Unsteady state conduction of heat in a cylinder 2

[DMull]

Hi everyone,

Onsite there is a fat maturation tank, which is jacketed and held at 32°C. Fat going into this tank is at around 22°C.

I am trying to derive an equation to work out the minimum temperature (at the centre of the tank) after a given time t. Unfortunately I’ve forgotten a lot of my college equations, and unsteady state conduction in a cylinder is giving me some trouble…

I am going to treat the fat as a solid cylinder at 22°C, which is exposed to a temperature of 32°C at time t = 0. I am assuming that there is no convection or radiation – the tank is not stirred.

I know the starting temperatures and the dimensions of the tank. I don’t know any of the heat transfer coefficients of the material – but before I go looking for these I’d like to have an equation to put them in.

Any suggestions on what equation I should use?

Any advice on where I can locate thermodynamic coefficients for vegetable oils/fats?

 

[peaktop]

Take a look at this:

http://www.mhtl.uwaterloo.ca/pdf_papers/mhtl96-8.pdf

Regards,

PT

Fudd's First Law of Opposition: "If you push something hard enough, it will fall over."

Teslacle's Deviant to Fudd's Law: "It comes in, it must go out."

 

[peaktop]

Also:

http://en.wikipedia.org/wiki/Heisler_Chart

Fudd's First Law of Opposition: "If you push something hard enough, it will fall over."

Teslacle's Deviant to Fudd's Law: "It comes in, it must go out."

 

[peaktop]

One more.

http://www.me.umn.edu/education/courses/me8341/modules/Mod-10.pdf

Also, lots of hits when googling 'Heisler Grober'.

Good luck.

PT

Fudd's First Law of Opposition: "If you push something hard enough, it will fall over."

Teslacle's Deviant to Fudd's Law: "It comes in, it must go out."

 

[zekeman]

"I am going to treat the fat as a solid cylinder at 22°C, which is exposed to a temperature of 32°C at time t = 0. I am assuming that there is no convection or radiation – the tank is not stirred."

Is the fat in solid state?

If so, then I suppose the problem becomes a solid cylinder with a surface temperature T=32 and you want the transient solution?

I would suggest Googling Schneider, heat transfer. His books have the transient solutions both graphically and in equation form ( an infinite series). Another source is
" Conduction of Heat in Solids", Carslaw and Jaeger,Oxford Clarendon Press, 1959.

 

[DMull]

Thanks for all of the help.

zekeman, the fat is in a pseudosolid state so I am treating it as a solid while it is in the tank.

Here’s where I am so far:

The radius, height, density and temperatures were easy to get:

Radius of tank 1m
Height of tank 2m
T0 22°C
Ti 30°C
T? 32°C
Density of fat 1035kg/m3

The others were a little more difficult. Fortunately I only want a rough estimate, so I’m really only concerned about order of magnitude figures here.

o The k value of the fat I got from

http://palmoilis.mpob.gov.my/Netacg...&u=/public/tascessearch.html&r=43&f=G&op1=ADJ

It’s for the liquid phase, but looking at the k values of various fats online it seems to be around 1.5 – 2.5, so this is close enough.

o The specific heat I got from a paper on “Determination of some physical and transport properties of palm oil and of its methyl esters”. Again, this was for the liquid phase – but it’s reasonably close to other figures for fats.

o These let me calculate the thermal diffusivity – the result is inline with other figures I’ve seen for vegetable oils.

The heat transfer coefficients are problematic. The figures are taken from papers on frying oils – I’m really not sure how applicable they are. Any ideas?

k value of fat 0.17 W/K m
Heat transfer coefficient 200 W/m2 K
Specific heat capacity 1746.8 J/kgK
Thermal diffusivity 9.4 × 10 -8 m2/s

Assuming that these are all ok though, I can now look at using the Heisler charts peaktop linked to:

Biot number 1176
1/Biot number 0.00085
(T0-T?) / (Ti - T?) 5

And here I run into problems because my values are way off the charts. Any ideas on how to proceed?

 

[IRstuff]

Heat transfer coefficient from what to where? The HTC represents a liquid convection, of which you have none.

A simple estimate should simply be something like lumping all the fat at the end of a thermal conductance equivalent to, say, 1/3rd of the radial thickness of the fat.

TTFN

FAQ731-376

 

[peaktop]

It appears that different authors have different nomenclature for Tinf, T0, Ti, etc., but all seem to agree that dimensionless temperature is final temperature excess divided by initial temperature excess. In your case:

T0-Tinf == Final temperature - Ambient temperature = 30 - 32

Ti-Tinf == Initial temperature - Ambient temperature = 22 - 32

(T0-Tinf)/(Ti-Tinf) = 0.2

Fudd's First Law of Opposition: "If you push something hard enough, it will fall over."

Teslacle's Deviant to Fudd's Law: "It comes in, it must go out."

 

[chicopee]

Your problem reminds me of a test I had in a college heat transfer test on how long would it take to warm up a can of beer in a room. There is good information presented above which at first appears unyieldy and a pain in the neck to handle.

Altho you classified fat as a pseudosolid, I would not be surprised if it had convective movement eventho it may have a high viscocity. If this is the case and for a ball park value why not try a time step analysis on a spread sheet of the equation of Qin(= hc*A*delta T1)= m*Cp*delta T2/delta theta where delta T1=(32-Tb1), delta T2=(Tb2-Tb1), delta theta is the time increment. Tb is bulk temperature,Tb2 & Tb1 being final and initial bulk temps during time increments; as Tb increase you may want to change Hc and Cp.

 

[zekeman]

In this problem, the only parameter of interest is the diffusivity which you state is 9.4*10^-8, a number that is probably close to the mark for solid fat.

Surprisingly (to me) I found from Schneider charts for your exact case that for a temperature rise (at the center) of 1/2 the total possible, i.e.
(T-Ti)/(Tamb-Ti)=0.5. or a rise of .5*(32-22)=5 C, it takes 400 hours.

This derives from the corresponding chart parameter
kappa*t/x^2=0.15, where

kappa=diffusivity
t= time
x= radius of cylinder

Substituting
x=1 m

kappa=9.4*10^-8

t=0.15/(9.4810^-8

which says that it takes about 1.5 million seconds or 400 hours,a number, if accurate, would suggest the answer to the ultimate insulator.Greenhouse problem solved!

 

[DMull]

Wow, thanks for all of the help everyone.

IRstuff: You’re right of course. I was trying to get the h value (or an estimate of it) so I could calculate the Biot number – but it isn’t applicable to this problem.

Peaktop: Thanks for that, I thought I might be missing something obvious when the charts all stopped at 1...

chicopee: thanks for the suggestion, but I don’t seem to have any practical way to measure the h value.

zekeman: Thanks for that, it seems to be exactly what I’m looking for!

Where did you get these Schneider charts from? I have access to Knovel so I might be able to view them.

I’d like to be able to view them so I can see how the time needed changes as the intial temperature varies.

 

[davefitz]

It would seem that one important parameter not defined is the outer cylinder boundary condition- likely the jacket uses condensing steam - so the steam temp of 32 C needs to be augmented with an effective outer convective heat transfer coeficient of condensing steam + fouling + liquid wetted layer resistance.

 

[zekeman]

"I'd like to be able to view them so I can see how the time needed changes as the intial temperature varies."


I found the socalled Schneider charts in a copy I have of "Handbook of Heat Transfer",Rohsenow and Hartnett,McGraw Hill, 1973, p 3-59, geometry f ( length =2*radius)

Fortuitously,it has your exact geometry and BTW, varying initial temperature should be easily handled since kappa*t/x^2 vs (T-Ti)/Tamb-Ti) are correlated , so varying Ti can be accommodated. I developed a table for a few values of kappa*t/x^2for your information


Kappa*t/x^2 (T-Ti)/(Tamb-Ti)

0.5 0.15

0.7 0.25

0.9 0.35

 

[DMull]

Thanks for that zekeman, that's exactly what I needed.

From playing about with the figures, it seems that any kind of temperature gradient between prodcut coming in and the tank will takes weeks to even out - we need to ensure that the inlet temperature is 32°C.

davefitz - thanks for the comment, but in this case I'm assuming that the resistance to convective heat transfer at the boundry is negligible compared to the resistence to conductive heat transfer within the body. Since the (very rough) Biot number I calculated above was so large this seems to be a reasonable assumption.

 

[zekeman]

D'Mull,

As you may have noticed, the headings in the table I posted should be inverted to read

(T-Ti)/(Tamb-Ti) Kappa*t/x^2

0.5 0.15

0.7 0.25

0.9 0.35

 

[DMull]

Oh yeah, forgot to mention that.

Thanks.