Author Topic: Talking Thermodynamics  (Read 122164 times)

Offline MJM460

  • Full Member
  • ****
  • Posts: 909
  • Melbourne, Australia
Re: Talking Thermodynamics
« Reply #1080 on: October 19, 2018, 12:42:02 PM »
Hi Willy, I couldn't agree more about the interesting descriptions of what all those early pioneers were discovering.  The names are familiar, many having been given as special names for units or observed phenomenon, but use of the name does not reveal much of what they actually did without further information.

I have done some calculations on the temperature profile of the jug with insulation.  A little more checking required, to easy to make an error in a spreadsheet and have a very accurate looking wrong answer.  Takes careful checking by a fresh mind, so perhaps tomorrow.

MJM460
The more I learn, the more I find that I still have to learn!

Offline MJM460

  • Full Member
  • ****
  • Posts: 909
  • Melbourne, Australia
Re: Talking Thermodynamics
« Reply #1081 on: October 20, 2018, 11:33:09 AM »
Hi Willy, good news at last.  I have been struggling with the analysis of the jug for way too long for a relatively simple concept.  I was getting unlikely results and had clearly made a mistake.  At last I have found it.  A case of systematically checking the formula in each cell until I found it.  So what did I conclude?

Start by remembering the basics - the total heat loss is determined by the resistance of the convection film coefficient between the water and the inside wall of the jug, the conductivity and thickness of your cork insulation, and then the convection film coefficient between the outside of the cork and the air in the room.  The lowest conductivity or highest thermal resistance has the highest effect.  In this case, the water coefficient is so high it is not a big influence on the total resistance.  However the cork and air are close enough in value that both are important.  The same rate of heat loss has to pass through all layers in series.  Unfortunately, we don't know with any accuracy the air side film coefficient.  My text book lists it as somewhere in the range of 5 - 25 W/m^2.K.  Read this as Watts per square meter per deg Kelvin, and of course one degree centigrade is the same magnitude as one Kelvin, so when temperature differences are involved, heat transfer problems for example, you can use either.

The effect of this coefficient is that apart from its effect on the overall rate of heat transfer, it can be directly seen in the temperature difference between the outside of the cork and the room air temperature.  A high film coefficient leads to a small temperature difference across the film, while a low film coefficient leads to a high temperature difference across that film. 

The published figures for the air side film coefficient lead to much higher film temperature differences than those you have observed.  However, if I assume the film coefficient in the range of 100 to 200, the film temperature difference is predicted to be 3 (for 100) or 2 (for 200).  This just seems a bit high compared with the textbook figures.

That outside cork temperature is in practice quite difficult to measure with a thermocouple as the thermocouple is influenced by the air temperature as well as the cork temperature, and the air temperature range occurs over a very small distance.  It is likely that your thermocouple reads low, but it is interesting to try.  An infrared device may be able to measure the temperature differences more accurately as the jug heats, but I haven't tried to compare it.

I would be interested to know if you tried touching the cork with your finger when the jug was hot, and whether the temperature seemed higher than what you were measuring.

Any way, with consideration of a reasonable error range, the calculation yields a heat loss somewhere around 20 Watts, quite small compared with the element rating of 1000 W.  If it is as low as 12, which would require a film coefficient of only 5, the cork surface temperature would be nearer 38 C above ambient, quite a lot higher than the measured value.  A value of 25 for the film coefficient would give a cork surface temperature about 12 C above ambient which would feel warm but definitely not hot.  A coefficient of between 100 and 200 would give a cork surface temperature rise of 2 - 3 C above ambient, which would definitely be in line with your measurement.

If it is of interest, perhaps I could write out the relevant equations and attach them rather than describe the whole procedure.

Just for interest I repeated the calculation for the uninsulated jug.  I don't really have any data on the thickness of the plastic jug walls so I guessed 2 mm, or 0.002 m.  For the plastic, I found a thermal conductivity for plexiglass 0.195 W/m^2.K, so, much less than glass, but high compared with cork.  I am guessing that it might be in the right range for the jug plastic (without any evidence however!)

The issue of what value to use for the air side film coefficient still influences the answers, so I  repeated the same calculations several times to cover the same range, the joy of using a spreadsheet.

The heat loss is is then likely to be in the range 100 to 200 W and the outside wall temperature 35 to 60 degrees above the room temperature, so 55 to 80 C, would definitely feel very hot, and could inflict a burn on your skin.  Now starting to become significant compared to the element, so probably slows that final approach to boiling point.  Of course, when you first switch on the jug, the temperature difference is very low, so the heat loss is also quite low for most of the heating time.

By comparing your readings with the calculation, you can see the influence of the film coefficients and material properties.  You can also see that while the electric power saved, probably would not cover the cost of the cork (in environmental or economic terms), it has real value in reducing the outside temperature of the vessel and so the likelihood hood of burn injuries.

Though of course, most of us learn early to expect the jug to be hot, so probably don't need it.  However when in a laboratory boiling liquids at much higher temperatures, or in some special cases where the occurrence of burns is a problem, including many industrial situations where piping carries hot fluids, the insulation can be useful, and is applied for personal protection.

I guess the last step is to revisit the convection calculations and see if the film coefficient can be estimated a little more closely.  That will have to wait for another day.

Thanks for looking in,

MJM460
The more I learn, the more I find that I still have to learn!

Offline steam guy willy

  • Full Member
  • ****
  • Posts: 2228
Re: Talking Thermodynamics
« Reply #1082 on: October 22, 2018, 12:04:46 AM »
Hi MJM , thanks for the reply...what i should have done as well was check on the cooling down time of the various states of the insulation breakdown ...that would have been much more interesting ,However that would have taken a lot of time clock watching !!..When i did the cooling down . Also the time cooling down with the boiler it took about 18 hours !! Also i did the kettle timing in reverse with the full insulation in reverse !!! :facepalm:

Willy

Offline MJM460

  • Full Member
  • ****
  • Posts: 909
  • Melbourne, Australia
Re: Talking Thermodynamics
« Reply #1083 on: October 22, 2018, 12:34:40 PM »
Hi Willy,

I don't think I would have waited for the cooling time either.  It is not necessary with an electric element as there is no air flow as an extra mechanism for heat loss like there is with a fired boiler.  In that case, the cooling curve allows an approximate measure of the heat loss through the insulation alone.

I am still puzzling over what you mean by reverse?  Obviously not the same as up side down.  The handle facing the other way?  Time backwards? Not usually, though the clock could count down instead of up, though the elapsed time would be the same either way.

But the big learning is about the temperature profile.  I will try and post a picture tomorrow.

MJM460
The more I learn, the more I find that I still have to learn!

Offline steam guy willy

  • Full Member
  • ****
  • Posts: 2228
Re: Talking Thermodynamics
« Reply #1084 on: October 23, 2018, 01:09:55 AM »
hi MJM,  By reverse i meant i put all the insulation on first  after cutting it out and making it all fit properly as each layer became larger. I then did the tests starting with all the layers and doing each consecutive test by removing each layer in turn ! Also some of the layers of cork were not completely touching everywhere due to the shape of the kettle, so there were layers of 'air' as well however that should possibly increase the insulation ?.. I can remember wearing an 'airtex' vest years ago but i don't know if they still make them. !


willy

Offline MJM460

  • Full Member
  • ****
  • Posts: 909
  • Melbourne, Australia
Re: Talking Thermodynamics
« Reply #1085 on: October 23, 2018, 12:14:50 PM »
Hi Willy,

Ok, it really doesn't matter which order you do the experiments, as I am sure you knew.  But you have made an important point about those air gaps.

When two surfaces touch, there is a contact resistance to heat flow, or a local lower heat transfer coefficient.  First there is a temperature drop across the contact faces, and second, the lower conductivity is in the line that all the heat must pass through so further reduces the overall heat transfer.  Air is not a very good conductor, and in your jug experiment, the transfer to air is nearly as important as the transfer through the cork in determining the overall result.  And the transfer that does take place to the air is significantly aided by convection, which keeps replacing warmed air with cooler air, so increases the average temperature difference.

In those little air gaps created by the insulation "fit", there is very limited scope for convection, so the actual conductivity of the air becomes more important as heating of the air reduces the temperature difference driving the heat transfer.  So the air gaps are only helpful in an insulation system.  (They do add a degree of complexity for trying to do a complete calculation, but really only important when you are trying to increase the heat transfer for heating or cooling purposes, so in this case I have ignored them.) The thermal conductivity of air is only 0.026 W/m.K at room temperature.  It increases with temperature, but at 120 C, it is still only 0.033 W/m.K, so lower conductivity than cork at the temperatures we are talking about.  My text book does not list anything better.  This contact resistance is part of the reason it is so hard to measure the outside cork temperature with your thermocouple.  Along with the fact that the thermocouple sheath lies in the thickness of that convection profile, where the temperature is changing rapidly with distance from the cork, so difficult to determine what it is actually measuring.

Unfortunately if you have a bigger air gap, convection is able to start, so increases the conduction.  But in closed cell foams, and those air-tex jackets, (I am not sure of the construction of those), the very popular and incredibly effective down jackets and doonas all work mainly due to the low conductivity and low mobility of the trapped air.

All these problems involving heat transmission through multiple layers are solved using the same equation, q= U x A x (T2 - T1), where U is an overall combined heat transfer coefficient.

For a system involving two convection layers (convection heat transfer coefficient is usually given the symbol h) and a solid material such as your cork, the overall heat transfer coefficient is calculated by combining the individual layer values as shown in the attached extract from the text book, which also show pictorially the temperature profile.  If you have more layers, like the glass wall of the jug, wood outer cladding, or perhaps a second insulating layer of different material, the extra L/k terms are easily added.  L is the thickness and k the thermal conductivity of the layer.

The inside film coefficient for the water is quite high, so really has negligible effect on U, especially once the boiling starts, as the vigorous agitation of the bubbling fluid increases the transfer coefficient to a much higher level again.  But the air side coefficient is just the level that gives it significant importance in our calculation.

So the last step in this little section is to do the calculation of the convection coefficient and see if we can narrow down that a bit.  That will allow us to predict the temperature of the outside of that cork.  I will have a go at that in the next few days.

Thanks for looking in,

MJM460
The more I learn, the more I find that I still have to learn!

Offline steam guy willy

  • Full Member
  • ****
  • Posts: 2228
Re: Talking Thermodynamics
« Reply #1086 on: October 24, 2018, 11:40:34 PM »
Hi MJM ,thanks for the info........Wondering about water evaporating ... can water evaporate from an enclosed vessel upside down ? I am thinking of a fish bowl that has a lip with a small cavity for some water .... but open at the bottom  Can water only evaporate upwards and what is the lowest temperature to enable evaporations to occur ??  I suppose the answers are available somewhere ..!! I was thinking that if it does evaporate and rises to the top of the bowl does it condense and run back down to the bottom ?
Willy
« Last Edit: October 24, 2018, 11:44:56 PM by steam guy willy »

Offline MJM460

  • Full Member
  • ****
  • Posts: 909
  • Melbourne, Australia
Re: Talking Thermodynamics
« Reply #1087 on: October 25, 2018, 11:51:28 AM »
Hi Willy, the nice thing about thermodynamics is that the very small number of laws cover most situations, even if sometimes it is a little obscure how they might operate in a particular situation. 

So the answer to your evaporation question comes back to that basic vapour pressure of the water which of course depends on temperature and is listed in the steam tables.  Water or any fluid really, evaporates when it's vapour pressure is equal or higher than the absolute pressure at the location.  Once the water vapour pressure exceeds the vapour pressure, it evaporates to form a gas phase.  Of course the gas phase has much lower density than the liquid, so under the influence of gravity, the vapour phase tends to separate from the liquid.  However if the pressure is higher than the vapour pressure at the prevailing temperature, the evaporation is suppressed.

Now in a container of liquid, the pressure at the liquid surface is the same as the air pressure, so about 101 kPa, depending on the weather.  At this pressure the equilibrium vapour pressure is reached at 100 deg C.  But below the surface, the local pressure increases due to the density of the water and the depth.  Because the surface pressure is normally the lowest pressure in the liquid, that is where evaporation starts.  In your kettle or in a boiler with a good heat input, the liquid is generally not uniform temperature throughout, and is hotter near the element.  Now as we know, hotter water has a lower density, so starts to rise so mixing up the water and evening out the temperature somewhat.  But it also has a higher vapour pressure, and with sufficient heat input the temperature, hence vapour pressure is high enough that it exceeds the local pressure even allowing for the pressure increase with depth, and so formation of a second phase or evaporation starts making a bubble right there on the element.  The expansion into vapour phase compared with the liquid specific volume means the bubble is quite large compared with the liquid from which it came and rapidly rises under the influence of gravity until it breaks out at the surface.  This is what we call boiling, as opposed to just evaporation.

Now in your fish tank, I am assuming you mean or at least have seen one of those fish tanks where there is a handle like loop filled with water, so fish can swim through the loop even though it is above the free surface at the opening of the tank.  The first thing to realise is that to determine the pressure everywhere in the tank, you start at the only point you know the pressure which is where it meets the atmosphere at that free surface.  As you go deeper in the tank the pressure increases, but equally, as you go above that free surface elevation in the handle, the pressure decreases, as does the temperature at which the vapour pressure reaches the local pressure.  Now in the dimensions of a typical fish tank, the height in meters is quite small, and one meter of water increases the pressure by 9.8 kPa or about 0.1 atmospheres, or 1.5 psi, so three or four inches does not make much difference.  And the vapour pressure increases in a similar manner towards the bottom of the tank.

So the changes in boiling point and vapour pressure above and below that free surface, directionally are lower pressure and boiling point above the level of the free surface, and higher pressure and boiling temperature below the free surface.  But in practical terms, the difference is insignificant.  However, if you warm up the water (preferably remove the fish first) to about 100, vaporisation occurs first at the lowest pressure point, the top of the handle even though there is no contact with air, a vapour bubble will form and gravity keeps it at the highest point.  More vaporisation will cause the bubble to enlarge.  It's expansion displaces the water down until the vapour space breaks through to the free surface.

If the temperature outside the tank is lower than the water temperature, there will be heat lost at the walls, water will condense, and gravity will make the water drops,once they are big enough to coalesce, run down to rejoin the liquid phase.

So evaporation does not require a free surface, though it will cause a very high pressure if there is nowhere for the water to go to make room for a lower pressure bubble.  It is the action of gravity, and the difference in density between the liquid and vapour phases that causes the vapour bubbles to rise.  If the astronauts had a vessel half full of water and heated it in space, beyond the influence of gravity, the bubbles would tend not to "rise", but would stay near where they formed on the element.  And as the heat transfer coefficient to a vapour is much less than to a liquid, the element would get much hotter to dissipate the heat (presumably from an electric element.). Of course it is pretty difficult to get far enough away from everything so the gravity is really zero.  Even beyond the earth's influence, the bubbles would probably tend to 'rise' away from the moon or sun or any other planetary bodies nearby, so to speak, but not so vigorously as in a stronger gravitational field.

As to the lowest temperature that water would evaporate, well it is pretty low.  There is a vapour pressure even over ice.  At about 0.01 deg C there is a point at which liquid vapour and solid can all exist together in equilibrium, called the triple point.  At this point, the vapour pressure is 0.61 kPa, so not very much, but it is the lowest temperature for liquid to exist.  Below that the vapour and solid continue to exist, and the solid can 'evaporate', though the vapour pressure is very low.  So I guess 0.01 degrees is about the lowest temperature at which liquid water can evaporate.

I hope that you can see the problem does not involve any new concepts, just understanding of the implication of the configuration of that fish tank.  And of course, keep separate the effects of phase change  and gravity.

Thanks for following along,

MJM460
The more I learn, the more I find that I still have to learn!

Offline steam guy willy

  • Full Member
  • ****
  • Posts: 2228
Re: Talking Thermodynamics
« Reply #1088 on: October 25, 2018, 11:36:44 PM »
Hi MJM. Thanks for the info and i will need to read through it a few times to get more info from it . I should have included a drawing of what i was getting at though and when i get home i shall include one...11 Ok here is the drawing ,it is a small circular fish bowl with a lip that is upside down with some water in the lip . so does the water just evaporate upwards and if so would it condense on the top and run down the sides to go back into the lip ? ...or can the water evaporate SIDEWAYS AND DOWN to escape to atmosphere. If it only evaporates upwards, would the water stay there for ever ??  or is this another unintuitive sillyish question ??!!!

thanks again
Willy
« Last Edit: October 26, 2018, 02:34:53 AM by steam guy willy »

Offline MJM460

  • Full Member
  • ****
  • Posts: 909
  • Melbourne, Australia
Re: Talking Thermodynamics
« Reply #1089 on: October 26, 2018, 12:46:16 PM »
Hi Willy, well, not what I was thinking of at all.  Not much room for fish in a bowl placed like that.  But the same physics applies, just a slightly different problem to the one I was thinking of.

I would look at that one in two stages.  First, look just near the surface of the liquid.  In terms of molecular sizes and distance between collisions even a few mm is quite a long way.  The liquid will evaporate at the surface providing only that the humidity of the air in the bowl is not too high, so that the vapour pressure of water vapour in the air is less than the equilibrium vapour pressure at the temperature.  The water will evaporate to make the water vapour pressure near the surface up to the equilibrium vapour pressure.  The random motion of the molecules means that once the molecules are in the vapour space, there is a tendency to move on average into space where the concentration is lower, so they do not stay close to the surface but drift further away throughout the whole space, and the evaporation tends to continue.

Now in an open dish, the slightest air current will tend to move the vapour away from the surface, so more has to evaporate to get to equilibrium, though it tends to spread purely due to the random motion of the molecules anyway.   But under your inverted bowl, movement of air is restricted.  There is still connection with the atmosphere, so the extra water vapour does not increase the pressure in the bowl, some of the air moves out to the atmosphere through the opening, whether large or small and regardless of orientation.  So the pressure does not change.

So we are left with just looking at how gases mix when occupying the same volume.

Water vapour, being a lighter gas than oxygen or nitrogen or most other components of air, tends to rise under the action of gravity.   It is not as simple as a bubble membrane filled with water vapour floating in air, as the vapour molecules of each gas in the mixture are free to move in all directions and tend to each fill the entire space regardless of the other components.  Gravity acts on all the molecules equally, but the lower mass of the water molecules tends to means that after a large number of random collisions, the lighter molecules tend to experience a slow drift upwards. 

This means there is a small tendency for the number air molecules moving out of the bowl at the bottom to be more in proportion to their total numbers and water.  Consequently, the humidity in the bowl will tend to increase and this might eventually inhibit further evaporation.

The heat necessary to evaporate the water comes from the remaining liquid which gets a bit cooler.  This means conditions inside the bowl are a bit cooler than outside so the wall of the bowl is relatively a bit warmer, and their would not cause condensation.  However, if the temperature falls in the room generally, perhaps as evening approaches, the heat loss from inside to outside would result in some condensation once the humidity gets near 100%, and eventually this condensate would run back down into the water.  It is all a question of a delicate heat balance, and which side on the bowl is warmer.  Then heat always travels from the warmer space to the cooler space.  Room temperature is rarely sufficiently constant that you would never get some condensate under some conditions.

Essentially, condensation requires a cooler surface to absorb the latent heat, and humidity sufficiently high that the equilibrium vapour pressure at the prevailing temperature is reached.  If the bowl is sitting on say a stone bench which is still warm from earlier in the day, your could possibly get a situation where the bench supplies heat to evaporate the water which then condenses on the glass wall, giving the heat to the glass and hence to the room air, while the condensed water runs down to join the water at the bottom.  For a while, until all the temperatures become essentially equal, it could look like a continuous process.

I hope that makes it a little clearer.

Thanks everyone for following,

MJM460
The more I learn, the more I find that I still have to learn!

Offline steam guy willy

  • Full Member
  • ****
  • Posts: 2228
Re: Talking Thermodynamics
« Reply #1090 on: October 27, 2018, 04:56:27 PM »
Hi MJM , thanks for that...thinking about your comments on condensation ...They say that in  a dessert you can get drink  ing water by placing large stones in hollows to collect the condensate...however the humidity must be really low in the desert. However i suppose this process uses a different way to p produce the moisture ???

Willy

Offline MJM460

  • Full Member
  • ****
  • Posts: 909
  • Melbourne, Australia
Re: Talking Thermodynamics
« Reply #1091 on: October 28, 2018, 07:13:26 AM »
Hi Willy, despite many trips to the desert, I have never tried that one.  Though I have heard of the principal being used in desert revegetation programmes.   Watering the young plants is not very practical so they place a rock by each one.  The idea is that the rock cools with the surrounding sand at night, but has a higher specific heat, so stays cooler than the surrounding sand well into the heat of the day.

Any moisture in the sand, perhaps from condensed dew overnight, is also heated during the day, becoming a bit more mobile as vapour, and tends to condense, or at least concentrate in the sand under the rock where it is cooler.  I am not sure if you would get enough to drink, but apparently the plant is able to take advantage of any higher moisture content.  Later in the day, the rock temperature catches up, but the plant has already taken what it can.  Overnight the whole process starts again.  Eventually the plant is big enough that it's own leaves are able to contribute to shading the roots, particularly if it is one of many.

It is difficult to believe it would always have enough effect in all circumstances, as you say the desert gets pretty dry.  But perhaps it does make enough difference in borderline circumstances.  Or it may be more effective than I might think.  And perhaps near the edge of vegetation surrounding a desert, the effect is enough to extend the viable growing area, and so gradually beat back the desert.

But at the end of the day, the principal is the same, molecular motion slows at lower temperature, so vapour is less able to escape from cooler areas, resulting in a slightly higher concentration.  Depending on the concentration reached, water will condense, or perhaps at least concentrate enough for the plant roots to take advantage.  Even the condition of water in soil may be a factor.  If it attaches in some loose chemical way to the molecules in sand, as it does in many compounds, it may not need much increase in concentration to become useful to a thirsty plant.  And may not behave as a simple binary liquid-vapour system.  Perhaps we need a botanist on the forum!  But we are getting away from engines.

Thanks for looking in,

MJM460
The more I learn, the more I find that I still have to learn!

Offline steam guy willy

  • Full Member
  • ****
  • Posts: 2228
Re: Talking Thermodynamics
« Reply #1092 on: November 03, 2018, 01:52:34 AM »
Hi MJM , I was reading my book and found these pages about Solar Chimneys I have not seen this before and it docent seem very efficient to just get 100 KW.  it would be interesting to find out how many watts you get for the square foot !!

so More calculations i'm afraid !!!

willy

Offline MJM460

  • Full Member
  • ****
  • Posts: 909
  • Melbourne, Australia
Re: Talking Thermodynamics
« Reply #1093 on: November 03, 2018, 09:51:49 AM »
Hi Willy, that's an interesting project, but a very high chimney.  I have been up one of 100 meters though much smaller diameter.  But 200 m!  You would have a good view from up there.  Not for cyclone or earthquake areas though.

The sun heats the air, which lowers the density of the air.  The hot air rises just like a boiler chimney, or a hot air balloon.  Amazing how much power is generated that way.  Solar panels are rated based on 1000 W/sq.m.  But of course the peak is only reached (or even exceeded in the desert or tropical areas) only for the few peak hours of the day.  The performance they quote as an average is much more important in terms of the total quantity of power generated.  And of course, they really need to be accompanied by storage system so the power generated can be dispatched at the time it is required.  Battery banks for short term issues or pumped hydro system for really large systems.

Driving a turbine with the chimney draft is a great idea, as all the moving parts are accessible from ground level for maintenance.

Other systems being tried around the world use reflectors on a ground level array to concentrate the energy on a heat exchanger to melt salts which in turn are used to generate steam.  The molten salts are also stored in well insulated pressure vessels to spread the time of generation. 

We will get there eventually, probably a mixture of different technologies, to make the most of natural resources such as sun light, wind, tide and underground thermal, in different locations and to get best efficiency in different locations.  And we have to upgrade the grid to cope with many generators spread far apart, with power flowing different directions at different times, rather than most existing grid designs which assume one way distribution from a large central power station.

It all makes for interesting reading and research.

MJM460
The more I learn, the more I find that I still have to learn!

Offline steam guy willy

  • Full Member
  • ****
  • Posts: 2228
Re: Talking Thermodynamics
« Reply #1094 on: November 14, 2018, 11:00:49 PM »
Hi MJM, a quick question ...In a steam engine we burn fuel in a boiler to generate steam to drive the engine to create power to do work .......... this fuel can be wood coal gas etc etc ... So in the human body we eat vegetation in the form of veg and this enables the body to function and work also keeping us warm and cool,  So how does the human body change the fuel/food into a functioning  work producing "engine"   ?? This may be more medical than technical, and is it still a thermodynamic Process ??

Willy