Viscous Flow in Pipes

Types of Engineering Problems How big does the pipe have to be to carry a flow of x m3/s? What will the pressure in the water distribution system be when a fire hydrant is open?

Example Pipe Flow Problem cs1 D=20 cm L=500 m Find the discharge, Q. 100 m valve cs2 Describe the process in terms of energy!

Viscous Flow: Dimensional Analysis Remember dimensional analysis? Two important parameters! R - Laminar or Turbulent e/D - Rough or Smooth Flow geometry internal _______________________________ external _______________________________ Where and in a bounded region (pipes, rivers): find Cp flow around an immersed object : find Cd

Laminar and Turbulent Flows Reynolds apparatus inertia damping Transition at R of 2000

Boundary layer growth: Transition length What does the water near the pipeline wall experience? _________________________ Why does the water in the center of the pipeline speed up? _________________________ Drag or shear Conservation of mass v v v Pipe Entrance Non-Uniform Flow Need equation for entrance length here

Laminar, Incompressible, Steady, Uniform Flow Between Parallel Plates Through circular tubes Hagen-Poiseuille Equation Approach Because it is laminar flow the shear forces can be easily quantified Velocity profiles can be determined from a force balance Don’t need to use dimensional analysis

Laminar Flow through Circular Tubes Different geometry, same equation development (see Streeter, et al. p 268) Apply equation of motion to cylindrical sleeve (use cylindrical coordinates)

Laminar Flow through Circular Tubes: Equations a is radius of the tube Max velocity when r = 0 Velocity distribution is paraboloid of revolution therefore _____________ _____________ average velocity (V) is 1/2 umax Q = VA = Vpa2

Laminar Flow through Circular Tubes: Diagram Shear Velocity Laminar flow Shear at the wall True for Laminar or Turbulent flow

The Hagen-Poiseuille Equation cv pipe flow Constant cross section h or z Laminar pipe flow equations CV equations!

Example: Laminar Flow (Team work) Calculate the discharge of 20ºC water through a long vertical section of 0.5 mm ID hypodermic tube. The inlet and outlet pressures are both atmospheric. You may neglect minor losses. What is the total shear force? What assumption did you make? (Check your assumption!)

Turbulent Pipe and Channel Flow: Overview Velocity distributions Energy Losses Steady Incompressible Flow through Simple Pipes Steady Uniform Flow in Open Channels

Turbulence A characteristic of the flow. How can we characterize turbulence? intensity of the velocity fluctuations size of the fluctuations (length scale) instantaneous velocity mean velocity velocity fluctuation t

Turbulence: Size of the Fluctuations or Eddies Eddies must be smaller than the physical dimension of the flow Generally the largest eddies are of similar size to the smallest dimension of the flow Examples of turbulence length scales rivers: ________________ pipes: _________________ lakes: ____________________ Actually a spectrum of eddy sizes depth (R = 500) diameter (R = 2000) depth to thermocline

Turbulence: Flow Instability In turbulent flow (high Reynolds number) the force leading to stability (_________) is small relative to the force leading to instability (_______). Any disturbance in the flow results in large scale motions superimposed on the mean flow. Some of the kinetic energy of the flow is transferred to these large scale motions (eddies). Large scale instabilities gradually lose kinetic energy to smaller scale motions. The kinetic energy of the smallest eddies is dissipated by viscous resistance and turned into heat. (=___________) viscosity inertia head loss

Velocity Distributions Turbulence causes transfer of momentum from center of pipe to fluid closer to the pipe wall. Mixing of fluid (transfer of momentum) causes the central region of the pipe to have relatively _______velocity (compared to laminar flow) Close to the pipe wall eddies are smaller (size proportional to distance to the boundary) constant

Turbulent Flow Velocity Profile Turbulent shear is from momentum transfer h = eddy viscosity Length scale and velocity of “large” eddies Dimensional analysis y

Turbulent Flow Velocity Profile increases Size of the eddies __________ as we move further from the wall. k = 0.4 (from experiments)

Log Law for Turbulent, Established Flow, Velocity Profiles Integration and empirical results Laminar Turbulent Shear velocity y x

Pipe Flow: The Problem We have the control volume energy equation for pipe flow We need to be able to predict the head loss term. We will use the results we obtained using dimensional analysis

Pipe Flow Energy Losses Horizontal pipe Dimensional Analysis Darcy-Weisbach equation

Friction Factor : Major losses Laminar flow Turbulent (Smooth, Transition, Rough) Colebrook Formula Moody diagram Swamee-Jain

Laminar Flow Friction Factor Hagen-Poiseuille Darcy-Weisbach Slope of ___ on log-log plot -1

Turbulent Pipe Flow Head Loss ___________ to the length of the pipe ___________ to the square of the velocity (almost) ________ with the diameter (almost) ________ with surface roughness Is a function of density and viscosity Is __________ of pressure Proportional Proportional Inversely Increase independent

Smooth, Transition, Rough Turbulent Flow Hydraulically smooth pipe law (von Karman, 1930) Rough pipe law (von Karman, 1930) Transition function for both smooth and rough pipe laws (Colebrook) (used to draw the Moody diagram)

Moody Diagram 0.10 0.08 0.06 0.05 0.04 friction factor 0.03 0.02 0.01 0.015 0.04 0.01 0.008 friction factor 0.006 0.03 0.004 laminar 0.002 0.02 0.001 0.0008 0.0004 0.0002 0.0001 0.00005 0.01 smooth 1E+03 1E+04 1E+05 1E+06 1E+07 1E+08 R

Swamee-Jain no f Each equation has two terms. Why? 1976 limitations /D < 2 x 10-2 Re >3 x 103 less than 3% deviation from results obtained with Moody diagram easy to program for computer or calculator use no f Each equation has two terms. Why?

Pipe roughness Must be dimensionless! pipe material pipe roughness  (mm) glass, drawn brass, copper 0.0015 commercial steel or wrought iron 0.045 Must be dimensionless! asphalted cast iron 0.12 galvanized iron 0.15 cast iron 0.26 concrete 0.18-0.6 rivet steel 0.9-9.0 corrugated metal 45 0.12 PVC

Solution Techniques find head loss given (D, type of pipe, Q) find flow rate given (head, D, L, type of pipe) find pipe size given (head, type of pipe,L, Q)

Minor Losses We previously obtained losses through an expansion using conservation of energy, momentum, and mass Most minor losses can not be obtained analytically, so they must be measured Minor losses are often expressed as a loss coefficient, K, times the velocity head. High R

Head Loss due to Gradual Expansion (Diffusor)  diffusor angle () 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 20 40 60 80 KE

Sudden Contraction V2 V1 flow separation losses are reduced with a gradual contraction

Sudden Contraction Cc A2/A1 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 0.2 0.2 0.4 A2/A1 Cc

Entrance Losses vena contracta Losses can be reduced by accelerating the flow gradually and eliminating the vena contracta

Head Loss in Bends High pressure Head loss is a function of the ratio of the bend radius to the pipe diameter (R/D) Velocity distribution returns to normal several pipe diameters downstream Possible separation from wall R D Low pressure Kb varies from 0.6 - 0.9

Head Loss in Valves Function of valve type and valve position The complex flow path through valves can result in high head loss (of course, one of the purposes of a valve is to create head loss when it is not fully open)

Solution Techniques Neglect minor losses Equivalent pipe lengths Iterative Techniques Simultaneous Equations Pipe Network Software

Iterative Techniques for D and Q (given total head loss) Assume all head loss is major head loss. Calculate D or Q using Swamee-Jain equations Calculate minor losses Find new major losses by subtracting minor losses from total head loss

Solution Technique: Head Loss Can be solved directly

Solution Technique: Discharge or Pipe Diameter Iterative technique Set up simultaneous equations in Excel Use goal seek or Solver to find discharge that makes the calculated head loss equal the given head loss.

Example: Minor and Major Losses Find the maximum dependable flow between the reservoirs for a water temperature range of 4ºC to 20ºC. 25 m elevation difference in reservoir water levels Water Reentrant pipes at reservoirs Standard elbows 2500 m of 8” PVC pipe Sudden contraction Gate valve wide open 1500 m of 6” PVC pipe

Directions Assume fully turbulent (rough pipe law) find f from Moody (or from von Karman) Find total head loss Solve for Q using symbols (must include minor losses) (no iteration required) Obtain values for minor losses from notes or text

Example (Continued) What are the Reynolds number in the two pipes? Where are we on the Moody Diagram? What value of K would the valve have to produce to reduce the discharge by 50%? What is the effect of temperature? Why is the effect of temperature so small?

Example (Continued) Were the minor losses negligible? Accuracy of head loss calculations? What happens if the roughness increases by a factor of 10? If you needed to increase the flow by 30% what could you do? Suppose I changed 6” pipe, what is minimum diameter needed?

Pipe Flow Summary (1) Shear increases _________ with distance from the center of the pipe (for both laminar and turbulent flow) Laminar flow losses and velocity distributions can be derived based on momentum and energy conservation Turbulent flow losses and velocity distributions require ___________ results linearly experimental

Pipe Flow Summary (2) Energy equation left us with the elusive head loss term Dimensional analysis gave us the form of the head loss term (pressure coefficient) Experiments gave us the relationship between the pressure coefficient and the geometric parameters and the Reynolds number (results summarized on Moody diagram)

Pipe Flow Summary (3) Dimensionally correct equations fit to the empirical results can be incorporated into computer or calculator solution techniques Minor losses are obtained from the pressure coefficient based on the fact that the pressure coefficient is _______ at high Reynolds numbers Solutions for discharge or pipe diameter often require iterative or computer solutions constant

Columbia Basin Irrigation Project The Feeder Canal is a concrete lined canal which runs from the outlet of the pumping plant discharge tubes to the north end of Banks Lake (see below). The original canal was completed in 1951 but has since been widened to accommodate the extra water available from the six new pump/generators added to the pumping plant. The canal is 1.8 miles in length, 25 feet deep and 80 feet wide at the base. It has the capacity to carry 16,000 cubic feet of water per second.

Pipes are Everywhere! Owner: City of Hammond, IN Project: Water Main Relocation Pipe Size: 54"

Pipes are Everywhere! Drainage Pipes

Pipes

Pipes are Everywhere! Water Mains

Glycerin

Example: Hypodermic Tubing Flow