منهج أ. خالد فاضل Ce325.pdf · Version 1.pdf

University of Benghazi Faculty of Engineering Civil Engineering Department

FLUID MECHANICS (CE 325)

Mr. Khalid Fadel

1

CHAPTER ONE INTRODUCTION

1.1 DEFINITIONS A fluid is a substance that deforms continuously when subjected to a shear stress, no matter how small that shear stress may be. A substance may be solid or fluid, a fluid may be liquid or gas. A fluid can flow and has no fixed shape. Fluid mechanics is that branch of applied mechanics that concerns with the study of fluids at rest and motion. The branches of fluid mechanics are: a- Fluid statics 1- Hydrostatics: study of incompressible fluids at rest. 2- Aerostatics: study of compressible gases at rest. b- Fluid dynamics Study of fluid, velocity and acceleration, etc. with force or energy causing flow. 1- Hydrodynamics: study of incompressible flow of liquids. 2- Aerodynamics: study of compressible and incompressible flow of gases. c- Fluid Kinematics Study of flow patterns and relation between velocity and acceleration of fluid without reference to the cause of motion (like force or energy). Hydraulics is the study of liquids at rest and motion. 1.2 SCOPE OF FLUID MECHANICS A knowledge of fluid mechanics is required: i- to understand the fluid behavior under all conditions of rest and motion. ii- to determine the pressure distribution on hydraulic structures like dams, spillways, and gates. iii- for flow classification and flow analysis in pipes, open channels, and groundwater flow. iv- for the design of ships and naval vessels. 1.3 DIMENSIONS AND UNITS Physical quantities such as length, time, mass, and temperature are referred as dimensions, whereas meter, second, kilogram, and Kelvin are referred as units. For example, Width and height have the dimension of length [L] and unit of (m), (cm), (mm), etc. Table 1.1 gives the main dimensions and units in the international system of units (SI System). CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 1: Introduction

2

Table 1.1 Dimensions and units in SI System Quantity

Dimension

Unit

Mass Length Time Temperature

M L T ө

Kilogram (kg) Meter (m) Second (s) Kelvin(K) Celsius (co)

Because Newton's 2nd law deals with force, mass and acceleration (F=ma), so there are two systems of dimensions (MLT) and (FLT). Some derived units in table 1.2. Table 1.2 Dimensions and units for some physical quantities in SI System

Dimension Quantity

SI Unit

Basic Units

F

Newton, N

kg m s-2

ML2T-2

FL

Joule, J (N.m)

kg m2 s-2

Power

ML2T-3

FLT-1

Watt, W (N.m/s)

kg m2 s-3

Pressure

ML-1T-2

FL-2

Pascal, Pa (N/m2)

kg m-1 s-2

(MLT)

(FLT)

Force

MLT-2

Energy

1.4 FLUID PROPERTIES 1.4.1 Density (ρ) The density, ρ, of a fluid is defined as its mass per unit volume.

where m is the mass of the fluid and V its volume. Fluid density is related to two properties, specific weight , and relative density S.

where w is the weight of the fluid. Because w=mg, where g is the acceleration of gravity (g=9.81 m/s2), Eq. (1.2) can be written as

The relative density S of a liquid (sometimes specific gravity) is the ratio of its mass to the mass of an equal volume of water at standard conditions. CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 1: Introduction

3

Example 1-1 2.5 m3 of a certain liquid weighs 15000 N. Determine: (i) specific weight, (ii) Density, and (iii) Relative density of the liquid. ( ) Solution (i)

Specific weight

(ii)

Density

or

(iii) Relative Density

or

1.4.2 Viscosity (µ) The viscosity, µ, is the property that causes existence of friction forces in flowing fluids result from cohesion and momentum interchange between molecules. In liquids viscosity decreases with increasing temperature, while in gases viscosity increases with increasing temperature. Consider the situation in Fig. 1.1, where a fluid is between to plates, the top plate is moved by a force F moving at a constant velocity v.

Moving plate of area A

h

Fluid Fixed plate

v

F

Velocity profile

Fig. 1.1 Definition of viscosity

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 1: Introduction

4

or

The constant of proportionality is known as the dynamic viscosity µ. Since shear stress as

can be written

then

This relation is right when the velocity profile is linear. In general the velocity profile is as shown in Fig. 1.2

This relation (Eq. 1.10) is known as Newton’s law of viscosity.

dy

The SI unit of is Pa.s (N.s/m ),it has no name. The common unit of is Poise.

Velocity profile

v

2

y

dv

Fig. 1.2 General case for velocity profile

The kinematic viscosity is the ratio of viscosity to density

The SI unit of is (m2/s), it has no name. The common unit of is Stoke, 1Stoke = 10-4 m2/s.

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 1: Introduction

5

1.4.2.1 Newtonian and Non-Newtonian Fluids

id Flu ew ton ian n-N

an ni o t ew

No

Shear Stress, 

Fluids that obey Newton’s law of viscosity are known as Newtonian fluids. For such fluids, doesn't change with velocity gradient dv/dy, i.e, is constant as shown in Fig. 1.3. In some cases the friction forces due to viscosity are very small and can be neglected, in this case the fluid is called ideal fluid ( =0). In reality, there is no existence of ideal fluid.

N



 tan 

d

ui Fl

Ideal Fluid 

Velocity Gradient, dv/dy Fig. 1.3 Relation between shear stress and velocity gradient

Example 1-2 A plate, 0.5mm distance from a fixed plate, moves at 0.25 m/s and requires a force per unit area of 2 Pa to maintain this speed, see Fig. 1.4. Determine the viscosity of the fluid between the plates, assuming linear variation of velocity.

Moving plate

v=0.25 m/s

 = F =2Pa A

0.5 mm Fixed plate

Solution Fig.1.4 EX 1-2

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 1: Introduction

6

Example 1-3

l

Oi

A flat plate of area 1m2 and weight 100 N slidesdown a 30o inclined plane lubricated with 1mm uniform layer of oil of viscosity 0.1 N.s/m2, see Fig. 1.5. Determin the steady velocity of the plate.

v=

?

1mm w=100N

30°

Fig. 1.5 EX 1-3 Solution Since the velocity is steady (v=constant), then the acceleration a=0, so

x

y

30°

F 30°

w

Fig. 1.6 EX 1-3

Example 1-4 The velocity distribution over a plate is given as v=2y-y2 in m/s, where y(m) is the distance above the plate, see Fig. 1.7. Determine the shear stress at the boundary and 0.15m above it. The viscosity of the fluid is 0.8 N.s/m2.

dy

v=2y-y

2

v y

dv

boundary y=0

Fig. 1.7 EX 1-4

Solution

at the boundary y=0:

[

at y=0.15m:

[

CE325 Fluid Mechanics, Mr. Khalid Fadel

] ]

Chapter 1: Introduction

7

Example 1-5 The velocity v at radius r in a pipe of radius r0 is given in terms of center-line velocity vc for laminar flow as ( ) If the center-line velocity in a pipe of 1 m diameter is 6 m/s and the viscosity is 0.002 N.s/m2, draw the velocity and shear stress profile for a cross section. What is the drag per km length of the pipe. Solution *

[

( ) +

(

) ]

Velocity profile is given by

Shear stress profile is given by

Table 1.3 Velocity and Shear stress distribution in laminar flow in a pipe

r (m) 0 0.1 0.2 0.3 0.4 0.5

(m/s) 6 5.76 5.04 3.84 2.16 0

(N/m2) 0 0.0096 0.0192 0.0288 0.0384 0.0480

Remarks At center -line 𝝉=0, v=max

At boundary 𝝉=max, v=0

Drag force per km length of pipe,

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 1: Introduction

8

0

0 v

r0

dv v+dv v

r

dy

 0

y v r 2 = 1- ( ) vc r0

Shear stress profile

Flow

vmax (vc ) r



velocity profile

velocity gradient =

y



dv dv =dy dr

r

(a) Velocity distribution

0 =

0 r0

(b) Shear stress distribution

Fig. 1.8 Laminar flow in a pipe

1.4.3 Surface Tension At the interface between a liquid and a gas, or two immiscible liquids, a film seems to form on the liquid owing to attraction of liquid molecules below the surface. The surface tension is then the stretching force required to form this film. Surface tension is the force per unit length, the SI unit of is N/m

Capillary attraction is caused by surface tension and by the relative value of adhesion between liquid and solid to cohesion of the liquid. In Fig. 1.9 (a) water wets the glass because force of adhesion is greater than force of cohesion that cause capillary rise. In Fig. 1.9 (b) mercury doesn't wet the glass because force of cohesion is greater than force of adhesion that cause capillary depression.  F

Glass Tube

h (Capillary Rise)

Water

Diameter d

(Capillary h Depression) Mercury F



(a)

(b) Fig. 1.9 Capillary rise and depression

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 1: Introduction

9

For water, see Fig. 1.10, from equilibrium

F



F

h w=  V Fig. 1.10

F cos

=w

Equating the right hand sides of Eq. 1.15 and Eq. 1.16 yields

If the tube is clean, the contact angle

CE325 Fluid Mechanics, Mr. Khalid Fadel

for water and

for mercury.

Chapter 1: Introduction

10

1.4.4 Vapor Pressure Liquids evaporate because of molecules escaping from the liquid surface. The vapor molecules exert a partial pressure in the space, known as vapor pressure Pv. The vapor pressure depends upon temperature and increase with it. When the pressure above a liquid equals the vapor pressure of the liquid boiling occurs, in this condition, Pv is called saturation vapor pressure SVP. Boiling of water, for example may occur at room temperature if the pressure is reduced sufficiently, see Fig. 1.11. Pv has the dimensions of pressure ( [Pv]= F/L2, SI unit is Pa or N/m2 ).

P=101.3 kPa o

P=101.3 kPa o

P=4.3 kPa o

Water 30 C Pv =4.3 kPa

Water 100 C Pv =101.3 kPa

Water 30 C Pv =4.3 kPa

(a)

(b)

(c)

Pv

Pv=P Boiling occurs

Pv=P Boiling occurs

Fig. 1.11 Boiling of water

1.5 Compressibility The compressibility is the capability of a substance of being reduced in volume by application of pressure. Compressibility of a fluid is expressed by means of its bulk modulus of elasticity K.

At normal temperature and pressure condition Kwater= 2.07 × 106 kpa and Kair= 101.3 kpa, that means, air (a gas) is 20,000 times more compressible than water (a liquid). For this reason, the density of liquids can be assumed to be constant without any serious loss in accuracy. On the other hand, gases are very compressible.

Example 1-6 Find the bulk modulus of elasticity of a liquid if its volume is decreased by 15% when the applied pressure is increased from 700 N/cm2 to 1300 N/cm2. Solution

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 1: Introduction

11

(

)

PROPLEMS 1-1

A right circular cone of diameter 15 cm and hight 25 cm. When filled with a liquid, it weighs 20 N. When empty, it weighs 8 N. Estimate the relative density of this liquid. (Ans. S=0.83)

1-2

A liquid of kinematic viscosity of 1000 cS, and relative density of 0.92. Find (i) kinematic viscosity in m2/s, (ii) viscosity in N.s/m2 and in Poise. (Ans. ν=10-3 m2/s, 𝝁=0.92 N.s/m2, 𝝁=9.2 Poise)

1-3

A vertical gap of 25 mm of infinite extent contains oil of viscosity 2.5 N.s/m2. A square metal plate of 1.5 m side and 1.5 mm thick weighing 50 N is to be lifted through the gap and at a constant speed of 0.1 m/s. Find the force and the power required to lift the plate. (Ans. F = 145.7 N, P = 14.57 W)

V = 0.1 m/s

F=?

Metal Plate

oil

 = 25 poise



1.5 mm 25 mm W = 50 N

1-4

1-5

A shaft 59 mm Dia. rotates inside a cylinder 60 mm Dia. Both the shaft and the cylinder are 80 mm long. Find the number of revolutions per minute (rpm) of the shaft if the space between the cylinder and shaft is filled with oil of viscosity of 3 Poise and a torque of 1.5 N.m is applied. What is the power required to rotate the shaft?. Hints: Torque T = F.r, Power P= T.ω, v= ω.r (Ans. N=1850 rpm, P= 290.6 W)

Derive an expression to find the capillary depression h, for mercury, shown in the figure. If the tube is 5mm diameter, determine h using θ=40o, σ =0.51N/m and S=13.56 (Ans. h=4σ cos θ/(γd), h= 2.3mm)

Cylinder 60mm 

oil

r

 = 3 poise



Shaft 59mm 

Diameter d

h Mercury F

 CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 1: Introduction

12

1-6

Estimate the pressure difference in (i) 1 cm diameter soap bubble in air, use σ =0.037 N/m. (ii) 1 mm diameter air bubble in water, use σ =0.073 N/m. (iii) 5 mm diameter rain droplet in air, use σ =0.073 N/m. (Ans. (i) 29.6 Pa, (ii) 292 Pa, (iii) 58.4 Pa)

1-7

If K = 2.2 GPa is the bulk modulus of elasticity for water, what pressure is required to reduce a volume by 0.6%? (Ans. 13.2 MPa)

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 1: Introduction

13

CHAPTER TWO FLUID STATICS

2.1 PRESSURE AT A POINT IN A STATIC FLUID Consider an element of fluid as shown in Fig. 2.1. Let p is the pressure at the center. The forces acting on the element are the force of gravity, w, and the surrounding fluid at right angles to the sides of the element. p z (p + z 2 )  x  y

z

(p -

p y y 2 ) x z

(p +

p x ) y  z x 2

y

(p -

p x ) y  z x 2

p z

x

p y (p + y 2 )  x  z

(p w

p z z 2 )  x y

=  ( x y  z) x

y

Fig. 2.1 Forces acting on an element of static fluid

Applying the equations of statics, (

)

(

)

(

)

(

)

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

14

(

)

(

)

From Eq.(2.1), (2.2), and (2.3),

Hence in a static fluid the pressure, p, is the same at every point in the horizontal plane, and decreases as z increases. For an incompressible fluid ( =constant), Fig. 2.2, let the pressure at point 1, h meters below the free surface be p1. From Eq. (2.6) ∫

∫

At the free surface p2=0 (atmospheric pressure) and z2-z1=h, hence

Point 1 and 3 are in the same horizontal plane, from Eq. (2.4) and (2.5),

p is the pressure and h is known as the head.

2

Free Surface Atm. pressure (p=0)

h

Liquid (  )

1

z2

3

z1 Datum Fig. 2.2 Pressure at a point in a static fluid (incompressible)

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

15

2.2 PASCAL’S LAW Consider a wedge shaped element at a point in a static fluid with thickness (perpendicular to the plane of paper) dy, Fig. 2.3. Since no tangential stresses can exist in a fluid at rest, let p be the pressure on a plane at angle  to the horizontal, and px, pz are the pressures on horizontal and vertical planes. Then resolving

p(dl dy)



px (dy dz) dl



dz

dx

pz (dx dy)

w

=  ( 12 dx dy dz)

Fig. 2.3 Pressure on a fluid element at rest

Neglecting the third term, and

Hence, at any point in a fluid at rest, the intensity of pressure is exerted equally in all directions, which is called Pascal’s law. Example 2-1 Determine the pressure at A, B, C, and D.

D

Air

Solution

Air 30 cm

Water

B

1

C

100 cm

60 cm 4

130 cm

100 cm 2

A

3

Water

Fig. 2.4 Ex. 2-1 CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

16

,

"Ignoring

"

Example 2-2 For the vessel containing glycerin (S=1.25) under pressure as shown in Fig. 2.5, Determine the pressure at the bottom of the vessel.

50 kPa

Solution

Glycerin

2m

Fig. 2.5 Ex. 2-2

Example 2-3 In the closed tank shown in Fig. 2.6, the pressure at point A is 98 kPa, what is the pressure at point B? What percent error results from neglecting the specific weight of the air? ( )

B

Air

Air

5m

3m

3 1

Solution i-

A

2

5m

Considering

3m Water Fig. 2.6 Ex. 2-3

ii-

Neglecting

Error = (78.405 – 78.38) / 78.405 = 0.00031, or

CE325 Fluid Mechanics, Mr. Khalid Fadel

0.031%.

Chapter 2: Fluid Statics

17

2.3 PRESSURE AND ITS MEASUREMENTS The atmosphere of Earth is the layer of gases surrounding the planet Earth that is retained by Earth's gravity. Atmospheric pressure is the total weight of the air above unit area at the point where the pressure is measured. Thus air pressure varies with location and weather. The average atmospheric pressure at sea level is called standard atmospheric pressure. The local atmospheric pressure may be lower or higher than the standard atmospheric pressure; lower if the place under question is higher than sea level and higher if the place is lower than sea level.

(+)ve gauge pressure Standard atmospheric pressure (at sea level) Local atmospheric pressure

1 atm 101.325 kPa 760 mm Hg 10.34 m water 1 bar (100 kPa)

(-)ve gauge pressure

Absolute Pressure Local barometer reading

Absolute Pressure Absolute zero (perfect vacuum)

Fig. 2.7 Absolute and gauge pressure

Pressure measured above the perfect vacuum (absolute zero) is called absolute pressure, Fig. 2.7. The perfect vacuum is the lowest possible pressure, therefore, an absolute pressure is always positive. Gauges record pressure above or below the local atmospheric pressure, since they measure the difference in pressure of the fluid to which they are connected and that of the surrounding air. Hence the gauge pressure may be positive or negative (vacuum pressure). Absolute pressure = Local atmospheric pressure + Gauge pressure

Pressure can be recorded by: 1- Barometer (for local atmospheric Pressure) 2- Bourdon gauge (pressure at a point) 3- Piezometer (pressure at a point) 4- Manometer (pressure at a point) 5- Differential manometer (pressure difference between two points) 6- Inclined manometer (for measurement of pressure to a great sensitivity) 7- Micro manometer (for measurement of pressure to a great sensitivity)

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

18

2.3.1 Barometer The mercury barometer is used to measure the local atmosheric pressure, It consists of a glass tube closed at one end, filled with mercury, inverted and submerged in mercury, Fig. 2.8

pv 0 (complete vacuum)

Mercury is used because it has relatinely big relative density (S=13.6), also the vapour pressure, pv, is very small, so it can be neglicted.

R patm

M

Mercury

Fig. 2.8 Barometer

Where R is the hight of mercury, or simply it is the reading of the barometer.

Example 2-4 Air

Determine the gage and the absolute pressure at A and B, if the barometer reading is 755 mm Hg. (SHg = 13.6) A

Oil S=0.8

0.9 m

Solution 2m

Water B

Fig. 2.9 Ex. 2-4

Example 2-5 At a base of a mountain, a mercury barometer reads Rb=740 mm, R whereas at the top it reads Rt=595 mm. If the density of air is 3 M 1.23 kg/m and is assumed constant, find the pheight of the mountain.

h air,t

(compl

atm

Mercury

h air,b

Solution

H Oil S = 0.85

Fig. 2.10 Ex. 2-5 CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

19

At the top:

At the base:

From Eq. (2) and Eq. (3) substitute in Eq. (1)

2.3.2 Bourdon Gauge Pressure at a point in fluid is commonly measured by Bourdon gauge, Fig 2.11. In this gauge a curved hollow matal tube, open to the fluid, of elleptical cross-section changes its curvature according to the pressure. Bourdon gauge records pressure above or below the local atmospheric pressure, since it measures the difference in pressure of the fluid to which it is connected and that of the surrounding air. A pressure and vacuum gauge can be combined into one called the compound gauge.

a a

Section a-a

deformed state

flexible metal tube (original state) Fluid under pressure

Fig. 2.11 Bourdon gauge

Example 2-6 Air

The Bouron gauge shown in Fig. 2.12 reads 34.5 kPa (gauge). What is the absolute air pressure in the tank? Assuming standard atmospheric pressure of 101.325 kPa.

30 cm

Solution

15 cm

G

Water 1

2

(gauge) Fig. 2.12 Ex. 2-6

(gauge) CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

20

(abs) Example 2-7 Bourdon gauge A inside a pressure tank reads 80 kPa. Another Bourdon gage B outside the pressure tank and connected with it reads 120 kPa, and an aneroid barometer reads 750 mm Hg. What are the absolute pressures measured by gauge A and gauge B.

B b A a

Solution

Fig. 2.13 Ex. 2-7

(gauge),

(gauge) A 80 kPa B

(gauge)

200 kPa 120 kPa Local Atm. Pr.

(abs)

300 kPa

220 kPa

100 kPa

Absolute 0

(abs) Fig. 2.14 Ex. 2-7

2.3.3 Piezometer A piezometer is used to measure moderate positive pressure. It is a simple device consists of a glass tube one end being connected to the point of pressure measurement and the other end open to atmosphere, Fig. 2.15.

Piezometers pA h= 

h

A pressure pipe

Fig. 2.15 Piezometer

Example 2-8 Liquid A will simply rise in piezometer a to the same elevation as liquid A in the tank (i.e., to elevation 2m). Determine the pressure at the bottom of the tank (El 0), and determine the elevation to which CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

21

a b

liquid B will rise in piezometer b.

El 2m

Solution Liquid A (S=0.72)

El=?

El 0.3m El 0

Liquid B (S=2.36)

Elevation of liquid B in piezometer b =0.819 m Fig. 2.16 Ex. 2-8

2.3.4 Manometer A manomete is a pressure measurement device uses the relationship between pressure change and elevation change in a static liquid. It is used to measure high pressures using a manometric fluid immiscible with the fluid in A and B. Basically there are two types of manometers; open manometer and differential manometer, Fig 2.17. Open manometer measures pressures at a point, while differential manometer measures difference in pressure between two point. B A

A

Manometer Liquid

(a) Open manometer

(b) Differential manometer Fig. 2.17 Manometers

The following steps are followed to solve manometers problems: 1- Start with one end or surface and write the pressure. 2- Add or subtract the change in pressure from the end or surface in step (1) to the next surface (add if the next surface is lower, and subtract if it is higher). 3- Continue in the procedure in step (2) till reaching the other end or surface then write equal sign and the pressure at this end or surface.

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

22

Example 2-9

S1

In the open manometer shown in Fig. 2.18; S1=1, S2=13.6, h1=300 mm, h2=200 mm. Find pA.

A h1 h2

Solution S2

Fig. 2.18 Ex. 2-9

Example 2-10 In the differential manometer shown in Fig. 2.19, S1=S3=0.83, S2=13.6, h1=150mm, h2=70mm, and h3=120 mm. (a) Find the difference (pA - pB). (b) Find pA if pB = 70 kPa (gage) (c) For pA = 140 kPa (abs) and a barometer reading of 720 mm, find pB in kPa (gage).

S1 A

S3

B

h3 h1 h2 S2

Fig. 2.19 Ex. 2-10

Solution

Substitute in Eq. (3)

Note that, Dividing Eq. (1) by

, Eq. (1) becomes

where hA and hB are the pressure heads in meter of water. CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

23

Example 2-11 In the differential manometer shown in Fig. 2.20, find the difference in pressure between A and B in meters of water.

S 2 = 0.96

S1 = 1.6 A

Solution

0.3 m

y

0.4 m

Dividing by

yields S1 = 1.6 B

Fig. 2.20 Ex. 2-11

Example 2-12 In the differential manometer shown in Fig. 2.21, find the difference in pressure between A and B.

S=0.8

Solution

B

0.7m Water

X+0.7-1.5=(X-0.8)m

Water 1.5m

X

A

Fig. 2.21 Ex. 2-12

Example 2-13 Find the difference in the mercury levels in the U-tube manometer, reading h, as shown in Fig. 2.22.

Vacuum Gauge -76 mm Hg

Gauge 20 kPa Air

Solution For the vacuum gauge,

Air

Water

2m Oil S = 0.8

Starting from the vacuum gage as follows

1m

1m

h=? Hg (S = 13.6)

Fig. 2.22 Ex. 2-13 CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

24

Example 2-14 A

In Fig. 2.23, A contains water. When the left meniscus is at zero on the scale, as shown in the figure, hA = 90 mm H2O. Find the reading of the right meniscus for pA = 8 kPa with no adjustment of the U tube or scale.

600 mm h1

Solution 0

First determine the reading h1 for hA=90 mm (Fig. 2.23)

0

S = 2.94

Fig. 2.23 Ex. 2-14

Now determine the reading h2 for pA=8 kPa (Fig. 2.24)

A

h

600 mm

h1 h 2 0

0

h Fig. 2.24 Ex. 2-14

Example 2-15 G1 =?

In Fig. 2.25, Find the gauge readings G1 and G2, if the barometric pressure of the place is 750 mm mercury.

1

Closed

Air

Solution 1.5m

Hg vapor

Oil S=0.8

(a) to find G1

0.6m c

1.8m

Water

G2 =? Hg S=13.6

Gauge G1 reads the difference in pressure between point 1 and the outside atmosphere Fig. 2.25 Ex. 2-15

or

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

25

(b) to find G2

2.4 FORCES ON PLANE AREAS The resultant force F due to fluid pressure acting on a plane immersed area A is given by ∫ and acts on a point on the area A normal to the surface, this point is called the center of pressure C.P, Fig. 2.26. To develop an expression for F and C.P, consider the plane of the immersed area making an angle  with the free surface, X axis is a line where the plane of the area intersects the free surface. F.S



h

h cp

X

h

F ycp

y 0

y dA

x' G

C.P y'

x xcp

Y Fig. 2.26 Hydrostatic force on an inclined surface

On an element area dA ∫

∫

CE325 Fluid Mechanics, Mr. Khalid Fadel

∫

∫

∫ Chapter 2: Fluid Statics

26

It is known that from the centroid of an area ∫

∫

From Eq. (2.19) substitute in Eq. (2.18)

To determine the center of pressure C.P. take the moment about X axis ∫ ∫

∫

∫

From the parallel axes theorem ̅ substituting in Eq. (2.24) yields ̅

Similarly to determine xcp take the moment about Y axis with an element area dA parallel to Y axis that yields ̅ Summary for an inclined surface:

̅

̅ Notes: -

is the vertical distance from the centroid of area, G to the free surface (real or imaginary). F is the resultant force, perpendicular to the surface. ̅ is the second moment of area about the centroidal axis x'. ̅ is the product of inertia. Note that, if either of the centroidal axes x' and y' is an axis of symmetry, then ̅ , also ̅ may be either (+)ve or (-)ve.

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

27

For a vertical area (=90o, Fig. 2.27),

X

F.S

y cp h cp

With the same formulae and notes in the inclined surface summary.

y

h

G

x' F

x

C.P y'

x cp Y

Fig. 2.27 Hydrostatic force on a vertical surface

For a Horizontal area, Fig. 2.28, the resultant force is perpendicular to the area and acts on the centroid G, i.e. G coincides with C.P.

F.S

F

h=h cp=h

Y

x

where is the vertical distance from the centroid of area, G to the free surface (real or imaginary).

C.P. G y

X

Fig. 2.27 Hydrostatic force on a horizontal surface

Note that, the resultant static fluid forces are perpendicular to the plane surfaces (inclined, vertical, and horizontal). CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

28

Fig. 2.28 shows the second moment of area for some common shapes about the cetroidal axes y'

y'

y' y'

d

d

x'

G

x'

G d/3 b

x'

G

x'

G r

4r

r



b

̅

̅

̅

̅

̅

Fig. 2.28 The second moment of area of some common shapes

The pressure prism In some cases, it is easy to calculate the hydrostatic force using the idea of the pressure prism. In Fig. 2.29 the acting force on an element area dA is , where dV is an element volume of area dA and height h, hence force F can be written as

where F is the resultant hydrostatic force and is the volume of the pressure prism. The line of action of F passes through the centroid of the pressure prism. F.S h2 F

h

X

h1 h1

Pressure prism 0 h2

dA

C.P

Y

Fig. 2.29 Pressure prism CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

29

Example 2-16 A cubic tank has a side of 2 m contains water as shown in Fig. 2.30, find the hydrostatic forces acting on the sides.

Water

1.5 m

2m

Fig. 2.30 Ex. 2-16

Solution i- Using the formulae (Fig. 2.31) 2m F.S

̅

⁄

X hcp h

y

ycp

G C.P.

x' F

̅

1.5 m x cp

because y' is an axis of symmetry y'

̅ Y

Fig. 2.31 Ex. 2-16

ii- Using the pressure prism (Fig. 2.32)

F acts at the center of gravity of the pressure prism 2m 1m

1m

1m h=1.5m

C.P.

F h

F

Pressure prism

h 3 =0.5m

F

0.5m

h Fig. 2.32 Ex. 2-16

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

30

Example 2-17 For the cubic tank Example 2-16, Fig. 2.30, find the hydrostatic force acting on the bottom.

F.S

F

Solution

1.5 m

i- Using the formulae (Fig. 2.33)

F acts at the centroid of the area. Y

1m C.P.

2m

G 1m X 2m

Fig. 2.33 Ex. 2-17 ii- Using the pressure prism (Fig. 2.34)

F acts at the center of gravity of the pressure prism F Pressure prism

1m

F 1m

C.P.

2m

h

2m

1m Fig. 2.34 Ex. 2-17

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

31

Example 2-18 Determine the moment required to hold the gate as shown in Fig. 2.35.

F.S

A

Water

M

X

3m 2m

Gate 1.3m wide

Fig. 2.35 Ex. 2-18

Solution

̅

X

⁄

h= y

1m A

M

ycp d

x'

F

G C.P. y'

Y

1.3m

Fig. 2.36 Ex. 2-18

Check using the pressure prism (Fig.2.37): F.S 1

To determine the centroid of the prism, take the first moments of area about a-a axis yields (

)(

)

a

a

d F

3 Fig. 2.37 Ex. 2-18 CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

32

Example 2-19 -15 kPa

Gate AB in Fig. 2.38 is 1.2 m wide, hinged at A. Find the force P, applied at B, for equilibrium of gate AB. Water

Solution

3.7m

For oil: ⁄

̅

A

⁄

Oil S=0.75 B

For water:

1.8m

P

Fig. 2.38 Ex. 2-19

To determine the imaginary free surface (I.F.S.), we will locate a surface at which the pressure = 0 as follows:

⁄

̅

1.53m

⁄

I.F.S. 2.17 w

2.17m

3.16m d

1.2m

Fw P 3.97 w

Check using the pressure prism:

Fo

1.8 o

0.6m

Fig. 2.39 Ex. 2-19

acts at (h/3)=(1.8/3)=0.6m from the bottom.

Fw acts at the prism centroid, taking ⁄

yields ⁄

⁄

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

33

Example 2-20 Gate AB in Fig. 2.40 is semi-circular and hinged at B. Find the force P, applied at A, for equilibrium of gate AB.

Oil S=0.8 7.5m

A

P

Water Gate side view

4m B

Fig. 2.40 Ex. 2-20

Solution Convert the pressure head of oil to equivalent pressure head of water as follows: X Water 6m

y

h

ycp P

̅

G

x'

4m F

C.P. 4r =1.698m 

Y

y'

Fig. 2.41 Ex. 2-20

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

34

Example 2-21 Determine the minimum reading of the gauge, G, for which the square gate shown in Fig. 2.42 will be closed.

Water

Solution

4m

For water:

hinge

G=?

=γ h

1m

Gas 1m Fig. 2.42 Ex. 2-21

y

h ̅

⁄

Water

ycp

For Gas:

3 acts through the centroid of the gate. Moment about the hinge is equal to zero.

p 0.5m FG

d Fw 4

p

Fig. 2.43 Ex. 2-21

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

35

Example 2-22 An inclined circular gate shown in Fig. 2.44. Determine the magnitude and the location of the acting force

60° 1.5m

Water

r = 0.5m

Fig. 2.44 Ex. 2-22

Solution

X

60°

⁄

h

h cp

y

̅

F ycp

x'

⁄

G C.P. y' Y

Fig. 2.45 Ex. 2-22

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

36

Example 2-23 Isosceles triangle gate shown in Fig.2.46 is hinged at AC and weighs 1500 N. Determine the horizontal force P required for equilibrium.

C

1m

Oil S=0.83

3m

A A

Solution

B

2m 50° P

̅ ̅

B Fig. 2.46 Ex. 2-23

̅ ̅ 2.611 =0.87m 3

1m 3.916

̅

G

̅ ̅ 2.611m

Fig. 2.47 Ex. 2-23

x y ycp

h G

h cp

A

x'

C.P

y

F 0.87 cos50=0.559m

P 1500 N

Fig. 2.48 Ex. 2-23 CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

37

Example 2-24 Rectangular gate AB in shown in Fig. 2.49 is 4 m wide. Determine the hydrostatic force acting on the gate if a- p is atmospheric pressure. b- p=26 kPa gage.

Air with pressure p

Water

4m

A B 2m

30°

Fig. 2.49 Ex. 2-24

Solution a- p=p0 (Fig. 2.50) X

30°

h hcp A

y

F ycp

B

x' G

C.P.

y' 4m

Y

Fig. 2.50 Ex. 2-24

̅ ̅ ̅

̅⁄

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

I.F.S.

̅

̅

38

⁄

2.65m

̅

6.65m 7.16m 4m

b- p=26 kPa (Fig. 2.51) To determine the imaginary free surface (I.F.S.), we will locate a surface at which the pressure = 0 as follows: ⁄

F

⁄

̅ Fig. 2.51 Ex. 2-24

̅ ̅

̅⁄ ⁄ 4m 5m 0.963m

Check using the pressure prism: Case a (Fig. 2.52):





F

F acts at the prism centroid, taking ⁄

about lower edge. Fig. 2.52 Ex. 2-24

⁄

⁄

Case b (Fig. 2.53):

6.65m 7.65m

F acts at the prism centroid, taking ⁄

about lower edge.

0.98m 

⁄

⁄

F



Fig. 2.53 Ex. 2-24 CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

39

Example 2-25 What is the depth of water required to tip the rectangular flash board in Fig. 2.54? Determine the hydrostatic force acting on the board. (Consider 1m width).

Water L h=?

Solution Hinge

AS the depth of water increases, the resultant hydrostatic force, F, moves upwards, and just before tipping, F acts at the hinge.

1m 50°

From Fig. 2.53

̅

Fig. 2.54 Ex. 2-25

̅ ̅ ⁄ ⁄

Substitute from Eq. (2) in Eq. (1)

X

50°

̅ ̅

hcp

⁄ ̅

h h

F y

̅

ycp

x' G 1m

C.P y'

1m Y

Fig. 2.55 Ex. 2-25

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

40

2.5 FORCES ON CURVED SURFACES Consider a curved surface AB immersed in a liquid, Fig. 2.56. The resultant hydrostatic force, F, acting on a curved surface has two components, the horizontal component, FH, has the same magnitude and line of action as the force computed on the projection of this surface on a vertical plane (A' B').

̅

D

h cp

E

F.S

C.G

h FV A'

C

A

AP



G

FH

F

B'

B

Fig. 2.56 Hydrostatic force on a curved surface

The vertical component, Fv, is equal to weight of the liquid extending above the area to the free surface (real or imaginary) and acts through the center of gravity, C.G, of this volume of liquid

where V is the volume EABCD in Fig. 2.56. The resultant hydrostatic forces √ Inclined at an angle

to the horizontal given by

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

41

Example 2-26

20 kPa

Determine the magnitude, direction, and location of the resultant hydrostatic force exerted by water on the curved surface AB which is a quarter of a circular cylinder, Fig. 2.57, the tank is 4 m long ( the paper).

Water

Solution

1.5m A

Determine the position of the imaginary free surface (I. F. S), i.e., a surface at which the gage pressure=0.

r=3m

⁄

B

⁄

Fig. 2.57 Ex. 2-26

I. F. S [

⁄

]

2.04m h h cp

C.G

1.5m

F

√



A FV

3m

√

FH (

FH

M

B

)

Fig. 2.58 Ex. 2-26

(

) 3.54 w A

F 6.54 w

M B Fig. 2.59 Ex. 2-26

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

42

Example 2-27 Determine the magnitude, direction, and location of the resultant hydrostatic force exerted by water on the gate shown in Fig. 2.60. Consider 1 m length ( the paper).

F.S

r= Water

6m

80 °

O

Solution ̅

Fig. 2.60 Ex. 2-27

V1: volume (A C C' A) V2: volume (A B C C' A) FV1

: volume (A B C A), "Sector – Triangle" (

C'

)

A

3.857

h

40°

C FH

√ (

40°

O

3.857

B

)

FV2

4.596

Fig. 2.61 Ex. 2-27



FH

O F

F



O

FV 7.714 w

Fig. 2.62 Ex. 2-27 CE325 Fluid Mechanics, Mr. Khalid Fadel

Fig. 2.63 Ex. 2-27 Chapter 2: Fluid Statics

43

Example 2-28 A cylindrical barrier holds water as shown in Fig. 2.64. Consider 1 m length ( the paper), determine (a)- the gravity force of the barrier. (b)- the force exerted by water against the smooth wall (horizontal force only).

2m

Water

Solution Fig. 2.64 Ex. 2-28

̅

FV1 I. F. S

(

2 w

W

) FH1

4 w

4 w

(

FH2

FV2

)

Fig. 2.65 Ex. 2-28

2 w

2 w

4 w Fig. 2.66 Ex. 2-28

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

44

Example 2-29 Determine the magnitude, direction, and location of the resultant hydrostatic force exerted by water on the curved surface which is a quarter of a circular cylinder, Fig. 2.67, consider 1 m long ( the paper).

Water 1m

Solution

0.5m

Fig. 2.67 Ex. 2-29

[

⁄

] I. F. S

√ w

√



(

FH

) 1.5 w

F (

FV

)

Fig. 2.68 Ex. 2-29

w

1.5 w

F Fig. 2.69 Ex. 2-29 CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

45

Example 2-30 A cylinder of radius 1.2 m and 1m length plugs a rectangular hole in a tank as shown in Fig. 2.70. With what force is the cylinder pressed against the bottom of the tank?

Water 3m

Solution

2m 1. 30°

Fig. 2.70 Ex. 2-30

[

⁄

[

]

]

2.4m

Fd 3m x2

x1

y

F u1

F u2

Fig. 2.71 Ex. 2-30

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

46

2.6 BUOYANCY FORCE A buoyancy force is defined as the net force exerted from fluid pressure on the surface of an object that is either completely or partially submerged in a fluid at rest. The buoyancy force and its line of action can be found by the principles just introduced for the calculation of forces on curved surfaces. Consider a body immersed in a liquid , Fig. 2.72. The horizontal components balance each other, the vertical components are , and , thus the forces acting on the body in vertical direction are: i- the weight W of the body ii- the net vertical component of the fluid pressure E

F

F.S

( ) FV1 B FH

G

A

V

C

FH

W D FV2 Fig. 2.72 Forces acting on submerged body

,

( is the volume EABCF),

acts through C.G of volume EABCF.

,

( is the volume EADCF),

acts through C.G of volume EADCF.

, ( is the volume ABCD, which is the volume of the body) is known as the buoyancy force FB.

= the weight of the liquid displaced. A submerged body or floating body is buoyed up with a force equal to the weight of the fluid displaced by the body and is called the force of buoyancy, FB, and acts through the center of gravity of the displaced volume and is called the center of buoyancy, B. If the fluid is homogeneous and the body is homogeneous, then the center of gravity of the body, G, and the center of gravity of the displaced fluid, B coincide. CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

47

When a body (stone or steel ball) is weighed in a liquid its effective weight is . Thus when a body is immersed in a liquid it suffers an apparent loss of weight equal to the weight of the liquid displaced, which is called the principle of Archimedes (250 B.C.).

F.S FB =  V

F.S

( )

V

G

( )

W V

G B

V is the volume of displaced fluid W (a) Immersed body

(b) Floating body

Fig. 2.73 Forces acting on immersed and floating body

If a body is submerged, i- if W = FB, the body is in equilibrium, which means the densities of the body and fluid are equal. ii- if W > FB, the body will sink. iii- if W < FB, the body will rise until the weight of displaced liquid equals the weight of the body (floating body like a ship).

Example 2-31 A body weighs 2080 N in air. When submerged in water it weighs 1000 N, Fig. 2.74. Determine the specific gravity of the body.

weight =2080N

weight =1000N

Balance String Air

Water V

V

Fig. 2.74 Ex. 2-31 CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

48

Solution

⁄

T= 2080 N

T=1000 N

Water

Air

V

V

W=mg=2080N W=mg=2080N

FB = wV

Fig. 2.75 Ex. 2-31

Example 2-32 A cube of timber 40 cm of each side floats in water as shown in Fig. 2.76. The specific gravity of the timber is 0.60. Find the submerged depth of the cube.

D=?

Water

Solution 40cm Fig. 2.76 Ex. 2-32

Weight of cube = weight of displaced water

Water

W

V

D

FB = wV Fig. 2.77 Ex. 2-32 CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

49

Example 2-33

F.S

A body having a volume of 0.20 m3 required a force of 250 N to keep it immersed in water and 160 N to keep it immersed in another liquid. Determine the weight of the body and the specific gravity of the body and the liquid.

G

Solution

( )

F

V

W In water:

FB =  V Fig. 2.77 Ex. 2-33

⁄

In liquid:

⁄

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

50

Example 2-34 A hydrometer weighing 0.1 N has a stem of 4 mm diameter. How much deeper will it float in oil of specific gravity 0.8 than in another liquid of specific gravity 0.84?

h=?

4mm

Solution Let a = the cross-sectional area of the stem

In liquid (S.G=0.84): Liquid (0.84)

Oil (0.8)

Fig. 2.78 Ex. 2-34

In oil (S.G=0.8):

From Eq. (1) V=1.2135 ×10-5 m3, substitute in Eq. (2) (

)

W FB =  V Fig. 2.79 Ex. 2-34

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

51

Example 2-35 To what depth will a 1 m diameter log of timber 4 m long and relative density 0.6 sink when floated in a river?

 Water

h

Solution

W

Weight of log = Weight of displaced water

FB = wV

Fig. 2.80 Ex. 2-35

(

)

* (

Using trigonometry,

(

+

, and simplifying yields

)

Using trial and error method to find (

)

as follows:

)



45

60

70

80

85

81

E

+ 0.242

+ 0.16

+ 0.09

+ 0.008

-0.035

-0.00064

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

52

Example 2-36 A block of steel (S=7.85) will float at a mercury-water interface as shown in the figure . what will be the ratio of distances a and b for this condition? The block has a width d perpendicular to the paper.

Water

h Steel block

a b

L Mercury (13.6) Solution Fig. 2.81 Ex. 2-36

[

]

FV1 [ [

] ]

w h w (h+a) W w (h+a) + Hgb

FV2 Fig. 2.82 Ex. 2-36

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

53

PROBLEMS 2-1

In the Fig., determine the pressure at point A, B, C, and D in pascals, ignoring air density. (Ans. pA = -5886 Pa, pB = pC = 5886 Pa, pD = 22661 Pa )

Air

A

Air

0.3 m

C

0.3 m

1

Oil S = 0.9

0.6 m

3

2

B

1m

Water

2-2

In the differential manometer shown in the figure h1=330mm, h2=480mm, and h3=230 mm, Find the difference (pA - pB). (Ans. pA – pB = 82.5 kPa)

D

Air B

h3

45°

Water A

h2 h1

Hg (S=13.6)

2-3

Oil (S=0.83)

For the inverted manometer shown in the figure, if (pB – pA)=97 kPa, what must the height h be in cm? (Ans. h = 22.5 cm)

18 cm

h A

Water 35 cm

B

Hg (S=13.6) CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

54

2-4

Find the gauge readings G1 and G2, if the barometric pressure of the place is 750 mm mercury. (Ans. G1=80.05 kPa, G2=109.48 kPa)

G1 =? 1

Open

Air

1.5m

Oil S=0.8

0.6m c

1.8m

Water

G2 =? Hg S=13.6

2-5

Circular gate ABC shown in the figure is 4m in diameter and hinged at B. Compute the force P just sufficient to keep the gate from the opening. (Ans. P = 61.64 kN )

Water 8m A

2m

B C

2-6

A dam retains 8 m of water, as shown in the figure. Find the total resultant force acting on the dam and the location of the center of pressure (Consider 1m width). (Ans. F = 362.48 kN, hcp = 5.333 m)

P

Water F

h cp 8m

60 °

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

55

2-8

2-9

A dam retains water, as shown in the figure. A strut is erected every 6 m to maintain the dam in the shown position. Determine the compressive force in the strut. Neglect the mass of the dam. (Ans. 602.51 kN)

The submerged sector gate AB shown in the figure has a radius of 6m. The length of the gate is 10 m. Determine the amount and location of the horizontal and vertical components of the total resultant force acting on the gate, also determine the amount and location of the total resultant force. (Ans. , )

The cylindrical gate shown in the figure has a radius of 1m, length of 1.5 m, and weight 200 kN. Determine the reactions at A and B, neglecting friction, also determine the resultant hydrostatic force acting on the gate. (Ans. )

Water

2m



5m

2-7

hinge

6m Strut

4m

B Water

60°

A

o r = 6m

1m

Oil (S=0.8)

A

W B

2-10 A cylindrical barrier holds water as shown in Fig. 2.64. Consider 1 m length ( the paper), determine (a)- the gravity force of the barrier. (b)- the force exerted by water against the smooth wall (horizontal force only). (Ans.

CE325 Fluid Mechanics, Mr. Khalid Fadel

Oil S=0.85

2m

Water

Chapter 2: Fluid Statics

56

2-11 The dome (hemisphere) is located below the water surface as shown. Determine the force components needed to hold the dome in place. (Ans. , )

Water

1m

1m

2-12 Determine the force components acting on the dome (hemisphere) shown in the figure. (Ans. , )

Oil S=0.8 1.5m

Water

2m

0. 5m

2-13 A body weighs 1000 N when immersed in water and when immersed in oil (S=0.85) it weighs 1150 N. Determine the volume, specific gravity and weight of the body in air. (Ans. , , )

weight =1150N

weight =1000N

Balance String

Water

Oil S=0.85 V

CE325 Fluid Mechanics, Mr. Khalid Fadel

V

Chapter 2: Fluid Statics

57

2-14 The cylindrical can in the figure floats in the position shown. What is its weight? (Ans. W = 5 N)

Water 8cm

D=9cm

2-15 To what depth d will this rectangular block (with density 0.8 times that of water) float in the two-liquid reservoir? Consider 1 unit length perpendicular to the paper. (Ans. d = 2.167 L )

L

S 1 =1.0

3L d=?

y

S 2 =1.2 6L

2 pulleys

2-16 A 3.5-cm-radius solid aluminum ball (Sp. Gravity, SA= 2.7) and a solid brass ball (Sp. Gravity, Sb=8.5) balance when submerged in water, as shown in the figure. What is the radius of the brass ball? (Ans. r = 2.13 cm )

Water

cm 3.5

? r= Brass

Aluminum

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 2: Fluid Statics

58

CHAPTER THREE BASIC CONCEPTS OF FLUID FLOW

3.1 INTRODUCTION The foundational axioms of fluid flow are the conservation laws, specifically, conservation of mass, conservation of energy (also known as first law of thermodynamics), and conservation of linear momentum (also known as Newton’s second law of motion). i. Conservation of Mass States that matter can be neither created nor destroyed. In fluid mechanics the rate of change of fluid mass inside a control volume must be equal to the net rate of fluid flow into the volume. ii. Conservation of Energy States that energy can be neither created nor destroyed, although it can be changed from one form into another. In an isolated system the sum of all forms of energy therefore remains constant. iii. Conservation of Linear Momentum States the fact that a body or system of bodies in motion retains its total momentum unless an external force is applied to it. The fundamental equations in fluid mechanics which are mathematical formulations of these conservation laws are: 1. Continuity Equation Implies to the law of conservation of mass. 2. Energy Equation and Bernoulli Equation Implies to the law of conservation of energy. 3. Momentum Equation Implies to the law of conservation of linear momentum. 3.2 STEADY AND UNSTEADY FLOW If the velocity (or any other variable) at a point remains constant with respect to time, the flow is said to be steady, otherwise unsteady (also called transient).

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

59

3.3 UNIFORM AND NON-UNIFORM FLOW If the velocity (or any other variable) at a point remains constant with respect to distance along a stream line, the flow is said to be uniform, otherwise non-uniform.

Steadiness refers to 'no change with time' and uniformity refers to 'no change with distance' therefore the flow can be steady or unsteady quite independent of its being uniform or non-uniform. Hence there can be four flow classifications as shown in Fig. 3.1.

Time (s) 1

i. Steady uniform 2

Constant

10 m 4

v

4

4

v

4

s1

s2

Constant 1

ii. Steady non-uniform 2

4

v

6

4

v

6 Variable

1

iii. Unsteady uniform 2

4

v

4

6

v

6 Variable

1

iv. Unsteady non-uniform 2

4

v

6

6

v

8

Fig. 3.1 Steadiness and uniformity

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

60

3.4 COMPRESSIBLE AND INCOMPRESSIBLE FLOW All fluids are compressible to some extent, that is, changes in pressure or temperature cause changes in density. However, in many situations the changes in pressure and temperature are sufficiently small that the changes in density are negligible, in this case the flow can be considered as an incompressible flow. Otherwise the more general compressible flow equations must be used.

3.5 STREAM LINE, STREAM TUBE, STREAK LINE, AND PATH LINE A stream line is an imaginary line within the flow for which the tangent at any point gives the velocity vector at that point and at that instant. Fig. 3.2(a) A stream line cannot intersect itself nor two stream lines can cross. Stream lines show the mean direction of a number of fluid particles at the same instant of time. Fig. 3.3(a) Stream lines v Stream line

(a) Stream line

(b) Stream tube Fig. 3.2 Stream line and stream tube

A stream tube is an imaginary tube of fluid bounded by closely spaced stream lines which enclose or confine the flow. Fig. 3.2(b). Streak lines are the lines traced by all the fluid particles at a given point at a particular instant. Fig. 3.3(b). A path line is the line traced by a single particle during successive instants of time and shows the directions of velocity (of the same fluid particle) at successive instants of time. A path line can intersect itself. Fig. 3.3(c). The stream line pattern remains same when the flow is steady. For unsteady flow, the stream line pattern changes from instant to instant. Stream lines, path lines and streak lines are all same for steady flow.

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

61

Dye injected

(a) Stream lines

(b) Streak lines

(c) Path lines

Fig. 3.3 Stream lines, streak lines and path line

3.6 FLOW RATE The discharge, Q, often called the volume flow rate, is the volume of fluid that passes through a cross-sectional area per unit time. The mass flow rate, ̇ , is the mass of fluid passing through a crosssectional area per unit time. Fig. 3.4. [ ⁄ ] [ ⁄ ] ̇

where v is the mean velocity A is the cross-sectional area (area perpendicular to the direction of v) is density of the fluid

A

v

Fig. 3.4 Definition of flow rate

3.7 CONTROL VOLUME A control volume is a volume located in space through which matter can pass. The selection of the control volume position and shape is problem-dependent. The control volume, CV, is enclosed by the control surface, CS, as shown in Fig. 3.5. Fluid mass enters and leaves the control volume through the control surface. The control volume can deform with time as well as move and rotate in space and the mass in the control volume can change with time. The control surface are represented by a vector, ⃗ , oriented outward from the control volume and with magnitude equal to the cross-sectional area. CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

62

A1

v1

Control surface

Control volume

v2 A2

Fig. 3.5 Definition of control volume

3.8 CONTINUITY EQUATION The law of conservation of mass, as mentioned before, states that the rate of change of fluid mass inside a control volume must be equal to the net rate of fluid flow into the volume. The general form of the continuity equation is as follows: ∫

∫

⃗

⃗

The first term represents the rate of decreasing of the mass inside the control volume. The second term represents the net mass outflow rate leaving the control volume (outflow – inflow). If the control volume has only a number of one-dimensional inlets and outlets, that is the velocities are uniform over the cross section, we can write ∑ ⃗ ⃗

∫

In case of steady state the first term in Eq. 3.5 equals to zero, hence we can write ∫

⃗

⃗

If the inlets and outlets are one-dimensional, we have for steady flow ∑ ⃗ ⃗ In case of incompressible flow ( is constant) the first term in Eq. 3.5 equals to zero, hence Eq. 3.5 becomes ∫ ⃗

⃗

or CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

63

∫ ⃗

⃗

If the inlets and outlets are one-dimensional, we have for steady flow ∑⃗ ⃗ Consider the one dimensional flow in the stream tube shown in Fig. 3.5, we can write ̇

̇

For incompressible flow

Because section 1 and 2 are arbitraries, therefore ̇ For incompressible flow

For incompressible fluids Eq. 3.15 is valid for steady and unsteady flow. In Fig. 3.5 if the velocities are uniform over the cross section, the net volume outflow rate can be written as ⃗ ⃗ ⃗ ⃗

Generally ∑⃗ ⃗ The net mass outflow rate can be written as ̇

⃗

⃗

⃗

⃗

Generally ̇

CE325 Fluid Mechanics, Mr. Khalid Fadel

∑ ⃗ ⃗

Chapter 3: Basic Concepts of Fluid Flow

64

Example 3-1 Water flows steadily through the nozzle in Fig. 3.6. Considering incompressible flow, compute Q1, Q2, v1, and

Water

̇.

1

Solution

⁄

2

D1 = 300 mm

D 2 = 450 mm

V1 = ?

V2 = 5 m/s

⁄ Fig. 3.6 Ex. 3-1

⁄

⁄

⁄

⁄

⁄ ⁄

̇ Example 3-2

Water flows through the wye fitting shown in Fig.

D 2 = 0.8 m

3.7. If Q2 = Q1/3 and the flow is steady and

V2 = ?

incompressible, compute Q1, v2, and D3. 1 2

Water

Solution

3

⁄ ⁄

D1 = 1.5 m

⁄

⁄

V1 = 1.92 m/s D3= ?

⁄

V3 = 2.5 m/s

⁄

⁄

⁄

⁄

Fig. 3.7 Ex. 3-2

∑⃗ ⃗

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

65

Or simply:

V2 A2

cs A1

V1

⁄

⁄

V3 A3

⁄

Fig. 3.8 Ex. 3-2 (Control Volume, CV)

⁄

Example 3-3

D 2 = 20 cm

Water flows through the tee fitting shown in Fig. 3.9. If

V2 = ?

the flow is steady and incompressible, compute the

2

volume flow rate and the mean velocity in section 2 and 1

show its direction. Water

Solution Assume ⃗ direction as shown in Fig. 3.10

D1 = 25 cm V1 = 4 m/s

⁄

3

⁄

D 3 = 20 cm

Fig. 3.9 Ex. 3-3 V3 = 10 m/s

⁄

⁄ V2

∑⃗ ⃗

A1

V1

A2

cs

Or simply

V3

⁄ CE325 Fluid Mechanics, Mr. Khalid Fadel

A3

Fig. 3.10 Ex. 3-3 Chapter 3: Basic Concepts of Fluid Flow

66

The negative sign of Q2 means that the direction of ⃗ is opposite.

⁄

⁄

⁄

⁄

Example 3-4 Water, Q 1 = 100 L/s

Water enters mixing device shown in Fig. 3.11 at a rate of 100 L/s through pipe 1, while alcohol with specific gravity 0.8 enters at 300 L/s If the flow is steady and incompressible, compute the volume flow rate and the density of the mixture.

Water-Alcohol Mix Q3 3

Solution For incompressible flow (Fig. 3.12): Alcohol (S=0.8), Q 2 = 300 L/s

∑⃗ ⃗

Fig. 3.11 Ex. 3-4

Or simply

A1

⁄

v1

⁄ v3

∑ ⃗ ⃗

cs

A3 v2

Or simply

A2

Fig. 3.12 Ex. 3-4

⁄

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

67

Example 3-5 A sprinkler shown in Fig. 3.13 has 25 nozzles, the

25 24 23

n

3

2 1 Q

velocity of water discharged from any nozzle is given by the relation

v25 v24 v23

vn

v3 v2 v1

⁄

Fig. 3.13 Ex. 3-5 Where n is the number of the nozzle. If the crosssectional area of each nozzle is 1 cm2, find the volume outflow rate from the sprinkler.

Solution (

)

∑⃗ ⃗

(

)

⁄

Example 3-6 Water is being added to the tank shown in Fig. 3.14 with a

Q 1 = 30 L/s

constant rate Q1= 30 L/s and drained with a constant rate Q2. Describe the following cases: i-

Q2 = 40 L/s

ii-

Q2 = 10 L/s

iii- Q2 = 30 L/s

Water Q2

Fig. 3.14 Ex. 3-6

Solution ∑⃗ ⃗ CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

68

Q 1 = 30 L/s

A1

Outflow > Inflow that means tank draining cs

v1

Outflow < Inflow that means tank filling

Water

CV

Q2

v2 A2

Outflow = Inflow that means steady state Fig. 3.15 Ex. 3-6

Example 3-7 Water is being added to the open tank with surface area AT = 100 m2 with a constant rate Q1= 80 m3/min and

dh dt Q1

drained with a constant rate Q2 = 50 m3/min. Determine

Water

the rate of rising of water level, dh/dt. Fig. 3.16.

h Q2

Fig. 3.16 Ex. 3-7

Solution ∑ ⃗ ⃗

∫

dh dt A1 cs

v1 CV v2 A2

Fig. 3.17 Ex. 3-7

⁄

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

69

Example 3-8 The pipe flow in Fig. 3.18 fills a cylindrical surge tank as shown. At time t = 0, the water depth in the tank is 30 cm.

D=75 cm

Estimate the time required to fill the remainder of the 1m

tank. 30 cm

Solution v1 = 2.5 m/s

∑ ⃗ ⃗

∫

Dia.= 12 cm

v 2 = 1.9 m/s

Fig. 3.18 Ex. 3-8

(

)

(

)

⁄ CV

∫

∫

[]

[ ]

cs

dh dt

v2 A1

v1 A2

Fig. 3.19 Ex. 3-8

Example 3-9 A 10 cm jet of water issues from a 1 m diameter tank as shown in Fig. 3.20. Assume that the velocity in the jet is

√

m/s,

how long will it take for the water surface in the tank to drop from h0= 2 m to hf= 0.50 m?

D =1 m

Water h

Dia.=10cm

Fig. 3.20 Ex. 3-9 CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

70

Solution ∑ ⃗ ⃗

∫

√

(

)

dh dt

⁄ CV

∫

h

cs

∫

Fig. 3.21 Ex. 3-9

[]

[ (

| )

3.9 BERNOULLI EQUATION Euler’s equation along a stream line in a flow of an ideal fluid is ( )

v 1

line

ds

p dA ) s + (p

dA

Stream

s

2

p dA

Fluid element

g dm (element weight) Fig. 3.22 Forces acting on an ideal fluid element

in the steady flow Euler’s equation becomes CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

71

( ) if the fluid is incompressible ( is constant), integrating Eq. (3.24) yields

Applying Eq. (3.25) in two points (1) and (2), Fig. 3.24, along the stream line yields

Eq. (3.26) is called the Bernoulli equation (Daniel Bernoulli, 1738). Fig. 3.23. 3.9.1 Restrictions on Bernoulli Equation Bernoulli equation is applied along a stream line and valid only for: i. Steady flow ii. Inviscid (frictionless) fluid iii. Incompressible fluid Piezometer

Pitot tube Total energy line (TEL) v22 2g

2

v1 2g

) e (HGL n i l e d a lic gr Hydrau

v12 2g p2



Constant Bernoulli head

p1



p1



2

Flow z2

1

z1

z1 Arbitrary datum (z=0)

Fig. 3.23 Frictionless flow in a duct (Bernoulli equation) CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

72

[ ] [ ] [ ] (

)

[ ]

[ ]

The law of conservation of energy, as mentioned before, states that energy can be neither created nor destroyed, although it can be changed from one form into another. In an isolated system the sum of all forms of energy remains constant. Therefore Bernoulli equation can be written as

where E is the energy per unit weight of the fluid. E has the dimension [L]. For an element has a mass m, it possess the following forms of energy: i. Potential energy (EP): Due to its elevation

ii. Kinetic energy(EK): Due to its velocity

iii. Flow energy(EF): Amount of work necessary to move the element distance L across a certain section against the pressure p.

Derivation of EF: To move the element (of volume V and weight W) distance L,

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

73

3.10 APPLICATIONS OF BERNOULLI EQUATION 3.10.1 Venturi Meter The smoothly necked-down system shown in Fig. 3.24 is called a venturi meter, it is used to determine the flow rate in a pipe.

TEL 2 1

v 2g

v22 2g

p1

HGL

p2





Flow

2

1 z2

z1

Datum

Fig. 3.24 Flow through a venturi meter

Example 3-10 Oil S=0.85

A venturi meter with a throat 10 cm diameter is installed in a pipe of 20 cm diameter as shown in Fig. 3.25. Determine the flow rate, Q, if the discharge coefficient Cd=0.98

Solution i-

D=10cm

2

30cm

Continuity Equation

1

D=20cm

y 20cm

S=13.6 Fig. 3.25 Ex. 3-10

ii- Bernoulli Equation

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

74

Take point 1 as the datum and substitute from Eq. (1) in Bernoulli Eq. (

)

(

)

iii- From the manometer:

substituting from Eq. (3) in Eq. (2) yields (

(

)

)

The theoretical flow rate Qth is

The actual flow rate Qa is

The fluid is assumed to be ideal in Bernoulli Eq. that means ignoring the shear stress and the velocity is distributed uniformly in the cross-sectional area, thus there will be a difference in the flow rate computed by (Qth=Av) and the actual flow rate Qa, This difference is considered using the correction factor Cd (discharge coefficient).

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

75

3.10.2 Flow through Small Orifices Consider the small orifice in the wall of the tank shown in Fig. 3.26. It is required to determine the velocity and the flow rate through the orifice.

1

applying Bernoulli Eq. (point 2 is the datum) Stream line h

Orifice diaeter = d

2 Fig. 3.26 Flow through a small orifice

The theoretical velocity at 2, v2th , is √ √

The actual velocity at 2, v2a , is

,

where Cv is the velocity coefficient.

Fig. 3.27 shows the stream lines in two different openings. Cc is the contraction coefficient

Stream line Stream line

Aa

A th

A th

Aa

2 2

(a) Sharp edge orifice

(b) Nozzle

Fig. 3.27 Flow through different openings

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

76

Summary:

Orifices and nozzles can be used to determine the flow rate in a pipe like a venturi meter. Fig. 3.28 Orifice

Pipe of diameter d

1

d

2 d0

y h

Diff. Manometer

(a) Sharp edge orifice Nozzle

Pipe of diameter d

1

d

2 d0

y h

Diff. Manometer

(b) Nozzle Fig. 3.28 Use of orifices and nozzles to determine the flow rate CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

77

Example 3-11 Water flows from water tank through an orifice in the wall of the tank to air tank as shown in

1

Fig. 3.29.

Determine the actual velocity and the discharge through

Water

the 5 cm diameter orifice if

G

h

a- G = 30 kPa Air

b- G=-30 kPa Assume a contraction coefficient Cc = 0.70 and a velocity coefficient Cv = 0.90

65 cm

2 50 cm

Hg (13.6)

Solution Fig. 3.29 Ex. 3-11

i-

From the manometer:

ii- Bernoulli Eq. (point1,2 datum point 2):

a- G=30 kPa:

The theoretical velocity at 2, v2th = 10.18 m/s The theoretical discharge Qth is

The actual velocity at 2, v2a , is

The actual discharge Qa is

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

78

b- G=-30 kPa:

The theoretical velocity at 2, v2th =

m/s

The theoretical discharge Qth is

The actual velocity at 2, v2a , is

The actual discharge Qa is

Example 3-12 A nozzle is installed in a pipe as shown in Fig. 3.28 (b). Derive an expression to find the actual discharge Q a as a function of the relative density S of the flowing fluid, relative density S o of the manometer fluid, Cd, d, d0, and h.

Solution iii- Continuity Equation

iv- Bernoulli Equation

substitute from Eq. (1) in Bernoulli Eq.

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

79

(

)

(

)

v- From the manometer:

(

)

(

)

substituting from Eq. (3) in Eq. (2) yields

(

)

(

(

)

(

[

( [

) )

]

) (

)

]

3.10.3 Flow through Large Orifices In small orifices the opening was small such that the can be considered constant, also the head h was constant. If the diameter or depth of the orifice is more as compared to the head h, neither the head nor the velocity is constant. For the large rectangular orifice shown in Fig. 3.30, consider an element strip (b dh) under a head h Area of the element: Velocity through the element:

√ √

Flow through the element: Flow through orifice: CE325 Fluid Mechanics, Mr. Khalid Fadel

∫ Chapter 3: Basic Concepts of Fluid Flow

80

∫ √

(

√

√

h

∫

√

⁄

(

⁄

⁄

⁄

) ⁄

)

H1

h

H2 d

dh

b

Fig. 3.30 Flow through large rectangular orifice

Example 3-13 For the large isosceles triangular orifice shown in Fig. 3.31, derive an expression to find the actual discharge Qa. Compute Qa if H1=60cm, H2= 120cm, b=90cm, Cd=0.7.

H1 H2



 b

Fig. 3.31 Ex. 3-14 CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

81

Solution Area of the element: H1 h

x h-H1

√

Velocity through the element:





dh

√

Flow through the element:

b

∫

Flow through orifice:

Fig. 3.32 Ex. 3-13

∫ √

[

√

√

⁄

⁄

⁄

[(

⁄

∫

√

⁄

)

| ⁄

(

⁄

)]

⁄

Substituting by H1=60cm, H2= 120cm, b=90cm yields ⁄

3.10.4 Flow through Notches and Weirs Notches are plates with sharp edged openings, they could be rectangular, triangular, trapezoidal, etc. used for flow measurements. Weirs are similar to notches usually provided in tanks and dams (spillway) to dispose off the flood waters (surplus works). In the rectangular notch (or weir) shown in Fig. 3.33, consider an element strip (b dh) under a head h √

Flow through the element: Flow through orifice:

∫

∫√ CE325 Fluid Mechanics, Mr. Khalid Fadel

√

∫

⁄

Chapter 3: Basic Concepts of Fluid Flow

82

h H dh

b Sharp-crested weir Fig. 3.33 Flow through rectangular weir

⁄

√

⁄

√

In the V – notch (or triangular notch) shown in Fig. 3.34, its left to the student to deduce that √

⁄

H



V- notch (or triangular notch) Fig. 3.34 Flow through V – notch (or triangular notch)

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

83

Example 3-14 For triangular notch shown in Fig. 3.34, compute Qa if H=20cm, = 90o, sand Cd=0.613.

Solution ⁄

√ (

)√

(

⁄

)

(

⁄

)

3.10.5 Cavitation In an enclosed flow system, Fig. 3.35, if the pressure at any point falls below the vapor pressure of the liquid, the liquid flashes into vapor and bubbles begin to form and travel along the flow till they reach a high pressure zone where they suddenly collapse. This phenomenon is known as cavitation. If the bubbles collapse on metallic surfaces, like impellers or blades of pumps or turbines, they cause damage called 'pitting'. Therefore cavitation must be avoided in the design of enclosed flow systems. Fig. 3.36 shows the condition of the negative pressure causing cavitation. TEL v12 2g

(+ve)

HGL

p1



v22 2g 1 (-ve)

z1

Flow

2 p2



z2 Datum

Fig. 3.35 Flow through a pipe of changing cross-sectional area (cavitation)

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

84

Local Atm. pressure

p2 p2

p0

p0

pv

pv

(a)

|

|

(b)

Boiling occurs then cavitation

Absolute zero

|

|

No boiling, no cavitation

Fig. 3.36 Condition of negative pressure causing cavitation

Example 3-15 A certain liquid has a relative density S=0.86 and vapor 2

7 N/cm2

pressure pv=2.7 N/cm . This liquid flows through the pipe shown in Fig. 3.37 with a discharge rate of 250 L/s. If the barometer

reading

is

758

mm

Flow

Hg (SHg = 13.6),

determine the minimum value of D2 at which no cavitation will occur.

1 D1 = 50 cm

2 D2 = ?

Fig. 3.37 Ex. 3-15

Solution

⁄

(

)⁄(

)

The maximum absolute value of p2 at which no cavitation will occur is |

|

(this value p2 is corresponding to the minimum value of D2)

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

85

i-

Continuity Equation ⁄

(

⁄

⁄

⁄

⁄

⁄

)⁄ ⁄

⁄

⁄

⁄

⁄

ii- Bernoulli Equation

⁄

3.10.6 Pitot Tube If a stream of uniform velocity flows around a body, the stream lines take a pattern as shown in Fig. 3.38. The stream line in the center is called dividing stream line, it goes to the tip of the body and stops because at this point, s, the velocity equals to zero. This point is known as a stagnation point.

Dividing stream line

s

Uniform flow Fig. 3.38 Streamlines around a body

Piezometer and Pitot tube, shown in Fig. 3.39, are used to determine the velocity of the flowing fluid in the horizontal pipe using the idea of the stagnation point. Applying Bernoulli Eq. between point 1 and the stagnation point s yields

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

86

Piezometer

Pitot tube

H

ps



p1



1

s

Flow

Fig. 3.39 Piezometer and Pitot tube installation in a pipe

Point 1 is located at the uniform flow, hence v1=v and p1=p

From Fig. 3.39

The theoretical velocity of the fluid, vth , is √ The actual velocity of the fluid, va , is √ From Eq. 3-35

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

87

Example 3-16 Water flows through the pipe shown in Fig. 3.40 with a discharge rate of 300 L/s.

Dia. = 30 cm Dia. = 20 cm 1

s

Q=300 L/s

2

Determine the manometer reading h.

y

Solution i-

Bernoulli Eq. (1 and s)

h SHg=13.6

Fig. 3.40 Ex. 3-16

ii- Manometer relation

Substitute from (1) in (2) for ps

iii- Bernoulli Eq. (1 and 2)

From (3) and (4)

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

88

⁄ ⁄

⁄

⁄

⁄

Substitute in (5) yields ⁄

3.10.7 Siphon The word siphon is used to refer to that involve the flow of liquids through tubes in an inverted 'U' shape, which causes a liquid to flow upward, above the surface of a reservoir, with no pump, but powered by the fall of the liquid

U-tube

Reservoir

as it flows down the tube under the pull of gravity, then discharging at a level lower than the surface of the reservoir it came from. Fig. 3.41.

Fig. 3.41 Siphon

Example 3-17 C

Water is siphoned through a 150 mm rubber tube as shown in . Fig. 3.42.

1

1.5m

B 1m

i-

If the depth of the outlet h=5m, determine the discharge through the siphon and the minimum

A h

Water

pressure in the siphon tube. What is the pressure at

D

B (inside the tube)? ii-

What is the depth of the outlet, h, at which no

Fig. 3.42 Ex. 3-17

discharge will occur? iii-

If the minimum pressure in the siphon should not fall 7.5 m of water below atmospheric pressure, what is the maximum possible depth of the outlet, h, below the water surface?

Solution i-

Bernoulli Eq. (1 and D, datum D):

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

89

⁄

√ Continuity Equation: ⁄

⁄

Since the area of the tube is constant, then The minimum pressure in the siphon is at the summit C. Bernoulli Eq. (1 and C, datum 1):

Bernoulli Eq. (1 and B):

ii- Bernoulli Eq. (1 and D, datum D):

No discharge will occur, that means Q=0 and v=0

iii- Minimum pressure in the siphon

⁄

:

Bernoulli Eq. (C and D, datum D):

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

90

3.11 MOMENTUM EQUATION The law of conservation of linear momentum, as mentioned before, states that a body or system of bodies in motion retains its total momentum unless an external force is applied to it. The momentum, ⃗⃗⃗ , for any moving body of mass m is a vector ⃗⃗⃗

⃗

Newton's II law is Force in any direction = Rate of change of momentum in that direction For steady flow, if the inlets and outlets of the control volume are one-dimensional, that is the velocities are uniform over the cross section, the momentum equation can be written as ̇ ⃗ Since ̇

⃗

, for incompressible flow equation 3.39 can be written as ⃗

⃗

Example 3-18 p2

The reducing pipe bend shown in Fig. 3.43 is in a horizontal plane. Water enters at section 1 with a

p1 =150 kPa

3

flow of 0.05 m /s and pressure p1=150 kPa. Section 2 100-mm Dia.

Determine the force exerted by water on the bend.

 =30°

Solution

Section 1 150-mm Dia.

i- Continuity Equation Fig. 3.43 Ex. 3-18

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

91

ii- Bernoulli Equation Horizontal pipe that means Z1=Z2

p2 A 2 v2

iii- Momentum Equation a- in x-direction ⃗

p1 A 1

⃗

(

v1

CV

FX

Fy Fig. 3.44 Ex. 3-18

)

(

)

b- in y-direction ⃗

⃗ FX 

(

) Fy

F

√ (

)

F

: is the force exerted by surface on water.

F

: is the force exerted by water on surface, same magnitude and opposite direction.

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

92

Example 3-19 Q 2 , D2

Fig. 3.45 depicts a horizontal wye-pipe. Water enters at section 1 with a flow Q1 = 60 L/s and pressure p1 = 35 kPa.

Q 3 , D3

If D1 = D2

3

2

=15 cm, D3 = 7.5 cm, and Q2 = 40 L/s, determine the force exerted by water on the pipe.

30°

Solution i- Continuity Equation

p1 = 35 kPa

1 Q1 , D1

Fig. 3.45 Ex. 3-19

ii- Bernoulli Equation Horizontal pipe that means z1 = z2 = z3 - Between 1 and 2 p2 A 2

v2

FX - Between 1 and 3

p3 A 3

v3

CV

Fy v1

p1 A 1 Fig. 3.46 Ex. 3-19

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

93

iii- Momentum Equation a- in x-direction ⃗

⃗

(

)

⃗

b- in y-direction (

⃗

⃗

(

)

)

(

⃗

(

)

)

(

)

√ Fy

(

)



FX

F

: is the force exerted by surface on water.

F

: is the force exerted by water on surface, same magnitude and opposite direction.

CE325 Fluid Mechanics, Mr. Khalid Fadel

F

Chapter 3: Basic Concepts of Fluid Flow

94

Example 3-20 For the flow of water over the spillway shown in Fig. 3.47 determine the thrust of water per meter length of the spillway.

Solution For the control volume bounded by 1, A

16 m Water

and 2, B and L =1m, Fig. 3.48

Spillway

i- Continuity Equation

1m

ii- Bernoulli Equation Two points 1 and 2 are on the free surface

Fig. 3.47 Ex. 3-20

both of them are on the same streamline.

1

v1

16 m

CV

F1

F v2 Spillway

 w

2 F2 B  w

A

1m

Fig. 3.48 Ex. 3-21 CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

95

Solving (1) and (2) yields v1 = 1.074 m/s,

v2 = 17.19 m/s ⁄

iii- Momentum Equation ⃗

⃗

,

(force exerted by water on the spillway)

Example 3-21 A vertical sluice gate spans a rectangular channel. When the gate is raised by 0.3 m above the floor, the upstream depth is 3m,

Water Sluice gate

Cc=0.6. Determine the horizontal force exerted by water per meter length of the sluice gate. Fig. 3.49.

3m

0.3 m

Fig. 3.49 Ex. 3-21

Solution For the control volume bounded by 1, A and 2, B and L =1m, Fig.3.50 Assume Cd = Cc = 0.6 (Cv=1)

i- Continuity Equation

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

96

v 2g

2 1

TEL 1 Sluice gate

Water v1

y1 = 3 m

v22 2g

F CV

F1

v2 2

0.3 m  w

B

A

F2 y 2 w

y2

Fig. 3.50 Ex. 3-21

ii- Bernoulli Equation 1 and 2 are two points on the free surface both of them are on the same streamline.

Solving (1) and (2) yields

v2 = 7.452 m/s,

v1 = 0.447 m/s ⁄

iii-

Momentum Equation ⃗

FWater

F gate

⃗ Pressure distribution on sluice gate

(Fgate) (force exerted by water on the spillway, Fwater) CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

97

D 3 = 6 cm

PROBLEMS 3-1

Three pipes steadily deliver water to a large exit pipe D 4 = 9 cm

shown in the figure. The velocity v2 = 5 m/s, and the exit flow rate Q4 = 33.33 L/s. Find (a) v1, (b) v3, and (c) v4 if it is known that increasing Q3 by 20 percent

D 2 = 5 cm

would increase Q4 by 10 percent. (Ans. v1 =5.45 m/s, v3=5.9 m/s, v4=5.24 m/s) D1 = 4 cm

3-2

Dia.=8 cm

The nozzle shown in the figure is conical. Find a

Dia.=3 cm

relation between the velocity along the nozzle axis Q = 10 L/s

with the distance x. (Ans.

⁄(

)

⁄ ) x L

3-3

Q 1 = 17.5 L/s

A jet of water discharges into an open tank at a rate of 17.5 L/s, and water leaves the tank through an

dh dt

orifice in the bottom at a rate of 3 L/s. At what rate is water accumulating in the tank? If the tank surface area AT = 9 m2, compute the rate of rising of water

Water

level. (Ans. Qo =14.5 L/s, dh/dt = 1.6 mm/s) Q 2 = 3 L/s

3-4

A river discharges into a reservoir at a constant rate of 7000 m3/s. The outflow

River

rate from the reservoir through the flow Dam

passages in the dam is 11000 m3/s. If the reservoir surface area is 100 km2, what is the rate of fall of water level in the

Reservoir Flow passage

reservoir? (Ans. dh/dt = - 3.456 m/day)

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

98

3-5

A venturi meter with a throat 15 cm diameter is installed in a pipe of 30 cm diameter as shown in the figure.

20 cm

y1

Determine the flow rate, Q, if the

1

discharge coefficient Cd=0.98 (Ans. Qa = 0.0355 m3/s)

y2

z1 -z 2 2

Water 3-6

The Figure shows an orifice in the wall of the tank.

62 kPa

Air

Determine the discharge through the orifice. Assume a

1

contraction coefficient Cc = 1 and a velocity coefficient Cv = 0.98 Water

(Ans. Qa = 15.7 L/s)

2m

Nozzle 4cm Dia. 2

3-7

Water is discharged from a dam through 10 cm diameter pipe, as shown in the figure. Determine the discharge through the pipe and the pressure at A in the pipe a- with the nozzle b- without the nozzle. What is the power available from the jet in the two cases? (Ans. (a) Q = 32.5 L/s, pA = 69.9 kPa, P = 4461 W )

1 Water 8m

Dam

Pipe 10 cm Dia. A 6m

Nozzle 5 cm v 2

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

99

3-8

Determine the discharge of water through the nozzle shown in the figure. (Ans. Q = 241 L/s)

Dia. = 15 cm

Dia. = 7.5 cm s

2

1

75 cm SHg=13.6

3-9

Determine the horizontal force required to fix the nozzle as shown in the figure. (Ans. Fx = 7.07 kN)

Dia. = 20 cm 1

Dia. = 10 cm 2

s

Oil (S=0.85)

20 cm

20 cm

SHg=13.6

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 3: Basic Concepts of Fluid Flow

100

CHAPTER FOUR VISCOUS FLOW IN CLOSED CONDUITS

4.1 INTRODUCTION A conduit is any pipe, tube, or duct that is completely filled with a flowing fluid. Examples include a pipeline transporting water or liquefied natural gas, and a duct transporting air for heating of a building. A pipe that is partially filled with a flowing fluid, for example a drainage pipe, is classified as an open-channel flow and will be analyzed separately. In ideal fluid flow there is no energy loss because shear stress is not considered. In reality there is no existence of ideal (or inviscid) fluid. In real (or technically viscous or viscid) fluid flow energy loss takes place because shear stress is considered. The main goal of this chapter is to describe how to predict head loss 4.2 ENERGY EQUATION The law of conservation of energy, as presented in chapter three, states that energy can be neither created nor destroyed, although it can be changed from one form into another. In an isolated system the sum of all forms of energy therefore remains constant. The general energy equation in the direction of flow in a pipe (Fig. 4.1) can be written as, Energy at Sec.1 + Energy added - Energy deleted = Energy at Sec.2 + Energy lost The above energy equation can be written as, (

)

(

)

Where Hp is the head added by pump (energy added per unit weight), [L] Ht is the head deleted by turbine or fluid motors (energy deleted per unit weight), [L] hL is the head losses (the summation of hf, major losses due to friction, and hm, minor losses due to fittings fixed on conduits, inlets, outlets, and valves), [L] the other terms were predefined in section 3.9

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

101

Where f:

friction factor

D:

internal diameter of pipe

L:

length of pipe

K:

minor loss coefficient

Total en ergy line

hL

(TEL) v22 2g

2 1

v 2g

) e (HGL

e lin lic grad u a r d y H

p2



p1



2

Flow z2

z1

1 Arbitrary datum (z=0)

Fig. 4.1 Viscous flow in a duct (E1 = E2 + hL)

Example 4-1 A pipe 200 m long shown in Fig. 4.2 carries 100 L/s of oil (S=0.85). Determine the velocities at the two ends and the pressure at end 2 assuming head losses hL=1 m.

Solution ⁄ CE325 Fluid Mechanics, Mr. Khalid Fadel

⁄ Chapter 4: Viscous Flow in Closed Conduits

102

50 kPa 200 m long

G=? Q=100 L/s

D 1 = 80 cm z1 = 1 m

D 2 = 40 cm z 2= 0

Fig. 4.2 Ex. 4-1

iv- Continuity Equation

⁄

⁄

⁄

⁄

v- Energy Equation

Note: p2 > p1 indicates that flow can take place from a section of lower pressure head to a section of higher pressure head. It is the total energy E which governs the flow direction, i.e. E1> E2 for flow from 1 to 2.

TEL

2

v1 2g =0.002 m

h L =1m 2

HGL

p1



v2 2g =0.032 m

=5.996 m

p2



50 kPa

=6.966 m

58.09 kPa

z 1 =2 m

Q=100 L/s

Fig. 4.3 Ex. 4-1 CE325 Fluid Mechanics, Mr. Khalid Fadel

z 2 =0 Chapter 4: Viscous Flow in Closed Conduits

103

Example 4-2 Water has to be conveyed at a rate Q=220 L/s from tank A to tank B through the pipe line as shown in Fig. 4.4. Assume constant friction factor f=0.012, determine the required power of the pump having an efficiency of 80%.

Level = 249 m B

K=0.35

K=0.5

Tank B Level = 220 m A

Tank A

Valve

Q=220 L/s

Pump

K=0.2

K=0.3

K=0.35

L1=500m, D1=40cm

L 2 =1000m, D 2 =30cm

Fig. 4.4 Ex. 4-2

Solution Applying the energy Eq. between point 1 and 2, point 1 is the datum

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

104

Substituting in (1) yields

TEL h f 2 +h m 2 elbows

HGL 2

v2 2g

Exit loss

Hp

Entrance loss h f 1 +h mvalve 2

v1 2g suction pipe

Pump

discharge pipe

Fig. 4.5 Ex. 4-2 (TEL and HGL)

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

105

Example 4-3 What power can be developed by the turbine shown in Fig. 4.6 if the turbine efficiency is 85%? Assume that f =0.015 for the two pipes and neglect the minor losses.

Elev. 93 m A

Reservoir

Dam

Elev. 50 m B

L 2 =20m, D 2 =60cm

L1=50m, D1=30cm

water 3

Q=0.28 m /s Turbine

Fig. 4.6 Ex. 4-3

Solution Applying the energy Eq. between point A and B, point B is the datum

Substituting in (1) yields

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

106

TEL 2 1

v

2g

h f1

HGL Ht

Reservoir

h f2

v22 2g water suction pipe

discharge pipe Turbine

Fig. 4.7 Ex. 4-3 (TEL and HGL)

4.3 LAMINAR AND TURBULENT FLOW Flow in a conduit is classified as being either laminar or turbulent, depending on the magnitude of the Reynolds number. The original research involved visualizing flow in a glass tube as shown in Fig. 4.7a. Reynolds in the 1880s injected dye into the center of the tube and observed the following: - When the velocity was low, the streak of dye (a streakline) will remain as a well-defined line as it flows along. - If velocity was increased, the dye streak fluctuates and intermittent bursts of irregular behavior appear along the streak. - If velocity was increased, the dye streak almost immediately becomes blurred and spreads across the entire pipe in a random fashion. These three characteristics, denotes as laminar, transitional, and turbulent flow, respectively, are illustrated in Fig. 4.8b. Dye

Laminar Q = vA

water

Transitional

D Turbulent

(a)

(b) Fig. 4.8 Reynolds’ experiments

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

107

Based on Reynolds’ experiments, engineers use guidelines to establish whether or not flow in a conduit will be laminar or turbulent. The guidelines used in this text are as follows: Re < 2000 2000 ≤ Re ≤ 4000 Re > 4000

Where D is the pipe diameter, respectively

laminar flow transitional flow turbulent flow

(4.5)

are the density, dynamic viscosity, and velocity of the fluid,

4.4 HEAD LOSSES 4.4.1 Head Loss Due to Friction The Darcy-Weisbach equation (Eq. 4.3) is the most general equation in the pipe flow application. It was obtained experimentally.

The friction relation depends on the state of flow, which is classified according to the Renolds number. For laminar flow (Re <2000), the friction factor is a function of the Renolds number only.

In the critical region of Re between 2000 and 4000 (transitional flow) the flow alternates between laminar and turbulent flow. Any friction factor relation cannot be applied with certainty in this region. In turbulent regime, the friction factor is a function of the Renolds number as well as the relative roughness of the pipe surface. During 1932 and 1933, Nikuradse published the results of now famous experiments on smooth (uncoated) and rough pipes coated with sand grains of uniform size. In contrast to the Nikuradse sand roughness, the roughness of the commercial pipe is not uniform, it is given in equivalent sand roughness. Table 4.1indicates equivalent roughness, , for pipe of different material. Prandtl and von Karaman established the following equations for smooth and fully rough pipes of. For smooth pipe in a turbulent regime:

√ CE325 Fluid Mechanics, Mr. Khalid Fadel

(

√

) Chapter 4: Viscous Flow in Closed Conduits

108

For fully rough pipes in a turbulent regime: (

√

⁄

)

In 1939, Colebrook and White established the following equation which covers the entire turbulent regime: (

√

⁄ √

)

In 1976, Jain has suggested the following explicit equation for the entire turbulent regime, which gives results within 1% of the Colebrook - White equation:

√

(

⁄

)

From the implicit relations of Prandtl and von Karaman and Colebrook – White, Moody (1944) prepared a diagram between the friction factor versus the Renolds number and the relative roughness as shown in Fig. 4.9. Table 4.1 Roughness Values for Pipes Pipe Material

Equivalent Roughness, (mm) Smooth Smooth

Brass, copper, aluminum PVC, plastic Cast iron 0.24 new old 0.15 Galvanized iron 0.12 Asphalted iron 0.046 Wrought iron 0.046 Commercial iron and welded steel 1.83 Riveted steel 1.22 Concrete 0.61 Wood stave Ram S. Gupta, hydrology and hydraulic systems

CE325 Fluid Mechanics, Mr. Khalid Fadel

Hazen-Williams Coefficient, C 140 150 130 100 120 120 110 130 120

Chapter 4: Viscous Flow in Closed Conduits

109

Fig. 4.9 Moody diagram

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

110

The turbulent flow is classified in three categories as follows: iFlow in smooth pipe, where the relative roughness is very small. Curve in Fig. 4.10a. ii- Flow in fully rough pipe. Zone illustrated in Fig. 4.10b. iii- Flow in partially rough pipe where both the relative roughness and viscosity are significant. It is transitional zone from smooth to rough pipes. Zone illustrated in Fig. 4.10c.

Transitional zone from smooth to rough

Rough turbulent zone

smooth pipes

4000

4000

(a)

4000

(b)

(c)

Fig. 4.10 Categories of turbulent flow

Summary: a- Friction factor for laminar can be found from Eq. 4.7 or from Moody diagram. b- Friction factor for turbulent flow can be found in three ways: i- Graphical solution: Moody Diagram, Fig. 4.9. ii- Implicit equation : Colebrook-White formula, Eq. 4.10. iii- Explicit equation: Jain formula, Eq. 4.11. 4.4.2 Types of Single Pipes Problems There are three types of single pipes problems: a. Type I, given: Q, L, D,

required: hf

b. Type II, given: hf , L, D,

required: Q

c. Type III, given: hf , Q, L,

required: D

Example 4-4 Determine the head loss due to friction for water flowing at 20oC at a rate of 2.8 L/s in a new cast iron pipe of diameter 5cm and length 10m. Solution (Type I problem) ⁄

⁄

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

111

⁄

⁄

⁄

At 20oC, kinematic viscosity,

f = 0.032

/D=0.0048 Re=7.125 10 4

Since R>4000, turbulent flow. From table, = 0.24 mm relative roughness ⁄

⁄

from Moody diagram f = 0.032 Alternative Solution: From Jain equation (4.11) (

√

⁄

(

√

)

)

Alternative Solution: From Colebrook and White equation (4.10) (

√

√

(

⁄ √

)

√

)

An iterative procedure to obtain f can be done as follows. We assume a value of ( f =0.02, for example), substitute in the right- hand side of Eq. (1) and calculate a new f from the left- hand side of Eq. (1) ( f = 0.0316 in this case). Since the two value do not agree, the assumed value is not the solution. Hence, we try again. This time we assume f = 0.0316 (the last value calculated) and calculate a new value as f =0.0313. One more iteration shows that the assumed and the calculated values converge to the solution: f =0.0313.

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

112

Example 4-5 Water flows at a rate 0.020 m3/s from reservoir A to reservoir B through a smooth pipe (PVC) as shown in Fig. 4.11. Find the difference in water surfaces elevation in the reservoirs, H, neglecting minor losses. ⁄ . Assume a

Reservoir A

H

PVC p

ipe, L =500 0m, 3 00 m m Di a.

b

Reservoir B

Fig. 4.11 Ex. 4-5

Solution smooth pipe

f = 0.0185 Re=8.49 10 4

Using Moody diagram, from smooth pipes curve f = 0.0185 Alternatively, using Jain equation with ⁄

√

√

(

(

for smooth pipe as follows

)

)

TEL HGL

2

v1 2g

Applying the energy Eq. between point a and b, point b is the datum

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

113

Example 4-6 Solve example 4-5 with flow rate Q=0.4 L/s Solution Q=0.0004 m3/s

Using Moody diagram, f = 0.038 Laminar flow

Alternatively, using the equation for laminar flow as follows

f = 0.038

Re=1698

Example 4-7 A swimming pool of 8 m length, 4 m width, and 2.5 m depth is to be filled from a garden hose (smooth interior) of length 30 m and diameter 2 cm diameter, Fig. 4.13. If the pressure at the faucet is 1 bar, how long will it take to fill the pool? Water exits the hose as a free jet at the same elevation of the faucet. ⁄ . Assume

1

, 2cm L=30m

Dia.

2 swimming pool 8 4 2.5m

Fig. 4.13 Ex. 4-7

Solution Applying the energy Eq. between point 1 and 2

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

114

v1 = v2 = v,

z1 = z2 = 0,

p2 = 0

hf , L, and D are known, Q is required

using Jain equation with ⁄ (

√

(Type II problem)

for smooth pipe as follows:

)

Using trial and error method to solve Eqs. (1) and (2) as follows: -Assume f =.02, from Eq. (1) v = 2.557 m/s, from Eq. (2) Re = 51147, from Eq. (3) -Assume f =.0206, from Eq. (1) v = 2.518 m/s, from Eq. (2) Re = 50351, from Eq. (3)

⁄

from Eq. (1) v = 2.514 m/s, V = Q t, ⁄

where t = filling time,

⁄

and V = pool volume.

⁄

Note that, alternatively, f can be determined using trial and error method using Moody diagram from smooth pipes curve as follows: -Assume f =.02, from (1) v = 2.557 m/s, from (2) Re = 51147, from Moody diagram CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

115

-Assume f =.0206, from (1) v = 2.518 m/s, from (2) Re = 50351, from Moody diagram

Example 4-8 Water flows at a rate 1.0 m3/s from reservoir A to reservoir B through a rough concrete pipe ( =3mm) as shown in Fig. 4.14. Determine the pipe diameter if the difference in water surfaces elevation in the ⁄ . reservoirs is 10 m and the pipe length is 1000 m. Neglect minor losses. Assume 1

Reservoir A

10 m

=3m m, L =100 0m, D=

2

?

Reservoir B

Fig. 4.14 Ex. 4-8

Solution Applying the energy Eq. between point 1 and 2

hf , L, and Q are known, D is required

(

(Type III problem)

)

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

116

using Jain equation: (

√

⁄

)

(

√

)

Using trial and error method to solve Eqs. (1), (2) and (3) as follows: -Assume f =0.02, from Eq. (1) D = 0.698 m, from Eq. (2) Re = 1.63×106, from Eq. (3) -Assume f =.029, from Eq. (1) D = 0.752 m, from Eq. (2) Re = 1.51×106, from Eq. (3) -Assume f =.0285, from Eq. (1) D = 0.749 m, from Eq. (2) Re =1.518×106, from Eq. (3)

, and D = 0.749 m ⁄

from Eq. (1) v = 2.514 m/s, V = Q t, ⁄

where t = filling time,

⁄

and V = pool volume.

⁄

Note that, alternatively, f can be determined using trial and error method using Moody diagram.

4.5 Noncircular Conduits Many round pipe results can be carried over, with slight modification, to flow in conduits of other shapes. Practical, easy-to-use results can be obtained by introducing the hydraulic diameter.

where Dh = hydraulic diameter A = cross-sectional flow area P = wetted perimeter The hydraulic diameter is used in the definition of the friction factor, CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

117

the Renolds number,

and the relative roughness, ⁄

Example 4-9 ⁄ flows Air at a temperature of 45oC from a furnace through an 20-cm-diameter pipe with an average velocity of 3 m/s. It then passes through a transition section and into a square duct whose side of length a. The pipe and duct surfaces are smooth ( =0). Determine the duct size, a, if the head loss per meter is to be the same for the pipe and the duct.

20 cm

a

(1)

(2)

Fig. 4.15 Ex. 4-9

Solution For circular pipe: ⁄

(

√

√

(

)

)

For the square duct:

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

118

From continuity

⁄

⁄

vs is the velocity in the square duct. Combining Eqs. 1 and 2 we obtain ⁄

⁄

⁄

using Jain equation: √

(

)

(Type III problem). We have three unknowns (a, f, and Re) and three equations (3), (4), and (5). Using trial and error method to solve these equations as follows: -Assume f =0.023, from Eq. (3) a = 0.181 m, from Eq. (4) Re = 29845, from Eq. (5)

, and a = 0.181 m

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

119

4.6 Hazen=Williams Equation for Fiction Head Loss Another common formula for head loss in pipes that has found almost exclusive usage on water supply engineering is the Hazen-Williams equation:

where v A R S

= = = =

mean velocity of flow, (m/s) Hazen-Williams coefficient of roughness given in table (4.1) hydraulic radius, (m) slope of energy gradient

where A P hf L

= = = =

cross-sectional flow area wetted perimeter friction head loss length of conduit

Hazen-Williams equation can be written as

( )

Note that for circular pipe as follows:

⁄

⁄ , Eq. (4.15) can be rearranged for circular pipe

Two sources of error in the Hazen-Williams equation formula are: (1)the multiplying factor 0.849 should change for different R and S, and (2) the Hazen-Williams coefficient C is considered to be related to the pipe material only, whereas it must also depend on pipe diameter, velocity, and viscosity similar to the friction factor of Darcy-Weisbach.

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits

120

Example 4-10 Solve Example 4-4 using Hazen-Williams equation. Solution From table 4.1 C=130

( )

(

)

(

)

or

CE325 Fluid Mechanics, Mr. Khalid Fadel

Chapter 4: Viscous Flow in Closed Conduits