15.1 Hydrostatics and Aerostatics

15.1.1 Hydrostatics: Incompressible Fluids

A particularly simple but very important case of fluid mechanics is the hydrostatic state. In hydrostatics, we assume the density of the fluid to be constant, thus we usually consider only incompressible fluids. Furthermore, changes of temperature are usually ignored. Because the fluid is not moving, we effectively have only one unknown field variable (see section 10.1.1), which is the pressure p. The increase of pressure can be induced either by external forces acting on the resting liquid or by the liquid’s weight itself. In the following, we consider a fluid column parallel to the z-axis (see Fig. 15.1). From our three fundamental laws of conservation, we will require only a single equation: the Navier-Stokes equation parallel to the gravitational vector (the z-axis). Because the fluid is not moving, the continuity equation (in the case of compressible fluids, Eq. 10.16; in the case of incompressible fluids, Eq. 10.17) cannot be used. If changes in energy are ignored, we also will not require the conservation of energy (see Eq. 12.21). Furthermore, if thermodynamic state changes are ignored, we also will not use thermodynamic equations of state (see Eq. 12.23).

There are two potential ways of finding the analytical solution to the pressure equation in this case: using a force balance and using the Navier-Stokes equation along the z-axis.

Force Balance. The first approach uses a force balance on the infinitesimal volume, which yields

$\begin{array}{ll}p-\left(p+\frac{dp}{dz}dz\right)dxdy-\rho gdxdydz\hfill & =0\hfill \\ \frac{dp}{dz}\hfill & =-\rho g\hfill \end{array}$

Derivation Using the Navier-Stokes Equation. We can also derive Eq. 15.1 from the Navier-Stokes equation directly. Here we consider only the balance along the z-axis. Using Eq. 11.39 and ignoring all terms related to momentum transport and fluid movement, we find

$\begin{array}{ll}-\frac{dp}{dz}+k_z\hfill & =0\hfill \\ \frac{dp}{dz}\hfill & =-\rho g\hfill \end{array}$

which is exactly Eq. 15.1. Here we used k_z = −ρ g. Eq. 15.1 is straightforward and easy to solve. However, we need to distinguish between two cases here: incompressible and compressible fluids.

Incompressible Fluids. In the case of incompressible fluids, the density ρ is not a function of z, and therefore Eq. 15.1 can be integrated to

$\begin{array}{l}p\left(z\right)=-\rho gz+c\hfill \\ p\left(z\right)={p|}_{z=0}-\rho gz\hfill \end{array}$

Eq. 15.3 is the hydrostatic pressure equation, which tells us that when z increases, the pressure gradually decreases.

Rate of Pressure Increase in Water. If we assume a density of around ρH₂O ≈ 1 kg m⁻³, we can derive the rate at which the pressure increases by decreasing z in Eq. 15.3 to be

$\begin{array}{l}-\rho g\approx 1\text{kg}{\text{m}}^{-3}9.81\text{m}{\text{s}}^{-2}\approx \frac{9.81}{1000}\text{kg}\text{m}{\text{s}}^{-2}{\text{m}}^{-3}\hfill \\ =9810{\text{Nm}}^{-2}{\text{m}}^{-1}=\frac{9810}{1000}\text{bar}{\text{m}}^{-1}\text{=}\text{0}\text{.981}\text{bar}{\text{m}}^{-1}\hfill \\ \approx 0.1\text{bar}{\text{m}}^{-1}\hfill \end{array}$

This relationship is important, e.g., during diving. It tells us that for every additional 10 m of depth, the pressure will increase by approximately 1 bar.

15.1.2 Aerostatics: Compressible Fluids

In the case of a compressible fluid, we need to express the density ρ as a function of the pressure p. To do so, we use the ideal gas equation (see Eq. 6.8). From this equation, we find

$\rho =\frac{p}{R_{S}T}$

in which case, Eq. 15.1 can be rewritten as

$\begin{array}{l}\frac{dp}{dz}=\frac{p}{R_{S}T}g\hfill \\ \frac{dp}{p}=\frac{g}{R_{S}T}dz\hfill \end{array}$

which can be integrated to

$\begin{array}{ll}{\int }_{po}^p\frac{dp}{p}\hfill & =-{\int }_{zo}^z\frac{g}{R_{S}T}dz\hfill \\ \ln p-\ln p_0\hfill & =-\frac{g}{R_{S}T}\left(z-z_0\right)\hfill \\ \ln \frac{p}{p_0}\hfill & =-\frac{g}{R_{S}T}\left(z-z_0\right)\hfill \\ \frac{p}{p_0}\hfill & ={\text{e}}^{-\frac{g}{R_{S}T}\left(z-z_0\right)}\hfill \\ p\left(z\right)\hfill & ={\text{p}}_0{\text{e}}^{--\frac{g}{R_{S}T}\left(z-z_0\right)}\hfill \end{array}$

Eq. 15.4 is the hydrostatic pressure equation for compressible fluids and is appropriate, e.g., for calculating the pressure drop at increased altitudes. We know from the ideal gas law that the term R_ST is constant. For example, for air we find R_S = 0.287 J kg⁻¹ mol⁻¹ (see Tab. 6.1). This allows us to calculate the pressure decrease in the atmosphere as we increase the height, e.g., when climbing a mountain or traveling in a plane. Fig. 15.2 shows the decrease in pressure as a function of the height z. At a height of 100 m, the atmospheric pressure will decrease to 98.8 % of the value at z = 0. At 500 m, it will decrease to 93.9 % and at 1000 m to 88.2 %. At a height of 3000 m (which is about the height of the Zugspitze), it will reach a value of around 68.7 %. On the Mont Blanc (around 5000 m), it will decrease to 53.5 % and on the Mount Everest (around 9000 m) to 32.6 %, which is almost too low to sustain breathing. Finally, at the typical traveling height of planes (around 10 000 m), the pressure will reach only 28.7 % of the ground pressure value.

15.1.3 Buoyancy

If a control volume of a given solid or fluid is immersed in a second fluid, the control volume will experience buoyancy forces if the densities of the two materials are different. This fact can be directly derived from Eq. 15.1. We assume two control volumes of equal height. For each of these control volumes, we find the forces acting on the infinitesimal cell to be

$\begin{array}{l}dF=\left(p\left(z\right)-p\left(z+dz\right)\right)dxdy\hfill \\ =\left({p|}_{z=0}-\rho gz-\left({p|}_{z=0}-\rho g\left(z+dz\right)\right)\right)dxdy\hfill \\ =\rho gdxdydz=\rho gdV\hfill \end{array}$

Now if we assume different densities of the two materials ρ₁ and ρ₂, we find a buoyancy force difference

$\begin{array}{l}d{F}_{\text{buoyancy}}=d{F}_2-d{F}_1=\left({\rho }_2-{\rho }_1\right)gdV\hfill \\ =\Delta \rho gdV\hfill \end{array}$

Therefore, if an object of lower density ρ₁, e.g., an air balloon is immersed in a liquid of higher density ρ₂, it will experience a positive buoyancy force pulling it upward. This force is proportional to the volume of the balloon and to the density difference.

15.2.1 Derivation

In the following, we will derive the velocity and the shear stress profile for the Couette flow. We assume the profiles to be stationary and fully developed. Furthermore, we ignore the influence of gravity and assume the system to be infinitely stretched along the axis parallel to the view. Therefore we essentially have a one-dimensional system and will consider only the Navier-Stokes equation in one direction, namely along the x-axis (see Eq. 11.37). There are no volume forces acting, and there is no flow along the z-axis or the y-axis. Eq. 11.37 can thus be simplified to

$\begin{array}{ll}\eta \frac{d^2{v}_x}{d{z}^2}\hfill & =0\hfill \\ \frac{d}{dz}\left(\frac{d{v}_x}{dz}\right)\hfill & =0\hfill \end{array}$

This is a homogenous second-order ODE, which can be integrated twice to result in

$v_x\left(z\right)=c_{1}z+c_2$

where the two integration constants c₁ and c₂ can be found by the boundary conditions v_x (z = 0) = 0 and v_x (z = h) = v₀. From the first boundary condition, we find c₂ = 0; from the second, we find that $c_1=\frac{v_0}{h}$.

Therefore the velocity profile for the Couette flow is given by

$v_x\left(z\right)=v_0\frac{z}{h}$

The profile is shown exemplarily in Fig. 15.3a.

15.2.2 Shear Stress Profile

The shear stress profile can be calculated from the velocity profile (see Eq. 15.7) using Eq. 9.5. In this case, we find

$\mathcal{T}\left(z\right)=\eta \frac{d{v}_x}{dz}=\frac{v_0}{h}$

As stated, the shear profile of the Couette flow is very simple: it is a constant. If the surface area of the top plate is known, the shear force can be easily calculated by multiplication from Eq. 15.8. If the force can be measured accurately, the viscosity of the fluid can be easily calculated from Eq. 15.8. This was the concept Couette suggested in 1887 to measure the viscosity of fluids, and it is still one of the easiest and most straightforward methods in use today.

15.2.3 Flow Rate

Having derived the velocity and the shear stress profile, we will now derive the flow rate for the Couette flow, which can be calculated according to Eq. 10.9 as

$\begin{array}{l}Q={\int }_{A}vdA\hfill \\ ={\int }_0^h{v}_0\frac{z}{h}wdz\hfill \\ ={\left[v_0\frac{z^2}{2h}w\right]}_0^h\hfill \\ =v_0\frac{wh}{2}\hfill \end{array}$

where we introduced the width of the system w.

15.4.1 Navier-Stokes Equation for Poiseuille Flows

We will start with the most basic case, i.e., the pressure-driven flow in a microfluidic channel with an arbitrary cross-section that is invariant along the x-axis (see Fig. 15.6). We will employ the most general form of the Navier-Stokes equation (see Eq. 11.40). We make a number of simplifications that will make solving the Navier-Stokes equation a little easier.

• The flow profile is fully developed; therefore any (partial) derivative with respect to time is zero: $\frac{\partial \dots }{\partial t}=0$

• The flow is parallel; therefore there is no velocity component in y- or z-direction: v_y = 0, v_z = 0

• The pressure driving the flow is a function of only x but not a function of y or z: p (x); therefore $\frac{\partial p}{\partial y}=\frac{\partial p}{\partial z}=0$, which allows to write the pressure gradient as total differential $\frac{\partial p}{\partial x}=\frac{dp}{dx}$

The Navier-Stokes equation can be simplified because we can assume flow along only one axis (usually the x-axis). From Eq. 11.37, we find

$\begin{array}{l}\frac{\partial p}{\partial x}=\eta \left(\frac{{\partial }^2{v}_x}{\partial z^2}+\frac{{\partial }^2{v}_x}{\partial y^2}\right)+k_x\hfill \\ \frac{dp}{dx}=\eta \left(\frac{{\partial }^2{v}_x}{\partial z^2}+\frac{{\partial }^2{v}_x}{\partial y^2}\right)+k_x\hfill \end{array}$

The pressure drop drives the flow and is maintained from the outside, which is why we can treat it as a constant. You will occasionally find the notations $\frac{\Delta P}{\Delta x}$ instead of $\frac{dp}{dx}$. In this work, we will stick to writing $\frac{dp}{dx}$. The individual terms refer to the following contributions:

• $\frac{dp}{dx}$ is the pressure driving the flow.

• $\eta \left(\frac{{\partial }^2{v}_x}{\partial z^2}+\frac{{\partial }^2{v}_x}{\partial y^2}\right)$ describes the viscous forces in the liquid, which are the reason for the formation of a flow profile.

• k_x describes volume forces, such as magnetic or electric forces, which are important, e.g., for electroosmosis and similar effects; in many applications, this term can simply be set to 0 if no volume forces are acting.

Eq. 15.14 is a second-order inhomogeneous PDE. We have already studied this case of differential equation, which is referred to as a Poisson equation (see section 8.1.5). The methods we can use to solve this equation depend strongly on the geometry C of the microfluidic channel. Because the Poiseuille flow is one of the most important flow cases in microfluidics, we will spend some time studying solutions to this PDE for the most relevant flow channel geometries, in the following section.