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Calculating Cooling Times 7
- Thread starter mmelville
- Start date
"New PostHelpful Member!IRstuff (Aerospace)
30 Mar 10 10:28
OK, you guys need to get a room winky smile '""
Or a gymnasium.
I'm not arguing Biot for radial heat flow , just the axial part of the problem; it does not readily lend itself to "lumping" as I have been yelling about.
30 Mar 10 10:28
OK, you guys need to get a room winky smile '""
Or a gymnasium.
I'm not arguing Biot for radial heat flow , just the axial part of the problem; it does not readily lend itself to "lumping" as I have been yelling about.
Eventually got a room!
Well it is obvious lumped model from a theoretical point of view gives the right answer only with Bi = 0, but as it was written above the door of my college: “Perfection does not exist”. From an engineering point of view (at least here we are all engineers and what matters is to find answers to real problems), the reliability of the lumped model depends on the accuracy one is looking for. My intent is never to be polemic or too picky, I have just reported the reference value I use as demarking threshold for Bi, and I think corus has done the same. The problem is not to take Bi <0.1 or Bi<0.2 as in this case both would imply a “h” value which is not realistic for a forced convection problem.
Hey Zekeman,
A blow-molded plastic tube 40 mm diameter and “200 feet long”. It could be interesting to see such a mould!
All my consideration descend from the OP first statement. I have considered this as 1-D problem (radial heat flow) since the very beginning. I now leave the OP free to take his/her conclusions about the most useful approach for his/her requirements
One last thing back to the OP’s issue (assuming he/she is still with us). I have found the following link on the web. There are some formulas which seem to be interesting for plastic items cooling (page 281).
Well it is obvious lumped model from a theoretical point of view gives the right answer only with Bi = 0, but as it was written above the door of my college: “Perfection does not exist”. From an engineering point of view (at least here we are all engineers and what matters is to find answers to real problems), the reliability of the lumped model depends on the accuracy one is looking for. My intent is never to be polemic or too picky, I have just reported the reference value I use as demarking threshold for Bi, and I think corus has done the same. The problem is not to take Bi <0.1 or Bi<0.2 as in this case both would imply a “h” value which is not realistic for a forced convection problem.
Hey Zekeman,
A blow-molded plastic tube 40 mm diameter and “200 feet long”. It could be interesting to see such a mould!
All my consideration descend from the OP first statement. I have considered this as 1-D problem (radial heat flow) since the very beginning. I now leave the OP free to take his/her conclusions about the most useful approach for his/her requirements
One last thing back to the OP’s issue (assuming he/she is still with us). I have found the following link on the web. There are some formulas which seem to be interesting for plastic items cooling (page 281).
ione,
Too late, the link's not working, and, it's responding in Italian... ;-(
The English version has a bunch of hits, but I don't know if the one you cited is there:
TTFN
FAQ731-376
Too late, the link's not working, and, it's responding in Italian... ;-(
The English version has a bunch of hits, but I don't know if the one you cited is there:
TTFN
FAQ731-376
![[wink]](/data/assets/smilies/wink.gif "[wink] [wink]")
Hey IRstuff,
Try the following link (if you want of course) and let me know
Try the following link (if you want of course) and let me know
OK, in this case, a small length of pipe, I would concede that the axial problem for a reasonable flow rate would not cause any significant temperature gradient, but it is not good practice to make general assumptions without proof.
So, if we now accept the 1 dimensionality of the problem, why are we having a conversation on whether or not we accept the radial lumped model, when the exact solution that is probably 100 years old is available, if needed.
The solution is plotted in the form of Biot number as the parameter and the independent variable kappa*t/l^2 , plotted against the dependent parameter (T-Tc)/(Ti-Tc),where
T= temperature of material
Ti= initial temperature of material
Tc= coolant temperature (assuming negligible axial gradient
kappa= diffusivity
l= thickness of material.
Two references come to mind:
Carslaw and Jaeger , "Conduction of Heat in Solids"Oxford Clarendon Press. 1959, pp119-127, and
Schneider (I don't have the exact reference for his book) but he has a chapter in " Handbook of Heat Transfer",McGraw-Hill,1973,chapter 3)
The important plots are for the center and the two surfaces.
The OP probably needs a conservative time it takes to cool the surface that is not exposed to the coolant air; he/she can first calculate the Biot number from the geometry and thermal/flow conditions,pick out the desired temperature that acceptable as the cooling temperature. The intersection of the ordinate and the Biot curve will give the abscissa,
kappa*t/l^2 from which the t,the cooling time is calculated.
If I could figure out how to post the plot(s) I would gladly do so, (if the OP is interested). But by now he/she must be fast asleep.
So, if we now accept the 1 dimensionality of the problem, why are we having a conversation on whether or not we accept the radial lumped model, when the exact solution that is probably 100 years old is available, if needed.
The solution is plotted in the form of Biot number as the parameter and the independent variable kappa*t/l^2 , plotted against the dependent parameter (T-Tc)/(Ti-Tc),where
T= temperature of material
Ti= initial temperature of material
Tc= coolant temperature (assuming negligible axial gradient
kappa= diffusivity
l= thickness of material.
Two references come to mind:
Carslaw and Jaeger , "Conduction of Heat in Solids"Oxford Clarendon Press. 1959, pp119-127, and
Schneider (I don't have the exact reference for his book) but he has a chapter in " Handbook of Heat Transfer",McGraw-Hill,1973,chapter 3)
The important plots are for the center and the two surfaces.
The OP probably needs a conservative time it takes to cool the surface that is not exposed to the coolant air; he/she can first calculate the Biot number from the geometry and thermal/flow conditions,pick out the desired temperature that acceptable as the cooling temperature. The intersection of the ordinate and the Biot curve will give the abscissa,
kappa*t/l^2 from which the t,the cooling time is calculated.
If I could figure out how to post the plot(s) I would gladly do so, (if the OP is interested). But by now he/she must be fast asleep.
- Thread starter
- #29
WOW! I never expected so much activity for this post. At this point there are 27 replies to my original question.
zekeman, you were right with the comment "The OP probably needs a conservative time it takes to cool the surface.." That is really what I was looking for.
I am not an engineer, though I took mechanical engineering technology classes in college, my major (B.S.) was computer graphics technology (CAD design); I am now designing assembly equipment.
I will provide a model of the pipe in question, if anyone really want to try to calculate it. (At this point, some may want it to prove/disprove their own formulas and methods!) I am not asking anyone to do my work for me, I am just looking for resources to get an approximate time to cool to a given temperature. Maybe someone can suggest a CFD program with decent thermal analysis that would the job. I primarily use Autodesk Inventor but have access to SolidWorks and Catia.
Again, THANK YOU to everyone who has posted thus far!
zekeman, you were right with the comment "The OP probably needs a conservative time it takes to cool the surface.." That is really what I was looking for.
I am not an engineer, though I took mechanical engineering technology classes in college, my major (B.S.) was computer graphics technology (CAD design); I am now designing assembly equipment.
I will provide a model of the pipe in question, if anyone really want to try to calculate it. (At this point, some may want it to prove/disprove their own formulas and methods!) I am not asking anyone to do my work for me, I am just looking for resources to get an approximate time to cool to a given temperature. Maybe someone can suggest a CFD program with decent thermal analysis that would the job. I primarily use Autodesk Inventor but have access to SolidWorks and Catia.
Again, THANK YOU to everyone who has posted thus far!
I have no access to the book quoted by Zekeman (Carslaw and Jaeger , "Conduction of Heat in Solids"Oxford Clarendon Press. 1959) but have found a link to another book (Coulson & Richardson's Chemical Engineering: Fluid flow, heat transfer, and mass transfer) which quotes the same book as Zekeman quoted.
At page 403 there is a diagram which could be useful.
Knowing ambient surrounding temperature, pipe starting temperature, pipe cooled temperature, pipe material diffusivity and geometry one could extrapolate the value Fourier number (Fo = diffusivity*t/(length of interest)^2) and solve for time t.
Unfortunately pipes are not mentioned amongst shapes tabulated.
At page 403 there is a diagram which could be useful.
Knowing ambient surrounding temperature, pipe starting temperature, pipe cooled temperature, pipe material diffusivity and geometry one could extrapolate the value Fourier number (Fo = diffusivity*t/(length of interest)^2) and solve for time t.
Unfortunately pipes are not mentioned amongst shapes tabulated.
Ione,
The figure 9.14 of your reference is approximately the Carslaw ans Jaeger plot we need for hottest part of the surface ( the outer one) cooling of the "slab". The reason we use the slab is that the pipe wall is usually thin in comparison to the hole size so a "slab" solution is OK.
Your description of how you get the time is better than my description for anybody trying to do it.
Mmellvile,
I am having trouble opening the file you posted. I wonder if that is just my limited software; if you could upload it with another format or give a physical description of the problem , I would appreciate that and could easily solve the problem with what I have. These cooling curves we discussed above have everything we need to solve it, without any further ado.
The figure 9.14 of your reference is approximately the Carslaw ans Jaeger plot we need for hottest part of the surface ( the outer one) cooling of the "slab". The reason we use the slab is that the pipe wall is usually thin in comparison to the hole size so a "slab" solution is OK.
Your description of how you get the time is better than my description for anybody trying to do it.
Mmellvile,
I am having trouble opening the file you posted. I wonder if that is just my limited software; if you could upload it with another format or give a physical description of the problem , I would appreciate that and could easily solve the problem with what I have. These cooling curves we discussed above have everything we need to solve it, without any further ado.
- Thread starter
- #32
electricpete
Electrical
No problem to attach jpegs. There is a step "Insert Link" which is easy to overlook.
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- Thread starter
- #39
OK, I don't have thermal data on your plastic, so as an example I will use the properties of brick, since it is closer to plastic than metal and you may be able to use it as a guide.
First the geometry
L=wall thickness= 3.5mm=0.01167 ft
Now the thermal properties of brick
K=0.5 BTU/HR-FT-DEGF
kappa=.0147 FT^2/HR
From the flow rate of 700CFM through the tube I used charts to get
h=100 BTU/HR-FT^2-degF
BIOT number=hL/K=100*.01167/.5=2.33
The problem is the time to cool to 25 C from 85 C with 23 C coolant.
A more realistic time is to cool within 10% of the total drop or .1*(85-23)=6 C or a final temperature of 29 C
From the C&J plot that I posted, set the horizontal coordinate =0.1, and move vertically until you reach the BIOT curve=2.3(or close to it), then go horizontall to pick off the log(kappa t/L^2) which I found to be 0.4.
Then
kappa t/L^2=10^.4=2.5
Solving for t I get
t=2.5*L^2/kappa=2.5*.01167^2/.0147= .0233 hr=1.4 minutes
If this was done using the lumped model, the answer would be about 0.7 minutes or an error of factor of 2.
Doing this for 25 degrees as you suggested doubles the time to cool to 2.8 minutes.
NOTES:
1- opening the C&J curve with MS PAINT, if possible, will much improve the details.
2- The horizontal coordinate of the plot , v/V, is actually (T-Tc)/(T(0)-Tc)
3- heat flow to ambient on the outside is ignored since it is negligible for the transient times of interest; its film coefficient is about h=3 or factor of 30 less than the internal cooling.
First the geometry
L=wall thickness= 3.5mm=0.01167 ft
Now the thermal properties of brick
K=0.5 BTU/HR-FT-DEGF
kappa=.0147 FT^2/HR
From the flow rate of 700CFM through the tube I used charts to get
h=100 BTU/HR-FT^2-degF
BIOT number=hL/K=100*.01167/.5=2.33
The problem is the time to cool to 25 C from 85 C with 23 C coolant.
A more realistic time is to cool within 10% of the total drop or .1*(85-23)=6 C or a final temperature of 29 C
From the C&J plot that I posted, set the horizontal coordinate =0.1, and move vertically until you reach the BIOT curve=2.3(or close to it), then go horizontall to pick off the log(kappa t/L^2) which I found to be 0.4.
Then
kappa t/L^2=10^.4=2.5
Solving for t I get
t=2.5*L^2/kappa=2.5*.01167^2/.0147= .0233 hr=1.4 minutes
If this was done using the lumped model, the answer would be about 0.7 minutes or an error of factor of 2.
Doing this for 25 degrees as you suggested doubles the time to cool to 2.8 minutes.
NOTES:
1- opening the C&J curve with MS PAINT, if possible, will much improve the details.
2- The horizontal coordinate of the plot , v/V, is actually (T-Tc)/(T(0)-Tc)
3- heat flow to ambient on the outside is ignored since it is negligible for the transient times of interest; its film coefficient is about h=3 or factor of 30 less than the internal cooling.
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