FEA consistant thermal units

[MITravis]

Currently working a model with standard US/Imperial units only.
I am trying to determine if the following are consistant units for the model?
1)Nodal lengths = inches
2)Thermal Conductivity = Btu-in/(hr-ft2-F)
3)Density = lb/in3
4)Surface Transfer Coefficient = Btu/hr-ft2-F
5)Internal Heat Source = Btu/in3 hr

thanks

 

[vpl]

Well first you have to assume that there is a standard ...

Over the years, I've found what keeps me out of trouble is to do a unit analysis -- write out the formula in terms of the units and make sure all units cancel out so that I'm left with just the ones I'm expecting.

With the US units, that may mean that you need to use conversion factors such as inches to foot, pounds mass (such as in density) to pounds force (such as in BTU).

Patricia Lougheed

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[ginsoakedboy]

These are not.

Express all quantities in terms of: inches, hours, BTU, lbm; and they will become consistent.

 

[electricpete]

A strict approach to guarantee consistent units is to select a minimal set of initial units which encompasses all the remaining units of the problem, then express all other units in terms of that minimal intial set.

There are only 4 independent units which can be expressed. After you pick those three, all others are "dependent" (i.e. can be expressed in terms of the first 4).

I would pick the set: mass, length, time, temperature.

The easy answer is kg, m, sec, deg C. Work everything in SI and you are good to go, guaranteed. In fact we know that's the case without ever looking for the minimal set of units, because SI is a consistent set of units.

It can certainly be done in US units. Whether expressing everything in inches, hours, BTU, lbm (and I assume degF) is guaranteed to work, I can't say. It doesn't follow the roadmap that I outlined above (it starts with 5 variables when only 4 independent variables exist).


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[electricpete]

Some corrections/clarifications in bold:

A strict approach to guarantee consistent units is to select a minimal set of initial units which encompasses all the remaining units of the problem, then express all other units in terms of that minimal intial set.

There are only 4 independent units which can be expressed. After you pick those four, all others are "dependent" (i.e. can be expressed in terms of the first 4).

I would pick the set: mass, length, time, temperature.The easy answer is kg, m, sec, deg C. Work everything in SI and you are good to go, guaranteed. In fact we know that's the case without ever looking for the minimal set of units, because SI is a consistent set of units.

It can certainly be done in US units. Whether expressing everything in inches, hours, BTU, lbm (and I assume degF) is guaranteed to work, I can't say. It doesn't follow the roadmap that I outlined above (it starts with 5 units when only 4 independent units exist).

By the way vpl's method of course works too. We can't do it here because we don't have the equations. It can be a little tricky when programming since the variables don't carry around the units. htlyst's method might work... I just can't follow it all the way through myself.

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[ginsoakedboy]

There is no "method".

As a matter of fact, FEA using ANSYS is blind to units. So, the user has to make sure that all the numbers (properties, loads, simulation time etc.) are in consistent units or the results will not make any sense. That's all.

I suggested the above system of units as MITravis had already mentioned some of them in his post. So, for completeness, I would use the following units:

inches, hours, lbm, °F

(1 BTU = 4.672e13 lb*in^2/hr^2)

That being said, I don't know the appropriate units that should be used for simulations of electromagnetic systems, consistent with above.

 

[vpl]

I may be getting a bit pedantic here, but I need to point out that there is a difference between pounds mass (which is a mass unit) and pounds force (which is not). Density is in terms of pounds-mass; the other terms, in which BTU appears, are in terms of pounds-force.

The two are connected by Newton's second law: F=ma. Or, to be specific, lb_f = lb_m*g_l/g_c.

g_l is the local gravitational constant. It may differ from the gravitational constant of g_c (which is = 32.174 lb_m.ft/lb_f.sec²). If accuracy is not too, too important, than g_l can be set equal to g_cor 32.2 (units of g_l are in ft/sec²). However, if you're not located at the 45° latitude and sea level, then the two values start differing at the second decimal point and the ratio is not equal to 1.

For clarity, when working with these two units, it is important to use the subscript to describe whether you have a force or a mass.

I can't comment on htlyst's conversion because I'm not sure what assumptions went into it. However, I would recommend using feet rather than inches.

Patricia Lougheed

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[electricpete]

I agree htlyst's most recent list: Expressing everything in inches, hours, lbm, F will work - that is a minimal set of independent units as I described earlier. But finding a way to express energy in terms of those will require some conversion work and will end up with a set of units that I dare say no-one is familiar with. But if all quantities are in imperial units to begin with and imperial results are required, it may be no more unit conversion work than my suggestions to convert input quantities to SI and then convert results to imperial at the end. That choice is a matter preference among minimal sets.

To expand a little bit on what I said beforeabout minimal sets, here's an example. Let's say I study the equation for distance travelled during uniform acceleration starting from 0 velocity:
distance = 0.5 * acceleration * time^2

What units can be used as consistent set of units for the above equation?
The minimal set of dimensions incldes two independent dimensions: the most natural are time and distance.

If I pick any unit for time and any unit for distance, then express all quantities in terms of the first two, I can't go wrong. Let's choose time in fortnights and distance in furlongs. I know right off the bat how to express distance and time in my equation. But what are the units for acceleration? They have to be expressed in terms of the minimal set.... fuglongs/fortnight^2. Any other unit for acceleration (such as g's or m/sec^2) would give the wrong answer. This is a simple example where the answer is obvious. It's not so obvious when you get more variables and more relationships, but follow the recipe and it'll work.

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[electricpete]

No-one asked, but I will continue a little.

It is instructive to work the problem save uniform acceleration problem with a slightly more bizarre set of units where the answer is not so obvious
(distance = 0.5 * acceleration * time^2)
Choose the minimal set of dimensions as distance and acceleration.
Choose distance in inches and acceleration in g's.

We need to express time in terms of the first two... i.e. the units of time in this system must be sqrt(inches/g)

Let's say using the above set of units (inches and g's) I want to find distance travelled in 5 seconds at 6'gs. acceleration expressed in g's is 6. I have to convert my time to the proper units:
5 seconds * sqrt((9.8*m)/(g * second^2))*sqrt(40*inch/m) =
(5 *second) * sqrt((9.8*m)/(g * second^2))*sqrt(40*inch/m)=100 sqrt(inch/g)

So we find
d = 0.5 * 6 * 100^2 = 30000 (distance in this system are in inches, by definition).

Try the same problem in your favorite unit system and convert to inches and you will get the same answer. The advantage of finding the consistent set of units up front is you know the units will work out as long as you convert your quantities to your consistent units up front... you don't have to wait to plug them into equations and see how they turn out. I could use these same set of units for any other problem involving time, acceleration, and quantities derivable fro those first two (distance, velocity, jerk, frequency etc).

The same advantage carries a step further with SI system. Use SI units and you have a consistent set of units every time, guaranteed. No up-front analysis required. No need to invent a system of units for every complex problem. The SI system follows the same appraoch I outlined above....starts with 7 basic indepednent units (meter kilogram, second, Amp, Kelvin, mole Candela). None can be expressed in terms of the other base units. Then all other SI units are derived in terms of the first 7. And by the way they are called "derived" SI units for that reason. The group of derived SI units is huge. Looking just at electromagnetics it encompasses pretty much everything you'd ever need:

http://online.physics.uiuc.edu/courses/phys435/fall06/Lecture_Notes/SI_Units_EM_Quantities.pdf

I think I'm done rambling. I am sure everyone participating in this forum already has their own strategy for units successfully and efficiently. I just felt like typing a little more.

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[electricpete]

The last 7 items at the previous link were constants, not derived SI units. But everything before that was derived SI units.

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[ginsoakedboy]

vpl,

The conversion I stated is consistent with pound-mass.

 

[FredRosse]

Keep in mind that every fundamental equation in all of engineering and science is dimensionally consistent, the units come out right every time, without exception. If the units are tracked throughout the calculations, then the result will be in well defined units. The same fundamental equations work in any consistent system of units, English, Metric, SI, etc.

Also remember that any term in any equation can be multiplied by (or divided by) a non-dimensional “1”, without changing the equality.

For example, a well known exact equality: 25.4 mm = 1 inch

Dividing both sides of the equation by 1 inch: 25.4 mm / inch = 1 (ND)

The term (ND) signifies Non-Dimensional, and this non dimensional “1” can be used in any fundamental equation without changing the equality. Unfortunately, there are many “bastardized” equations floating out in the industry with “conversions” built in, which require the use of various inconsistent units. Better to use only the fundamental equations, and multiply by non-dimensional “1” as required to arrive at the desired set of units for any given problem.

Another relevant point, as mentioned in previous replies is to differentiate between “Mass” and “Force”. For example, one should never label something as “Pounds”, it is not clear if the unit is Pounds force or Pounds Mass. Always and without exception label a force as a force, and a mass as a mass within calculations.

The two are related by Newton’s equation: F = M * A

In the English system, a 1 Pound force will result from 1 Pound mass under the influence of standard gravity, 32.174 ft / sec^2

1 LBf = 1 LBm * 32.174 ft / sec^2

again, this equation can be turned into a “Non-Dimensional One”, dividing both sides by 1 LBf:

1 (ND) = 32.174 ft – LBm / LBf – sec^2,

or, inverting:

1 (ND) = LBf – sec^2 / 32.174 ft – LBm

Either of these can be used as required, in all equations as required. It is common practice to think of these values as “conversions”, they are really just the value of unity, with null dimensions.

Keeping track of the units avoids much needless uncertainty (and errors) in engineering problem solving.