F lu id Mechanics I I(BDA 30203)

Chapter 1: Viscous Flow in Pipes

Part 1

Objectives

1. Have a deeper understanding of laminar and turbulent flow inpipes and the analysis of fully developed flow

2. Calculate the major and minor losses associated with pipe flow inpiping networks and determine the pumping power requirements

Introduction

• Piping systems are encountered in almost every engineering area.

• Problems are related to flow in ducts or pipes with various velocities, fluids, duct and pipe shapes and sizes.

• When ‘real world’ (viscous effect) effects are important, it is difficult to use theoretical method to obtain the desired result.

• A combination of experimental data with theoretical considerations and dimensional analysis provide the desired results.

3

Pipeline

Industrial Pipes

4

Pipe Flow Characteristics

• Not all conduits used to transport fluid are round in cross section.

• Heating and air conditioning ducts are often of rectangular cross section. Why?

• For heating and air conditioning, pressure difference between inside and outside is relatively small and basic principle involved are independent of the cross-sectional shape.

5

Assumption:

Assumption involved in this chapter:

• The pipe is round in cross section

• The pipe is completely filled with fluid

• Viscous fluid μ ≠ 0

• Incompressible fluid, 𝜌=constant

• Steady Flow, 𝜕

𝜕𝑡= 0

6

𝐹𝑙𝑢𝑖𝑑𝜌=constant𝜇

7

Steady and unsteady flow

• Steady flows occur when flow parameters such as pressure, velocity, temperature etc. do not vary with time.

• If flow parameters vary with time, it is called unsteady.

Compressible and incompressible

• Fluid is incompressible when its density does not depend on pressure. (Volume does not change when pressure is applied).

• When density changes when pressure is applied, it is called compressible.

8

9

Reynolds number

for pipe flow

10

Laminar and Turbulent Flow

Re ≤ 2100 Laminar Flow

2100 < Re < 4000 Transitional Flow

Re ≥ 4000 Turbulent Flow

Example 1.1

Water at temperature 16oC flows through a pipe of diameter

0.018 m.

a. Determine the minimum time taken to fill a 3.54 × 10-4 m3 glass

with water if the flow in pipe is to be laminar.

b. Determine the maximum time taken to fill a 3.54 × 10-4 m3 glass

with water if the flow in pipe is to be turbulent.

Solution:

Given, D = 0.018 m

V = 3.54 × 10-4 m3

11

a. Minimum time occur when Reynolds number is maximum

allowed for laminar flow (maximum velocity)

v = Re (µ)/ D

= 2100 (1.12 × 10-3)/[1000(0.018)] = 0.131 m/s

Q = vA

= 0.131[π(0.018)2/4] = 3.33 × 10-5 m3/s

t = V/Q

= 3.54 × 10-4/3.33 × 10-5

= 10.63 s

12

b. Maximum time occur when Reynolds number is minimum allowed for

turbulent flow (minimum velocity)

v = Re (µ)/ D

= 4000 (1.12 × 10-3)/[1000(0.018)] = 0.249 m/s

Q = vA

= 0.249 [π(0.018)2/4] = 6.33 × 10-5 m3/s

t = V/Q

= 3.54 × 10-4/6.33 x 10-5

= 5.59 s

13

14

The Entrance Region

• Entrance region - the region of flow near where the fluid

enters the pipe.

• The fluid enters the pipe with nearly uniform velocity profile

[section (1)].

• As the fluid move through the pipe, viscous effects cause it to stick to the pipe wall. Thus boundary layer is produced along the pipe such that the initial velocity profile changes with distance along the pipe until the fluid reaches the end of the entrance length [section (2)].

• From section (2) to section (3), the velocity profile does not vary with pipe length and the boundary layer is fully developed (fully developed flow).

15

• The shape of velocity profile and the dimensionless entrance length, le/D

depends on whether the flow is laminar or turbulence.

• le/D = 0.06 Re for laminar flow (1.1)

• le/D = 4.4 (Re)1/6 for turbulent flow (1.2)

• Because of the character of the pipe changes from section (3) to section (4), the flow gradually begin its return to its fully developed character (section (5)).

16

Example 1.2

Water flows through a 15m pipe with 1.3 cm diameter at 20 l/min.

Determine the length of entrance region, le?

Solution:

Given, L = 15m

D = 1.3 cm = 0.013 m

Q = 20 l/min = 20/(1000 × 60)m3/s = 3.33 × 10-4 m3/s

v = Q/A

= 3.33 × 10-4/[π(0.013)2/4] = 2.50 m/s

Re = v D/µ

= 1000(2.5)(0.013)/1 × 10-3 = 32500 (> 4000, turbulent flow)

therefore

le/D = 4.4 (Re)1/6

le = 4.4 (Re)1/6D

le = 4.4(32500)1/6(0.013) = 0.32 m

17

• Pressure difference between 2 pointsforces the fluid through the horizontalpipe and viscous effects provide therestraining force that exactly balancesthe pressure force.

• In the entrance region, fluid accelerateor decelerate as it flows, thus there is abalance between pressure, viscous andinertia (acceleration) forces

• The magnitude of the pressure gradientp/x is larger in the entrance regionthan in the fully develop flow, where it isa constant,

p /x = -p / l 0

18

Pressure and Shear Stress

Equation for Fully Developed Laminar Flow in Pipes

• Fully developed laminar flow - velocity profile is the same at any cross section of the pipe.

• From the velocity profile, we can get other information regarding the flow such as pressure

drop, flow rate, shear stress etc.

• 3 methods can be used to derive equations pertaining to fully developed laminar flow in

pipes.

▪ Applying F = ma to a fluid element.

▪ Dimensional analysis.

▪ Navier-Stokes equation of motion.

19

Applying F = ma to a Fluid Element

• Consider fluid element at time t – circular cylinder of fluid of length l and radius r.

• Even though the fluid is moving, it is not accelerating, so ax = 0

Apply F = ma

( p 1 ) π r 2 – ( p 1 - p ) π r 2 – ( ) 2 π r l = 0

(1.3)

is depending on r

= C r

where C is a constant

at r = 0, there is no shear stress ( = 0) at r = D/2, the shear stress is maximum ( =

w)

C = 2 w / D

therefore

(1.4)

and

(1.5)

These equation (1.3, 1.4 and 1.5) valid for both laminar and turbulent flow.

20

How shear stress related to velocity?

Two governing laws for fully developed laminar flow:

(for laminar flow of a Newtonian fluid, shear stress simply proportional to the velocity gradient)

Combine these two equations

velocity profile

21

at r = D/2, u = 0 and C1 = (p / 1 6 µ l ) D 2

at r = 0, centerline velocity, Vc

V c = (p D 2 / 1 6 µ l )

(1.6)

combine equations 1.5 and 1.6, and D/2 = R

22

Flowrate

or

knowing that average velocity, V = Q/A = Q/π R 2

(1.7)

and

(1.8)

Equation 1.8 is commonly referred to as Poiseuille’s law which is

valid for laminar flow only.

23

For a horizontal pipe, the

flowrate (Q) is:

1. directly proportional to the

pressure drop, (Δp)

2. inversely proportional to the

viscosity, (µ)

3. inversely proportional to the

pipe length, (l) and

4. proportional

to the pipe diameter to the

fourth power, (D4)

Summary - Flow properties for horizontal pipe

Flow PropertiesEquation Remarks

Entrance Length, le/Dle/D = 0.06 Re

le/D = 4.4 (Re)1/6

Laminar flow

Turbulent flow

Pressure drop per unit length p/l = 2/r Valid for both laminar and turbulent flow

Shear stress = 2wr/D Valid for both laminar and turbulent flow

Pressure drop p = 4lw/D Valid for both laminar and turbulent flow

Velocity profile ur = Vc 1 – (2r/D) 2 Laminar flow

Average velocity

V = (π R2 Vc/2)/ πR2

V= Vc/2

V = pD2/32µl

Laminar flow

Flowrate Q = πD4p/128µl Laminar flow

24

non-horizontal pipe – gravity effect

θ - angle between pipe centerline axis and horizontal axis

Apply F = ma

( p + p ) π r 2 – ( p ) π r 2 – mg sin θ – ( ) 2 π r l = 0

( p + p ) π r 2 – ( p ) π r 2 – ( π r 2) l g sin θ – ( ) 2 π r l = 0

(p – γ l sin θ ) / l = 2 / r (1.9)

effects of non-horizontal pipe

p →→ (p – γ l sin θ )

therefore

V = (p – γ l sin θ ) D 2 / 3 2 µ l (1.10)

and

(1.11)

25

Summary - Flow properties for non-horizontal pipe

26

Flow Properties Equation Remarks

Entrance Length, le/D

l e / D = 0 . 0 6 R e

l e / D = 4 . 4 ( R e ) 1 / 6

Laminar flow

Turbulent flow

Pressure drop per unit length ( p – γ l sin θ ) / l = 2 / r Valid for both laminar and turbulent flow

Shear stress = 2 w r / D Valid for both laminar and turbulent flow

Pressure drop p – γ l sin θ = 4 l w / D Valid for both laminar and turbulent flow

Velocity profile u r = V c 1 – ( 2 r / D ) 2 Laminar flow

Average velocity

V = ( π R 2 V c / 2 ) / π R 2

V = V c / 2

V = ( p – γ l sin θ ) D 2 / 3 2 µ l

Laminar flow

Flowrate Q = π D 4 ( p – γ l sin θ ) / 1 2 8 µ l Laminar flow

Example 1.3

An oil with a viscosity of µ = 0.40 N.s/m2 and density = 900 kg/m3 flows in pipe of

diameter D = 0.020 m.

a) What pressure drop is needed to produce a flowrate of Q = 2.0 × 10-5 m3/s if the pipe

is horizontal and x1 = 0 m and x2 = 10 m.

b) How steep a hill, θ, must the pipe be on if the oil is to flow at the same rate as in part

(a) but with p1 = p2.

c) For a condition of part (b), if p1 = 200 kPa, what is the pressure at x3 = 5m.

27

Solution:

a) Given, µ = 0.40 N.s/m2, = 900 kg/m3 D= 0.020 m Q = 2.0 x 10-5 m3/s, x1 = 0 m x2 = 10 m.

from equation Q = π D 4 p / 1 2 8 µ l

p = Q ( 1 2 8 µ l ) / π D 4

= [128(2.0 x 10-5)(0.40)10]/[3.14(0.02)4]

= 20400 N/m2

b) Given, µ = 0.40 N.s/m2 , = 900 kg/m3 D = 0.020 m, Q = 2.0 × 10-5 m3/s x1 = 0 m, x2 = 10 m, p= 0

from equation Q = π D 4 ( p – γ l sin θ ) / 1 2 8 µ l

p – γlsin θ = Q ( 1 2 8 µ l ) / π D 4

sin θ = - 1 2 8 Q µ / π D 4 γ

= - 128(2.0 × 10-5)(0.40)/[3.14(0.02)4(900)(9.81)]

θ = sin-1[[- 128(2.0 × 10-5)(0.40)]/[3.14(0.02)4(900)(9.81)]]

= -13.340

c) Condition as part (b), pressure different along the pipe, p = 0 and p1 = p2 = p3,

therefore, at x3 = 5 m, p3 = 200kPa

28

Thank You

29