Linus School

A roast does not develop heat momentum, aka thermal momentum, because this doesn’t exist. As far as I can tell from what has been written about it by Rob Hoos and others (including some on this forum), the idea is that the rate of rise (RoR) of the bean mass has its own inertia that must be accounted for when making changes to roast conditions.

The purpose of this post is to apply some fundamental thermodynamics to this and hopefully make it disappear in a puff of logic.

RoR is the rate of change of bean mass temperature with time, in normal notation this is δT

_{b }/ δt ; T is in Kelvins, t is in seconds. If you are not familiar with Kelvins, the more familiar Celsius scale is derived from the Kelvin scale by

^{ o}C = K – 273.1.

There is a good reason for using this unit which will become obvious later.

We know that δT

_{b }/ δt = Q

_{b}/ C

_{b}

Where Q

_{b}is the net heat input to the bean mass and C

_{b}is the heat capacity of the bean mass. C is in Joules per kg.

This is what’s known as a lumped parameter model: the parameter C

_{b}incorporates such things as the heat required to vaporise the water content and the heat required to drive the reactions while the parameter Qb allows for heat losses outside the beans (eg to ambient). I will tease out some of these factors later but for now we’ll stick with lumped parameters, hence the partial differential indicated by the lower case deltas.

The fundamental rule of heat transfer is

Q = U . A . ΔT

Where U is the thermal transmittance*, A is the system boundary area and ΔT is the temperature difference across that boundary. U is in watts/ m

^{2}/K, A is in m

^{2.}. In this case the area with which we will be concerned is the net area of the air layer next to the skin of the beans, this will again become obvious later. ΔT = T

_{g}– T

_{b}where T

_{g}is the temperature of the heating gas ( air) and T

_{b}is the bean mass temperature as above, so

Q

_{b}= U . A . (T

_{g}– T

_{b})

Putting this together, we have

δT

_{b }/ δt = U . A . (T

_{g}– T

_{b}) / C

_{b}or

δT

_{b }/ δt = U . A . T

_{g}/ C

_{b}- U . A . T

_{b}/ C

_{b}

Letting U . A / Cb = k, this is

δT

_{b }/ δt = k . T

_{g}- k . T

_{b}

For constant T

_{g}this means the rate of change of T

_{b}is proportional to T

_{b}, so we have an exponential** decay curve. Over small time scales where k can be considered as a constant we can convert to a full differential equation

dT

_{b }/ dt = k . T

_{g}- k . T

_{b}so we can solve to get

T

_{b }(t)= (k . T

_{g}).t - T

_{b}(0)e

^{-kt}

which has a characteristic time constant = 1/k which from the above = Cb / U . A

Since Cb and A depend on the coffee bean while U is a function of the roaster design, the value of this time constant varies with roaster and bean and can be between about 30 seconds and a few minutes. Here’s what a time constant of 60 seconds and an input temperature of 240

^{o}C looks like:

Exponential decay

This time constant is important because it controls the rate of system response to an input: the larger the value of the time constant, the slower the system responds.

This result was derived with the assumption of the constituent factors being constant over small time scales. Over longer time scales the individual factors vary. The boundary area (A) increases as the beans expand during the roast, the thermal transmittance (U) decreases for more or less the same reasons. The heat capacity Cb changes as the roast develops as will be teased out later. Taken together these mean that the time constant reduces somewhat over the history of the roast but the changes are small enough that it is still a useful guide.

If you are measuring your roast profile with a temperature probe it will also have a finite lag and thus its own time constant. These two are additive in terms of system control.

The analysis so far assumes that the heating medium is entirely gas, as is the case for a fluidised bed roaster. If there are significant amounts of metal in contact with the gas and the beans, as is the case with a drum roaster, the analysis is more complex and includes another time constant for the response of the metal structures.

The important thing to note here is that the rate of rise is entirely passive, it is an output from the roasting system, not an input to it. It indicates the balance between the heat input into the roasting system and the ability of the bean mass to absorb that input. There is no need to invoke the mythical quantity of thermal momentum.

The existence of the thermal time constant means that when something changes the system will continue on its exising path for some time before the effect of the change become apparent: this can look like the rate of rise has a life of its own but that isn't the case.

Quote Rob Hoos (Modulating the Flavour Profile of Coffee p 16): “If the turn-around happens earlier than planned, you are entering the roast with more thermal momentum than you may have expected and should lower your heat application”.

This can be replaced by:

if the turnaround happens earlier than planned the total energy in the system is higher than expected so energy input can be reduced to compensate.

Quote Paul Songer Coffee Thermodynamics

"The change in the internal energy of the system at this stage is the sum of the heat entering the system from the environment and the heat being created by the sugar browning reactions. Like a snowball rolling downhill, the sugar browning reactions develop a momentum of their own and the increase in internal energy depends less and less on the heat in the environment as the roast progresses and more on the heat already taken on, the buildup of pressure within the bean, and the heat released by sugar browning."

He is wrong about almost every point here. The browning reactions are endothermic so there is no internal heat generation. The heat already taken on has increased the temperature of the bean mass so there’s no need to account for it again. The buildup of pressure within the bean probably actually decreases reaction rate by itself: the pressure is however a symptom of increased temperature which will increase reaction rate but again we’ve already accounted for that.

To be continued.

* U is the inverse of the R value used in insulation, so U * R = 1

** If you really care, visit Khan academy for tutelage

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