Engineered coiled condenser

[engunear]

This thread is about building a coiled condenser that is as light and efficient as possible. The plan is to measure and so optimise the heat transfer properties of the coil, build a condenser and see if it works as designed. This is in contrast to the proven method of guessing and adding a safety margin.

It started as a rat-hole in a thread "Do turbulators really work" http://homedistiller.org/forum/viewtopi ... 87&t=57690

[pfshine]

There is a calculator for that on the parent site.

[rad14701]

There is a calculator for that on the parent site.

That calculator has been proven to be more than a bit inaccurate, which is most likely why this discussion was started, and the one engunear linked has progressed... Nobody runs liebigs as long as the calculator indicates as being required...

[Edwin Croissant]

This thread is about building a coiled condenser that is as light and efficient as possible. The plan is to measure and so optimise the heat transfer properties of the coil, build a condenser and see if it works as designed. This is in contrast to the proven method of guessing and adding a safety margin.

Like this?

[engunear]

Yeah the calculator is interesting. Although I don't think it says so on the page, the calculator is workable for coiled condensers; the maths is the same. But the catch is the HTC value. It talks about ranges of HTC from 150 to 850 W/sq m/C; already a ~ 6:1 range, and my measurements may be screwed, but the lowest measurement I've made is 2,000 W/sq m/C and the highest is 20,000. So its garbage in, garbage out on the calculator. We need values of HTC we trust so we can then use the calculator with confidence. Does HTC vary with: pipe diameter? flow? the material used in the pipes? If there are multiple pipes are they better in parallel or series?

The other problem with the calculator is that it gives no thought to safety margin. You can have safety margin in flow, and you can have safety margin in length. Which is better and why?

Anyone trying to make sense of this thread should read this http://homedistiller.org/forum/viewtopi ... urbulators

At this point, in summary, to measure HTC I have piece of ~5mm OD copper pipe poked through a ~30mm PVC pipe that is driven from a boiler. 29mm of the pipe is exposed to the steam. You can calculate the power absorbed as the temperature difference x flow x specific heat water. You can calculate the average temp along the pipe as the average of the inlet and outlet temp, and knowing the surface area of the pipe calculate the HTC. (BTW, there is an approximation in the maths here.) By varying the flow I can measure HTC vs flow.

So two experiments this weekend. One was to flatten the pipe so its cross sectional area dropped by to about 65% of its previous value. My figuring was that would move the water closer to the walls and increase the velocity - both good. The second to run the flow down lower to see what is down there, after Martimer's interesting question about zero flow.

See graph. So the flattening made no difference.

But what is weird is the set of low flow points. What the hell are they doing? It was all nice and linear ... and then this. I'm a bit stumped. I've measured these points with three different setups and thats what they are. At first I had trouble getting the flow that low, and they are below the point where all the steam condenses, so it was venting and I thought that maybe I was getting steam on my thermometer. So I set up a second flow system just for the experimental pipe (a bucket and a siphon) so I could run without venting steam and put an insulating sleeve on the thermometer ... and it didn't make a damn bit of difference. I don't believe those points, so there is something screwed somewhere. The temperature difference along the pipe is 25 degrees for the lowest flow, so maybe that is having an effect (conduction down copper?). Or maybe my Python is wrong. Measurement is a bitch, anyone tells you its easy, they are a liar or a fool.

[raketemensch]

God I love this place. engunear, if you'd like a second set of eyes, I'm also a python guy.

[engunear]

Hi raketemensch, I nearly posted the raw data (and if anyone wants it let me know). But first I plotted it without manipulation (see graph) and thought "that looks remarkably sensible". If the line for the black, red and blue points had continued it would make no sense, because at zero flow the pipe will heat to the boiling point, but the extrapolation pegs out at 7.5 degrees or something.

So then I wrote up the maths behind the calculator (t was on scraps of paper, a minefield of possible errors), to see what it says. Its attached FYI (definitely only for consenting geeks in private). It points out that the equations are not defined for flow=0, or maybe converge to some shitty limit that is beyond my time budget to care about. Its odd how it looks like there is an angle in the data rather than a smooth transition between modes. (By the way, these equations match the website calculator when plotted together.)

In conclusion, I now trust the measurements at higher flow rates. Don't know where the trust/distrust boundary should be and am probably going to pick a putative operating point and start analyzing. Its only a frigging pipe, and the worst that can happen is I get the length wrong.

An interesting thing in that plot is that the points for the thick and thin pipes are indistinguishable. So presumably the increase in flow velocity offsets the smaller area. This is definitely useful as it says that if you can keep the flow up (against water friction) make the pipe as small as possible.

[Maritimer]

Hi engunear,

The limit of (e^(-1/f))/f as f->0 is 0. I used Mathematica. If you are interested in using math to explore things, I can't stop recommending using a computer algebra system. Impossible math becomes trivial. When I first started using it (after retiring), I thought that it was like having a graduate student at my beck and call to do all the hard math. So the temperature gradient along the pipe at zero flow, is, as expected, zero.

Sorry if I've set you off on a wild goose chase, but, on the other hand, you have really come to a good understanding of your system!

M

[LBHD]

So, assuming you can get 1/4 inch to work with your pump/tap pressure, it is not only the easiest to bend, but cheapest to buy and most effective for heat distro.

Sounds like I will be picking up 1/4 instead of 3/8 for my coil

[engunear]

Hey Martimer, thanks for that advice. I'm too much and old fart and head to the paper and pencil rather than the internet. The gradient in practical terms is non-zero though as we start with the water coming in at Ti and it leaves at Tb. What is happening is that we have to measure over a piece of pipe where temperature change is small (2-3 degrees I guess) to approximate the differential so the maths works. But the 29mm piece with its 25C difference is too long.

Rather than build a new test setup I just assumed the linear equation for HTC worked to zero flow.

Combining the equation for HTC with the exponential equations for the water in the coolant pipe heating (the website calculator equations) leads to the attached graph.

In summary the website calculator is fine, but it needs an HTC value that is right. HTC varies linearly with flow. A good method for measuring HTC with a Leibig would be helpful, but thats off topic for here.

Time to calculate the operating parameters methinks.

[Maritimer]

Hi engunear,

After eq. 10 in your pdf, you write:

This looks complicated but it has a simple interpretation. The water starts at Ti and absorbs heat from the surroundings at a rate proportional to the difference in temperature between itself and the steam. As it heats it absorbs less as the difference decreases. It asymptotically approaches the boiling point at which it absorbs no more. As the temperature rises exponentially, the heat absorbed falls exponentially. The hot end of the condenser is less effective than the cold end.

I wonder about this. You are using steam as your power source, so the heating of the coolant would be gained at the expense of condensation of the steam, which would involve changing its latent heat of vapourization into sensible heating of the coolant. There would also be an additional heating, which is described by eq. 10. This is the cooling of the condensed vapour by the coolant, which would depend on the temperature difference. But this temperature difference would be much less, because the heating of the coolant due to the condensation would be considerably more.

(I'm just trying to keep the conversation going!)

M

[engunear]

No problem about keeping this running. Its got a way to go yet. I'm still working through what the linear HTC relationship does to things overall. What does it mean? On the (off) topic of equation solvers, we had a lecturer once who, halfway through some hairy derivation, would choose a student at random and ask them what the line actually meant. Rather than algebraic manipulation being a sausage machine that produces an answer, he took the view that every step was a piece of its own logic with its own conclusion and its own lesson. It was scary being in the firing line, but we learnt a lot. You can take it too far, but there are deep lessons that a solver skips.

Your comment is right, the cooling of the condensate does add an extra term that I'm ignoring in this analysis. A Leibig often uses a single unit for both, and it is not well designed for liquid cooling, so they get kinda long to do a good job. If you look at some offset designs, e.g. Ian Smiley's, he has a short tube-in-tube system to cool the distillate. I've built two of those neither of which work very well. So I'm thinking about designs for a compact cooler. My current thinking is a small bottom-fed, top-bled reservoir with a cooling coil will work well.

[Maritimer]

Hi engunear,

The first difficulty I experienced in using Mathematica was using the screen and keyboard instead of paper. After a lifetime of writing expressions without any thought of what and how to write, the keyboard and screen became an impediment to clear thinking.

But putting up with that problem revealed the wonder of having so much power. Mathematica isn't just a problem solver, it actually contains vast mathematical knowledge. And attached to this is "Wolfram Alpha", a curated computational knowledge engine.

This is sounding too much like a sales pitch. But if you really are an old fart (I'm 68, figure), and you'd like to massage the brain synapses, there's nothing quite like playing around with Mathematica.

M

[engunear]

I can imagine a few readers of this thread wondering where the hell are we going. Oh well. But I'd like guessless model of a condenser, preferably closed form. Feel like working some math?

There are three problems I'm mucking with.

1) The linear approximation of the HTC is probably good enough, but the H0 H1 values are wrong. The data has got that kick in the tail at low flow which is messing with the H0 and H1 approximation. So how should we find H0 and H1? Do a least-squares fit of the data with f(H0,H1)? Here is the raw data. These are temp time to fill a bottle vol seconds, with Ti inlet temp. These are Python lists (if you don't know Python or an object-oriented language, thats another thing to sharpen those grey things. Python is way cool. Numpy is used by the US met office. Its free.)

2) Since the points for different diameters seem to fit on the same line, what is the general equation? It won't be right for all, but we can cover the range that people are using.

3) How does one best add safety margin? If you have an infinitely long condenser, you can work out the flow below which the thing will vent. But when it is shorter, how does one solve it? For a finite length, what is the flow that is exactly 2X the flow at which it vents? What is the solution for NX?

Since time is money and money is power, you can measure time in dB. I'm less than 0.5dB younger than you, so in real terms we are the same.

vol=0.75
Ti=18.2
dia=5.5e-3
points=[[21.2,26],[20.7,22],[22.8,60],[22.7,62],[21.7,41.6],[24.6,108],[23.9,74],[22.6,52],[22,45],[21.5,34]] # Temp time

vol=0.75
Ti=18.2
dia=2.7e-3
points=[[20.9,44.0],[23.7,68.0],[22.5,72.0],[22.0,43.0],[21.0,20.6],[20.6,19.0],[22.0,89],[23.2,118],[21.3,29],[22.0,49],[21.8,46]] # Temp time

vol=0.75
dia=2.7e-3
Ti=19.3
points=[[22.3,28.6],[22.1,32],[22.3,40.9],[22.9,61],[23.9,98],[26.2,173],[24.6,132],[25.8,132],[42.1,330]] # Flattened.

vol=0.19
dia=2.7e-3
area=math.pi*length*dia
#points=[[30.2,72.2],[32.4,79],[39,108]] # Flattened, slow. I rejected these because I didn't believe them at first.
points=[[28.9,49.0],[45.1,107],[41,98],[33.8,73]] # Flattened, slow
Ti=19.7

Pipe length is 29mm.

[Maritimer]

Hi engunear,

I'll look into it.

61? That's a little young for old-fartness.

What are H0 and H1?

Did you download A Heat Transfer Textbook?

M

[engunear]

Having had a pick at Heat Transfer Textbook: a bit of math is is fun, but this is too much effort without getting paid. He takes to opposite approach to textbooks to Feynmann whose books are short on maths and long on what it means. Maybe you can find some pearls in there?

However, the example on P351 for flow in a 1mm dia pipe is spot in the middle of what we are talking about. He gets an HTC of 2,778 W/m2/K for a 1mm tube with a flow of 0.2m/s. This has a Reynold's number of 360 (laminar). For 2.7mm dia pipe, minimum flow for a 2kW condenser has a flow velocity of 1.6m/s and so a Reynolds number of 7,600. HTC should increase over this range so I'm much more comfortable with my results. I would keep thinking "I'm must be missing a constant here, maybe the speed of light or something" as the numbers seemed a bit out there.

He also makes a tantalising comment about micro-channels: 100uM channels in the surface to get HTC to 10,000 W/m2/K. Anyone out there with milling equipment that can try this? Its more Leibig/turbulator territory rather than coiled condenser. I did an experiment with a rough file but got no benefit.

BTW, I suggest most reading this are not maths perverts, so we sort that out in another thread, or via email. I'd like to take this thread back to practical stuff, with an improved calculator coming in from "out there" when we have it.

[Maritimer]

Chapter 8 is about condensation. I'm using the paper edition; my page numbers are a little different, so look for Fig. 8.14. Fig. 8.13 is even more disheartening. On the other hand, Chapter 3 on heat exchangers is accessible.

I still don't know what H0 and H1 are.

[engunear]

Sorry: HTC=H0+H1.f

Thats example 7.1 I referred to. 8.15 and 8.15 are... yes, disheartening. Maybe I was getting a little hopeful with my desire for closed form solutions. A graph we can read off is fine.

I'm just thinking to assume HTC is a straight line (everything is linear to first order). Even the values we already have are pretty good, its likely gilding the lily to go too much farther. I mean the current 850 W/m2/K is probably 10X off. We only need to be much better than 10X error and we have made a step forward.

I don't have a measure for pressure changes through the day, or when the shower goes on. We are going to have to have a safety margin for flow and for length. I'm used to safety margins like 2X and 3X so it probably won't matter all that much if we are off a bit here and there.

But meanwhile, I'm getting itchy to cut some metal, so have been trying to catalog the ways of making a coiled condenser with common pipe fittings. I'm leaning towards the side fed at the moment, though it will need brazing or hard soldering to make it strong enough.

[Bagasso]

You left out concentric.

[Maritimer]

The standard VM reflux coil is missing, too.

This post is a trial to see if the readers want math.

So, HTC = H0 + H1*f. (* means times)

Engunear is afraid that math will scare off readers. I can't think of anything more attractive than math! Let's examine that equation.

In high-school algebra the equation of a straight line is:

y = m x + b (nothing between m and x also means times)

y is the vertical coordinate, x is the horizontal. So if x=0, y = m 0 + b = 0 + b = b.

That means that b is the value of the equation when x=0. The line starts at x=0, y=b, and then goes either up or down from there. The slope of the line is b. If b is a large positive number, the line goes up quickly; if it is a small negative number, the line goes down slowly-- both starting at b on the vertical axis.

If we go back to HTC = H0 + H1*f, this is saying that the Heat Transfer Coefficient starts at H0 and then goes up at the rate of H1 times the rate of flow of the coolant, f.

This equation characterizes the condenser.

You could have a catalogue of condensers with H0 and H1 values at various orientations and configurations of the condenser, so you could compare how they do in the application you are considering.

With static coolant, the condenser takes heat away from the source at H0 watts/square meter/degree Celsius. (Read "/" as "per".) Think about that. Watts is the stuff that heats up the boiler charge (and reciprocally, is absorbed by the condenser), square meters is the area of the condenser, and "per degree Celsius" is the difference between the inside and outside walls of the condenser. So, if you increase the flow rate of the coolant, the Heat Transfer Coefficient also increases.

So, do you want math?

M

[engunear]

This is where the theory side of this thread is headed:

There is a graph that tells us how to design a condenser. It works for both coiled and reflux condensers and it looks like the picture. We can't yet draw this graph properly, but it is where we are headed. This picture packs a lot of meaning. There may be one picture for each pipe diameter, or a way to make one work for all pipe diameters. This is still TBD.

Some starting points: it is obvious what the flow is; the active length is the length over which condensation occurs. A condenser has an actual length (how big you made it) and an active length. If the active length exceeds the actual length then vapour is vented. This is a fire hazard, and besides losing good whiskey, an explosion will ruin your whole day. So we need to know what these are.

There is also a "minimum flow". For a given power, you can't run below that or the condenser will again vent. This will happen even if you had a huge condenser. This comes from the principle of "Conservation of Energy" - the cooling water can only heat up so much, so for a give flow there is a limit to how much it can absorb. So we need to know the minimum flow as well. Its easy to work out.

So imagine you are wanting to build a condenser for a 1000W still. The middle line applies to you. We have a (dotted) line of actual flow which comes across and tells you the actual active length for that flow (drop down from the intersection to the point). The difference between the actual flow and the minimum flow is referred to as the "flow safety margin". It is most useful to express this as a multiple e.g. 2 times.

There is also safety margin for length. By the way, it is common to refer just to "flow" and "active length" meaning "actual flow" and "actual active length" because the graph's x and y axes shows all possible values of these parameters, but at any moment we are only talking about one value for each.

For high flows, the active length decreases slightly as flow increases. This slope is controlled by the parameter H1. There is a point where the graph turns a corner. This is controlled by the parameter H0. So we care about H0 and H1 because they tell us how to build this graph.

For me, the maths in this thread will be done when we can draw that line from the math, and plot measured points on this graph that line up with the theory, and do it for all common sizes of pipe. That way we can build condensers knowing in advance how the design will work.

On the questions ... is there a concentric coiled condenser? There is one with a bottom feed and a rain hat that I'll add to the list.

[DeepSouth]

Yeah the calculator is interesting. Although I don't think it says so on the page, the calculator is workable for coiled condensers; the maths is the same. But the catch is the HTC value. It talks about ranges of HTC from 150 to 850 W/sq m/C; already a ~ 6:1 range, and my measurements may be screwed, but the lowest measurement I've made is 2,000 W/sq m/C and the highest is 20,000. So its garbage in, garbage out on the calculator. We need values of HTC we trust so we can then use the calculator with confidence. Does HTC vary with: pipe diameter? flow? the material used in the pipes? If there are multiple pipes are they better in parallel or series?

Engunear, you are missing one very important thing when calculating the overall HTC for condensers. The actual heat transfer mediums and their phases play a huge role. You are solving for the HTC for a steam condenser using water as the cooling medium. Water has a thermal conductivity of 0.609 W/m K and alcohol has a thermal conductivity of 0.171 W/m K. For this reason, a water to steam condenser will have a much larger HTC than a water to alcohol condenser. During engineering school, the heat transfer textbook we used was Heat Transfer by Cengel. Chapter 13 deals with heat exchangers. This book lists overall heat transfer coefficients for a water cooled steam condenser as 1000-6000 W/m^2 C and water cooled alcohol condensers as 250-700 W/m^2 C. The measurements you are finding for a water cooled steam condenser are pretty well inline with already known values. This is great that your experiments match up with other prior, well established data. However, the overall HTC that you will see for an ALCOHOL condenser will be much less than what you are calculating right now for a STEAM condenser.

[DeepSouth]

One more quick thing to add. The reason there is such a wide range of overall heat transfer coefficients is because it is very dependent on the flow regimes of the fluids. The mass flow rates of the fluids as well as the physical geometry of the heat exchanger will determine the Reynolds number. Knowing the Reynolds number, you can have an idea of whether the flow is turbulent or laminar. There is also the Nusselt number, which can be empirically calculated from the Reynolds number and Prandtl number. Knowing Reynolds number and Prandtl number, there are empirical calculations that can give you an idea of the convective heat transfer coefficient for a certain flow. Tons of heat transfer work comes down to curve fits and empirical calculations because closed form solutions for fluid flows and turbulence and heat transfer don't exist. A man could spend his entire life doing research on this subject alone.

[Brutal]

I'm not qualified to be posting here. Got that out of the way, but I do enjoy reading about it!

But meanwhile, I'm getting itchy to cut some metal, so have been trying to catalog the ways of making a coiled condenser with common pipe fittings. I'm leaning towards the side fed at the moment, though it will need brazing or hard soldering to make it strong enough.

I want to point out an easy version of figure (1). You can use 2" or similar throughout with just one elbow, no need to close in the end. Just make sure the pipe housing the coiled condenser is angled down a little. Using this you could easily slide coils of different types up the output end for testing.

[engunear]

Hey DeepSouth, thanks for clearing that up. I've been hoping an adult would find this and either agree or disagree with the direction. Naturally I would have preferred be right from the outset, but this is better than continued error. Need a bit of time to think and reframe where to go from here. Yep I should read the textbooks, but they are so heavy and it seems easier to shove a thermometer in something.

What is happening here? What is the limiting factor on HTC? It sounds like it is not the water/copper interface but the alcohol/copper interface - that this is building a layer of fluid which is impeding heat transfer, and providing the additional resistance. As I understand it the Nuessel, Prandtl and Reynolds numbers all apply to the water side of the interface, but if its overall HTC is 10X less than this, and due to the low thermal conductivity of ethanol then this is the other side of the pipe.

So does this mean that a horizontal coil condenser, where the alcohol drips out of the way quickly is superior to a vertical one, where it runs down over the coils?

[DeepSouth]

There are lots of similarities between electrical resistance networks and thermal resistance networks. Take for example, 3 electrical resistors wired in series. The overall resistance of that system is the sum of the 3 individual resistances. Similarly, thermal resistance networks are thought of the same way. In a house for example where you have a wall constructed of sheetrock, studs, insulation, sheetrock, brick or other siding etc. heat transfer through that wall will encounter a series of thermal resistances. The thickness and material properties of each layer can be used to create an overall thermal conductivity of the entire wall. There are three basic modes of heat transfer, conduction, convection, and radiation.

Let's look at your coiled condenser inside of a water jacket. At the interface between the alcohol vapor and the inside wall of the tubing, there is convection. The convective heat transfer coefficient at this interface is dependent on the mass flow rate of the vapor, the size of the tube, the surface roughness, (these things will determine whether the flow is turbulent or laminar and how well the fluid can transfer it's heat), the thermal conductivity of the vapor, the surface roughness of the pipe etc. The next resistance in the series is the conduction through the wall of the pipe. This thermal resistance is easily calculated as it is just a simple conduction problem. The thermal resistance here is dependent on the material of the pipe, (copper, stainless, etc. which gives the thermal conductivity) and the wall thickness. The next thermal resistance is the interface between the external water and the outside wall of the coil. This could either be a conduction or a convection problem. For something like a flake stand, where the water isn't circulating, you have basic conduction. For a liebig or other external jacket where the cooling water is flowing, you will have convection. The convective heat transfer coefficient at this surface will be determined by the overall geometry of the outer jacket, the mass flow rate of cooling water, etc., etc. which can determine whether the flow is laminar or turbulent. Now, at the outer surface of your cooling jacket, the outer pipe can radiate some additional heat into the room. There can also be the case of natural convection if the outer jacket is warm enough, where the outer jacket heats a thin layer of air, changing it's buoyancy and causing it to rise and flow over the surface further aiding in cooling. The sum of the thermal resistances at all of the interface points yields an overall heat transfer coefficient for the entire system.

When you see in a heat transfer textbook or heat exchanger design book a range of overall heat transfer coefficients for the entire system, it has been empirically calculated for a wide range of materials, geometries, and flow rates. We haven't even discussed that in a condenser a portion of the condenser is used to actually condense the vapor and the rest of the condenser is used to further cool down the condensed liquid. This would be the case for a still condenser where it is desirable that the product is cooled to around room temperature. Heat transfer is a really complicated subject, but can be pretty fun. If you are really interested in doing more experiments, you should pick up some old textbooks at the library. Make sure on your experiments that you are measuring temperatures at the right locations and mass flow rates correctly in your experiment. Also, you'll see in textbooks that it isn't as simple for calculating heat transfer coefficient with inlet and outlet temperatures. A lot of equations depending on the problem use bulk temperatures and log mean temperature differentials. Just make sure you have your experiments set up properly and are using the right calculations is what I'm saying.

[Maritimer]

For those who would like to have a look, you can download A Heat Transfer Textbook here: http://web.mit.edu/lienhard/www/ahtt.html" onclick="window.open(this.href);return false;" rel="nofollow . It's also available from Dover (or Amazon, etc) for about $25.

As for spending a lifetime learning about heat transfer, notice that A Heat Transfer Textbook has two authors, John H. Lienhard V and John H. Lienhard IV. That's a good two lifetimes to write the book!

M

[engunear]

So in both a heat transfer problem and an electrical circuit resistance is the reciprocal of conductance. And we find the total resistance by finding each of the conductances, taking their reciprocal and adding them. Usually one dominates the rest, and the problem reduces to determining which one it is in our case, and what is its value.

Since there is a lot of energy transferred when vapour condenses, I don't think the vapour movement, or the vapour condensation will be part of the limit. Physical systems move towards equilibrium, and the more energy transfer on the table, the more violent the move. I believe a thumper thumps because they create a vapour/water interface that sucks the coolant water up the vapour feed pipe. This is violent, aggressive.

The cooling of liquid is a real but small contribution - as can be worked out by comparing the latent heat of vapourization (energy to go from gas to liquid) to specific heat (energy to change temperature, once a liquid).

The thermal conductivity of the copper is so high, its resistance is so small, that it does not materially contribute to the rest. Forget it.

Comparing alcohol and water, the inside of the pipe does not know what is happening outside. So the existing measurements put an upper bound on what we'll get for alcohol.

So my current understanding is it boils down to the layer of fluid on the surface of the pipe (pun intended).

I have the gear, the will and the technique to measure this, just not the time ... yet. When you don't have time to measure a good guess is all you have, so its worth practicing. So a challenge to you all: What are you guessing is the value, based on what I've measured, what DeepSouth and Martinez have said, and gut feel? I now there is at least these two other techies reading this and probably a lurker or two. Pick a number between 250 and 4,000, it will be in there. My guess is 1500, far enough out of the textbook range to be interesting and down on the water measurements. Units of course are W/m2/C. This is a thin horizontal pipe, 29mm long, slightly flattened.

This really puts a nail in the question about turbulators - my measurements overstate the effect of turbulent cooling water, and I struggled to see any benefit from it. If the performance is dominated by the alcohol, then the effect is even further down.

And it makes me laugh, its just a frigging pipe in another frigging pipe. Whoda thunk it was so hard?

[DeepSouth]

http://www.rshanthini.com/tmp/PM3125/RM ... oblems.PDF" onclick="window.open(this.href);return false;" rel="nofollow

Check out page 1036 and 1037 in the document attached in the link. It is an entire chapter on heat exchangers from the textbook I mentioned previously. It describes in detail exactly what you just hit on, that in the case of a series of thermal conductances, if one value is much smaller than the other, a bottleneck effect is created. The thermal conductivity of a network in series is limited by the smallest thermal conductivity value. For a condenser when you are not only condensing the vapor but also cooling down the condensed liquid, the overall heat lost from the condensing liquid is equal to the mass flow rate of the vapor times its latent heat of vaporization, plus the mass flow rate of the newly condensed liquid times the specific heat capacity of the liquid times the temperature change. I'm pretty sure you already knew that though.

[DeepSouth]

The cooling of liquid is a real but small contribution - as can be worked out by comparing the latent heat of vapourization (energy to go from gas to liquid) to specific heat (energy to change temperature, once a liquid).

I decided to actually plug in some values to see how much of the overall heat transfer in the condensing process is attributed to actually condensing the liquid and how much is attributed to further cooling down the liquid.

In the case of a steam condenser, steam has a latent heat of vaporization of 2264 kJ/kg and specific heat capacity of 4.183 kJ/kg K at 1 atmosphere. If you cooled down the condensing steam from 100 C to 20 C ( or a delta T of 80 C or 80 K), the condensing portion gives up 2264 kJ/kg and the further cooling down gives up 334 kJ/kg. Cooling the condensed liquid is only about 13% of the overall heat transfer in this process. The actual condensing portion is where all of the energy is transferred.

In the case of an alcohol condenser, ethanol has a latent heat of vaporization of 855 kJ/kg and a specific heat capacity of 2.44 kJ/kg K at 1 atmosphere. If you cooled down the condensing alcohol from 78.4 C to 20 C (or delta T of 58.4 C or 58.4 K), the condensing portion gives up 855 kJ/kg and the further cooling down gives up 142 kJ/kg. Cooling the condensed liquid is only about 14% of the overall heat transfer process. The actual condensing portion is where all of the energy is transferred.