Viscous Flow in Pipes CEE 331 Fluid Mechanics February 13, 2020 Monroe L. Weber-Shirk School of Civil and Environmental Engineering 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? Can we increase the flow in this old pipe by adding a smooth liner? Viscous Flow in Pipes: Overview Boundary Layer Development Turbulence Velocity Distributions Energy Losses Major Minor Solution Techniques Laminar and Turbulent Flows Reynolds apparatus VD inertia

Re damping Transition at Re of 2000 Boundary layer growth: Transition length What does the water near the pipeline wall experience? _________________________ Drag or shear Why does the water in the center of the pipeline speed Conservation of mass up? _________________________ Non-Uniform Flow v v v Entrance Region Length le f Re D le 0.06 Re D le 1/ 6

4.4 Re D 100 Distance for velocity profile to develop l e /D 10 1 Shear in the entrance region vs laminar shear in long pipes? Re turbulent 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 uniform _______velocity (compared to laminar flow)

Close to the pipe wall, eddies are smaller (size proportional to distance to the boundary) Log Law for Turbulent, Established Flow, Velocity Profiles yu* u 2.5ln 5.0 Dimensional analysis and measurements u* yu* 20 Valid for 0 Turbulence produced by shear! u* Shear velocity Velocity of large eddies rough smooth h d Force balance 0 f 4l ghf d u* 4l y f u* V 8 u/umax

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 Viscous Flow: Dimensional Analysis Remember dimensional analysis? D Cp f l , Re D 2p VD and C p Where Re V 2 Two important parameters! Re - Laminar or Turbulent /D - Rough or Smooth Flow geometry

internal _______________________________ in a bounded region (pipes, rivers): find Cp external _______________________________ flow around an immersed object : find Cd Pipe Flow Energy Losses D f C p f , Re L D 2p Cp V 2 2 ghf D f 2 V L L V2 hf f D 2g u*2 f=8 2 V Dimensional Analysis ghl p g z ghl p

Cp 2 ghl V2 More general Assume horizontal flow Always true (laminar or turbulent) Darcy-Weisbach equation L u*2 hf 8 D 2g Friction Factor : Major losses Laminar flow Turbulent (Smooth, Transition, Rough) Colebrook Formula Moody diagram Swamee-Jain Laminar Flow Friction Factor gD 2 hl V 32 L 32 LV hf gD 2 L V2 hf f D 2g 32 LV LV2

f 2 D 2g gD 64 64 f VD Re Hagen-Poiseuille hf V D 4 ghl Q 128 l Darcy-Weisbach f independent of roughness! Slope of ___ -1 on log-log plot L V2 hf f D 2g Turbulent Flow: Smooth, Rough, Transition Hydraulically smooth pipe law (von Karman, 1930) Rough pipe law (von Karman, 1930) Transition function for

both smooth and rough pipe laws (Colebrook) f u* V 8 Re f 1 2 log 2.51 f 1 3.7 D 2 log f 1 2.51 D 2 log 3.7 f Re f (used to draw the Moody diagram)

Moody Diagram 0.10 0.08 friction factor D f C p l 0.06 0.05 0.04 0.03 0.05 0.02 0.015 0.04 0.01 0.008 0.006 0.004 0.03 laminar 0.002 0.02

0.001 0.0008 0.0004 0.0002 0.0001 0.00005 0.01 1E+03 smooth 1E+04 1E+05 1E+06 Re 1E+07 1E+08 D Swamee-Jain 1976 limitations f /D < 2 x 10-2 Re >3 x 103

less than 3% deviation from results obtained Q D 5 / 2 2 with Moody diagram easy to program for computer or calculator use 0.25 5.74 log 3.7 D Re0.9 ghf log 2.51 L 3.7 D Colebrook 1.25 LQ D 0.66 ghf

2 4.75 2 no f L 2 ghf D 3 5.2 L Q ghf 0.04 9.4 Each equation has two terms. Why? Pipe roughness pipe material glass, drawn brass, copper commercial steel or wrought iron asphalted cast iron galvanized iron cast iron concrete rivet steel

corrugated metal PVC pipe roughness (mm) 0.0015 0.045 0.12 d Must be 0.15 dimensionless! 0.26 0.18-0.6 0.9-9.0 45 0.12 Solution Techniques find head loss given (D, type of pipe, Q) 0.25 2 8 LQ f 4Q 2 hf f 2 Re 5

5.74 g D D log 3.7 D Re0.9 find flow rate given (head, D, L, type of pipe) Q D5 / 2 2 find ghf log 2.51 L 3.7 D L 2 ghf D 3 pipe size given (head, type of pipe,L, Q) 2

LQ 1.25 D 0.66 gh f 4.75 5.2 L Q ghf 9.4 0.04 Example: Find a pipe diameter The marine pipeline for the Lake Source Cooling project will be 3.1 km in length, carry a maximum flow of 2 m3/s, and can withstand a maximum pressure differential between the inside and outside of the pipe of 28 kPa. The pipe roughness is 2 mm. What diameter pipe should be used? Minor Losses: Expansions! 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. hex High Re C p f geometry, Re 2p Cp V 2 2 ghex Cp 2 V V2 K 2g V2 hex C p 2g Venturi hex 2 in V

2g Sudden Contraction Ain 1 A out EGL Cc 2 1 V2 2 hc 1 C 2g c HGL Ac A2 2

1 0.95 0.9 0.85 Cc 0.8 0.75 0.7 0.65 0.6 0 0.2 0.4 0.6 0.8 A2/A1 V1 1 c 2 V2 vena contracta Losses are reduced with a gradual contraction Equation has same form as expansion equation!

Entrance Losses V2 he K e Losses can be 2g K e 1.0 reduced by accelerating the flow Estimate based on contraction equations! gradually and K e 0.5 eliminating the vena contracta K e 0.04 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 D returns to normal several pipe diameters hb downstream Kb varies from 0.6 - 0.9

high low R Possible separation from wall n 2 V p dn z C R Low pressure K b V2 2g 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) Yes!

Can Kvbe greater than 1? ______ What is V? hv K v V2 2g 8Q 2 hv K v g 2 D 4 Solution Techniques Neglect minor losses Equivalent pipe lengths Iterative Techniques Using Swamee-Jain equations for D and Q Using Swamee-Jain equations for head loss Assume a friction factor Pipe Network Software Solution Technique: Head Loss Can be solved explicitly hminor Re V2 K 2g 4Q D

hminor 8Q 2 g 2 K D4 0.25 f 5.74 log 3.7 D Re 0.9 hl hf hminor 2 hf f 8

g 2 LQ 2 D5 Find D or Q Solution Technique 1 Assume all head loss is major head loss Calculate D or Q using Swamee-Jain equations 8Q 2 hex K Calculate minor losses g 2 D 4 Find new major losses by subtracting minor hf hl hex losses from total head loss LQ D 0.66 1.25 gh f 2 Q D5/ 2 2 ghf

log 2.51 L 3.7 D L 2 ghf D 3 4.75 5.2 L Q ghf 9.4 0.04 Find D or Q Solution Technique 2: Solver Iterative technique Solve these equations Re 4Q D hminor K f

0.25 5.74 log 0.9 3.7 D Re 8Q 2 g 2 D 4 hl hf hminor 2 hf f 8 g 2 LQ 2 D5 Use goal seek or Solver to find discharge that makes the calculated head loss equal the given head loss.

Spreadsheet Find D or Q Solution Technique 3: assume f The friction factor doesnt vary greatly If Q is known assume f is 0.02, if D is 1 3.7 D 2 log known assume rough pipe law f Use Darcy Weisbach and minor loss equations Solve for Q or D Calculate Re and /D Find new f on Moody diagram Iterate Example: Minor and Major Losses Find the maximum dependable flow between the reservoirs for a water temperature range of 4C to 20C. Water 25 m elevation difference in reservoir water levels Reentrant pipes at reservoirs Standard elbows 2500 m of 8 PVC pipe 1500 m of 6 PVC pipe

Spreadsheet Sudden contraction Gate valve wide open Directions Example (Continued) What are the Reynolds numbers in the two pipes? 90,000 & 125,000 /D= 0.0006, 0.0008 Where are we on the Moody Diagram? What is the effect of temperature? Why is the effect of temperature so small? What value of K would the valve have to produce to reduce the discharge by 50%? 140 friction factor 0.1 0.02 0.015 0.01 0.008 0.006 0.004 laminar 0.01 1E+03 Spreadsheet 0.05

0.04 0.03 1E+04 1E+05 1E+06 Re 1E+07 1E+08 0.002 0.001 0.0008 0.0004 0.0002 0.0001 0.00005 smooth Example (Continued) Were the minor losses negligible? Yes Accuracy of head loss calculations? 5% What happens if the roughness increases by f goes from 0.02 to 0.035 a factor of 10? If you needed to increase the flow by 30% what could you do? Increase small pipe diameter 0.1

0.05 0.04 0.03 friction factor 0.02 0.015 0.01 0.008 0.006 0.004 laminar 0.002 0.001 0.0008 0.0004 0.0002 0.0001 0.00005 0.01 1E+03 smooth 1E+04 1E+05

1E+06 Re 1E+07 1E+08 Pipe Flow Summary (1) Shear increases _________ linearly 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 (Navier Stokes) and energy conservation Turbulent flow losses and velocity distributions require ___________ experimental results 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 _______ constant at high Reynolds numbers Solutions for discharge or pipe diameter often require iterative or computer solutions Pressure Coefficient for a Venturi Meter Cp 10 2 p Cp V 2 1 1E+01 1E+02 1E+03 1E+04 1E+05 1E+06 Re

Vl Re 0.1 0.05 0.04 0.03 0.02 0.015 friction factor 1E+00 0.01 0.008 0.006 0.004 laminar 0.002 0.001 0.0008 0.0004 0.0002 0.0001 0.00005 0.01 1E+03 smooth 1E+04

1E+05 1E+06 Re 1E+07 1E+08 Moody Diagram 0.1 D f C p l 0.05 0.04 0.03 friction factor 0.02 0.015 0.01 0.008 0.006 0.004 D

laminar 0.002 0.001 0.0008 0.0004 0.0002 0.0001 0.00005 0.01 1E+03 smooth 1E+04 1E+05 1E+06 Re 1E+07 1E+08 Minor Losses LSC Pipeline z=0 cs1 cs2

0 Ignore minor losses p1 V12 p2 V22 z1 1 z2 2 hl 2g 2g KE will be small Q 2m3 / s -2.85 m 28 kPa is equivalent to 2.85 m of water 4.75 5.2 0.04 2 LQ L

D 0.66 1.25 Q 9.4 gh f gh f V22 D 154 . m V 1.07 m / s 2 0.06 m 2g 10 6 m 2 / s L 3100m 0.002m h f 2.85m Directions Assume fully turbulent (rough pipe law) find f from Moody (or from von Karman) Find total head loss (draw control volume) Solve for Q using symbols (must include minor losses) (no iteration required) hl hf hminor Solution

0.1 Water 0.05 0.04 0.03 friction factor 0.02 0.015 0.01 0.008 0.006 0.004 laminar 0.002 0.001 0.0008 0.0004 0.0002 0.0001 0.00005 0.01 1E+03 smooth 1E+04 1E+05

1E+06 Re 1E+07 1E+08 Pipe roughness Find Q given pipe system hminor K 8Q 2 2 g D 4 h f f 8 g 2 hl hf hminor 8Q 2 L K hl f 5 4 2

g D D Q ghl L K 8 f 5 4 D D Water LQ 2 D5