Hydrodynamics

Navier-Stokes equation

For an incompressible Newtonian fluid with constant viscosity, in vector form:

$$\rho\left(\frac{\partial\vec V}{\partial t} + (\vec V\cdot\nabla)\vec V\right) = \rho\vec g - \nabla p + \mu\nabla^2\vec V$$

x-component in Cartesian form:

$$\rho\left(\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} + w\frac{\partial u}{\partial z}\right) = \rho g_x - \frac{\partial p}{\partial x} + \mu\left(\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} + \frac{\partial^2u}{\partial z^2}\right)$$

(The y and z equations are similar.) Together with continuity $\nabla\cdot\vec V = 0$ it describes the flow.

Bernoulli's equation

For steady, incompressible, frictionless flow along a streamline:

$$\frac{p}{\gamma} + \frac{V^2}{2g} + z = \text{constant}$$

Each term has the unit of length (metre of fluid). Multiplying by $\gamma$ gives pressure, static + dynamic + hydrostatic: $p + \tfrac12\rho V^2 + \gamma z =$ constant.

Navier-Stokes equation (3-D, incompressible, constant $\mu$)

$$\begin{aligned} \rho\left(\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} + w\frac{\partial u}{\partial z}\right) &= \rho g_x - \frac{\partial p}{\partial x} + \mu\nabla^2 u\\ \rho\left(\frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} + w\frac{\partial v}{\partial z}\right) &= \rho g_y - \frac{\partial p}{\partial y} + \mu\nabla^2 v\\ \rho\left(\frac{\partial w}{\partial t} + u\frac{\partial w}{\partial x} + v\frac{\partial w}{\partial y} + w\frac{\partial w}{\partial z}\right) &= \rho g_z - \frac{\partial p}{\partial z} + \mu\nabla^2 w \end{aligned}$$

with $\nabla^2 = \partial^2/\partial x^2 + \partial^2/\partial y^2 + \partial^2/\partial z^2$ and continuity $\partial u/\partial x + \partial v/\partial y + \partial w/\partial z = 0$.

Simplification

Assumptions: steady ($\partial/\partial t = 0$); no z-variation and $w = 0$; fully developed ($\partial/\partial x = 0$ for velocity). Continuity then gives $\partial v/\partial y = 0$, and with $v = 0$ at a wall, $v = 0$ everywhere. Take $x$ along the flow and gravity along $-y$ ($g_x = 0$, $g_y = -g$).

• Left-hand side: all inertia terms vanish ($u\,\partial u/\partial x = 0$, $v = 0$).
• x-momentum:

$$0 = -\frac{\partial p}{\partial x} + \mu\frac{d^2u}{dy^2}$$

• y-momentum:

$$0 = -\rho g - \frac{\partial p}{\partial y}$$

• z-momentum: $0 = -\partial p/\partial z$.

So the pressure is hydrostatic across the flow and $dp/dx$ is constant, giving the simplified model:

$$\mu\frac{d^2u}{dy^2} = \frac{dp}{dx}$$

Velocity distribution

Yes. Integrating twice:

$$\frac{du}{dy} = \frac{1}{\mu}\frac{dp}{dx}y + C_1,\qquad u = \frac{1}{2\mu}\frac{dp}{dx}y^2 + C_1y + C_2$$

For Couette-type flow between plates at $y = 0$ (fixed) and $y = h$ (moving at $U$), with $u(0) = 0$, $u(h) = U$:

$$u = \frac{U y}{h} + \frac{1}{2\mu}\frac{dp}{dx}\left(y^2 - hy\right)$$

If $dp/dx = 0$, this reduces to the linear profile $u = Uy/h$ (simple Couette flow). If both plates are fixed ($U=0$), it gives the parabolic Poiseuille profile $u = \dfrac{1}{2\mu}\dfrac{dp}{dx}(y^2 - hy)$.

Navier-Stokes equations (2-D, incompressible, constant viscosity)

$$\begin{aligned} \rho\left(\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y}\right) &= \rho g_x - \frac{\partial p}{\partial x} + \mu\left(\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2}\right)\\ \rho\left(\frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y}\right) &= \rho g_y - \frac{\partial p}{\partial y} + \mu\left(\frac{\partial^2v}{\partial x^2} + \frac{\partial^2v}{\partial y^2}\right) \end{aligned}$$

Euler equations (2-D, inviscid)

Set $\mu = 0$ in the above:

$$\begin{aligned} \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} &= g_x - \frac{1}{\rho}\frac{\partial p}{\partial x}\\ \frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} &= g_y - \frac{1}{\rho}\frac{\partial p}{\partial y} \end{aligned}$$

Definition of terms

Together with the continuity equation $\partial u/\partial x + \partial v/\partial y = 0$ these form a closed system.

Applications

• Navier-Stokes: laminar flow in pipes and between plates (Hagen-Poiseuille and Couette flow), boundary layer analysis, flow around bodies, lubrication in bearings, weather and ocean modelling, CFD for turbulence (RANS) and aerodynamics.
• Euler: flow outside boundary layers where viscosity is negligible, derivation of Bernoulli's equation, potential flow around airfoils and bodies, flow through nozzles and open-channel waves, high-speed gas dynamics, and for a quick estimate of pressure distribution in non-viscous flow.

Given: $Q = 0.12$ m³/s, $d = 0.15$ m, $h_f = 27V^2/2g$, efficiency 75 %, lower reservoir surface at 120 m, upper reservoir surface at 150 m. The pump discharges to the upper reservoir, and both surfaces are open to atmosphere with negligible velocity.

Velocity and losses

$$\begin{aligned} A &= \frac{\pi}{4}(0.15)^2 = 0.017671\ \text{m}^2\\ V &= \frac{Q}{A} = \frac{0.12}{0.017671} = 6.791\ \text{m/s}\\ \frac{V^2}{2g} &= \frac{6.791^2}{19.62} = 2.350\ \text{m}\\ h_f &= 27\times 2.350 = 63.46\ \text{m} \end{aligned}$$

Pump head (energy equation between the two surfaces)

$$\begin{aligned} z_1 + H_p &= z_2 + h_f\\ H_p &= (150 - 120) + 63.46 = 93.46\ \text{m} \end{aligned}$$

Power

$$\begin{aligned} P_{water} &= \gamma Q H_p = 9.81\times 0.12\times 93.46 = 110.02\ \text{kW}\\ P_{input} &= \frac{110.02}{0.75} = 146.69\ \text{kW}\\ &= \frac{146.69\times 1000}{746} = 196.6\ \text{hp} \end{aligned}$$

Answer: pump head 93.46 m; power to drive the pump $\approx$ 146.7 kW, i.e. about 197 hp (water power 110 kW or 147.5 hp).

TEL and HGL

 El.                                     ________ 150 m
 (m)                           _________/  upper reservoir
 213.46 ......... TEL ........./\ jump at pump (Hp = 93.46)
                              /  \     friction line (TEL falls
                    ...HGL  /     \    63.46 m over pipe length)
 150  ................/ ....       \
                 jump |
 120 __________| pump  P              
  lower reservoir

• Both TEL and HGL start at 120 m in the lower reservoir (free surface; no velocity).
• In the suction pipe, the TEL falls slowly with friction; at the pump the TEL rises suddenly by $H_p = 93.46$ m.
• After the pump, the TEL slopes down uniformly with friction and reaches 150 m at the upper reservoir (total fall 63.46 m).
• The HGL lies below the TEL by $V^2/2g = 2.35$ m throughout the pipe (uniform diameter) and meets the water surface at 150 m at the upper reservoir, after an entry/exit effect.
• The TEL in the delivery pipe starts at 150 + 63.46 = 213.46 m just after the pump (if all the friction is in the delivery pipe).

Euler's equation to Bernoulli's equation

Euler's equation of motion along a streamline $s$ for a steady flow of an ideal fluid:

$$\frac{dp}{\rho} + V\,dV + g\,dz = 0$$

Integrating along the streamline for an incompressible fluid ($\rho$ constant):

$$\frac{p}{\rho} + \frac{V^2}{2} + gz = \text{constant}$$

Dividing by $g$:

$$\frac{p}{\gamma} + \frac{V^2}{2g} + z = H = \text{constant}$$

This is Bernoulli's equation: the sum of pressure head, velocity head and elevation head is constant along a streamline (steady, incompressible, frictionless flow).

With head loss, turbine and pump

Between points 1 and 2, with energy supplied by a pump $H_p$ (added to the fluid), energy extracted by a turbine $H_T$ and head loss $h_L$:

$$\frac{p_1}{\gamma} + \frac{V_1^2}{2g} + z_1 + H_p - H_T - h_L = \frac{p_2}{\gamma} + \frac{V_2^2}{2g} + z_2$$

• $h_L$ = head lost to friction and minor losses between 1 and 2 (always positive).
• $H_p$ = head added to the flow by the pump (positive).
• $H_T$ = head taken from the flow by the turbine (positive).

Power: pump $P = \gamma Q H_p/\eta_p$ (input), and turbine $P = \eta_T\gamma Q H_T$ (output).

Bernoulli's equation from Euler's equation

Consider a small fluid element along a streamline of length $ds$ and area $dA$, in steady flow of an ideal fluid. Newton's second law along $s$ (pressure, weight component, no friction):

$$-\frac{\partial p}{\partial s}\,ds\,dA - \rho g\,ds\,dA\cos\theta = \rho\,ds\,dA\,V\frac{\partial V}{\partial s}$$

With $\cos\theta = dz/ds$, this gives Euler's equation:

$$\frac{dp}{\rho} + V\,dV + g\,dz = 0$$

For incompressible flow, integrate along the streamline:

$$\frac{p}{\rho} + \frac{V^2}{2} + gz = \text{constant}\quad\Rightarrow\quad\frac{p}{\gamma} + \frac{V^2}{2g} + z = \text{constant}$$

Assumptions: steady flow, incompressible, frictionless (inviscid), along a streamline, no energy added or removed.

Four applications

1. Venturimeter: a pipe contraction raises velocity and lowers pressure. From the pressure difference, $Q = C_d\dfrac{A_1A_2}{\sqrt{A_1^2 - A_2^2}}\sqrt{2g\,h}$. Used to measure discharge in pipelines.
2. Pitot tube: the stagnation point converts velocity head into pressure; $V = \sqrt{2g\,h}$ gives the velocity in pipes, channels and aircraft airspeed indicators.
3. Orifice / Torricelli's theorem: water issuing from a tank under head $H$ has velocity $V = \sqrt{2gH}$; used for orifices, sluice gates and discharge estimates.
4. Siphon: Bernoulli's equation gives the discharge velocity $\sqrt{2gh}$ and the pressure at the crest. The crest must stay above vapour pressure (maximum height about 7 to 8 m of water at sea level) to avoid cavitation and breaking of flow.

Other uses include the pump and turbine head calculation, hydraulic jump estimates, aerofoil lift and spray nozzles.