange groves in a cold snap. The heat of fusion of water is high, so it takes a lot of energy for the phase change, and until all the water freezes, the temperature remains just above 0 deg C.
(I don't know if that works with bananas. Even close to zero is probably too cold for them.)
This is also how they fly supplies into north and south pole settlements. They put a large bag of water in the airplane cargo hold. If they don't do that, everything will fall well below zero and the equipment and food would be damaged.
-
The trick is to make them very, very wet. Spray water all over them, all night. That's what they do with orange groves in a cold snap. The heat of fusion of water is high, so it takes a lot of energy for the phase change, and until all the water freezes, the temperature remains just above 0 deg C.
(I don't know if that works with bananas. Even close to zero is probably too cold for them.)
This is also how they fly supplies into north and south pole settlements. They put a large bag of water in the airplane cargo hold. If they don't do that, everything will fall well below zero and the equipment and food would be damaged.
The trick is to dry them out a bit. Bananas are mostly water. What you want is to increase the sugar and electrolyte level of the water already in the stalks to lower the freezing point. Rain water soaking into the stalk while dormant seems to cause the most damage, hence the upside down bucket or coffee can on top. (All leaves chopped off for winter). In spring the new growth lifts the top cover, and then you know it is time to unwrap. Mine are already 12 feet high, starting from 5 feet of stalk overwintering. It is too cold here to grow store size bananas (the fruit). That takes 3 years with the leaves on all the time. I do get the giant flowers and finger size bananas sometimes
-
But if you put the bananas in well insulated thermos with a thermoelectric heater, do the bananas with a higher electrolyte level keep longer than those in a poorly insulated thermos, and can either be used as a perpetual motion machine? 🙄
-
MR3 Cal2 has started. A new diaphragm-sealed valve is installed on the gas manifold port.
-
The resistance heaters are set at the same power level. They deliver the same amount of energy to both. They have to; energy is power over time. The insulated vessel gets hotter inside, with the same power level, for the same reason you get hotter under a heavier blanket. You are not producing any more heat. The outside of the cylinder (or the top of the blanket) radiates exactly the same amount of heat eventually, after plateauing.
I agree. *If* both cylinders received the same amount of energy they must ultimately emit the same amount of energy, insulated or not, just at different timescales.
You may have changed my mind, but frankly I'm not sure, because I have something a little different in mind. Here's a thought experiment to show what I mean.
It's cold outside.
Scenario 1 : A battery provides 1 kilo joule of energy into each of two identical bananas. Each banana gets heated up by the energy, by the same amount. Total energy inventory imparted to the bananas is 2 kJ. Since both bananas are at the same energy, there can be no transfer of heat between them; No more work can be done.
Scenario 2: A batter provides 2kJ of energy into one banana, but none into the other, identical banana. One banana gets heated up, the other not. Because one banana is hotter than the other, there can be a transfer of heat between bananas; additional work can be done. The total energy imparted to the bananas now appears to be greater than 2kJ!
But how? It must be that the battery *actually* exhausted more energy to heat up one banana 2kJ than two bananas 1kJ each. In other words, in a cold environment it requires more energy to heat an already hot object than a cold object. So : It takes more energy to heat the hot core of an insulated cylinder than the cooler core of an uninsulated cylinder.
Or am I drunk on fermented banana juice?
-
Because one banana is hotter than the other, there can be a transfer of heat between bananas; additional work can be done.
Nope. No additional work is done. The thermoelectric device converts some the heat back into high grade energy (electricity). That's all there is to it. It does not not add any energy to the system. It just converts some fraction of the heat energy to higher grade energy (and a much larger fraction of it to lower grade energy, or it would violate thermodynamics). You don't need the second banana; you can let the heat radiate from the cool side of the thermoelectric device into the atmosphere.
Any heat engine that can work at a given temperatures will do that. In actual practice, with a test of this nature, I think only a thermoelectric device would produce a measurable flow of electricity. A small steam engine would never even turn.
Here is a small steam engine, by the way. The piston is 2.5 cm. It is sitting on my bookshelf. It only works at combustion temperatures. Like any heat engine, it converts heat into work, following the laws set forth by M. Carnot.
But how? It must be that the battery *actually* exhausted more energy to heat up one banana 2kJ than two bananas 1kJ each
Nope. 2 kJ is 2 kJ. One banana or a hundred, it is the same amount of energy. If you release the same amount of energy inside identical bodies, the surface must eventually radiate that same amount of energy. If it is brief pulse, and neither banana comes to a stable temperature, the one with less internal insulation will get hotter at the surface for a while, while the better-insulated one will stay hot for a longer time. If the power is left on until they settle at a fixed temperature, with output balancing input, then the surface temperatures have to be the same.
The point is, output must balance input, eventually. No matter how well insulated the vessel is, eventually the heat coming out must balance the resistance heater electricity going in. Either that, or the vessel melts or ruptures, I suppose.
This is how isoperibolic calorimetry works. Here is a formal definition from Hemminger and Hohne, p. 99:
6.2.2 Isoperibol Calorimeters
In isoperibol calorimeters (see Section 5.2) the measuring system is coupled via defined heat conduction paths with surroundings which have constant temperature and have good thermal conductivity. In such calorimeters all processes can with some approximation be reduced to a heat exchange between the system to be investigated and an isothermal heat reservoir of infinite capacity. Fig. 6-10 shows the construction of such a calorimeter and a simplified representation of the heat conduction path between the isothermal surroundings and the sample being examined. If an abrupt heat-generating event occurs in the sample at the moment to, the temperature T of the calorimeter vessel undergoes the change shown in Fig. 6-11 if both the vessel (2) and the sample (3) have a high thermal conductivity. The temperature jump DT is proportional to the quantity of heat produced: DT = a • Q. The constant a (the reciprocal heat capacity of the device) is best determined by calibration using electric heating. The function f l (t) according to which the temperature of the system approaches that of the surroundings depends on the nature and magnitude of heat transport between the system and the isothermal surroundings; in other words it is a function of the apparatus used. For pure heat conduction it is an exponential function (see Section 8.2.1). The relationship between the temperature of the system and the heat flux to the surroundings is in any case defined; consequently the area F under the curve T = f1(t) is proportional to the heat produced at t0.
-
Jed, it seems you are missing my point. My (perhaps unorthodox?) point is this: It requires energy to send energy. Specifically, (in a cold, low energy environment) it seems to require more energy to send energy to a high energy object that to a low energy object. That's all. Perhaps you think that's bananas. Which is OK.
-
Jed, it seems you are missing my point. My (perhaps unorthodox?) point is this: It requires energy to send energy. Specifically, (in a cold, low energy environment) it seems to require more energy to send energy to a high energy object that to a low energy object. That's all. Perhaps you think that's bananas. Which is OK.
A few years ago I tried something like you described. I inserted an additional heater into the inside of the same heater coil I use now. (The coil is on the outside of that heater). After some calibration I determined the temperature each could reach, in or outside of the ceramic tube, separately. Then I heated the outer coil, and set the internal one so it could not get as hot. The idea was that it could not add heat to the hotter coil, since that is counter to thermodynamics. It seemed like the power was sent in, and never materialized. Just lost, gone to nowhere. Weird.
Unfortunately, at the time, I did not have enough measurement equipment to do a good job measuring power in to both coils at the same time. It may be worth trying that again. However, experience tells me that somehow the heat did go in from the internal coil anyway. Or the power to each dropped accordingly. But maybe not. That was first experience of Insufficient Heat (the opposite of Excess Heat).
-
Jed, it seems you are missing my point. My (perhaps unorthodox?) point is this: It requires energy to send energy. Specifically, (in a cold, low energy environment) it seems to require more energy to send energy to a high energy object that to a low energy object. That's all.
That is not the situation we are describing here. No energy is going from low energy to a high energy (hotter) object. In all cases, the heat flows from the warmer body to the colder body, as it must according to the second law of thermodynamics.
When the test begins, the poorly insulated cell surface gets warmer at first, and heat flows from that surface to the cooler surface of the well insulated cell. If you leave the power on, after a while, they reach equal temperatures, and no heat flows either way. It all radiates into the surroundings. When you turn off the cells, the poorly insulated cell cools faster, and heat flows in the other direction, from well-insulated to poorly-insulated. Overall there is no net difference. The exact same amount of heat comes out of both, and there is no net increase anywhere just because some of it flows in one direction, and then the same amount flows in the other direction. The fact that you can put a heat engine in between them and concentrate some of the heat into high-grade energy is the same for all heat flows. Whenever there is a temperature difference and a heat flow, you can run a heat engine (if the difference is large enough to work with that particular heat engine type). The heat engine outputs high grade mechanical or electrical energy, but it also always degrades much more of the heat into lower grade waste heat. That's Carnot's law.
You are also wrong because heat can, in fact, be sent from a cooler body to a hotter body. With a heat pump, which is a heat engine run in reverse. Such as a refrigerator. It moves heat from the cold air inside the refrigerator to the warmer outside in your kitchen. Of course it takes more energy to do that than the energy content (enthalpy) of the warm air in the kitchen. Otherwise it would be a perpetual motion machine.
The second law is that heat cannot of itself go from one body to a hotter body. It can go, but not of itself. Meaning you can push it. But in the example of the two cells, well insulated and poorly insulated, there is no heat pump and no violation of the second law. Heat always flows from the hotter to the cooler in this example.
Perhaps you are thinking the thermoelectric device moves heat from the colder to the hotter. It does not, overall. Here is how you can think of this. 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. It was somewhat similar to this gadget, which lifts a small amount of water while being pushed by a much larger amount of water going downhill:
A thermoelectric device can produce a very small amount of high grade energy (electricity) but to do this it must create a far larger amount of degraded, colder, waste heat. Overall it creates no energy at all.
You could use the electrical output from a thermoelectric device to run a laser that heats a very small object up to much higher temperatures than the temperature of surface driving the thermoelectric device. You could vaporize a target. The temperature would be higher, but the overall energy release is far smaller. This is similar to using the energy from a 300 deg C fission reactor core to boil water, to generate electricity, which you use to run a 3000 deg C arc welder. The welder is hotter than the core of the fission power plant. But the fission power plant produces much more waste heat than useful, high grade electricity. It has to. That's Carnot's law.
