11 Ideal fluid
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11.1 The Euler equations for an ideal fluid
Euler wrote down the partial differential equations describing an ideal fluid already in 1755. This was one of the first partial differential equations systems to be studied.
An ideal fluid has two simplifying properties:
Constant density: The density is constant throughout the fluid at all times, so the density that in a general fluid could be a function of space and time is a constant: \(\rho(\underline{x},t)=\rho\), i.e., \[ \partial_t\rho=0=\underline{\nabla}\rho. \tag{11.1}\] This implies that also the material derivative is zero: \[ \frac{D\rho}{Dt}=\partial_t\rho+\underline{u}\cdot\underline{\nabla}\rho=0. \tag{11.2}\] We had seen already that this implies that the fluid is incompressible: \(\underline{\nabla}\cdot\underline{u}=0\). But the condition on an ideal fluid is stronger than incompressibility, because it demands that the density is not only constant as experienced by each fluid particle, but that it is also the same across different fluid particles.
No viscosity: The force that the fluid exerts on any infinitesimal surface element is solely in the direction of the normal vector to the surface element. Thus this surface force contribution can be written as \[ d\underline{F}^{(s)}=p\, \underline{n}\, dS \tag{11.3}\] where \(p=p(\underline{x},t)\) is called the pressure, \(\underline{n}\) is the normal vector to the surface, and the factor of \(dS\) expresses that the force is proportional to the area of the surface element.
Thus in an ideal fluid there are no surface forces acting tangentially to the surface, also called shear forces. Such a fluid is also called inviscid. The opposite would be a viscous fluid. In reality every fluid has at least a little bit of viscosity, but in circumstances where these viscous forces are negligible compared to the other forces we can neglect them and treat the fluid as inviscid.
There are phenomena that we will not be able to model correctly when we neglect viscosity. For example, when we will study the flow of air over an aerofoil, a thin layer of fluid near the surface experiences a high level of shear stress due to the no-slip condition at the surface. This results in the formation of a boundary layer, which is a region of fluid where the velocity gradients are large and the viscosity of the fluid plays a critical role in determining the behaviour of the flow. Luckily the details of the flow in this boundary layer does not affect the lift force that we want to calculate.
Another phenomenon we can not describe with inviscid flows is turbulence. Turbulent flows arise for example at the rear of real-world aerofoils when the boundary layer detaches from the aerofoil. There the viscosity of the fluid plays a critical role in dissipating the kinetic energy of the turbulent motion into heat and this produces the drag on the airplane. But again this does not affect the lift force, just the drag force.
The equation of motion for the fluid is simply derived from Newton’s equation \(m\mathbf{a}=\underline{F}\). We consider a small volume \(V\) inside the fluid. We sum up the mass multiplied by the acceleration for each fluid particle inside the volume, which due to our continuity approximation becomes an integral of the density multiplied by the acceleration vector field \(D\underline{u}/Dt\) over the volume \(V\): \[ m\underline{a} = \int\limits_{V} \rho \, \frac{D\underline{u}}{Dt} \, dV . \tag{11.4}\]
As discussed above, for an ideal fluid the only force acting on the surface of the volume \(V\) is the force due to the pressure in the surrounding fluid. The total force exerted on the volume \(V\) is obtained by integrating the contributions from all infinitesimal surface elements: \[ \underline{F}^{(s)} = -\oint\limits_{\partial V} p \, \underline{n} \, dS . %\quad \text{or, in components,} \quad \underline{F} = -\oint\limits_{\partial V} p \, n_i \, dV \ \ \text{for} \ \ i=1,2,3. \tag{11.5}\] The reason for the minus sign is that by convention \(\underline{n}\) denotes the outwards normal, but we are interested in the force acting on the volume from the outside.
We now use a form of the divergence theorem that you may not have met yet:
Theorem 11.1 For any scalar function \(f(\underline{x})\) \[ \oint\limits_{\partial V} f \, \underline{n} \, dS = \int\limits_{V} \underline{\nabla} f \, dV. \tag{11.6}\]
Proof. The \(i\)th component of the above vector equation can be obtained by setting \(\underline{F}=f\underline{e}_i\) in the divergence theorem Theorem 10.1.
Applying this to the expression for the surface force \(\underline{F}^{(s)}\) by setting \(f=p\) allows us to express the force as a volume integral: \[ \underline{F}^{(s)} = -\int\limits_{V} \underline{\nabla} p\, dV. \tag{11.7}\]
We added the superscript \((s)\) to the pressure force to indicate that this is a force acting on a surface – a so-called surface force. An ideal fluid can also be subject to other forces, called body forces, that act at every point in the fluid, like the gravitational force or possibly electromagnetic forces. We will here consider the gravitational force. We obtain the gravitational force acting on our volume \(V\) by summing the gravitational force (mass times gravitational acceleration \(\underline{g}\)) on all particles, which in our continuum approximation becomes an integral: \[ \underline{F}^{(g)}=\int_V\rho\,\underline{g}\,dV. \tag{11.8}\]
So, Newton’s equation \(m\underline{a}=\underline{F}\) applied to our volume V of fluid takes the form \[ \int\limits_{V} \rho \, \frac{D\underline{u}}{Dt} \, dV = -\int\limits_{V} \underline{\nabla} p\, dV + \int_V\rho\,\underline{g}\,dV. \tag{11.9}\] This must hold for any volume \(V\). Therefore, we conclude that \[ \rho \, \frac{D\underline{u}}{Dt} = -\underline{\nabla} p + \rho\,\underline{g}. \tag{11.10}\] This equation, together with the incompressibility condition Eq. 10.30 are called Euler’s equations for an ideal fluid.
\[\begin{split} \frac{D\underline{u}}{Dt} &= -\frac{1}{\rho}\underline{\nabla} p + \underline{g}\\ \underline{\nabla}\cdot\underline{u} &= 0 \end{split} \tag{11.11}\]
11.2 Water in a rotating bucket
Example 11.1 We will now use the steady flow from Example Example 10.1: \[ \underline{u}=(-y, x,0) \tag{11.12}\] to illustrate the Euler equations. We have seen that this is a circular flow. This describes for example the flow in a rotating bucket of water. 1
1 The reason why the water in a rotating bucket rotates along with the bucket is that shear forces act until the steady rotating flow is achieved. By neglecting viscosity we can not model this initial phase of the flow but we can model the steady flow that is achieved eventually.
We now want to show that \(\underline{u}\) satisfies the Euler equations and find the pressure \(p\). We will then see that Euler’s equations can tell us the shape of the surface of the water in the rotating bucket. You probably know from personal observation that you expect the surface to be higher towards the rim of the bucket than in the middle because the centrifugal force presses water outwards. We will easily find the exact shape.
We have written Euler’s equations in vector notation. Let’s expand them in components. Let the \(z\) axis be directed vertically up, then \(\underline{g}=(0,0,-g)\). The equations for the material derivatives of the components of \(\underline{u}\) are. \[\begin{split} &\partial_t u_x + \left(u_x \partial_x + u_y \partial_y + u_z \partial_z\right) u_x = -\frac{1}{\rho} \, \partial_x p ,\\ &\partial_t u_y + \left(u_x \partial_x + u_y \partial_y + u_z \partial_z\right) u_y = -\frac{1}{\rho} \, \partial_y p ,\\ &\partial_t u_z + \left(u_x \partial_x + u_y \partial_y + u_z \partial_z\right) u_z = -\frac{1}{\rho} \, \partial_z p - g. \end{split} \tag{11.13}\] Also, we have the incompressibility condition \[ \partial_x u_x + \partial_y u_y + \partial_z u_z =0. \tag{11.14}\] Thus, we have 4 equations for 4 unknowns: three components of the velocity, \(u_x\), \(u_y\) and \(u_z\), and pressure, \(p\).
The pressure cannot be uniquely determined from Eq. 11.13, because if \(\underline{u}(\underline{x},t)\) and \(p(\underline{x},t)\) represent a solution, then \(\underline{u}(\underline{x},t)\) and \(p(\underline{x},t)+f(t)\) for arbitrary function \(f\) is also a solution. To determine the pressure uniquely, we heed to impose some additional condition. For example, if we consider a flow in the whole space, we may require that the pressure at infinity is a given constant: \(p(\underline{x},t)\to p_0=const\) as \(\vert\underline{x}\vert\to \infty\).
First we check that the flow with \(\underline{u}=(-y,x,0)\) is incompressible, i.e., that Eq. 11.14 is satisfied. We find \[ \partial_x u_x + \partial_y u_y + \partial_z u_z =\partial_x (- y) + \partial_y x + \partial_z 0 =0. \tag{11.15}\] So the flow is incompressible and thus can be the flow of an ideal fluid.
Substituting \(\underline{u}=(-y,x,0)\) into Eq. 11.13, most terms are zero and we get \[ \left\{\begin{array}{l} x \partial_y(- y) = -\frac{1}{\rho} \, \partial_x p , \\ - y\partial_x\, x = -\frac{1}{\rho} \, \partial_y p , \\ 0 = -\frac{1}{\rho} \, \partial_z p - g \nonumber \end{array}\right. \quad \Rightarrow \quad \left\{\begin{array}{l} \partial_x p = \rho\,x, \\ \partial_y p = \rho\,y, \\ \partial_z p = -\rho \, g \end{array}\right. \tag{11.16}\] We see that these equations for the pressure do indeed have a solution: \[ p=\rho \left(\frac{1}{2} \, (x^2 + y^2) - g z\right) +\text{ constant}. \tag{11.17}\] This shows that this velocity field satisfies Euler’s equations and thus describes an ideal fluid with the given pressure.
At the surface of the water in the bucket the pressure is constant – equal to the atmospheric pressure. From that we can deduce that \[ z = \frac{1}{2g}(x^2+y^2) +\text{ constant}. \tag{11.18}\] Thus the surface of the water is a perfect paraboloid.