How would I go about comparing the efficiency of different heater block shapes for the outlined system?

So, in the attached diagram there is a cylindrical heater cartridge (A) that is used to transfer heat to the cylindrical polymer (B) through the heater block(C).

enter image description here

(A): Heater Cartridge

(B): Polymer liquid flow, carrying heat out

(C): Heater block, aluminium

(D): Temperature Sensor

This setup is for the heater block of a 3D printer. The aim being to provide constant heat to the polymer (B) that flows through the system. The aluminium block acts as a thermal storage to reduce temperature fluctuation in the polymer filament as the flow rate changes. The temperature sensor (D) reacts to changes and controls the heater cartridge in order to try and maintain constant temperature.

I wish to redesign this heater block to find the most efficient shape for this function.

So, my question is how would I calculate the efficiency of different shapes for this purpose? Losses to the surrounding air will have to be taken into account as the surface area will change with different shapes. Different shapes having different surface area to volume ratios.

I must admit I don't know where to start with this problem, it has been quite some time since I've had any formal thermodynamics education. Any answers, pointers or links to resources are greatly appreciated. Thank you!

  • $\begingroup$ You seem to be confusing temperature and heat. Your sketch doesn't show a temperature sensor mounting but presumably you will have one and run the block at constant temperature. Efficiency probably won't be important as it is likely to be a low-power heater powered from the mains. A sphere will give you the lowest surface area for a given volume but is unlikely to be of any use in this application. I think your challenge is to make the filament path long enough that you heat through it thoroughly during transit but not too much while it is stationary. $\endgroup$
    – Transistor
    Commented Dec 31, 2019 at 15:34
  • $\begingroup$ @Transistor Thank you, I've now updated the diagram to include a temperature sensor and more accurate representation of the setup. Why would the sphere not be of use in this application? My aim is to be able to reduce the footprint of the heater block in order to allow greater tool access of the nozzle (part of a bigger project). Is the ideal design simply the one with lowest surface area to volume ratio? (Ignoring melt zone length for now). I'm struggling to find any quantifiable way of comparing designs. $\endgroup$
    – FEA42
    Commented Dec 31, 2019 at 16:01
  • 1
    $\begingroup$ A sphere is difficult to make, to mount, to insulate. "Is the ideal design simply the one with lowest surface area to volume ratio?" No. The ideal design will be the one that gives you the best balance between function, cost, reliability, running cost, cleanability, etc. A regular cuboid shape will be the easiest to insulate. I would rate heatloss among the lowest of your concerns. $\endgroup$
    – Transistor
    Commented Dec 31, 2019 at 16:12
  • $\begingroup$ @Transistor From the perspective of heater block efficiency though? This solution will be bespoke and a prototype, most likely manufactured with a CNC mill and/or lathe. The rest of the hot end is bespoke also as the whole assembly is being redesigned. I can certainly appreciate why the factors you list are important in any engineering problem. However, at this stage I am trying to design the optimum solution purely from a thermodynamics aspect. $\endgroup$
    – FEA42
    Commented Dec 31, 2019 at 16:20
  • 1
    $\begingroup$ How are you calculating efficiency? $ \frac {Heat\ into\ polymer}{Heat\ into\ block} $? $\endgroup$
    – Transistor
    Commented Dec 31, 2019 at 16:26

1 Answer 1


Here is a first order analysis.

Consider the input heat flow $\dot{q}_{in}$ is distributed between heating the polymer $\dot{q}_{p}$ and loss to the air $\dot{q}_{air}$. We can write a basic energy balance as below.

$$ \dot{q}_{in} = \dot{q}_{p} + \dot{q}_{air}$$

The air loss is controlled by the convection coefficient and the area of the block as $\dot{q}_{air} = h_{air}A\Delta T$. Use this to write efficiency as the ratio of heat to the polymer divided by heat input.

$$\varepsilon = \dot{q}_{p}/\dot{q}_{in} = 1 - \left( h_{air} A \Delta T / \dot{q}_{in}\right)$$

The system will be more efficient to heat the polymer when the convection coefficient around the block is decreased. It will also be more efficient to heat the polymer when its area is smaller.

The shape of the block appears in a calculation of the Biot number factor.

$$Bi = h_{air} (V/A) / k $$

This has the ratio volume to surface area as well as the thermal conductivity of the block. The smaller the value of Bi, the closer the system is to a uniform temperature profile throughout. Values of $Bi \approx 0.1$ or less is the typical cutoff for lumped analysis.

This all suggests that smaller is better. However, this neglects the heat transfer from the block to the polymer. In this case, smaller is not necessarily better. To achieve the same desired temperature change of the polymer in a shorter channel, a slower mass flow rate of the polymer will be required.

Where you go next depends on the details.

Hope this sets a starting point.

  • $\begingroup$ I suspect I have gone wrong in attempting this... When calculating the heat loss to air I got a value in the region of 8*10^-3 W for a standard size of aluminium heater block at 473.15k in a 293.15k environment. A typical heater cartridge is 40W. So I end up with an answer of 99.97% efficiency... $\endgroup$
    – FEA42
    Commented Jan 15, 2020 at 4:52

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