Author Topic: Talking Thermodynamics  (Read 105121 times)

Offline MJM460

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Re: Talking Thermodynamics
« Reply #870 on: April 30, 2018, 11:16:26 AM »
Hi Willy, thank you for some feedback on my answers.  As you get to know enough to understand  which topics to look for in text books, and how to use the information, it becomes easier, and quicker.  Hence my reluctance to give a yes or no answer, I feel it is important to understand the why, rather than just learn facts.  Of course long winded answers are no use if they are not understood, so I do hope that my answers are helpful in increasing understanding.

Your bath problem is a typical heat transfer problem which shows some of the complexity in arriving at an exact answer.  The basic approach is to start with the usual equation, that says heat loss is proportional to area, temperature difference and heat transfer coefficient.

Each litre of water, (close enough to 1 kg), has to lose 4.185 Joules of energy to cool one degree.  So if you release half the volume of water down the drain, you have only to wait for half the heat loss required to be lost from the full tub, but that is not the same as half the time.

Cooling is an unsteady process, meaning that the temperature is changing as the process proceeds, so the temperature difference reduces during cooling, which is the reason your coffee cup cooling and your boiler cooling curves have their particular shape.  As you will not want to reduce the temperature all the way to room temperature you will be operating on the steeper part of the curve.

Heat is lost from your bath by a combination of convection from the walls and bottom of the bath tub, and evaporation from the surface.  How much each contributes depends mostly on the tub installation.  If the bath is built in with a wooden frame, some insulation around the tub and perhaps tiles to finish the now straight and regular shape, convection may not be very important due to the low transfer through that construction.   However, if it is an old fashioned galvanised steel tub, on cast iron feet in the centre of the room, convection would be much more important.  The heat lost from the surface by evaporation would be similar regardless of the installation, but would be significantly influenced by any draft which increased air circulation.  So the heat transfer coefficient is not really the same when you lower the level.

Now a bath tub is not usually exactly rectangular, but it is reasonable to assume the vertical sides area would be approximately halved in height, so that halves the heat loss in that area.  Again, the heat transfer area is not the same as you lower the level.  The bottom is not changed.  On the other hand the area for evaporation is not significantly changed, though the surface is a little more sheltered from air movement. 

So you can see the area for heat loss is reduced bit not exactly proportional, and the overall effect depends on the bath installation.  I would be reasonably confident that the remaining water would take less time to cool, but not as little as half.

Finally, it is worth remembering that the heat in that water has been paid for through your utility bill.  If you run the excess water down the drain, that heat is lost with the water.  If you wait for the bath to cool, the heat is retained in the house, and contributes to your comfort.  If you have a thermostat controlling the heating, the excess heat in the water would definitely tend to be off set by a reduction in heat used for house heating. 

If we delve a little deeper into theoretical considerations, cast your mind back to your cup of our coffee and the teaspoons, you could use a large number of teaspoons, or perhaps a lump of cast iron to lower the temperature, but retain the total amount of heat, (providing the bath did not overflow!) but probably not worth the risk of injuring your back or chipping the bath enamel.  Better to lower the level by pumping the excess water to a tub in the lounge room, (obviously using a steam engine driven pump) where it would help heating the house, avoid the overflow risk, and by using a pump and hose, minimise the risk of spoiling the floor by spilling water.  Ok,  getting silly again, but I hope that helps with understanding the concepts of heat compared with temperature, and how the various factors affect heat transfer and heat transfer rates.

With regard to your question about flywheels, the answer is very clearly yes.  I am still admiring your lateral thinking in using a bar for extra inertia to get our engine running.  It did not add to the energy driving the engine, but encouraged you to tweak the timing and eventually get it running without the bar.  Extending that thought to a practical method of determining the required flywheel for a new engine is the reason I have continued discussing that idea.

This is how it works.  By drilling and reaming a rectangular bar to suit your engine shaft right at it's centre of mass, and adding a set screw to hold it in place, you can quickly produce a flywheel of any desired moment of inertia.  Not too much of a chore to produce a larger one if it seems too small, or a smaller one if that seems desirable.

I would suggest a starting point is to sketch a flywheel that "looks about right", and calculate the moment of inertia of the rim.  This is a simple calculation using one formula twice.  First to calculate the moment of inertia of a solid disk the same o.d. as you assumed flywheel, then the same formula a second time for the inertia of a disk the same diameter as the i.d. of your rim and subtract this from the the first answer.  Don't worry about spokes and hub.  The disk formula is
  I = 1/2 x M x R^2.  R is the radius of the rim, or half the diameter. 

For mass, calculate the volume of the disk, and assume a density of about 7200 kg per cubic meter for C.I. or 7800 kg per cubic metre for steel.  Even  use 2800 kg per cubic meter for aluminium.  Use metres for the diameter and width measurements.

Now calculate the size of a rectangular bar of the same moment of inertia.  Use the formula for a bar about its centroid, I = 1/12 x M x L^2.  L is the length of the bar, M is the mass.  Again for the mass, calculate the volume of the bar, and use the appropriate density.  Cut a suitable size bar from whatever you have available, drill and ream at its centre, use a sander or file to balance the bar on your shaft, and weigh it to check the density.  Use the actual weight to update your bar calculation.  Then install it on your engine. 

The flywheel is large enough if the engine runs smoothly to your satisfaction, at the slowest speed you require.  Remember, the slower you want to run, the more moment of inertia you need. 

You can easily increase the moment of inertia of your bar by bolting a short length to each end, using bolts in shear (parallel to the shaft) with nuts or tapped holes, and making sure you keep the assembly in balance.  Each add on mass adds M x R^2 to the inertia, where R is the radius to the centre of mass of the added mass.  Or if necessary you can start again with a longer or wider bar.

If the engine runs very smoothly you may want to try a lighter flywheel, so start again with a shorter bar.  However, as you started with a size for a flywheel that looked right, there is probably no need to go smaller, simply demonstrate that it is adequate.

I hope you can see that this procedure gives a quick and cheap way to determine the moment of inertia you need.  If you intend to drive a load, you may need to try another temporary flywheel to provide more moment of inertia, depending on the load characteristic, if the load device was not available for your initial testing.  I know they I always want to prove I have a runner at the earliest opportunity.

I hope that answers the question in a way that you can use.

Thanks for following along,

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 #871 on: May 01, 2018, 02:19:16 AM »
Hi MJM, just a quick question   (and is the length of the answer always inversely proportional to the question!!!) does anything connected to the flywheel {large gears} ? add to the MoI ?.........
Willy

Offline MJM460

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Re: Talking Thermodynamics
« Reply #872 on: May 01, 2018, 01:22:34 PM »
Hi Willy, now that is two questions, surely you don't expect both answers to be short?  Oh well, perhaps I could try, so here goes.

The answers don't have to be long, but you have a real talent for asking complex questions, at least more complex than they look.  But don't give up, simple questions would make for a boring thread.

Yes, every part of the rotating system has its own moment of inertia, and they all add up to contribute to the total.  The crank shaft, balance weights, couplings, the whole lot.  If there are gears in the train with a ratio other than 1:1, allowance has to be made for the speed.  Alternators normally run much slower than a turbine driving them, but the low speed means their moment of inertia has less influence on the high speed turbine, as less energy is stored at low speed.  But a high speed centrifugal compressor driven by a normal electric motor has much more influence relative to its mass and moment of inertia.

I do hope the answers are both short enough, and complete enough, but please ask if there is something I seem to have left out, or if the short answer prompts more questions.  But in the spirit of the questions, I will give everyone's brain a rest by making it a short night.  Might even get to bed a bit earlier myself.

Thanks for looking in,

MJM460


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Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #873 on: May 01, 2018, 03:14:57 PM »
Hi MJM , I don't mind long answers as everything is relevant !  I was just making an observation !!!! ;D  also i usually make these posts just before i go to bed so am a bit tired and things just pop out of my brain and get typed/triped out randomly !!! So todays question does everything in the motion train have a positive MoI ? as you say  'add up  to' so does that actually mean an increase?
Willy

Offline MJM460

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Re: Talking Thermodynamics
« Reply #874 on: May 02, 2018, 08:35:13 AM »
Hi Willy, I may be back to inversely proportional on this one.  I hope I am passing the relevance test.

First, the simple answer is yes, you simply add the moment of inertia for each component to get the total, so long as they all rotate at the same speed.  Just the same as you add the mass of all the components to calculate the total mass in a linear system.

In fact, the moment of inertia of any object is calculated by adding the moment of inertia of each tiny element of the object, down to crystals or even atoms if you must, but the process of addition is "simplified" by the mathematical process of integration.  I am sure you don't want me to go into integral mathematics.  However, in simple words, it is a process of adding the contribution of many small components.

The formula for every component, no matter how small is M x R^2, where R is the distance from the axis of rotation.  The formula for a solid disk I = 1/2 x M x R^2 is simply the result of that integration performed over the whole disk.

For our purposes, rather than perform the integration for a complex shape, we can often use known results for a few simple shapes and add them all up.  The known formulae are normally about the centre of mass (which is always the smallest moment of inertia in that plane of rotation) or some other defined axis of rotation.  For a different parallel axis, you add the moment of inertia about its centre, and the moment of inertia of that mass about the required axis, M x s^2.  Where s is the perpendicular distance between the axis through the centre and the required centre.

If you want to spin your flywheel like a coin on edge, there is a different formula for that axis.

Now that formula always results in a positive answer, there is no negative moment of inertia.  Mass is always positive, and even if your chosen reference point for zero results in a negative R, R^2 is always positive, so M x R^2 is always positive.

However, the easiest way to calculate the moment of inertia for a rim is to imagine you first make a solid disk, then remove the centre.  The moment of inertia of the centre you remove can be subtracted from the larger disk to leave the moment of inertia of the rim.

You continue building up the moment of inertia of a flywheel by adding on the spokes and the hub, or subtracting the moment of inertia of any part you machine away from your initial solid disk.

You would use a similar procedure if you wanted to make a novelty engine with your set square for a flywheel, but including a guard would be a good idea to save fingers.

Note also that that formula does not include g, or any contribution from gravity.  Your flywheel has the same moment of inertia on the moon or Mars or wherever.  In fact it may work a little better on the moon, as there would be little or no air resistance, and the lower gravity means less force on the bearings due to the mass, so less friction.  Give it a spin and it should go for a very long time.

By the way, the discussion since I first looked at this topic has prodded me into going back to my post #849 on 21 April, (on page 57 with my forum settings) on calculating flywheel moment of inertia, and editing it to try and clarify a few points that were badly or incompletely expressed, if not just wrong.  I feel it is better to go back and edit, rather than leave errors stand with corrections inserted many pages later.

Thanks for following along,

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

Offline derekwarner_decoy

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Re: Talking Thermodynamics
« Reply #875 on: May 02, 2018, 12:16:36 PM »
Just a small observation on the comment.....

"if not just wrong.  I feel it is better to go back and edit, rather than leave errors stand with corrections inserted many pages later"

Absolutely......100 years ago we read, studied & learnt by fixed paper text....today with the advantage of live electronic documents, any revision as deemed necessary is to the advantage or the new reader if revised in the original body of the text

This could be simply achieved by inserting the revised text as...eg., jhbs  dvhdv  bsdbhb  Color or Font change

This revision need not reference date or change to the original text, however is a backstop to alert any new reader accessing an older version of the document that may have been stored on an external facility...[zip or disc drive]

Derek
Derek Warner - Honorary Secretary [Retired]
Illawarra Live Steamers Co-op - Australia
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Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #876 on: May 02, 2018, 11:50:04 PM »
Hi MJM, wondering about the correlation between the flywheel MoI helping the engine go over top dead centre and the revolving load 's MoI trying to stop it ?? or is that incorrect /silly ??? I am thinking this should be a relevant point to consider ?..thanks....Would it be helpful/quicker to use a planimeter to do some of these calculations ?
Willy,
« Last Edit: May 02, 2018, 11:55:28 PM by steam guy willy »

Offline MJM460

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Re: Talking Thermodynamics
« Reply #877 on: May 03, 2018, 11:43:22 AM »
Hi Derek, it is quite a good feature to be able to go back and modify a post when necessary.  Then anyone who jumps in using the search box gets the latest information instead of having to wonder if any corrections were ever posted.  So long as they bookmark the post and do not print or save the text.  Another world from where you and I grew up, with only paper books.  I put a lot of effort into avoiding errors, and correct them when I notice them, however, sometimes the wording could have been clearer when read in the light of a new day.  If you see any other posts which need updates please let me know.

Hi Willy, right on topic and clearly describes the issue you are grappling with, so not silly at all.  Let me try and make it all clearer.  It is not as easy as the short question might make it look.

First it does not matter whether the moment of inertia is part of the engine, the flywheel, or the driven load, it is only the total moment of inertia that matters. 

It is the same in a linear system, one that is more familiar to most of us.  Suppose you are pushing a load on a cart along a smooth level path.  If we assume you have nice ball bearing wheels with negligible friction, only the total load that matters.  It does not matter if the load is a heavy iron cart with a load of a few aluminium blocks, or an aluminium cart with a load of iron, (providing the cart is strong enough!), or even a long plank with a weight on each end.  The equation is F = M x a, or force equals mass times acceleration.  There is no velocity term, velocity does not matter, only acceleration or change of velocity.

You have to push to get the cart moving, because that involves changing the cart's momentum, but once it is moving at your walking pace, you only have to overcome that negligible friction to keep it going.

If you are feeling fit and decide to break into a run, you have to push again to increase the speed and momentum of the cart to your running speed, but then, to keep it going at constant speed, again you only have to overcome that negligible friction.  Do young mum's push those jogging prams in your area?  They make it look effortless, on the flat!   But stay off hills and corners.  We will get back to corners.

You can analyse the system by looking at kinetic energy, 1/2 x mass x velocity squared, this tells you how much work that kinetic energy can do before the system comes to a stop, but it does not tell you much about the time it will take to stop. 

You can also analyse the system by looking at momentum.  Momentum = Mass times velocity.  It is not hard to demonstrate that change of momentum involves force.  In fact change of momentum per unit time equals force.  If you measure the time in seconds that the known mass takes to stop from a known speed, you can calculate the magnitude of the resistance to motion.  Conservation of momentum and conservation of energy provide two independent equations which can be used to analyse a linear dynamic system.

A rotating system is very similar, quite accurately described as analogous.  If you replace mass with Moment of Inertia, and linear velocity with angular velocity, then replace force by torque the same equations apply.  Of course if you use a plank with a mass attached to each end, it is still only the moment of inertia that counts, but in this case the change will be much higher in proportion to the mass of the plank and weights.

So, to try and keep to your specific question, the moment of inertia of the whole system contributes to the angular momentum and kinetic energy, and is important only while the rotational speed is changing.  It stores energy and increases in momentum when there is excess torque available, and returns energy and decreases in momentum when the engine torque is insufficient.  The driven load is trying to slow the whole system all the time, not just at the dead centres, unless the load is also a reciprocating machine. 

The moment of inertia only resists the increase of velocity as it builds angular momentum and rotational kinetic energy when the engine has enough torque to accelerate the system.  But, at the top and bottom dead centres, the system is actually slowing down, so all that the momentum and kinetic energy stored by that moment of inertia is actually applying the positive torque to carry the engine through those dead centres.  The more moment of inertia, the greater the positive torque and less slowing down. 

Now I said earlier that we would get back to corners, but that requires quite a bit more information, so it would be a very long post.  I will save that topic for tomorrow.

You also asked about planimeters.  Integration gives the area under a line in a manner sort of the reverse to differentiation giving the slope of a line.  One way of doing integration would be to plot the curve of the line on graph paper and determine the area under it.  I did learn to use a planimeter, but they were too expensive for me to own, so I generally counted squares on the graph paper.  Where the equation to the line is known, the answers to the integration process are also well known, so the mathematical method would not only be quicker, but more accurate.  A planimeter would come into its own if the equation of the line is not known, say for an indicator diagram, though these days a planimeter is superseded by computer programs which calculate the area enclosed by a line traced out on a digitising pad or perhaps even a touch screen, or better still a photograph.  Or, in absence of a suitable program, just by counting squares.

I hope that answers your question without being too long,

Thanks for following along,

MJM460


« Last Edit: May 05, 2018, 12:00:13 PM by MJM460 »
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Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #878 on: May 04, 2018, 03:05:01 AM »
Hi MJM, thanks for the latest posts, a bit busy with the council elections and leafleting and the allotment as it has stopped raining at last...so no questions tonight !! I don't mind long answers actually !! I did ask a question a few posts ago about a steam engine that has the fire heating the cylinder and water being injected, so dispensing with the boiler ?? i have not heard of one ,but would this work ??
Willy

Offline MJM460

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Re: Talking Thermodynamics
« Reply #879 on: May 04, 2018, 12:51:25 PM »
Hi Willy, just as well there is no new question, as I was going to continue on the effect on a spinning system of going around corners.

That engine, are you thinking of the Hargreaves engine?  Or a different one.  It actually had the combustion gases entering the cylinder if I remember, so not quite what you asked.  If you think about the heat transfer area necessary to raise enough steam in a boiler, it is obvious that there is not enough area on the outside of a cylinder to raise enough steam.  Was it the Trevanick(?) that had the cylinder inside the boiler.  That would give a suitable amount of area and give an idea of the proportions needed, but could be described as applying the fire to the outside of a (jacketed) cylinder.

Yesterday I said I would return to the effect of going around corners, or turning the axis of a rotating system.

Velocity is a vector, which means it has both magnitude and direction.  So a change of direction is a change of velocity, even when it's magnitude, the speed, stays constant.  Force is also a vector, so also has magnitude and direction.   So to continue moving in the new direction, the force also has to change direction.  Fortunately, the direction of force and velocity is quite obvious.

Now, angular velocity is also a vector and has an associated direction.  Similarly torque is a vector, as is an angle of rotation, and each not only has magnitude, but also has an associated direction.  Of course you will immediately think of rotating in a clockwise or an anticlockwise direction, but those actually are different magnitudes.  If you take clockwise as the positive direction, then anticlockwise is just a negative magnitude, but the direction of the axis of rotation is still the same.  This is not easy to imagine.  The direction of a rotation is defined as the direction of the axis of rotation.  The positive direction is defined by a "right hand rule".  If you wrap your right hand fingers around the shaft in the direction the object is spinning, the positive direction is defined by the direction your thumb is pointing.  Similarly for a torque, the direction is the direction of the axis about which the torque applies.  And again the right hand rule defines the positive direction of a torque.

These definitions are somewhat arbitrary.  Other definitions could be used, so long as the same definition is applied everywhere the direction applies.

If we look then at momentum, momentum equals mass times velocity.  Mass is not a vector and only has magnitude.  But momentum is a vector like the velocity and has the same direction as the velocity.  However, when we look at Newton's law, and conservation of momentum, if we apply a force to something moving the direction of the force and the velocity interact to change the direction of the momentum in the direction of the force.  The equation involves the cosine of the angle between the direction of the force and the velocity, but the result is reasonably intuitive.  It is a two dimensional system which can be solved by arranging the vectors in a triangle.


So how does this work in a rotating system?  Suppose your flywheel rotates on an axis that can change direction.  Think of the front wheel of a bicycle.  The direction of the angular momentum of the wheel is to the left for forward motion.  If you apply a torque to change this direction using the handlebars (assume the stem is close enough to vertical).  Using that right hand rule, the direction of the torque (for turning left) is vertical upwards.  The torque will cause a change of direction of the angular momentum of the wheel.   It becomes a three dimensional problem.  (Another right hand rule is used, involving the thumb, first and second fingers to show the direction of that rotation, but here it gets more complicated, because you have to multiply the torque and angle turned in the right order, to get the right direction.  That requires delving further into vector maths than I intended, so I hope it sufices to say that this process is the mathematics behind precession of a gyroscope.)

The simplest way to visualise and then solve the problem is probably to take the front wheel off your bicycle, hold the ends of the axle with the wheel between your outstretched arms.  Start the wheel turning, then try to change its direction by turning the axle.  Even a slow turning wheel will tell you the answer very quickly.  That is a very minimal description of the procedure, but is probably more than enough vector maths for most of us.

Ok, I got a bit of the subject of yesterday's question, but I mentioned gyroscopic effects in a recent post, so I wanted to touch on the topic somewhere.  But it is about change of momentum!  I hope you found it of interest.

Thanks for looking in,

MJM460

PS Some aspects of this post as originally made did not stand up to re-reading the next day.  So I have removed the unhelpful wording and had another go at discussing these ideas in the next post.

MJM460
« Last Edit: May 05, 2018, 12:16:29 PM by MJM460 »
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Offline MJM460

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Re: Talking Thermodynamics
« Reply #880 on: May 05, 2018, 12:32:08 PM »
I often think of this thread as a build log, as in building a knowledge base.  Well, what is a build log without the odd day where the parts so carefully crafted are consigned to the scrap bin?  Yesterday seems to be one of those days.  It was a bit like engaging the cross feed instead of the longitudinal feed, but was revealed much more like slow motion.  Amazing how the subconscious works while you are asleep.  I hope no one has spent too much time trying to follow the detail of yesterday's post.  Please bear with me while I have another go.

Basically, if you apply a torque to a spinning object there are two possibilities.  If the torque and angular momentum of the object both have the same direction, then the rotational analogy of Newton's law applies and torque equals moment of inertia times angular acceleration.  The object spins faster or slower, depending on whether the torque is positive (accelerating) or negative (retarding) relative to the spinning object.

However, if you apply a torque that is at right angles to the axis of rotation, the speed of rotation of the object does not change, but only its direction.  Newton's law still applies, but must be expressed in vector terms.  The direction of angular acceleration in this case is at right angles to both the torque and angular momentum.  The spinning wheel moves in response to the torque in an unexpected direction. 

There are two ways of doing vector multiplication.  One is usually called a dot product, A . B = scalar result.  For example work (not a vector) is the dot product of two vectors Force and distance.  This product obeys the normal rules, A . B = B . A.  The other way is called a cross product and results in a vector answer.  Unfortunately A x B = - B x A, so the multiplication has to be done in the correct order.

There is a second right hand rule, involving the thumb, forefinger and middle finger, which gives the direction for that angular acceleration in accordance with the rules of vector multiplication.  Torque and angular momentum use the cross product and result in a vector result, angular acceleration.   A simple experiment with a bicycle wheel is probably the easiest way to find the correct direction.  Then, remember whether the torque is the thumb or forefinger, the other is the direction of the angular momentum, while the middle finger then points in the direction of the angular acceleration, so tells you which way the wheel will move.

This is the basis for the unexpected actions of a gyroscope or a spinning bicycle wheel held by the ends of the axle.  You can also feel it with a large portable circular or angle grinder.  The unexpected reaction to movement of spinning wheels adds to the danger of handling them.

If the applied torque is neither parallel or at right angles to the angular momentum, you can use a force or torque triangle to resolve the torque vector into two components, one in the direction of the angular momentum and one at right angles.  You can then apply the effects of each component separately.  The parallel component will change the spin speed, while the component at right angles will change the direction of the spin axis.

I hope that makes yesterday's topic a little clearer.  I have gone back and edited yesterday's post to  improve it a little, and referred to today's post for additional explanation.  It wasn't too bad.  I have removed a doubtful clause and a couple of sentences which did not help convey anything useful.  I hope that will be adequate to convey the best description I can, for a topic that is often seen as mysterious, and to avoid sending anyone the wrong direction.

Thanks for following along,

MJM460


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Offline Admiral_dk

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Re: Talking Thermodynamics
« Reply #881 on: May 05, 2018, 08:42:48 PM »
You talking about dot products and the hand finger method for the resulting vector, really reminds about how much I have forgotten of my study years  ::)

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #882 on: May 05, 2018, 11:09:44 PM »
Hi MJM talking about  'right hand rules'   when i studied electronics we also had a "Flemmings right hand and left hand rules' !! So there must be quite a lot more in different  sciences !!! No questions today just an observation about locomotives using propellers for the drive ??? and Los angeles painting the roads white to stop the black tarmac absorbing the suns heat and giving it back strait away to make everybody feel very hot ??!!!  Vectors and convection ...any correlation ??  Oh dear i seem to have posted a few questions actually !!!!!
Willy
« Last Edit: May 05, 2018, 11:13:09 PM by steam guy willy »

Offline MJM460

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Re: Talking Thermodynamics
« Reply #883 on: May 06, 2018, 12:45:21 PM »
Hi Admiral DK, I think most of us are in that category, but it is surprising what will rise again to the surface when suitably prompted.  We are all taught about vectors having magnitude and direction, and we learn from Newton's law, F = m x a.  I often wondered how to apply this to a change of direction but never had time to really sort it out.  Easy for the linear case, but harder to grasp for the rotational analogy.  More recently, thinking about vectors, I made some connections.  Retirement has given me time to assimilate some of the information I gained from education and experience, and make some appropriate connections.  Definitely not there yet, plenty still to keep me going a long time.

Hi Willy, quite a collection of thoughts in different areas today.  My textbooks only have a Fleming's right hand rule.  And it is definitely used in the context of the vector cross product in relation to forces on current carrying conductors in magnetic fields.   The right hand rule goes with the right handed x-y-z coordinate system.  I have not heard of a left hand rule, but that does not mean there isn't one.

Propellor driven locomotives are not exactly intuitive, but some say neither is flying heavier than air aeroplanes, especially those stunt planes that can fly vertically upwards.  A propellor is a thrust producing device.  The momentum of the air is increased by the propellor which applies a force to the air, and the propellor experiences an opposite thrust from the air.  So if the propellor can change the velocity of a great enough mass of air, the thrust is enough to drive a boat, hovercraft, aeroplane or, I guess, even a locomotive.  But you would need to keep well clear of that propellor.  However, thrust is a force, it has direction so is a vector.

Painting roads?  A white reflective coating would reduce the heat absorption by the asphalt, which in return would reduce the convection heating of air.  The reflected radiant heat is not so well absorbed by the air, so some eventually gets back into space.  Even if it is all trapped in the atmosphere, it is better distributed, so you don't get quite such high localised temperatures in the ground level air we occupy.  So it would add to comfort, though it only helps lessen global warming to the extent that some of the reliant heat is reflected back into space.  You might be interested to know that despite us using high temperature formulations for road making here, it actually starts to melt on a sunny day, so perhaps painting it white would help.  But the glare would be horrific, especially early and late in the day.  But I don't believe temperature and heat are classified as vectors, so another topic really, even though heat does transfer in the direction of the temperature gradient.

The interaction of torque and angular momentum produces an angular rotation at right angles to both the torque and the angular momentum suggested by the cross product of the vectors, 

If we use the equation for change of momentum per unit time (=T) and the equation of motion, change of angular velocity equals angular acceleration times time, we can use a bit of algebra to demonstrate that Torque is proportional to the angular velocity times angular acceleration.  In vector terms, the multiplication using the cross product definition seems to give the correct direction of precession providing you multiply omega cross alpha, and not the other way around.  (Remember alpha is the rate of change of angular velocity, or angular acceleration.)

For those interested, using w as omega or angular velocity, and subscripts 2 and 1 for final and initial values.  Then a as alpha for rate of change of angular velocity, T for torque and t for time, I for moment of inertia and use * for multiply to avoid confusion with the two vector operations.  We have the following equations-

 w2 = w1 + a * t

T = I /t * (w2 - w1)

If we substitute the first equation in the second, we find T = I/t w1* a

As w and a are vectors we need to use vector operations, either dot product or cross product.  The physical phenomena seems to require the cross product option.  So T = I/t * w1 x a.  With the operation in this order, the thumb points in the direction of w1 the index finger the direction of the applied torque and middle finger indicates the direction of the torque.  Sorry about the w and a, instead of omega and alpha, it may make more sense if you write it out with a pencil using your preferred notation.

For those shy of a little maths, that last little section can be skipped over.

Thanks for following along.

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 #884 on: May 06, 2018, 11:04:23 PM »
Hi MJM, I found a rule at the car boot sale today and have learnt a new word that i have never seen before  Intumescent  But i don't know what the compound actually is ??
Perhaps you could en light en us !!!!!!
Willy