Fluid Mechanics : Manometry and Buoyancy & Hydrostatic Forces on Surface Notes

Manometry and Buoyancy

Manometry

The pressure is proportional to the height of a column of fluid.
Manometry is the field of science which deals with the evaluation of the pressure of the fluid.
The instrument used to carry out the complete process is termed a Manometer.
Types of Manometers: Barometer, Piezometer and U-tube Manometer.

Manometers use the relationship between pressure and head to measure pressure.

Relation between Hydrostatic pressure & Head

We have the vertical pressure relationship p = ρgz + constant measuring z from the free surface so that z = -h and surface pressure is atmospheric,p_atm

We generally assume atmospheric pressure as the datum,

Gauge pressure, p_g = ρgh

The lower limit of any pressure is the pressure in a perfect vacuum. Pressure measured above a perfect vacuum (zero) is known as absolute pressure.

Absolute pressure, p_a = ρgh+ p_atmospheric

Absolute pressure = Gauge pressure + Atmospheric

Piezometer Tube Manometer

The simplest manometer is an open tube. This is attached to the top of a container with liquid at pressure. containing liquid at a pressure.
The tube is open to the atmosphere, The pressure measured is relative to atmospheric so it measures gauge pressure.
Pressure at A = pressure due to column of liquid h₁

p_a = ρgh₁

Pressure at B = pressure due to column of liquid h₂

P_b = ρgh₂

Limitations of Piezometer:

Can only be used for liquids
Pressure must above atmospheric
Liquid height must be convenient i.e. not be too small or too large

U-tube Manometer

It consists of a U shaped bend whose one end is attached to the gauge point ‘A’ and the other end is open to the atmosphere.
It can measure both positive and negative (suction) pressures.
“U”-Tube enables the pressure of both liquids and gases to be measured “U” is connected as shown and filled with manometric fluid.

Note:

The manometric fluid density should be greater than of the fluid measured, ρ_man > ρ
The two fluids should not be able to mix they must be immiscible.

Pressure in a continuous static fluid is the same at any horizontal level, pressure at B = pressure at C

P_B = P_C

For the left-hand arm pressure at B
- pressure at A + pressure of height of liquid being measured

P_B = P_A + ρgh₁

For the right-hand arm pressure at C =
- pressure at D + pressure of height of manometric liquid

P_C = P + ρ_manogh₂

We are measuring gauge pressure we can subtract p_atmospheric giving

P_B = P_C

P_A = P +ρ_manogh₂- ρgh₁

Differential U-Tube Manometer

A U-Tube manometric liquid is heavier than the liquid for which the pressure difference is to be measured and is not immiscible with it.

The pressure difference between A and B is given by equation

P_A – P_B = ρ₂h₂ + ρ₃h₃ – ρ₁h₁

Inverted U-Tube Manometer

Inverted U-Tube manometer consists of an inverted U Tube containing a light liquid.
This is used to measure the differences of low pressures between two points where better accuracy is required.
It generally consists of an air cock at top of the manometric fluid type.

Pressure difference can be calculated from equation:

P₁ – ρ₁gH₂ – ρ_mg(H₁– H₂)=P₂ – ρ₂gH1

Micro Manometer

Micro Manometer is the modified form of a simple manometer whose one limb is made of larger cross sectional area.
It measures very small pressure differences with high precision.

Let ‘a’ = area of the tube, A = area of the reservoir, h₃ = Falling liquid level reservoir,

h₂ = Rise of the liquid in the tube,

By Volume Equality, Ah₃ = ah₂
Equating pressure heads at datum,

P₁ = (ρ_m – ρ₁)gh₃ + ρ_mgh₂ – ρ₁gh₁

Inclined Manometer

An inclined manometer is used for the measurement of small pressures and is to measure more accurately than the vertical tube type manometer.
Due to inclination, the distance moved by the fluid in manometer is more.

Pressure difference between A and B is given by equation

P_A – P_B = ρ₂_LsinΘ + ρ₃h₂ – ρ₁h₁

Buoyancy

Buoyancy is also known as buoyant force. It is the force exerted on an object that is wholly or partly immersed in a fluid.

Concept of Buoyancy: When a body is immersed in a fluid, an upward force is exerted by fluid on the body which is equal to weight of fluid displaced by body. This acts as upward.

Archimedes’ Principle: It states, when a body is immersed completely or partially in a fluid, it is lifted up by a force equal to weight of fluid displaced by the body.

Buoyant force = Weight of fluid displaced by body

Buoyant force on cylinder =Weight of fluid displaced by cylinder

V_{Sm =}Value of immersed part of solid or Volume of fluid displaced

F_B = P_water X g X Volume of cylinder immersed inside the water

= (w = mg = pVg)

Principle of Flotation: According to this principle, if weight of body is equal to buoyant force then, body will float.

F_B = mg

The factors that affect buoyancy are: the density of the fluid, the volume of the fluid displaced, and the local acceleration due to gravity.
The buoyant force is not affected by the mass of the immersed object or the density of the immersed object.

Center of Buoyancy: The point at which force of buoyancy acts is called center of buoyancy. It lies on center of gravity of volume of fluid displaced or center of gravity of the part of the body which is inside the water. Point B is the center of buoyancy.

Buoyancy on a submerged body:

The Archimedes principle states that the buoyant force on a submerged body is equal to the weight of liquid displaced by the body, and acts vertically upward through the centroid of the displaced volume.
Thus the net weight of the submerged body, (the net vertical downward force experienced by it) is reduced from its actual weight by an amount that equals the buoyant force.

Buoyancy on a partially immersed body:

According to Archimedes principle, the buoyant force of a partially immersed body is equal to the weight of the displaced liquid.
Therefore the buoyant force depends upon the density of the fluid and the submerged volume of the body.
For a floating body in static equilibrium and in the absence of any other external force, the buoyant force must balance the weight of the body.

Metacentre of a Floating Body: If a body that is floating in liquid is given a small angular displacement, it starts oscillating about some point M. This point is called the metacentre.

The equilibrium of a submerged body in a liquid requires that the weight of the body acting through its centre of gravity should be colinear with equal hydrostatic lift acting through the centre of buoyancy. Let us suppose that a body is given a small angular displacement and then released. Then it will be said to be in distance MG is called metacentric height (it is the distance between the gravity centre and metacentre)

Stability of Submerged Body: It is classified into three groups.

Stable Equilibrium: When the centre of buoyancy lies above the centre of gravity, the submerged body is stable.

Unstable Equilibrium: When B lies below G, then body is in unstable equilibrium.

Neutral Equilibrium: When B and G coincide then, body is in neutral equilibrium.

Stability of Floating Bodies: When the body undergoes an angular displacement about a horizontal axis, the shape of the immersed volume changes and so the centre of buoyancy moves relative to the body.

Stale Equilibrium: When a body is given a small angular displacement by external means and if body comes to its original position due to internal forces then, it is called stable equilibrium.

It occurs, when metacentre lies above centre of gravity.

Unstable Equilibrium: In the above case, if body does not come in its original position and moves further away then, it is known as unstable equilibrium. M lies below centre of gravity.

Neutral equilibrium: When a body is given a small angular displacement and it sets on new position then, body is called in neutral equilibrium. In this, M and G coincide.

Relation between B,G and M is

GM = I/V - BG

Here, l = Least moment of inertia of plane of body at water surface

G = Centre of gravity

B = Centre of buoyancy

M = Metacentre

V is volume submerged inside the water can be given as

Where b,d and x are the length, width and depth of the section or body.

BG is the distance between centre of gravity and centre of buoyancy. (In other words, BG=distance between centre of gravity of whole body and centre of gravity of submerged part of body)
When we find out GM then, we can determine the status of body as
- GM > 0 (stable equilibrium),
- GM < 0 (unstable equilibrium),
- GM = 0 (neutral equilibrium )

Hydrostatic Force on Surfaces

Fluid Statics

Fluid Statics deals with fluids at rest while Fluid Dynamics studies fluids in motion.
Any force developed is only due to normal stresses i.e, pressure. Such a condition is termed the hydrostatic condition.
Fluid Statics is also known as Hydrostatics.
A static fluid can have no shearing force acting on it, and that any force between the fluid and the boundary must be acting at right angles to the boundary.
For an element of fluid at rest, the element will be in equilibrium. The sum of the components of forces in any direction will be zero. The sum of the moments of forces on the element about any point must also be zero.
Within a fluid, the pressure is same at all the points in all the directions.
Pressure at the wall of any vessel is perpendicular to the wall
Pressure due to depth is P = ρgh, and is the same at any horizontal level of connected fluid.

Fluid Pressure at a Point

If a fluid is Stationary, then force acting on any surface or area is perpendicular to that surface.
If the force exerted on each unit area of a boundary is the same, the pressure is said to be uniform

Pascal’s Law for Pressure At A Point

It states that pressure or intensity of pressure at a point in a static fluid (fluid is in rest) is equal in all directions. If fluid is not in motion then according to Pascal’s law,

p_x = p_y = p_z

where, p_x,p_y and p_z are the pressure at point x,y,z respectively.

General Equation For Variation Of Pressure in a Static Fluid

(A cylindrical element of fluid at an arbitrary orientation)

Vertical Variation Of Pressure in a Fluid Under Gravity

Taking upward as positive, we have

Vertical cylindrical element of fluid cross sectional area = A

mass density = ρ

The forces involved are:

Force due to p₁ on A (upward) = p_1.A
Force due to p₂ on A (downward) = p_2.A
Force due to weight of element (downward) = mg

= mass density x volume x g

= ρ.g.A.(z₂ - z₁)

Thus in a fluid under gravity, pressure decreases linearly with increase in height

p₂- p₁ = ρgA(z₂- z₁)

This is the hydrostatic pressure change.

Equality Of Pressure At The Same Level In A Static Fluid

Horizontal cylindrical element cross sectional area = A

mass density =ρ

left end pressure = p_l

right end pressure = p_r

For equilibrium, the sum of the forces in the x direction is zero= p_l.A = p_r.A

p_l= p_r

So, Pressure in the horizontal direction is constant.

As we know, p_l= p_r

Thus, pressure at the two equal levels is the same.

Total Hydrostatic Force on Plane Surfaces

For horizontal plane surface submerged in liquid, or plane surface inside a gas chamber, or any plane surface under the action of uniform hydrostatic pressure, the total hydrostatic force is given by

F = p. A

where p is the uniform pressure and A is the area.

In general, the total hydrostatic pressure on any plane surface is equal to the product of the area of the surface and the unit pressure at its center of gravity.

F = p_cg.A

where p_cg is the pressure at the center of gravity.

For homogeneous free liquid at rest, the equation can be expressed in terms of unit weight γ of the liquid.

F=γh'A

where h' is the depth of liquid above the centroid of the submerged area.

Derivation of Formulas(Not required for exam)

The figure shown below is an inclined plane surface submerged in a liquid. The total area of the plane surface is given by A, cg is the center of gravity, and cp is the center of pressure.

(Forces on a inclined plane surface)

The differential force dF acting on the element dA is

$d F = p d A$

$d F = γ h d A$

From the figure

$h = y \sin θ$ ,
$d F = γ (y \sin θ) d A$

Integrate both sides and note that γ and θ are constants,
$F = γ \sin θ \int y d A$

$F = γ \sin θ \int y d A$

Recall from Calculus that

$\int y d A = A \bar{y}$

$F = (γ \sin θ) A \bar{y}$

$F = γ (\bar{y} \sin θ) A$

From the figure, $\bar{y} \sin θ = \bar{h}$ thus,

F = γh¯A

The product $γ \bar{h}$ is a unit pressure at the centroid at the plane area, thus, the formula can be expressed in a more general term below:

F = p_cg.A

Location of Total Hydrostatic Force (Eccentricity)

From the figure above, S is the intersection of the prolongation of the submerged area to the free liquid surface. Taking moment about point S.

$F y_{p} = \int y d F$

Where

$d F = γ (y \sin θ) d A$
$F = γ (\bar{y} \sin θ) A$

$[γ (\bar{y} \sin θ) A] y_{p} = \int y [γ (y \sin θ) d A]$

$(γ \sin θ) A \bar{y} y_{p} = (γ \sin θ) \int y^{2} d A$

$A \bar{y} y_{p} = \int y^{2} d A$

Again from Calculus, $\int y^{2} d A$ is called moment of inertia denoted by I Since our reference point is S,

$A \bar{y} y_{p} = I_{S}$

Thus,

$\bar{h}$

By transfer formula for moment of inertia $I_{S} = I_{g} + A {\bar{y}}^{2}$ the formula for y_p will become

y_p $y_{p} = \frac{I_{g} + A {\bar{y}}^{2}}{A \bar{y}}$ or

_yp=y¯+I_g/Ay¯

From the figure above, y_p $y_{p} = \bar{y} + e$ thus, the distance between cg and cp is

Eccentricity, $e = \frac{I_{g}}{A \bar{y}}$

Example 1

An opening in a dam is covered with a plate of 1 m square and is hinged on the top and inclined at 60 ⁰ to the horizontal. If the top edge of the gate is 2 m below the water level what is the force required to open the gate by pulling a chain set at 45 ⁰ angle with the plate and set to the lower end of the plate. The plate weighs 2200 N.

Solution:

Total Hydrostatic Force on Curved Surfaces:

In the case of curved surface submerged in liquid at rest, it is more convenient to deal with the horizontal and vertical components of the total force acting on the surface. Note: the discussion here is also applicable to plane surfaces.
Horizontal Component: The horizontal component of the total hydrostatic force on any surface is equal to the pressure on the vertical projection of that surface.

F_H=p_cg.A

Vertical Component: The vertical component of the total hydrostatic force on any surface is equal to the weight of either real or imaginary liquid above it.

F_V=γ. V

Total Hydrostatic Force:

F=√(F_H²+F_V²)

Direction of $F$

$F$

$F$

The vertical component of the hydrostatic force is going upward and equal to the volume of the imaginary liquid above the surface.

Example 2

The length of a tainter gate is 1 m perpendicular to the plane of the paper. Find out the total horizontal force on the gate and the total hydrostatic force on the gate.

Solution:

Example 3:

A quarter circle (10 m diameter) gate which is 10 m wide perpendicular to the paper holds water as shown in the figure. Find the force required to hold the gate. The weight of the gate can be neglected.

Solution:

Fluid Mechanics : Manometry and Buoyancy & Hydrostatic Forces on Surface Notes

Manometry and Buoyancy

Manometry

Buoyancy

Hydrostatic Force on Surfaces

Equality Of Pressure At The Same Level In A Static Fluid

_yp=y¯+I_g/Ay¯

Eccentricity, $e = \frac{I_{g}}{A \bar{y}}$

F_H=p_cg.A

F_V=γ. V

F=√(F_H²+F_V²)

Comments

AE & JE Exams

Featured Articles

Fluid Mechanics : Manometry and Buoyancy & Hydrostatic Forces on Surface Notes

Manometry and Buoyancy

Manometry

Buoyancy

Hydrostatic Force on Surfaces

Equality Of Pressure At The Same Level In A Static Fluid

yp=y¯+Ig/Ay¯

Eccentricity, e=Ig/Ay¯

FH=pcg. A

FV=γ. V

F=√(FH2+FV2)

Comments

AE & JE Exams

Featured Articles

_yp=y¯+I_g/Ay¯

Eccentricity, $e = \frac{I_{g}}{A \bar{y}}$

F_H=p_cg.A

F_V=γ. V

F=√(F_H²+F_V²)