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Calculating Cooling Times(7)

I am looking for a formula to determine the cooling time of a blowmolded plastic tube with forced cold air. The pipe is 40mm o.d. with a 34mm wall thickness. I appreciate any help offered. Thanks in advance! 

IRstuff (Aerospace) 
25 Mar 10 14:53 
There's pretty much only two equations that are ever used in heat transfer, Q = htc*area*deltaT Q = emiss*sigma*area*(T_1^4T_2^4) You calculate the latent heat, the heat transfer coefficient (htc), emissivity, and determine what kind of lumped model you're going to use to model where the heat energy resides in the material. TTFN
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(2) corus (Mechanical) 
25 Mar 10 15:33 
As a simple solution you could assume that the pipe wall acts as a lumped mass with a negligible temperature gradient across the wall thickness as it cools. You then solve V.p.Cp.dT/dt = h.A.(TTa) to get the temperature T as an exponential of time t. Values of the heat transfer coefficient, h, you'd have to calculate based upon the air speed, and loss of heat to the outer ambient. excorus (semidetached) 

(3) ione (Mechanical) 
26 Mar 10 12:00 
Mmelville, You've already got valid inputs from the posts above. The solution of the Newton's cooling equation (the one reportes in corus' post) will lead to an exponential timedecaying profile of the temperature. T(t) = Tamb + (Ti  Tamb)*exp[A*h*t/(m*cp)] Where: T(t) = wall temperature at time t Tamb = ambient temperature Ti = starting wall temperature at time t = 0 s (as noted by corus with assumption of constant temperature through the pipe wall) A = surface area h = convection coefficient m = pipe mass cp = pipe material specific heat t = time If you want to go any further take a look at the link below (besides the Newton's equation this paper takes into account a more sophisticated model, which also considers radiation heat transfer ). http://www.loreto.unican.es/Carpeta2007/00TorreonCartes2007/CoolingNewton02.pdf 

The only part I am having trouble with is determining h. I understand that as a general rule for natural convection, h=10. Not sure what to use for forced air convection though. I would be using a fan moving 725 cfm. Any thoughts? 

zekeman (Mechanical) 
26 Mar 10 16:28 
This problem recurs from time to time and I believe no one has gotten it right. It turns out to involve solving a very difficult partial differential equation so the best thing to do is solve it numerically. The last time I saw this problem I used EXCEL and broke up the pipe into discrete sections of delta x length and deltat=deltax/V from the known flow velocity V moving the gas along the pipe iteratively,exchanging heat and repeating the process until "equilibrium" is obtained.With EXCEL, you could write a MACRO to do this so that only one row could be updated with each increment of time, delta t.Using a counter you determine the time t=delta t*N If the OP would furnish the geometry of the pipe and the air mass flow rate I might take a crack at it in my spare time.


Spare time!! Whats that?
Unfortunately, I work for an automotive supplier and cannot really give out the 3d data.
I really just want to "ballpark" the cooling time required to achieve a given temperature. The equation ione gave got me close, i just need an approximate "h" value.
Until I get samples of the parts I am designing around, I cannot perform experiments and must rely on calculations to get a rough idea of the cooling time. 

zekeman (Mechanical) 
26 Mar 10 17:03 
hc, convection coefficient depends on flow velocity, diameter and more weakly on gas temperature; hr, the equivalent radiation coefficient depends on emissivity and temperature of pipe and gas; so unless you can "ballpark" the geometry, ( e.g. a pipe with diameter of x opening up to a diameter of y over a distance z) we can't ballpark the h.


ione (Mechanical) 
27 Mar 10 9:17 
mmelville, The equation given in my previous post was just the solution of the differential equation suggested by corus. If you are looking for a correlation to calculate h in regime of forced convection (assuming the pipe is cooled just from the outside) you can use one of the correlations in the attached paper (from 9.13 to 9.18). I also suggest you to give a shot to the calculator at the link below. http://www.thermalwizard.com/tmwiz/default.htm 

If you check the internet or heat transfer books you'll find heat transfer convective coefficients (h's) values and formulae for forced air. You will need to determine the inside velocity under the 750 cfm air flow and probably other parameters such as bulk viscocity and air density. 

zekeman (Mechanical) 
27 Mar 10 16:22 
Following up on my objection to using lumped parameters without proof, I think you could make that determination by sending a slug of cool air through the pipe and if the drop in temperature is small next to the overall temperature difference, then you could write the 1st order linear differential eqaution rho*A*c*dT/dt=hi*l*(TcT)+ho*l*(T0T) whose solution is T=T(0)exp(t/tau)+hi/(h1+h0)*Tc+h0/(hi+h0)*To T(0) initial temperature pipe tau= time constant= rho*A*c/(hi+ho)l h1,ho inside and outside film coefficients rho = density coolant c specific heat l = perimeter A cross sectional area of pipe To test for the slug of air, assume that the pipe remains at the initial temperature. T(0) for the time it takes the slug of cools air to pass through the pipe. Then the delta temperature of the coolant is delta Tc=to* hi*l*(T(0)Tc)/(rho*A*c) to= transit time thru pipe If the temperature change is small, then all subsequent changes will be even smaller, thus making the case for the lumped model.


ione (Mechanical) 
29 Mar 10 4:45 
To check whether the heat transfer lumped model is acceptable, it is necessary to evaluate the Biot number:
Bi = (h* V/A)/k
Where:
h = convective heat transfer coefficient V = volume of the mass to be cooled A = surface of the mass to be cooled k = thermal conductivity of the mass
For low Biot number (Bi<0.1) it is acceptable to adopt lumped model and Newton's cooling equation works fairly well. A low Biot number implies that the thermal resistance between mass/cooling fluid (related to convection) exceeds the thermal resistance through the pipe wall (related to conduction) and so the pipe wall can be considered constant. In the case of a plastic pipe the thermal conductivity is low but the wall thickness is low as well, so the lumped model should be applicable. 

Thanks to everyone that posted. I think I can now get close at estimating a cooling time using the calculator link and the formulas provided. 

zekeman (Mechanical) 
29 Mar 10 14:47 
"To check whether the heat transfer lumped model is acceptable, it is necessary to evaluate the Biot number:
Bi = (h* V/A)/k..................."
Not that easy. That is true of radial conduction part of the problem. I had assumed that but it does not answer the flow problem here which must address the longitudinal gradient of gas temperature so that one could assume an "average" gas temperature over the length of pipe, reducing it to the total lumped model. Without that, you have no validity in doing it as a lumped model.


ione (Mechanical) 
30 Mar 10 3:28 
If the cooling air moves orthogonally to the pipe (outer diameter 40 mm), the fluid temperature should not be that affected, and evaluation of Biot number should give a reliable answer.
So while I still consider evaluation of Biot number a valid approach to define whether a lumped model is applicable or not for the transient analysis, I am now prone to say that the lumped model (Newton cooling law) is not applicable here (or at least it should give misleading results).
Despite of my previous statement ("In the case of a plastic pipe the thermal conductivity is low but the wall thickness is low as well, so the lumped model should be applicable"), here we are dealing with forced convection and so assuming the lumped model valid for Bi < 0.1:
h = Bi * k * A/V = 0.1 * 0.2 / 0.004 = 5 W/(m^2 * K)
Being:
k = 0.2 W/(m*K) plastic thermal conductivity (it could be also greater but the order of magnitude should be right)
V/A = s = 4 mm (pipe wall thickness)
The value of h, computed as shown above, undoubtedly reminds values more typical of natural convection than forced convection. 

corus (Mechanical) 
30 Mar 10 4:12 
ione, I'm not sure of your limit of 0.1 for the Biot modulus as I have seen elsewhere that 0.2 should be a limit, and hence double your value of h. In either case the lumped mass model is only intended to give an approximate answer that can be derived fairly easily if no other methods were avaliable. If it were possible, I'd suggest using finite difference methods or finite elements if the OP were familiar with those. For this 1D case there might be free software available that could do the job. excorus (semidetached) 

ione (Mechanical) 
30 Mar 10 4:41 

corus (Mechanical) 
30 Mar 10 9:53 

IRstuff (Aerospace) 
30 Mar 10 10:28 
OK, you guys need to get a room There are few things in nature that are discontinuous; near as I can tell, the Biot number is a guide, no more. I mean, what happens if the Biot number is 0.11, or 0.21, for Corus? You can still do the calculation, but the accuracy simply gets worse. It's not like it goes completely wonkers, is it? TTFN
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zekeman (Mechanical) 
30 Mar 10 10:34 
"agree that lumped model could give an approximate solution, and the approximation is as rougher as the more prevailing is the convection heat transfer at the interface pipe/fluid over conduction through the wall pipe (increasing Biot number). For a metal pipe the lumped model would have worked fine."
It will NOT work fine for a long pipe like 200 feet. If you think so , then what is the "average" fluid temperature over that distance. You have a 2 dimensional heat equation and you have only gotten the radial part approximately right, not the axial. So, unless that you can prove that there is an "average" temperature over the transient time, you have no valid solution. Try doing the classic problem of how long it takes for hot water to reach a faucet 200 feet away from the the source. What will you take for the average temperature of the water. To solve that accurately or even reasonably accurately, you can't, in general, lump the model. 

zekeman (Mechanical) 
30 Mar 10 10:46 
"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. 

ione (Mechanical) 
30 Mar 10 11:53 

IRstuff (Aerospace) 
30 Mar 10 12:32 

ione (Mechanical) 
30 Mar 10 13:05 
IRstuff, Are you pulling my legs? Just tried the link and it works! Anyway the book is "How to make injection molds" by Mengens/Michaeli/Mohren (page 281) 

ione (Mechanical) 
30 Mar 10 13:16 

IRstuff (Aerospace) 
30 Mar 10 17:12 

zekeman (Mechanical) 
30 Mar 10 17:52 
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 (TTc)/(TiTc),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, pp119127, and Schneider (I don't have the exact reference for his book) but he has a chapter in " Handbook of Heat Transfer",McGrawHill,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. 

zekeman (Mechanical) 
30 Mar 10 22:42 
"The OP probably needs a conservative time it takes to cool the surface.."
Edit to read:
The OP probably needs a conservative temperature at the surface...... 

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! 

ione (Mechanical) 
31 Mar 10 11:21 

zekeman (Mechanical) 
31 Mar 10 14:31 
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.


Attached is the pipe in an IGES format. The parameters are: Tube temperature: 85C Forced air temperature: 23C Target pipe temperature: 25C Time to cool: TBD Air flow inside pipe: 700CFM 

zekeman (Mechanical) 
31 Mar 10 21:07 
Can you give the developed length, diameter and wall thickness? 

zekeman (Mechanical) 
31 Mar 10 22:03 
I'm trying to post the C&J plot that applies to this transient and is similar to the one in fig 9.14 as I noted above. I think it is much clearer. If it posts, I will comment later on how to handle the specific problem posed by the OP 

zekeman (Mechanical) 
31 Mar 10 22:17 
I guess it didn't go thru. It's a JPEG file. Is that the problem, or is it the size ? Any help on this please? 

Quote:I guess it didn't go thru. It's a JPEG file. Is that the problem, or is it the size ? Any help on this please?
No problem to attach jpegs. There is a step "Insert Link" which is easy to overlook. ===================================== Engtips forums: The best place on the web for engineering discussions. 

Thanks guys, I'll give it another shot now. 

Length of pipe is around 48 inches. OD of pipe is 40mm, with wall thickness of 3.5mm
Thanks! 

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/HRFTDEGF kappa=.0147 FT^2/HR From the flow rate of 700CFM through the tube I used charts to get h=100 BTU/HRFT^2degF 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*(8523)=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 (TTc)/(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. 

zekeman (Mechanical) 
2 Apr 10 16:19 
It gets worse...
I found that the thermal conductivity of most plastics is more like 0.1 BTU/HRFT^2_DEGF as opposed to 0.5 for the example case. I redid the problem (assuming a similar reduction in diffusivity, kappa) and found the cooling time to be 7 minutes for the "exact" solution vs the 0.7 minutes using the lumped method or a factor of 10.
So much for using "lumped" solutions for this type of problem. 

zekeman (Mechanical) 
2 Apr 10 16:21 
Correction units of thermal conductivity should be BTU/HRFTDEG F 

ione (Mechanical) 
3 Apr 10 8:15 
Hey Zekeman,
The fact the lumped model was not applicable here (or at least the consideration it gives misleading results) was already stated in one of my previous posts (30 Mar 10 3:28). The Biot number is definitely too high (the poor thermal conductivity of plastic makes void the lumped model). I have to reassert that with a metal pipe and standing the same other conditions, the lumped model would have worked fine! 

zekeman (Mechanical) 
3 Apr 10 11:25 
Hey Ione, "...........In the case of a plastic pipe the thermal conductivity is low but the wall thickness is low as well, so the lumped model should be applicable."
Your quote of Mar 29.
I only commented because someone ( like you at that writing and me for accepting it at the time) may think that a "thin" section meets the "lumped" requirement and also that the magnitude of the the error could be game changing.
So we now all agree that Biot number MUST be small for the lumped model but how small is still at issue; so why not use the more exact solution found in the plots I found. It only takes a few more minutes of work to get an accurate answer, since the "error" in using the "lumped" solution cannot be readily quantified for Biot values that are "small". 

ione (Mechanical) 
6 Apr 10 8:22 

zekeman (Mechanical) 
6 Apr 10 14:40 
"'fathers' of the graphical approach for this kind of problem are Heisler and Grobler (Heisler charts and Grobler charts)"
Good references, but "fathers". That's a stretch. Similar graphs in C&J date back to the early 30's, long before Heisler and Grober were even out of grade school. 

ione (Mechanical) 
7 Apr 10 4:26 
Ok, someone here is related to Coulson and/or Jaeger and can't resign to terms that someone else has hijacked their work and got the glory.... Zekeman probably you are right (Heisler's work was presented only in 1947 and Grobler improved it in 1961), but the charts I have quoted have become part of the scientific literature as Heisler charts and Grobler charts. 

zekeman (Mechanical) 
7 Apr 10 11:10 
Yes, I admit that I am Carslaw's second cousin twice removed. 



