## Calculating Cooling Times

## Calculating Cooling Times

(OP)

I am looking for a formula to determine the cooling time of a blow-molded plastic tube with forced cold air. The pipe is 40mm o.d. with a 3-4mm wall thickness. I appreciate any help offered. Thanks in advance!

## RE: Calculating Cooling Times

Q = htc*area*deltaT

Q = emiss*sigma*area*(T_1^4-T_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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## RE: Calculating Cooling Times

ex-corus (semi-detached)

## RE: Calculating Cooling Times

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 time-decaying 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://ww

## RE: Calculating Cooling Times

## RE: Calculating Cooling Times

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.

## RE: Calculating Cooling Times

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.

## RE: Calculating Cooling Times

## RE: Calculating Cooling Times

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.thermal-wizard.com/tmwiz/default.htm

## RE: Calculating Cooling Times

## RE: Calculating Cooling Times

rho*A*c*dT/dt=hi*l*(Tc-T)+ho*l*(T0-T)

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.

## RE: Calculating Cooling Times

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.

## RE: Calculating Cooling Times

## RE: Calculating Cooling Times

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.

## RE: Calculating Cooling Times

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.

## RE: Calculating Cooling Times

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.

ex-corus (semi-detached)

## RE: Calculating Cooling Times

My source concerning Bi<0.1 is "Introduction to thermodynamics and heat transfer" by Yunus A. Cengel (Italian edition), but the same statement is easily found with a Google search ("Technology of thermoforming" )

http://

I 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.

## RE: Calculating Cooling Times

says "If the Biot modulus os less than 0.2 no significant gradient is expected". So the answer is you takes your pick what would be acceptable.

ex-corus (semi-detached)

## RE: Calculating Cooling Times

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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## RE: Calculating Cooling Times

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.

## RE: Calculating Cooling Times

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.

## RE: Calculating Cooling Times

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).

h

## RE: Calculating Cooling Times

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: http://

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## RE: Calculating Cooling Times

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)

## RE: Calculating Cooling Times

Try the following link (if you want of course) and let me know

http

## RE: Calculating Cooling Times

TTFN

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## RE: Calculating Cooling Times

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.

## RE: Calculating Cooling Times

Edit to read:

The OP probably needs a conservative temperature at the surface......

## RE: Calculating Cooling Times

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!

## RE: Calculating Cooling Times

http:

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.

## RE: Calculating Cooling Times

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.

## RE: Calculating Cooling Times

Tube temperature: 85C

Forced air temperature: 23C

Target pipe temperature: 25C

Time to cool: TBD

Air flow inside pipe: 700CFM

## RE: Calculating Cooling Times

## RE: Calculating Cooling Times

I think it is much clearer.

If it posts, I will comment later on how to handle the specific problem posed by the OP

## RE: Calculating Cooling Times

## RE: Calculating Cooling Times

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## RE: Calculating Cooling Times

TTFN

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## RE: Calculating Cooling Times

## RE: Calculating Cooling Times

OD of pipe is 40mm, with wall thickness of 3.5mm

Thanks!

## RE: Calculating Cooling Times

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.

## RE: Calculating Cooling Times

I found that the thermal conductivity of most plastics is more like 0.1 BTU/HR-FT^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.

## RE: Calculating Cooling Times

units of thermal conductivity should be

BTU/HR-FT-DEG F

## RE: Calculating Cooling Times

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!

## RE: Calculating Cooling Times

"...........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".

## RE: Calculating Cooling Times

These curves could be found at the link below from page 194 to 205 (for given geometries).

h

I have also found another link which gives a more direct access to the charts aforementioned:

ht

## RE: Calculating Cooling Times

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.

## RE: Calculating Cooling Times

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.

## RE: Calculating Cooling Times

## RE: Calculating Cooling Times

h