Funny, I was just looking at those, since we used to use those in spacecraft when I worked in aerospace a million years ago. I think that probably answers my question.
Unless I’m misreading the article those are about the opposite phenomenon, i.e. producing a heat differential by applying a voltage. But if Wikipedia is choosing to represent the names as refering to just one direction and they’re often used both ways I wouldn’t be surprised.
These analogies easily break down, but to make an equivalent model for the system using an electric generator you have to introduce the same kind of energy reservoair and sink as you have in the water dam or thermoelectric couple. And you have to think about what role internal resistance plays. And it’s possible an electric generator is just a bad comparison overall. They’re not my field of expertise (not that anything in this thread really is).
A short circuit is a light load. It means using the energy faster and wasting more of it on the loads in the system itself. It’s equivalent to the freely spinning water wheel.
You can remove the hot/cold distinctions, because a thermocouple works either way
You need to flow current around the loop if you want to take heat energy away from the hot junction.
If the other ends of the two dissimilar wires are open-circuit, then no current flows, and there’s no cooling effect except for ordinary thermal conduction; you just get a voltage across the two leads.
If the other ends of the two dissimilar wires are connected to each other, then you have a short-circuit, and current will flow around the loop, with magnitude subject to the electrical resistance of wires. There will be ohmic heating all along the wires due to the current, but the cold junction will also warm up as electrons deposit their energy there, dropping to a lower voltage as the flow across the interface between the two dissimilar metals. The energy to warm that junction comes from the cooling of the hot junction in your coffee cup: it will be cooler than the coffee, drawing heat from your coffee (separate from thermal conduction down the wires) to keep the electrical current going.
If the other ends of the two dissimilar wires are connected across an electrical load, then the heat from your coffee manifests as electrical power delivered to that load. As above, the hot junction will be cooler than your coffee, and it will pull heat from your coffee (separate from thermal conduction down the wires) to power that electrical load. This is the basic design for a thermoelectric generator:
In answer to your original question, yes, if your cup of coffee is powering a thermoelectric generator that’s delivering electrical power, it will cool your coffee faster than if the electrical load is removed. “Electrical load” here refers to not just your cell phone, but also the charger itself, i.e. any electronic components comprising the switched-mode power supply that is required in order to supply your phone with a well-regulated charging voltage/current.
I’m pretty sure this is wrong.
A short-circuit in an electrical system is a low-resistance connection, resulting in the circuit delivering the maximum electrical current it’s capable of providing. It’s the heaviest load the circuit can possibly deliver, and it manifests as heat in the circuit wiring itself due to ohmic heating, which is why short-circuits are dangerous.
The water wheel is a questionable analogy, as its upper speed will be limited to the speed of the falling water, and when the speed of the water and the speed of the wheel are matched, then there’s no mechanical power being delivered to the wheel’s spindle. Energy isn’t being deposited anywhere in the system at that point.
I interpreted @naita 's comment to mean that a short circuit may induce excessive currents and Ohmic heating in your wiring, but is not delivering any useful power to a load (which in certain situations requires impedance matching, or think how much power is delivered by a DC motor that is spinning freely or stuck fast, versus a situation with intermediate RPMs)
Yeah, this is how I messed up describing the load. Most electrical systems operate close to voltage constant, in which case a short circuit represents the maximum use of energy.
Conversely, if you attach a Peltier element to a battery, without any heat sink, coffee, etc attached, I imagine that it will fry itself and self-destruct rather quickly.
Here’s a technical spec sheet for a peltier-seebeck element by the way. Note figure 5. which shows the response with a constant 100 C temperature difference. Maximum power is at 1.8 ohm. Lower than that and the voltage drops faster than current increases, and vice versa.
If you short circuit it you top out at 3A, but no power out (except the element itself will heat up), with an open circuit you get north of 5 V, but of course no current. In the first case you will still have a flow of heat, even if you’re not getting much electrical power, in the second the element isn’t contributing to the transfer of heat, except for it being a chunk of material and not a reflective vacuum bubble.
What you need is a small demon to sit on the rim of the cup and block the fast moving molecules from escaping while allow all of the slow moving molecules to leave. Get back to me when you’ve figured out how to conjure that and I’ll write up the patent application.
Stranger
It sure is. I had a constant temperature viscosity bath that I used a Peltier cooler to control. It was originally designed to use tap water for cooling, but the damn tap water was over 25 deg C during some of the year so I had to readjust. Peltier coolers are good for that, they don’t move a lot of joules, but the bath was close enough to room temperature to only need a few removed. I just set the Peltier to cool as much as possible, then back titrated the temperature in a manner of speaking with a microprocessor controlled heater.
Yes, for the reason I stated. Charging a phone needs energy which has to come from somewhere and in this case it’s the hot liquid. That means that it will cook faster than if the energy drain wasn’t there. adding a metal spoon to the example is an unnecessary complication.
You could argue about the thermal conductivity of the container too, but the point is that yes, the liquid will cool down and yes, it will cool down faster if something like a thermocouple takes energy from it.
I will have to go back and check, but one thing I remember is that dual metal (thermocouples) had low internal resistance and could give high currents with low voltages (the voltage from temperature is very small).
I think you’re missing the point. I get from other explanations that it does speed the cooling of the coffee in this case. But no one was proposing that it was a case of getting energy from nowhere. It is clear that it comes from the coffee.
There are plenty of examples of capturing energy that would otherwise be lost. It doesn’t mean the energy comes from nowhere. Regenerative braking, e.g., where the recaptured energy occurs as part of the desired effect. Another example I recently saw is micro-hydroelectric turbines installed in a water treatment plant as part of the pressure-reduction step. The effect of the turbines of slowing the flow is a desired effect. That isn’t energy coming from nowhere, it’s energy that would otherwise just be lost to the environment.
Saying the energy has to come from somewhere is missing the point. The whole point of the question was whether the energy was all or mostly coming from what would have otherwise been lost to the environment, but could be recaptured (or, more accurately, not lost but redirected – a desirable effect) or whether the thermocouple would be extracting heat from the coffee faster, which would be undesirable to most coffee-drinkers.
You seem to be responding to a different, more ignorant question.
No one is saying you can come out ahead energy wise. But you can create a more efficient system by capturing energy that would otherwise be lost.
Physicists / Scientists / Engineers has the same confusion with early electrical systems: maximum power transfer is when the load impedance is matched to the source impedance, right? So the best a power generation system can do is 50% efficiency? Part of Edison’s success was the insight that he wasn’t trying to get the maximum power out of the Coffee into the Phone, he was trying to get power into the Phone at minimum cost.
A generator is easiest to turn when open circuit. It transfers at least cost when there is a very large impedance connected, almost like an open circuit. It transfers at maximum rate when half the energy goes to the load, and half lost to local heating. All of the energy is lost to local heating when the load is short circuit. It would be back to easiest to turn when the load is s/c and the generator is also perfect, lossless and s/c, with no energy storage, but that’s not a real thing.
If we had a perfect thermo-electric generator, energy would flow from the coffee to the phone as fast as the phone would accept it, and the coffee would get cold as fast as the phone charged.
If we have a normal / lossy TEG, energy will flow from the coffee to the phone as fast as the generator will send and the phone accept. Losses in the TEG will just make it warm, which heat will just run back to the coffee (I assume lossless transmission, pre-heated insulated cup).
Since loss in the TEG has no effect on the Coffee, you should run the TEG in lossy mode, exporting power to the Phone as fast as possible, ignoring TEG loss which is retained in the Coffee.
Sending power to the Phone as fast as possible makes the Coffee cold.
Of course, sending any power to the Phone makes the coffee cold, but I’m talking about the design of the system: the faster it makes the coffee cold, the faster it charges the phone.
In a real system, you’d also have to worry about warming the cup and TEG, insulating the cup and TEG, losses in the transmission system (wires) and in the charger, which would change the optimum operating point, you’d want to increase the impedance of the load to exceed that of the loss elements outside the cup.
@eschrodinger, yes, exactly. Everyone else – thanks for your patience with me, and it’s clear to me now that the coffee will cool faster, since the thermocouple will actually extract heat faster than a regular thermal conductor if it is powering an electric load.
Resistive heating in the system would partly be going back into the coffee where it’s doing no harm, but it would also be partly going into the outside of the cup. And heat going to the outside of the cup is just a loss all around: It’s cooling down the insides, but not charging the phone at all.
I did a very quick back of the envelope estimate.
It takes 4.2 Joules to raise the temperature of 1 gram of water 1 degree Celsius
A pint of water is about 474 grams.
Let’s assume a big cup of coffee (say…400 grams of water) near boiling (100 C) in a cool room (20 C). So, about 80 C of temperature difference.
So there is 4.2*80*400 Joules available to charge a phone. About 135,000 Joules.
So now we need to take the end to end efficiency of conversion of the energy in the hot water to energy into the cell phone battery. I’m sure someone can provide a detailed end to end analysis, but I’ll just guesstimate a best case efficiency of 10% (even laboratory best thermoelectric junctions have <50% efficiency IIRC, especially in a temperature range as low as this case; there will be loss in the regulation of the charging; and the loss of energy to anything other than the junction, like cooling of the surface of the liquid exposed to the room). So maybe 14,000 Joules available for charging.
So not enough to charge a cell phone from zero to full (~30,000 Joules), but enough to get 30-40% charge.
But…as the coffee cools, the efficiency of the thermoelectric junction will fall, so my estimate may be an absolute best.
So, my final guess is less than 10% charge available from a cup of hot coffee, probably single digit percentages of charge.