Basic Equations for Fluid Flow

Statement

The continuity equation expresses conservation of mass: the net rate of mass flowing out of a control volume equals the rate of decrease of mass inside it.

Derivation

Consider a fixed elemental control volume $dx\,dy\,dz$ in a flow with velocity $\vec V = u\hat i + v\hat j + w\hat k$ and density $\rho$.

          y
          ^      _______
          |    /|      /|
          |   /_|_____/ |   dz
          |  |  |____|__|
          |  | /     | /  dy
          |  |/______|/
          +--------------> x
               dx

Mass inflow through the face perpendicular to $x$ (area $dy\,dz$) is $\rho u\,dy\,dz$. Mass outflow through the opposite face is

$$\left[\rho u + \frac{\partial(\rho u)}{\partial x}dx\right]dy\,dz$$

Net outflow in the $x$-direction is $\dfrac{\partial(\rho u)}{\partial x}dx\,dy\,dz$. Similarly for $y$ and $z$. Total net outflow:

$$\left[\frac{\partial(\rho u)}{\partial x} + \frac{\partial(\rho v)}{\partial y} + \frac{\partial(\rho w)}{\partial z}\right]dx\,dy\,dz$$

Rate of decrease of mass in the element: $-\dfrac{\partial\rho}{\partial t}dx\,dy\,dz$.

Equating:

$$\frac{\partial\rho}{\partial t} + \frac{\partial(\rho u)}{\partial x} + \frac{\partial(\rho v)}{\partial y} + \frac{\partial(\rho w)}{\partial z} = 0$$

or in vector form, $\dfrac{\partial\rho}{\partial t} + \nabla\cdot(\rho\vec V) = 0$.

Special cases

• Steady flow ($\partial\rho/\partial t = 0$):

$$\frac{\partial(\rho u)}{\partial x} + \frac{\partial(\rho v)}{\partial y} + \frac{\partial(\rho w)}{\partial z} = 0$$

• Incompressible flow ($\rho$ constant, steady or unsteady):

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0 \quad \text{i.e. } \nabla\cdot\vec V = 0$$

• For 2D incompressible flow: $\partial u/\partial x + \partial v/\partial y = 0$.

(a) Continuity equation in cylindrical coordinates

With velocity components $v_r$ (radial), $v_\theta$ (tangential) and $v_z$ (axial):

$$\frac{1}{r}\frac{\partial (r v_r)}{\partial r} + \frac{1}{r}\frac{\partial v_\theta}{\partial \theta} + \frac{\partial v_z}{\partial z} = 0$$

• First term: net radial outflow per unit volume. The factor $r$ appears because the flow area $r\,d\theta\,dz$ grows with $r$.
• Second term: net tangential outflow. The arc length $r\,d\theta$ appears in the denominator.
• Third term: net axial outflow, same as in Cartesian form.

For compressible unsteady flow the term $\partial\rho/\partial t$ is added and $\rho$ is included inside the derivatives. For 2D axisymmetric flow, $\partial/\partial\theta = 0$.

(b) Radial flow between discs

Given $b = 0.02\ \text{m}$, $Q = 0.04\ \text{m}^3/\text{s}$. The flow is radial and symmetric, so $v_\theta = 0$, $v_z = 0$, and the flow area at radius $r$ is $2\pi r b$.

From the continuity (volume balance), $v_r = \dfrac{Q}{2\pi r b}$, so $r v_r = \dfrac{Q}{2\pi b} = \text{constant}$.

Hence

$$\frac{1}{r}\frac{\partial (r v_r)}{\partial r} = \frac{1}{r}\frac{\partial}{\partial r}\left(\frac{Q}{2\pi b}\right) = 0$$

and the cylindrical continuity equation is satisfied.

Velocities:

$$\begin{aligned} v_r(0.10) &= \frac{0.04}{2\pi \times 0.10 \times 0.02} = 3.18\ \text{m/s} \\ v_r(0.25) &= \frac{0.04}{2\pi \times 0.25 \times 0.02} = 1.27\ \text{m/s} \end{aligned}$$

The velocity falls inversely with radius, as the flow area increases.

Answer: $v_r = 3.18\ \text{m/s}$ at $r = 0.10\ \text{m}$ and $1.27\ \text{m/s}$ at $r = 0.25\ \text{m}$

Derivation

Newton's second law for a system: the resultant external force equals the time rate of change of linear momentum of the system.

$$\sum \vec F = \frac{d(m\vec V)}{dt}\bigg|_{system}$$

Using the Reynolds transport theorem to convert the system to a control volume (CV):

$$\sum \vec F = \frac{\partial}{\partial t}\int_{CV}\rho\vec V\,dV + \int_{CS}\rho\vec V(\vec V\cdot\hat n)\,dA$$

For steady flow the first term is zero. For a CV with one inlet (1) and one outlet (2), with uniform velocity over each area:

$$\sum \vec F = \dot m(\vec V_2 - \vec V_1) = \rho Q(\vec V_2 - \vec V_1)$$

In components:

$$\sum F_x = \rho Q(V_{2x} - V_{1x}), \qquad \sum F_y = \rho Q(V_{2y} - V_{1y})$$

$\sum \vec F$ includes pressure forces on the control surface, body forces (weight) and the force $\vec R$ exerted by the solid boundary on the fluid.

Assumptions: steady flow, uniform velocity at the inlet and outlet sections, a fixed control volume, and an inertial frame.

Applications

1. Pipe bend: take the CV as the fluid inside the bend. Forces are the end pressure forces $p_1A_1$ and $p_2A_2$ and the reaction $R$ from the bend. The change in direction and speed of flow gives the force on the bend, which is used to design anchor blocks and thrust blocks.

2. Jet on a vane: the vane changes the direction of the jet. For a smooth vane deflecting the jet through $\theta$, the force in the jet direction is $F = \rho a V^2(1-\cos\theta)$ for a fixed vane and $F = \rho a (V-u)^2(1-\cos\theta)$ for a single vane moving at speed $u$, found from the change of momentum using the relative velocities. Used in impulse turbines (Pelton wheel).

3. Jet propulsion: a fluid is accelerated backward and the reaction thrust pushes the body forward. Thrust $F = \dot m(V_{jet} - V_{body})$ for ships with water jets and aircraft engines. For a rocket, $F = \dot m V_e + (p_e - p_a)A_e$.

Other applications: force on a nozzle, thrust on a sluice gate, and force in a hydraulic jump.

Given data

$d_1 = 0.3\ \text{m}$, $d_2 = 0.15\ \text{m}$, $Q = 0.15\ \text{m}^3/\text{s}$, $p_1 = 200\ \text{kPa}$ (gauge), $\theta = 45^\circ$, $\rho = 1000\ \text{kg/m}^3$. Take $x$ along the inlet flow and $y$ perpendicular to it, towards the side to which the flow turns.

Areas and velocities

$$\begin{aligned} A_1 &= \frac{\pi}{4}(0.3)^2 = 0.07069\ \text{m}^2, & A_2 &= \frac{\pi}{4}(0.15)^2 = 0.01767\ \text{m}^2 \\ V_1 &= \frac{0.15}{0.07069} = 2.122\ \text{m/s}, & V_2 &= \frac{0.15}{0.01767} = 8.488\ \text{m/s} \end{aligned}$$

Outlet pressure (Bernoulli, horizontal, no loss)

$$p_2 = p_1 + \frac{\rho}{2}(V_1^2 - V_2^2) = 200\,000 + 500(4.503 - 72.05) = 166\,226\ \text{Pa}$$

Pressure forces

$$p_1A_1 = 200\,000 \times 0.07069 = 14\,137\ \text{N}, \qquad p_2A_2 = 166\,226 \times 0.01767 = 2937\ \text{N}$$

Momentum equation (forces on the fluid)

Let $R_x$ and $R_y$ be the forces exerted by the bend on the fluid. $\rho Q = 150\ \text{kg/s}$.

In the $x$-direction:

$$p_1A_1 - p_2A_2\cos45^\circ + R_x = \rho Q (V_2\cos45^\circ - V_1)$$ $$\begin{aligned} 14\,137 - 2077 + R_x &= 150(6.002 - 2.122) = 582 \\ R_x &= -11\,478\ \text{N} \end{aligned}$$

In the $y$-direction:

$$\begin{aligned} -p_2A_2\sin45^\circ + R_y &= \rho Q V_2\sin45^\circ \\ R_y &= 2077 + 150 \times 6.002 = 2977\ \text{N} \end{aligned}$$

Force of water on the bend

By Newton's third law, the force on the bend is equal and opposite:

$$F_x = 11\,478\ \text{N}, \qquad F_y = -2977\ \text{N}$$ $$\begin{aligned} F &= \sqrt{11\,478^2 + 2977^2} = 11\,858\ \text{N} \\ \alpha &= \tan^{-1}\frac{2977}{11\,478} = 14.5^\circ \end{aligned}$$

$F_x$ acts in the direction of the incoming flow, and $F_y$ acts away from the side to which the flow turns.

Answer: $F = 11.86\ \text{kN}$ at $14.5^\circ$ to the inlet flow direction, tilted away from the side to which the water turns ($F_x = 11.48\ \text{kN}$, $F_y = 2.98\ \text{kN}$)

Given data

$d = 0.05\ \text{m}$, $V = 25\ \text{m/s}$, deflection angle $\theta = 165^\circ$, $u = 10\ \text{m/s}$.

Jet area: $a = \dfrac{\pi}{4}(0.05)^2 = 1.9635\times10^{-3}\ \text{m}^2$

For a smooth vane with deflection $\theta$, the change of velocity component in the jet direction (relative to the vane) is $V_r(1 - \cos\theta)$.

$$1 - \cos165^\circ = 1 + 0.9659 = 1.9659$$

(a) Fixed vane

All the jet strikes the vane, $\dot m = \rho a V$ and $V_r = V$:

$$\begin{aligned} F &= \rho a V^2 (1 - \cos\theta) \\ &= 1000 \times 1.9635\times10^{-3} \times 625 \times 1.9659 \\ &= 2412.6\ \text{N} \end{aligned}$$

(b) Moving vane (single vane)

For a single moving vane the mass striking the vane per second is only $\rho a (V - u)$, because the vane is moving away from the nozzle. The relative velocity is $V - u = 15\ \text{m/s}$.

$$\begin{aligned} F &= \rho a (V - u)^2 (1 - \cos\theta) \\ &= 1000 \times 1.9635\times10^{-3} \times 15^2 \times 1.9659 \\ &= 868.5\ \text{N} \end{aligned}$$

Work done per second (power) = force $\times$ vane velocity:

$$P = F u = 868.5 \times 10 = 8685\ \text{W} = 8.685\ \text{kW}$$

Kinetic energy of the jet per second:

$$\text{KE} = \tfrac12 \rho a V^3 = 0.5 \times 1000 \times 1.9635\times10^{-3} \times 15\,625 = 15\,340\ \text{W}$$ $$\eta = \frac{8685}{15\,340} = 0.566 = 56.6\%$$

Answer: (a) $F = 2.41\ \text{kN}$; (b) $F = 868.5\ \text{N}$, $P = 8.69\ \text{kW}$, $\eta = 56.6\%$

(a) Hydraulic jump

A hydraulic jump is the sudden rise in water surface when a rapid (supercritical, $Fr > 1$) flow changes to a slow (subcritical, $Fr < 1$) flow. It is a turbulent phenomenon with large energy dissipation.

Derivation (horizontal channel, unit width, neglect friction on the bed): apply the momentum equation to a control volume enclosing the jump.

Hydrostatic forces: $F_1 = \tfrac12\rho g y_1^2$ and $F_2 = \tfrac12\rho g y_2^2$. Discharge per unit width: $q = V_1y_1 = V_2y_2$.

$$\tfrac12\rho g y_1^2 - \tfrac12\rho g y_2^2 = \rho q (V_2 - V_1)$$

Using $V_1 = q/y_1$, $V_2 = q/y_2$:

$$\frac{g}{2}(y_1^2 - y_2^2) = q^2\left(\frac{1}{y_2} - \frac{1}{y_1}\right) = \frac{q^2 (y_1 - y_2)}{y_1y_2}$$

Dividing by $(y_1 - y_2)$ and rearranging:

$$\frac{q^2}{g} = \frac{y_1y_2(y_1 + y_2)}{2}$$

With $q = V_1y_1$ and $Fr_1 = V_1/\sqrt{gy_1}$, this gives the quadratic in $y_2/y_1$:

$$\left(\frac{y_2}{y_1}\right)^2 + \frac{y_2}{y_1} - 2Fr_1^2 = 0 \quad \Rightarrow \quad \frac{y_2}{y_1} = \frac12\left(\sqrt{1 + 8Fr_1^2} - 1\right)$$

Head loss: $h_L = E_1 - E_2$, where $E = y + \dfrac{V^2}{2g}$. Substituting from above:

$$h_L = \frac{(y_2 - y_1)^3}{4y_1y_2}$$

(b) Numerical

Given $y_1 = 0.3\ \text{m}$, $V_1 = 6\ \text{m/s}$.

$$\begin{aligned} q &= 0.3 \times 6 = 1.8\ \text{m}^2/\text{s} \\ Fr_1 &= \frac{6}{\sqrt{9.81 \times 0.3}} = 3.497 \\ y_2 &= \frac{0.3}{2}\left(\sqrt{1 + 8(3.497)^2} - 1\right) = 1.341\ \text{m} \\ V_2 &= \frac{q}{y_2} = \frac{1.8}{1.341} = 1.342\ \text{m/s} \\ Fr_2 &= \frac{1.342}{\sqrt{9.81 \times 1.341}} = 0.370 \end{aligned}$$

Head loss and power:

$$\begin{aligned} h_L &= \frac{(1.341 - 0.3)^3}{4 \times 0.3 \times 1.341} = 0.702\ \text{m} \\ P &= \rho g q h_L = 1000 \times 9.81 \times 1.8 \times 0.702 = 12\,390\ \text{W/m} = 12.4\ \text{kW per m width} \end{aligned}$$

Since $Fr_2 < 1$, the downstream flow is subcritical, as expected.

Answer: $y_2 = 1.34\ \text{m}$, $V_2 = 1.34\ \text{m/s}$, $Fr_2 = 0.37$, $h_L = 0.70\ \text{m}$, $P = 12.4\ \text{kW per metre width}$

Given data

$Q = 2.5\ \text{m}^3/\text{s}$, $V_j = 15\ \text{m/s}$ (relative to the ship, astern), $V_s = 6\ \text{m/s}$, $\rho = 1000\ \text{kg/m}^3$.

Principle

Relative to the ship, water enters the system with velocity $V_s$ (the ship's speed) and leaves with $V_j$. The thrust equals the rate of change of momentum:

$$F = \rho Q (V_j - V_s)$$

Thrust

$$F = 1000 \times 2.5 \times (15 - 6) = 22\,500\ \text{N} = 22.5\ \text{kN}$$

Useful power

$$P_{out} = F V_s = 22\,500 \times 6 = 135\,000\ \text{W} = 135\ \text{kW}$$

Power input to the jet (rate of kinetic energy gain of the water)

$$\begin{aligned} P_{in} &= \tfrac12 \rho Q (V_j^2 - V_s^2) \\ &= 0.5 \times 1000 \times 2.5 \times (225 - 36) \\ &= 236\,250\ \text{W} = 236.25\ \text{kW} \end{aligned}$$

Propulsive efficiency

$$\eta_p = \frac{P_{out}}{P_{in}} = \frac{135}{236.25} = 0.571 = 57.1\%$$

Check with the formula $\eta_p = \dfrac{2V_s}{V_j + V_s} = \dfrac{12}{21} = 0.571$.

The efficiency rises as the jet velocity approaches the ship speed, which means a larger mass flow at a smaller jet velocity is more efficient.

Answer: $F = 22.5\ \text{kN}$, $P_{out} = 135\ \text{kW}$, $P_{in} = 236.25\ \text{kW}$, $\eta_p = 57.1\%$

Basis

The Navier-Stokes (N-S) equations are Newton's second law written for a fluid element, with the viscous stresses related to velocity gradients by Stokes' law for a Newtonian fluid:

$$\tau_{xy} = \mu\left(\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\right), \qquad \sigma_{xx} = -p + 2\mu\frac{\partial u}{\partial x}$$

Steps

1. Take a fluid element $dx\,dy\,dz$ with normal and shear stresses on its faces.
2. Net surface force in the $x$-direction $=$ (net normal stress gradient + shear stress gradients) $\times\ dx\,dy\,dz$.
3. Add the body force $\rho g_x\,dx\,dy\,dz$.
4. Equate the sum to mass $\times$ acceleration, $\rho\,\dfrac{Du}{Dt}\,dx\,dy\,dz$.
5. Substitute the Newtonian stress relations, and use incompressible continuity ($\nabla\cdot\vec V = 0$) with constant $\mu$ to simplify.

x-component (incompressible, constant viscosity)

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

In vector form: $\rho\dfrac{D\vec V}{Dt} = \rho\vec g - \nabla p + \mu\nabla^2\vec V$.

The y and z components are written in the same way.

Assumptions

Newtonian fluid, incompressible, constant viscosity, continuum, and laminar or turbulent instantaneous flow (the equation is exact).

Reduction

• Euler's equation: for an inviscid fluid, $\mu = 0$:

$$\rho\frac{D\vec V}{Dt} = \rho\vec g - \nabla p$$

• Hydrostatic equation: for a fluid at rest, $\vec V = 0$, and with $z$ upward, $\partial p/\partial z = -\rho g$ and $\partial p/\partial x = \partial p/\partial y = 0$.

The N-S equations have exact solutions only for simple cases, such as laminar flow between parallel plates and in circular pipes (Hagen-Poiseuille flow).