- How cold does your freezer need to be for freezing that piece of fish in less than 2 minutes?
- How fast does your cooling water need to flow to keep a pipe full of soup at temperature?

If you’re designing food production processes, controlling heat is essential for the process to succeed. You don’t want the soup to burn or the milk to freeze. Designing your production process well, so temperature is controlled, is essential. Even on a smaller scale, in your kitchen, temperature control is important, you don’t want to burn or accidently freeze your food.

At a small scale, such as your kitchen, temperature control is pretty straightforward. The feedback you get from the process is almost instantaneous. If you see your food is starting to turn brown or remains too hot, you can quickly adjust. Seeing as how you’ve got pretty small quantities you’re working with, they can change temperature quick enough. But once you start working with bigger quantities, it becomes more complex. This is where some basic calculations will come in handy.

In this article we’ll zoom in on two simplified examples of cooling food down: a perfectly stirred pot of food and a fish we’re planning to freeze.

### The science of temperature control

Before digging into the calculations you need to be aware of some basic principles of energy and heat transfer that will always be applicable:

**Conservation of energy**– Energy cannot be lost within a system. The type of energy may change though, so movement (kinetic energy) may turn into heat.**Temperature is a measure for energy**– The warmer the object (or food) the more energy it contains.**Heat flows from hot to cold**– The fast moving (hot) molecules bump into slower ones, transferring their energy over.**Systems always aim for a minimal energy state**– As a result, systems will want to reach equilibrium. For heat, this means that everything has the same temperature.

The field of study that focuses on this types of challenges is the field of transport phenomena. In transport phenomena you study the flow of mass and energy (and several others).

### Types of heat transfer

We now know that energy can’t get lost and that if it is not in equilibrium energy will flow from hot to cold places. When heat moves, it can do so in three different ways:

**Conduction**: two objects touch one another and transfer heat, this could be a pan standing on top of a hot plate. The objects themselves don’t move, the internal movements of molecules is what transfers the energy.**Convection**: heat in transferred via flows of material. Movement of molecules ensures that heat is moved around, this could be air, steam or a vigorously boiling water for instance.**Radiation**: there does not have to be a medium that actually moves the heat here, the sun is an example of radiant heat transfer. We won’t further focus on this scenario.

For each of these three mechanisms, the way to calculate heat transfer is very different! For conduction, the properties of the material through which the heat travels is very important. Whereas for convection the speed of the flow of materials may be essential.

In most real life cases, it is actually really hard to calculate the heat transfer in any of these scenarios. Generally several mechanisms are at play at once. That hot plate on the stove for instance, may be impacted by what is inside the pot, as well as the air around it.

That said, by simplifying the situations and making some assumptions, we can get some decent calculations going. For large scale processing where the system is well controlled with clearly set conditions, this makes it possible to still do the required calculations. Here, we’ll focus on one scenario: cooling down a well stirred pot of food.

### Cooling down a well stirred pot of food

So let’s have a look at a hot pot of food that’s on your stove. You want to know how fast that pot of food cools down once you’ve turned off the heat on the stove. For this we can use Newton’s law of cooling, but we do have to make a few assumptions:

**Heat travels throughout your food more easily than it leaves the food**. So, we’re assuming that the food is one homogeneous temperature (scientist use the Biot number for this, see below). By stirring something very well you can approach this ideal situation.**The temperature outside of your food is constant and completely homogeneous**. In order for this to be true, you either need a lot of space around your pot of food (e.g. a kitchen at room temperature) or use tricks to keep the surrounding temperature constant (e.g. a continuous flow of cooling water around the pot).**The temperature of your food is not that different from whatever you’re cooling it with**, luckily food generally does not get that hot.**All the processes of transferring heat are the same whatever the temperature**. That is, during cooling of the product, all physical mechanisms stay the same. As such, we’re assuming that your food being at a lower or higher temperature, does not make it more or less good at transferring heat.

In this specific instance we can use a simplified formula for calculating the temperature changes in our food: Newton’s law of cooling. We can write this as:

energy flow over time (in Joule/s, Watt)

=

dQ/dt

=

– h * A * (T_{product} – T_{environment})

=

-k * (T_{product} – T_{environment})

Where:

- h = heat transfer coefficient (
*W/(m*^{2}*K)*) - A = surface area over which the heat transfer takes place (
*m*^{2}) - T = temperature in either the product of the surrounding environment (
*K*) - k = sometimes k is used as a constant value, which is a combination of the effect of surface area and the heat transfer coefficient (
*W/k*)

Even without filling in any numbers into this formula, we can draw a few conclusions. Most of these will sound very intuitive for sure!

- A larger surface area of our pot of food cools down more quickly (A is larger)
- A larger temperature difference between our food and the surroundings results in faster cooling down.

Let’s apply this to a possible scenario: a hot cup of tea.

#### Example 1: cooling down a cup of tea

A cup of hot tea is a decent example where we can apply Newton’s law of cooling. By putting the cup of tea in a large room, we can assume that the cup of tea will not affect the temperature of the room. By stirring the cup we keep the temperature homogeneous. You do have some evaporating water which will affect the changes in temperature, which can complicate calculations, but we’ll neglect those for now.

We brought some water to the boil, poured in a cup of tea and then looked at how that water temperature was dropping down over time.

In the graph below you can see our observations of this cup of tea while cooling down. You can see that it cools down fast at first, but then levels off. In other words, when the driving force (the temperature difference of the cup and the surrounding air) is larger, more heat is transferred. The colder the cup of tea, the smaller the driving force, the slower further heat reduction goes.

#### Example 2: Cooling your drinks

Cooling you wine, soda, beer or water to the exact right drinking temperature can be challenging. You might be too late, wanting to drink it soon, or you may be pretty early, being afraid of overcooling your drink. That’s why two students (Álvaro Díez and Tibor Pal) made a calculation guide for you. It uses the Newton’ law of cooling as well. Again, this is not a perfect scenario. The drinks aren’t perfectly homogeneous in temperature. That said, it is close enough to get some decent approximations with Newton’s law (and in all reality, a few degrees differences tends to not matter as much!).

### Some further theory

There is a lot more theory to heat transfer phenomenon than we could ever cover in just one article. We did mention the Biot number earlier. This is not essential to understand the calculations, but it is a helpful concept to understand what is going on.

#### Biot number

Engineers tend to use a lot of ‘numbers’ to help them determine what type of scenario they’re dealing with. These numbers are ratios of properties of the system you’re evaluating, often called dimensionless numbers (since they don’t have a dimension).

One of these numbers, the Biot number tells you whether the rate limiting step of you cooling down the food sits within the food or outside of the food. The lower the Biot number (generally <0.1) the simpler the math for making calculations.

As an example, let’s assume you’ve got a huge pot of chili that you’re not stirring. You can probably imagine that it takes a long time for the heat from the inside of that pot to move to the outside. Once the heat is on the outside, it can leave easily. However, if you now take a small pot that you’re stirring all the time, it will be a lot easier for that heat to get to the outside. The large chili pot will have a large Biot number, the small well stirred one a lot smaller.

##### The formula

Engineers will use the following formula to calculate the Biot number (Bi).

Bi = L_{c} * h / k

Where:

- L
_{c}= characteristic length of the product - h = heat transfer coefficient (measure for the amount of heat that is transported over a surface area per temperature degree difference)
- k = thermal conductivity of the product (how well the product conducts heat)

Without even fully understanding the formula you can learn a few things:

- A bigger product will have a larger Biot number is everything else is the same
- If it is hard to transfer heat from the product to the outside world (represented in h) the Biot number is higher
- A product through which heat cannot travel easily (e.g. that unstirred chilli) has a higher Biot number (since the value for k will be low)

### Sources

Alfa laval, The theory behind heat transfer, link

Nave, R., Heat transfer, Hyperphysics, link

Wikipedia, Newton’s law of cooling, link

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