cooling of barrels with hot tomato paste 5

[CARF]

Dear all,

Our filling line fills large barrels (300 liter) with hot tomato paste (90 gradC). We want to cool these drums by putting them in cold water (20 gradC)(and perhaps apply a high water flow).

Does anybody has experience on this?
How long will it take to cool the paste down to say 30 gradC?
Models?
Spreadsheets?

Thanks for sharing knowledge !

 

[Maui]

We will need some more information in order to generate an answer for you. What are the physical dimensions of the cans? Diameter, height, and wall thickness? What material are they made from? The thermal conductivity of the can material is important. Will the cans be completely covered by the flowing water? Will they be placed in the water in large groups, or will they be coooled individually? What is the temperature and flow rate of the water? All of these parameters must be specified in order to calculate an accurate time.


maui

 

[tiwanaku]

You may want to consult the classic book "Heat Conduction in Solids" by Carslaw and Jeager (Cambridge Press or a Dover book)to look at a simplified solution to your problem as a conduction-only problem involving a hot cylinder (tomato paste)losing heat by conduction to the surrounding medium (water). A summed series solution involving Bessel functions is provided in the text provided a surface heat transfer coefficient(h)is specified on the container surface(you may want to use a free convection h value here if no water circulation is assumed to exist). A more precise solution that allows for convection currents developed in the water and (nonNewtonian) tomato paste during cooling can be easily found by transient FLUENT or FLOW-3D solutions if you have these CFD codes available.
tiwanaku

 

[Montemayor]

MVD (Industrial):

In order to stimulate efficient heat transfer, you should provoke good convection currents in order to accentuate the controlling film coefficient (presumed to be that of your hot tomato paste). However, I envision that it would be next to impossible to expect an agitator within the drum being filled. Any attempt to cool the paste after it is inside the drums, without any stirring or agitation is going to be very time consuming and inefficient.

My first attempt to cool the paste would be to do it while the paste is flowing from the filling machine and into the drum. I visualize a cooled coil as an extended spout from the machine, leading to the drum to be filled. This method allows you use the fill machine's pressure as the driving force to stimulate a large Reynolds number within the cooled coil, giving a better convection film coefficient. You may have to increase the machine's fill pressure due to the high viscosity of the paste and the added resistance of the coil. But this is what is usually the case: a higher driving force is typically used to generate a better film coefficient by increasing the Reynolds number.

I would use a helical coil (with an external cooling jacket - a "double tube" affair) because it allows for more cooling area in a reduced volume and it accentuates internal eddies and added turbulence to the paste. The paste would flow in the internal tube while the water would be in the outer tube. Spiral and Helical coils are known and proven to yield much better film coefficients because of the added internal eddies (HTRI reports). The natural configuration of the coil also lends itself well to thermal expansion and mechanical vibrations. A trial run on an experimental setup would not be expensive to try - at least on one drum filling line. I would recommend a trial setup first, because you and I know from the outset that the viscosity of the paste is going to change differentially as the paste is cooled within the coil. This viscosity change will affect the driving force required to fill the drum in the time you require. I presume the paste is NOT a newtonian fluid - am I correct?

Depending on your filling rates, I would try a 3/4" SS tube inside a 1" SS tube. The water connections on the outer tube would be counter-current and be done with Tygon tubing.

There are assembled double-tube suppliers in the market place. I have seen them advertised in Thomas Register. You may have a simple application here if you can contact one of them.

Hope this helps.


Art Montemayor
Spring, TX

 

[CARF]

Dear Montemayor,

Thank you for spending so much time on my problem. I shall explain it a bit more. The system you describe (tube in tube heat exhangers often with helical static mixers) we have in use now. The problem is that the product is so viscous that we apply lots of shear and pressure on the product in the tube in tube pasteurisation unit (heating - holding - cooling). This damages the product. Yes it's non-Newtonian => shear thinning. So what we want is to cool it down without this lengthy tube systems. So yes it maybe inefficient as a cooling method; it might keep product quality at a higher level.

So we want to get an idea what the maximum cooling rate by barrel cooling will be.

Barrel diameter = 500 mm
Barrel height = 1100 mm
(So 216 liter instead of 300 liter)
The paste is really thick, like *putty*
Barrel wall thickness = 2 mm, carbon steel
They will be totally flooded in cooling water
Water flowrates roughly = 1 m/s
Water temps = 20 gradC
Paste start temp = 90 gradC

Cooling of an individual barrel as a first trial.

Hope you understand the problem better now, thanks for helping!

MVD

 

[cbiber]

I would start the analysis by treating the paste as a solid object. You'll need the thermal mass (mass * specific heat) and conductivity, or thermal diffusivity.

Look in any heat transfer text to find a convection coefficient for the water around the barrel. It will be in the form of a Nusselt number as a function of Reynolds number (for forced convection). You will want the Reynolds number based on the diameter of the barrel. Look for correlations that apply to "flow over a cylinder". If you don't have a book handy, use something in the middle of a typical range -- maybe 1000 W/m^2K.

Then look up Heisler-Grober charts for cylinders with external convection. These charts describe the time behavior of a solid, and avoid having to deal with Bessel series (although you can't put them in a spreadsheet). You should be able to find this in a good thermal text along with the convection coefficient.

Failing that, take a guess at the cooling time using a thermal time constant of the form RC, where R=1/(h*A), h being the heat transfer coefficient and A the submerged surface area of the barrel, and C being the total thermal mass of the paste in the barrel. For the guess, don't worry about the barrel material -- it's probably not the dominant parameter for the type of analysis I think you're asking for.

Biber Thermal Design

http://www.biberthermal.com

 

[pmureiko]

I recommend you solve this using the transient cooling charts for solids objects. The paste in the drum will act as a solid because of its very high viscosity. The heat transfer resistance from the metal walls will be negligible. The natural convection coeficient for water can be approximated for a liquid on a vertical plate since the water is not controlling the cooling process.

A rough estimate a cooling time for the center of the drum to reach 30 oC is 86 hours. It may be quicker since I treated the drum as a cylinder and neglected heat transfer from the top and bottom. I also assumed the tomato paste had the same heat capacity and thermal conductivity as water, with a specific gravity of 2.

You'll probably want to repeat the calculation with more accurate physical property values and perhaps a more rigorous method.

Good luck.

 

[CARF]

Hi pmureiko,

The paste in the drum will act as a solid because of its very high viscosity. => Agree

86 hrs ?

Hmmm that's long

Here some more properties:

Cp = 2000 [J/kg.gradC]
rho = 1250 [kg/m3]
labda = 0.2 [W/(mK)] (rough guess)

Could I have an example of the calcs?

Thanks !

 

[Maui]

As a rough estimate for the time required to cool one can to your aim temperature, the following formula should get you in the ballpark:

t = [2.3*D*Rho*Cp/Lambda]*ln{[To - Tw]/[Ts - Tw]}

where t = time in seconds
D = diameter of the can in meters
Rho = density in kg/m^3
Cp = the specific heat in J/kg.gradC
Lambda = heat transfer coefficient in W/gradC
To = initial temperature of the paste in gradC
Tw = temperature of the cooling water in gradC
Ts = final temperature of the paste in gradC

Your actual cooling time should be slightly less than the value calculated from this formula because heat loss through the top and bottom of the can was not incorporated in the derivation. I hope this helps.


Maui

 

[25362]

Others use refrigerated chilled air-cooling tunnels for similar purposes. I've seen them used on continuous wax slabbing.

 

[CARF]

OK, here's my speadsheet:

=(2.3*B3*B4*B5/B6)*LN((B7-B8)/(B9-B8))

INPUTS

D 0.5 diameter drum meter (B3)
Rho 1250 density kg/m3
Cp 2000 specific heat J/kg.C
lambda 0.2 heat transfer coeff W/m2.C
To 80 initial temp. of paste gradC
Tw 20 temp. coolingwater gradC
Ts 30 final temp. of paste gradC

OUTPUT

t_cool 25756542.37 sec
7154.595103 hr

Well, my lambda is prop. wrong? Any ideas?

ThX,
MVD

 

[Maui]

MVD, my mistake for leaving out the m^2 in the denominator of the units for the heat transfer coefficient. In a previous post you listed the value for lambda as 0.2 W/mK, but in your last post where you listed your calculations you used 0.2 W/m^2-gradC for lambda. If the units differ but a factor of 1000 in the denominator, this could be the source of your problem.


Maui

 

[Maui]

MVD, I interpreted the lambda value you originally posted, 0.2 W/mK, as "0.2 Watts per milliKelvin". According to "Transport Phenomena in Metallurgy" by Geiger and Poirier, page 259, the heat transfer coefficient for heat flow between steel and water with no agitation is 240 Btu/hr-ft^2-gradF, which converts to 1,363 W/m^2-gradK. By substituting this value into the equation in my prior post, you get t = 3779 seconds. It should take approximately one hour for the can to cool to 30 degrees C.


Maui

 

[25362]

Thermal conductivities are expressed in W/(m.K), or Watts per meter per Kelvin.
Heat Transfer Coefficients, in W/(m².K)