Unit 1 INTRODUCTION 1.1.Introduction 1.2.Objectives

Unit 1

INTRODUCTION

1.8.Surface Tension and Capillarity 1.9.Capillarity

1.10. Summary 1.11. Keywords 1.12. Exercise

1.1.Introduction

Fluid mechanics that branch of science which deals with the behavior of the fluids (liquids orgases) at rest as well as in motion. Thus this branch of science deals with the static, kinematics and dynamic aspects of fluids. The study of fluids at rest is called fluid statics. The study of fluids in motion, where pressure forces are not considered, is called fluid kinematics and if the pressure forces are also considered for the fluids in motion, that branch of science is called fluid dynamics.

1.2.Objectives

After studying this unit we are able to understand − Properties of Fluids

− Viscosity

− Types of Fluids.

− Thermodynamic Properties − Compressibility

− Surface Tension and Capillarity − Capillarity

1.3.Properties of Fluids Density or Mass Density

Density or mass density of a fluid is defined as the ratio of the mass of a fluid to its volume. Thus mass per unit volume of a fluid is called density. It is denoted the symbol p (rho). The unit of mass density in SI unit is kg per cubic meter, i.e. kg/m3. The density of liquids may be considered as constant while that of gases changes with the variation of pressure and temperature.

Mathematically, mass density is written as

The value of density for water is 1 gm/cm3 or 1000 kg/m3.

Specific Weight or Weight Density

Specific weight or weight density of a fluid is the ratio between the weights of a fluid to its volume. Thus weight per unit volume of a fluid is called weight density and it is denoted by the symbol w.

Specific Volume

Specific Volume of a fluid is defined as the volume of a fluid occupied by a unit measure volume per unit mass of fluid is called specific volume. Mathematically, it is expressed as specific volume

Thus specific volume is the reciprocal of mass density. It is expressed on m3/kg. It is commonly applied to gases

Specific Gravity

Specific gravity is defined as the ratio of the weight density (or density) of fluid to the weight density (or density) of a standard fluid. For liquids, the standard fluid is taken water and for gases, the standard fluid is taken air. Specific gravity is also called relative density. It is dimension less quantity and is denoted by the symbol S.

Problem 1.1.

Calculate the specific weight, density and specific gravity of one liter of a liquid which weighs 7 N.

Problem 1.2.

Calculate the density, specific weight and weight of one liter of petrol of specific gravity = 0.7.

1.4.Viscosity

Viscosity is defined as the property of a fluid which offers resistance to the movement of one layer of over another adjacent layer of the fluid. When two layers of a fluid a distance ‘dy’ apart, move one over the other at a different velocities, say u and u+du as shown in Fig. 1.1, the viscosity together with relative velocity causes a shear stress acting between the flid layers

Fig.1.1 Velocity variationnear a solid boundary

The top layer causes a shear stress on the adjacent lower layer while the lower layer causes a shear stress on the adjacent top layer. This shear stress is proportional to the rate of change of velocity with respect to y. It is denoted by symbol t called Tau.

• Units of viscosity

The units of viscosity is obtained by putting the dimensions of the quantities in equation (1.3)

In MKS system, force is represented by kgf and length by meter (m), in CGS system, force is represented by dyne and length by cm and in SI system force is represented by Newton (N) and length by meter (m).

Thus for solving numerical problems, if viscosity is given in poise, it must be divided by 98.1 to get its equivalent numerical value in MKS.

Note. (i) In SI units second is represented by 's' and not by 'sec'.

(ii) If viscosity is given in poise, it must be divided by 10 to get its equivalent numerical value in SI units. Sometimes a unit of viscosity as centipoise is used where

The viscosity of water at 20°C is 0.01poise or 1.0centipoise. • Kinematic Viscosity

It is defined as the ratio between the dynamic viscosity and density of fluid. It is denoted by the Greek symbol (v) called 'nu'. Thus, mathematically,

In MKS and SI, the unit of kinematic viscosity is metre2/sec or m2/sec while in CGS units it is written as cm2/s. In CGS units, kinematic viscosity is also known stoke.

• Newton's Law of Viscosity

It states that the shear stress (T) on a fluid element layer is directly proportional to the rate of shear strain. The constant of proportionality is called the co-efficient of viscosity. Mathematically, it is expressed as given by equation (1.2) or as

Fluids which obey the above relation are known as Newtonian fluids and the fluids which do no: obey the above relation are called Non-newtonian fluids.

• Variation of Viscosity with Temperature

Temperature affects the viscosity. 'The viscosity of liquids decreases with the increase of temperature while the viscosity of gases increases with the increase of temperature. This is due to reason that in liquids the cohesive forces predominates the molecular momentum transfer, due to closely packed molecules and with the increase in temperature, the cohesive forces decreases with the result of decreasing viscosity. But in case of gases the cohesive force are small and molecular momentum transfer predominates. With the increase in temperature, molecular momentum transfer increases and hence viscosity increases. The relation between viscosity and temperature for liquids gases are:

Fig. 1.2 Type of fluids

The fluids may be classified into the following five types:

(1). Ideal fluid, (2) Real fluid, (3) Newtonian fluid, (4) Non-Newtonian fluid, and (5) Ideal plastic fluid.

1. Ideal Fluid : A fluid, which is incompressible and is having no viscosity, is known as an ideal fluid. Ideal fluid is only an imaginary fluid as al1 the fluids, which exist, hive some viscosity.

2. Real Fluid. A fluid, which possesses viscosity. is known as real fluid. All the fluids, in actual practice., are real fluids.

3. Newtonian Fluid. A real' fluid, in which the shear stress is directly, proportional to the rate of shear strain (or velocity gradient) is known as Newtonian fluid.

4. Non-Newtonian Fluid. A real fluid, in which the shear stress is not proportional to the rate of shear strain (or velocity gradient) is known as a Non-Newtonian fluid.

5. Ideal Plastic Fluid. A fluid,. in which shear stress is more than the yield value and shear stress is proportional -Velocity gradient (&) to the rate of shear strain (or velocity gradient)is, known as d Y ideal plastic fluid.

Problem: 1 .3.

If the velocity distribution over a plate is given by u = 2/3y y2- in which u is tile velocity in meter per second at a distance y meter above the plate, determine the shear stress at y = 0 and y = 0.15 m. Take dynamic viscosity of fluid as 8.63 poises.

Problem 1.4.

A plate, 0.025 mm distant from a fixed plate, moves at 60 cm/s and requires a force of 2N per unit area i.e. 2N/m2 to maintain this speed Determine the fluid viscosity between the plates.

Problem 1.5

A flat plate of area 1.5 x lo6 mm2 is pulled with a speed of 0.4 m/s relative to another plate located at a distance of 0.15 mm from it. Find the force and power required to maintain this speed, if the fluid separating them is having viscosity as 1poise

Problem 1.6.

Determine the intensity of shear of an oil having viscosity = 1 poise. The oil is used for lubricating the clearance between a shaft of diameter 10 cm and its journal bearing. The clearance is 1.5 mm and the shaft rotates at 150 r.p.m

.

Problem 1.7. Calculate the dynamic viscosity of an oil, which is used for lubrication between a square plate of size 0.8 m x 0.8 m and an inclined plane with angle of inclination 3(P as shown in Fig. 1.4. The weight of the square plate is 300 N and it slides down the inclined plane with a uniform velocity of 0.3 m/s. The thickness of oil film is 1.5 mm.

1.6.Thermodynamic Properties

Fluids consist of liquids or gases. But gases are compressible fluids and hence thermodynamic properties play an important role. With the change of pressure and temperature, the gases undergo large variation in density. The relationship between pressure (absolute), specific volume and temperature (absolute) of a gas is given by the equation of state as

1.5.1. Dimension of R. the gas constant R depends upon the particular gas. The dimension of R is obtained from equation (1.5) as

1.5.2. Isothermal Process. If the changes in density occurs at constant temperature, the process is called isothermal and relationship between (p) and density (P) is given by

1.5.3. Adiabatic Process. If the change in density occurs with no heat exchange to and may the gas, the process is called adiabatic. And if no heat is generates within the gas due to friction, the relationship between pressure and density is given by

Where k = Ratio of specific heat of a gas at constant pressure and constant volume = 1.4 for air.

1.5.4. Universal Gas Constant. Let m = Mass of a gas in kg

V = Volume of gas of mass m p = Absolute pressure

T = Absolute temperature Then, we have

Where R = Gas constant.

Equation (1.8) can be made universal, i.e. applicable to all gases if it is expressed in mole-basis.

Let n = Number of moles in volume of a gas V = Volume of the gas

M= Mass of the gas molecules Mass of a hydrogen atom m = Mass of a gas in kg P

Substituting the value of m in equation (1.8), we get

The product M x R is called universal gas constant and is equal to

In MKS units and 8314/kg-mole K in SI units.

One kilogram mole is defined as the product of one kilogram mass of the gas and its molecular weight Problem 1.2 0.A gas weighs 16 N/m3 at 25oC and at an absolute pressure of 0.25 N/mm2. Determine the gas constant and density of the gas.

Sol. Given :

Weight density, w = 16 N/m2 Temperature, t = 25°C

T=273 +t=273 +25 =288o K

p = 0.25 N/mm2 (abs.) = 0.25 x l06 N/m2 = 25 x l04 N/m2 (i) Using relation w = pg, density is obtained as

(ii) Using equation (1.5)

Problem 1.8 .A cylinder of 0.6 m3 in volume contains air at 50°C and 0.3 Nlmm2 absolute pressure. The air is compressed to 0.3 m3. Find (i) pressure inside the cylinder assuming isothermal process and (ii) pressure and temperature assuming adiabatic process. Take k = 1.4.

Problem 1.9 Calculate the pressure exerted by 5 kg of nitrogen gas at a temperature of 10°C if the volume is 0.4 m3. Molecular weight of nitrogen is 28. Assume, ideal gas laws are applicable. Sol. Given : Mass of nitrogen =5kg Temperature, t = 10°C . . T = 273 + 10 = 283°K Volume of nitrogen, V = 0.4 m3 Molecular weight = 28

1.7.Compressibility

Compressibility is the reciprocal of the bulk modulus of elasticity, K which is defined as the ratio of compressive stress to volumetric strain. Consider a cylinder fitted with a piston as shown in Fig. 1.3

Fig. 1.3

Consider a cylinder fitted with a piston as shown in Fig. 1.3. Let ∀ = Volume of a gas enclosed in the cylinder

p = Pressure of gas when volume is ∀

Let the pressure is increased top + dp, the volume of gas decreases from ∀ to ∀ - d ∀ Then increase in pressure = dp kgf/m2

Decrease in volume =d ∀

Relationship between K and Pressure (p) for a Gas

The relationship between bulk modulus of elasticity (K) and pressure for a gas for two different

processes of compression are as :

(i) For Isothermal Process. Equation (1.6) gives the relationship between pressure (p) and density (P) of a gas as

Problem 1.10. Determine the bulk modulus of elasticity of a liquid, if the pressure of the liquid is Increased from 70 Nlcm2 to 130 Nlcm2 The volume of the liquid decreases by 0.15per cent

Sol. Given :

Initial pressure = 70 N/cm2 Final pressure = 130 N/cm2

Therefore dp = Increase in pressure = 130 - 70 = 60 N/cm2 Decrease in volume = 0.15%

Problem 1.11. What is the bulk modulus of elasticity of a liquid which is compressed in a cylinder from Compressibility is given by - -1 a volume of 0.0125 m3 at 80 N/m3 pressure to a volume of 0.0124 m3 at 150N/cm3 pressure.

1.8.Surface Tension and Capillarity

Surface tension is defined as the tensile force acting on the surface of a liquid in contact with a gas or on the surface between two immiscible liquids such that the contact surface behaves like a membrane under tension. The magnitude of this force per unit length of the free surface will have the same value as the surface energy per unit area. It is denoted by Greek letter a (called sigma). In MKS units, it is expressed as kgf/m while in SI units as N/m. The phenomenon of surface tension is explained by Fig. 1.4. Consider three molecules A, B, C of a liquid in a mass of liquid. The molecules A is attracted in all directions equally by the surrounding molecules of the liquid. Thus the resultant force acting on the molecule A is zero. But the molecule B, which is situated near the free surface, is acted upon by upward and downward forces which are unbalanced. Thus a net resultant force on molecule B is acting in the downward direction. The molecule C, situated on the free surface of liquid does experience a resultant downward force. All the molecules on the free surface experience a downward force. Thus the free surface of the liquid acts like a very thin film under tension of the surface of the liquid acts as though it is an elastic membrane under tension.

• Surface Tension on Liquid Droplet

Consider a small spherical droplet of a liquid of radius 'r'. On the entire surface of the droplet, the tensile force due to surface tension will be acting.

Let a = Surface tension of the liquid

p = Pressure intensity inside the droplet (in excess of the outside pressure intensity) d = Dia. of droplet.

• Surface Tension on a Hollow Bubble.

A hollow bubble like a soap bubble in air his two surfaces in contact with air, one inside and other outside. Thus two surfaces are subjected to surface tension. In such case, we bave

1.63. Surface Tension on a Jet. Consider a liquid jet of diameter 'd' and length 2' as shown in Fig. 1.5.

Let

p = Pressure intensity inside the liquid jet above the outside pressure a = Surface tension of the liquid.

Consider the equilibrium of the semi jet, we have Force due to pressure = p x area of semi jet

= p x L x d Force due to surface tension = a x 2L. Equating the forces, we have

Problem 1.12.

The surface tension of water in contact with air at 20°C is 0.0725 N/m. The pressure inside a droplet of water is to be 0.02 N/cm2 greater than the outside pressure. Calculate the diameter of the droplet of water.

Problem 1.13. The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm2 (atmospheric pressure). Calculate the pressure within the droplet if surface tension k given as 010725 N/m of water.

1.9.Capillarity

Capillarity is defined as a phenomenon of rise or fall of a liquid surface in a small tube relative to the adjacent general level of liquid when the tube is held vertically in the liquid. The rise of liquid surface is known as capillary rise while the fall of the liquid surface is known as capillary depression. It is expressed in terms of cm or mm of liquid. Its value depends upon the specific weight of the liquid, diameter of the tube and surface tension of the liquid.

Expression for Capillary Rise Consider a glass tube of small diameter ‘d’ opened at both ends and is inserted in a liquid, say water. The liquid will rise in the tube above the level of the liquid.

Let h = height of the liquid in the tube. Under a state of equilibrium, the weight of liquid of height h is balanced by the force at the surface of the liquid in the tube. But the force at the surface of the liquid in the tube is due to surface tension.

Fig.1.6 Capillary rise Let σ = Surface tension of liquid

Θ = Angle of contact between liquid and glass tube. The

weight of the liquid of height h in the tube

=(Area of the tube x h) x p x g

=π/4 d2 x h x p x g ……….(1.17) Where p = Density of liquid

The value of θ between water and clean glass tube is approximately equal to zero and hence cosθ is equal to unity of water is given by

Expression for Capillaty Fall. If the glass tube is dipped in mercury, the level of mercury in the tube will be lower than the general level of the outside liquid as shown in Fig. 1.6.

Let h = Height of depression in tube.

Then in equilibrium, two forces are acting on the mercury inside the tube. First one is due to surface tension acting in the downward direction and is equal to a x πd x cosθ

Second force is due to hydrostatic force acting upward and is equal to intensity of pressure at a depth 'h' x Area

Problem 1.14. Calculate the capillary rise in a glass tube of2.5 mm diameter when immersed vertically in (a) water and (b) mercury. Take surface tensions o = 0.0725 N/m for water and O=0.52N/m for mercury in contact with air. The specific gravity mercury is given as 13.6 and angle of contact = 130".

Problem 1.15. Calculate the capillary effect in millimeters in a glass tube of 4 mm diameter, when immersed in (i) water, and (ii) mercury. The temperature of the liquid is 20°C and the values of the surface tension of water and mercury at 20°C in contact with air are 0.073575 N/m and 0.51 N/m respectively. The angle of contact for water is zero that for mercury 1.30". Take density of water at 20°C as equal to 998 kg/m3.

Problem 1.16: The capillary rise in the glass tube is not to exceed 0.2 mm of water. Determine its minimum size, given that surface tension for water in contact with air = 0.0725 N/m.

1.10. Summary

In this unit we have studied − Properties of Fluids − Viscosity

− Types of Fluids.

− Thermodynamic Properties − Compressibility

− Surface Tension and Capillarity − Capillarity 1.11. Keywords Viscosity Kinematic Viscosity Ideal Fluid Compressibility Capillarity Real Fluid Newtonian Fluid Non-Newtonian Fluid

Ideal Plastic Fluid

1.12. Exercise

1. Define the following fluid properties Density, Weight density, Specific volume and specific gravity of a fluid.

2. Differentiate between : (i) liquids and gases, (ii) Real fluids and ideal fluids, (iii) Specific weight and specific volume of a fluid.

3. What is the difference between dynamic viscosity and kinematic viscosity? State their units of measurements.

4. Explain the terms: (i) dynamic viscosity and (ii) kinematic viscosity. Give their dimensions.

6. Enunciate Newton's law of viscosity. Explain the importance of viscosity in fluid motion. What is the effect of temperature on viscosity of water and that of air?

7. Define Newtonian and Non-Newtonian fluids.

8. What do you understand by terms : (i) Isothermal process, ( i i ) Adiabatic process, and (iii) Universal-gas constant.

9, Define compressibility. Prove that compressibility for a perfect gas undergoing isothermal compression is P while for a perfect gas undergoing isentropic compression is WP

10. Define surface tension. Prove that the relationship between surface tension and pressure inside a droplet of liquid in excess of outside pressure is given byp = 4σ/d

11. Explain the phenomenon of capillarity. Obtain an expression for capillary rise of a liquid. 12. Distinguish between ideal fluids and real fluids. Explain the importance of compressibility in fluid flow.

13. Define and explain Newton's law of viscosity. 14. Convert 1 kg/s-m dynamic viscosity in poise.

15. Why does the viscosity of a gas increases with the increase in temperature while that of a liquid deceases with increase in temperature?

NUMERICAL PROBLEMS

1. One liter of crude oil weighs 9.6 N. Calculate its specific weight, density and specific gravity.

2. The velocity distribution for flow over a Flat plate is given by u = 3/2.y-y3/2, where u is the point velocity in meter per second at a distance y meter above the plate. Determine the shear stress at y = 9 cm. Assume dynamic viscosity as 8 poise.

3. A plate, 0.025 mm distant From a fixed plate, moves at 50 cm/s and requires force of 1.471 N/m2 maintain this speed. Determine the fluid viscosity between the plates in the poise. 4. Determine the intensity of shear of an oil having viscosity = 1.2 poise and is used for lubrication in the clearance between a 10 cm diameter shaft and its journal bearing. The clearance is 1.0 mm and shaft rotates at 200 r.p.m.

5. Two plates 2 are placed at a distance of 0.15 mm apart. The lower plate is f i e d while the upper plate having surface area 1.0 m is pulled at 0.3 m/s. Find the force and power required to maintain this speed, if the fluid separating them is having viscosity 1.5 poise.

6. An oil film of thickness 1.5 mm is used for lubrication between a square plate of size 0.9m x 0.9 m and an inclined plane having an angle of inclination 20'. The weight of the square

plate is 392.4 N and it slides down the plane with a uniform velocity of 0.2 m/s. Find the dynamic viscosity of the oil.

7. In a stream of glycerin in motion, at a certain point the velocity gradient is 0.25 meter per sec per meter. The mass density of fluid is 1268.4 kg per cubic meter and kinematic viscosity is 6.30 x 104square meter per second. Calculate the shear stress at the point.

Unit 2

BERNOULLI EQUATION AND APPLICATIONS

Structure

2.1. Introduction 2.2. Objectives

2.3.Forms of Energy Encountered in Fluid Flow

2.4.Variation in the Relative Values of Various Forms of Energy During Flow 2.5.EULER’S Equation of Motion for Flow along a Stream Line

2.6.Bernoulli Equation for Fluid Flow

2.7.Energy Line and Hydraulic Gradient Line 2.8.Volume Flow through a Venturimeter

2.9.Euler and Bernoulli Equation for Flow with Friction

2.10. Concept and Measurement of Dynamic, Static and Total Head 2.11. Pitot Tube 2.12. Solved Problems 2.13. Summary 2.14. Keywords 2.15. Exercise 2.1.Introduction

In chapter five flow of ideal fluids was discussed. The main idea was the study of flow pattern. The determination of equal flow paths and equal potential lines was discussed. No attempt was made to determine the numerical value of these quantities. In this chapter the method of determination of the various energy levels at different locations in the flow is discussed. In this process first the various forms of energy in the fluid are identified. Applying the law of conservation of energy the velocity, pressure and potential at various locations in the flow are calculated. Initially the study is limited to ideal flow. However the modifications required to apply the analysis to real fluid flows are identified. The material discussed in this chapter are applicable to many real life fluid flow problems. The laws presented are the basis for the design of fluid flow systems.

Energy consideration in fluid flow:

Consider a small element of fluid in flow field. The energy in the element as it moves in the flow field is conserved. This principle of conservation of energy is used in the determination of flow parameters like pressure, velocity and potential energy at various locations in a flow. The concept is used in the analysis of flow of ideal as well as real fluids.

Energy can neither be created nor destroyed. It is possible that one form of energy is converted to another form. The total energy of a fluid element is thus conserved under usual flow conditions.

If a stream line is considered, it can be stated that the total energy of a fluid element at any location on the stream line has the same magnitude.

2.2.Objectives

After studying this unit we are able to understand − Forms of Energy Encountered in Fluid Flow

− Variation in the Relative Values of Various Forms of Energy During Flow − EULER’S Equation of Motion for Flow along a Stream Line

− Bernoulli Equation for Fluid Flow

− Energy Line and Hydraulic Gradient Line − Volume Flow through a Venturimeter

− Euler and Bernoulli Equation for Flow with Friction

− Concept and Measurement of Dynamic, Static and Total Head − Pitot Tube

2.3.Forms of Energy Encountered in Fluid Flow

Energy associated with a fluid element may exist in several forms. These are listed here and the method of calculation of their numerical values is also indicated.

Kinetic Energy

This is the energy due to the motion of the element as a whole. If the velocity is V, then the kinetic energy for m kg is given by

(…2.1.1) The unit in the SI system will be Nm also called Joule (J) {(kg m2/s2)/(kg m/N s2)}

The same referred to one kg (specific kinetic energy) can be obtained by dividing 2.1.1 by the mass m and then the unit will be Nm/kg.

….(2.1.1b)

In fluid flow studies, it is found desirable to express the energy as the head of fluid in m. This unit can be obtained by multiplying equation (6.1.1) by go/g.

..(2.1.2)

Apparently the unit appears as metre, but in reality it is Nm/N, where the denominator is weight of the fluid in N.

The equation in this form is used at several places particularly in flow of liquids. But the energy associated physically is given directly only be equation 2.1.1.

The learner should be familiar with both forms of the equation and should be able to choose and use the proper equation as the situation demands.

When different forms of the energy of a fluid element is summed up to obtain the total energy, all forms should be in the same unit.

Potential Energy

This energy is due to the position of the element in the gravitational field. While a zero value for KE is possible, the value of potential energy is relative to a chosen datum. The value of potential energy is given by

PE = mZ g/go Nm (2.1.3)

Where m is the mass of the element in kg, Z is the distance from the datum along the gravitational direction, in m. The unit will be (kg m m/s2) × (Ns2/kgm) i.e., Nm. The specific potential energy (per kg) is obtained by dividing equation 2.1.3 by the mass of the element. PE = Z g/g0 Nm/kg (2.1.3. b)

This gives the physical quantity of energy associated with 1 kg due to the position of the fluid element in the gravitational field above the datum. As in the case of the kinetic energy, the value of PE also is expressed as head of fluid, Z.

PE = Z (g/go) (go/g) = Z m.

This form will be used in equations, but as in the case of KE, one should be familiar with both the forms and choose the suitable form as the situation demands.

Pressure Energy (Also Equals Flow Energy)

The element when entering the control volume has to flow against the pressure at that location. The work done can be calculated referring Fig. 2.1.1.

Fig.2.1: Flow work calculation

The boundary of the element of fluid considered is shown by the dotted line, Force = P1 A, distance to be moved = L, work done = P1AL = P1 mv as AL = volume = mass × specific volume, v.

∴ flow work = P mv. The pressure energy per kg can be calculated using m = 1. The flow energy is given by

FE = P.v = P/ρ, Nm/kg

As in the other cases, the flow energy can also expressed as head of fluid.

(2.1.5) As specific weight γ = ρ g/go, the equation is written as,

(2.1.5b)

It is important that in any equation, when energy quantities are summed up consistent forms of these set of equations should be used, that is, all the terms should be expressed either as head of fluid or as energy (J) per kg. These are the three forms of energy encountered more often in flow of incompressible fluids.

Internal Energy

This is due to the thermal condition of the fluid. This form is encountered in compressible fluid flow. For gases (above a datum temperature) IE = cv T where T is the temperature above the datum temperature and cv is the specific heat of the gas at constant volume. The unit for internal energy is J/kg (Nm/kg). When friction is significant other forms of energy is converted to internal energy both in the case of compressible and incompressible flow.

Electrical and Magnetic Energy

These are not generally met with in the study of flow of fluids. However in magnetic pumps and in magneto hydrodynamic generators where plasma flow in encountered, electrical and magnetic energy should also be taken into account.

2.4.Variation in the Relative Values of Various Forms of Energy During Flow

Under ideal conditions of flow, if one observes the movement of a fluid element along a stream line, the sum of these forms of energy will be found to remain constant. However, there may be an increase or decrease of one form of energy while the energy in the other forms will decrease or increase by the same amount. For example when the level of the fluid decreases, it is possible that the kinetic energy increases. When a liquid from a tank flows through a tap this is what happens. In a diffuser, the velocity of fluid will decrease but the pressure will increase. In a venturimeter, the pressure at the minimum area of cross section (throat) will be the lowest while the velocity at this section will be the highest.

The total energy of the element will however remain constant. In case friction is present, a part of the energy will be converted to internal energy which should cause an increase in temperature. But the fraction is usually small and the resulting temperature change will be so small that it will be difficult for measurement. From the measurement of the other forms, it will be possible to estimate the frictional loss by difference.

2.5.EULER’S Equation of Motion for Flow along a Stream Line

Consider a small element along the stream line, the direction being designated as s.

The net force on the element are the body forces and surface forces (pressure). These are indicated in the figure. Summing this up, and equating to the change in momentum.

PdA – {P + (∂P/∂s} dA – ρg dA ds cos θ = ρ dA ds as (2.3.1) where as is the acceleration along the s direction. This reduces to,

(2.3.2)

(Note: It will be desirable to add go to the first term for dimensional homogeneity. As it is, the first term will have a unit of N/kg while the other two terms will have a unit of m/s2. Multiplying by go, it will also have a unit of m/s2).

and as cos θ = dz/ds, equation 6.3.2 reduces to,

(2.3.2 a) For steady flow ∂V/ ∂t = 0. Cancelling ∂s and using total derivatives in place of partials as these are independent quantities.

(2.3.3) (Note: in equation 6.3.3 also it is better to write the first term as go.dp/ρ for dimensional

homogeneity). This equation after dividing by g, is also written as,

(2.3.4) which means that the quantity within the bracket remains constant along the flow. This equation is known as Euler’s equation of motion. The assumptions involved are: 1. Steady flow

2. Motion along a stream line and 3. Ideal fluid (frictionless)

In the case on incompressible flow, this equation can be integrated to obtain Bernoulli equation.

2.6.Bernoulli Equation for Fluid Flow

Euler’s equation as given in 2.3.3 can be integrated directly if the flow is assumed to be incompressible.

2.4.1

The constant is to be evaluated by using specified boundary conditions. The unit of the terms will be energy unit (Nm/kg).

In SI units the numerical value of go = 1, kg m/N s2. Equation 2.4.1 can also be written as to express energy as head of fluid column.

(2.4.2)

(γ is the specific weight N/m3). In this equation all the terms are in the unit of head of the fluid. The constant has the same value along a stream line or a stream tube. The first term represents (flow work) pressure energy, the second term the potential energy and the third term the kinetic energy. This equation is extensively used in practical design to estimate pressure/velocity in flow through ducts, venturimeter, nozzle meter, orifice meter etc. In case energy is added or taken out at any point in the flow, or loss of head due to friction occurs, the equations will read as,

(2.4.3)

where W is the energy added and hf is the loss of head due to friction. In calculations using SI system of units go may be omitting as its value is unity.

Example.1 A liquid of specific gravity 1.3 flows in a pipe at a rate of 800 l/s, from point 1 to point 2 which is 1 m above point 1. The diameters at section 1 and 2 are 0.6 m and 0.3 m respectively. If the pressure at section 1 is 10 bar, determine the pressure at section 2. Using Bernoulli equation in the following form (2.4.2)

Taking the datum as section 1, the pressure P2 can be calculated. V1 = 0.8 × 4/π × 0.62 = 2.83 m/s, V2 = 0.8 × 4/π × 0.33 = 11.32 m/s P1 = 10 × 105 N/m2, γ = sp. gravity × 9810. Substituting.

Solving, P2 = 9.092 bar (9.092 × 105 N/m2).

As P/γ is involved directly on both sides, gauge pressure or absolute pressure can be used without error. However, it is desirable to use absolute pressure to avoid negative pressure values (or use of the term vacuum pressure).

Example 2 Water flows through a horizontal venturimeter with diameters of 0.6 m and 0.2 m. The guage pressure at the entry is 1 bar. Determine the flow rate when the throat pressure is 0.5 bar (vacuum). Barometric pressure is 1 bar.

Using Bernoulli’s equation in the form,

and noting Z1 = Z2, P1 = 2 × 105 N/m2 (absolute) P2 = 0.5 ×105 N/m2 (absolute), γ = 9810 N/m3

V1 = Q × 4/(π × 0.602) = 3.54 Q, V2 = Q × 4/(π × 0.202) = 31.83Q

Solving, Q = 0.548 m3/s, V1 = 1.94 m/s, V2 = 17.43 m/s.

2.7.Energy Line and Hydraulic Gradient Line

The total energy plotted along the flow to some specified scale gives the energy line. When losses (frictional) are negligible, the energy line will be horizontal or parallel to the flow direction. For calculating the total energy kinetic, potential and flow (pressure) energy are considered.

Energy line is the plot of P/γ+ Z + V2g2 along the flow. It is constant along the flow when losses are negligible.

The plot of P/γ+ Z along the flow is called the hydraulic gradient line. When velocity increases this will dip and when velocity decreases this will rise. An example of plot of these lines for flow from a tank through a venturimeter is shown in Fig. 2.3

The hydraulic gradient line provides useful information about pressure variations (static head) in a flow. The difference between the energy line and hydraulic gradient line gives the value of dynamic head (velocity head).

Fig.: 2.3 Energy and hydraulic gradient lines 2.8.Volume Flow through a Venturimeter

Example Under ideal conditions show that the volume flow through a venturimeter is given by

This is a general expression and can be used irrespective of the flow direction, inclination from horizontal or vertical position. This equation is applicable for orifice meters and nozzle flow meters also. In numerical work consistent units should be used.

Pressure should be in N/m2, Z in m, A in m2 and then volume flow will be m3/s. A coefficient is involved in actual meters due to friction.

Fig 2.4

2.9.Euler and Bernoulli Equation for Flow with Friction

Compared to ideal flow the additional force that will be involved will be the shear force acting on the surface of the element. Let the shear stress be τ, the force will equal τ 2πr ds (where r is the radius of the element, and A = π r2)

Refer Para 2.3 and Fig. 2.3.1. The Euler equation 2.3.3 will now read as

ds can also be substituted in terms of Z and θ

Bernoulli equation will now read as (taking s as the length)

The last term is the loss of head due to friction and is denoted often as hL hf in head of fluid in metre height (check for the unit of the last term).

Example A vertical pipe of diameter of 30 cm carrying water is reduced to a diameter of 15cm. The transition piece length is 6 m. The pressure at the bottom is 200 kPa and at the top it is 80 kPa. If frictional drop is 2 m of water head, determine the rate of flow.

2.10. Concept and Measurement of Dynamic, Static and Total Head

In the Bernoulli equation, the pressure term is known as static head. It is to be measured by a probe which will be perpendicular to the velocity direction. Such a probe is called static probe. The head measured is also called Piezometric head. (Figure 2.5 (a))

The velocity term in the Bernoulli equation is known as dynamic head. It is measured by a probe, one end of which should face the velocity direction and connected to one limb of a manometer with other end perpendicular to the velocity and connected to the other limb of the manometer. (Figure 2.5 (b))

The total head is the sum of the static and dynamic head and is measured by a single probe facing the flow direction. (Figure 2.5 (c))

The location of probes and values of pressures for the above measurements are shown in Fig. 2.5.

Fig.2.5: Pressure Measurement 2.11. Pitot Tube

The flow velocity can be determined by measuring the dynamic head using a device known as pitot static tube as shown in Fig. 2.6. The holes on the outer wall of the probe provides the static pressure (perpendicular to flow) and hole in the tube tip facing the stream direction of

flow measures the total pressure. The difference gives the dynamic pressure as indicated by the manometer. The head will be h (s – 1) of water when a differential manometer is used (s > 1).

The velocity variation along the radius in a duct can be conveniently measured by this arrangement by traversing the probe across the section. This instrument is also called pitot– static tube.

Fig. 2.6: Pitot-Static tube

Example : The dynamic head of a water jet stream is measured as 0.9 m of mercury column. Determine the height to which the jet will rise when it is directed vertically upwards. Considering the location at which the dynamic head is measured as the datum and converting the column of mercury into head of water, and noting that at the maximum point the velocity is zero,

0.9 × 13.6 + 0 + 0 = 0 + 0 + Z ∴ Z = 12.24 m

Note. If the head measured is given as the reading of a differential manometer, then the head should be calculated as 0.9 (13.6 – 1) m.

Example A diverging tube connected to the outlet of a reaction turbine (fully flowing) is called ‘‘Draft tube’’. The diverging section is immersed in the tail race water and this provides additional head for the turbine by providing a pressure lower than the atmospheric pressure at the turbine exit. If the turbine outlet is open the exit pressure will be atmospheric as in Pelton wheel. In a draft tube as shown in Fig. Ex. 2.7, calculate the additional head provided by the draft tube. The inlet diameter is 0.5 m and the flow velocity is 8 m/s. The outlet diameter is 1.2 m. The height of the inlet above the water level is 3 m. Also calculate the pressure at the inlet section.

Fig. Ex.2.7 Draft tube Considering sections 1 and 2

Considering tail race level, 2 as the datum, and calculating the velocities

Additional head provided due to the use of draft tube will equal 6.16 m of water

Note: This may cause cavitation if the pressure is below the vapor pressure at the temperature condition. Though theoretically the pressure at turbine exit can be reduced to a low level, cavitation problem limits the design pressure.

2.12. Solved Problems

Problem 6.1 A venturimeter is used to measure the volume flow. The pressure head is recorded by a manometer. When connected to a horizontal pipe the manometer reading was h cm. If the reading of the manometer is the same when it is connected to a vertical pipe with flow upwards and (ii) vertical pipe with flow downwards, discuss in which case the flow is highest.

Consider equation 6.6.2

As long as ‘h’ remains the same, the volume flow is the same for a given venturimeter as this expression is a general one derived without taking any particular inclination. This is because

of the fact that the manometer automatically takes the inclination into account in indicating the value of (Z1 – Z2).

Problem 6.2 Water flows at the rate of 600 l/s through a horizontal venturi with diameter 0.5 m and 0.245 m. The pressure gauge fitted at the entry to the venturi reads 2 bar. Determine the throat pressure. Barometric pressure is 1 bar.

Using Bernoulli equation and neglecting losses

Problem 6.3 A venturimeter as shown in Fig P. 6.3 is used measure flow of petrol with a specific gravity of 0.8. The manometer reads 10 cm of mercury of specific gravity 13.6. Determine the flow rate.

= 4.245 × 10–3 m3/s or 15.282 m3/hr or 4.245 l/s or 15282 l/hr or 3.396 kg/s Problem 6.4 A liquid with specific gravity 0.8 flows at the rate of 3 l/s through a

venturimeter of diameters 6 cm and 4 cm. If the manometer fluid is mercury (sp. gr = 13.b) determine the value of manometer reading, h.

Using equation (6.6.2)

2.13. Summary In this unit we have studied

− Forms of Energy Encountered in Fluid Flow

− Variation in the Relative Values of Various Forms of Energy During Flow − EULER’S Equation of Motion for Flow along a Stream Line

− Bernoulli Equation for Fluid Flow

− Energy Line and Hydraulic Gradient Line − Volume Flow through a Venturimeter

− Euler and Bernoulli Equation for Flow with Friction

− Pitot Tube 2.14. Keywords Kinetic Energy Potential Energy Internal Energy Pitot Tube 2.1.Exercise

Q. 2.1. Fill in the blanks:

1. Kinetic energy of fluid element is due to its –––––––––––––––––.

2. The amount of kinetic energy per kg is given by the expression ––––––––––––––––– the unit used being head of fluid.

3. The kinetic energy in the unit Nm/kg is given by the expression –––––––––––––––––. 4. Potential energy of a fluid element is due to its –––––––––––––––––.

5. Potential energy of a fluid element in head of fluid is given by –––––––––––––––––. 6. Potential energy of a fluid element in Nm/kg is given by –––––––––––––––––. 7. Pressure energy or flow energy of a fluid element is given in head of fluid by the expression –––––––––––––––––.

8. Pressure energy or flow energy of a fluid element in the unit Nm/kg is given by the expression –––

9. Internal energy is due to ––––––––––––––––.

10. In the analysis of incompressible fluid flow, internal energy is rarely considered because –––––––.

11. Electrical and magnetic energy become important in the flow of –––––––––––––––––. Q. 2.2. Fill in the blanks:

1. Eulers equation is applicable for flow along a –––––––––––––––––. 2. Bernoulli equation is applicable for flows which are –––––––––––––––––. 3. Bernoulli equation states that the total head –––––––––––––––––.

4. Total head in a steady incompressible irrotational flow is the sum of –––––––––––––––––. 5. In steady flow along a horizontal level as the velocity increases the pressure –––––––––. 6. The pressure along the diverging section of a venturi –––––––––––––––––.

7. Cavitation will occur when the pressure at a point –––––––––––––––––.

8. Draft tube ––––––––––––––––– the available head in the case of reaction turbines. 9. Energy line along the flow ––––––––––––––––– if there are no losses.

10. Hydraulic grade line represents the sum of ––––––––––––––––– along the flow. 11. If a pump supplies energy to the flow the energy line ––––––––––––––––– . 12. If there are frictional losses the energy grade line will ––––––––––––––––– Q. 2.3. Indicate whether the statement is correct or incorrect.

1. Energy line along the direction of flow will dip if there are losses.

2. When a pump supplies energy to a flow stream, the energy line will decreases by a step. 3. For ideal flows the energy line will slope upward along the flow.

4. If velocity increases, the hydraulic grade line will dip along the flow direction.

5. If the differential manometer reading connected to a venturimeter is the same, the flow will be independent of the position or flow direction.

6. For the same reading of the differential manometer connected to a vertical venturimeter, the flow rate will be larger if flow is downwards.

7. A pitot probe connected perpendicular to flow will indicate the total head. 8. A pitot probe facing the flow will indicate the dynamic head.

9. A pitot probe facing the flow will indicate the total head.

10. A pitot-static tube has probes both facing the flow and perpendicular to flow. 11. Flow will take place along hydraulic gradient.

12. Flow will take place along energy gradient. Q. 2.4 Exercise Problems

2.1. A pipe inclined at 45° to the horizontal converges from 0.2 m dia to 0.1 m at the top over a length of 2 m. At the lower end the average velocity is 2m/s. Oil of specific gravity 0.84 flows through the pipe. Determine the pressure difference between the ends, neglecting losses. If a mercury manometer (specific gravity 13.6) is used to measure the pressure, determine the reading of the manometer difference in m of mercury. Oil fills the limbs over mercury in the manometer. (36.854 N/m2, 0.201m)

2.2. Oil of specific gravity of 0.9 flows through a venturimeter of diameters 0.4 and 0.2 m. A Utube mercury manometer shows a head 0.63 m. Calculate the flow rate. (0.105 m3/s)

2.3. Water flows from a reservoir 240 m above the tip of a nozzle. The velocity at the nozzle outlet is 66 m/s. The flow rate is 0.13 m3/s. Calculate (1) the power of the jet. (2) the loss in head due to friction. (283.14 kW, 17.98 m)

2.4. Water flows in the middle floor tap at 3 m/s. Determine the velocities at the taps in the other two floors shown in Fig. E. 2.4.

2.5. Oil flows through a horizontal pipe will line which has a diameter of 0.45 m at the start. After some distance the diameter of reduces to 0.3 m at which point the flow divides into pipes of 0.15 m and 0.225 m diameter. The velocity at the beginning is 1.8 m/s. The velocity in the pipe line of 0.225 m dia is 3.6 m/s. If the pressure at the start is 20 m head of oil and the specific gravity of the oil is 0.91 determine the pressure at the fork and also at the end of the two branch pipes. Neglect losses.

(V at fork = 4.05 m/s, V0.15 = 8.1 m/s, Pfork = 13.33 m, P0.15 = 16.821 m, P0.225 = 19.5 m)

2.6. A nozzle of 25 mm dia. directs a water jet vertically with a velocity of 12 m/s. Determine the diameter of the jet and the velocity at a height of 6 m. (38.25 mm, 5.13 m/s)

2.7. A pipe line is 36 m above datum. The pressure and velocity at a section are 410 kN/m2 and 4.8 m/s. Determine the total energy per kg with reference to the datum. (774.7 Nm.kg)

2.8. The supply head to a water nozzle is 30 m gauge. The velocity of water leaving the nozzle is 22.5 m/s. Determine the efficiency and power that can be developed if the nozzle diameter is 75mm. (84.3%, 25.2 kW)

2.9. The suction pipe of a pump slopes at 1 m vertical for 5 m length. If the flow velocity in the pipe is 1.8 m/s and if the pressure in the pipe should not fall by more than 7 m of water, determine the maximum length. (35.8 m)

2.10. The pressure at the entry to the pipe line of 0.15 m dia. is 8.2 bar and the flow rate at this section is 7.5 m3/min. The pipe diameter gradually increases to 0.3 m and the levels rises by 3 m above the entrance. Determine the pressure at the location. Neglect losses. (8.14 bar)

2.11. A tapering pipe is laid at a gradient of 1 in 100 downwards. The length is 300 m. The diameter reduces from 1.2 m to 0.6 m. The flow rate of water is 5500 l/min. The pressure at the upper location is 0.8 bar. Determine the pressure at the lower location. (0.73 bar)

2.12. A horizontal pipe carrying water is gradually tapering. At one section the diameter is 150 mm and flow velocity is 1.5 m/s. (i) If the drop in pressure is 1.104 bar at a reduced

section determine the diameter at the section. (ii) If the drop in pressure is 5 kN/m2, what will be the diameter? Neglect losses. (47.6 mm, 100 mm)

2.13. The diameter of a water jet at nozzle exist is 75 mm. If the diameter at a height of 12 m is 98.7mm, when the jet is directed vertically, determine the height to which the jet will rise. (18 m)

2.14. Calculate the height to which the jet, issuing at 18.8 m/s will rise when (i) The jet is directed vertically (ii) when it is directed at 45°. Also find the horizontal distance travelled in this case. (18 m, 9 m, 9 m)

2.15. A jet directed at 30° reaches a maximum height of 3 m at a horizontal distance of 18 m. Determine the issuing velocity of the jet. (16.9 m/s)

2.16. Determine the flow rate of a fluid of specific gravity 0.83 upward in the set up as shown in Fig. E. 2.16.

Fig.2.16

2.17. Determine the flow rate and also the pressure at point 2 in the siphon shown in Fig. E. 2.17. Diameter of the pipe is 2.5 cm. (4.2 l/s, 45.1 kPa ab)

Fig. E.2.17 Fig. E.2.18

2.19. Derive an expression for the variation of jet radius r with distance y downwards for a jet directed downwards. The initial radius is R and the head of fluid is H.

2.20. For the venturimeter shown in Fig. E. 2.24, determine the flow rate of water.

2.21. A horizontal pipe divides into two pipes at angles as shown in Fig. E. 2.25. Determine the necessary forces along and perpendicular to the pipe to hold it in place. Assume that these are no losses.

Unit 3

EQUILIBRIUM OF FLOATING BODIES

Structure 3.1.Introduction 3.2.Objectives 3.3.Buoyancy 3.4.Center of Buoyancy 3.5.Meta-Center 3.6.Meta-Centeric Height

3.7.Analytical Method for Meta-Centre Height

3.8.Conditions of Equilibrium of a Floating and Sub-Merged Bodies 3.9.Experimental Method of Determination of Metacentric Height 3.10. Oscillation (Rolling) of a Floating Body

3.11. Summary 3.12. Keywords 3.13. Exercise

3.1. Introduction

In this chapter, the equilibrium of the floating and sub-merged bodies will be considered. Thus the chapter will include : 1. Buoyancy, 2. Center of buoyancy, 3. Metacentre, 4. Metacentric height, 5. Analytical method for determining metacentric height, 6. Conditions of equilibrium of a floating and sub-merged body, and 7. Experimental method for metacentric height.

3.2. Objectives

− Buoyancy

− Center of Buoyancy − Meta-Center

− Meta-Centeric Height

− Analytical Method for Meta-Centre Height

− Conditions of Equilibrium of a Floating and Sub-Merged Bodies − Experimental Method of Determination of Metacentric Height − Oscillation (Rolling) of a Floating Body

3.3. Buoyancy

When a body is immersed in a fluid, an upward force is exerted by the fluid on the body. This upward force is equal to the weight of the fluid displaced by the body and is called the force of buoyancy or simply buoyancy.

3.4. Center of Buoyancy

It is defined as the point, through which the force of buoyancy is supposed to act. As the force of buoyancy is a vertical force and is equal to the weight of the fluid displacement by the body, the center of the buoyancy will be the centre of gravity of the fluid displaced.

Problem 1

Find the volume of the water displaced and position of center of buoyancy for a wooden block of width 2.5mand of depth 1.5m, when it floats horizontally in water. The density of wooden block is 65kg/m and its length 6.0m.

Problem 2:

A wooden log of 0.6m diameter and 5m length is floating in river water. Find the depth of the wodden log in water when the sp. Gravity of the log is 0.7.

Problem 3:

A float valve regulates the flow of oil of sp. gr. 0.8 into a cistern. The spherical float is 15 cm in diameter. AOB is a weightless link carrying the float at one end, and a valve at the other end which closes the pipe through which oil flows into the cistern. The link is mounted in a frictionless hinge at O and the angle AOB is 135". The length of OA is 20 cm, and the distance between the centre of the float and hinge is 50 cm. When the flow is stopped A0 will be vertical. The valve is to be pressed on to the seat with a force of 9.81 N to completely stop the flow of oil into the cistern. It was observed that the flow of oil is stopped when the free surface of oil in the cistern is 35 cm below the hinge. Determine the weight of the float.

3.5. Meta-Center

It is defined as the point about which a body stark oscillating when the body is tilted by a small angle. The meta-centre may also be 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.

Consider a body float& in a liquid as shown in Fig. 3.1 (a). Let the body is in equilibrium and G is the centre of gravity and B the centre of buoyancy. For equilibrium, both the points lie on the normal axis, which is vertical.

Let the body is given a small angular displacement in the clockwise direction as shown in Fig. 3.1 (b). The centre of buoyancy, which is the centre of gravity of the displaced liquid or centre of gravity of the portion

Fig. 3.1 Meta-cenre

of the body sub-merged in liquid, will now be shifted towards right from the normal axis. Let it is at B1 as shown in Fig. 3.1 (b). The line of action of the force of buoyancy in this new

position, will intersect the normal axis of the body at some point say M. This point M is called Meta-centre.

3.6. Meta-Centeric Height

The distance MG, i.e., the distance between the meta-centre of a floating body and the centre of gravity of the body is called meta-centric height.

3.7. Analytical Method for Meta-Centre Height

Fig. 3.2 (a) shows the position of a floating body in equilibrium. The location of centre of gravity and centre of 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 Fig. 3.2 (b). The new centre of buoyancy is a B1. The vertical line through Bl cuts the normal axis at M. Hence M is the meta-centre and GM is meta-centric height.

Fig. 3.2: Meta-centre height of a floating body

The angular displacement of the body in the clockwise direction causes the wedge-shaped prism BOB’1 on the right of the axis to go inside the water while the identical wedge-shaped prism represented by AOA' emerges out of the water on the left of the axis. These wedges represent a gain in buoyant force on the right-side and a corresponding loss of buoyant force on the left side. The gain is represented by a vertical force df1 acting through the C.G. of the prism BOB’ while the loss is represented by an equal and opposite force dFB acting vertically downward through the centroid of AOA'. The couple due to these buoyant forces dFB, tends to rotate the ship in the counter clockwise direction. Also the moment caused by the

displacement of the centre of buoyancy from B to Bl is also counter clockwise direction. Thus these two couples must be equal.

Couple Due to Wedges. Consider towards the right of the axis a small strip of thickness dx & at a distance x from 0 as shown in Fig. 3.2. The height of strip x x LBOBr = x x 0.

Problem 1

A uniform body of size3 m long x2 m wide x1 m deep floats in water. What is the weight of the body if depth of immersion is 0.8 m? Determine the meta-centric height also.

Problem 2

A block of wood of specific gravity 0.7 floats in water. Determine the meta-centric height of the block if its size is 2m x Im x 0.8m.

The meta-centric height is given by equation (4.4) or

Problem 3

A body has the cylindrical upper portion of 3 m diameter and 1.8 m deep. The lower portion is a curved one, which displaces a volume of 0.6 m3 of water. The centre of buoyancy bf the curved portion is at distance of 1.95 m below the top of the cylinder. The centre of gravity of the whole body is 1.20 m below the top of the cylinder. The total displacement of water is 3.9 tonnes. find themeta-centric height of the body.

Sol. Given :

Dia. of body = 3.0m Depth of body = 1.8 m

Volume displaced by curved portion = 0.6 m3 of water.

Let B1 is the centre of buoyancy of the curved surface and G is center of gravity of the whole body

Then CB1 = 1 .95 m CG = 1.20 m

Total weight of water displaced by-body = 3.9 tonnes = 3.9 x 1000 = 3900 kgf

= 3900 x 9.81 N = 38259 N Find meta-centric height of the body.

Let the height of the body above the water surface x m. Total weight of water displaced by body

= Weight density of water x [Volume of water displaced] = 1000 x 9.81 x [Volume of the body in water]

3.8. Conditions of Equilibrium of a Floating and Sub-Merged Bodies

A sub-merged or a floating body is said to be stable if it comes back to its original position after a slight disturbance. The relative position of the centre of gravity (G) and centre of buoyancy (83 of a body determines the stability of a sub-merged body.

Stability of a Sub-merged Body

The position of centre of gravity and centre of buoyancy in case of a completely sub-merged body an fixed. Consider a balloon, which is completely sub-merged in air. k t the lower portion of the balloon contains heavier material, so that its centre of gravity is lower than its

centre of buoyancy as shown in Fig. 3.4 (a). Let the weight of the balloon is W. The weight W is acting through G1 vertically in the downward direction, while the budyant force FB i s acting vertically up, through B For the equilibrium of the balloon W = FB. If the balloon is given an angular displacement in the clockwise direction as shown in Fig. 3.4 (a), then Wand FB constitute a couple acting in the anti-clockwise direction and brings the balloon in the original position, Thus the balloon in the position, shown by Fig. 3.4 (a) is in stable equilibrium.

Fig. 3.4: Stabilities of sub-merged bodies

(a) Stable Equilibrium When W = FB and point B is above G, the body is said to be in stable equilibrium.

(b) Unstable Equilibrium. If W = FB, but the centre of buoyancy (B) is below centre of gravity (G), the body isinunstableequilibrium as shown in Fig. 3.4 (b). A slight displacement to the body, in the clockwise direction, gives the couple due to W and FB also in the clockwise direction. Thus the body does not return to its original position and hence the body is in unstable equilibrium. (c) Neutral Equilibrium. If FB = Wand B and G are at the same point, as shown in Fig. 3.4 (c), the body is said to be in Neutral Equilibrium.

Stability of Floating Body

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

(a) Stable Equilibrium. If the point M is above G, the floating body will be in stable equilibrium as shown in Fig. 3.5 (a). If a slight angular displacement is given to the floating body in the clockwise direction, the centre of buoyancy shifts from B to B1 such that the vertical line through B1 cuts at M. Then the buoyant force FB through B1 and weight W

through G constitute a couple acting in the anti-clockwise direction and I thus bunging the floating body in the original position.

Fig. 3.5: Stability of floating bodies

(b) Unstable Equilibrium. If the point M is below G, the floating body will be in unstable equilibrium j as shown in Fig. 3.5 (b). The disturbing couple is acting in the clockwise direction. The couple due to buoyant force FB and W is also acting in the clockwise direction and thus overturning the floating body

(c) Neutral Equilibrium. If the point M is at the centre of gravity of the body, the floating body will be in neutral equilibrium.

Problem 4

Solid cylinder of diameter 4.0 m has a height of 4.0 m. Find the meta-centric height I I 11 of the cylinder if the specific gravity of the material of cylinder = 0.6 and if is floating in water with its axis vertical. State whether the equilibrium is stable or unstable.

- ve sign means that the meta-centre (M) is below the centre of gravity (G). Thus the cylinder is in unstable equilibrium. Ans.

Problem

Solid cylinder of 10cm diameter and 40cm long, consist of two parts made of different The first part at the base is 1.0 cm long and of specific gravity = 6.0. The other part of the cylinder is made of the material having specific gravity 0.6. State, if it can float vertically in water. Sol. Given : D= 1Ocm,

Length, L=40cm,

Length of 1st part, L1 = 1.0 cm

Density of 1st part, pl = 6 x 1000 = 6000 kg/m3

Length of 2nd part, 12= 40- 1.0 = 39.0 cm

sp. gr., S2 = 0.6

Density of 2nd part, p2 = 0.6 x 1000 = 600 kg/m3

The cylinder will float vertically in water if its meta-centric height GM is positive. To find meta-centric height, find the location of centre of gravity (G) 1.0 and centre of buoyancy (B) of the combined solid cylinder. The distance of the centre of gravity of the solid cylinder from A is given as

AG = [(Weight of 1st part x Distance of C.G. of 1st part from A)

+ (Weight of 2nd part of cylinder x Distance of C.G. of 2nd part from A)]

Problem

A rectangular pontoon 10.0 m long, 7 m broad and i.5 m deep weighs 686.7 kN. It carries on its upper deek an empty boiler of 5.0 m diameter weighing 588.6 kN. The centre of gravity of the boiler and the pontoon are at their respective centers along a vertical line. Find the meta-centric height. Weight density of sea water is 10.104 kN/m3.

Sol. Given : Dimension of pontoon = 10 x 7 x 2.5

Weight of pontoon, Wl = 686.7 kN

Dia. of boiler, D = 5.0m I

Weight of boiler, W2 = 588.6 kN

w for sea water = 10.104 kN/m3

To find the meta-centric height, first determine the common centre of gravity G and .common centre of buoyancy B of the boiler and i A pontoon. Let GI and G2 are the centre of gravities of pontoon and boiler respectively. Then

Let h is the depth of immersion. Then

Total weight of pontoon and boiler = Weight of sea water displaced

or (686.7 + 588.6) = w x Volume of the pontoon in water

= 10.104 x L x b x Depth of immersion

. . 1275.3 = 10.104 x 10 x 7 x h

Therefore The distance of the common centre of buoyancy B from A is

Meta-centric height is given by GM = I/V - BG

= Volume of the body in water

=L*b*h =10.0 *7*1.857

Therefore Meta-centric height of both the pontoon and boiler = 0.12 m. Ans.

Problem

Show that a cylindrical buoy of 1 m diameter and 2.0 m height weighing 7.848 W will j not float vertically in sea water of dens@ 1030 kg/m3. Find the force necessary in a vertical chain attached t at the centre of base of the buoy that will keep it vertical.

As the meta-centric height is –ve, the point M lies below G and hence the cylinder will be in unstable equilibrium and hence cylinder will not float vertically.

Part II. Let the force applied in a vertical chain attached at the centre of the base of the buoy is T to keep the buoy vertical.

Now find the combined position of centre of gravity (G') and centre of buoyancy (B'). For the combined centre of buoyancy, let h' = depth of immersion when the force T is applied.

Then Total downward force = Weight of water displaced

Or (7848 + T ) = Density of water x g x Volume of cylinder in water

The combined centre of gravity (G') due to weight of cylinder and due to tension T in the chain from A is