# Watts of filament extrusion

I'd like to calculate the wattage 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).

• Watts is defined 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 Wattage 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?

• 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. – Carl Witthoft Jun 4 '19 at 13:52

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.

also:

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.

• Good answer, thank you! Does that mean the formula W = (C_filament_at_nozzle_temp * T_nozzle - C_filament_at_env_temp * T_environment) * r ? – user1282931 Jun 4 '19 at 11:53
• @user1282931 No that does not work like that. When I taught gas turbine thermodynamics we used to let the students calculate with a mean specific heat (for air and combustion gasses) to do their calculations (which work because they are conveniently chosen values), but in fact (complicated) thermodynamic software integrate the specific heat over the temperature rise; the latter is exact, while the first is an assumption. – 0scar Jun 4 '19 at 11:59
• @user1282931 Even with the changed formula, there are lots of factors overlooked in the assumption. Only if one assumes equilibrium the problem becomes solvable at all, and even then there are more factors... – Trish Jun 4 '19 at 12:16
• But if the block is defined to lose heat only through the extrusion, then for every volume-unit of filament the system loses a defined ammount of energy and it gains another defined amount of energy via the volume of filament entering the system. The difference between these amounts of energies is what must be supplied through the heater cartridge so that the energy of the system doesn't change. It does not seem like integration would be required for this at all. – user1282931 Jun 4 '19 at 12:34
• @user1282931 your formula is still WAY off. I tossed you a somewhat more suitable estimation at the end of my wall of physics. Integration is not required, but your base assumptions are lacking. – Trish Jun 4 '19 at 12:40

# No.

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

## Watt

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 change over time in $$\text W$$ and one has to derive as $$\dot Q=\frac {\delta Q} {\delta t}$$.

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

Now, mass and specific energy don't change over time, but we can substitute the absolute temperature change for a temperature change over time. $$\Delta T =\dot T t + T_0$$ where $$T_0$$ is an integration constant and the temperature offset at the start of the experiment.

This gives us $$\Delta Q=m c (\dot T t + T_0)=m c \dot T t +Q_0$$ where $$Q_0=mc T_0$$ is the heat energy at the start of the experiment and allows us to pull $$\dot Q=\frac {\delta Q} {\delta t}=m c \dot T$$.

That is for the total machine.

## Problem in question

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

• melting plastic
• 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 $$P_\text{total}=P_\text{heating}+Q_\text{melting}+Q_\text{extrusion loss}+Q_\text{convection}+P_\text{bb}=0$$.

## 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 $$U$$ 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: $$P_\text{bb}=A \sigma T^4$$ 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

The change of heat energy via heat convection is roughly defined as $$\dot Q=\theta A (T-T_f)$$ which brings us another coefficient $$\theta$$ about how good the block heats the air and the Temperature of the medium (air) around $$T_f$$. We can just multiply by the time to get back to the overall power of this process.

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 number, for the other, we will get into problems.

## extruded heat

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 $$c$$ 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. So we need to somehow add the extrusion rate over time $$e$$. What factor is changing? It is the mass $$m=r\times t$$ where $$r=\frac {\text kg} {\text s}$$. So $$Q_\text{extrusion loss}=rtcT_\text{extrusion}$$

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:

$$P_\text{total}=P_\text{heating}+Q_\text{melting}+Q_\text{extrusion loss}+Q_\text{convection}+P_\text{bb}=0$$

$$P_\text{heating}-|Q_\text{convection}|-|P_\text{bb}|-Q_\text{extrusion loss}=|Q_\text{melting}|$$

$$\epsilon \frac {U^2}{R}-\theta A(T-T_f)t-A \sigma T^4-rtcT_\text{extrusion}=|Q_\text{melting}|$$

Remember, that $$U$$ is a time-dependent factor (it's most easy to assume U is a sine curve and even it over time into the $$\epsilon$$ for a ballpark number), $$T_f$$ is also not a steady thing and changes depending on the airflow, $$T$$ itself might change over time as a result. $$T$$ is not equal to $$T_\text{Extrusion}$$ but the temperature of the heater block system as a whole while the plastic filament does usually not have the same temperature on extrusion!

## Conclusion tl;dr

When eliminating the loss via convection and black body radiation and assuming them 0 or neglectable, we are left with this equilibrium situation, assuming that $$\epsilon$$ achieves a perfect smoothing of the heating:

$$\epsilon \frac {U^2}{R}-rtcT_\text{extrusion}=|Q_\text{melting}|$$

The sum of the thermal energy used to melt the filament and the energy stored in the extruded filament is equal to the energy deposited into the hotend (minus losses)