Subfield: Optics

Thermoacoustic Engine

A thermoacoustic engine is a device that converts heat into acoustic power, or sound waves - a form of mechanical work. Like many other heat machines, it can be operated in reverse to become a refrigerator, using sound to pump heat from a cold to a hot reservoir. The high operating frequencies reduce heat conduction and eliminate the need for any working chamber confinement. Unlike many other engine types, the thermoacoustic engine has no moving parts except the working fluid itself.

The efficiencies of thermoacoustic machines are typically lower than other engine types, but they have advantages in set up and maintenance costs. This creates opportunities for renewable energy applications, such as solar-thermal power plants and utilization of waste heat. Our analysis will focus on the creation of acoustic energy within the system, ignoring the extraction or conversion for powering external devices.

Part A: Sound wave in a closed tube

Consider a thermally insulating tube of length $L$ and cross-sectional area $S$, whose axis lies along the $x$ direction. The two ends of the tube are located at $x=0$ and $x=L$. The tube is filled with an ideal gas and is sealed on both ends. At equilibrium, the gas has temperature $T_{0}$, pressure $p_{0}$ and mass density $\rho_{0}$. Assume that viscosity can be ignored and that the gas motion is only in the $x$ direction. The gas properties are uniform in the perpendicular $y$ and $z$ directions.

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Figure 1
Context question:
A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by

$$
u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t)
\tag{1}
$$

(please note the $u$ here describes the displacement of a gas element)

where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system?
Context answer:
\boxed{$\lambda_{\max }=2 L$}


Extra Supplementary Reading Materials:

We will assume throughout the question an oscillation mode of $\lambda=\lambda_{\max }$.

Now, consider a narrow parcel of gas, located at rest between $x$ and $x+\Delta x(\Delta x \ll L)$. As a result of the displacement wave of Task A.1, the parcel oscillates along the $x$ axis and undergoes a change in volume and other thermodynamic properties.

Throughout the following tasks assume all these changes to the thermodynamic properties to be small compared to the unperturbed values.
Context question:
A.2 The parcel volume $V(x, t)$ oscillates around the equilibrium value of $V_{0}=S \Delta x$ and has the form

$$
V(x, t)=V_{0}+V_{1}(x) \cos (\omega t) .
\tag{2}
$$

Obtain an expression for $V_{1}(x)$ in terms of $V_{0}, a, k$ and $x$.
Context answer:
\boxed{$V_{1}(x)=a k V_{0} \cos (k x)$}


Question: A.3 Assume that the total pressure of the gas, as a result of the sound wave, takes the approximate form

$$
p(x, t)=p_{0}-p_{1}(x) \cos (\omega t) .
\tag{3}
$$

Considering the forces acting on the parcel of gas, compute the amplitude $p_{1}(x)$ of the pressure oscillation to leading order, in terms of the position $x$, the equilibrium density $\rho_{0}$, the displacement amplitude $a$ and the wave parameters $k$ and $\omega$.

Solution:
We use Newton's Second Law $\rho_{0} \ddot{u}=-p^{\prime}$ to deduce $p^{\prime}=-\rho_{0} \ddot{u}=\rho_{0} a \omega^{2} \sin (k x) \cos (\omega t)$, so that

$$
p(x, t)=p_{0}-a \frac{\omega^{2}}{k} \rho_{0} \cos (k x) \cos (\omega t) \quad \Rightarrow \quad p_{1}(x)=a \frac{\omega^{2}}{k} \rho_{0} \cos (k x) .
$$

Final answer: $p_{1}(x)=a \frac{\omega^{2}}{k} \rho_{0} \cos (k x)$