# Thermodynamics and conservation of energy

• Roy Green
In summary: The 100 kW is thermal energy moved from cold to hot. it costs 25 kW to transfer the energy because it is increasing the thermal gradient. If what you said was true, heat pumps would not exist as you may as well use a convection heater.Normally if you then run a heat engine you'd get back less than the 25 kW, but because I am adding extra heat from the industrial process the heat engine efficiency increases.
Roy Green
I have a question about conservation of energy in my system

A heat pump allows you to move a quantity of heat with a smaller energy input. It is most efficient for small temperature differences.

A heat engine allows you to convert a temperature difference into mechanical work. It is most efficient for large temperature differences.

Since their efficiencies are inversely proportional, if you use a heat pump to create a temperature gradient you won't be able to get as much energy back out with a heat engine.

Now imagine we have 500 kJ of high temperature thermal energy from an industrial process. The heat was 'created' through regular heating, let's assume at 100% efficiency for simplicity. The industrial heat drives a heat engine and the temperature difference is such that we get a 50% efficiency, i.e. 250 kJ of mechanical power out.

The heat from the industrial process is added via a heat exchanger.

Let's say we preheat the heat exchanger via heat pump, but by only a small amount and therefore at high efficiency e.g. COP=4. e.g. we have added 100 kJ heat at a cost to us of 25 kJ .I'm guesstimating the numbers for the purpose of the explanation, but we now add 400 kJ from the industrial process so that there is 500 kJ in the heat exchanger and get the same power output of 250 kJ, of which 20% is attributed to the preheating.

In the latest scenario instead of getting 250 kW mechanical power out at a cost of 500 kW, we achieve the same 250 kW at a cost of 425 kW. We have therefore increased efficiency. What I am worried about is that when I look at the preheat on its own, it feels like we have violated the conservation of energy. Can someone please provide explain either where the concept breaks the laws of thermodynamics or if not, where does the energy come from?

Don't get too stuck up on why we are doing this, I've left out the application because of IP owned by someone else.

You want to "fire" a boiler with a heat pump, and you are calling the "exhaust" heat from the heat pump a "free" energy source. It does NOT work that way. The heat pump will move heat from low T to high T, but if you do not furnish it a low T energy source, it cannot move anything.

Please read again, you've not understood it. I didn't mention free energy once. I am trying to figure out if I can increase efficiency by preheating the heat exchanger with a pump because I can add 100 kW of heat at a cost of 25 kW.

I'll then be adding heat from a higher temperature industrial process, it's not waste heat but it can be thought of on a similar basis, though I can't elaborate on why I am doing this.

Since the heat engine is running between ambient and a high temperature (much higher than that of the preheat), I get 50% efficiency. By first preheating with a heat pump rather than doing all the heating with the heat from the industrial process, it would appear that I increase efficiency above 50% (not sure where you got the idea that I was trying to get it above 100%, I am trying to reduce losses).

Obviously the basis for the operation of the heat engine is that it takes energy out of the temp gradient between the hot heat exchanger and ambient, so the heat exchanger cools down.

I'm asking because I may have missed something, as something about it feels odd to me, but you've not explained why it won't work.

Roy Green said:
because I can add 100 kW of heat at a cost of 25 kW.
No. You have to provide the other 75 kW the heat pump is pumping; your 100 kW preheat costs you to 25 kW to run the heat pump plus the 75 kW it pumps.

Bystander said:
No. You have to provide the other 75 kW the heat pump is pumping; your 100 kW preheat costs you to 25 kW to run the heat pump plus the 75 kW it pumps.
The 100 kW is thermal energy moved from cold to hot. it costs 25 kW to transfer the energy because it is increasing the thermal gradient. If what you said was true, heat pumps would not exist as you may as well use a convection heater.

Normally if you then run a heat engine you'd get back less than the 25 kW, but because I am adding extra heat from the industrial process the heat engine efficiency increases. The conversion of heat from the industrial process to mechanical work is 50% efficient. The preheat is meant to increase efficiency slightly.

Roy Green said:
The 100 kW is thermal energy moved from cold to hot.
You meant "75 kW" rather than 100. What happens to your cold reservoir as you pump from it at 75 kW?

Bystander said:
You meant "75 kW" rather than 100. What happens to your cold reservoir as you pump from it at 75 kW?

The heatpump compresses gas, which heats it. the heat is exchanged with the heat exchanger. Gas is then expanded, it cools to below ambient. The surrounding air heats it. The air can be considered an infinite source for practical purposes.

Let's back up and start again. This looked to me (not to anyone else that I'm aware of) on first (and second, and third reading like a Rube Goldberg end run).
Roy Green said:
Let's say we preheat the heat exchanger via heat pump
Then the outlet T of the unspecified process is elevated by however much results from the preheat.
Roy Green said:
where does the energy come from?
The heat that has not been removed from that original process stream is the unaccounted heat in your heat balance/conservation of energy, which on "nth" reading is what bothered you enough to request another view of the idea. I was answering an entirely different question from what you asked --- my bad.

The "industrial process" is waste heat, right? So instead of getting 500 kJ of free energy you are getting 400 kJ of free energy and paying for the other 100 kJ. That's not better, it is worse.

Also, and maybe this is more germane to the question, you are not adding the heat energies at the same temperature, so you can't take credit for the same efficiency of the output from them. You've kept the efficiencies separate on the input, so keep them separate at the output: what happens to the efficiency of your use of the "industrial process" energy if you reject the heat form it at a higher temperature?

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Bystander said:
Let's back up and start again. This looked to me (not to anyone else that I'm aware of) on first (and second, and third reading like a Rube Goldberg end run).

Then the outlet T of the unspecified process is elevated by however much results from the preheat.

The heat that has not been removed from that original process stream is the unaccounted heat in your heat balance/conservation of energy, which on "nth" reading is what bothered you enough to request another view of the idea. I was answering an entirely different question from what you asked --- my bad.
Ok thanks, I can see how you misunderstood, I asked the question because I was concerned there might be a Rube Goldberg paradox and was trying to see if someone else saw one.

Did I figure out correctly what your question was?

russ_watters said:
The "industrial process" is waste heat, right? So instead of getting 500 kJ of free energy you are getting 400 kJ of free energy and paying for the other 100 kJ. That's not better, it is worse.

Also, you are not adding the heat energies at the same temperature, so you can't take credit for the same efficiency of the output from them.
But its not waste heat, I said you could treat it as if were but I can't say exactly what the system is used for and therefore using 400 kJ instead of 500 kJ of the 'waste heat' is a saving.Could you elaborate on your second point. In my design the preheat is added at a lower temperature than that of the final temperature and so both the heat pump and heat engine will be efficient. Unfortunately as the heat engine runs and the heat exchanger cools, efficiency will fall from 50%, but this will happen by the same amount regardless of whether i preheat since the target temperatures are still the same.

Bystander said:
Did I figure out correctly what your question was?

I think so but by this stage I'm getting confused

Roy Green said:
I think so but by this stage I'm getting confused
I think what Russ was pointing out is the following:
If the "waste" heat from the heat exchanger would have been dumped into a cooling tower/whatever, literally wasted, you pick up whatever you get from your heat engine for "free." If you throw in another 25 kW to drive a heat pump that in some ways degrades performance of the heat exchanger, you still get the heat engine output, but you've invested another 25 kW to power it.

Roy Green said:
In my design the preheat is added at a lower temperature than that of the final temperature and so both the heat pump and heat engine will be efficient. Unfortunately as the heat engine runs and the heat exchanger cools, efficiency will fall from 50%, but this will happen by the same amount regardless of whether i preheat since the target temperatures are still the same.
The output is the same total, but the outputs from each process are different. You are assuming each is 50% efficient in providing you output, but they aren't the same. If you keep them separate on the input you must keep them separate on the output and calculate your efficiencies for them separately. And you haven't calculated the effect of rejecting heat from the high temp process at a higher temperature -- that's where your extra loss is that you haven't accounted for.

I suggest that the OP should back off a bit and consider what he is trying to accomplish.

If you are trying to analyze a complex system, you should assume that energy is conserved, and use that principle to aid in your analysis. Anywhere where energy conservation appears to be violated means that the analysis must be in error. That applies to the system as a whole, and to subsets of the system if you draw imaginary boundaries around them.

If you are trying to prove that energy conservation really works, then you should reduce your experiments or analysis to the smallest simplest system you can imagine. Indeed, a single particle's interactions. Then you can use higher level of abstractions to simplify larger systems. If energy is conserved in every possible elementary particle comprising a system, then it is conserved in the composite system.

But it makes no sense to dream up a system so complex that I confuse myself, then to post a non-mathematical non-graphical verbal description of it to a PF forum, and ask others to point out my error.

## 1. What is the definition of thermodynamics?

Thermodynamics is the branch of science that deals with the relationship between heat and other forms of energy, and how it affects matter.

## 2. What is the first law of thermodynamics?

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only transferred or converted from one form to another.

## 3. How does the second law of thermodynamics relate to entropy?

The second law of thermodynamics states that the total entropy of a closed system will always increase over time. This means that energy tends to disperse and become more disordered, leading to a decrease in usable energy.

## 4. Can energy be considered a conserved quantity?

Yes, energy is considered a conserved quantity in thermodynamics. This means that in any physical process, the total amount of energy remains constant, even if it may change forms.

## 5. How does thermodynamics apply to everyday life?

Thermodynamics is present in many aspects of everyday life, from the functioning of household appliances to the processes that occur in our bodies. It helps us understand how energy is transferred and transformed, and allows us to make predictions about the behavior of systems.

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