A submerged or a floating body is said to be stable if it comes back to its original position after a slight disturbance.

Stability of a submerged body:

The position of center of gravity and center of buoyancy in case of completely submerged body are fixed. Consider a balloon which is completely submerged in air. Let, the lower portion of balloon contains heavier material so that its center of gravity is lower than its center of buoyancy as shown in figure a. Let, the weight of the balloon is W. The weight W is acting through G vertically in the downward direction while the buoyant force F_B is acting vertically up, through B. For the equilibrium of the balloon W= F_B. If the balloon is given an angular displacement in the clockwise direction as shown in figure a, then W and F_B constitute a couple acting in the anti- clockwise direction and brings the balloons in the original position. Thus the balloon in the position shown in figure a is in stable equilibrium.

• Stable Equilibrium :

When W =F_B and center of buoyancy (B) is above center of gravity (G), the body is said to be in stable equilibrium.

• Unstable Equilibrium:

If W= F_B, but the center of buoyancy (B) is below center of gravity (G) the body is in unstable equilibrium as shown in figure b. A slight displacement to the body in the clockwise direction gives the couple due to W and F_B also in the clockwise direction. Thus the body does not return to its original position and return to its original position and hence the body is in unstable equilibrium.

• Neutral Equilibrium:

If F_B =W and center of buoyancy (B) and center of gravity (G)  are at the same point as shown in figure c the body is said to be in neutral equilibrium.

The stability of a floating body is determined from the position of metacenter (M). In case of floating body the weight of the body is equal to the weight of liquid displaced.

• Stable Equilibrium:

If the point M is above G, the floating body will be stable equilibrium as shown in figure a. If a slight angular displacement is given to the floating body in the clockwise direction the center of buoyancy shifts from B to B₁such that the vertical line through B₁ cuts at M. Then the buoyant force F_B through B₁ and weight W through G constitute a couple acting in the anticlockwise direction and thus bringing the floating body in the original position.

• Unstable Equilibrium:

If the point M is below G, the floating body will be in unstable equilibrium as shown in figure B. The disturbing couple is acting in the clockwise direction. The couple due to buoyant force F_B  and W is also acting in the clockwise direction and thus overturning the floating body.

• Neutral Equilibrium:

If the point M is at the center of gravity of the body, the floating body will be in neutral equilibrium.

Summary:

Metacenter:

Metacenter is defined as the point at which the line of action of the force of buoyancy will meet the normal axis of the body when the body is given a small angular displacement.

Metacentric Height:

The distance between the metacenter of a floating body and the center of a floating body and the center of gravity of the body is called metacentric height.

Experimental Method for determination of metacentric height:

Let,

W= Weight of vessel including w₁

G = Center of gravity of the vessel

B = Center of buoyancy of the vessel

The weight w₁   is moved across the vessel towards right through a distance x as shown in figure B. The vessel will be tilted. The angle of heel θ is measured by means of a plumbline and a protractor attached on the vessel.The new center of gravity of the vessel will be G₁ as the weight w₁ has been moved toward the right.

Above the center of buoyancy will change to B₁ as the vessel has tilted.

Under equilibrium the moment caused by the movement of load w₁ through a distance x must be equal to the movement caused by the shift of center of gravity from G to G₁ .

Thus,

The moment due to change of G

= GG₁ * W

= W * GMtanθ

 The moment due to movement of w₁

= w_{1 *} x

From above equation,

w_{1 *} x = W * GMtanθ

or, GM = (w_{1 *} x)/ Wtanθ

Analytical Method for Metacentric height:

In figure A, it shows the position of a floating body in equilibrium. The location of center of gravity and center buoyancy in this position is at G and B. The floating body is given a small angular displacement in the clockwise direction. This is shown in figure B. The new center of buoyancy is B_1. The vertical line through B₁ cuts the normal axis at M. Hence, M is the metacenter and GM is metacentric height.

Couple due to wedges

Consider towards the right of the axis a small strip of thickness dx at a distance x from O as shown in figure B.

The height of strip x ∠ BOB^’ = x* θ [∴∠BOB^’ = ∠AOA^’ = ∠BMB₁^’=θ]

∴Area of strip = Height * Thickness

                            = x* θ*dx

If L is the length of the flaoting body,then

Volume of strip = Area * L

                          = x* θ*L*dx

∴Weight of strip = ρg*Volume

                            = ρg *x* θ*L*dx

Moment of the couple = Weight of each strip * Distance between these weigth

                                    = ρg x θLdx*[x+x]

                                    = 2ρg x² θLdx

∴Moment of couple for the whole wedge=∫2ρg x² θLdx ——(i)

Moment of couple due to shifting of center of buoyancy from B to B₁

= F_B * BB₁

= F_B * BM* θ [∴ BB₁ = BM* θ , if θ is very small]

= W * BM* θ [∴ F_B = W] ————(ii)

Hence,equating equation (i) and (ii)

W * BM* θ =∫2ρg x² θLdx

Or, W * BM* θ = 2ρgθ∫ x² Ldx

Or, W * BM = 2ρg∫ x² Ldx

But,

Ldx = dA [∴ Elemental area on the water line shown in figure C]

∴W*BM=2ρg∫ x² dA

But, figure C it is clear that 2∫ x² dA is the second moment of area of the plane of the body at water surface about the axis Y-Y.

Therefore,

W*BM = ρgI [∴ where I = 2∫ x² dA]

∴ BM = (ρgI)/ W

But,

W= Weight of the body

    = ρg * Volume of the body submerged in water

    = ρg * ∀

∴ BM = (ρgI)/ ρg * ∀

Or, BM = I/∀

Now,

GM = BM – BG

        = (I/∀)  – BG

∴Metacentric height = GM = (I/∀)  – BG

Total pressure: Total pressure is defined as the force exerted by a static fluid on a surface either plane or curved when the fluid comes in contact with the surface. This force always acts normal to the surface.

Centre of pressure: Centre of pressure is defined as the point of application of the total pressure on the surface.

The fluid element is of very small dimensions i.e. dx, dy, and ds.

Consider an arbitrary fluid element of wedge shape in a fluid mass at rest as shown in figure. Let, the width of the element perpendicular to the plane of the paper is unity and P_x, P_y and P_z are the pressure or intensity of pressure acting on the force AB, AC and BC respectively.

Then,

Force on side AB = P_x* Area of face AB

Or, Force on side AB = P_x*dy*1

Or, Force on side AB = P_x dy

Similarly,

Force on side AC = P_y dx

Force on side BC = P_z ds

Weight (W) = Area of triangular element*depth*Specific weight

Or, W = 1/2* AB * AC *1*

Or, W = * dxdy * ρg

Resolving the forces in x-direction

P_x dy = P_z ds sin (90- θ)

Or, P_x dy = P_z ds cos θ

Also from figure,

dy = ds cos θ

Now,

P_xdy = P_z dy

Or, P_x = P_z              ………………….             (i) 

Similarly,

Resolving the forces in y-direction

Pydx = Pzds cos(90- θ) + 1/2* dxdy * ρg

Also,

dscos θ = dx {also the element is very small and hence weight is negligible}

Now,

P_ydx = P_zdx

Or, P_y = P_z             ………………..               (ii)

From equation (i) and (ii), we have

P_x = P_y = P_z

The above equation shows that pressure at any point in x, y and z direction is equal.

Absolute pressure is a pressure that is relative to the zero pressure in the empty, air-free space of universe. This reference pressure is the ideal or absolute vacuum. Also, it is the sum of gauge pressure and atmospheric pressure.

i.e. P_ab= P_gauge+P_atm

Where,

Absolute pressure = P_ab

Gauge pressure = P_gauge

Atmospheric pressure = P_atm

Gauge pressure is defined as the difference between an absolute pressure and atmospheric pressure.

i.e. P_gauge = P_ab – P_atm

Where,

Absolute pressure = P_ab

Gauge pressure = P_gauge

Atmospheric pressure = P_atm

Atmospheric pressure is also known as barometric pressure, it is the pressure within the atmospheric of earth.

i.e. P_atm= P_ab – P_gauge

Where,

Absolute pressure = P_ab

Gauge pressure = P_gauge

Atmospheric pressure = P_atm

A pressure reading below the atmospheric pressure is known as a vacuum pressure.

The graphical representation of relationship between Absolute, gauge and atmospheric pressure at a point:

Let, B is the point at which pressure is to be measured, whose value is P. The datum line is A-A.

Let,

h₁= Height of light liquid above the datum line.

h₂= Height of heavy liquid above the datum line.

s₁= Specific gravity of light liquid.

s₂= Specific gravity of heavy liquid.

ρ₁ = Density of light liquid. = 1000 * s₁

ρ₂  = Density of heavy liquid. = 1000 * s₂

As the pressure is the same for horizontal surface. Hence, pressure above the horizontal datum line A-A in the left column and in right column of U-tube manometer should be same.

Pressure above A-A in the left column = P + ρ₁ g h₁

Pressure above A-A in the right columns = ρ₂ g h₂

Hence, equating the two pressure

P + ρ₁ g h₁= ρ₂ g h₂

 Or,    P =    ρ₂ g h_{2 –} ρ₁ g h₁

For measuring vacuum pressure the level of the heavy liquid in the manometer will be as shown in figure.

Let,

h₁= Height of light liquid above the datum line.

h₂= Height of heavy liquid above the datum line.

s₁= Specific gravity of light liquid.

s₂= Specific gravity of heavy liquid.

ρ₁ = Density of light liquid. = 1000 * s₁

ρ₂= Density of heavy liquid. = 1000 * s₂

Then,

Pressure above A-A in the left column = P + ρ₁ g h₁+ ρ₂ g h₂

Pressure above A-A in the right columns = 0

Hence, equating the two pressure

              P + ρ₁ g h_{1 +} ρ₂ g h₂= 0

       Or,   P = – (ρ₁ g h_{1 +} ρ₂ g h₂)

h₁= Height of light liquid above the datum line.

h₂=Height of heavy liquid above the datum line.

 s₁= Specific gravity of light liquid.

s₂= Specific gravity of heavy liquid.

ρ₁= Density of light liquid.

ρ₂= Density of heavy liquid.

△h= Fall of heavy liquid.

P = Pressure at A.

A= Cross- sectional area of the reservoir.

a = Cross-sectional area of the right limb.

Fall of heavy liquid in reservoir will cause a rise of heavy liquid level in the right limb

A*△h = a * h₂

 Or, △h = (a * h₂) / A                                     (a)

Now, consider the datum line Y-Y as shown in figure.

Then,

 Pressure in the right limb above Y-Y = ρ₂g (△h + h₂)                     (b)

Pressure in the left limb above Y-Y = ρ₁g (△h + h₁) + P                      (c)

Equating equation (b) and (c) , we have

ρ₂g (△h + h₂) = ρ₁g (△h + h₁) + P                      

Or, P = ρ₂g (△h + h₂) – ρ₁g (△h + h₁)

Or, P = △h [ρ₂g – ρ₁g] + h₂ ρ₂g – h₁ ρ₁g

From equation (a)

P =  [(a * h₂) / A ] * [ρ₂g – ρ₁g] + h₂ ρ₂g – h₁ ρ₁g

As, the area A is very large compared to a, hence ratio a/A becomes very small and can be neglected.

Thus,

P= h₂ ρ₂g – h₁ ρ₁g

Above figure shows the inclined single column manometer. This manometer is more sensitive. Due to inclination the distance moved by the heavy liquid in the right limb will be more.

Let,

L= length of heavy liquid moved in right limb from X-X

θ = Inclination of right limb with horizontal

h₂ = Vertical rise of heavy liquid in right limb from X-X

    h₂  = Lsinθ

Now,

From equation of pressure at A

P= h₂ ρ₂g – h₁ ρ₁g

Substituting the value of   , we get

P=Lsinθ ρ₂g – h₁ ρ₁g

In above figure the two points A and B are at different level and also contain liquids of different specific gravity. These points are connected to the U- tube differential manometer. Assume the pressure at A and B are

P_A and P_B respectively.

Let,

h= Difference of mercury level in U-tube

y = Distance of the center of B from the mercury level in the right limb

x= Distance of the center of A from the mercury level in the right limb

ρ₁= Density of liquid at A

ρ₂= Density of liquid at B

ρ_m= Density of heavy liquid or mercury

Pressure above X-X in the left limb = ρ₁g(h+x) + P_A

Pressure above X-X in the right limb = ρ_m gh + ρ₂gy + P_B

Equating the two pressure, we have

ρ₁g(h+x) + P_A = ρ_m gh + ρ₂gy + P_B

 Or, P_A – P_B = ρ_m gh + ρ₂gy – ρ₁g(h+x

Or, P_A – P_B = hg(ρ_m – ρ₁ )+ ρ₂ gy – ρ₁gx

It is defined as the ratio of mass of fluid to the volume of fluid. Its SI unit is kg/m³

Mathematically,

Density (ρ) =(Mass of fluid (m) )/(Volume of fluid (V))

Where,

Density = ρ

Mass of fluid= m

Volume of fluid = V

Note: Density of Water = 1000 kg/ m³

 It is defined as the ratio of weight of fluid to the volume of fluid. Its SI unit is N/ or KN/

Mathematically,

Weight Density (γ) = (Weight of fluid(W))/(Volume of fluid(V))

Then,

         γ =   W/V                                 

Or, γ = (mg)/V                               

γ = ρ g

Where,

Weight Density or Specific weight= γ

Gravity= g

Density= ρ

Mass= m

Volume = V

It is defined as the ratio of volume per unit mass.

Its SI unit is m³ /kg.

Mathematically,

Specific Volume (Ѵ) =  (Volume of fluid(V))/(Mass of fluid(m))

Then,

   Ѵ=  (V)/m

       Ѵ=     (1)/(ρ)

It is defined as the ratio of weight density of liquid to the weight density of standard liquid like water.

Mathematically,

Specific Gravity(S) =   (Weight Density of Liquid)/(Weight Density of Standard Liquid)                                 

Also,

Specific Gravity(S) =  (Mass Density of Liquid)/(Mass Density of Water)                             

Note: Specific Gravity of Mercury is 13.6

If k_n=λ /L is less than 0.01 then it is continuum.

A continuous and homogenous fluid media is known as continuum. Analysis of fluid flow problems is made by a concept that treats fluid as continuous media. All cavities (voids) between the molecules are neglected.

The physical properties like density, temperature, velocity, pressure etc. are the define as continuous function of space and time within the matter or at any time we define properties or parameter as continuous function of space within the matters. So, it assume that each and every point in the matter there is a molecule (because all matter are composed of molecules) and hence, there is no discontinuity. This continuous and homogenous fluid media is called continuum. This is highly true for solid and liquid because molecule are highly pack but for gases if the pressure is very low, distance between the molecule increases cohesive force decreases and the concept of continuum may not be void.

• Based on Knudsen number the validity of continuum concept is describe as if k_n = λ /L < 0.01 then the concept of continuum is valid.
• Where,
• Distance between the molecule = λ
• Characteristics dimension of problem = L

Newton’s Law of viscosity states that the shear stress in a flowing fluid is directly proportional to the rate of shear strain.

Mathematically,

Ʈ ∝(du)/(dy)

Ʈ = µ(du)/(dy)

Where, 

Dynamic Viscosity = µ

Rate of shear stress = Ʈ

Velocity gradient = du/dy

Then,

µ= (Ʈ)/(du/dy)

Significance:

The fluid which follow Newton’s law of viscosity are called Newtonian fluids.

Weight (W) = Mass (m)*Gravity (g)

Or, W = ρVg

Or, W= ρ*(A*h)*g

Where,

Density = ρ

Volume = V

Area = A

Height of rise = h

Tensile Force = Surface tension * Circumference * cosθ

Or, Tensile Force = σ*πd*cosθ

At equilibrium condition,

ρ *(A*h)*g = σ*πd*cosθ

Or, ρ *(πd² /4 * h)*g = σ*πd*cosθ

Or, h = (4σ * cosθ)/ρgh

If Ɵ =0

Then,

 h =4σ/ρgh

Weight (W) = Mass (m) * Gravity (g)

Or, W= ρVg

Or, W= ρ (A*h) g

Or, W= ρ (L*t) h g

Tensile force= σ* 2L *cosθ

At equilibrium condition,

ρ (L*t) h g = σ* 2L*cosθ

Or, h =(σ* 2L*cosθ)/ρ (L*t) g

 h =(2σcosθ)/ρtg