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School of Mechanical and Chemical Engineering

SEMESTER 1, 2012 EXAMINATIONS

ENSC3003 Fluid Mechanics

FAMILY NAME: ____________________________ GIVEN NAMES: ______________________

STUDENT ID: SIGNATURE: ________________________

This Paper Contains: 18 pages (including title page) Time allowed: 3 hours 10 minutes

INSTRUCTIONS: This paper contains 5 questions, a table of constants and conversions (1 page), a general equation sheet (3 pages), and the equations of motion in Cartesian and Cylindrical coordinates (4 pages). Students should attempt all 5 questions Each question is worth 30 marks, for a total of 150 marks

PLEASE NOTE

Examination candidates may only bring authorised materials into the examination room. If a supervisor finds, during the examination, that you have unauthorised material, in whatever form, in the vicinity of your desk or on your person, whether in the examination room or the toilets or en route to/from the toilets, the matter will be reported to the head of school and disciplinary action will normally be taken against you. This action may result in your being deprived of any credit for this examination or even, in some cases, for the whole unit. This will apply regardless of whether the material has been used at the time it is found. Therefore, any candidate who has brought any unauthorised material whatsoever into the examination room should declare it to the supervisor immediately. Candidates who are uncertain whether any material is authorised should ask the supervisor for clarification.

Supervisors Only – Student left at:

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Question 1 (30 points total) (a) (15 points) Gas enters a pipe system at section 1 at a velocity of 5.2 m/s and a temperature

of 125˚C, and leaves at section 2, where the temperature is 80˚C. The pipe diameter at section 1 is 200 mm, and the pipe diameter at section 2 is 250 mm. The gauge pressures at sections 1 and 2 are 600 kPa and 400 kPa respectively. The system is at steady state. Determine the following;

The mass flow rate (in kg/s) at section 1 The mass flow rate (in kg/s) at section 2 The flow velocity (in m/s) at section 2 You may assume that at 300 K and standard atmospheric pressure, the density of the gas is

0.7 kg/m3. (b) (10 points) A body of water is held back behind a 5 metre long diamond shaped wall, as

illustrated in the diagram below (the diamond is a square rotated to a 45 degree angle). The water surface is level with the top of the wall, and atmospheric pressure acts at the surface. Calculate the horizontal and vertical components of the pressure force (in Newtons) acting on the dam wall. Your answer must identify the direction in which the force acts. You may use "shortcuts" in determining your answers, and assume that the density of water is 1000 kg/m3.

(c) (5 points)

The temperature of a liquid undergoing laminar flow in a horizontal circular pipe is decreased. If the pressure gradient is unchanged, how will this affect the volume flow rate in the pipe? Justify your answer.

2 metres

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Question 2 (30 marks total)

A liquid is transported through a horizontal circular pipe of radius R via pressure-driven flow, as illustrated below. The pressure at the upstream end of the duct (z=0) is Po, and at the downstream end (z=L) is PL. The flow may be regarded as laminar and isothermal, and the oil is a Newtonian liquid; the duct may be regarded as stationary. (a) (20 marks) Starting with the appropriate form of the Continuity Equation and Equations of

Motion (attached at the end of the examination paper), derive an equation for determining the z component of velocity in the pipe at steady state. Be sure to identify all assumptions and boundary conditions. You may use the attached equation sheets – submit any marked equation sheets with your examination booklet.

(b) (5 marks) Derive an equation for the volume flow rate of oil through the pipe at steady state.

You must show all working for full credit. (c) (5 marks) In a modified version of this system, the pipe radius tapers from R at z=0 to a

radius of 0.5R at z=L. While continuing to assume isothermal laminar flow, re-derive the differential forms of the continuity equation and z-direction momentum balance for the tapered pipe (NOTE – do not integrate the resulting equations)

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Question 3 (30 marks total)

A centrifugal mixing device, as illustrated in the diagram below, is being developed to stir small quantities of blood for experimental procedures. The proposed device is confined within a cylinder 2 cm in diameter and 10 cm tall, and uses a stirrer 1 cm in diameter. The proposed operating speed for the stirrer is 20 rpm. Blood is very viscous, and significant surface vortex can develop. If this vortex becomes large enough to reach the stirrer, the delicate cells within the blood will be severely damaged by the extreme shear gradients arising. It is therefore proposed to build a large scale model using an alternative fluid in order to investigate the phenomenon. Due to material restrictions, it has been established that the model cylinder diameter (DCM) must be 10 cm. If the specific gravity of blood is 1.1, and the viscosity of blood at room temperature is 2.0 cP, determine (i) The other dimensions (height HM and stirrer diameter DSM) of the model (ii) The kinematic viscosity required of the model fluid (in m2/s) (iii) The speed (in rpm) of the stirrer in the model system

DS

DC

H

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Question 4 – 30 Marks Total A centrifugal pump is to be installed to transfer water (ρ=1000 kg/m3, μ=0.001 Pas) between two reservoirs via the system illustrated in the diagram below. (a) (16 marks) The pump curve for the proposed pump is provided overleaf. Calculate the

system curve, plot it on the chart provided, and determine the duty point for the proposed system. To plot the system curve, calculating 3 points on the curve will be sufficient. (NB please detach the finished chart and submit it with your solutions). Charts for the friction factor and fitting loss factors are provided.

(b) (6 points) It is proposed to increase the flow delivered through the system by operating a

second identical pump with the original pump. Using the system curve you have calculated and the pump curve provided, determine the optimal arrangement of the two pumps (in terms of maximising flow rate) and the resulting duty point.

(c) (3 points) If the pump efficiency at the duty point is 76%, calculate the pump power (in

watts) that must be supplied by the motor when the pump is operating at the regular (single pump) duty point.

(d) (5 points) At the duty point, the NPSHR for the proposed pump is 9.2 metres. Under

extreme conditions, the water level in the supply reservoir may fall to as low as 1.5 metres above the pump inlet centreline. Assuming that the friction losses in the suction side pipework can be considered to be negligible (for the purpose of NPSH calculation only), is the pump safe from cavitation ? Show working to support your answer. The vapour pressure (head) of water at 293 K is 0.238 m (H2O)

PPipework

Length 675 metresDiameter 450 mmCommercial Steel

90 degreeLong radius

Fully OpenGate Valves

Swing CheckValve

11.5 metres

90 degreeLong radius

90 degreeLong radius

Sharp entrance4.3 metres

(to pump centreline)

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Pump Curve for Question 4

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9

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Question 5 - 30 marks total The pipe section shown below is incorporated in an oil delivery system (for the oil, SG=0.9, μ=2.5 centipoise). The flow velocity at section 1 is 1.5 m/s. The absolute pressures at section 1 is 450 kPa. The system is at steady state, and the pipe section is horizontal (so that body forces may be neglected). It may be assumed that β=1.05 for this system. (a) (14 points) Calculate the absolute pressure (in Pa) at section 2. For the purposes of this

calculation, it may be assumed that the straight pipe losses in the pipe are negligible. Fitting losses, however, must be taken into account. The loss factor chart was provided for question 4.

(b) (16 points) Determine the magnitude (in Newtons) and direction of the force exerted on the

pipe section by the flowing oil

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Constants Gravity 9.81 m/s2 Density of Water 1000 kg/m3 Viscosity of Water 0.001 Pas Atmospheric Pressure 1.015 x 105 Pa

Common Unit Conversions

Density

1 lb/ft3 16.02 kg/m3

Force 1 lbf 4.448 N 1 dyn 1 x 10-5 N

Length 1 foot 0.3048 m 1 inch 0.0254 m

Mass 1 pound (lbm) 0.04536 kg 1 ton (2000 lb) 907.0 kg

Pressure 1 atmosphere 1.015 x 105 Pa 1 mm H2O 9.806 Pa 1 mm Hg 133.3 Pa 1 psi 6895 Pa

Viscosity

1 poise 0.1 Pa.s 1 lbf.s/ft2 47.88 Pa.s 1 lbm/ft.s 1.488 Pas

Volume

1 litre 0.001 m3 1 gallon 0.003785 m3

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EQUATION SHEET : ENSC 3003 – SEMESTER 1, 2012

CONSERVATION OF MASS (MACROSCALE)

General Conservation of Mass for a System dMdT

= ˙ m in − ˙ m out = ρQ( )in

− ρQ( )out

Ideal Gas Law PV = nRT = mR 'T

VISCOSITY

Shear Stress & Absolute Viscosity - Newtonian Fluids

τ xy = −μdux

dy

Kinematic Viscosity

ν =μρ

Shear Stress & Absolute Viscosity - Bingham Fluids

τ xy = −μ0

dux

dy±τ o if τ xy ≥ τ o;

dux

dy= 0 if τ xy ≤ τ o

Shear Stress & Absolute Viscosity - Power-Law Fluids

τ xy = −ηdux

dy

n−1dux

dy

Chapman-Enskog Equation: Viscosity of Gases

μ = 2.6693×10−6 MWT( )1

2

σ 2Ωμ

Viscosity of Mixtures of Gases

μMIX =xiμi

x jΦij

j=1

N

∑

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

i=1

N

∑

Where

Φij =1

81+

Mi

M j

⎛

⎝⎜⎜

⎞

⎠⎟⎟

−12

1+μi

μ j

⎛

⎝⎜⎜

⎞

⎠⎟⎟

12 M j

Mi

⎛

⎝⎜

⎞

⎠⎟

14⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

2

Temperature Dependence of Viscosity in Liquids

μ = AeB

T Falling Ball Viscometer

μ =ρS − ρF( )gD2

18U

Couette Viscometer

T = 4πμω0R2Lk2

1− k2

⎛

⎝⎜

⎞

⎠⎟

HYDROSTATICS

Specific Gravity

SG =ρ

ρwater

Pressure Gradient dp

dx3

= −ρg

Gauge Pressure Pgauge = Pabsolute − Patmosphere

Pressure Head

H =P

ρg

Force Exerted on a Surface by a Pressure Distribution F = PdA

Area

∫

Centre of Pressure

LC =

PLdAArea

∫PdA

Area

∫=

dMArea

∫dF

Area

∫

Archimedes’ Principle FB = ρ f gVD

SHELL MOMENTUM BALANCES

Steady state momentum balance for a shell element {The rate of momentum flow in}

-{The rate of momentum flow out}

+{The sum of all the forces acting on the system} =0

DIMENSIONAL ANALYSIS

Reynolds Number (General)

Re =ρUL

μ

Froude Number (General)

Fr =U 2

gL

Weber Number (General)

We =ρU 2 L

σ

Buckingham-Pi Equation No. of dimensionless groups

= No. of parameters - No. of fundamental dimensions

BOUNDARY LAYER ADJACENT TO A FLAT PLATE

Local Reynolds Number at Laminar to Turbulent Transition

Re x

Trans =ρUx

μ= 5 .0 x 105

Drag Coefficient - Definition

CD =τ o

12

ρU 2

Boundary Layer Thickness - Laminar Boundary Layer

δ =4 .91x

Re x

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Local Surface Shear Stress - Laminar Boundary Layer

τ o =0.332μU Re x

x

Local Drag Coefficient - Laminar Boundary Layer CD =

0 .664

Rex

Mean Drag Coefficient - Laminar Boundary Layer

C D =1.328

Rex

Surface drag force due to laminar boundary layer FD = 0 .664μUW Re x

Boundary Layer Thickness - Turbulent Boundary Layer

δ =0.377xRe

x

0. 2

Local Surface Shear Stress - Turbulent Boundary Layer

τ o =0.0587ρU 2

2 Rex

0. 2

Mean Drag Coefficient - Turbulent Boundary Layer

C D =0.455

logRe x( )2. 5 8

Mean Drag Coefficient - Laminar and Turbulent Boundary Layer

C D =0.455

logRe x( )2. 5 8 −1700Re

x

DRAG FORCE ON BODIES IMMERSED IN A FLOW

Reynolds Number for Spheres

Re =ρUD

μ

General Equation: Drag Force Exerted on a Submerged object

FD =12

ρ fU2A CD

General Equation: Drag Force Exerted on a Sphere

FD =π8

ρ FU2D

2CD

Stokes Law : Total Drag Force Exerted on a Sphere for Creeping Flow FD = 3πμD - Valid for Re < 0.1

Drag Coefficient - Creeping Flow Around a Sphere

CD =24

Re - Valid for Re < 0.1

Terminal Velocity of a Falling Sphere

U =4

3

ρS − ρF( )ρF

gD

CD

⎡

⎣⎢

⎤

⎦⎥

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CONSERVATION OF MOMENTUM (MACROSCALE)

General Equation ∂∂t

ρu.dVVol�∫

⎛

⎝⎜

⎞

⎠⎟ = ρ1Q1β1U1n1( ) − ρ2Q2β2U2n2( )

+ P1G A1n1( ) + P2G A2 -n2[ ]( ) + FR + Mg

Conservation of Momentum - Jet of Incompressible Fluid deflected by a Moving Blade 0 = ρQSuJ S

ˆ n 1 − ρQS uE Sˆ n 2 + F

CONSERVATION OF MECHANICAL ENERGY

General “Pressure” Equation (Incompressible Fluid, Steady State)

0 =ΔU 2

2+

ΔP

ρ+ gΔx3 + EV +W

General “Head” Equation (Incompressible Fluid, Steady State)

0 =ΔU 2

2g+

ΔP

ρg+ Δx3 +

EV

g+

W

g

General Equation (Isothermal Ideal Gas, Steady State)

ΔU 2

2

⎛

⎝⎜

⎞

⎠⎟+

RT

Mln

p2

p1

⎛

⎝⎜

⎞

⎠⎟+ gΔx3 +W + Ev = 0

Friction Loss – Straight Pipe – Darcy-Wessbach equation EVSP

g∑ = ∑

2 fFU2L

gD

⎛

⎝⎜

⎞

⎠⎟ = ∑

fMU 2L

2gD

⎛

⎝⎜

⎞

⎠⎟

Friction Loss – Fittings EVF

g∑ ≡ ∑

kU 2

2g

⎛

⎝⎜

⎞

⎠⎟

INCOMPRESSIBLE FLOW IN A CIRCULAR PIPE

Average Flow Velocity U =

QA

Reynolds Number for Flow in a Circular Pipe

Re =ρUD

μ

Fanning Friction Factor

f =TR

12

ρU 2

Moody Friction Factor fM = 4 fF

Fanning Friction Factor for Laminar Flow

fF =16

Re

Blausius Equation : Fanning Friction Factor for Turbulent Flow in Hydraulically Smooth Pipes f = 0 .079Re −0. 2 5 Valid for 4000 < Re < 100000

Prandtl Resistance Law : General Equation for Fanning Friction Factor for Turbulent Flow in Hydraulically Smooth Pipes

1

f= 4 log Re f( ) −0 .4 Valid for 4000 < Re < 3000000

Colebrook Equation: General Equation for Fanning Friction Factor for Turbulent Flow in Rough Pipes

1

f= −4 log

ε3.7D

+1 .255

Re f

⎛

⎝ ⎜

⎞

⎠ ⎟

Dynamic Pressure Loss in Laminar Flow in a Straight Circular Pipe: Hagen-Poiseuille Equation

ΔPD =8μQLπR4

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PUMPS AND PIPING SYSTEMS

Hydraulic Power ˙ W H = −ρgQΔHP

Change In Dynamic Head

ΔH P =Δpρg

+ΔU 2

2g

Pump Efficiency

η =˙ W H˙ W P

Net Positive Suction Head Available

NPSHA =P0

ρg+

U02

2g⎛

⎝ ⎜ ⎞

⎠ ⎟ − Δx3 −

ˆ E VSP

g+

ˆ E VF

g

⎛

⎝ ⎜

⎞

⎠ ⎟ − HVAP

System Head Curve

ΔHP =ΔP

ρg+ Δx3

⎛

⎝⎜

⎞

⎠⎟+

ΔU 2

2g+

EVSP

g∑ +

EVF

g∑

⎛

⎝⎜

⎞

⎠⎟

FLOW IN POROUS MEDIA

Reynolds Number for Flow Through a Packed Bed of Spheres

Re PM =ρUo Dp

μ 1 −ε( )

Superficial Velocity

U0 =QA

Definition - Friction Factor for Flow in a Packed Bed of Spheres

f =ε 3 Dp ΔPD

ρU0

2L 1− ε( )

Blake-Kozeny Equation - Friction Factor for Laminar Flow in a Packed Bed of Spheres

f =150μ 1 −ε( )

ρU0D

p

Valid for RePM < 10

Burke-Plummer Equation - Turbulent Flow, Packed Bed of Spheres

ε 3DpΔPD

ρU0

2 L 1 −ε( )=1.75 Valid for RePM >1000

Ergun Equation - Transition Flow in a Packed Bed of Spheres

f =1.75 +150μ 1 −ε( )

U0D

pρ

Valid for 10 < Re PM < 1000

Darcy's Law - Cartesian Co-ordinates

U0 = −K D

μdPdx

Minimum Fluidization Velocity

U0F =ρP − ρF( )gε 3DP

2

150μ 1 −ε( )

COMPRESSIBLE FLOW Speed of Sound in an Ideal Gas

c =RG TγM

W

⎛

⎝ ⎜ ⎞

⎠ ⎟

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Mach Number

Ma =vc

Mach Cone - Internal Angle

sinα =cv

=1

Ma

Adiabatic Compressible Flow with Friction - Pipe Length Required to Achieve Sonic Velocity 1− Ma 2

γMa2 +1+ γ2γ

ln1 +γ( )Ma 2

2 + γ −1( )Ma 2

⎛

⎝ ⎜

⎞

⎠ ⎟ =

4 fLs

D

Adiabatic Compressible Flow with Friction - Ratio of Initial Pressure Divided by the Sonic Pressure

pp* =

1Ma

1 +γ( )2 + γ −1( )Ma 2

⎛

⎝ ⎜

⎞

⎠ ⎟

12

Adiabatic Compressible Flow with Friction - Ratio of Initial Temperature Divided by the Sonic Temperature TT * =

1 +γ( )2 + γ −1( )Ma 2

⎛

⎝ ⎜

⎞

⎠ ⎟

Adiabatic Compressible Flow with Friction - Ratio of Initial Density Divided by the Sonic Density

ρρ* =

1 +γ( )2 + γ −1( )Ma2

⎛

⎝ ⎜

⎞

⎠ ⎟

−12

Isothermal Compressible Flow with Friction - Pipe Length Required to Achieve Maximum Theoretical Velocity 1− γMa2

γMa 2+ ln γMa 2( )=

4 fLT

D

Where the maximum theoretical velocit

Ma * =1

γ

⎛

⎝ ⎜ ⎞

⎠ ⎟

12

y

Isothermal Compressible Flow with Friction - Ratio of Initial Pressure Divided by the Pressure at Ma*

pp

T

=1

γ Ma

Adiabatic Compressible Frictionless Flow - Ratio of Entrance Pressure to Reservoir Pressure p1

p0

= 1 +γ −1

2Ma

2⎛ ⎝

⎞ ⎠

−γγ −1

Adiabatic Compressible Frictionless Flow - Ratio of Entrance Temperature to Reservoir Temperature T1

T0

= 1 +γ −1

2Ma

2⎛ ⎝

⎞ ⎠

−1

Adiabatic Compressible Frictionless Flow - Ratio of Entrance Density to Reservoir Density ρ1

ρ0

= 1 +γ −1

2Ma

2⎛ ⎝

⎞ ⎠

−1γ −1

Reynolds Number for Compressible Flow in a Pipe

Re =4 ˙ m

πDμ

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The Equations of Motion The Continuity Equation (Conservation of Mass)

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The Equation of Motion (Momentum Balance) Cartesian Coordinates (x1, x2, x3)

General form x1 component

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17

The Equation of Motion (Momentum Balance) Cylindrical Coordinates (r, θθ , z) General form r component

ρ∂ur

∂t+ ur

∂ur

∂r+

uθ

r

∂ur

∂θ−

uθ

2

r+ uz

∂ur

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟ = −

1

r

∂∂r

rTrr( )+1

r

∂Trθ

∂θ+

∂Trz

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟ + ρgr

θ component

ρ∂uθ

∂t+ ur

∂uθ

∂r+

uθ

r

∂uθ

∂θ−

uruθ

r+ uz

∂uθ

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟ = −

1r2

∂∂r

r2Tθr( )+1r

∂Tθθ

∂θ+

∂Tθz

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟ + ρgθ

z component

ρ∂uz

∂t+ ur

∂uz

∂r+

uθ

r

∂uz

∂θ+ uz

∂uz

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟ = −

1r

∂∂r

rTzr( )+1r

∂Tzθ

∂θ+

∂Tzz

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟ + ρgz

In terms of velocity gradients, for an incompressible Newtonian fluid (i.e. with constant viscosity), under isothermal conditions r component

ρ ∂ur

∂t+ ur

∂ur

∂r+

uθ

r∂ur

∂θ−

uθ

2

r+ uz

∂ur

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟

= −∂p

∂r+ μ ∂

∂r

1

r

∂∂r

rur( )⎛ ⎝

⎞ ⎠

+1

r2

∂ 2ur

∂θ 2−

2

r2

∂uθ

∂θ+

∂2ur

∂z 2

⎛

⎝ ⎜ ⎞

⎠ + ρgr

θ component

ρ ∂uθ

∂t+ ur

∂uθ

∂r+

uθ

r∂uθ

∂θ−

ur uθ

r+ uz

∂uθ

∂z⎛ ⎝

⎞ ⎠

= −1

r

∂p

∂θ+ μ

∂∂r

1

r

∂∂r

ruθ( )⎛ ⎝

⎞ ⎠

+1

r 2

∂ 2uθ

∂θ 2 −2

r 2

∂ur

∂θ+

∂ 2uθ

∂z 2

⎛

⎝ ⎜ ⎞

⎠ + ρgθ

z component

ρ ∂uz

∂t+ur

∂uz

∂r+

uθ

r∂uz

∂θ+ uz

∂uz

∂z⎛ ⎝

⎞ ⎠

= −∂p

∂z+ μ

1

r

∂∂r

r∂uz

∂r

⎛ ⎝

⎞ ⎠

+1

r2

∂ 2uz

∂θ 2 +∂ 2uz

∂z2

⎛

⎝ ⎜ ⎞

⎠ + ρgz

18

Components of the stress tensor, for an incompressible Newtonian fluid (i.e. with constant viscosity), under isothermal conditions

Trr = p − 2μ∂ur

∂r

Tθθ = p − 2μ1

r

∂uθ

∂θ+

ur

r

⎛ ⎝

⎞ ⎠

Tzz = p − 2μ∂uz

∂z

Trθ = Tθr = −μ r∂∂r

uθ

r⎛ ⎝

⎞ ⎠ +

1

r

∂ur

∂θ⎡ ⎣ ⎢

⎤ ⎦ ⎥

Tθz = Tzθ = −μ∂uθ

∂z+

1

r

∂uz

∂θ⎡ ⎣

⎤ ⎦

Tzr = Trz = −μ∂uz

∂r+

∂ur

∂z⎡ ⎣

⎤ ⎦