Author Topic: Talking Thermodynamics  (Read 5665 times)

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #135 on: July 13, 2017, 03:11:53 PM »
Thanks for that.......also in my book..There are tables that mention the Latitude at Dublin.... Is this significant ?? Do thermodynamic tables alter at the equator  from the poles ??

Offline Maryak

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Re: Talking Thermodynamics
« Reply #136 on: July 13, 2017, 09:19:24 PM »
Hi, Also in an infernal combustion engine the inlet/exhaust ports are different sizes and the cams can be set differently . so .could a steam engine have four ports  2 for the inlet cycle and two (larger)? for the exhaust cycle ? My brain is really starting to hurt now.!! Is it because i am an autodidact ??................

Some steam engines were fitted with double ported slide valves.


Если вы у Тетушки были яйца, она была бы Дядюшкой

Offline MJM460

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Re: Talking Thermodynamics
« Reply #137 on: July 14, 2017, 02:22:31 PM »
Continuing on Willy's questions from post #130 -

The next question was whether the mks system we may have been used to is the same as SI.  This was a very contentious issue at the time of adoption of SI metric, as it seemed to some that everyone was being made to change so no one was disadvantaged, and no one had a head start.  However this was a spurious argument, put up by people who did not understand physics.

The fundamental huge leap forward inherent in the SI metric system is the definition of the unit of force by application of Newton's first law and the associated equation, F= ma.  This step provides a clearly defined unit of force (the Newton) and removes the necessity to use the arbitrary constant equal in magnitude of the acceleration due to gravity, and instead allowing the constant to be one, and removing the confusion around force and mass.  This is a major contribution to a rational set of units.  Previously all systems defined force in terms of the effect of gravity on a mass, giving rise to the imperial unit of pound force, the mks system with its kgf, cgs with gmf, and probably others.  So mks system is a sort of inferial metric system using sensible size units for large industrial equipment (the chemical engineer's test tubes capacity measured in tons) and cgs is good for laboratory size equipment where grams and centimetres are more convenient.  The systems arose from historical understanding of mass and force.  Newton's law is now recognised as a fundamental law of physics, and now more eloquently expressed as the law of conservation of momentum, which applies to bodies moving in a straight line.  A very similar universal law which is totally analogous, expressed as the law of conservation of angular momentum, applies to torque, moment of inertia and acceleration of angular rotation.  Using this basic law of physics to define force results in a very rational set of units, and clearly defines the relationship between mass and force in a universal way, it applies equally on the moon, on Mars or anywhere else in the universe.  So please remember these legacy unit systems only for the purpose of working out conversion factors, and then continue in SI metric.

The question on steam enclosed in a cylinder with one end a piston is very interesting and the answer might not be what you would expect.  When you move the piston to reduce the volume, the steam is compressed.  If you have adiabatic compression, that is no heat flow in or out, the pressure increases, along with the temperature.  So the properties of the steam change toward superheated, and yes, you can compress steam.  The work input adds more than enough energy to avoid condensation.  However you asked what if you remove heat as you compress so the temperature does not rise.  Now for a permanent gas, such as air, isothermal compression means less work to do the compression.  However with steam, removing heat means the properties change toward condensation.  I have not yet found an example to rely on, to say what happens if you try isothermal compression on steam, but I would tackle the problem this way.  Using steam tables, if you use the superheat section, and look up the properties at a temperature like 150 deg C, starting from say 0.01 MPa, you will see that as you move to the next pressure and reduce the volume (your cylinder contains a constant mass, so reducing volume means reducing the specific volume), while the pressure increases, the saturation temperature also increases, so by about 0.48 MPa, the saturation temperature exceeds your 150 deg C, so steam will start condensing.  It seems likely that if you start at the saturation temperature, then the reduction in volume will cause some condensation immediately.  But if you do not remove all the heat added by the work you do in moving the piston, you can keep compressing and avoid condensing.

Your text book is up to date for its time, and much of the maths is still current, though the units are more difficult to deal with than SI.  The main trick to following the text is to keep track of which assumptions are current at any point in the text.  The pages you have shown talk about isothermal and adiabatic processes, and calculate the work done as the process proceeds.  It was already understood that you had to know the process to calculate the work done.  Modern terminology is that work done is a process dependent quantity, unlike enthalpy and enthalpy, which purely depend on the measured pressure, temperature, and specific volume, not the process followed to get there.  Personally, I find a modern text much easier to read, and it is definitely worth getting an SI version of the steam tables.

I am not sure about the Dixon table.  It looks like an early version of a steam table.  I note that the pressure is measured in inches of mercury, which is of course dependent on the local value of gravity.  Now the gravitational acceleration changes from the equator to the poles, and also with height above sea level.  Hence the use of the defined latitude (of Dublin in this case).  The last column looks like enthalpy with a reference temperature of 32 deg F, but I am not sure what the H and L columns are.  Modern steam tables use absolute pressure, and there is no dependence on latitude.

I hope that adds a bit more clarity, and a couple more blocks to our knowledge base.  I will check on what I have missed from your post for next time.

Thanks for following along

MJM460

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Offline MJM460

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Re: Talking Thermodynamics
« Reply #138 on: July 15, 2017, 10:35:31 AM »
Hi Willy, a few more answers for you in the hope of dealing adequately with the outstanding issues.

While I basically leave valves to Dan and Maryak, I was going to add a couple of comments.  If I remember history correctly, the first engines were controlled by a boy on the taps, opening and closing them as required.  Obviously a real impediment to high speed, continuous operation.  Then some lever systems were developed to let the boy off, but still pretty clumsy.  Jobs sacrificed to technology even then!  So the slide valve was a huge step forward.  Then, I suspect that, just as increases in boiler pressure had to wait for pressure vessel technology to advance, I suspect that valve design had to wait for advances in machining technology, and even metallurgy.  The slide valve worked well, but there was a lot of development in the linkages which drive the valve to allow early cut off and reversing.  That valve of Maryaks would be a real trick to machine, perhaps requiring soldering some layers together.  Would be interesting to understand that one better.

Now post #133.  Some real confusion there.  I will try and make things a little clearer.  First introducing steam into a closed container.  You did not say if the container already had air in it, and you did not make any comment on heat transfer during the operation.  Now steam is normally hotter than atmospheric pressure, unless the steam itself is at quite low pressure.  So you can expect heat to flow from high temperature to low temperature, unless you make specific provisions.  You might for example insulate the vessel very well so heat transfer is prevented.  Let's start with that.  We assume no heat transfer in or out.  You introduced the steam to the container, rather than evaporated steam in the container.  So not a constant volume process, and no mention of moving pistons for external work.  So let's just assume some steam in the container, and the steam definitely has mass.  Steam will fill the entire space.  Pressure works in all directions, so pressure is experienced on the walls, but they do not move, so no work done.  As you imply, the steam "has weight", but weight is simply the result of gravity acting on a mass, there is no conflict there.  But I expect that you are alluding to the normal assumption of equal pressure in all directions, so what is weight doing?  Basically, just as we saw in your water column experiment, weight means that the steam pressure is higher at the bottom of the container than at the top.  Let's see if we can calculate the pressure difference.  Basically, that conservation of energy equation gives us the necessary clue.  We saw before that the pressure energy due the height if the fluid, in this case steam, is calculated as P= density x g x h.

If our steam is at 100 kPa (absolute), we have looked up the specific volume (tells us density) for that before.  I think the density was about 0.6 kg/m^3.  The value of g in SI units is near enough to 9.8 m/s^s, and as we are mostly interested in models, let's assume the height of our container is just 1 metre.  It is then easy to scale for other heights.  So P = 0.6 x 9.8 x 1 = 5.9, so what are the units?  Using our dimensional analysis, we see we have kg/(m.s^2) which is the dimensions of pressure in Newton/m^2.  So we have nearly 6 Newton's per square meter which has the name Pascal. A Pascal is a very small unit.  We have 1000 Pa = 1 kPa.  And standard atmospheric pressure is about 101.3 kPa.  So 6 Pa is 6 / 100000 times 1 atmosphere.  Or 6/100000 x 14.7 psi or 0.0009 psi.  You can see why it is normally considered not important.  The more so in most of our model sizes?

Condensing requires the steam to lose heat.  So with our assumption of no heat transfer, and no work done, no condensing.

Oh, and in case you were assuming some air in the container to which the steam was added.  So we can continue to assume no heat transfer, the air and the container and the air have to be at steam temperature, otherwise irreversible mixing occurs until the temperature is uniform.  Then Dalton's law of partial pressures tells us that the air and steam each fill the space as if the other was not there.  So it makes no difference to the steam, however, the total pressure in the container is the sum of the pressure due the air plus the pressure to the steam.  Each called a partial pressure in this case.  Dalton's law is close enough until quite high pressure.

And a water powered rocket?  Some one has no doubt tried it.  However rockets are about thrust to mass ratio, which of course gives acceleration.  You would need a large mass of water, plus fuel to heat it, then most of the heat goes into evaporating the steam before there is energy to accelerate the steam out of the rocket nozzles at high velocity.  And the latent heat is lost with escaping steam.  Conservation of momentum then tells us the resulting propulsive force.  Modern rockets whether solid fuel or liquid fuel usually employ chemical reactions to release much higher amounts of energy, and give much higher thrust to mass ratio, so make a more efficient rocket.  However, you might have seen those soda pop rockets where the pop bottle part full of water is pumped up with air, inverted so the neck is at the bottom, and fitted with a nozzle with a means of sudden release.  They go up in a spectacular manner, but the next continent is not in any danger.  No doubt something similar could be done with steam, but with a high probability of scolding the rocketeer, definitely not recommended.

So not silly questions, but questions which highlight the significant consequences of physics and thermodynamics.  I hope I have been able to provide at least a little clarity.  More building blocks in place.

Thanks to all who are looking in

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

Offline Stuart

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Re: Talking Thermodynamics
« Reply #139 on: July 15, 2017, 11:17:32 AM »
By ek my brain hurts in a good way

All this has reminded me of when I worked in HVAC and because we used chillers 4 500 ton units ( with now banned R11 ) I had to do a refrigeration theory course amongst other at Manchester Uni.
All the talk of  e’s ( sorry my dyslexia is bad today) made me think how much I learned in the past but now do not use ,it’s been over 22 years since I was in work but we used the SI system for the HVAC calcs

Thank for taking the time to explain the behaviour of gases

A add on topic could be the explanation as to why it takes so much energy input/extraction to change the state of , gas ,fluid or solid into the next state

Stuart
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Offline MJM460

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Re: Talking Thermodynamics
« Reply #140 on: July 16, 2017, 12:31:32 PM »
Thanks for dropping in, Stuart, I'm glad the brain hurt is in a good way, and I appreciate the thanks.  You will understand my granddaughter who recently told her mother that "Grandpa really is hard work!"

Those 500 ton chillers should be able to keep a lot of beer cold.  I never had much to do with any of the freons, went straight to the hard stuff, like propane, propylene and ethylene.  They are very good refrigerants but there is an issue of flammability that makes them unsuitable for most uses.  Same principal though.  Refrigeration is great for getting to understand the principals, and always useful.

The issue of learning all this good stuff, then forgetting overtime due to not using it often enough, I think is pretty normal.  I don't think any theory will help you find and cure a leak, or install a unit in insufficient space, and when you have the pressure of jobs to complete there is not much time for contemplation.  But over the years I have come to realise that my career was a little unusual, in that I frequently needed to use the theory, all through my career.  And so many questions are asked in the model engineering community that are not really answered, so I was prompted to have a go.  Vixens comment on 4000 strong in Hugh Currin's thread struck a note though.  I know we have a few student members of the forum, and I would be delighted if I could help them in getting started too.  This is intended to be for everyone interested, not just the old hands. 

Now your question on the amount of energy required to cause a change of state, solid to liquid or liquid to gas.  This is an interesting question which leads into some of the fundamental ideas in physics, atoms molecules and the forces between these tiny building blocks of the universe.  Just what forces hold the separate atoms together whether in a fluid state such as a liquid, or in a solid state where the strength can literally be like steel?

In a gas, the individual molecules are moving rapidly in all directions and they are a big distance apart relative to their size, apart from the occasional collision.  Conservation of momentum applies to each molecule, so they travel in a straight line until a force makes them change, either collision with another molecule, or the wall of the vessel which contains them.  So gas molecules always fill the space, and they are so far apart relative to their size that they each act independently.  But there are forces between them none the less.  These are gravitational forces, not gravity in the sense of gravity causing weight on earth, though essentially the same force in action.  Basically, any two masses are attracted to wards each other by a gravitational force which is proportional to the product of their masses and inversely proportional to the distance between them.  Symbolically F is proportional to m1 x m2 / d^2.  And there is a universal gravitational constant which completes this equation.  You can look it up if you like.  Now gravity is classed by physicists as a weak force.  The gravity that holds us on earth does not feel weak, but then compared with an atom, the masses are large, and the distance is about the same magnitude as the size of the objects, and if we have a rocket with enough energy, it can escape.  If m1 is the earth, the mass is obviously large, and m2 our body mass (even without being political about it) is also very large compared with atoms.  You can see why gravity varies from place to place, the distance between centre of mass of the earth and the centre of our body mass changes of we climb a mountain, descend to the bottom of the sea, or travel towards the poles from the equator, ( the earth is not a perfect sphere but technically an oblate spheroid, sort of like a slightly squashed ball.

But in molecular dimensions, the masses are extremely small, and the distances large compared with the size of the molecules.  Also gravity acts over very large distances, just getting weaker as the distance increases.  It is the Suns gravitational force which holds Jupiter and Pluto in their orbits, not to mention Earth.  So the gravitational forces exist between molecules in a gas, but are very small, and are classed as weak forces.  The energy in the fast moving molecules allows them to very easily escape the attractive forces between them.

Now this may not seem relevant to your question, but it is necessary to have an understood starting point.  That's probably more than enough for one session.  I will have a go at moving from this starting point to understanding latent heat of liquids next time.

Thanks for dropping in,

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

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #141 on: July 16, 2017, 01:37:25 PM »
more fascinating info and a whole new set of questions............. so in an enclosed vessel half full of air and water at sea level the water sinks to the bottom. if this vessel were taken to outer space how would the water know where to go ??  How does gravity get through the walls of the vessel ??What is the temperature of outer space and would the water be frozen ?   Just wondering really...........

Offline Stuart

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Re: Talking Thermodynamics
« Reply #142 on: July 16, 2017, 07:34:42 PM »
Thanks MJM

Ever whished you had not asked :lolb:

Now my brain is even more tasked

Yes the chillers were pretty big all to keep some IBM  main frames cool long time ago they needed chilled water to cool them
‘Bluechip “ was at the same complex looking after bit on the computor floor

The most interesting session I did a Manchester uni was on vibration analysis of machines using fast Fournier transformation but at that time the resultant graphs had to be manually interpretated hence the course , it main use was to predict a machine failure before it failed , but in practice some times it beat you the failure   :censored:

Thanks again I will follow with interest
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Offline Jo

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Re: Talking Thermodynamics
« Reply #143 on: July 16, 2017, 08:56:28 PM »
Some where you lost me: Please send beer  :noidea: I feel the need for experimentation   :naughty:

Jo
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Offline crueby

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Re: Talking Thermodynamics
« Reply #144 on: July 16, 2017, 10:30:56 PM »
more fascinating info and a whole new set of questions............. so in an enclosed vessel half full of air and water at sea level the water sinks to the bottom. if this vessel were taken to outer space how would the water know where to go ??  How does gravity get through the walls of the vessel ??What is the temperature of outer space and would the water be frozen ?   Just wondering really...........

Well, I remember watching astronauts (different from us wingnuts) up in space playing with water - when squeezed out of the bottle into the air, the surface tension kept it in a sphere bobbling and wobbling about. So, that container would likely have a blob of water floating around in it. Depending on the surface of the inside of the tank, anyway... I think.... Jo is right, we need some beer to help the thought experiment get forgotten!!
  :cheers:

Offline MJM460

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Re: Talking Thermodynamics
« Reply #145 on: July 17, 2017, 12:12:31 PM »
I had intended to continue to discussing latent heat for condensation and freezing, but not before I acknowledge today's contributions, they are all thought provoking and very welcome. 

Willy, I think Chris has answered your question about water in space.  Many thanks Chris.  To analyse that case, the energy equation would need to include a term for energy in surface tension.  Gravity is described by physicists as a field which acts through space, and difficult to detect unless you are working on those gravity wave detectors, apart from its interaction with mass.  A bit like magnetic and electric fields, though those are more obviously affected by solid materials.  When you find the complete explanation, you should check that your suit is good enough for the ceremony. 

Empty space does not have a measurable temperature.  Temperature results in matter as a result of the balance between energy in and energy out.  So analysis of the water temperature would depend on how the space ship external temperature is transferred to things inside.  The side of the craft facing the sun would be quite hot, no clouds to shield it, while the other side I expect would be cold, as it effectively radiates to absolute zero. Presumably there are heat insulating layers and clever tricks to make sure the inside surface temperature are acceptable to the astronauts.  We will eventually look at means of heat transfer either in the condenser topic or when we get to boilers.  Outside the space ship the total vacuum means all the water would evaporate and the resulting vapour would attempt to fill space.  Before long it would all be gone, unless the gravity of a nearby planet or other space body captured it.

Stuart, you are welcome, it's good to have you on board, and please don't go away, there is more.  I have set the scene, but still have not answered your question on latent heat.  That fast Fourier transformation technique (FFT) is interesting to apply to a frequency spectrum, but I suspect that there is something futile about trying to predict random events.  However the analysis can help us see them coming, if we can interpret the signs.  Basically outside my expertise, I will leave any more details to the maintenance experts, and I suspect that few if any of us have the equipment necessary to apply this to our models.

Sorry Jo, Ade has not yet included the "send beer" button on the forum.  I suspect he is waiting for Bill Gates to get his act together on incorporating a missing sub routine from Windows, so please don't overwhelm him with requests.  But I can provide a little more information for those who have not come across the tons unit in refrigeration.  Will that help?

A ton of refrigeration is the heat necessary to produce a ton of ice at 32 deg F, starting from water  at 32 deg F.  Please correct me if I am wrong, Stuart.  Obviously another infernal unit, but four times 500 ton units should be able to cool an awful lot of beer, especially if you don't actually want to freeze it.  Though I understand that you guys don't actually chill your beer, but serve it tepid.  Strange!  By the way, great to see you back fondling castings, we don't want them getting lonely, and best wishes for your upcoming post op assessment.  We all hope the reports will be good.

Now let's try and make a little progress on that latent heat question.  I described the molecules in a gas state last time.  When heat is lost from the gas, the molecules slow down.  We sense it as cooling.  As a gas cools, the molecular speeds slow, all relatively predictable and uniform. 

But as well as the weak force there are also forces classed as strong forces. (I should possibly be saying weak forces, not sure for the moment if there are other forces also classed as weak forces.  Perhaps we have a physicist reading who can chip in.)  Now the strong forces are very strong, even beside the force of gravity that holds the universe together.  But the strong forces act over a much smaller distance.  It is one of the strong forces, or perhaps several that, when the energy of the molecules gets sufficiently low, the strong attractive force captures any that come close but do not have sufficient energy to escape, and holds the atoms closer together.  This is when liquid droplets start to form.  They still have a large range of movement relative to their size, and they all still move in a random manner, though loosely connected way that we recognise as liquid, by those strong forces.  In this state, the random movement allows the boundary to be flexible and to conform to the shape of their container. Gravity over the whole mass causes liquid to rest in the bottom of a container, or a river to run down hill.
 
We observe that  heat travels from higher temperature to a lower temperature.  In molecular terms, the faster ones keep loosing more than they gain from their collisions with the slower ones while the slow ones are kept moving by the energy they gain from the faster ones. So there is no further temperature change with cooling until all the molecules join the liquid.  Then the liquid can continue to cool.  So the latent heat is the heat that must be lost by the gas molecules before they can no longer escape that close range where the strong attractive forces take hold.  Similarly, to evaporate the liquid, energy must be supplied to give the molecules in the liquid enough energy to escape the strong forces.  The ones that escape are the higher energy ones, so their escape cools the liquid.  Again, they all have to evaporate before the temperature can rise.  I hope I have explained it well enough to explain the constant temperature part of the observation.

 As the liquid molecules are cooled, they continue to get slower.  There comes a point where the free movement within the liquid becomes limited by the close proximity of the surrounding molecules, and the molecular motion becomes more of a vibration.  Molecules tend to jiggle until they drop into a matrix where they still all fit, and the constraints to movement by the surrounding molecules causes the mass to take on the properties of a solid.  But the molecules are still moving within that matrix.  They do not actually stick together.  If they get too close, they actually repel with enough force to resist the attractive force.  As solidification involves losing enough heat that the molecules drop into this regular pattern characteristic of a solid, melting involves adding enough energy for all the molecules to "pop out" or escape that matrix.

I am sorry if my terminology is a bit woolly here, but I am not a physicist, this is just a mechanical engineers understanding of the mechanism, but I hope that is sufficient to answer the satisfy curiosity for now.

Thank you to everyone who is following along, I hope not too much brain pain.

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

Offline Stuart

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Re: Talking Thermodynamics
« Reply #146 on: July 17, 2017, 01:57:32 PM »
sounds about right ,

note the chillers where at the bank about 22 years ago mainframe du dads are now better

we only chilled down to 42  deg F ( they were USA machines ) and the controls were also made by Jonhsons ) we did get them to change to a better system

the analysis was used to spot trends and to diagnose actual faults e.g. a peak at twice mains frequency would indicate a motor over load

still looking in  :)
My aim is for a accurate part with a good finish

Offline MJM460

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Re: Talking Thermodynamics
« Reply #147 on: July 18, 2017, 01:14:50 PM »
Hi Stuart, clearly you would not want to actually freeze the water in your computer cooling application.  This highlights the point that a ton of refrigeration is a measure of the amount of heat removal capacity in terms of the amount if ice that could be produced, not necessarily the amount that is produced.  Obviously a historical unit from the early days when the main use of industrial refrigeration was to produce ice for food preservation.  There were no domestic refrigerators, the ice man came around with a horse drawn trailer and deliver ice to each household, closely followed by the milk man.  I am sure that I am not the only one who can remember the family ice chest.

To finish off your question on latent heat, it is interesting that as the solid cools further, there can be further phase changes which X-Ray diffraction reveals as different crystal structures.

We are perhaps getting into deeper theory than we need to understand and build an engine, but if would like to read more about it I would suggest a little book called Six Easy Pieces, a small collection from the Richard Feynman lectures in physics. It has been very helpful to me in developing my understanding.  His Nobel prize should be enough authority for us to rely on.  And a very interesting, readable and readily available little book.  If you get hooked, move on to Six Not So Easy pieces, which are, well, not so easy, but very interesting and worth having a go at.

A shorter post this time, I need to get the average down.  Tomorrow, I will get back to heat transfer and condensers.

Thanks for dropping in.

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

Offline MJM460

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Re: Talking Thermodynamics
« Reply #148 on: July 19, 2017, 12:31:50 PM »
The heat transfer equation

To design our condenser, we need to look at the factors which are relevant to heat transfer.  The primary factors are the temperature difference, heat transfer is faster with a higher temperature difference, area available for heat transfer, a direct measure of the necessary physical size of the exchanger.  It turns out that these factors are related in a deceptively simple looking equation.  The equation that applies to all heat transfer problems is written as follows:-

Q = U x A x (delta T).

In the equation, delta is the Greek capital letter usually written like an isosceles triangle with the base flat.  However I haven't conquered the technique on my iPad, so I will abbreviate delta T to dT, it means temperature difference.

Q is the heat transfer rate in Watts (or J/s)

U is called the overall heat transfer coefficient.  The units are w/(m^2.K)

A is the heat transfer area in m^2

Delta T is the temperature difference driving the heat transfer.  Technically the units are the absolute temperature units, Kelvin, but as it is normally in the context of a temperature difference, so you can also use deg C.  However radiation heat transfer calculations, for example, must use K.

Looks like a simple equation, and it would be if any of the terms were as simple as they appear.   The equation does tell us that the heat transfer rate is dependant on the heat transfer area, the the temperature difference, and a heat transfer coefficient.  Time to examine each of these in more detail.

The simplest to understand is the heat transfer area, A.  Measured as you would expect in square meters.  In model sizes we will mostly measure in mm, so we have to deal with a lot of zeros.  One square meter = 1,000,000 square mm, which can be conveniently written as 10E6.  You might read this as 10 exponential 6, or 10 to the power 6.  When dividing is required, you can instead multiply by 10E-6, ten to the power -6.  I believe excel allows this, and has engineering and scientific notation, unfortunately not Numbers for iPad.  I tend to use 10^6 (10 raised to the power of 6) for this reason.

The other wrinkle with area, particularly when tubes or pipes are involved, is that the area based on the outside diameter is larger than the area based on the inside diameter, so you need to keep track of which is the relevant area.

The temperature difference in dT is a bit more complicated.  In our condenser, the steam side of the heat exchanger is essentially constant temperature while the latent heat is transferred.  However if we are cooling with water, the water temperature is rising as is moved through the exchanger and takes up heat.  Similar considerations if the heat transfer is between two liquids except that then both fluid temperatures are changing.  It is necessary in these conditions to use a log mean temperature difference.  It involves the temperature difference on each end as well as the natural logarithm of the ratio of the inlet and outlet temperature differences.  I can write it out if anyone is really interested, but I don't think it will be useful for most readers.

The final factor, U is the really difficult one, a look at that next time.

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

Online Kim

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Re: Talking Thermodynamics
« Reply #149 on: July 19, 2017, 03:37:22 PM »
The temperature difference in dT is a bit more complicated.  In our condenser, the steam side of the heat exchanger is essentially constant temperature while the latent heat is transferred.  However if we are cooling with water, the water temperature is rising as is moved through the exchanger and takes up heat.  Similar considerations if the heat transfer is between two liquids except that then both fluid temperatures are changing.  It is necessary in these conditions to use a log mean temperature difference.  It involves the temperature difference on each end as well as the natural logarithm of the ratio of the inlet and outlet temperature differences.  I can write it out if anyone is really interested, but I don't think it will be useful for most readers.

So, why would we use natural Log of the ratio rather than just the average of the delta temperatures between the inlet and outlet?

I'm finding this thread quite interesting. Thanks MJM!
Kim