1. Continuum and statics

The question

A tank is motionless, yet the pressure on its wall grows with depth. Before simulating moving water, predict that pressure and the force on a vertical gate. This establishes the sign and units of the stresses we will later compute.

A small region containing many molecules

A continuum model assigns density $\rho$, velocity $\boldsymbol u$ and pressure $p$ to points. A point represents an average over a region containing many molecules, small compared with the length on which the flow varies. The model loses this interpretation if molecular scales become comparable to the geometry.

For the initial route, assume a single Newtonian fluid with constant density and viscosity, no free-surface motion and no temperature dependence. These assumptions are suitable for studying the equations; they must be reconsidered for a boiling liquid, a rarefied gas or a strongly compressible flow.

Quantity	Meaning	SI unit
$\rho$	Mass per volume	$\mathrm{kg/m^3}$
$p$	Isotropic compressive stress	$\mathrm{Pa=kg/(m\,s^2)}$
$\mu$	Dynamic viscosity	$\mathrm{Pa\,s}$
$\nu=\mu/\rho$	Momentum diffusivity	$\mathrm{m^2/s}$

Pressure and viscosity are different physical quantities. Pressure can exist at rest. A Newtonian viscous stress requires a velocity gradient, as we derive in Chapter 2.

Balance a horizontal slice

Choose $z$ upward and gravitational acceleration $\boldsymbol g=-g\boldsymbol e_z$. Take a stationary slice of area $A$ and height $\mathrm dz$. Pressure below pushes upward, pressure above pushes downward, and weight acts downward:

$$p(z)A-p(z+\mathrm dz)A-\rho gA\,\mathrm dz=0.$$

Dividing by $A\,\mathrm dz$ and taking the limit gives

$$\frac{\mathrm dp}{\mathrm dz}=-\rho g.$$

If the free surface is at $z=H$ and has pressure $p_{\rm atm}$, then

$$p(z)=p_{\rm atm}+\rho g(H-z).$$

surface z = H: p = p_atm
     │
     │ depth h = H − z       gravity ↓
     │
point  z:     p = p_atm + ρgh
     │
bottom z = 0: p = p_atm + ρgH

Depth increases downward while the chosen coordinate increases upward.

For an illustrative constant density $1000\,\mathrm{kg/m^3}$ and $g=9.81\,\mathrm{m/s^2}$, the gauge pressure one metre below the surface is $9810\,\mathrm{Pa}$. This value follows from the balance and chosen inputs; it is not a numerical-solver result.

Integrate the load

Consider a vertical rectangular gate of width $b$, extending from the surface to depth $H$, with atmospheric pressure on its other side. At depth $h$, the net force on a strip of height $\mathrm dh$ is $\mathrm dF=\rho ghb\,\mathrm dh$. Therefore

$$F=\int_0^H\rho ghb\,\mathrm dh=\frac{\rho gbH^2}{2}.$$

The depth of the resultant follows by equating moments about the surface:

$$h_{\rm cp}=\frac{\int_0^H h\,\mathrm dF}{F}=\frac{2H}{3}.$$

For $b=H=1\,\mathrm m$, the load is $4905\,\mathrm N$, acting two thirds of a metre below the surface. The resultant lies below the midpoint because the lower strips carry larger loads.

Pressure reference and physical pressure

Gauge pressure subtracts a reference. When both sides of the gate see the same atmosphere, that contribution cancels in the net force. It would not cancel if the back side were evacuated. A constant shift in a mathematical pressure unknown is harmless only when the problem’s boundary data permit the same shift. Absolute pressure also matters for physical questions such as cavitation, outside this incompressible teaching model.

The cylinder workflow later uses a prescribed outlet traction. Do not add an arbitrary pressure shift while holding that traction fixed: you would change the boundary-value problem.

Try it and diagnose a failure

1. Double the liquid depth. Predict the bottom gauge pressure and gate force.
2. Integrate the pressure over a gate extending from depths $h_0$ to $h_1$.
3. Replace the upward coordinate with downward depth. Derive the pressure derivative again before writing a sign.
4. Someone multiplies bottom pressure by the whole gate area. Why is the result wrong, and by what factor for a gate starting at the surface?

Check your reasoning. In exercise 1, pressure doubles and force quadruples. In exercise 4, the result is twice the correct force because the mean gauge pressure is half the bottom value. These are checks of the model, independent of mesh or software.

For the continuum framework and equilibrium reasoning, consult the instructor’s fluid mechanics notes.

Book map · Next: Mass, momentum and stress