cve 372 hydro mechanics - 2. flow in closed conduits 1

CVE 372 Hydromechanics 1/20

Dr. Bertuğ AkıntuğDepartment of Civil EngineeringMiddle East Technical University

Northern Cyprus Campus

CVE 372CVE 372HYDROMECHANICSHYDROMECHANICS

FLOW IN CLOSED CONDUITSFLOW IN CLOSED CONDUITSII


2. FLOW IN CLOSED CONDUITS2. FLOW IN CLOSED CONDUITS

Overview

2.1 General Characteristics of Flow in Closed Conduits

2.1.1 Definition of Laminar and Turbulent Flows

2.1.2 Entrance Region and Fully Developed Flow

2.1.3 Head Losses in Pipes



General Characteristics of Flow in Closed Conduits

The governing equations for mass, momentum and energy transfer are developed in CVE 371 Fluid Mechanics.

Now, we will apply them to viscous, incompressible fluids in pipes or ducts

Typical pipe system components




We assume that pipe is fully filled with the fluid.Main driving force is likely to be a pressure gradientalong the pipe.If the pipe is not full, it is not possible to maintain this pressure difference, p1-p2.

(a) Pipe flow, (b) Open-channel flow




2.1.1. Definition of Laminar and Turbulent FlowOsborne Reynold (1842-1912), a British scientists and mathematician was the first to distinguish the difference.In pipe flow:

(a) Experiment to illustrate type of flow. (b) Typical dye streaks.

Large Re

Small Re

Intermediate Re

µρ DVa=Re

Re ≤ 2100 Laminar2100 < Re < 4000 TransitionRe ≥ 4000 Turbulent




2.1.1. Definition of Laminar and Turbulent Flow

Time dependence of fluid velocity at a point



The time-averaged, ū, and fluctuating, u', description of a parameter for tubular flow.



∫=T

dtuT

u0

1

u(t): instantaneous velocity in the x-directionu’(t): fluctuating part of u(t)

)()( tuutu ′+=





Example 1:Water at a temperature of 10°C flows through a pipe of diameter D=1.85 cm. a) Determine the minimum time taken to fill a 355 cm3 glass with

water if the flow in the pipe is to be laminar.b) Determine the maximum time taken to fill the same glass if the

flow is to be turbulent.Repeat the calculation if the water temperature is 60°C.

Solved in the class room




Example 2:Water is flowing through capillary tubes A and B into tube C. IfQA=2x10-3 lt/s in tube A, what is the largest QB allowable in tube B for laminar flow in tube C? If the water has a temperature of 40°C with the calculated QB, what kind of flow exists in tubes A and B?(DA= 5 mm, DB= 4 mm, and DC= 6 mm)

Solved in the class room

A

B

C




2.1.2. Entrance Region and Fully Developed Flow

Entrance length Re06.0=Dle

Entrance length

Laminar flow

Turbulent flow6/1(Re)4.4=Dle

accelerated decelerated

u(r), du/dr, and τ are fixed




2.1.2. Entrance Region and Fully Developed FlowPressure and Shear Stress

If viscosity was zero, pressure would not vary along x direction.The physical properties of the shear stress are quite different for laminar flow then for turbulent flow.

Pressure distribution along a horizontal pipe

( )dydu

dydu

tµµτ

µτ

+=

=

turbulent

laminar

µ = molecular viscosityµt = turbulent viscosity




2.1.3. Head Losses in PipesWe need to determine the head loss. For convenience, we will consider two types of energy losses; minor and major loss

Total head loss, hL

hf : Friction (viscous, major) losshm :Local (minor) loss

Note that major and minor do not necessarily reflect the magnitude of the energy losses

mfL hhh +=




2.1.3. Head Losses in PipesDarcy – Weisbach Equation for friction loss

ifgV

DLfh

2

2

f =

22

2

5f 216 KQ

gQ

DLfh ==

π

( ) 42

2

222

2

2

22 16

4/ DQ

D

QAQV

ππ===

f: Darcy-Weisbach friction factorL: pipe length (m)D: Pipe diameter (m)V: average velocity (m/s)Q: Discharge (flow rate) (m3/s)

where

528DgfLK

π=




2.1.3. Head Losses in PipesHazen – Williams Equation for friction loss

or85.1

165.185.1f8.6 VDL

Ch =

C: Hazen-Williams coefficient of roughness (unitless)L: pipe length (m)D: Pipe diameter (m)V: average velocity (m/s)Q: Discharge (flow rate) (m3/s)

85.187.485.1f

6.10 QDL

Ch =




2.1.3. Head Losses in PipesMajor losses are due to energy loss in long straight sections of pipe which can be calculated by use of the friction factor.However, most pipe systems are not straight and have valves, bends, etc. which result in additional losses. These energy losses are collectively called minor losses.Typically, minor losses are evaluated from

where KL is the loss coefficient which depends on both the geometry of the system and the Reynolds number. For large Re, it simply becomes a function of the geometry.

gVKh m 2

2

m =




2.1.3. Head Losses in PipesLocal (minor) LossEmpirical equation (except for sudden enlargement)

gVKh m 2

2

m =

K: constantV: average velocity (m/s)

Entrance flow conditions and loss coefficient. (a) Reentrant, Km = 0.8, (b) Sharp-edged, Km = 0.5, (c) Slightly rounded, Km = 0.2, (d) Well-rounded, Km = 0.04




2.1.3. Head Losses in PipesLocal (minor) Loss

Loss coefficient for a sudden contraction.



2

2

1 )1(AAKL −=





Character of the flow in a 90° bend and the associated loss coefficient



cve 372 hydro mechanics - 2. flow in closed conduits 1

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