Fluid flow measurement

This chapter covers the following:

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

Momentum equations

Definitions

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:

Water flows through a pipe bend with:

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

Then,

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

Dimensional analysis and similitude

Definitions

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 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

Methods of dimensional analysis

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 $L$:

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

Use dimensional homogeneity to find $a,b,c$.

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

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.

Types of similarity

• 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:

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

Distorted vs undistorted models

• Undistorted model: All similarities are preserved.

Definitions

A model in which geometric, kinematic, and dynamic similarities are all preserved.

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

Definitions

A model in which one or more similarity conditions (usually geometric) are intentionally violated.

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

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

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

Example:

Given:

• $A=0.01{\text{m}}^2$
• $H_1=3\text{m},H_2=1\text{m}$
• $C_d=0.6$

Then:

$$Q=0.6\times 0.01\times \sqrt{2\times 9.81\times (3-1)}\approx 0.0265{\text{m}}^3/\text{s}$$

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:

Given:

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

Then:

$$Q=0.62\times 0.005\times \sqrt{2\times 9.81\times 2}\approx 0.0136{\text{m}}^3/\text{s}$$

Pitot tubes

Definitions

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

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:

Given:

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

Then:

$$Q=0.98\times 0.02\times \sqrt{\frac{2\times 9.81\times 0.5}{1-(0.02/0.05{)}^2}}\approx 0.154{\text{m}}^3/\text{s}$$