During the initial stages of the growth of the nanowire of...

During the initial stages of the growth of the nanowire of Problem 3.109, a slight perturbation of the liquid catalyst droplet can cause it to be suspended on the top of the nanowire in an off-center position. The resulting nonuniform deposition of solid at the solid-liquid interface can be manipulated to form engineered shapes such as a nanospring, that is characterized by a spring radius $r$, spring pitch $s$, overall chord length $L_{c}$ (length running along the spring), and end-to-end length $L$, as shown in the sketch. Consider a silicon carbide nanospring of diameter $D=15 \mathrm{~nm}, r=30 \mathrm{~nm}, s=$ $25 \mathrm{~nm}$, and $L_{c}=425 \mathrm{~nm}$. From experiments, it is known that the average spring pitch $\bar{s}$ varies with average temperature $\bar{T}$ by the relation $d \bar{s} / d \bar{T}=0.1 \mathrm{~nm} / \mathrm{K}$. Using this information, a student suggests that a nanoactuator can be constructed by connecting one end of the nanospring to a small heater and raising the temperature of that end of the nano spring above its initial value. Calculate the actuation distance $\Delta L$ for conditions where $h=10^{6} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, T_{\infty}=T_{i}=25^{\circ} \mathrm{C}$, with a base temperature of $T_{b}=50^{\circ} \mathrm{C}$. If the base temperature can be controlled to within $1^{\circ} \mathrm{C}$, calculate the accuracy to which the actuation distance can be controlled. Hint: Assume the spring radius does not change when the spring is heated. The overall spring length may be approximated by the formula,

Key Concepts

Convection Heat Transfer in Nanosystems

Heat transfer by convection is a critical concept in managing the thermal environment of nanoscale devices. The heat transfer coefficient characterizes the efficiency of energy exchange between a nanosystem and its surroundings. In the context of nanoactuation, accurate modeling of this process enables prediction and control of temperature distributions, which in turn affect physical deformations.

Nanostructure Growth Mechanisms

Nanostructure growth methods, especially those involving a catalyst such as in the vapor–liquid–solid (VLS) process, rely on precise control of the deposition process. Slight deviations in the position or behavior of the catalyst droplet during growth can lead to asymmetries and enable the formation of engineered nanostructures with unique shapes. Understanding these mechanisms is fundamental to tailoring the properties and functionalities of nanomaterials.

Thermal Actuation at the Nanoscale

Thermal actuation uses localized heating to induce mechanical changes in nanoscale structures. By applying heat, small changes in geometry—such as variations in spring pitch—can produce measurable movement, which is key for designing functional nanoactuators. This concept is essential for integrating thermal control with mechanical responses in advanced nanodevices.

Precision Temperature Control in Nanoscale Devices

Precision temperature control is vital in nanoscale engineering because even small variations in temperature can cause significant changes in the behavior and performance of nanomaterials. Tight thermal management ensures that nanoactuators operate reliably, with the accuracy of actuation being directly related to how precisely the temperature can be controlled. This level of control is essential for applications that demand high repeatability and fine-tuning at the nanometer scale.

Transcript

00:01 So we have two strips that are connected together.

00:10 Let's see.

00:18 And each one has a different alpha value.

00:21 So give green alpha 2, red will be alpha 1.

00:26 And the length will call l.

00:35 Now when the strip is heated, it's going to bend.

00:54 And what we'll say is that alpha 2 is larger than then alpha 1 so that had been to the left and it's going to create this angle here which we call theta and then we have r1 and r2 which are measured to the center of the strip so r1 is going right to the center of this strip r2 is going to the center of this strip and now what we want is theta as a function of the original length alpha 1 alpha 2 the change in temperature and delta r and delta r is defined as r2 minus r1 so on our diagram that would be right here delta r well we can first define how much each of these how much the lengths of each of these changes so the change in the length of the first strip is just going to be alpha one times its original length times delta t for the second strip it's the same same thing l2 equals alpha 2 times l delta t you see the only thing that's different in these two formulas are the alphas so we still have an introduced angle which we need and so to do that i'm going to introduce arc length so the arc length is s and the function for that is r theta and so in this case the arc length of each of these is the length of the arc through the center of the strip when it bends.

03:03 And so i'm going to call each of these arc lengths s1, and this one is going to be s2.

03:16 Now s1, which the arc length is equal to the original length plus, help them using the wrong color.

03:32 S1 is equal to the original length plus the change in the length, which is equal to r1 times data, for the second strip s2 equals the original length plus the change in the length of that strip which is equal to r2 theta now we could solve one of these equations for theta but we don't we don't want r2 or r1 so to eliminate those i'm going to write r2 theta minus r1 theta which is just going to equal l plus delta l2 minus l plus delta l1.

04:30 On the right hand side i can factor out the theta.

04:34 So r2 minus r1 times theta.

04:37 We can see that these ls will cancel out.

04:40 So we just get delta l2 minus delta l1 and we can plug in what those are.

04:46 So delta l2 is alpha 2 times l times delta t.

04:54 Minus alpha 1 times l times delta t and then on this left side here we can call r2 minus r1 this is just delta r so we have delta r times theta equals we can factor out the l and the delta t so we get alpha 2 minus alpha 1 times l delta t and finally we're solving for theta equals alpha 2 minus alpha 1 times l delta t all divided by delta r now we can look at this equation to make sure it makes sense so first of all is it positive well the only thing that can be well delta t we know we said it's increasing in temperature so delta t is positive l is always positive r is always positive and alpha 2 minus alpha 1 is positive because we said alpha 2 is greater than alpha 1.

06:06 So that checks out.

06:08 Part b asks to make sure that theta goes to 0 as delta t goes to zero.

06:13 So as delta t goes to zero, not theta, as delta t goes to zero, it is true that theta also goes to zero.

06:25 So that makes sense.

06:27 And then asks if alpha 2 equals alpha 1, does it go to zero? and the answer is yes, theta does go to zero? which is what we expect.

06:43 Now for part c, i'm going to look at these two.

06:49 So i'm going to redraw our two strips.

07:00 They're going to be the same two strips.

07:01 So we still have alpha 1 and alpha 2...