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ENG 213: Fundamentals of Fluid Mechanics

Learning Outcomes

• Explain the properties of fluids.
• Determine force in static fluids and fluids in motion.
• Determine stability of floating bodies.
• Determine the effects of various components (valves, orifices, bends, and elbows) on fluid flow in pipes.
• Measure flow parameters with venturi meters, weirs, and other meters.
• Perform calculations based on principles of mass, momentum, and energy conservation.
• Perform dimensional analyses and simple fluid modeling problems.
• Specify the type of pumps and turbines for engineering applications.

Course Outline

• Fluid Properties
• Hydrostatics
• Fluid dynamics using principles of mass, momentum, and energy conservation from a control volume approach.
• Flow measurements in pipes
• Dimensional analysis and similitude
• 2-dimensional flows
• Hydropower systems

Chapter 1: Fluid Properties

(1.0) Introduction to Fluid Mechanics

• Fluid mechanics is the discipline within applied mechanics that is concerned with the behavior of liquids and gases at rest or in motion.
• It covers a wide variety of phenomena that occur in nature, biology, and numerous engineered or manufactured situations.
• Fluids are involved in almost every aspect of our lives directly or indirectly.

Characteristics of Fluids

• Solid: Densely spaced molecules with large intermolecular cohesive forces that allow the solid to maintain its shape and not be easily deformed.
• Liquid: Molecules are spaced farther apart with smaller intermolecular cohesive forces than solids. Molecules have more freedom of movement. Liquids can be easily deformed but not easily compressed.
• Gas: Greater molecular spacing and freedom of motions with negligible cohesive intermolecular forces. Easily deformed and compressed and will completely fill the volume of any container. Liquids and gases are fluids.

Definition of a Fluid

• A fluid is defined as a substance that deforms continuously when acted upon by a shearing stress.
• Stress (force per unit area) is created whenever a tangential force acts on a surface.

Properties of Fluids

2.1 Density

• Density (\rho) of a fluid is the mass of fluid contained within a unit volume.
• Formula: \rho = \frac{m}{V} where:
  • \rho = density,
  • m = mass (kg),
  • V = volume (m^3).
• Units: kg/m³

2.2 Specific Weight

• Specific weight (\gamma) is the weight of fluid per unit volume.
• Formula: \gamma = \rho g where:
  • \gamma = specific weight (N/m³),
  • \rho = density (kg/m³),
  • g = acceleration due to gravity (approximately 9.81 m/s²).

2.3 Specific Gravity

• Specific gravity (S) of a fluid is the ratio of its specific weight to the specific weight of water.
• Formula: S = \frac{\gamma{\text{fluid}}}{\gamma{\text{water}}} = \frac{\rho{\text{fluid}}}{\rho{\text{water}}}

Course Outline: Fundamental Notions and Definitions

• Continuum property
• Density
• Pressure
• Specific Volume
• Viscosity
• Compressibility

Fluid Statics

• Hydrostatic Forces and Submerged Surfaces
• Pressure Variations in Static Incompressible fluids
• Floatation
• Stability Considerations of floating bodies

Fluid Dynamics

1.2.2 Density

(i) Mass Density

• Mass per unit volume.
• Represented by the symbol \rho (Greek letter rho).
• SI unit: kg/m³ (i.e., \rho = \frac{m}{V}).
• fps absolute unit: lb-mass/ft³
• technical unit: slug-mass/ft³

(ii) Weight Density

• Also known as specific weight and is the weight per unit volume at the standard temperature and pressure.
• w = \rho g
• SI unit: N/m³
• fps absolute unit: poundal/ft³
• technical unit: lb-wt/ft³

(iii) Specific Gravity

• Specific gravity or relative density, S, is the ratio of the weight of a substance to the weight of an equal volume of water at 4°C.
• S = \frac{w{\text{substance}}}{w{\text{water}}}

(iv) Specific Volume

• The reciprocal of density is what is referred to as the specific volume.
• Thus, V_s = \frac{1}{\rho}. SI unit: m³/kg.

1.2.3 Surface Tension (\tau, Greek Sigma)

• Within the body of a liquid, a molecule is attracted equally in all directions by other molecules surrounding it, but at the surface between liquid and air, the upward and downward attractions are unbalanced.
• Surface tension is caused by the force of cohesion at the surface.
• The liquid behaves as if it were an elastic membrane under tension.
• This surface tension is the same at every point on the surface.
• Surface tension is not affected by the curvature of the surface, and it is constant at a given temperature for the surface of separation of two particular substances.
• Increase of temperature causes a decrease of surface tension, which causes drops of liquid to tend to take a spherical shape and is responsible for capillary action.
• Capillary action: a liquid to rise in a fine tube when its lower end is inverted in a liquid which wets the tube. If the liquid does not wet the tube, it will be depressed in the fine tube below the free surface outside.

1.2.4 Compressibility

• For liquids, the relationship between change of pressure and change of volume is given by the bulk modulus, K.

• Bulk Modulus, K = \frac{\text{Change in pressure intensity}}{\text{Volumetric strain}} = \frac{\text{Change in pressure intensity}}{\frac{\text{Change in Volume}}{\text{Original Volume}}}

• The relation between pressure and volume for a gas can be found from the gas laws.

• For all perfect gases, PV = RT where:

  • P = absolute pressure,
  • V_s = Specific Volume = \frac{1}{\rho},
  • T = absolute temperature,
  • R = gas constant.

• If changes occur isothermally:

  • PV = \text{Constant} (at constant temperature).

• If changes occur adiabatically (without gain or loss of heat):

  • PV^{\gamma} = \text{Constant}
  • where \gamma = ratio of specific heat at constant pressure & specific heat at constant volume.

Viscosity

• Viscosity of the fluid determines its ability to resist shearing forces. Affects motion when shearing forces are set up between layers of fluid moving at different velocities.

• Viscosity of fluids is due to cohesion and interaction between molecules.

• Consider two layers of fluid, at distance 'dy' apart, moving one over another at different velocities, say u and u + du; the viscosity together with relative velocity causes a shear stress acting between the third layers.

• 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 \tau (called Tau).

• Mathematically, \tau = \mu \frac{du}{dy}

  • where \mu = constant of proportionality, known as the coefficient of dynamic viscosity.

• The coefficient of Dynamic Viscosity, \mu, is defined as the shear force per unit area required to drag one layer of fluid with unit velocity past another layer unit distance from it in the fluid.

• SI unit is N-s/m² (Pascal-second) or kg/m-s.

• (fps absolute unit lb-mass/ft-s; Technical unit slug/ft-sec)

• The unit of viscosity in C.G.S. is also called poise.

  • 1 poise = \frac{dyne \cdot sec}{cm^2} = \frac{1}{10} Ns/m^2

Kinematic Viscosity

• \nu (Greek nu) is the ratio of dynamic viscosity to mass density.
• \nu = \frac{\mu}{\rho}
• Note that if \mu is in kg/m-s, \rho must be in kg/m³. Thus, the units of \nu are independent of mass.
• The SI unit is m²/s.
• (FPS unit: ft²/s). In the cgs system the unit is the Stoke which is divided into 100 centistokes.

Newton's Law of Viscosity:

• This law states that the shear stress (\tau) in a fluid element layer is directly proportional to the rate of shear strain.
• Mathematically, \tau = \mu \frac{du}{dy}.
• The fluids which follow this law are known as Newtonian fluids.

Examples

1. Determine the mass density, specific volume, and specific weight of a liquid whose specific gravity is 0.85.

• Solution:
  • S = \frac{w{\text{liquid}}}{w{\text{water}}} = 0.85
  • w{\text{liquid}} = 0.85 \times w{\text{water}} = 0.85 \times 9.81 = 8.3385 \text{ kN/m}^3
  • \text{Mass density } \rho = \frac{w}{g} = \frac{8.3385 \times 1000}{9.81} = 850 \text{ kg/m}^3
  • \text{Specific Volume } V_s = \frac{1}{\rho} = \frac{1}{850} = 0.00118 \text{ m}^3/\text{kg}
  • \text{Specific weight } w_l = 8.3385 \times 1000 = 8339 \text{ N/m}^3

2. A liquid has a specific gravity of 1.9 and kinematic viscosity of 6 stokes. What is its dynamic viscosity?

• Solution:
  • Given: S = 1.9, \nu = 6 Stokes = 6 \times 10^{-4} m^2/s
  • \mu = ?
  • S = \frac{wl}{ww}
  • wl = S \times ww = 1.9 \times 9.81 = 18.639 \text{ kN/m}^3
  • \rho = \frac{w}{g} = \frac{18.639}{9.81} = 1900 \text{ kg/m}^3
  • \text{Kinematic viscosity } \nu = \frac{\mu}{\rho} \implies \mu = \nu \times \rho = 6 \times 10^{-4} \times 1900 = 1.14 N.S/m^2 = 11.4 \text{ Poise}

3. A plate 0.05 mm distant from a fixed plate moves at 1.2 m/s and requires a force of 2.2 N/m² to maintain this speed. Find the viscosity of the fluid between the plates.

• Solution:
  • Velocity of the moving plate, u = 1.2 m/s
  • Distance between the plates, dy = 0.05 mm = 0.05 \times 10^{-3} m
  • Force on the moving plate, F = 2.2 N/m²
  • Viscosity of the moving fluid \mu = ?
  • \tau = \mu \frac{du}{dy}
  • where \tau = shear stress or force per unit area = 2.2 N/m²
  • du = Change of Velocity = 1.2 - 0 = 1.2 m/s
  • dy = Change of distance = 0.05 mm = 0.05 \times 10^{-3} m
  • 2.2 = \mu \frac{1.2}{0.05 \times 10^{-3}} \implies \mu = \frac{2.2 \times 0.05 \times 10^{-3}}{1.2} = 9.16 \times 10^{-5} N \cdot s/m^2 = 9.16 \times 10^{-4} Poiss

Questions

1. The velocity distribution for flow over a plate is given by u = 2y - y² where u is the velocity in m/s at a distance y meters above the plate. Determine the velocity gradient and shear stress at the boundary and 0.15m from it. Take dynamic viscosity of fluid as 0.9 N.S/m².

• (Ans: 25, 1.8 N/m²; 1.75, 1.53 N/m²)

CHAPTER 2: FLUID STATICS

2.0 STATIC PRESSURE AND HEAD

• Pressure is the force exerted by a fluid at on any surfaces with which it is in contact, or part of the fluid on the adjoining part.
• The intensity of pressure at a point is the force exerted on a unit area at that point and is measured in newtons per square meter in SI units (pounds per square foot in fps technical units).
• An alternative metric unit is the bar, which is 10^5 N/m^2.

2.1 PRESSURE HEAD

• Consider a vessel containing a liquid.
• Let h = Height of liquid in the Cylinder
• A = Area of the Cylinder base
• \gamma = Specific weight of the liquid
• P = Intensity of pressure
• Total pressure on base of Cylinder = Weight of liquid in the Cylinder
• P \cdot A = w A h
• P = w h the intensity of pressure in a liquid due to it's depth will vary directly with depth.
• From equation 2.1, h = \frac{P}{w}, where h is known as static head.
• Here, the intensity of pressure can be expressed as:
  • In force per unit area (N/m², N/mm²)
  • As equivalent static head (metres, mm or cm of fluid)

Hydrostatic Law:

• The rate of increase of pressure in a vertically downward direction must be equal to the specific weight of the fluid at that point.

2.2 PRESSURE VARIATION IN FLUID AT REST

• Let:

  • p = Intensity of Pressure on face LM
  • \Delta A = Cross-sectional area of the element
  • z = Distance of the fluid element from the free surface, and
  • \Delta z = Height of the fluid element

• The forces acting on the element are:

  • Pressure force on the face LM = P \Delta A (acting downwards)
  • Pressure force on the face ST = P + \frac{\partial P}{\partial z} \Delta z) \Delta A (acting upwards)
  • Weight of the fluid element = weight density x volume = w (\Delta A \Delta z).
  • Pressure forces on the vertical surfaces MT and LS are equal and opposite.

• For equilibrium of the fluid element, we have:

  • P \Delta A - (P + \frac{\partial P}{\partial z} \Delta z) \Delta A + w(\Delta A \Delta z) = 0
  • P \Delta A - P \Delta A - \frac{\partial P}{\partial z} \Delta z \Delta A + w \Delta A \Delta z = 0
  • -\frac{\partial P}{\partial z} \Delta z \Delta A + w \Delta A \Delta z = 0
  • -\frac{\partial P}{\partial z} = -w \implies \frac{\partial P}{\partial z} = w

• \frac{dP}{dz} = \rho g (since w = \rho g).

• On integrating equation 2.3:

  • \int dP = \int \rho g dz
  • P = \rho g z = wz (2.4)

• Equation 2.3 states that the rate of increase of pressure in a vertical direction is equal to weight density of the fluid at that point. This is hydrostatic law.

Pascal's Law

• The intensity of pressure at any point in a fluid at rest, is the same in all directions.

2.3 ABSOLUTE AND GAUGE PRESSURES

Atmospheric Pressure

• Also known as Barometric pressure, it is the normal pressure which air exerts on all surfaces with which it is in contact.
• The atmospheric pressure at sea level (above absolute zero) is called standard atmospheric pressure.

Gauge Pressure

• This is measured with instrument in which the atmosphere pressure is taken as datum.
• Gauges record pressures above or below atmospheric pressure. If the pressure of the liquid is below the local atmospheric pressure, then the gauge is designated as Vacuum gauge and the recorded value indicates the amount by which the pressure of the liquid is below local atmospheric pressure, i.e., negative pressure

Absolute Pressure

• Is any pressure measured above the absolute zero of pressure.
• Mathematically,
  • Absolute Pressure = Atmospheric pressure + gauge pressure
  • Absolute pressure = Atmospheric pressure - Vacuum pressure

2.4 PRESSURE MEASUREMENT

2.4.1 U-TUBE MANOMETER

• A U-tube manometer consists of a glass tube bent in U-shape, one end of which is connected to a point at which pressure is to be measured and the other end remains open to the atmosphere.

(1) For positive pressure:

• Let A be the point at which pressure is to be measured.

• X-X is the datum line.

• Let:

  • h = Height of the light liquid in the left limb above the datum line.
  • h_2 = Height of the heavy liquid in the right limb above the datum line.
  • h = pressure in pipe, expressed in terms of head.
  • S_1 = Specific gravity of the light liquid
  • S_2 = specific gravity of the heavy liquid.

• Pressures in right and left limbs above datum line x-x are equal (pressures at the same level in a homogeneous liquid are equal).

• Pressure head above X-X in the left limb = h + h1 S1

• Pressure head above X-X in the right limb = h2S2

• Equating, h + h1 S1 = h2 S2

• h = h2 S2 - h1 S1

E. For negative pressure!

• Pressure head above X-X in the left limb = h + h1S1 + h2S2
• Pressure head above X-X in the right limb = 0
• Equating, h + h1S1 + h2S2 = 0
• h = -(h1S1 + h2S2)

Example 2.1

• A U-tube manometer is used to measure the pressure of oil of specific gravity 0.85 flowing in a pipeline. Its left end is connected to the pipe and the right limb is open to the atmosphere. The center of the pipe is 100mm below the level of mercury (specific gravity = 13.6) in the right limb. If the difference of mercury level in the two limbs is 160mm, determine the absolute pressure of oil in the pipe.

• Solution:

  • Given: Specific gravity of oil, S_1 = 0.85
  • Specific gravity of mercury S_2 = 13.6
  • Height of oil in the left limb, h_1 = 160 - 100 = 60mm = 0.06m
  • Difference of mercury level h_2 = 160mm = 0.16m
  • Let h = Gauge pressure in pipe in terms of head of water P = Gauge pressure in terms of KN/m².

• Equating pressure heads above datum,xx,

  • h + h1S1 = h2S2
  • h= h2S2 - h1S1 = 0.16 \times 13.6 - 0.06 \times 0.85 = 2.125m

• Pressure P = w h = 9.81 \times 2.125 \text{ KN/m}^2 = 20.85 \text{ KPa} (w = 9.81 \text{ kN/m}^2 in SI units)

• Absolute pressure of oil,

  • P{\text{abs}} = P{\text{atm}} + P_{\text{gauge}} = 100 + 20.85 = 120.85 \text{ KPa}.

Assignment:

1. A U-tube manometer containing mercury was used to find the negative pressure in the pipe, containing water. The right limb was open to the atmosphere. Find the Vacuum pressure in the pipe, if the difference of mercury level in the two limbs was 100mm and height of water in the left limb from the Centre of the pipe was found to be 40 mm below. (13.73 kPa) (Vacuum)

2. A pipe contains Carbon tetrachloride of specific gravity 1.594 under a pressure of

2. The pressure of water in a Pipeline was measured by means of a simple manometer containing mercury. The reading of the manometer is shown in fig 2.5. Calculate

3. If a mercury barometer reads 700mm and a Bourdon gauge at a point in a flow system reads 500kN/m², what is the absolute pressure at the point? (393.4 kN/m² abs).

4. A simple manometer (U-tube) containing mercury is Connected to a pipe in which an out of specific gravity 0.8 is flowing. The pressure in the pipe is Vacuum. The other end of the manometer is open to atmosphere. Find the vacuum pressure in pipe, if the difference of mercury level in the two limbs is 200mms and height of oil in the left limb from the Centre of the pipe is 150mm below. Show the arrangement schematically. (-27.86 kPa)

2.4.2 DIFFERENTIAL MANOMETERS

• A differential manometer is used to measure the difference of pressures between two points in a pipe, or in two different pipes.

CASE I

• Shows a differential manometer whose two ends are connected with two different points A and B at the Same Level and Containing same liquid.

• Let:

  • h = Difference of the mercury level in the U-tube.
  • h_1 = Distance of the centre of B, from the mercury Level in the right limb.
  • S1 = S2 = Specific gravity of liquid at the two points A and B.
  • S = Specific gravity of heavy liquid or mercury in the U-tube.
  • h_A = pressure head at A, and
  • h_B = pressure head at B.

• Pressures in left limb & right lims are equal above the datuon level.

• Pressure head in left Limb = hA + (h1 + h)S_1

• Pressure head in right lims = hB + h1 S_1 + h S

• hA + (h1 + h) S1 = hB + h1 S1 + h S

• hA - hB = h1S1 + hS - (h1 + h) S1

• hA - hB = h1S1 + hS - h1S1 - hS_1

• hA - hB = h (S-S_1)

CASE II

• Shows a differential manometer whose two ends are connected to two different pes points A and B at different levels and containing different liquids.

• Let:

  • h = Difference of mercury level in the tube.
  • h_1 = Distance of the Centre of A from the mercury Level in the right limb.
  • h_2 = Distance of the Centre of B from the mercury Level in the right limb-
  • S_1 = Specific gravity of liquid in pipe A
  • S_2 = Specific gravity of liquid in pipe B
  • S = Specific gravity of heavy liquid or mercury

• h_A = pressure head at A, and

• h_B = pressure head at B-

• Pressures above datuon line, are equal

• Pressure head in left limb = hA + (h1 + h)S_1

• Pressure head in right limb = hB + h2x S_2 + hxS

• Equating,

• hA + (h1 + h)S1 = hB + h2S2 + hS

• hA - hB = h2S2 + hS - (h1 + h) S1

• hA - hB = h2S2 + hS - h1S1 - hS_1

• hA - hB = h(S - S1) + h2S2 - h1S_1

Example 2.2

• A differential manometer Connected at the two Points A and B in a pipe Containing an oil of specific gravity 0.9, Shows a difference in mercury levels as 150mm. Find the difference in pressures at the two points.
• Solutions
  • Specific gravity of oil, S_1 = 0.9
  • Specific gravity of mercury S = 13.6
  • Difference of mercury levels h = 150mm
• Let
  • hA - hB = Difference of pressures between A & B in terms of head of water
  • PA - pB = Difference of pressures between A &B.
• Now hA - hB = h (s-S_1)
  • = 150 (13.6-0.9) = 1905 mm
  • = 1.905m
• Difference of pressures between A & B are;
  • PA - PB = w h = 9.81 \times 1.905 = 18.68 KN/m^2

Questions

1. A manometer consists of two tubes A and B, with Vertical axes and uniform cross-sectional areas 500 mm² and 800 mm² respectively, connected by a U-tube C of cross-sectional area 70mm² throughout. Tube A Contains a liquid of relative density 0.8; tube B contains one of relative density 0.9. The Surface of Seperation between the two liquids is Vertical side of C connected to tube A. Determine the additional pressure which, when applied to the tube B, will cause the surface of Seperation to rise to 10 mm in the tube C.

• Ans: (121 Pa)

2. A pressure gauge Consists of a U-tube with equal enlarged ends and is filled with water on one side and oil of specific gravity 0,99 on the other, the Surface of seperation being in the tube below the enlarged ends. Calculate the diameter of each enlarged ends if the tube diameter is 3 mm and the Surface seperation moves 25mm for a difference in pressure head of 1mm of Water. (Ans: 70mm)

3. A differential manometer Consists of a U-tube Connected to two Pipes A and B. The liquid in pipe A is water while the liquid in Pipe B is oil of specific gravity 0.8. Type A is higher than B and the distance of seperation of the water and mercury is 80mm. The distance of seperation of the oil and mercury in the right lime is 120mm while the difference of mercury level in the U-tube is 100mm. If at A air pressure is 78.5 KN/m², find the absolute Pressure at B. (Ans: 69.1 KN/m²)

CHAPTER THREE HYDROSTATIC PRESSURE ON SURFACES

3.1 INTRODUCTION

• A fluid in contact with a solid surface will exert a force on every small area of the surface equal to the product of the Pressure P on the small element and its area \delta a. Usually the pressure P will vary from point to point

• Total pressure on Solid Surface = sum of the forces on all these elements P = \sum P \delta a

• Since there are no shear stresses in a fluid at rest the force on each element is at right angles to the Solid surface. The Combined action of all these elementary forces can be represented by a single resultant force called the resultant pressure, acting at a point on the Surface Called the Centre of pressure.

• If the solid surface is a Plane Surface all the elementary forces are parallel, so that Resultant Pressure = Total Pressure.

• If the surface is curved the elementary forces will not be parallel and will have components opposing each other so that the resultant pressure will be less than the total pressure.

• In terms, total pressure and resultant pressure, the word pressure is used to mean force not intensity of Pressure

3.2 HORIZONTALLY IMMERSED SURFACE

• Let
  • A = Area of the immersed Surface
  • x = Depth of horizontal Surface from the liquid Surface,
  • w = Specific weight of the liquid
• Total pressure, P = Weight of liquid above immersed surface = Specific weight of liquid x volume = Specific weight of liquid x area of Surface x depth of liquid = w A x.

3.3 VERTICALLY IMMERSED SURFACE

• Let
  • A = Total area of the Surface
  • \bar{x} = Depth of Centre of area from Free Surface of liquid and
  • h = Distance of Centre of pressure from free Surface of liquid

(i) Total Pressure (P)

• Let there be a thin horizontal strip of thickness dx and breadth b.

Let x = depth of strip
P = Intensity of pressure on strip
Then P= wx, where w = specific weight of the liquid

• Total Pressure on strip = P \delta a = w x b dx
• Total pressure on the whole area P = \int w x b dx = w \int b x dx. But \int b x dx = moment of Surface area about Liquid level
  \int b x dx= A\bar{x}
  \therefore P= w A \bar{x}

(ii) Centre of pressure (h)

The resultant pressure on any immersed Surface will act at some point, below the Centre of gravity of the immersed Surface and towards the Lower edge of the figure. The point through which this resultant pressure acts is known as "Centre of pressure" and is always expressed in terms of depth from the water surface.

• Let C_p be the Centre of pressure of the immersed figure. Then the resultant Pressure P will act through this point.
• Let h = Depth of Centre of pressure below free liquid surface, and
• I_o = Moment of inertia of the surface about free Surface

Now total pressure on any Strip = wx.b.dx

• Moment of this pressure about free surface = (w x b dx) x = w.x^2 b.dx
• Total moment of all such pressures for whole area, M= \int w x^2.b.dx = w \int x^2. b.dx
• But \int x^2. b.dx = moment of inertia of the surface about the free Surface 00 (or second moment of area).
• M = wI_o
  The sum of the moments of the Pressures is equal to P x h
• Equating equations (2) and (3), P h = wI_o
• w A\bar{x}h= w I_o
• h = \frac{I_o}{A \bar{x}}
• Also $$I_o = I