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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

CE154

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)

CE154

concepts froude number
Concepts – Froude Number
  • Froud Number (ratio of inertia force to gravitational force)
  • V = velocityg = gravitational accelerationh = depth of water

CE154

concepts turbulence
Concepts - Turbulence
  • Turbulent vs. laminar flow

CE154

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

CE154

concepts laminar flow
Concepts – laminar flow
  • Turbulent and Laminar flows

CE154

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)

CE154

conservation of mass
Conservation of Mass

1

Control Volume

2

CE154

conservation of mass1
Conservation of Mass
  • Consider the control volume

CE154

conservation of mass2
Conservation of Mass
  • For steady & incompressible flow, dS/dt = 0I = OV1A1 = V2A2 ViAi =  VoAo

CE154

conservation of mass3
Conservation of Mass
  • Apply to a pipe junction, Q1+Q2 = Q3+Q4

CE154

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

CE154

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

CE154

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

CE154

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

CE154

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

CE154

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:

CE154

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

CE154

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?

CE154

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

CE154

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

CE154

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

CE154

darcy weisbach
Darcy-Weisbach 
  • Laminar flow (Re<2000)Turbulent flow in smooth pipes (Re>4000)

CE154

darcy weisbach cont d
Darcy-Weisbach  (cont’d)
  • Turbulent flow in rough pipesTransition between turbulent smooth and rough pipes

CE154

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):

CE154

minor loss
Minor Loss
  • Use minor loss coefficient (k) in this form

CE154

minor loss2
Minor Loss
  • For abrupt expansion, from D1 to D2, the loss coefficient may be estimated by

CE154

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 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.

CE154

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

CE154

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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