I'd like to calculate the power lost through the filament being extruded (or in other words, at how many Watts I'd have to run an ideal heater that loses heat ONLY through filament so that it stays at constant temperature).

  • Power is defined in Watts as $\text W =\frac{\text J}{\text s}$
  • Specific heat capacity of a material is defined as $C =\frac{\text J}{ \text K \times\text{kg}}$

When extrusion happens, the filament of higher temperature leaves the hotend while the same weight of filament of lower temperature enters the hotend.

Let's say the specific heat capacity of the filament is $C$ and the extrusion rate $r$ is given with units kg/s. The temperatures are $T$.

Is it correct then to say that the power consumption of filament extrusion is $W = (c\times T_\text{Nozzle} - c\times T_\text{Environment}) \times r$

meaning that if I would run a heater cartridge at exactly "W" watts and extrude filament with rate "r" and the block would not loose heat through any other means than through the filament extrusion, then the nozzle temperature would stay constant?

  • $\begingroup$ Specific heat of a material is drastically different in solid and liquid phase. In addition, there's the heat of transition: the energy required to change from solid to liquid (with no change in temperature). And remember that the hotend heats the liquid phase to well above the transition temperature. $\endgroup$ – Carl Witthoft Jun 4 '19 at 13:52
  • $\begingroup$ @CarlWitthoft: The materials in question don't have discrete solid and liquid phases involved (no discrete phase change and thus, at least AIUI, no jump in energy without a change in temperatures) but a glass transition temperature at which they begin to soften. $\endgroup$ – R.. GitHub STOP HELPING ICE Apr 16 at 18:44

This is very simply stated, in fact the specific heat is a function of temperature and state of the material (liquid or solid). Also you need to consider which type of specific heat you use, e.g. the one for constant volume $C_V$ or for constant pressure $C_P$. Constant pressure is probably preferred considering the mechanics of the printer (pressing filament into the nozzle-heatbreak assembly).

A very interesting source of information is the PolymerDatabase.com.

This source confirms that:

In the case of polymers, we have to distinguish between the heat capacity of liquid, rubbery and glassy polymers. The heat capacity increases with increasing temperature, therefore, a liquid or rubbery polymer can hold more energy than a solid polymer. All materials show this increase in heat capacity with temperature.


Specific heat capacities as a function of temperature have been published for only a limited number of polymers. In many cases, the heat capacity (at constant pressure) as a function of temperature can be approximated by straight lines.

In such cases you can use the value of the specific heat at a predefined temperature (in thermodynamics that is frequently 298 K) to get approximations for your thermoplastic material. These formulae can then be used to integrate over the temperature rise.

Please remember that a cartridge is of a certain value of Watt; to have a lower power, the cartridge modulates voltage to keep the heating block within a predefined temperature range.



Your formula is quite off, and it starts with the nomenclature:


Watt is the unit of energy transfer which equals power.

The commonly used term "wattage" does not exist in science. It is a very despised shorthand only used in terms of electric power $P=UI$.

Both power $P$ (like work over time) and heat energy transfer $\Delta Q$ (which is one variant of power) use the unit $\text W=\frac {\text J} {\text s}$, which is confusing but a necessary distinction. Always remember that $P_\text{total}=\sum_{i=1}^nP_i$ - the total power in and out of an object is the sum of all partial powers!

Heat Energy transfer

The Heat energy transfer through an object is defined as the change of the heat energy $Q$ stored inside an object. $Q$ is given in $\text J$, so its change $\Delta Q$ is given in $\text J$ too. To get to the power, the energy change needs to be measured at several spots in time, so we make the derivate over time and get the power in $\text W$. We're looking at $\dot Q=\frac {\delta Q} {\delta t}$.

The absolute change of heat energy of an item is defined as $\Delta Q(t)=m(t) c \Delta T$: Increasing the temperature $T$ of an object with the mass $m$ and specific heat capacity $c$ by $\delta T$ (between times $t_0$ and $t$) results in a change of the stored energy by $\Delta Q(t-t_0)$.

So, we know $Q=c m \Delta T$ and $P=\dot Q=\frac \delta {\delta t} c m \Delta T$

Problem in question

We know that the drain (loss) of thermal energy from the system is via three ways:

  • melting plastic (phase transition)
  • extruding heated plastic
  • convective heat loss to the air
  • black body radiation of the heater block

We know that the total balance in equilibrium should be $P_\text{total}=P_\text{heating}+P_\text{melting}+P_\text{extrusion}+P_\text{convection}+P_\text{bb}=0$.

heat deposited into the system

Let's start at the simplest: we simply know the nominal heating power of the cartridge, it is usually written upon the cartridge itself, usually something in the area of 20 to 40 W. In praxis, it is not exactly that, but the ballpark fits. Otherwise, we'd plug in $P_\text{heating}=\epsilon \frac {U^2} {R}$ for our specific resistor, where $\epsilon$ is a coefficient between 0 and 1 telling us how good it is in converting electric to heat energy. Remember that since $U$ is technically a function of time (it is modulated to control heating behavior), our heating power also is, even though not explicit!

black body radiation loss $P_\text{bb}$

Black body radiation: $P_\text{bb}=A \sigma T^4$ where $A$ is the surface area of the object, $\sigma$ is a constant called Stefan-Boltzmann Constant. That much thermal energy is just lost due to radiation via photons, even if we don't see it glowing.

convection loss $P_\text{convection}$

The change of heat energy via heat convection is roughly defined as $H=\theta A (T-T_f)$ which brings us another coefficient $\theta$ about how good the block heats the air and the temperature of the medium (air) around $T_f$ - which we can replace as $(T-T_f)=\Delta T_a$.

And then we get to the biggest can of worms: the thermal heat transfer for melting the plastic and how much thermal energy is extruded from the system. For one of them, we can estimate some ballpark numbers, for the other, we will get into problems.

extrusion loss $P_\text{extrusion}$

The heat energy removed from the system by extruding plastic we can estimate from what we already established about thermal energy back in the Heat Energy transfer paragraph: $Q=mc(T_0+\Delta T)$ using the specific heat capacity $c [\frac {\text{J}}{\text kg K}] $ of the molten plastic as it is extruded (more about that later). But that's not the loss per time, but the heat energy stored in it in Joules. What factor is changing? In this case, it is the mass $m=r\times t$ where $r=\frac {\text kg} {\text s}$ is the extrusion rate. So $Q_\text{extrusion}=rtc\ \Delta T_\text{extrusion}$ and subsequently $P_\text{extrusion}=rc\ \Delta T_\text{extrusion}$

This leaves us with the big problem: as 0scar correctly pointed out by directing to the PolymerDatabase the specific heat capacity is not a constant and not linear but changes depending on the aggregate of the substance. We can make some estimate about it though from how we formulated the total power and adding a few absolutes for convenience:



$$\epsilon \frac {U^2}{R}-\theta A \Delta T_a-A \sigma T^4-rc\ \Delta T_\text{extrusion}=P_\text{melting}$$

Remember, that $U$ is a time-dependent factor (because of the control board activating it or disabling it), $T_f$ is also not a steady thing and changes depending on the airflow (though we can just pin it for our thought experiment) and thus $T$ itself might change over time as a result. $T$ is not equal to $\ \Delta T_\text{Extrusion}$ but is the temperature of the heater block system as a whole. $\ \Delta T_\text{Extrusion}$, on the contrary, is the temperature increase of the filament and not necessarily the same $\Delta T_{air}$, the differential between the heater block and the air. Why this differentiation is necessary becomes apparent if one realizes that the path of the filament might benefit from the heat that is lost from the heater block along that path, pre-heating the filament.

Phase Transition $P_\text{melting}$

$\propto$ is the proportionality sign and indicates that I might skip factors or constants.

What is that last part? That $P_\text{melting}$? It is the power of the Phase Transition. Matter does not shift between phases freely. There is energy stored in the state itself! So when transitioning from one phase to the other, that energy either has to be added (when going from solid to liquid or liquid to gas) or removed (when going the other way).

The "heat of fusion" is a material constant. For this look, I'll call it $\phi [\frac {\text{J}} {\text {g}}]$. We can make an estimation for the power that is put into melting the filament: there's an amount of filament that gets an amount of heat and undergoes the phase transition per time increment $$P_\text{melting}\propto \frac \delta {\delta t}\phi m_\text{melting}=\phi*\dot m$$ Now, we have the product of specific "latent heat" and melting mass derived over time... We had earlier the flow rate of material $m=r\times t$ and the "latent heat" is a constant. So, we pull out $\dot m=r$ again. So in the end we get that the power that is needed to melt our filament is proportional to the flow of the material and the material constant. $$P_\text{melting}\propto \phi r$$

Conclusion tl;dr

When eliminating the loss via convection and black body radiation and assuming them 0 or neglectable, we assume our heater is packed in perfect isolation - and call them losses. Assuming $\epsilon=1$ for a perfect heater, we are left with this equilibrium situation:

$\frac {U^2}{R}-P_\text{losses}=P_\text{melting}+rc\ \Delta T_\text{extrusion}$

The sum of the power of the phase transition (melting of the filament) and the energy stored in the extruded filament per time $(\frac{dQ}{dt})$ is equal to the energy deposited into the hotend over time [minus losses over time]

  • $\begingroup$ Explanation: is Textrusion changing over time? no, it's the same final value. Therefore if's a constant, and derivative does not touches it. Is Tinitial changin over time? no, as well. Conclusion: deltaT is a multiplicative factor, like c, and is not affected by the derivative. Is m changing over time? yes, the total mass that has passed through the extruder is increasing over time (or you would not be extruding at all), therefore you can derive over time that one, and get m'. $\endgroup$ – FarO Apr 16 at 11:50
  • $\begingroup$ @FarO Correct, I think I found the last errors now. $\endgroup$ – Trish Apr 16 at 12:28
  • $\begingroup$ Pmelting is a bit hanging there without much explanation (the only one without a section for it), but in general it's ok. $\endgroup$ – FarO Apr 16 at 16:15
  • $\begingroup$ @FarO Not sure if I got it down to the last factors, but I get something akin to $P_\text{melting}\propto \phi r$ where $\phi$ is energy that is needed/released for one gram doing the phase transition and $r$ is the flow rate of our filament. I am not entirely sure if I might have missed some offset or other factor, that's why I only point to the proportionality. $\endgroup$ – Trish Apr 16 at 18:34

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