1

CE 642

HYDRAULICS

Dr. Emre Can

� Tentative Course Outline

� Introduction � Pipe Flow� Open Channel Flows

� Uniform Flow� Non-Uniform Flow� Local Changes in Water Levels

� Channel Controls� Sedimentation in Open Channels and Rivers� Dimensional Analysis & Theory of Models

HYDRAULICS

EXAM SCHEDULE

� 31 March 15:00 Midterm exam 1� 12 May 15:00 Midterm exam 2� The exams will always be closed book,

(however formula sheets will be provided)

� Questions will be in English and there will be no translation of questions into Turkish,

� Answers to all the questions should be in English.

HYDRAULICS

REFERENCES:

� Chow, V.T., Open Channel Hydraulics, , Mc Graw Hill, New York, 1959.

� Henderson, F.M., Open Channel Flow, Macmillan Co, 1966.

� Vennard, J.K. & Street, R.L., Elementary Fluid Mechanics,John Wiley & Sons, 1977.

� Linsley, R.K. & Franzini, J.B., Water Resources Engineering, McGraw Hill, Newyork, 1972

HYDRAULICS

REFERENCES:

� Sümer, B.M, Ünsal, İ. & Bayazıt M. Hidrolik,Birsen yayınevi

� Yanmaz, A. Melih, Applied Water Resources Engineering, Metu Press, 3rd edition, 2006

� CE 372 Hydromechanics Lecture Notes, Middle East Technical University, Civil Engineering Department

� UTAH STATE UNIVERSITY Open Coursewarehttp://ocw.usu.edu/Civil_and_Environmental_Engineering/Fluid_Mechanics

Scope of the Course

� In many water systems, transportation of water from one location to another is the main concern.

� Two main modes of transportation are:

� Closed conduits with pressurized flow inside� Open conduits with free surface flow inside

� The main objective in this course is to study the flow in closed conduits (mainly pipes) and in open channels

Examples include:

� Water distribution networks in urban areas

� Water transmission line from Çamlıdere Dam to İvedik Water Treatment Plant

(φφφφ = 3.4 m, L = 15.5 km)

� Urfa Tunnels from Atatürk Dam to Harran Plain (φφφφ = 7.62 m, L = 2 x 26.4 km)

� Main irrigation canal in Harran Plain

� (L=118 km, Q = 80 m3/s)

The View of Atatürk Dam

GAP WATER RESOURCES ROJECTS

Total 22 dams, 19 HPP

1.7 million ha, 7485 MW, 27 billion kWh

Urfa Tunnels from Atatürk Dam to Harran Plain

� φφφφ = 7.62 m,

� L = 2 x 26.4 km

� Q=80 m3/s

Main irrigation canal in Harran Plain (L=118 km, Q = 80 m3/s)

Before 1995

HARRAN PLAINHARRAN PLAIN

YEYEŞİŞİLLÇÇAY SYSTEMAY SYSTEM

BLACKSEAYEŞİLÇAY REG.

SUNGURLUDAMİSAKÖY

DAM

KABAKOZDAM

DARLIKDAM

EMİRLİ TREATMENT

ÖMERLİDAM

AĞVA

M A R M A R A SEA

STORAGE

YEŞİLÇAY SYSTEM CHARACTERISTICS

Length of transmission lines: 723 712 m

Length of water Network : 11 738 km

Volume of water reservoir : 914 000 m3

Water Supplied (2003) : 920 million m3/year

Water treatment capacity : 3.5 million m3/day

Ø3 000 mm Prestressed Concrete Cylinder Pipes

BLACKSEACumhuriyet

Pompa İstasyonu

Cumhuriyet Pompa İstasyonu

Hüseyinli Su Arıtma Tesisi

700 000 m³/gün

Hüseyinli Su Arıtma Tesisi

700 000 m³/günŞile-Alaçalı

Tünel 3.5 km

Şile-AlaçalıTünel 3.5 km

Melen Pompa İstasyonu

Melen Pompa İstasyonu

Melen Regülatörü

8.5 m3/s

Melen Regülatörü

8.5 m3/s

Melen Barajı(ileri aşama)

Melen Barajı(ileri aşama)

Melen-Alaçalıİsale Hattı

131 km

Melen-Alaçalıİsale Hattı

131 km

AlaçalıBarajı

AlaçalıBarajı

Hamidiye Tüneli 5.2

km

Hamidiye Tüneli 5.2

km

Ömerli Barajı(mevcut)

Ömerli Barajı(mevcut)

Alaçalı-Ömerli Hattı

Alaçalı-Ömerli Hattı

Bekleme Tüneli 1.3 km

Bekleme Tüneli 1.3 km

Beykoz Tüneli2.6 km

Beykoz Tüneli2.6 km

Ortaçeşme Tüneli0.8 km

Ortaçeşme Tüneli0.8 km

AyazağaTüneli2.8 km

AyazağaTüneli2.8 km

Osmankuyu Su deposu

Osmankuyu Su deposu

Boğaz Tüneli5.5 km

Boğaz Tüneli5.5 km

Boğaz Tüneli Profili

Boğaz Tüneli Profili

Boğaz TüneliBoğaz Tüneli

Boğaz Tüneli Ø=4.0-3.6 m

L=5.5 km

Boğaz Tüneli Ø=4.0-3.6 m

L=5.5 km

MARMARA SEA

GREATER MELEN PROJECT OF ISTANBUL

GREATER MELEN PROJECT OF ISTANBUL

Great Melen Project Technical Specifications

System Length : 185 600 m

Ø 2 500 mm Steel Pipe : 163 950 mØ 4 500 mm tunnel length : 8 700 mØ 4 000 mm tunnel length : 11 550 mØ 3 600 mm tunnel length : 1 400 m

Examples of Fluid Mechanics System

Physical Properties of Fluids

� Density

� Specific weight

� Specific Gravity

� Specific Volume

� Viscosity

� Surface Tension

� Vapor Pressure

� Compressibility

Density, ρ

� Mass per unit volume

� ρ = m/∀

� [ρ]=ML-3

Specific Weight, γ:

� Weight per unit volume

� γ = W/∀

� [γ]=FL-3

γ = ρg

Specific Gravity, SG

� The ratio of the density of the fluid to the density of water (or air) at standard conditions

w

liquid)SG(ρ

ρ=

air

gas)SG(ρ

ρ=

Density and Specific Weights of some fluids (g=9.81m/s2)

Gases

Fluid Temperature°C

Densitykg/m3

Specific WeightN/m3

Water 4.0 1000. 9810.

Mercury 20.0 13600. 133416.

Gasoline 15.6 680. 6671.

Alcohol 20.0 789. 7740.

Air 15.0 1.23 12.0

Oxygen 20.0 1.33 13.0

Hydrogen 20.0 0.0838 0.822

Methane 20.0 0.667 6.54

Liq

uid

sG

ase

s

Deformation of fluid for a short time interval ∆t

τ=∝ hA

hFUp F

Up∆S

h

A

B B’

∆θy

u(y)

τ

x

h

Up∝τ

dt

dθ∝τ

Shear stress is proportional to the rate of angular deformation

Newton’s Law of Viscosity

For the linear velocity profile

y

)y(u

h

Up=

yh

U)y(u p

=

dt

d

h

U

dy

du p θ==

dy

du∝τ

Newton’sLaw of

viscositydy

duµ=τ The proportionality constant µ is known as

dynamic viscosity of the fluid.

Dynamic and Kinematic Viscosity

Viscosity can be made independent of fluid density; kinematic viscosityis defined as the ratio

ρ

µ=ν

µµµµ Dynamic Viscosity : N⋅⋅⋅⋅s/m2 N (Mass/Length/Time)

νννν Kinematic Viscosity : (m2/s) (Length2/Time)

Viscosities of air and water

Fluid Temperature(°°°°C)

µµµµ

(N⋅⋅⋅⋅s/m2)νννν

(m2/s)

Water 20 1.00E-03 1.01E-06

Air 20 1.80E-05 1.51E-05

Reynolds Experiment

Q=VA Dye streak

Dye

pipe

Smooth well-rounded entrance

D

Characteristics of Turbulent Flow

Velocity components in a turbulent pipe flow: (a) x-component velocity; (b) r-component velocity; (c) θ-component velocity.

Type of Flow

� Re…Dimensionless numberf( velocity, diameter, viscosity)

� Laminar flow: Re < 2000

� Transitional flow: 2000 < Re < 4000

� Turbulent flow: Re > 4000

νπ=

ν D

Q4VD = Re