Author Topic: Talking Thermodynamics  (Read 194710 times)

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #375 on: October 19, 2017, 06:12:55 PM »
Hi MJM, first, i was going to incorporate it into the heater sheath if i made a new boiler...and if it only requires 50 watts could i make it so the saturated steam pipe only goes in and out of the sheath for  1/10 of the length. this would require some cunning metal work !! Secondly. would it be possible to calculate the temperatures in the different places on the drawing when the boiler is at full steam production  A to G ??

Offline MJM460

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Re: Talking Thermodynamics
« Reply #376 on: October 20, 2017, 01:08:16 PM »
Superheater design continued -

Hi Willy, so I was not too far out on your intention for incorporating the superheater in the boiler, but first the temperature questions.  Easy ones first.

H, atmospheric temperature, I am assuming 15 or 16 deg, i.e. your summer, not mine!

B, C and D, water and steam in the boiler, reasonable to assume all three are equal, and equal to the saturation temperature for the pressure, especially once the air is mostly gone.  The heat transfer within the boiler is largely facilitated by the mass transfer associated with vigorous boiling, which does not look much like the benign situation with a clear liquid interface like you have drawn.  The huge volume expansion when a little bit of water flashes into vapour and the associated huge density difference mean it is very turbulent, vapour quickly rises and is replaced by more water.  This is why the boiling heat transfer coefficient is so high, and the temperature is reasonably uniform throughout.

E, steam in the outlet pipe to the engine, same as B, C and D above, providing you have some insulation wrapped around it to minimise heat loss before the engine.

Calculation of F, the outside temperature of copper is a fairly straightforward variation of a problem illustrated in most heat transfer texts  In reality, it is easier because earlier we did calculate the total heat loss from your boiler from your test run.  If we proportion this heat loss to the area of the ends and the cylindrical part of the shell or put extra insulation on the ends and ignore the ends, we can use the standard formula to calculate the corresponding outside temperature.  The formula is a bit more complicated than for conduction through a flat plate because as the heat travels outward the area is increasing due to the increased radius.  It is getting late here and I have an early morning coming, so I will give you the formula, and you can easily work it out.

The basic formula is q = 2 x Pi x k x L x (T1-T2) / ln (R2/R1)

Here, q is the heat loss in Watts which in your case we already know, k is the conductivity of copper, use 399 W/m.K, which is actually for pure copper, and small amounts of alloy reduce it markedly, worth trying how much difference it would make if it was only 200, and brass is only 111.  T1 is your steam temperature, T2 is the outside temperature you are looking for.  L is the length of the boiler in m, R1 and R2 are the inside and outside radii of the shell, also on m.  The function ln means log to base e, or natural logarithm, and is one of the standard functions provided in any spreadsheet program or scientific calculator.

So you can manipulate the formula to find T2, using a calculator, or just use trial and error in a spreadsheet to find the temperature that gives the right answer.  That is the joy of a spreadsheet, but that formula is not difficult to manipulate to find T2 directly.  You might also know that if R1=R2, R2/R1=1, and ln(1) = 0, so T2-T1 =0 as you would expect, and it is still very small when the difference between R1 and R2 is small, as in a thin boiler shell.  But that ln term starts making a difference for a thick shell.

Having found T2 you can apply the same formula to the insulation, using k = about 0.035, to find the temperature difference across the insulation as q is the same for both insulation and shell.  This figure will be a little smaller than the difference between the copper temperature and the atmospheric temperature as there is a film resistance between the insulation surface and the air, and a contact resistance between the copper and the insulation.  There are standard formulae which enable us to calculate these film resistances now your test run has given us a heat loss based on the heat up phase, (as I have mentioned before they are hard to predict purely on theory).   You have already done the experimental determination, we just have to find our way through the calculations, which I will have a go at tomorrow.

Still have to tackle the temperatures A and B at the heater, and then look closer at your superheater design.  I need to get a couple of sketches into small files for that, so it won't happen tonight.

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 #377 on: October 20, 2017, 05:11:01 PM »
Hi MJM , thanks for this ....ok .....a thermodynamic conundrum.....If the element is rated at 500 watts and it is 100 mm long does this workout at 5 watts per mm  ?  the total heat given off is  X degrees but at each mm is the heat given off only X ÷ 500 degrees ?  the heat given off is uniform from every square mm so perhaps not ?? Am i being a bit silly here or is this a valid question......and it  relates to only getting a 50 watts input for the superheater input from the existing 100 mm long element. !!

Offline MJM460

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Re: Talking Thermodynamics
« Reply #378 on: October 21, 2017, 10:32:04 AM »
Hi Willy, not silly, but a little bit of wholly thinking there.  Remember to be consistent with the units.  You can add apples and oranges to make fruit salad, but you can't multiply them, and you never get pears.  If you stick to consistent units, the issues will be easier to understand.  The total heat from the element is 500 watts, the name given to Joules/sec.  Degrees is not part of it.  Then heat transfer comes in, a bit like flow of energy, and the flow requires a temperature difference to drive it, just like flow of electrical energy requires volts.  You can even use current as an analogy for the energy flow.  Strictly, physicists don't like to call it flow, but it's a bit like flow for most of us.  Your heater generates an amount of energy, or power, which you can calculate as V x I, or V^2/R, by conversion of electrical energy and it will be uniform along the length of the wire, just as the resistance is uniform along the length of the wire.  We don't know how the wire is arranged inside the insulation in the sheath, but reasonable to assume it is distributed uniformly, probably in a coil, possibly even on a ceramic former, we just don't know.  If you are picky, the spec sheet you posted a while back says the is a short non heating bit on each end within the 100 mm, so probably only about 85 mm, but as some heat will flow axially to the ends of the sheath 100 mm is near enough for our purposes.

Now all the heat generated by the electrical energy must escape.  If the heat flow out of the element is less than that generated, the extra heat will be stored in the form of higher temperature of the element.  So the element temperature goes up, and eventually the temperature difference between the wire and the water is enough to drive heat transfer equal the heat generated, and we at last get a stable temperature for the wire.  The temperatures are determined, not by the heater energy input, but by the heat balance, and the law of conservation of energy.

Now the heat must flow through the insulation within the sheath, the spec sheet says it is magnesia, which has a conductivity of about 2.68 W/m.K, and we don't know the thickness, then through the manufacturers sheath which I think is stainless steel, conductivity 14.4 W/m.K, again thickness unknown, then through the contact resistance between the stainless steel and your brass component, on through your pressure containing sheath, brass, conductivity 111 W/m.K, you know the thickness of that, and finally through the boiling liquid film into the water, which is very well stirred up, so a uniform temperature except for a very thin film very close to the the surface of your sheath.  All those thermal resistances need a temperature difference to drive the heat transfer.

We can actually make an estimate of the film coefficient using the general equations and your boiler test results.  I can go through it if you wish, but for the moment I will just tell you that from your boiler test, it is in the range of 5300 W/m^2.K, (which is very modest for boiling water, for which it can range from 3,000 to 60,000) based on the outside area of your pressure containing part, and this means the outside surface temperature of the brass will be in the range 10 to 100 deg hotter than the water, I suspect more likely less than 50.  The temperature gradient through the brass will depend on the thickness, but it is of the order of 3 degrees.  The temperature of the wire looks like it will be 30 - 200 degrees above the stainless steel sheath temperature depending on the thickness of the magnesia, possibly only 0.5 to 1 mm thick in your 10 mm diameter heater.

When you look at all those components, clearly your high conductivity brass is the least of the problem, and with the high film coefficient to the water, no fins are necessary.  Not sure about that contact resistance, but the spec sheet says you need H7 fit, and a design that would accommodate that heat transfer grease would be a good idea.  Clearly the highest resistance is the magnesia insulation, which by the way, allows enough current leakage to trip an earth leakage relay, so you need a good earth so that your equipment cannot build up a voltage, and I don't know what you have to do about the earth leakage protection.

The whole arrangement so far is all symmetrical and radially uniform around the centre line of the element, and it is easily analysed with simple one dimensional techniques only slightly modified.  You can make a simple electrical analogy with temperatures and thermal resistance analogous to voltages and electrical resistance.  With constant temperature water and constant heat generation along the length, it is all very uniform. 

 When you add your superheater tube, all that changes.  If you make a cross section through your pressure containment, you now have three specified temperatures, the element, which must be the highest temperature if heat is to flow the right way, and the superheater must be well above the water temperature as we saw in the engine performance discussion.  You now have three paths instead of just one, you have heater to superheater, heater to water, and superheater to the water.  The geometry of these paths is quite irregular, even though the high conductivity brass is the main heat conductor.  It can be solved, but requires finite element techniques.  Even then, it is not simple as it is a three dimensional problem.  As the steam temperature rises in the superheater, the temperature distribution has to change, so that the heater is still the highest temperature, and so that enough heat is transferred into the superheater despite much lower film coefficients.  This means there is no constant temperature along the length of the element.  Unfortunately I don't have a suitable programme.  I know we have some students on the forum and I hope they are reading.  They may be able to help.  These days students seem to use finite element like we used a slide rule, and it is good they spend their time that way, it is a very powerful tool once you know how to drive it, and enables solving of problems that can never be tackled with a slide rule, a calculator, or even a spreadsheet.

Generally, I would expect the placement of the superheater would be quite critical, it has to be close enough to the heater to get enough heat and force the heater temperature high enough, while leaving an adequate heat path direct to the water so the element cools enough.  In addition,  the rising superheater temperature would mean there are longitudinal heat flow as well as radial so not at all easy to analyse.  Any required adjustment after a test run has to be changed by changing the geometry, so not an easy exercise, but is your only available variable to control the heat flows.  Your separate superheater behaviour is much more predictable, and more controllable as you can adjust the superheater energy input independently of the boiler energy input.  I know I am a bit conservative, but in the oil industry they do not like the crash and burn approach, and even on model scale, that hot steam is very dangerous if you do not totally contain it.

I hope that helps to clarify a few more issues, and perhaps gives a slightly clearer picture of what happens inside your boiler, and how the element transfers energy to the water.

Thanks for reading,

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 #379 on: October 21, 2017, 02:08:57 PM »
Hi MJM ,once again thanks and i was a bit confused with the "Degrees is not part of it" ! . so i wondering how much heat(temperature) you can get from a 500Watt heater given enough seconds to do it ? In my boiler if there was no safety valve ,and the boiler was strong enough, what is the maximum temp/ pressure that one could attain ? and would a time/temp graph of this show a continuous line to infinity ? Also i can see now that an entirely separate superheater would be required as in a Loco firebox, I am now feeling a bleet sheepish with my wooly thinking !!!,And this is why we need people like you to help with our knowledge and prevent unnecessary contraptions from being built !! there have been quite a few expensive constructions in the past that have been failures, as depicted in The Engineer over the previous centuries .

Offline MJM460

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Re: Talking Thermodynamics
« Reply #380 on: October 22, 2017, 12:19:54 PM »
Heat and temperature -

Hi Willy, I hope my terminology did not offend you.  And the "wooly" term refers as much to my explanation as anything else as it was obviously not sufficiently clear.  So thank you for you kind words.  We are very fortunate to have access to information and tools that Watt and his colleagues could hardy dream of.  We can calculate a likely result and narrow down to those experiments with a reasonable chance of success.  They mostly had only trial and plenty of errors, and really amazing intuition. 

I was also trying to emphasise that all those words about dimensional analysis are not just esoteric waffle, dimensional analysis is a very practical approach to every day problems.  If you check the dimensions of each part of your problem, and they are the same, then you are likely to have all the elements of the problem and that is a good start.  Especially when working in SI units, when there are very few constants which have units associated them.  Your question, as usual, puts your finger right on the area that is still not adequately explained.  So let's try again.

The electrical power input from your heating elements can be calculated from the electrical formula, P= V x I, and the answer is in watts, or J/sec.  The big advantage of SI units is that mechanical power is also measured in J/sec, and they really are equal.  You only need talk about J/s, you don't have to qualify it as to whether is is mechanical or electrical joules.  For historical reasons, the unit J/s is given the name Watt.  Nothing in your boiler is able to provide any "push back" to change the electrical energy flow, no matter what temperature it reaches.  That is why I said temperature has nothing to do with it.

Now the fundamental law of physics comes into the picture, conservation of energy.  Energy does not disappear, it can always be accounted for, even if sometimes it is hard to know where to look.  But in your boiler with the electric heating element, we know the energy input, and the total energy output, or heat losses, plus any energy stored by increase in temperature of the materials in the system is always equal to the input.  When the heat losses equals the heat input, there is no more storage, so we have a steady temperature.  Now you will remember that I divided your operation into two phases, first a heat up process then steam production.  In the heat up phase, there is no steam production, there is some heat loss to the air from the shell which you can detect by holding your hand close, and all the rest of the heat input is stored in the water, the water vapour in equilibrium with the liquid, the copper shell, and even a little in the insulation.  And we did a calculation of how much energy was required for the storage.  We can calculate this using the mass of the materials, specific heat of copper, and the steam tables for the water.  Probably should have allowed a little for the heat stored in the element which also got hotter.  Then I assumed that the difference between the heat required for the observed heating of the materials and the input from the element was lost to the atmosphere.  It seemed like a lot and I suggested some insulation which would reduce the losses.  It did not reduce it as much as I expected, and I mentioned that we only know the rated output of the element, but did not have facilities to measure it, a potential source of error.  We will get back to that.

Once the boiler is up to your selected pressure, and you open the steam regulator, it runs at essentially constant temperature, so the losses continue at a constant rate.  We should have measured the steam consumption to check the losses at this temperature, because they are not constant during the heat up period, but vary with the temperature.  Now we have constant temperature, no more storage, so there are losses to atmosphere, and heat carried out with the steam.  We can measure the steam production, the tables tell us the energy carried out in the steam, and calculate the heat losses by subtraction from the energy input.

The point is that the energy equation is a complete explanation of the energy flow, and while the balance between storage, atmospheric losses and steam production changes with temperature, you only use the temperature as a measure of the amount of energy stored, and the properties of the steam, but nothing to do with the total amount of energy input.  And all because of your electric heating element which is ideal for demonstrating these things for that reason.

Now if we look at the heat transfer, the process which describes the flow of heat we do need temperature.  The equation is basically q = U x A x temperature difference.

Now let's look at the dimensions, q is J/s.  U has the dimensions of W/m^2.K, and of course W is the same as J/s.  A is the area in square metres, or m^2, and temperature difference, whether in C or K, has the dimensions of temperature or K.

The right hand side of the equation then has the units of J/(s. m^2 .K) times m^2 times K.  You can see the metres and temperature units "cancel" out, leaving only J/s, the same as the left hand side of the equation.  So we are off to a good start in understanding heat transfer, as our units are dimensionally consistent.

So on to look at your question about the ultimate temperature and/ or pressure.  I am glad that you are aware that the boiler could bust, this must remain a purely hypothetical thought experiment.  I am pretty confident that the boiler will fail before the heating element, not a good result.  You definitely need that safety valve and a high temperature switch, both properly set.

First you insulate the boiler, right over the top, with say 3 inches of good insulation such as rock wool tightly packed, or any one of several others with a similar conductivity, around 0.04 W/m.K, so the heat loss is minimal.  And then we start the heat up process.  Initially nearly all the heat goes into the water, the copper and a little into the insulation.  And as the temperature rises, the heat loss to the atmosphere increases.  We don't set a limit when we open the regulator, we want to see how high it goes, at least before bedtime!

Now that law of conservation of energy comes in.  The energy input is constant.  If there are no losses, all the energy goes into storage, and so the temperature just rises, as the energy is stored as sensible heat (higher temperature) plus all the ways energy is stored in steam.  It is not possible to eliminate all losses, some heat will be lost to the atmosphere, and as the temperature rises this heat loss will increase, it will increase until it equals the heat input, or until something breaks.

Now I did a few calculations with increasing insulation thickness from practically none and calculated the temperature difference required to dissipate the full 1000 J/s of the heater input.  It takes very little insulation to get over 1000 deg C, way above any safe temperature for your silver soldered copper boiler.  Something will definitely break, and no safe way of doing the experiment.

So in summary, energy input is not related to temperature in your electric boiler.  Temperature is a property of steam that we can measure, and if we have two independent such properties, the steam tables can tell us all the properties of the steam.  Temperature is also the measure of sensible heat stored in the copper, and other materials as energy is stored in them.  But temperature is not a quantity that has to be defined to determine energy input.

A fired boiler has a fundamental difference which sets it apart from your electric heated boiler.  The energy input comes from the mass of fuel burned and the fuel properties, but to release the chemical energy contained in the fuel, you have to burn it which requires oxygen, which usually only comes in a mixture with four times as much nitrogen which absorbs heat but makes no contribution to the amount of heat, and the combustion products have to be allowed to escape for the fire to continue.  They take a considerable amount of heat with them, and can if necessary take it all.  And it always takes away all the heat that is not transferred into the boiler water and copper.  It is never negligible.

I hope that makes it a little clearer.  But I noticed something interesting in the process of formulating an answer.  I will send you a pm after I have posted this, please let me know if you don't receive it.

Thanks to all for looking 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 #381 on: October 23, 2017, 12:18:30 PM »
Unfortunately I have not much to add today.  I spent the available time reviewing where we are up to, and doing some calculations in preparation for the next step.

I hope to have something useful to discuss tomorrow.

Apologies to all the regular readers.

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 #382 on: October 24, 2017, 02:27:54 AM »
Hi MJM i have checked out the currant going into the boiler and it looks like just over 4 amps  "Virtual" ?  so this may help with the calculations !

Offline MJM460

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Re: Talking Thermodynamics
« Reply #383 on: October 24, 2017, 12:01:40 PM »
On Electric heating elements -

Hi Willy, that is a beautiful meter, and I am guessing it is at least as good as your Avometer.  With the zero clearly correct when power is off, I would say it reads pretty close to 4.0 amps.  I am guessing that "virtual amps" is the terminology of the time for what we now call RMS amps.  That stands for root mean square, and without diverging too deeply into the maths, it gives us the effective average of a sine wave.  For a pure resistive load with AC voltage we can apply ohms law that we are familiar with for DC, if we use RMS volts and amps, not the maximum voltage of the sine wave.  So it could be described as the effective volts and current that give us the same heating effect as a DC voltage of that value.  Perhaps "virtually the same as as a DC current of that value"? 

So let's look at what all that can tell us about the electric heater.  It is rated at 240 volts and 500 watts.  If we use the definition of power, power = volts times amps, we can say rated amps = power divided by volts, or 500/240 = 2.08 amps. With two elements in parallel, a total of 4.16 amps.  I am guessing you would see that 0.16 easily on that meter scale.  We can also use ohms law to calculate the expected resistance of the element, based on 240 volts rating with rated current of 2.08 amps, R = V/ I = 240/2.08 = 115 Ohms.


On the other hand, with the measured current of 4 amps, and assuming the voltage to be the rated voltage, 240 V, we get power = V x I = 240 x 4 = 960 watts, or two times 480 watts, compared with the specification 500 watts.  This time, using ohms law, we find R = V/R = 240 / 2 = 120 ohms for each element.

Already, we can see an inconsistency in the data.  The manufacturer has to be reasonably careful about the published ratings, so I would expect that data to be accurate, with the understanding that it is probably assuming all at a standard temperature of 20 deg C.  These ratings are at the specified voltage of 240 V mentioned on the data sheet.  The discrepancy in calculated resistance values would be expected if the actual voltage was less than the rated voltage.  Now I want an alternative to measuring the voltage, as mains voltage is too dangerous for unqualified people to mess with, meaning most of us, including me.  However, with an element removed from the boiler, and no power connected, we can safely measure the resistance of the element.  Now the actual resistance should be the same in both cases.  We have assumed the voltage is the same as the rated voltage, 240 V.  Now many sources I have seen mention 220 V as the normal European voltage, and there is some tolerance allowed to the electricity authorities, so it is reasonable to rate equipment for 240 V, so long as it can also operate properly at the minimum allowable supply and at the expected normal voltage.  We can assume the resistance is the same, whether connected to 240 V or passing 4 amps, so ohms law tells us the voltage must be different between the two cases.  If we measure the resistance of an element, we can calculate the voltage at 4 amps, then with voltage and current we can calculate the actual power output at our actual voltage.

Now if we do a little algebra on Ohms law, V= I x R, and on the definition of electrical power, P = V x I, we can derive P = I^2 x R, or P = V^2 / R.  Note P = power in Watts or J/s, I is in amps and V in Volts.

That second form is particularly interesting as it tells us that the power is proportional to voltage squared, which means a small difference in voltage has a big effect on power.  If the actual voltage is nearer 220 V, it would make a significant difference to the heater output.  If the resistance is 115 ohms, with 220 V, the power output would be only 420 watts each element.

There is a little more we need to understand about a resistive heater.  I have mentioned a few times that resistance increases with temperature, but I didn't have the relevant data for the wire nominated on the heater spec sheet, chrome-nickel resistive wire.  Well, I have found it.  That hyper physics site from gsu is a really great resource.  Should have thought of it earlier as I use it often.  The paid hyper physics Ap on the Ap store is money well spent, but it often turns up in response to a normal web search.  So I assumed the resistance of the element might increase by say 6 ohms when up to temperature for generating steam.  It turns out that the increase in temperature to cause this change from the specified 20 degrees would result in an element temperature of 158 degrees.  Now this is a little above our steam temperature, but probably not enough to give the required heat flow.  Hence it is likely that 6 ohms rise is a low estimate.  If we look at that voltage form of the power equation, P = V^2/R, we can see that as the increase in temperature causes the resistance to increase, the power output decreases.  This tells us that in our particular application, the temperature increase of the wire in the element and any amount of reduced voltage compared with the rated 240 V, both cause the power output of the element to be below the rated power, and that will affect our heat balance.

We need to return now to that heat balance to see how this all fits together, so that will be the topic for another post.  I hope this is all making sense so far.

Thanks for reading,

MJM460
« Last Edit: October 24, 2017, 12:16:47 PM by MJM460 »
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Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #384 on: October 25, 2017, 03:02:20 AM »
Hi MJM ,i have tested another element not connected up and it reads 112.9 ohms so your calculation is pretty close !!Thanks for all the latest info and the theoretical and the actual figures depend on so many other factors that we  are not usually aware of .!! and the temp when i took the reading was 19 C....
« Last Edit: October 25, 2017, 03:25:37 AM by steam guy willy »

Offline MJM460

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Re: Talking Thermodynamics
« Reply #385 on: October 25, 2017, 12:16:27 PM »
Electric heaters continued -

Hi Willy, measured resistances provide hugely useful additional data for out investigation, thank you.  I think that makes two you have measured now, one was 109, and this one say 113.  This gives us some idea of the variation, and it is probably within reasonable manufacturing tolerance. 

Last time, I used the measured current, and assuming the rated voltage, calculated the resistance.  But now with that measured resistance we can manipulate ohms law again to  V = I x R.  Let's use our measured value of current, divided by two for each of two parallel elements, so 2 amps each element.  In addition let's use the average of the two resistances you have measured, or 111 ohms.  We put both of those into ohms law, so V = I x R = 2 x 111 = 222 Volts.

Now with better estimate of the voltage, we can use the power definition P = V x I = 222 x 2 = 444 watts.  Now this is quite a bit below the rated value of 500 watts for each element, and that will increase our heat up time.  The data sheet is still correct in specifying 500 watts, because it also specifies the voltage as 240.  When we use the element, we have to use the data sheet specifications and in addition apply the actual voltage we have in our location and equipment in order to understand the heat we will get from the element.

We also need to be aware of the fact that we are unlikely to have just picked out two elements one above the average and the other exactly the same amount below the average of all the elements that have been made.  In the vague shadows of my memory there were lectures on statistics, and the average called an estimate of the average based on a sample of two, and there was a method of calculating a standard deviation and with it, the likely spread.  But this thread is about thermodynamics, not statistics, so I won't go there, apart from mentioning that the difference between those two measured values gives us an idea of the variation between different elements.  To get any more accurate values, we have to measure the actual ones we intend to use.

Then there is that question of the increase in resistance with temperature.  I finally found the value of temperature coefficients resistance and the formula to calculate the resistance at different temperatures.  For Nichrome resistor wire the coefficient is 0.0004 per deg C. 

If we measure the resistance R0 at any temperature T0, then the resistance R at another temperature, T, is found using the following formula:

R = R0 x (1 + 0.0004 x (T - T0))

If we apply this to your element with R0 measured at 19, and we assume the heater wire starts at 17, and in order to transfer its output to the water it gets say 50 degrees above the water, we can calculate the resistance at each of your measured temperature points up to 135 deg C where the element would be 135 + 50 = 185.  Over this temperature range the element which measured our assumed average for the elements of 111 at 19 degrees, ranges from approx.110.9 at 17 to 117 at 185 degrees, and the power output of the element ranges from 440 watts at the start to 412 watts when it is at 185 , generating steam at 135 degrees with a constant voltage of 220 V.  Now this only 82% of the output we all assumed based on the data sheet.  It still requires that estimate of the temperature difference between the wire and the water, but it is enough to give you the idea.

Now I used a spreadsheet for all those calculations, which, because of the repetitive nature, saved quite a bit of time.  I was able to develop the formula in one cell or row, then simply copy and paste the formulae and the computer did the rest.  Actually the iPad did it.  Just a matter of learning how to use those absolute and relative references.  Then, as extra data was discovered, a simple substitution, and the spreadsheet programme updated everything in an instant.

I think that just about covers the performance of electric heaters, but if I have missed anything please ask, and I will try and cover it before returning to the calculation of those temperatures you asked for a few days back.

Thanks for following along,

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

Offline Admiral_dk

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Re: Talking Thermodynamics
« Reply #386 on: October 25, 2017, 08:09:36 PM »
Just as a little info - EU standardized the mains voltage to 230V +10/-6% one phase and 400V +10/-6% three phase more than 20 year ago (instead of the 220V / 240V in different member countries).

In practicality I actually measure between 227V and 231V with a very precise True-RMS Fluke meter at any given day over the last 22 years - and I often do as part of fault analyse on all the gear I've repaired over the years .....

Hmm - just checked WiKi to confirm the date 1995 - and discovered that there are places in the UK (a few) that still uses some very old power stations that supply 250V (still within the +10% limit) and some more that supply 240V .... Here in Denmark we adjusted our power stations from 220V to 230 within a few months after the new (back then) regulation was activated.

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #387 on: October 25, 2017, 11:58:53 PM »
Hi all, i have just checked my mains voltage and it is 240.3 volts.......!Thanks for more info on the boiler heating front ...Due to to all the disparate parameters that can occur with all the info it is quite a formidable task to actually come up with  exact figures. ! I suppose that is why when Rolls Royce is asked about power and performance , the reply is "adequate" !!

Offline crueby

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Re: Talking Thermodynamics
« Reply #388 on: October 26, 2017, 12:07:55 AM »
Hi all, i have just checked my mains voltage and it is 240.3 volts.......!Thanks for more info on the boiler heating front ...Due to to all the disparate parameters that can occur with all the info it is quite a formidable task to actually come up with  exact figures. ! I suppose that is why when Rolls Royce is asked about power and performance , the reply is "adequate" !!

When the Rolls Royce test driver comes back from the track all wide-eyed and quietly saying 'wow' over and over, do they translate that to 'adequate' for the brochure?!

Offline steam guy willy

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Re: Talking Thermodynamics
« Reply #389 on: October 26, 2017, 02:14:55 AM »
Hi Chris , possibly ...It is the british stiff upper lip and "reserve ' that made us what we are/were.!  Also MJM does the specific gravity and the hard/softness of the water have any bearing on the steam ability in the boiler ??  Should i assemble for the next boiler test Hydrometer,Barometer,Ammeter, Voltmeter, temperature gauge ,pressure gauge , measuring jug etc etc etc

 

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