Viscous Flow in Pipes

CHAPTER 1

VISCOUS FLOW IN PIPES

Why this chapter is so important?

Why study this topic?

Fluid problem – flow in pipes Viscous fluid

Shear stress and friction

uniform velocity profile to parabolic velocity profile other flow properties also changed

CHAPTER 1: VISCOUS FLOW IN PIPES 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 or experimental data with theoretical considerations and dimensional analysis provide the desired results.

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.

� Assume involved in this chapter

- The pipe is round in cross section

- The pipe is completely filled with fluid - Viscous fluid

- Incompressible fluid

� Pipe flow vs Open channel flow

- Pipe flow – pressure gradient in the driving force (gravity may be important).

- Open channel flow - gravity is the driving force.

� 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.

� Laminar and turbulent flow

- Flow is said to be laminar when adjacent fluid layers move at same velocity and paths of

individual particles of fluid do not cross each other. Occur at low velocities and high viscosity (Re ≤ 2100).

- Flow is turbulent when streamlines cross each other and mixing of fluid flow occur. Occur at high velocities and low viscosity (Re ≥ 4000). - 2100  Re  4000 ?

Reynold’s Experiment : Dye Streaks

Fluid velocity at a point � 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.

� Example : Water flows through a pipe of diameter 0.018 m.

a. Determine the minimum time taken to fill a 3.54 x 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 x 10-4 _m3 _{glass with water if the flow}

in pipe is to be turbulent. Solution :

Given, D = 0.018 m

V = 3.54 x 10-4 _m3

a. Minimum time occur when Reynolds number is maximum allowed for laminar flow (maximum velocity)

v = Re (µ)/D = 2100 (1.12 x 10-3_{)/[1000(0.018)]} = 0.131 m/s Q = vA = 0.131[π(0.018)2_/4] = 3.33 x 10-5 _m3_/s t = /Q = 3.54 x 10-4_{/3.33 x 10}-5 = 10.63 s

b. Maximum time occur when Reynolds number is minimum allowed for turbulent flow (minimum velocity) v = Re (µ)/D = 4000 (1.12 x 10-3_{)/[1000(0.018)]} = 0.249 m/s Q = vA = 0.249 [π(0.018)2_/4] = 6.33 x 10-5 _m3_/s t = /Q = 3.54 x 10-4_{/6.33 x 10}-5 = 5.59 s

Entrance Region and Fully Developed Flow

� 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)].

Entrance region, developing flow and fully developed flow � 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).

� 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)).

� Example : 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 x 60)m3_{/s = 3.33 x} 10-4 _m3_/s v = Q/A = 3.33 x 10-4_/[π(0.013)2_/4] = 3.33 x 10-4_{/1.33 x 10}-4 = 2.50 m/s Re = vD/µ = 1000(2.5)(0.013)/1 x 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

Pressure and Shear Stress

�Pressure different between 2 section/point forces the fluid through the horizontal pipe and viscous effects provide the restraining force that exactly balances the pressure force.

� In the entrance region, fluid accelerate or decelerate as it flow, thus there is a balance between pressure, viscous and inertia (acceleration)

� The magnitude of the pressure gradient p/x is

larger in the entrance region than in the fully develop flow, where it is a constant,

p/x = -p/l  0

Pressure distribution along the pipe

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 method can be used to derive equations pertaining to fully developed laminar flow in pipes.

- Applying F = ma to a fluid element. - Dimensional analysis.

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

Free body diagram of fluid element

� Apply F = ma

(p1)πr2 – (p1 - p)πr2 – ()2πrl = 0

p/l = 2/r (1.3) This equation represents the basic balance in force needed to drive each fluid particle along the pipe with constant velocity.

� is depending on 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  = 2wr/D (1.4) and p = 4lw/D (1.5)

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



� How shear stress related to velocity?

Two governing laws for fully developed laminar flow - p/l = 2/r

-  = -µu/r

Combine this two equation

u/r = - (p/2µl)r velocity profile  u = - (p/2µl)r r u = - (p/4µl)r 2 _{+ C} 1 at r = D/2, u = 0 and C1 = (p/16µl)D 2

ur = (pD2/16µl)1 – (2r/D) 2 at r = 0, centerline velocity, Vc Vc = (pD2/16µl) therefore ur = Vc 1 – (2r/D) 2 (1.6)

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

ur = w D/4µ 1 – (r/R) 2 �Flowrate Q =  u dA Q =  u 2πr dr Q = 2π Vc 1 – (r/R) 2r dr Q = π R2 _V c/2

knowing that average velocity, V = Q/A = Q/πR2

V = (π R2 _V

c/2)/ πR2 = Vc/2 = pD2/32µl

(1.7) and

� Summary - Flow  properties  for horizontalpipe

Flow

Properties Equation Remarks

Entrance Length, le/D le/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  = 2wr/D Valid for both

laminar and turbulent flow Pressure

drop p = 4lw/D Valid for bothlaminar and turbulent flow Velocity profile ur = Vc 1 – (2r/D) 2_{ Laminar flow} Average velocity V = (π R 2 _V c/2)/ πR2 V= Vc/2 V = pD2_/32µl Laminar flow

Flowrate Q = πD4p/128µl _{Laminar flow}

� Adjustment to account for non horizontal pipe – gravity effect

Free body diagram of fluid element for non horizontal pipe

θ - angle between pipe centerline axis and horizontal axis

(p + p)πr2 _{– (p)πr}2 _{– mgsinθ – ()2πrl = 0}

(p + p)πr2 _{– (p)πr}2 _{– (πr}2_{)lgsinθ – ()2πrl = 0}

(p – γlsin θ)/l = 2/r

(1.9)

effects of non horizontal pipe

p  (p – γlsin θ) therefore V = (p – γlsin θ)D2_/32µl (1.10) and Q = πD4_{(p – γlsin θ)/128µl} (1.11)

� Summary - Flow  properties  for nonhorizontalpipe

Flow

Properties Equation Remarks

Entrance Length, le/D - le/D = 0.06 Re - le/D = 4.4 (Re)1/6 Laminar flow Turbulent flow Pressure

drop per unit length

(p – γlsin θ)/l = 2/r Valid for both laminar and turbulent flow Shear stress  = 2wr/D Valid for both

laminar and turbulent flow Pressure

drop p – γlsin θ = 4lw/D

Valid for both laminar and turbulent flow Velocity

profile ur = Vc 1 – (2r/D)

Average velocity V = (π R 2 _V c/2)/ πR2 V= Vc/2 V = (p–γlsin θ)D2_/32µl Flowrate Q = πD4_{(p – γlsin} θ)/128µl

� Example : 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 x 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 = 5 m.

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 = πD4_p/128µl p = Q(128µl)/πD4 = [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 x 10-5 _m3_/s x1 = 0 m x2 = 10 m p= 0

from equation Q = πD4_{(p – γlsin θ)/128µl}

p – γlsin θ = Q(128µl)/πD4 sin θ = - 128Qµ/πD4_γ = - 128(2.0 x 10-5_{)(0.40)/[3.14(0.02)}4₍₉₀₀₎ (9.81)] θ = sin-1_{[[- 128(2.0 x 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

Pressure Drop and Head Loss

� Important of pressure drop – it is related to power required by pump or fan to maintain fluid flow.

� Power, W = gQhL

� From energy equation

for horizontal pipe, v1 = v2 and z1 = z2, = kinetic

energy coefficient and for uniform flow 1 = 2 = 1

hL = p/g

(1.12)

from previous researcher, for laminar flow

hL = f(l/D)(v2/2g)

(1.13)

combine equation (1.12) and (1.13), and f = 64/Re p/g = (64/Re)(l/D)(v2_/2g)

p = 32µlv/D2

(1.14)

combine equation (1.3)and (1.12)

hLg = 2l/r

hL = 2l/gr

hL = 4wl/gD (1.15)

� Example : Water with a viscosity of µ = 1.545 x 10-3

kg.s/m and density  = 998 kg/m3 _{is flowing}

through 0.003 m diameter 9 m long horizontal pipe steadily at an average velocity of 0.9 m/s. Determine

a. the head loss

b. the pressure drop

c. the pumping power requirement to overcome this pressure drop.

a. head loss?

Re = vD/µ

= 998(0.9)(0.003)/ 1.545 x 10-3

= 1744 ( 2100, laminar flow) for laminar flow

f = 64/Re = 64/1744 = 0.0367 and hL = f(l/D)(v2/2g) = (0.0367)(9/0.003)[0.92_/2(9.81)] = 4.545 m b. pressure drop? for laminar flow

p/g = (64/Re)(l/D)(v2_/2g) p = (64/Re)(l/D)( v2_/2) = 0.0367(9/0.003)[998(0.92_)/2] = 44.5 kPa c. power required? P = gQhL and p/g = hL therefore P = Qp

= 0.9π(0.0032_/4)(44500)

= 0.283 W

Concept of Turbulent Flow in Pipes

�Characteristics of Turbulent Flow in Pipes - Re > 4000.

- Random movements of eddies which mixes up the layers of fluid.

- Particle path is irregular. - Most common type of flow.

- Difficult mathematical analysis to describe the flow.

� Important of turbulent flow - mixing process

- heat and mass transfer process

axial velocity measured at a given location

Turbulent Shear Stress

� Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow.

� 3-dimensional eddies conveys mass with average velocity ū. Therefore flow momentum exists. The result of this momentum transfer is shear force. � Shear stress in pipe is given as the summation of

laminar shear stress and turbulent shear stress

τ = τlam + τturb

� Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe.

� The region where laminar shear force dominates is called the viscous sublayer or the viscous wall

layer.

� The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer

layer.

� There is also a region where both laminar and shear are important. This region is called the overlap

region.

� The character of the each layers such as their velocity are different, so we need different equations to describe them

Turbulent Velocity Profile

� Velocity profile for viscous sublayer

ū/u* = yu*/ (1.16) where ū = average velocity

y = distance measured from wall = R – r

u* _{= friction velocity = (τ}

w /)1/2

 = kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 ≤ yu*/ ≤ 5

� Velocity profile for overlap region

ū/u* = 2.5 ln (yu*/) + 5.0

where 2.5 and 5.0 are constants determined by experiments

�For the outer layer, the Power Law is used from the following expression.

ū/Vc = [1 – (r/R)]1/n

(1.18)

The value of n which indicates the “power” of the equation is a function of Re and determined

experimentally.

exponent, n, for power laws velocity profiles

� The relationship between average velocity, V, volume flowrate, Q, and centerline velocity, VC can be

obtained by integrating the power law velocity profile.

Q =  Vc [1 – (r/R)]1/n A Q =  Vc [1 – (r/R)]1/n 2πr r Q = 2πR2 _V c n2/[(n + 1)(2n + 1)] (1.19) since Q = πR2_V πR2_{V/ πR}2 _V c = 2n2/[(n + 1)(2n + 1)] V/Vc = 2n2/[(n + 1)(2n + 1)] V = 2n2_V c /[(n + 1)(2n + 1)] (1.20)

� Example : Water at 200_{C (ρ = 998 kg/m}3 _{, and ν = 1 x}

10−6 _m2_{/s flows through a horizontal pipe of}

D = 0.1 m diameter with a flowrate of Q = 4 x 10−2 _m3_{/s and a pressure gradient of ∆p/l =}

2.59kPa/m. Determine

a. the approximate thickness of the viscous sublayer.

b. the approximate Darcy friction coefficient f.

c. approximate centerline velocity.

Head Loss

� Energy equation for steady incompressible flow in horizontal pipes,

p1/g + 1v12/2g + z1 = p2/g + 2v22/2g + z2 + hL+ w - q

where hL= head loss

w = turbine head q = pump head

� Head loss

- major loss - minor loss

Major Losses

� Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll, rub and slide each other.

� Losses cause by doing work against friction

hL = flv2/2gD

where f = friction factor

l = length v = velocity

d = gravity acceleration

� For laminar flow

f = 64/Re

� For turbulent flow

1/f0.5 _{= - 1.8 log [(6.9/Re) + (/3.7D)}1.1_]

� Surface Roughness

Pipe Equivalent Roughness, 

(ft) (mm)

Concrete 0.001 – 0.01 0.3 – 3.0 Wood Stave 0.0006 – 0.003 0.19 – 0.9 Cast Iron 0.00085 0.26 Galvanized Iron 0.0005 0.15 Commercial Steel 0.00015 0.045 Drawn Tubing 0.000005 0.0015 Plastic, glass 0.0 (smooth) 0.0 (smooth)

� Normally for pipe analysis, we obtain the dependence of friction factor on Re & ε/D through the Moody

Chart.

� To construct this chart the equivalent roughness ε is usually obtained for ‘clean’ and new pipes because after considerable use, most pipes may have increased roughness.

� For high Re flows, the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls.

� For smooth pipes (ε = 0), we notice that friction factor (f) is not zero because there is still head loss. These pipes are called “hydraulically smooth”.

� The moody chart offers the relationships between ε/D,

f and Re for a very wide range of pipe flows including

that for laminar flows as long as the flow is steady, fully developed and incompressible.

� The Moody chart is useful because in real applications, a large variety of D, V, ρ and μ exists but only for small ranges win the Moody chart.

� The Moody chart is valid for all steady, fully developed, incompressible pipe flows.

Non Circular Conduits

� Air Conditioner Ducting System

� Hydraulic Radius

RH = Area/ Circumference

Circle cross section area

RH = Area/ Circumference

= [πD2_/4]/πD

= D/4 D = 4 RH

therefore hf = flv2/2gD = flv2_/8gR H /D = /4 RH vD/µ =4v RH/µ Minor Losses

� Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows, tees, bends etc.

�In equation form

hL = KLv2/2g

where KL is loss coefficient

KL = hL2g/v2 = 2p/v2

� KL vary depending on the shapes involved

Entrance loss

� Head loss when liquid enters pipe from a large tank/reservoir

Exit Loss

� Head loss produced when liquid leaves pipe and enters a large tank/reservoir

� The entire kinetic energy of exiting fluid v1 is

dissipated through viscous effects and eventually becomes v2 = 0

� Exit loss from (1) to (2) is equivalent to one velocity head

� KL = 1.0

� Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

� Loss coefficient is a function of are ratio A2/A1.

� For sudden expansion

KL = [1 – (D1/D2)2]2

� A2/A1 = 0 KL = 0.5 - extreme sharp edge entrance

� A2/A1 = 1 KL = 0 - no area change

Vena Contracta

� Fluid entering a sharp corner.

� Fluid cannot through sharp corners. At a sharp corner, the flow will separate and reattaches at the pipe wall.

� This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area.

� This causes the velocity of fluid passing through this small area to increase

� Maximum velocity exists at section with minimum area called the vena contracta.

� Because high speed flows cannot slow down efficiently, the kinetic energy could not be fully converted into pressure.

Pump

� Pumps are used to increase energy of the fluid (liquid).

� 40% - 50% of industrial energy is used to drive pumps and compressors.

� Proper design construction and selection of pumps are economically significant.

� One of the most common pump is the centrifugal pump.

� It consist of rotating elements called impeller which is contained within the pump housing.

� The shaft transfers mechanical energy to the impeller. � A system of bearings and seals are required to

� Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward.

� Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades. � Fluid discharges into the housing which is designed

to reduce velocity. � Types of pumps

- Reciprocating pistons or plunger - Gear pump

- Double screw pump - Sliding vane

- Lobe pump

- Differential piston - Flexible squeegee

� In reality, pumps cause losses.

� Normally pumps are driven by electric motors, IC engines etc.

� In short we can say that pump draws kinetic energy and delivers it to the fluid.

� If we include energy of pump in the energy equation, we get:

p1/g + v12/2g + z1 + hP= p2/g + v22/2g + z2 + hL

where hP is the energy added to fluid and hL is the

Power Required by Pump

� Power is the rate of work or the rate of which energy is being transferred and is given by:

� If pump efficiency is 100%, power required by the pump is equal to power added to fluid:

� Power added to fluid:

Power = pghPQ

Unit: Watt, Nm/s or J/s

� Example : A tank of water empties by gravity through a horizontal pipe into another tank. There is a sudden enlargement in the pipe. At a certain time, the difference in level is 3 m. Each pipe is 2 m long and has a friction coefficient, f = 0.005. The inlet loss coefficient is 0.3. Calculate the flowrate at this point. Solution : 3 m 60 mm 20 mm

energy equation p1/g + v12/2g + z1 = p2/g + v22/2g + z2 + hL p1 = p2 = 0, v1 = v2 = 0 hL = z1 - z2 = 3 m major loss hf 1 = f1lv2/2gD = 0.005(2) [Q2_/π2_(0.022_/4)2_{]/[2(9.81)(0.02)]} = 258471Q2 hf 2 = f2lv2/2gD = 0.005(2) [Q2_/π2_(0.062_/4)2_{]/[2(9.81)(0.06)]} = 1064Q2 minor losses for entrance hL1 = KL1v2/2g = 0.3[Q2_/π2_(0.022_/4)2_]/[2(9.81)] = 155083Q2

for sudden enlargement

KL2 = [1 – (D1/D2)2]2 = [1– (0.02/0.06)2_]2 = 0.79 therefore hL2 = KL2v2/2g = 0.79[Q2_/π2_(0.022_/4)2_]/[2(9.81)]

= 408384Q2

for exit

hL3 = KL3v2/2g

= 1[Q2_/π2_(0.062_/4)2_]/[2(9.81)]

= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 +

408384Q2 _{+ 6382Q}2

Q2 _{= 3/596754}

Q = 2.242 x 10-3 _m3_/s

� Example : A vented tanker is to be filled with fuel oil with ρ = 920 kg/m3 _{and μ = 0.045 kg/m.s}

from an underground reservoir using a 20 m long, 5 cm diameter plastic hose with a slightly rounded entrance (KL = 0.12) and

two 90° smooth bends (KL = 0.3). The

elevation difference between the oil level in the reservoir and the top of the tanker where the hose is discharged is 5 m. For the flow rate 0.01 m3_{/s and the overall pump}

efficiency is 82%, calculate the power required by the pump.

� Example : In a chemical processing plant, benzene at 50o_{C must be delivered to point B with}

pressure of 550 kPa. A pump is located at point A, 21 m below point B and the two points are connected by 240 m of stainless steel pipe having an inside diameter of 40 mm as in figure below. If the volume flow rate is 1.833 x 10-3 _m3_{/s, calculate the}

required pressure at the outlet of the pump. Given:

 benzene = 860 kg/m3

μ benzene = 4.2 x 10-4 Pa.s

ε pipe = 0.002 mm

Pump P_A = ? A B A 21m Flow P_A = 550 kpa

Stainless steel pipe D = 40mm