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# Force Convection Heat Transfer2

## Force Convection Heat Transfer

(OP)
Hello, It's my first time posting in a blog...I was hoping to get some help from you smart engineers.

EXISTING SCENARIO:
-Biscuits have left the oven and are 45 deg C. Travelling on the conveyor at 0.28 m/s. Density of biscuit is 900 kg/m^3. Biscuit is 60mm diameter. There are approx 160 biscuits per square meter. (picture attached).

PROBLEM:
- I need to install industrial fans above the product to cool the biscuits down to 25 degrees.
- The fans will be using ambient air temp at 20 deg C, and a velocity of 3 m/s.
- I do not know how many fans I need to install.

QUESTION:
- I know the forced convective heat transfer coefficient increases the longer the biscuit is exposed to the forced air, but how long does it need to be exposed to this forced air to drop from 45 deg C to 25 deg C.
- I would greatly appreciate any help on this, I have spent a long time on it but I seem to be going in circles. Thanks again for any help.

Cheers,

Steve

### RE: Force Convection Heat Transfer

"I know the forced convective heat transfer coefficient increases the longer the biscuit is exposed to the forced air"

This is incorrect. The heat transfer coefficient is solely from mechanical aspects of the moving air; time is not a factor. Is this for school? Student posting is not allowed.

TTFN
I can do absolutely anything. I'm an expert!
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers

### RE: Force Convection Heat Transfer

This is obviously not a school question. OP gave the density of the biscuit because they were trying to provide helpful info.

IRstuff is correct that the heat transfer coefficient doesn't change - what changes is your temperature differential.
The amount of heat being transferred can be summarized (in a general case) as
Heat_Transferred = (SOMETHING NASTY TO CALCULATE) * (Temp_Hot - Temp_Cold)
The temperature of the biscuit is given by
Temp_New = Temp_Old - Heat_Transferred_Out * Heat_Capacity_of_Biscuit
So you can see that as the temperature of the biscuit you're trying to cool decreases, the amount of heat transferred out of it decreases as well. And as the amount of heat transfer out of the biscuit decreases, the temperature change decreases. This is the backbone of Newton's Law of Cooling. The result of this is that heat transfer for cooling an object looks like:
T(t) = T_a + (T_0 - T_a) * e^(-k * t)
Where
T = temperature of the thing (biscuit)
t = time
T_a = temperature of the atmosphere (in your example, 20C)
T_0 = starting temperature of the thing (in your example, 45C)
e = Euler's number (2.71828ish)
k = the bane of our existence. k is sort of a "summary factor" in this case. Now, what you could do is set up a tray of these biscuits (if you have access to it) and put one of the fans you'll be using on them. Every 10 seconds or so, record the temperature (preferably of one near the middle). Using your data and the starting conditions, you should be able to calculate your k for this system.

Given the k determined above, you can, calculate the amount of time required to cool the biscuits to your desired temperature under the forced air you describe. Now you just need to ensure that the biscuits spend enough time under these air conditions (e.g. enough fan coverage). You have your time to cool and a linear velocity of the biscuits as they move through the "cooling" zone - use it to determine how much linear distance the "cooling zone" must cover. I'd recommend multiplying this by a small factor of safety, maybe 1.2 or 1.3 (as I assume 25C is your max temperature, and going slightly under won't hurt). Now, determine how much "length" your fan can cover, and divide to get the number of fans needed.

I'd offer a guess at your k factor to help you along, but there's not a lot of tabulated data for "biscuit" - plus I'd need some more information. This is an experimental method, which (if you only have one shot to get it right) will give you the most accurate results.

Godspeed and good luck.

### RE: Force Convection Heat Transfer

Are the biscuits on trays or mesh? Trays will make it more difficult.
You need some idea of the heat transfer within a biscuit.
If you assume that you can cool the surface to 20C, then how long will it take to cool the interior to 25C (hint, use Heisler charts).
Then work on the external side of it.
You could do some searching to see what the coolers on commercial ovens look like, this is common.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube

### RE: Force Convection Heat Transfer

(OP)
Thank you everybody for your quick response and helpful answers. No this is not a student posting, it is a real problem I'm facing at work.

Our laboratory does not have data on thermal conductivity and there isn't much info on the net regarding related food products. But based on NATEPIERCY's excellent recommendation I have calculated the 'k' value for the existing process to be k=1.18

This value is without forced convection (the only forced convection is the air velocity from the conveyor belt moving). The fans will increase this air velocity by a factor of 10. I will take NATEPIERCY's suggestion and do an experiment on the biscuit under a fan (to simulate the proposed solution) and calculate the 'k' value. With this information I should be able to reasonably determine the number of fans I need to install.

Thanks again for everybody's help.

### RE: Force Convection Heat Transfer

Well, if there is "no data" ... GO MAKE SOME BISCUITS.

Seriously. GO MAKE SOME Biscuits of the approximate size and height with the same bread dough you use to make the real product.

Put them on a sheet (tray) or on a screen if that type of support will be used.
With 8x cooking thermometers, measure the temperature
1. at the left surface (an IR thermometer will be good enough). Record temperature every 15 seconds from "hot" to room temperature in still air.
2. 1/4 inch (the crust thickness, right?) below the left surface. Use a stick-in meat type thermometer.
3. 1/2 inch in
4. 3/4 inch in.
5 At the middle.
6 At the 3/4 inch depth in from right side.
7. 1/4 inch in from right side
8. At the right surface.

Now, plot those temperatures over time. You are looking for simple convective and radiation loss (and rad. gain from the tray) so you can get the surface heat transfer coefficients (used in the real problem solution above) AND the internal heat transfer coef of your biscuit from the internal dough out to the crust then through the crust (which is thermally different and chemically different and has a different density than the softer dough in the center, AND the thermal mass of the real biscuit.

### RE: Force Convection Heat Transfer

Yep. Too complex to solve analytically. Couple of other complicating factors:
1. If the biscuits are on a metal tray that has also been in the oven, much of the cooling will be required by the tray.
2. Biscuits contain moisture (steam). There will be a considerable quantity of latent heat to be removed as well.

A thermal imaging camera would be useful.

je suis charlie

### RE: Force Convection Heat Transfer

Eh... well seeing as how the fans and biscuits are pretty inexpensive, and your time is not... buy a bunch of fans.

You could indeed solve this analytically, but cooling a of a cooked biscuit means that biscuit has nice scattered heated pockets of air, making it a matrix material of no consistency.

Now you could use a thermal camera, I have experience with them, they only tell you what's on the surface, if you've "cooled" a biscuit too fast, it still has a lot of internal heat, and the thermal camera will just confirm that the biscuit is indeed still hot, or warms back up after you cool it. Alternative, less expensive method; stick finger in biscuit.

### RE: Force Convection Heat Transfer

I will also agree that this is one that is better solved experimentally. One very simple method of calculation, and it MIGHT be okay for initial estimates is to use the lumped capacitance method. Basically the lumped capacitance method assumes that for the transient heating or cooling of a solid object immersed in fluid, the change in temperature is due solely to convection at the fluid-solid interface and (this is an important one) the solid is assumed to be at a constant. The lumped capacitance equation is: (T-Tinf)/(Tinitial-Tinf)=exp(-((h*As)/(rho*V*c))*t where T is the surface temperature at some time t, Tinf is the ambient temperature, T initial is the initial surface temperature, h is the convective heat transfer coefficient, rho is the density of the solid, As is the surface area of the solid, V is the volume of the solid, and c is the specific heat of the solid. On the surface this seems pretty easy, but if you don't know the density and specific heat of the biscuits you are off to a bad start. Also, determining h is not a trivial thing and could be a pretty big source of error in your calculations. And if that is not enough, the lumped capacitance method may lead to very inaccurate results if there is a temperatue gradient within the biscuits (which I suspect there would be) Typically for the lumped capacitance method to be valid, the thermal conductivity (k) of the solid must be very high and/or the length scale of the solid object must be very small (think a 1" diameter copper sphere immersed in a liquid). The Biot number is typically used to determine the validity of the lumped capacitiance method. The Biot number is simply the ratio of convective resisitance to conductive resisntance- Bi=(hL)/K. Where Bi<<<1, it is pretty safe to assume that the solid has a uniform temperature distribution. If you can't use the lumped capacitiance method, you will have to solve the transient conduction problem in the solid as well as the transient convection at the surface, and that complicates things. Even if you were to make a CFD model, how would you define the material porperties of a biscuit? So like the others, I say that your only way to solve this is through testing. One thing that could be helpful to you is that if you look at the term on the right side of the lumped capacitiance equation it can be rewritten as 1/tau, where tau=(rho*V*c)/(h*A) is the thermal time constant. The time constant is a measure of how quickly the solid will respond to a change in thermal environment. The time constant can be determined experimantally and in theory an empirical relationship could be developed to solve your problem. But again, to echo the others, I don't think this is one you can solve accurately strictly on paper.

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