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Mathematical verification of heat flow 1
- Thread starter 82james82
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There are so many variables that the simpliest way would be to mount multiple RTDs in a grid pattern on the plate, heat it up, and generate the results. Once you know your pattern, any future calculations would be easy. I imagine that you are looking at temp rise, vs time, vs distance, vs ambient.
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- #3
Thanks
Is there no calculation that I can do having made some assumptions and neglecting plate thickness and time etc.
I have already used the Laplace FD equation but this relies on the temperature at each side of the plate being constrained
thanks again
Is there no calculation that I can do having made some assumptions and neglecting plate thickness and time etc.
I have already used the Laplace FD equation but this relies on the temperature at each side of the plate being constrained
thanks again
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- #5
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1
- #6
Is this a steady state problem or transient?
How is the heat supplied to the end?
Are you maintaining a fixed temp on the end?
Is the plate being held horizontally or vertically?
Before this boundary value problem can be tackled you must furnish the answers to these.
An example for fixed temp on one end and convective or radiative cooling from the surface of the plate involves
finding a solution to the eq
Kwdel^2(T)=2h(T-T0) with insulated boundaries on three sides
where
K= thermal conductivity
w= thickness
del^2= second partial derivative of T in x and y directions
h= film coefficient
T0=ambient temp
A closed form solution to this and others including transients for thin rectangular plates can be found (e.g.,p.169) in "Conduction of heat in Solids", Carslaw and Jaeger,Oxford at the clarendon press, 1959.
If you must, numerical methods are available from many other sources but closed forms are always more powerful.
How is the heat supplied to the end?
Are you maintaining a fixed temp on the end?
Is the plate being held horizontally or vertically?
Before this boundary value problem can be tackled you must furnish the answers to these.
An example for fixed temp on one end and convective or radiative cooling from the surface of the plate involves
finding a solution to the eq
Kwdel^2(T)=2h(T-T0) with insulated boundaries on three sides
where
K= thermal conductivity
w= thickness
del^2= second partial derivative of T in x and y directions
h= film coefficient
T0=ambient temp
A closed form solution to this and others including transients for thin rectangular plates can be found (e.g.,p.169) in "Conduction of heat in Solids", Carslaw and Jaeger,Oxford at the clarendon press, 1959.
If you must, numerical methods are available from many other sources but closed forms are always more powerful.
- Thread starter
- #7
This company makes a nice software program to do just that
If you don't want or need to own the software, this guy is a contributor to this site and apparently owns the software for consulting purposes.
"Venditori de oleum-vipera non vigere excordis populi"
If you don't want or need to own the software, this guy is a contributor to this site and apparently owns the software for consulting purposes.
"Venditori de oleum-vipera non vigere excordis populi"
Sounds like a very easy thermal model, but depending on the degree of accuracy required, a software model may not be necessary.
There are 3 forms of heat transfer involved: conduction, convection and radiation. The conduction formula is very straightforward. IF the following are true, you may be able to ignore convection & radiation:
1. plate is thick relative to the surface area
2. plate is thermally conductive (alum or copper)
3. air is still or plate is insulated
4. plate has low emissivity
5. heat source is fairly uniform along one edge
Conduction equation:
dT = QL/KA
dT = temperature change across L (deg C)
Q = power of heat source (W)
L = length of heat travel along the plate (m)
K = thermal conductivity of the plate (W/mK)
A = cross-sectional area of plate perpendicular to heat movement (m2)
Of course, there will still be some loss due to convection and radiation. At steady state, the energy must leave the plate at the same rate it enters and these are the only two ways to leave.
ko (
There are 3 forms of heat transfer involved: conduction, convection and radiation. The conduction formula is very straightforward. IF the following are true, you may be able to ignore convection & radiation:
1. plate is thick relative to the surface area
2. plate is thermally conductive (alum or copper)
3. air is still or plate is insulated
4. plate has low emissivity
5. heat source is fairly uniform along one edge
Conduction equation:
dT = QL/KA
dT = temperature change across L (deg C)
Q = power of heat source (W)
L = length of heat travel along the plate (m)
K = thermal conductivity of the plate (W/mK)
A = cross-sectional area of plate perpendicular to heat movement (m2)
Of course, there will still be some loss due to convection and radiation. At steady state, the energy must leave the plate at the same rate it enters and these are the only two ways to leave.
ko (
A thorough solution to your problem requires solving the heat diffusion equation. You probably need to solve the time dependent equation. The easiest way of getting a useful solution is a numerical solution, hence using computers. Someone mentioned that there is software to buy and that may be agood but often expensive solution. One also has to learn a new package. Another solution is to write some code. In any case, the plate will be losing heat with time due to convection and perhaps radiation if the temperature is high. The loss of heat requires knowledge of boundary conditions of your plate. One way to get those is to apply a sufficient number of temperature gauges along the surface but a better way may be to look at the plate with a thermal camera. Knowing what the temperature is along your surface will allow you to get a solution to the numeric problem and you end up knowing the temperature for throughout the volume of the plate. Finally, your problem may involve some simplifications like that one side is thermally insulated, there's no heat convection, no radiation etc. and that will in turn make the problem easier but in general you must know what the temperature is on the boundaries of the plate to be able to compute temperatures in the interior.
I agree janne. I was trying to answer james' question regarding an approximate calculation that might be possible if certain assumptions are made. I should mention that I also assumed a steady state condition.
ko (
ko (
IT DEPENDS ON THE RESOLUTION NEEDED.
IF THERE IS NO METHOD TO "SINK" OR REMOVE
THE HEAT AT THE END OF THE PLATE, IT WILL HAVE
A UNIFORM TEMP WITHIN 5 OR 10 DEGREES.
IF THE TEMP AT BOTH ENDS ARE KNOWN, IT
IS A SIMPLE RATIO OF TEMP TO LENGTH PROBLEM.
THE MORE YOU CAN VARIABLES YOU CAN GET RID OF
THE EASIER IT IS.
IN GENERAL THE MECHS ARE CORRECT, ONCE YOU ARE
PAST THESE BASIC CONSTRANTS, THINGS GET VERY COMPLEX.
I JUST HAD A PHD RUN A PROGRAM THAT TOOK OVER
40 HRS ON A 4G MACHINE, THAT TOOK THREE WEEKS TO SET UP,
AFTER THE MODEL WAS BUILT.
LETS SEE ABOUT 8 WEEKS *40 HRS *100 = 32000 +
30000 FOR THE PROGRAM = 62K, KIND OF STEEP FOR A HOME
PROJECT.
IF THERE IS NO METHOD TO "SINK" OR REMOVE
THE HEAT AT THE END OF THE PLATE, IT WILL HAVE
A UNIFORM TEMP WITHIN 5 OR 10 DEGREES.
IF THE TEMP AT BOTH ENDS ARE KNOWN, IT
IS A SIMPLE RATIO OF TEMP TO LENGTH PROBLEM.
THE MORE YOU CAN VARIABLES YOU CAN GET RID OF
THE EASIER IT IS.
IN GENERAL THE MECHS ARE CORRECT, ONCE YOU ARE
PAST THESE BASIC CONSTRANTS, THINGS GET VERY COMPLEX.
I JUST HAD A PHD RUN A PROGRAM THAT TOOK OVER
40 HRS ON A 4G MACHINE, THAT TOOK THREE WEEKS TO SET UP,
AFTER THE MODEL WAS BUILT.
LETS SEE ABOUT 8 WEEKS *40 HRS *100 = 32000 +
30000 FOR THE PROGRAM = 62K, KIND OF STEEP FOR A HOME
PROJECT.
Yes, some heat transfer problems can take many hours on a supercomputer but not this one.
I estimate less than an hour to build the model (steady-state conduction, convection, and radiation) and less than 20 minutes to solve it on a my PC. Transient would take longer: maybe 10-20 minutes per time step.
ko (
I estimate less than an hour to build the model (steady-state conduction, convection, and radiation) and less than 20 minutes to solve it on a my PC. Transient would take longer: maybe 10-20 minutes per time step.
ko (
THE PROBLEM I REF. WAS VERY COMPLEX INVOLVNG ABOUT 10
MATERIALS, ABOUT 30 HEAT SOURCES, AND FOUR PATHS,
AND A VERY SPECIFIC TOTAL POWER USAGE.
I THINK THAT ("82JAMES82"?) IS MISSING AT MIN TWO DIM
OF VARIBLES, ONE WAS IMPLIED BUT NOT STATED, (PLATE END TEMP.) THE OTHER, ?. (THICKNES OF THE PLATE, TIME, ETC.).
MY OBJECTIVE WAS TO SHOW RELEVANCE TO
A TYP EVERYDAY PROBLEM (SIMPLE ANSWERS) VERESES
MONEY FOR COMPLEX PROBLEMS.
ALL I WAS TRYING TO DO WAS HELP.
IF I AM IN ERROR PLEASE LET ME KNOW.
IF YOU HAVE BETTER ANSWERS, LET EVERY ONE KNOW.
NOT HOW COMPLEX BUT HOW SIMPLE CAN YOU MAKE THE PROBLEM.
MATERIALS, ABOUT 30 HEAT SOURCES, AND FOUR PATHS,
AND A VERY SPECIFIC TOTAL POWER USAGE.
I THINK THAT ("82JAMES82"?) IS MISSING AT MIN TWO DIM
OF VARIBLES, ONE WAS IMPLIED BUT NOT STATED, (PLATE END TEMP.) THE OTHER, ?. (THICKNES OF THE PLATE, TIME, ETC.).
MY OBJECTIVE WAS TO SHOW RELEVANCE TO
A TYP EVERYDAY PROBLEM (SIMPLE ANSWERS) VERESES
MONEY FOR COMPLEX PROBLEMS.
ALL I WAS TRYING TO DO WAS HELP.
IF I AM IN ERROR PLEASE LET ME KNOW.
IF YOU HAVE BETTER ANSWERS, LET EVERY ONE KNOW.
NOT HOW COMPLEX BUT HOW SIMPLE CAN YOU MAKE THE PROBLEM.
wrong100,
Are you permanently angry and shouting, or does your computer have a Caps Lock key which you could perhaps use? It would make your posts much more readable.
----------------------------------
If we learn from our mistakes,
I'm getting a great education!
Are you permanently angry and shouting, or does your computer have a Caps Lock key which you could perhaps use? It would make your posts much more readable.
----------------------------------
If we learn from our mistakes,
I'm getting a great education!
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