Heat Pumps

Jerramy · Feb 5, 2008

So I've been learning about Heat Pumps. By the investment of some work (in the classical sense), you can sort the entropy, with the added work landing in the temperature of the warmer object. The efficiency of this is maxed at (T1-T2)/(T1+T2). The closer the two temperatures are, the more efficient it is. Scale is in Kalvin, of course. So take a heat pump. When it is at room temperuature, it's efficiency is near infinite. At +100 degrees, it's still 2/3 efficient. Invest 1 joule of energy, and create a seperation of 100 degrees, cold on the outside, warm on the inside. Like a fridge, inside out. Heat is energy. Forget about the cold parts for now, and just look at the hot bit in the middle. If I throw water at that, it will make a hella lot of steam. Remember, I've already taken a fair bit of heat/energy out of the surrounding mass (many people bury these in the ground). All that heat is converted to energy. Steam generation can be up to 99% efficent, What comes out in the end is water droplets, at a degree above room temperature. So my question is, if I've taken all that heat out of the surrounding land, and converted it to electricity via steam generation, and all that energy flows down the wire, have I added to the total electricity that I got for throwing fossil fuel at it, or have I simply wasted it on my glorified resistor? It seems to me, that if the fuel is burnt, and the land is cold, that all that energy had to go -somewhere-. I think I organized it and cooled my environment to boot. A generator that gets too cold instead of too hot. Have I either: A. Violated a physical law, B. Invalidated a physical "law" (note the quotes), or C. Fooled my self into thinking something is possible when it ain't.

Jerramy · Feb 5, 2008

(I appologise for cross-posting. This belongs here.) So I've been learning about Heat Pumps. By the investment of some work (in the classical sense), you can sort the entropy, with the added work landing in the temperature of the warmer object. The efficiency of this is maxed at (T1-T2)/(T1+T2). The closer the two temperatures are, the more efficient it is. Scale is in Kalvin, of course. So take a heat pump. When it is at room temperuature, it's efficiency is near infinite. At +100 degrees, it's still 2/3 efficient. Invest 1 joule of energy, and create a seperation of 100 degrees, cold on the outside, warm on the inside. Like a fridge, inside out. Heat is energy. Forget about the cold parts for now, and just look at the hot bit in the middle. If I throw water at that, it will make a hella lot of steam. Remember, I've already taken a fair bit of heat/energy out of the surrounding mass (many people bury these in the ground). All that heat is converted to energy. Steam generation can be up to 99% efficent, What comes out in the end is water droplets, at a degree above room temperature. So my question is, if I've taken all that heat out of the surrounding land, and converted it to electricity via steam generation, and all that energy flows down the wire, have I added to the total electricity that I got for throwing fossil fuel at it, or have I simply wasted it on my glorified resistor? It seems to me, that if the fuel is burnt, and the land is cold, that all that energy had to go -somewhere-. I think I organized it and cooled my environment to boot. A generator that gets too cold instead of too hot. Have I either: A. Violated a physical law, B. Invalidated a physical "law" (note the quotes), or C. Fooled my self into thinking something is possible when it ain't.

richalex · Feb 5, 2008

Steam generation may be 99% efficient.. at producing steam! Even that, I really doubt. But, regardless, you are not going to get more than about half the energy out in electricity that you put in as fuel in a steam electric power plant. I'm saying that off the top of my head, based on my background as a steam propulsion tech (that was my job in the U.S. Navy; the title was "Machinest Mate"). The following link is to some global warming nonsense. I am including it here because it cites power generation efficiency numbers that are about what I am used to seeing: Global Warming from Electric Power Plants

richalex · Feb 5, 2008

I provided an answer for this over in Space Science & Astronomy. Someone else can nag you for posting the same post to multiple forums.

Jerramy · Feb 5, 2008

You're discounting it, offhandedly saying that I couldn't get more than half the energy out of it that I put into it. My question is, if it chilled the land around it, taking all the thermal/kinetic energy and converting it into turbine turning, can't I essentially "cheat", and take more power out than I put in? See here: http://en.wikipedia.org/wiki/Heat_pump My formula was wrong, the heat moved is related by the pump's COP, which can have values much higher than one (at infinity when they are the same temperature). This means that by the usage of 1 watt-hour, I've made something up to 4x as hot as if I had simply wrapped a resistor around it. So if I take that hot spot, which contains all the energy of the surrounding area, and generate power from it, such that at the end of the power extraction, I have a room temperature center, and cold surroundings, don't I have to say that if my general location is -colder- than it started, that that energy must have been added to the total amount that went down the line? This is a very different situation from when I burn coal to drive a steam engine, and end up with a burning hot engine and rather warm surroundings.

richalex · Feb 5, 2008

<blockquote>In reply to:<hr />You're discounting it, offhandedly saying that I couldn't get more than half the energy out of it that I put into it.<hr /></blockquote>You are the one who claimed that "Steam generation may be 99% efficient." On what do you base that claim? Where did you get that number? Much work has been done by several nations to extract as much power as possible from steam generated electricity. Some Russian invented a 7-stage process that extracts considerable work from the original system. But, the best systems don't produce anywhere near 99% efficiency in converting fuel to electricity! <blockquote>In reply to:<hr />My formula was wrong, the heat moved is related by the pump's COP, which can have values much higher than one (at infinity when they are the same temperature). <hr /></blockquote> COP = |Q| / W where Q is the useful heat supplied by the condenser and W is the work consumed by the compressor. If the temperature is the same on both sides, then you get a "0" on at least one side of the equation (both sides if the heat pump is turned off). This is scarcely impressive. <blockquote>In reply to:<hr />This means that by the usage of 1 watt-hour, I've made something up to 4x as hot as if I had simply wrapped a resistor around it. <hr /></blockquote>So, you expect that by putting 1 W*h in, you would get 4 W*h out? That only holds true if the heat pump can grab 4 units of heat from some place and move it somewhere else. The article you linked also mentions that the greater the temperature difference, the harder it becomes to pump the heat. If you want to extract energy from this heat, you might as well cut out the middle man (the heat pump) and just use the heat where the pump was getting it. Don't confuse temperature and energy. You may be able to move heat around, but you are not creating energy wit

wick07 · Feb 5, 2008

The closer the two temperatures are, the more efficient it is. Scale is in Kalvin, of course. I think you are incorrect here. As I recall from my thermo in college, efficiency increases as the difference between the two temperatures increases. I think you used the wrong equation, try: Theoretical Efficiency = {(Th - Tc)/Th} ×100% Th = temp of the heat source Tc = temp of the heat sink As you can see if the Th = Tc the the efficiency will be 0. Oh, and BTW, it's Kelvin. <div class="Discussion_UserSignature"> _______________________________ "If you are surrounded by those who constatly agree with you, then you're in an intellectual vacuum. If you feel like trying to make a difference, you have to BE different. How can you do that without interacting with those who are different from yourself?"- a_lost_packet_ </div>

Jerramy · Feb 5, 2008

Wow, you guys just jumped all over my a**. Guess you're right. So the formula I see on the wiki page is: COPheating = dQ/dA <= Thot/(Thot - Tcool) dQ is heat delivered to the hot reservoir. dA is dissipated work, hopefully dissipated to the reservoir. Also, I see: "a typical heat pump has a COP of three to four, whereas a typical electric resistance heater has a COP of 1.0" So for 1 Watt hour, my reservoir gets 4x as hot as if I had simply wrapped a resistor around it. I have a container, which is cold on the outside, and hot on the inside. I drop it in a bath, which warms up the outside, making the bath cold. That was easy. Now I have a cool (not cold) object, which is considerably warm on the inside. Take that away (and use it somewhere). The bath of water. It's cold. It used to be warm, but now it's cold. It has less energy than it did before. I know you're saying don't confuse energy with temperature, but there is less molecular kinetic energy in the bath. If my heat pump is as perfectly efficient as it can be, then I have to ask, have I not stolen energy from the background (room temperature), and displaced it to be used elsewhere?

Jerramy · Feb 5, 2008

"Theoretical Efficiency = {(Th - Tc)/Th} ×100% " Oh, I had to correct someone on that just the other day. Think about it. It's easy for a heat pump to take two things that are the same temperature and make them slightly different in temperature. It's really HARD to make one side hot as the blazing sun, and the take the other down to absolute zero. That efficiency relates to how efficient it is to -remove- energy from a heat source. It's very EASY to boil a cup of tea with a red hot poker, it's hard to boil it with your hand. The poker is more efficient at heating the tea than your hand.

origin · Feb 5, 2008

If my heat pump is as perfectly efficient as it can be, then I have to ask, have I not stolen energy from the background (room temperature), and displaced it to be used elsewhere? Sure, but you have not violated any laws of nature. You have put work into a system to move heat from one area to another. Think of a refrigerator. You are removing the heat from the low temperature of the inside and transfering it to the outside of the refrigerator using a heat pump. No magic here. A typical refrigerator has a COP of 5 or 6. The Clausius Statement: "It is impossible for a refrigerator (heat pump) working in a a cycle to produce no other effect than the transfer of heat from a colder body to a hotter body." In other words you always have to put energy into the system (the compressor (W)) so you cannot have a COP that is infinite. If the COP was infinite that would mean in principle you could have your refrigerator cooling with out plugging it in. Heat pumps are very good method of heating a house (unless it gets too cold) but they are very well understood and do not in any way violate any natrual laws. <div class="Discussion_UserSignature"> </div>

MeteorWayne · Feb 5, 2008

More correctly, a heat pump can move heat from one location to another. My heat pump heats us in winter when it's cold outside, and cools us in summer when it's hot outside. This shows that it can move heat (using power) when it is moving heat from a cold spot to a warm spot. If the temperatures are close (within 20C/36F) it's efficient. For +/- 30 C (54F), it's about breakeven, but runs a long time to move the heat in the wrong direction. Considering our electric prices, it's probably very low BTUs for the buck. It would intuitively be extremely efficient at moving heat from a warm spot to a cold spot. In any case juice runs the compressor (which moves the heat) and the fan (which distributes it). <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

origin · Feb 5, 2008

A heat pump by definition moves heat from a cold area to a warm area. In summer it moves the heat from your (relatively) cool house to the hot outside and in winter it moves the heat from the cold out side to the warm inside. If you want to move heat from a hot area to a cold area you don't need a heat pump just open a window. <div class="Discussion_UserSignature"> </div>

MeteorWayne · Feb 5, 2008

Really, it's a heat pump. It can pump in either direction. I didn't say moving heat in the ambient direction would be a good idea, just that it would be extremely efficient <img src="/images/icons/smile.gif" /> My point is that it's a pump that can move heat in either direction. <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

richalex · Feb 5, 2008

<blockquote>In reply to:<hr />Also, I see: "a typical heat pump has a COP of three to four, whereas a typical electric resistance heater has a COP of 1.0" So for 1 Watt hour, my reservoir gets 4x as hot as if I had simply wrapped a resistor around it.<hr /></blockquote>The resistor converts one form of energy (electricity) to another (heat). The heat pump (mostly) is just moving energy, not converting energy from one type to another. Of course, the heat pump has to use energy to move energy. <blockquote>In reply to:<hr />If my heat pump is as perfectly efficient as it can be, then I have to ask, have I not stolen energy from the background (room temperature), and displaced it to be used elsewhere?<hr /></blockquote>You appear to consider objects merely as "hot" or "cold," with no consideration of the quality of that heat or cold. You did this with your original steam example, which is why I wanted to revisit this thread (I feel I did not explain that aspect). Not all steam is equal in power. Steam that forms over boiling water at sea level (say 200° C) is very weak; it does not have very much energy in it, so cannot do very much work. If you want to get more work out of it, you need to put more energy into it. The Navy uses steam that is at least 400° C and 4000 kPa, and sometimes double that. Just barely getting a phase change or a small temperature difference is not sufficient to perform large amounts of work.

origin · Feb 6, 2008

MeteorWayne, we appear to be in violent agreement. <img src="/images/icons/smile.gif" /> <div class="Discussion_UserSignature"> </div>

siarad · Feb 6, 2008

Without wading through the topic, a rule of thumb is you get 3x the energy out you put in. A compound steam steam engine can have an efficency of 44% equal to the best Diesel & a dynamo 90%, varies. So you appear to be back to square one after taking into account mechanical & heat losses. Far better to use a Stirling engine driven by the heat. I seem to racal a company does this try Google <img src="/images/icons/smile.gif" />

wick07 · Feb 6, 2008

Sorry, you are correct. Looking over what you are trying to do I see that I was looking at the wrong way. I was giving the thermodynamic efficiency of a heat engine using a differnce in heat to create work, you are talking about moving heat from one location to another. <div class="Discussion_UserSignature"> _______________________________ "If you are surrounded by those who constatly agree with you, then you're in an intellectual vacuum. If you feel like trying to make a difference, you have to BE different. How can you do that without interacting with those who are different from yourself?"- a_lost_packet_ </div>

siarad · Feb 7, 2008

A B C No Look at this patent <blockquote>In reply to:<hr />so as to produce mechanical and/or electrical energy without consuming fuel, by extracting thermal energy from the surrounding medium - air or water<hr /></blockquote>

billslugg · Feb 7, 2008

Unless he has someplace cooler than the surrounding medium, in which to reject the heat, then he is just fooling himself. <div class="Discussion_UserSignature"> </div>

MeteorWayne · Feb 7, 2008

Again that's not quite true. A heat pump can pump against the gradiant, using energy. My heat pump is taking heat from outside (0C) and moving it to inside (17C). <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

billslugg · Feb 7, 2008

Wayne No, I am referring to the patent cited above, in which the inventor claims that a mixture of heat pumps, heat exchangers, motors and fans can run over unity by extracting heat from the ambient. In other words, he has a machine that sits there, extracts energy from the environment and pumps electricity back into the power system. You can't do that unless you have somewhere colder to reject the heat to. EDIT: In the case of your heat pump, you take some energy from the power company and push a bunch of BTU's uphill into your house. If you then let those BTU's fall back down hill, you can regain exactly the energy you got from the power company, but no more. This guy claims he can get more and do it without a cold sink. This is not possible. <div class="Discussion_UserSignature"> </div>

MeteorWayne · Feb 7, 2008

Right. We've kind of had two heat pump threads going on here, one real and one imaginary <img src="/images/icons/smile.gif" /> <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

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