Lesson 2 - Principles of Heat Transfer and Conduction

As mentioned earlier, latent heat is the energy absorbed or released from a substance during a phase change.

An easy example of latent heat to visualise is the boiling of water into water vapour or the freezing of water into ice. The amount of energy required to vaporise a liquid or solid or to condense a vapour is a standard quantity called the heat of vaporisation (ΔH_vap). Conversely, the amount of energy required to melt a solid or freeze a liquid is known as the heat of fusion (ΔH_fus). The SI unit for latent heat is in J Kg^-1. Each substance has its own latent heat values.

Let us put it another way. When we go from a liquid to a vapour or vice versa, it is called ‘vaporisation’. When we go from a liquid to a solid or vice versa, it is called ‘fusion’. Every substance has a known heat of fusion and heat of vaporisation.

Let us use the wort boiling example from the brewery to illustrate these concepts further. Heat is being supplied to the wort kettle continuously. We know that it is boiling because we see vapour exiting the kettle. If we monitor the temperature of the liquid wort during boiling, we will notice the temperature does not exceed the boiling temperature of the wort, which is around 100°C. So where does the heat energy go? The heat that is being transferred from the heat source to the wort is being used to vaporise the wort from a liquid to a vapour.

A substantially larger amount of thermal energy is required to change the phase of a substance compared to the sensible heat required to merely change its temperature.

As we know, most of wort is just water. When we boil wort, water is vaporised, leaving behind the sugars and other components of wort. For reference, the heat of water vaporisation is below:

To calculate the energy associated with vaporising a substance, we will use the following equation:

Where

Q is the required amount of heat, measured in kJ

M is the mass of the substance being heated, measured in kg

ΔH_vap is the heat of vaporisation, measured in KJ Kg^-1.

Let us now do a worked example to put this equation into practice.

We will calculate how much energy is required to boil the wort from the previous worked example. Let us say we want to boil 5% of our wort.

From our earlier example, we have 7,860 kg wort. As mentioned earlier, when we boil wort, we are vaporising water.

The mass of water we want to boil would then be:

Recall the equation just introduced:

And enter our values:

So, the energy needed to boil 5% of our wort is 887,001 kJ.

Let us quickly look at what we calculated in these two examples. We determined the amount of thermal energy required to bring the wort to nearly boiling, and the amount of thermal energy required to boil 5% of the wort. Therefore, if we add these two values, we will know roughly how much thermal energy is required to bring the wort to a boil and to boil 5% of the wort. So, we will need about 628,800 kJ + 887,001 kJ = 1,515,801 kJ of thermal heat supplied to the kettle during the boil.

Q and q

Throughout this unit we will use equations to calculate heat as Q or q. Q is the total amount of heat in joules which is a unit of energy. q is the rate of heat transfer in watts (joules per second). The difference between these is the element of time. Q has units of joules while q has units of joules per second. The two are interchangeable by using mass as a mass flow rate in kilogram per second. In other words, Q = MC∆T translates to q = mC∆T where m is the mass flow rate. Similarly, Q = MH_vap translates to q = MH_vap/t where t is the time taken for the known mass to change phase.