9.1 Fluid statics

Achievable FE Civil

9. Fluid mechanics

Fluid statics

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This chapter covers the following:

Definitions
Manometers and barometers
Forces on submerged surfaces and the center of pressure
Buoyancy and Archimedes principle

Fluid properties: definitions

In fluid mechanics, you’ll use a few core properties again and again. Three of the most common are density, specific weight, and specific gravity. We’ll define each one and show how to use the key equations.

Density ( $ρ$ )

Definitions

Density: Density is the mass of a fluid per unit volume. It indicates how much matter is packed into a given volume.

Equation:

$ρ = \frac{m}{V}$

Where:

$ρ$ = density ( $k g / m^{3}$ or $s l ug s / f t^{3}$ )
$m$ = mass ( $k g$ or $s l ug s$ )
$V$ = volume ( $m^{3}$ or $f t^{3}$ )

Example:

The density of water at $4° C$ is:

$ρ_{w a t er} = 1000 k g / m^{3}$

Specific weight ( $γ$ )

Definitions

Specific weight: Specific weight is the weight of a fluid per unit volume. It is a force quantity and depends on the local gravitational acceleration.

Equation:

$γ = ρ g$

Where:

$γ$ = specific weight ( $N / m^{3}$ or $l b / f t^{3}$ )
$ρ$ = density ( $k g / m^{3}$ or $s l ug s / f t^{3}$ )
$g$ = acceleration due to gravity ( $9.81 m / s^{2}$ or $32.2 f t / s^{2}$ )

Example:

The specific weight of water at $4° C$ is:

$γ_{w a t er} = 1000 k g / m^{3} \times 9.81 m / s^{2} = 9810 N / m^{3}$

Note the unit reduction: $k g / m^{3} \times m / s^{2} = k g / (m^{2} \cdot s^{2}) = N / m^{3}$ , since $1 N = 1 k g \cdot m / s^{2}$ .

Specific gravity (SG)

Definitions

Specific gravity: Specific gravity is the ratio of the density (or specific weight) of a fluid to the density (or specific weight) of a reference substance (usually water for liquids).

Equation (based on density):

$SG = \frac{ρ _{f l u i d}}{ρ _{w a t er}}$

Equation (based on specific weight):

$SG = \frac{γ _{f l u i d}}{γ _{w a t er}}$

Note: Since $g$ is constant, both definitions give the same value.

Example:

If a fluid has a density of $850 kg/m^{3}$ , then its specific gravity is:

$SG = \frac{850}{1000} = 0.85$

Pressure

Definitions

Pressure: Pressure is defined as the normal force exerted per unit area on a surface. It is a scalar quantity and acts equally in all directions at a point in a fluid at rest.

Mathematical expression:

$P = \frac{F}{A}$

Where:

$P$ = Pressure ( $P a$ or $N / m^{2}$ )
$F$ = Normal force acting perpendicular to the surface ( $N$ )
$A$ = Area over which the force is applied ( $m^{2}$ )

Example:

Suppose a force of $100 N$ is applied on an area of $0.5 m^{2}$ . The pressure is:

$P = \frac{100}{0.5} = 200 Pa$

Viscosity

Definitions

Viscosity: Viscosity is the property of a fluid that resists the relative motion between adjacent layers. It is a measure of the internal friction within the fluid.

There are two types:

Dynamic (or absolute) viscosity ( $μ$ ) - measured in $P a \cdot s$ or $N \cdot s / m^{2}$ .
Kinematic viscosity ( $ν$ ) - the ratio of dynamic viscosity to density, measured in $m^{2} / s$ .

$ν = \frac{μ}{ρ}$

Where:

$ν$ = Kinematic viscosity
$μ$ = Dynamic viscosity
$ρ$ = Fluid density

Newton’s law of viscosity states that shear stress between adjacent fluid layers is proportional to the velocity gradient perpendicular to flow, $τ = μ \frac{d u}{d y}$ ; fluids that follow this relationship (water, air, most common oils) are called Newtonian fluids.

Surface tension

Definitions

Surface tension: Surface tension is the property of a liquid that allows it to resist an external force, due to the cohesive nature of its molecules. Molecules at the surface experience a net inward force because they are not surrounded by similar molecules on all sides.

Mathematical expression:

Surface tension $σ$ is defined as the force per unit length acting along the surface of a liquid at rest:

$σ = \frac{F}{L}$

Where:

$F$ = Force acting along the surface ( $N$ )
$L$ = Length over which the force acts ( $m$ )
$σ$ = Surface tension ( $N / m$ )

For example, surface tension lets water form spherical droplets and allows a water strider to walk on the surface.

Capillarity (capillary action)

Definitions

Capillarity: Capillarity is the rise or fall of a liquid in a small diameter tube (capillary tube) caused by the interaction of adhesive and cohesive forces.

Adhesion: Attraction between liquid and tube wall
Cohesion: Attraction between liquid molecules

Mathematical expression:

The height of capillary rise (or fall) $h$ is given by:

$h = \frac{2 σ cos θ}{ρ g r}$

Where:

$h$ = Capillary rise or fall ( $m$ )
$σ$ = Surface tension of the liquid ( $N / m$ )
$θ$ = Contact angle between liquid and solid surface
$ρ$ = Density of the liquid ( $k g / m^{3}$ )
$g$ = Acceleration due to gravity ( $9.81 m / s^{2}$ )
$r$ = Radius of the capillary tube ( $m$ )

For example, water rises in a narrow glass tube because adhesion exceeds cohesion, while mercury is depressed because cohesion dominates.

The pressure field in a static fluid

In a static fluid, pressure increases with depth. The pressure difference between two points is given by $P_{2} - P_{1} = - γh$ , where point $2$ is located a vertical distance $h$ above point $1$ , and $γ$ is the specific weight of the fluid.

Absolute pressure = atmospheric pressure + gauge pressure reading

Absolute pressure = atmospheric pressure - vacuum gauge pressure reading

Watch out: manometer and other pressure-balance problems only work when both sides of the equation use the same reference. A pressure worked out from a mercury (or other) column is a gauge pressure unless atmospheric pressure is explicitly added to it. Before combining a manometer reading with another pressure value, confirm that both are gauge or both are absolute - mixing the two is one of the most common sources of a wrong-but-plausible answer.

Manometers and barometers

Manometer problem-solving procedure

The following is a general procedure for solving all manometer problems:

Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit or in an appropriate symbol if it is unknown.
Add the pressure change, in the same unit, from one meniscus to the next:
- Add if the next meniscus is lower.
- Subtract if the next meniscus is higher.
Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point, whether known or unknown.

The expression will contain one unknown for a simple manometer or will give a difference in pressure for a differential manometer.

Example:

A U-tube manometer containing Hg (specific gravity = 13.6) has its right limb open to the atmosphere. The left limb is full of water and connected to a pipe containing water under pressure. Task: Find the pressure of water in the pipe above atmospheric pressure, given the manometer readings as shown in the figure.

U-tube manometer
Achievable

Given

Vertical distance from the pipe centerline (point 1) down to the mercury surface in the left limb (point 2): $0.05 m$

Difference between the mercury surface in the left limb (point 2) and the mercury surface in the right limb (point 3): $h_{H g} = 0.15 m$

Specific gravity of mercury: $S G_{H g} = 13.6$

Solution

Let $h_{1}$ represent the gauge pressure of the water in the pipe, expressed as an equivalent height of water. Because every term is expressed as an equivalent height of water, each column’s height is multiplied by its own specific gravity relative to water (1 for water, 13.6 for mercury) - this puts the mercury and water columns on the same water-based scale so they can be added and subtracted directly. Equating the pressure at points 2 and 3, measured from the same datum, we have:

$h_{1} \times 1 + 0.05 \times 1 = 0.15 \times 13.6$

or,

$h_{1} = 0.15 \times 13.6 - 0.05 \times 1 = 1.99 m of water$

Therefore,

$p_{1} = γ h_{1} = 9.81 \times 1.99 = 19.52 kN/m^{2}$

Barometer

Another device that works on the same principle as the manometer is the simple barometer.

$p_{a t m} = p_{A} = p_{v} + γh = p_{B} + γh$

Where:

$p_{A}$ = pressure at the level of the reservoir’s free surface, which is open to the atmosphere
$p_{B}$ = pressure in the space at the top of the column, which holds only the fluid’s vapor, so $p_{B} = p_{v}$
$p_{v}$ = vapor pressure of the barometer fluid
$γ$ = specific weight of the barometer fluid
$h$ = height of the liquid column above the reservoir’s free surface

Forces on submerged surfaces and the center of pressure

A diagram illustrating hydrostatic forces on an inclined submerged surface and the location of the resulting center of pressure relative to the centroid. — Forces on a submerged surfaces and the center of pressure

The pressure at a point a vertical distance $h$ below the surface is:

$P = P_{atm} + γh$

where:

$P$ = pressure
$P_{a t m}$ = atmospheric pressure
$P_{C}$ = pressure at the centroid of area
$P_{CP}$ = pressure at the center of pressure
$y_{C}$ = slant distance from liquid surface to the centroid of area, $y_{C} = \frac{h _{C}}{s i n θ}$
$h_{C}$ = vertical distance from liquid surface to centroid of area
$y_{CP}$ = slant distance from liquid surface to center of pressure
$h_{CP}$ = vertical distance from liquid surface to center of pressure
$θ$ = angle between liquid surface and edge of submerged surface
$I_{x}$ = moment of inertia about the centroidal x-axis

Exam tip: the FE Reference Handbook (Fluid Mechanics section) already lists the center-of-pressure, resultant-force, and buoyancy equations below - you don’t need to memorize them, just practice finding them and matching the variables to the problem. Before substituting numbers, convert all lengths, densities, and forces to a single consistent unit system (SI or USCS).

If atmospheric pressure acts above the liquid surface and on the nonwetted side of the submerged surface:

$y_{CP} = y_{C} + \frac{I _{x}}{y _{C} A}$

or,

$y_{CP} = y_{C} + γ sin θ \cdot \frac{I _{x}}{P _{C} A}$

Wetted side force:

$F_{R} = (P_{atm} + γ y_{C} sin θ) A$

If $P_{a t m}$ acts on both sides:

$F_{n e t} = (γ y_{C} sin θ) A$

Example: Force on an inclined submerged gate

A rectangular gate holds back water in an open channel. The gate is $2 m$ wide and $3 m$ long (measured along its slant), inclined at $θ = 60°$ to the water surface. The centroid of the gate is $4 m$ below the surface, and atmospheric pressure acts on both sides of the gate. Find the resultant hydrostatic force $F_{n e t}$ and the slant distance to the center of pressure $y_{CP}$ .

Given

Width: $b = 2 m$ , slant length: $L = 3 m$

$θ = 60°$ , $h_{C} = 4 m$

$γ_{w a t er} = 9810 N / m^{3}$

Solution

First find the slant distance to the centroid and the area:

$y_{C} = \frac{h _{C}}{sin θ} = \frac{4}{sin 60°} = 4.62 m$

$A = b L = 2 \times 3 = 6 m^{2}$

Since atmospheric pressure acts on both sides, the net resultant force is:

$F_{n e t} = (γ y_{C} sin θ) A = γ h_{C} A = 9810 \times 4 \times 6 = 235, 440 N \approx 235.4 k N$

To find the center of pressure, first compute the moment of inertia of the rectangular gate about its own centroidal axis:

$I_{x} = \frac{b L ^{3}}{12} = \frac{2 \times 3 ^{3}}{12} = 4.5 m^{4}$

Therefore,

$y_{CP} = y_{C} + \frac{I _{x}}{y _{C} A} = 4.62 + \frac{4.5}{4.62 \times 6} = 4.78 m$

So $F_{n e t} \approx 235.4 k N$ , acting at $y_{CP} = 4.78 m$ (slant distance from the surface) - below the centroid, as expected.

Buoyancy and Archimedes principle

The buoyant force exerted on a submerged or floating body is equal to the weight of the fluid displaced by the body.
A floating body displaces a weight of fluid equal to its own weight; i.e., a floating body is in equilibrium.
The center of buoyancy is located at the centroid of the displaced fluid volume.

$F_{B} = γ V_{f}$

where:

$F_{B}$ = buoyancy force ( $N$ or $l b f$ )
$γ$ = specific weight of fluid ( $N / m^{3}$ or $l b f / f t^{3}$ )
$V_{f}$ = volume of displaced fluid ( $m^{3}$ or $f t^{3}$ )

Fluid properties: definitions

Density ( $ρ$ ): mass per unit volume, $ρ = \frac{m}{V}$
Specific weight ( $γ$ ): weight per unit volume, $γ = ρ g$
Specific gravity (SG): ratio of fluid density to water density, $SG = \frac{ρ _{f l u i d}}{ρ _{w a t er}}$
Pressure ( $P$ ): normal force per unit area, $P = \frac{F}{A}$
Viscosity:
- Dynamic viscosity ( $μ$ ): internal friction, units $P a \cdot s$
- Kinematic viscosity ( $ν$ ): $ν = \frac{μ}{ρ}$
Newton’s law of viscosity: $τ = μ \frac{d u}{d y}$ (shear stress proportional to velocity gradient)
Surface tension ( $σ$ ): force per unit length at liquid surface, $σ = \frac{F}{L}$
Capillarity: rise/fall in tube, $h = \frac{2 σ c o s θ}{ρ g r}$
Pressure in static fluid: increases with depth, $P_{2} - P_{1} = - γh$
- Absolute pressure = atmospheric pressure + gauge pressure

Manometers and barometers

Manometer problem-solving:
- Start at one end, assign pressure
- Add pressure if moving down, subtract if up
- Continue until circuit closes, solve for unknown
U-tube manometer: measures pressure difference using fluid column heights and densities
Barometer: measures atmospheric pressure using height of liquid column, $p_{a t m} = p_{v} + γh$

Forces on submerged surfaces and the center of pressure

Pressure at depth: $P = P_{a t m} + γh$
Center of pressure (vertical distance): $y_{CP} = y_{C} + \frac{I _{x}}{y _{C} A}$
- $y_{C}$ : centroid distance; $I_{x}$ : moment of inertia; $A$ : area
Hydrostatic force on surface: $F_{R} = (P_{a t m} + γ y_{C} sin θ) A$
Net force (if $P_{a t m}$ acts both sides): $F_{n e t} = (γ y_{C} sin θ) A$

Buoyancy and Archimedes principle

Buoyant force equals weight of displaced fluid, $F_{B} = γ V_{f}$
Floating body: displaces fluid weight equal to its own weight (equilibrium)
Center of buoyancy: centroid of displaced fluid volume

Next | 9.2 Mass and energy conservation

Fluid statics

This chapter covers the following:

Definitions
Manometers and barometers
Forces on submerged surfaces and the center of pressure
Buoyancy and Archimedes principle