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Closed Conduit Hydraulics. CE154 - Hydraulic Design Lecture 6. Hydraulics of Closed Conduit Flow. Synonyms - closed conduit flow - pipe flow - pressurized flow Objectives – to introduce - basic concepts of closed conduit flow, - its hydraulics, and - design method. Concepts.

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Closed conduit hydraulics

Closed Conduit Hydraulics

CE154 - Hydraulic Design

Lecture 6

CE154


Hydraulics of closed conduit flow
Hydraulics of Closed Conduit Flow

  • Synonyms- closed conduit flow- pipe flow- pressurized flow

  • Objectives – to introduce- basic concepts of closed conduit flow, - its hydraulics, and - design method

CE154


Concepts
Concepts

  • Closed Conduit vs. Open Channel

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Concepts reynolds number
Concepts – Reynolds Number

  • Reynolds Number (ratio of inertia force to viscous force)V = velocity (ft/sec)D = pipe diameter (ft) = density of fluid (lbm/ft3) = dynamic viscosity of fluid (lbm/ftsec or lbfsec/ft2) = kinematic viscosity (ft2/sec)

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Concepts froude number
Concepts – Froude Number

  • Froud Number (ratio of inertia force to gravitational force)

  • V = velocityg = gravitational accelerationh = depth of water

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Concepts turbulence
Concepts - Turbulence

  • Turbulent vs. laminar flow

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Concepts turbulent flow
Concepts – turbulent flow

  • Turbulent flow - Critical Re (laminar to turbulent) in the order of 1000

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Concepts laminar flow
Concepts – laminar flow

  • Turbulent and Laminar flows

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Concepts uniform steady flow
Concepts – uniform & steady flow

  • Uniform flow – constant characteristics with respect to space

  • Steady flow – constant characteristics with respect to time. Often adopted when establishing pipe system design parameters (pressure & flow at certain locations). Consider unsteady (transient) phenomena to refine design (pipe pressure class and thickness)

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Conservation of mass
Conservation of Mass

1

Control Volume

2

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Conservation of mass1
Conservation of Mass

  • Consider the control volume

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Conservation of mass2
Conservation of Mass

  • For steady & incompressible flow, dS/dt = 0I = OV1A1 = V2A2 ViAi =  VoAo

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Conservation of mass3
Conservation of Mass

  • Apply to a pipe junction, Q1+Q2 = Q3+Q4

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Conservation of momentum
Conservation of Momentum

  • Newton’s 2nd law – the resultant of all external forces on a system is equal to the time rate of change of momentum of this system

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Conservation of momentum1
Conservation of Momentum

  • Consider this control volume (CV) of fluid in a pipe elbow

x1=v1t

1

1’

2

2’

x2=v2t

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Conservation of momentum2
Conservation of Momentum

  • In a time t the fluid originally at Section 1 moves to 1’, and that at Section 2 moves to 2’

  • The control volume lost momentum equal to that of the fluid contained between 1 and 1’(A1x1)V1 = A1V12t = (QV1)tAt the same time it gained momentum (QV2)t

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Conservation of momentum3
Conservation of Momentum

  • The time rate of change of momentum is (QV2)- (QV1)

  • Hence, the 2nd Law becomes

  • This is the momentum equation for steady flow. Use this convention:

  • QVx1 Fx = QVx2

  • QVy1 Fy = QVy2

  • Where  depends on the direction of the force w.r.t. the coordinate system

CE154


Application of momentum eq
Application of Momentum Eq.

  • Forces on a pipe elbow:Taking momentum balance in the x direction,QV1 + (PA)1 – Fx = Q(0)Fx = (PA)1 + QV1

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Application of momentum eq1
Application of Momentum Eq.

  • Taking momentum balance in the y direction,External y force = (PA)2 - FyRate of change of momentum = QV2 (where V2 is in the negative direction) (PA)2 - Fy = QV2 Fy = (PA)2 - QV2 = (PA)2 + QV2

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Conservation of energy
Conservation of Energy

  • In pipeline design, most often consider steady state – flow not varying with time - first

  • Steady state (SS) Bernoulli Equation along a streamline:

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Conservation of energy1
Conservation of Energy

  • Pressure head p/

  • Elevation head z

  • Velocity head V2/2g

  • Piezometric head p/ + z

    (hydraulic grade line)

  • Total head p/ + z + V2/2g

    (energy grade line)

  • Head Loss h

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Example 3 1
Example 3-1

  • A plane jet of unit discharge q0 strikes a boundary at an angle of 45, what will be the ratio of q1/q2 for the divided flow?

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Head losses
Head Losses

  • Include mostly 2 types of losses:

  • Friction Loss- resulting from friction between the fluid and pipe wall

  • Minor Loss- resulting from pipe entrance, transition, exit, valve and other in-line structures

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Friction loss
Friction Loss

  • Most useful head loss equation for closed-conduit flow – Darcy-Weisbach equation

Pipe length

Friction head loss

Pipe velocity

Dimensionless Friction coefficient

Gravitational acceleration

Pipe diameter

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Friction loss1
Friction Loss

  • Darcy-Weisbach equation- derived from basic relationships of physics -  dimensionless, app. to all unit systems-  determined from experimental data

  • Other friction loss relationships – Hazen-Wiliams, Manning, Chezy, etc. – are also used in the industry, but are less accurate and will not be discussed here

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Darcy weisbach
Darcy-Weisbach

  • Laminar flow (Re<2000)Turbulent flow in smooth pipes (Re>4000)

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Darcy weisbach cont d
Darcy-Weisbach  (cont’d)

  • Turbulent flow in rough pipesTransition between turbulent smooth and rough pipes

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Darcy weisbach1
Darcy-Weisbach

  • Most recent development of Darcy Weisbach coefficient - Explicit equation [Swamee and Jain, 1976] applicable to entire turbulent flow regime (smooth, transition and rough pipes):

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Minor loss
Minor Loss

  • Use minor loss coefficient (k) in this form

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Minor loss1
Minor Loss

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Minor loss2
Minor Loss

  • For abrupt expansion, from D1 to D2, the loss coefficient may be estimated by

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Minor loss3
Minor Loss

  • American Water Works Association – Steel Pipe, A guide for design and installation, Manual of Water Supply Practices, M11, 4th Edition, 2004

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Minor loss4
Minor Loss

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Minor loss5
Minor Loss

  • Valve manufacturer has loss curves typically presented in terms of Cv vs. valve opening degrees. Cv is defined as the flow rate in gallons per minute of 60 water that flows through the valve under 1 psi of head loss.

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Example 3 2
Example 3-2

  • p. 2.24 of Mays’ Hydraulic Design Handbook – Calculate f and e/D from given discharge

V2/2g=1.21 m

Atmospheric Pressure

P=3MPa

L=2500 m

El. 200 m

D=27 in

El. 100 m

Q=1.8 cms

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Example 3 3
Example 3-3

  • Same problem but now we have an 20” in-line ball valve with a 20” bore opened at 70 from closed position, a contraction and expansion section each connected to the valve, and 2 90 elbows with r/D=2. What is the f now?

CE154


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