Fluid flow measurement

This chapter covers the following:

• Momentum equations
• Dimensional analysis and similitude
• Fluid flow measurement

Momentum equations

Definitions

Impulse-momentum principle

The impulse-momentum principle states that the net external force acting on a control volume equals the rate of change of momentum of the fluid.

The impulse-momentum principle connects the net external force on a control volume to how quickly the fluid’s momentum changes as it flows in and out.

Equation:

$$\sum \vec{F}=\frac{d\vec{P}}{dt}=\dot{m}({\vec{V}}_{\text{out}}-{\vec{V}}_{\text{in}})$$

Where:

• $\sum \vec{F}$: Net external force acting on the control volume
• $\dot{m}=\rho Q$: Mass flow rate
• ${\vec{V}}_{\text{in}},{\vec{V}}_{\text{out}}$: Average inlet and outlet velocity vectors

Example: Force on a straight reducing section

Water flows through a straight reducing section, so the inlet and outlet velocities lie along the same axis and the vector equation reduces to scalar subtraction along that axis:

• $V_1=3\text{m/s}$
• $V_2=5\text{m/s}$
• $\dot{m}=10\text{kg/s}$

$$F=\dot{m}(V_2-V_1)=10(5-3)=20\text{N}$$

Answer: $F=20\text{N}$

For a bend, the inlet and outlet velocities point in different directions, so apply the equation separately in each direction ($\sum F_x=\dot{m}(V_{2x}-V_{1x})$ and $\sum F_y=\dot{m}(V_{2y}-V_{1y})$) and combine the components into the resultant; subtracting speeds directly is valid only when the flow does not change direction.

Dimensional analysis and similitude

Definitions

Dimensional analysis

Dimensional analysis is a method used to simplify physical problems by expressing variables in terms of fundamental dimensions such as mass, length, and time.

Definitions

Similitude

Similitude is the concept of ensuring similarity between a model and its prototype so that experimental results can be reliably scaled.

Many fluid mechanics problems involve several physical variables at once. Dimensional analysis helps you organize those variables using fundamental dimensions (mass, length, time), which often reduces the number of independent quantities you need to work with. That same idea supports similitude, where you test a scaled model and use dimensionless relationships to predict how the full-size prototype will behave.

Dimensional analysis

Two methods are commonly used to perform dimensional analysis: Rayleigh’s method and the Buckingham π theorem.

Rayleigh’s method

Rayleigh’s method starts by assuming a power-law relationship and then uses dimensional consistency to solve for the unknown exponents.

Example:

If the drag force $F$ depends on velocity $V$, fluid density $\rho$, and length $L$:

$$F=k\cdot {\rho }^a\cdot V^b\cdot L^c$$

Matching dimensions on both sides ($F$ is $ML{T}^{-2}$, $\rho$ is $M{L}^{-3}$, $V$ is $L{T}^{-1}$, and $L$ is $L$) gives three equations:

$$\begin{array}{rl}M& :1=a\\ T& :-2=-b\\ L& :1=-3a+b+c\end{array}$$

Solving gives $a=1$ and $b=2$; substituting into the $L$ equation gives $c=1+3a-b=1+3(1)-2=2$. So $F=k\cdot \rho \cdot V^2\cdot L^2$.

Buckingham $\pi$ theorem

If a problem has $n$ variables and $r$ fundamental dimensions, it can be reduced to $(n-r)$ dimensionless groups ($\pi$ terms).

Steps:

1. List all variables
2. Write dimensions for each
3. Determine number of $\pi$ terms: ${\pi }_1,{\pi }_2,\dots ,{\pi }_{n-r}$
4. Choose repeating variables (must cover all dimensions)
5. Form dimensionless groups
6. Express the relationship using π terms

Example: Applying the Buckingham π theorem

Drag force $F$ depends on velocity $V$, density $\rho$, viscosity $\mu$, and length $L$ - 5 variables and 3 fundamental dimensions ($M$, $L$, $T$), so there are $5-3=2$ dimensionless $\pi$ terms. Choosing $\rho$, $V$, and $L$ as repeating variables gives:

$${\pi }_1=\frac{\rho VL}{\mu }=Re,{\pi }_2=\frac{F}{\rho V^2{L}^2}$$

Answer: The relationship reduces to $\frac{F}{\rho V^2{L}^2}=\varphi (Re)$.

Common dimensionless numbers

• Reynolds number (Re):

$$Re=\frac{\rho VL}{\mu }$$

• Froude number (Fr):

$$Fr=\frac{V}{\sqrt{gL}}$$

• Mach number (Ma):

$$Ma=\frac{V}{c}$$

• Weber number (We):

$$We=\frac{\rho V^{2}L}{\sigma }$$

Similitude

Similitude means setting up a model so it behaves like the prototype in the ways that matter for the problem you’re studying.

Similitude requires three types of similarity between model and prototype:

• Geometric similarity: Same shape, consistent scale ratios.
• Kinematic similarity: Similar velocity and acceleration fields.
• Dynamic similarity: Corresponding forces are in the same ratio.

Model laws

To maintain similarity, key dimensionless numbers must be equal for both model and prototype:

• Reynolds model law:

$$R{e}_{model}=R{e}_{prototype}$$

• Froude model law:

$$F{r}_{model}=F{r}_{prototype}$$

• Mach model law:

$$M{a}_{model}=M{a}_{prototype}$$

• Weber model law:

$$W{e}_{model}=W{e}_{prototype}$$

Which law governs depends on the dominant force - use Reynolds similarity when viscous forces dominate (submerged bodies, pipe flow) and Froude similarity when gravity/free-surface forces dominate (spillways, open channels, ships).

Example: Froude scaling

A model spillway is built at a length scale of $L_m/L_p=1/25$. If the prototype flow velocity is $V_p=4\text{m/s}$, find the model velocity required for dynamic similarity.

Setting $F{r}_{model}=F{r}_{prototype}$ gives $\frac{V_m}{\sqrt{g{L}_m}}=\frac{V_p}{\sqrt{g{L}_p}}$, which simplifies to $V_m=V_p\sqrt{L_m/L_p}$.

$V_m=4\times \sqrt{1/25}=4\times 0.2=0.8\text{m/s}$

Answer: $V_m=0.8\text{m/s}$

Distorted vs undistorted models

• Undistorted model: All similarities are preserved.
• Distorted model: One or more similarities (usually geometric) are intentionally violated to emphasize certain effects.

If complete similarity isn’t achieved, scale effects occur, meaning the model and prototype won’t match perfectly.

Fluid flow measurement

Flow measurement with orifices

Definitions

Orifice meter

An orifice meter measures flow rate by relating the pressure drop across a thin plate orifice to the discharge.

An orifice meter estimates flow rate by measuring the pressure drop created as fluid passes through a thin-plate orifice.

Equation:

$$Q=C_d{A}_o\sqrt{\frac{2\Delta P}{\rho }}$$

Where:

• $C_d$: Discharge coefficient
• $A_o$: Area of the orifice
• $\Delta P$: Pressure difference across the orifice
• $\rho$: Fluid density

Watch out: These flow-measurement equations are given in the FE Reference Handbook’s fluid mechanics section, so focus on knowing which variable goes where rather than memorizing the formulas. Keep every quantity in one unit system (this chapter uses SI throughout) before you plug into any of them.

Submerged orifice

Both sides of the orifice are submerged.

Equation:

$$Q=C_{d}A\sqrt{2g(H_1-H_2)}$$

Where:

• $H_1,H_2$: Upstream and downstream water levels
• $A$: Area of orifice

Orifice discharging freely in the atmosphere

Downstream is open to atmospheric pressure.

Equation:

$$Q=C_{d}A\sqrt{2gH}$$

Where:

• $H$: Height of water above orifice center

Example: Free-discharge orifice

Given:

• $A=0.005{\text{m}}^2$
• $H=2\text{m}$
• $C_d=0.62$

$$Q=0.62\times 0.005\times \sqrt{2\times 9.81\times 2}\approx 0.0194{\text{m}}^3/\text{s}$$

Answer: $Q\approx 0.0194{\text{m}}^3/\text{s}$

Pitot tubes

Definitions

Pitot tube

A Pitot tube measures local flow velocity using the difference between stagnation and static pressure.

A Pitot tube measures the local velocity at a point by comparing stagnation pressure to static pressure.

Velocity equation:

$$V=\sqrt{\frac{2\Delta P}{\rho }}$$

Or using head difference:

$$V=C\sqrt{2gh}$$

Where:

• $h$: Height difference in manometer
• $C$: Coefficient

Venturi meters

Definitions

Venturi meter

A Venturi meter measures flow rate using pressure differences across a converging-diverging section.

A Venturi meter measures flow rate using the pressure change between the inlet and the throat of a converging-diverging section.

Equation:

$$Q=C_d{A}_2\sqrt{\frac{2gh}{1-{\left(\frac{A_2}{A_1}\right)}^2}}$$

Where:

• $A_1,A_2$: Inlet and throat areas
• $h$: Differential head
• $C_d$: Discharge coefficient

Example: Venturi meter flow rate

Given:

• $A_1=0.05{\text{m}}^2$
• $A_2=0.02{\text{m}}^2$
• $h=0.5\text{m}$
• $C_d=0.98$

$$Q=0.98\times 0.02\times \sqrt{\frac{2\times 9.81\times 0.5}{1-(0.02/0.05{)}^2}}\approx 0.0670{\text{m}}^3/\text{s}$$

Answer: $Q\approx 0.0670{\text{m}}^3/\text{s}$