Throughout its history, humanity has built many different heat engines that operate on completely different cycles. In this problem, we will consider the Fluidyne – a type of Stirling engine where the working substance is a gas and the pistons are liquid.
The Fluidyne consists of three vertical tubes filled with water. They have the same cross-sectional area $A$, and the height of the water level is $L$. A pair of tubes are connected at the top, and inside the system, there is air with a volume of $2AL$ at atmospheric pressure $p_0$. The water is incompressible and its density is $\rho$.
We will analyze a mechanical system whose behavior is described by a system of linear differential equations. Such a system of equations has a remarkable property: if all quantities depend on time according to the law $x(t)= X_0 e^{\lambda t}$, then the system of differential equations reduces to a system of algebraic equations for $X_0$ and $\lambda$.
The simplest example of such a system is the equation of a spring pendulum:
\[ma=-\alpha v -kx \quad \Rightarrow \quad \ddot{x} + \frac{\alpha}{m} \dot{x} +\frac{k}{m} x = 0.\]
If we substitute the proposed dependence into it, we obtain an algebraic equation for $\lambda$:
\[X_0 \lambda^2 e^{\lambda t} + X_0 \frac{\alpha}{m} e^{\lambda t} + X_0 \frac{k}{m} e^{\lambda t} = 0 \quad \Rightarrow \quad \lambda^2 + \frac{\alpha}{m} \lambda + \frac{k}{m} = 0.\]
We will solve this equation using the discriminant:
\[\mathcal{D}=\frac{\alpha^2}{m^2} - \frac{4k}{m}.\]
If $\mathcal{D}<0$, then the equation for $\lambda$ has no real roots, but here a mathematical trick comes to our rescue. Suppose there exists a number $i$ such that $i^2 = -1$. Then we can say that
\[\lambda = \frac{-\frac{\alpha}{m} \pm \sqrt{\mathcal{D}}}{2} = \frac{-\frac{\alpha}{m} \pm \sqrt{-i^2\mathcal{D}}}{2} = \frac{-\frac{\alpha}{m} \pm i\sqrt{-\mathcal{D}}}{2}. \]
and
\[\tilde{x}(t) = X_0 e^{\left(-\frac{\alpha}{2m} \pm i \frac{\sqrt{-\mathcal{D}}}{2}\right)t} = X_0 e^{-\frac{\alpha}{2m}t} \cdot e^{\pm i \frac{\sqrt{-\mathcal{D}}}{2} t}.\]
The tilde over $x$ is used to indicate that the trick with $i$ has been applied. But from $\tilde{x}(t)$ we can move to the real $x(t)$ if we replace the exponential, whose exponent contains $i \cdot \omega t$, with the cosine $\omega t$:
\[x(t) =X_0 e^{-\frac{\alpha}{2m}t} \cdot \cos \left( \pm \frac{\sqrt{D}}{2} t \right)\]
Keep in mind that if the oscillations are shifted in time, i.e., $x(t)=X_0 \cos (\omega t + \phi)$, then the quantity $\phi$ is called the phase, and when written in exponential form we have $\tilde{x}(t)=X_0 e^{i\omega t + i\phi}$. This can be very convenient, since instead of the cosine of a sum, the phase manifests itself as a factor.
Suppose this system is disturbed from equilibrium and the water levels in the tubes shift downward by small amounts $z_1$, $z_2$ and $z_3$.
If the system is disturbed from equilibrium, the pressure at all points in the system may change. We will assume that inside the air cavity, the pressure is uniform and equal to $p = p_0 + \Delta p$, with $|\Delta p| \ll p_0$. The pressure at point $w$ is denoted as $p_w$, and this pressure also differs slightly from its equilibrium value: $p_w = p_0 + \rho gL + \Delta p_w$, with $|\Delta p_w| \ll p_0$. Neglect the influence of the connecting tubes on the mechanics of the system.
The following forces act on the water inside each of the tubes: gravity, the difference in pressure forces, and the viscous friction force.
Express your answer in terms of $\rho$, $g$, $A$, $z_1$ and $L$.
Express your answer in terms of $\rho$, $g$, $L$, $A$, $\Delta p$ and $\Delta p_w$.
There is an expression known as the Poiseuille formula. If water flows through a horizontal tube of length $l$ and radius $r$, and the pressure difference between the ends of the tube is $\Delta P$, then the volumetric flow rate $Q = \dot{V}$ of water through it is expressed as
\[Q = \frac{\Delta P \pi r^4}{8 \eta l},\]where $\eta$ is the viscosity of water.
Express your answer in terms of $\eta$, $L$, and $\dot{z}_1$.
\[ a \ddot{z} + b \dot{z} + c z = \dots\]
Let us eliminate $\Delta p_w$ from the system of equations. To do this, we will subtract the equations from one another.
\[
a(\ddot{z}_1 - \ddot{z}_2) + b (\dot{z}_1 - \dot{z}_2) + c (z_1 - z_2) = \dots
\]
\[
a(\ddot{z}_1 - \ddot{z}_3) + b (\dot{z}_1 - \dot{z}_3) + c (z_1 - z_3) = \dots
\]
Let all quantities vary according to the law $e^{\lambda t}$:
\[z_1 = z_{1,0} e^{\lambda t}, \quad z_2 = z_{2,0} e^{\lambda t}, \quad \Delta p= \Delta p_0 e^{\lambda t}\]
We are looking for an operating mode in which oscillations occur in the system. In this case, the air pressure inside the cavity is
\[ p(t) = p_0 + \Delta p(t) = p_0 + \Delta p_0 \cos \omega t,\]and the gas volume inside the cavity is
\[ V(t) = 2 AL + \Delta V(t) = 2 AL + \Delta V_0 \cos (\omega t + \varphi). \]We introduce
\[ K = \frac{\Delta p_0}{p_0} \frac{V_0}{\Delta V_0} > 0\]
Express your answer in terms of $\Delta p_0$, $\Delta V_0$, $\omega$ and $\varphi$.
The maximum power is achieved at $\varphi = \pi/2$. In this case, if we replace the cosines with exponentials, then
\[ \Delta \tilde{p}(t) = \Delta p_0 e^{i \omega t}, \quad \tilde{\Delta V(t)} = \Delta V_0 e^{i \omega t + i \varphi}, \quad \frac{\Delta \tilde{p}(t)}{\Delta \tilde{V}(t)} = \frac{\Delta p_0}{\Delta V_0} e^{-i \varphi}. \]It can be shown that $e^{-i \pi /2} = -i$, and then
\[\frac{\Delta \tilde{p}(t)}{\Delta \tilde{V}(t)} = -i \frac{\Delta p_0}{\Delta V_0}.\]In other words, assuming the presence of oscillations such that $\varphi = \pi/2$, in the equations for question A9 one can write that
\[\Delta p_0 = -i \frac{p_0}{2 L} (z_{1,0} + z_{2,0}) \cdot K.\]
\[ (z_{1,0} + z_{2,0}) \cdot \left( a \lambda^2 + b \lambda + c \right) = -iK (z_{1,0}+z_{2,0}) \]
In this equation, one can cancel $(z_{1,0}+z_{2,0})$ and obtain a quadratic equation for $\lambda$:
\[a \lambda^2 + b \lambda + c + i K = 0\]For further analysis, it will be convenient to switch to a dimensionless quantity: $x = \lambda \sqrt{L/g}$.
We will use the following values: $p_0 = 100 ~\mathrm{kPa}$, $\rho=1.00~\mathrm{g}/\mathrm{cm}^3$, $L=10.0~\mathrm{cm}$, $A=1~\mathrm{cm}^2$, $\eta=1.00 \cdot 10^{-3}~\mathrm{Pa} \cdot \mathrm{s}$, $g=9.8~\mathrm{m}/\mathrm{s}^2$.
It is known that for $u, v > 0$, the square root of $-u-iv$ is calculated using the formula:
\[ \sqrt{-u-iv} = \sqrt{u^2 + v^2} \left[ \cos \left( \frac{\pi}{2} - \frac{\arctan \frac{v}{u} }{2}\right) + i \sin\left( \frac{\pi}{2} - \frac{\arctan \frac{v}{u} }{2} \right) \right]\]