Advanced Fluid Mechanics
(PhD) Assistant Professor,
Mechanical Power Engineering Department,
Faculty of Engineering, Tanta University
Chapter Three
Applications of Equations of Motion
1
- About the Course ❑ Fluid Kinematics
❑ Differential Analysis of Fluid Flow
❑ Applications of Equations of Motion
❑ Bearings
❑ Turbulent Flow
References:
Munson, R. et. al. “Fundamentals of Fluid Mechanics”, US: John Wiley & Sons,
Inc., 2002 (Main)
Cengel, A. and Cimbala, M. “Fluid Mechanics: Fundamentals and Applications”,
US: McGraw-Hill Series, 2006
2
Applications of Equations of Motion
➢ Some Simple Solutions for Viscous incompressible Fluids
➢ Steady, Laminar Flow Between Fixed Parallel Plates
“Channel Flow”
➢ Flow Between Two Parallel Plates Bottom-Fixed, UpperMoving with Constant Velocity 𝑢 “Couette Flow”
➢ Steady, Laminar Flow in Circular Tubes “Hagen-
Poiseuille Flow”
➢ Steady, Axial, Laminar Flow in an Annulus
3
𝐍𝐚𝐯𝐢𝐞𝐫 − 𝐒𝐭𝐨𝐤𝐞𝐬 and 𝐂𝐨𝐧𝐭𝐢𝐧𝐮𝐢𝐭𝐲 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧𝐬
𝑁𝑎𝑣𝑖𝑒𝑟 − 𝑆𝑡𝑜𝑘𝑒𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
𝐼𝑛 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠
Expressing Continuity & Navier-Stokes equations in cylindrical coordinates
(r, 𝜃, z) for incompressible, Newtonian fluids.
𝐼𝑛 𝑟 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝐼𝑛 𝑧 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝐼𝑛 𝜃 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝐶𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (incompressible)
4
𝐬𝐧𝐨𝐢𝐭𝐚𝐮𝐪𝐞 𝐲𝐭𝐢𝐮𝐧𝐢𝐭𝐧𝐨𝐂 𝐍𝐚𝐯𝐢𝐞𝐫 − 𝐒𝐭𝐨𝐤𝐞𝐬 and
𝜃 𝑠𝑜𝑐 𝑟 = 𝑥
𝜃 𝑛𝑖𝑠 𝑟 = 𝑦
𝑟 = 𝑥2 + 𝑦2
)𝑥𝜃 = tan−1 (𝑦Τ
5
❑ للتحويل من اإلحداثيات
الكارتيزية إلى اإلحداثيات
القطبية (اإلسطوانية).
Steady, Laminar Flow in Circular Tubes
Hagen-Poiseuille Flow
❑ Probably the best-known exact solution to the Navier–Stokes
equations is for steady, incompressible, laminar flow through a
straight circular tube of constant cross section. This type of flow is
commonly called Hagen-Poiseuille flow, or simply Poiseuille flow.
❑ Consider the flow through a horizontal circular tube of radius R as
is shown in Fig. 3.4a. Because of the cylindrical geometry it is
convenient to use cylindrical coordinates. We assume that the flow
is parallel to the walls so that (𝑣𝑟 = 0.0 and 𝑣𝜃 = 0.0), and from the
continuity equation (𝜕𝑣𝑧 Τ𝜕𝑧 = 0.0). Also, for steady, axisymmetric
flow, 𝑣𝑧 is not a function of 𝑡 or 𝜃 so the velocity, 𝑣𝑧 , is only a
function of the radial position within the tube that is, 𝑣𝑧 = 𝑣𝑧 (𝑟) .
Under these conditions the Navier–Stokes equations reduce to
6
Steady, Laminar Flow in Circular Tubes
❑ Assumptions:
➢ Laminar flow & Steady flow (𝜕Τ𝜕𝑡 = 0.0).
➢ Incompressible, Newtonian fluid (𝜌, 𝜇 = Const.)
➢ Parallel flow (𝑣𝑟 = 0.0 and 𝑣𝜃 = 0.0) السريان في إتجاه مواز لمحور اإلسطوانة
➢ No swirl flow → 𝑣𝜃 = 0.0
➢ Axisymmetric flow (𝜕Τ𝜕𝜃 = 0.0)
➢ at tube-centerline (𝑟 = 0.0) → (𝜕𝑣𝑧 Τ𝜕𝑟 = 0.0), 𝑣𝑧 = 𝑣𝑧−𝑚𝑎𝑥
Fig. 3.4 The viscous flow in a horizontal, circular tube: a) coordinate system
and notation used in analysis; b) flow through differential annular ring.
7
Hagen-Poiseuille Flow
8
Hagen-Poiseuille Flow
➢ From Continuity equation for an incompressible fluids
𝑣𝑟 = 0.0 No swirl flow
1 𝜕(𝑟𝑣𝑟 )
1 𝜕𝑣θ
𝜕𝑣𝑧
+
+
𝑟 𝜕𝑟
𝑟 𝜕θ
𝜕𝑧
𝜕𝑣𝑧
= 0.0
𝜕𝑧
= 0.0
Parallel flow 𝑣θ = 0.0
𝑣𝑧 = 𝑓𝑛 (𝑟)
𝑣𝑧 ≠ 𝑓𝑛 (𝑡)
Steady flow
𝑣𝑧 ≠ 𝑓𝑛 (𝜃)
Axisymmetric flow
➢ 𝑁𝑎𝑣𝑖𝑒𝑟 − 𝑆𝑡𝑜𝑘𝑒𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
SteadyParallel flow
𝑣𝑧 ≠ 𝑓𝑛 (𝑧)
Fully developed flow
𝐼𝑛 𝑟 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
No swirl flow
𝑣𝜃2
Fully developed
flow
سرعة السريان عند أي
) ثابتة ال تتغيرz( نقطة
Parallel flow
𝜕𝑣𝑟
𝜕𝑣𝑟
𝑣𝜃 𝜕𝑣𝑟
𝜕𝑣𝑟
𝜌( + 𝑣𝑟
+
− + 𝑣𝑧 )
Parallel flow
𝜕𝑡
𝜕𝑟
𝑟 𝜕𝜃
𝑟
𝜕𝑧
𝜕𝑃
1 𝜕
𝜕𝑣𝑟
𝑣𝑟
1 𝜕2 𝑣𝑟
2 𝜕𝑣𝜃
𝜕2 𝑣𝑟
= − + 𝜌g𝑟 + 𝜇
(𝑟
)− 2+ 2 2 − 2
+ 2 (
𝜕𝑟
𝑟 𝜕𝑟
𝜕𝑟
𝑟
𝑟 𝜕𝜃
𝑟 𝜕𝜃
𝜕𝑧
g𝑟 = − gsin𝜃
Parallel flow
No swirl flow
9
Hagen-Poiseuille Flow
𝜕𝑃
0.0 = − 𝜌gsin𝜃
𝜕𝑟
𝜕𝑃
= −𝜌gsin𝜃
𝜕𝑟
𝑁𝑎𝑣𝑖𝑒𝑟 − 𝑆𝑡𝑜𝑘𝑒𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
𝐼𝑛 𝜃 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
No swirl flow
Steady Parallel flow
No swirl flow
𝜕𝑣𝜃
𝜕𝑣𝜃
𝑣𝜃 𝜕𝑣𝜃
𝑣𝑟 𝑣𝜃
𝜕𝑣𝜃
𝜌(
+ 𝑣𝑟
+
+
+ 𝑣𝑧
)
Parallel flow
𝜕𝑡
𝜕𝑟
𝑟 𝜕𝜃
𝑟
𝜕𝑧
1 𝜕𝑃
1 𝜕
𝜕𝑣𝜃
𝑣𝜃
1 𝜕2 𝑣𝜃
2 𝜕𝑣𝑟
𝜕2 𝑣𝜃
=−
+ 𝜌g𝜃 + 𝜇
(𝑟
)− 2+ 2 2 − 2
+ 2 (
𝑟 𝜕𝜃
𝑟 𝜕𝑟
𝜕𝑟
𝑟
𝑟 𝜕𝜃
𝑟 𝜕𝜃
𝜕𝑧
g𝜃 = − g cos 𝜃
1 𝜕𝑃
0.0 = −
− 𝜌g cos 𝜃
𝑟 𝜕𝜃
No swirl flow
No swirl flow
No swirl flow
𝜕𝑃
= − 𝑟𝜌g cos 𝜃
𝜕𝜃
𝑃 = −𝜌g 𝑟 sin 𝜃 + 𝑓1 (𝑧)
(𝑦 = 𝑟 𝑠𝑖𝑛 𝜃( ❖ بالتكامل بالنسبة ل 𝜃 حيث أن
𝑃 = −𝜌g 𝑦 + 𝑓1 (𝑧)
𝜕𝑓 )𝑧(
𝜕𝑃
= 0.0 + 1 ≠ 𝑓𝑛(𝑟,𝜃)
𝜕𝑧
𝜕𝑧
.) فقطBody force( ) و يكون تحت تأثيرالHydrostatic pressure( z ❑ الضغط عند أي مقطع
10
Hagen-Poiseuille Flow
𝑁𝑎𝑣𝑖𝑒𝑟 − 𝑆𝑡𝑜𝑘𝑒𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
𝐼𝑛 𝑧 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
Steady Parallel No swirl
Fully developed
𝜕𝑣𝑧
𝜕𝑣
𝑣 𝜕𝑣
𝜕𝑣
+ 𝑣𝑟 𝑧 + 𝜃 𝑧 + 𝑣𝑧 𝑧 )
Fully developed
𝜕𝑡
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
𝜕𝑃
1 𝜕
𝜕𝑣𝑧
1 𝜕2 𝑣𝑧
𝜕2 𝑣𝑧
= − + 𝜌g𝑧 + 𝜇
(𝑟
) + 2 2+ 2 (
𝜕𝑧
𝑟 𝜕𝑟
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
𝜌(
Axisymmetric
g𝑧 = 0.0
𝜕𝑃
1 𝜕
𝜕𝑣𝑧
0.0 =− + 𝜇 [
(𝑟
)]
𝜕𝑧
𝑟 𝜕𝑟
𝜕𝑟
since 𝑣𝑧 = 𝑓𝑛(𝑟) only
1 𝜕𝑃
1 𝜕
𝜕𝑣𝑧
=
(𝑟
)
𝜇 𝜕𝑧
𝑟 𝜕𝑟
𝜕𝑟
𝑑
𝑑𝑣𝑧
𝑟 𝜕𝑃
(𝑟
)=
𝑑𝑟
𝑑𝑟
𝜇 𝜕𝑧
𝑑𝑣𝑧
𝑟 2 𝜕𝑃
𝑟
= ( ) + 𝐶1
𝑑𝑟
2𝜇 𝜕𝑧
𝑑𝑣𝑧
𝑟 𝜕𝑃
𝐶
= ( ) + 𝑟1
𝑑𝑟
2𝜇 𝜕𝑧
𝑟 2 𝜕𝑃
𝑣𝑧 = ( ) +𝐶1 ln 𝑟 + 𝐶2
4𝜇 𝜕𝑧
(𝜋)
𝑟 ❖ بالتكامل بالنسبة ل
( ∅)
𝑟 ❖ بالتكامل بالنسبة ل
11
Hagen-Poiseuille Flow
➢ Boundary conditions:
(1) at 𝑟 = 𝑅 → 𝑣𝑧 = 0.0
𝑧𝑣𝑑
→ )𝑤𝑜𝑙𝑓 𝑐𝑖𝑟𝑡𝑒𝑚𝑚𝑦𝑠𝑖𝑥(2) at 𝑟 = 0.0 (A
= 0.0
𝑟𝑑
❖ بالتعويض بإستخدام ال ( )B. C.رقم ( )2في المعادلة (∅)
𝑧𝑣𝑑
𝑃𝜕 0.0
𝑟
= 0.0 = ( ) + 𝐶1
𝑟𝑑
𝑧𝜕 𝜇2
𝐶1 = 0.0
❖ بالتعويض بإستخدام ال ( )B. C.رقم ( )1في المعادلة (𝜋)
𝑃𝜕 𝑅 2
0.0 = ( ) + 𝐶2
𝑧𝜕 𝜇4
𝑃𝜕 𝑅2
) 𝑧𝜕 ( 𝜇𝐶2 = − 4
𝑃𝜕 𝑟 2
𝑃𝜕 𝑅 2
) ( 𝑣𝑧 = ( ) −
𝑧𝜕 𝜇4
𝑧𝜕 𝜇4
𝑃𝜕 1
) 𝑣𝑧 = ( )(𝑟 2 − 𝑅2
𝑧𝜕 𝜇4
𝑃𝜕
=−ve
𝑧𝜕
12
𝑃𝜕
𝑃∆
=−
𝑧𝜕
𝑙
𝑣𝑧 =+ve
𝟐
𝒍
𝟏
Hagen-Poiseuille Flow
➢ A relation for the flow rate:
2𝜋𝑟
𝑑𝑟
𝑑𝑄 = 𝑣𝑧 𝑑𝐴 = 𝑣𝑧 2𝜋𝑟𝑑𝑟
𝑑𝐴 = 2𝜋𝑟𝑑𝑟
𝑅
𝑄 = 0 𝑣𝑧 2𝜋𝑟𝑑𝑟
𝑅 1 𝜕𝑃
𝑄 = 0 4𝜇 ( 𝜕𝑧 )(𝑟 2 − 𝑅2 )2𝜋𝑟𝑑𝑟
𝑅
𝜋 𝜕𝑃
𝑄 = 2𝜇 ( 𝜕𝑧 ) 0 (𝑟 3 − 𝑅2 𝑟) 𝑑𝑟
𝜋 𝜕𝑃
𝑟4
𝑟2 𝑅
𝜋 𝜕𝑃
𝑅4
𝑅4
𝜋 𝜕𝑃
2
𝑄 = 2𝜇 ( 𝜕𝑧 )[ 4 − 𝑅 2 ]0 = 2𝜇 ( 𝜕𝑧 )[ 4 − 2 ] = − 8𝜇 ( 𝜕𝑧 )𝑅4
𝜕𝑃
∆𝑃
where, = −
𝜕𝑧
𝑙
𝜋
∆𝑃
𝑄 = 8𝜇 𝑅4 ( 𝑙 )
➢ The mean velocity:
2
𝑄
𝜋 4 ∆𝑃
1
𝑉𝑚𝑒𝑎𝑛 = = 𝑅 ( ) 2
A
8𝜇
𝑙
𝜋𝑅
𝑉𝑚𝑒𝑎𝑛 =
1 2 ∆𝑃
𝑅 ( )
8𝜇
𝑙
13
Hagen-Poiseuille Flow
➢ The maximum velocity:
𝜌𝑉𝑚𝑒𝑎𝑛 𝐷
𝑅𝑒 =
𝜇
at 𝑟 = 0.0 → 𝑣𝑧 = 𝑣𝑧,𝑚𝑎𝑥
1 ∆𝑃
𝑣𝑧,𝑚𝑎𝑥 = ( ) 𝑅2
4𝜇 𝑙
&
1 2 ∆𝑃
𝑉𝑚𝑒𝑎𝑛 = 𝑅 ( )
8𝜇
𝑙
𝑣𝑧,𝑚𝑎𝑥 = 2 𝑉𝑚𝑒𝑎𝑛
➢ The velocity distribution can be written in terms of 𝑣𝑧,𝑚𝑎𝑥 as:
𝑟2
= [1 − ( 2 )]
𝑣𝑧,𝑚𝑎𝑥
𝑅
𝑣𝑧
➢ The shear stress:
𝜏 = −𝜇
𝑑𝑣𝑧
𝑑𝑟
= −𝜇
1 𝜕𝑃
Remember 𝑣𝑧 = ( )(𝑟 2 − 𝑅2 )
4𝜇 𝜕𝑧
𝑑 1 𝜕𝑃
1 𝜕𝑃
[ ( )(𝑟 2 − 𝑅2 )] = −𝜇 2[ ( )(2𝑟)]
𝑑𝑟 4𝜇 𝜕𝑧
4𝜇 𝜕𝑧
1 𝜕𝑃
1 𝜕𝑃
𝜏 = − ( ) 𝑟 ➢ Wall shear stress: (𝑟 = 𝑅) → 𝜏𝑤 = − ( )𝑅
2 𝜕𝑧
2 𝜕𝑧
➢ Tube centerline shear stress: (𝑟 = 0.0) → 𝜏𝑤 = 0.0
14
Hagen-Poiseuille Flow
➢ The drag coefficient (𝐶𝐷 ):
1 𝜕𝑃
−2( 𝜕𝑧 )𝑅
𝜏𝑤
8𝜇
𝐶𝐷 = 1 2 = 1
=
1
𝜕𝑃
𝜌𝑉𝑅
𝜌𝑉
𝜌𝑉 [− 𝑅2 ( )]
2
2
8𝜇
Remember 𝑉𝑚𝑒𝑎𝑛 = 𝑉
𝜕𝑧
8𝜇
2
16𝜇
16𝜇
𝐶𝐷 =
x =
=
𝜌𝑉𝑅 2
𝜌𝑉(2𝑅)
𝜌𝑉(𝐷)
&
𝜌𝑉𝐷
𝑅𝑒 =
𝜇
16
𝐶𝐷 =
𝑅𝑒
15
Applications of Equations of Motion
❑ Some Simple Solutions for Viscous incompressible
Fluids
❑ Steady, Laminar Flow Between Fixed Parallel Plates
“Channel Flow”
❑ Flow Between Two Parallel Plates Bottom-Fixed, UpperMoving with Constant Velocity U “Coutte Flow”
❑ Steady, Laminar Flow in Circular Tubes “HagenPoiseuille Flow”
❑ Steady, Axial, Laminar Flow in an Annulus
16
Steady, Axial, Laminar Flow in an Annulus
❑ Assumptions:
➢ Laminar flow & Steady flow (𝜕Τ𝜕𝑡 = 0.0).
➢ Incompressible, Newtonian fluid (𝜌, 𝜇 = Const.)
➢ Parallel flow (𝑣𝑟 = 0.0 and 𝑣𝜃 = 0.0) السريان في إتجاه مواز لمحور اإلسطوانة
➢ No swirl flow → 𝑣𝜃 = 0.0
➢ Axisymmetric flow (𝜕Τ𝜕𝜃 = 0.0)
Fig. 3.4 The viscous flow between two coaxial cylinders: a) coordinate
system and notation used in analysis; b) flow through an annulus.
17
Steady, Axial, Laminar Flow in an Annulus
❑ An exact solution can be obtained for axial flow in the annular space
between two fixed, concentric cylinders
➢ Boundary conditions:
❑ The differential equations (1, 2, 𝑎𝑛𝑑 3) used in the preceding section for
flow in a circular tube also apply to the axial flow in the annular space
between two fixed, concentric cylinders (Fig. 3.4).
𝜕𝑃
0.0 = − 𝜌gsin𝜃 −
(1)
𝜕𝑟
𝜕𝑃
1 𝜕
𝜕𝑣
0.0 =− + 𝜇 [
(𝑟 𝑧 )]
𝜕𝑧
𝑟 𝜕𝑟
𝜕𝑟
1 𝜕𝑃
0.0 = −
− 𝜌g cos 𝜃
𝑟 𝜕𝜃
(2)
(3)
❑ Equation (𝜋) for the velocity distribution still applies, but for the
stationary annulus the boundary conditions become at (at 𝑟 = 𝑟𝑖 → 𝑣𝑧 = 0.0
& at 𝑟 = 𝑟𝑜 → 𝑣𝑧 = 0.0).
𝑟 2 𝜕𝑃
𝑣𝑧 = ( ) +𝐶1 ln 𝑟 + 𝐶2
4𝜇 𝜕𝑧
(𝜋)
(1) at 𝑟 = 𝑟𝑖 → 𝑣𝑧 = 0.0
(2) at 𝑟 = 𝑟𝑜 → 𝑣𝑧 = 0.0
18
Steady, Axial, Laminar Flow in an Annulus
➢ Starting from Eq. (𝜋) and substituting with the two B. C. s
(1) at 𝑟 = 𝑟𝑖 → 𝑣𝑧 = 0.0
𝑟 2 𝜕𝑃
𝑣𝑧 = ( ) +𝐶1 ln 𝑟 + 𝐶2 (𝜋)
4𝜇 𝜕𝑧
(2) at 𝑟 = 𝑟𝑜 → 𝑣𝑧 = 0.0
0.0 =
𝑟𝑜 2 𝜕𝑃
4𝜇
( ) +𝐶1 ln 𝑟𝑜 + 𝐶2 (4)
𝜕𝑧
− 𝑟𝑖 2 𝜕𝑃 −
−
0.0 = ( ) +𝐶1 ln 𝑟𝑖 +
𝐶2
4𝜇 𝜕𝑧
)4( ) من المعادلة5( ❖ بطرح المعادلة
(5)
1 𝜕𝑃
0.0 = ( ) (𝑟𝑜 2 − 𝑟𝑖 2 )+𝐶1 (ln 𝑟𝑜 − ln 𝑟𝑖 )
4𝜇 𝜕𝑧
1 𝜕𝑃 (𝑟𝑜 2 − 𝑟𝑖 2 )
𝐶1 = − ( )
4𝜇 𝜕𝑧 ln(𝑟𝑜 Τ𝑟𝑖 )
(ln 𝑟𝑜 − ln 𝑟𝑖 ) = ln(𝑟𝑜 Τ𝑟𝑖 )
)𝐶2 ( 𝐶) بهدف إيجاد1 ( ) عن قيمة4( ❖ بالتعويض في المعادلة
𝑟𝑜 2 𝜕𝑃
1 𝜕𝑃 (𝑟𝑜 2 − 𝑟𝑖 2 )
0.0 =
( )+[− 4𝜇 ( 𝜕𝑧 ) ln(𝑟 Τ𝑟 )] ln 𝑟𝑜 + 𝐶2
4𝜇 𝜕𝑧
𝑜 𝑖
(𝑟𝑜 2 − 𝑟𝑖2 )
1 𝜕𝑃
2
𝐶2 = − ( )[𝑟𝑜 −
ln 𝑟𝑜 ]
4𝜇 𝜕𝑧
ln(𝑟𝑜 Τ𝑟𝑖 )
19
Steady, Axial, Laminar Flow in an Annulus
➢ Substituting in Eq. (𝜋) with the values of 𝐶1 & 𝐶2
𝑟 2 𝜕𝑃
(𝑟𝑜 2− 𝑟𝑖2)
1 𝜕𝑃 (𝑟𝑜 2 − 𝑟𝑖 2 )
1 𝜕𝑃
2
𝑣𝑧 = ( ) − 4𝜇 ( 𝜕𝑧 ) ln(𝑟 Τ𝑟 ) ln 𝑟 − 4𝜇 ( 𝜕𝑧 )[𝑟𝑜 − ln(𝑟 Τ𝑟 ) ln 𝑟𝑜 ]
4𝜇 𝜕𝑧
𝑜 𝑖
𝑜 𝑖
➢ The velocity distribution becomes:
1 𝜕𝑃
(𝑟𝑜 2 − 𝑟𝑖 2) 𝑟
2
2
𝑣𝑧 = ( )[𝑟 − 𝑟𝑜 − ln(𝑟 Τ𝑟 ) ln 𝑟 ]
4𝜇 𝜕𝑧
𝑜
𝑜 𝑖
❑ The corresponding Flow Rate (𝑄) :
𝑑𝑄 = 𝑣𝑧 𝑑𝐴 = 𝑣𝑧 2𝜋𝑟𝑑𝑟
𝑟𝑜
𝑄 = 𝑧𝑣 𝑟2𝜋𝑟𝑑𝑟
𝑖
𝑟𝑜
2𝜋𝑟
𝑑𝑟
𝑑𝐴 = 2𝜋𝑟𝑑𝑟
2
2
𝜋 𝜕𝑃
(𝑟
−
𝑟
)
𝑟
𝑜
𝑖
3
2
𝑄 = ( ) න [𝑟 − 𝑟𝑟𝑜 −
𝑟𝑙𝑛 ] 𝑑𝑟
Τ
2𝜇 𝜕𝑧
𝑙𝑛(𝑟𝑜 𝑟𝑖 )
𝑟𝑜
𝑟𝑖
20
Steady, Axial, Laminar Flow in an Annulus
𝑟𝑜
2 − 𝑟 2)
𝜋 𝜕𝑃
(𝑟
𝑟
𝑜
𝑖
3
2
𝑄 = ( ) න [𝑟 − 𝑟𝑟𝑜 −
𝑟𝑙𝑛 ] 𝑑𝑟
2𝜇 𝜕𝑧
𝑙𝑛(𝑟𝑜 Τ𝑟𝑖 )
𝑟𝑜
𝑟𝑖
𝑟𝑜
න 𝑟𝑙𝑛(𝑟Τ𝑟𝑜 ) 𝑑𝑟 ➢ Integration by part
𝑟𝑖
❖ إستخدام التكامل بالتجزيئ
𝑣𝑢 = 𝑣𝑑𝑢 − 𝑢𝑑𝑣
𝑢 = 𝑙𝑛(𝑟Τ𝑟𝑜 )
𝑑𝑢 = [1Τ(𝑟Τ𝑟𝑜 )](1Τ𝑟𝑜 ) 𝑑𝑟 = (𝑑𝑟Τ𝑟)
𝑣 = 𝑟( = 𝑟𝑑𝑟 = 𝑣𝑑 2 Τ2)
𝑑𝑣 = 𝑟𝑑𝑟
𝑟𝑜
𝑟𝑜
𝑟𝑜
2
Τ
Τ
Τ
𝑟( = 𝑟𝑑𝑟 ) 𝑜𝑟 𝑟(𝑛𝑙 𝑟 = 𝑣𝑑𝑢 2) 𝑙𝑛(𝑟 𝑟𝑜 )]𝑟𝑖 − 𝑟 ( 𝑟2 Τ2 )( 𝑑𝑟Τ𝑟 )
𝑖
𝑖
𝑟
𝑟
𝑟( = 𝑣𝑑𝑢 2 Τ2) 𝑙𝑛(𝑟Τ𝑟𝑜 )]𝑟𝑜𝑖 − ( 𝑟 2 Τ4 )]𝑟𝑜𝑖
𝑟𝑖2
𝑟𝑖2
𝑟𝑜2
𝑟𝑜
𝑟𝑜2
𝑟𝑖
([ = 𝑣𝑑𝑢 2 ) 𝑙𝑛(𝑟 ) − ( 4 )] − [( 2 ) 𝑙𝑛(𝑟 ) − ( 4 )]
𝑜
𝑜
0.0 1
21
Steady, Axial, Laminar Flow in an Annulus
𝑟𝑜
𝑟𝑖2
𝑟𝑖2
𝑟𝑜2
𝑟𝑜
𝑟(𝑛𝑙 𝑟 = 𝑣𝑑𝑢 Τ𝑟𝑜 ) 𝑟𝑑𝑟 = −( 4 ) +( 2 ) 𝑙𝑛( 𝑟 ) + ( 4 )
𝑖
𝑖
𝑟𝑜
𝑟𝑖2
𝑟𝑖2
𝑟𝑜2
𝑟𝑜
Τ
𝑢𝑑𝑣
=
𝑙𝑛(
𝑟
𝑟
)
𝑟𝑑𝑟
=
−(
)
+(
)
+(
)
𝑙𝑛(
)
𝑟
𝑜
4
4
2
𝑟
𝑖
𝑖
𝑟𝑜
2
2
𝜋 𝜕𝑃
(𝑟
−
𝑟
𝑟
𝑜
𝑖 )
3
2
𝑄 = ( ) න [𝑟 − 𝑟𝑟𝑜 −
𝑟𝑙𝑛 ] 𝑑𝑟
2𝜇 𝜕𝑧
𝑙𝑛(𝑟𝑜 Τ𝑟𝑖 )
𝑟𝑜
𝑟𝑖
(𝑟𝑜2 −𝑟𝑖2)
𝑟𝑖2
𝑟𝑖2
𝜋 𝜕𝑃 𝑟 4
𝑟 2 𝑟𝑜2 𝑟𝑜
𝑟𝑜2
𝑟𝑜
𝑄 = ( )[ −
]𝑟𝑖 −
[−(
)
+(
)
+(
)
𝑙𝑛(
)]
Τ
2𝜇 𝜕𝑧
4
2
4
4
2
𝑟
𝑙𝑛(𝑟𝑜 𝑟𝑖 )
𝑖
𝑟𝑖4
𝑟𝑖2 𝑟𝑜2
𝑟𝑖2 𝑟𝑜2
𝑟𝑖4
(𝑟𝑜2 −𝑟𝑖2 ) 2
𝜋 𝜕𝑃
𝑟𝑜4
𝑟𝑜4
2 )]
𝑄 = ( )[( − − +
)−(
− )+
(𝑟
−
𝑟
𝑜
𝑖
2𝜇 𝜕𝑧
4
2
4
2
2
2
4𝑙𝑛(𝑟𝑜 Τ𝑟𝑖 )
𝑟𝑖4
(𝑟𝑜2 −𝑟𝑖2 )2
𝜋 𝜕𝑃
𝑟𝑜4
𝑄 = ( )[− + +
]
2𝜇 𝜕𝑧
4
4
4𝑙𝑛(𝑟𝑜 Τ𝑟𝑖 )
22
Steady, Axial, Laminar Flow in an Annulus
➢ The corresponding volume rate of flow is:
(𝑟𝑜2−𝑟𝑖 2)2
𝜋 𝜕𝑃
4
4
𝑄 = − ( )[𝑟𝑜 − 𝑟𝑖 −
]
8𝜇 𝜕𝑧
𝑙𝑛(𝑟𝑜 Τ𝑟𝑖 )
➢ or in terms of the pressure drop, ∆𝑃, in length, 𝑙, of the annulus,
𝜕𝑃
∆𝑃
where;
=−
𝜕𝑧
𝑙
(𝑟𝑜2 −𝑟𝑖 2 )2
𝜋∆𝑃
4
4
𝑄=
[𝑟 − 𝑟𝑖 −
]
8𝜇𝑙 𝑜
𝑙𝑛(𝑟𝑜 Τ𝑟𝑖 )
➢ The velocity at any radial location within the annular space can be
obtained from the following Eq.:
1 𝜕𝑃
(𝑟𝑖2 − 𝑟𝑜 2) 𝑟
2
2
𝑣𝑧 = ( )[𝑟 − 𝑟𝑜 + ln(𝑟 Τ𝑟 ) ln 𝑟 ]
4𝜇 𝜕𝑧
𝑜
𝑜 𝑖
23
Steady, Axial, Laminar Flow in an Annulus
❑ The mean velocity:
𝑄
1
𝑉= = ( ) 𝑄
A
A
and
𝑉𝑚𝑒𝑎𝑛 ≡ 𝑉
A = 𝜋(𝑟𝑜2 − 𝑟𝑖 2 )
(𝑟𝑜2 −𝑟𝑖2 )2
−𝜋 𝜕𝑃
2
2
2
2
𝑉= ( 2 2 ) ( )[(𝑟𝑜 − 𝑟𝑖 )(𝑟𝑜 + 𝑟𝑖 ) −
]
𝜋(𝑟𝑜 −𝑟𝑖 ) 8𝜇 𝜕𝑧
𝑙𝑛(𝑟𝑜 Τ𝑟𝑖 )
1
(𝑟𝑜2 −𝑟𝑖 2 )
−1 𝜕𝑃
2
2
𝑉= ( )[𝑟𝑜 + 𝑟𝑖 −
]
8𝜇 𝜕𝑧
𝑙𝑛(𝑟𝑜 Τ𝑟𝑖 )
𝑑𝑣𝑧
obtained at
= 0.0
𝑑𝑟
❑ The maximum velocity:
1 𝜕𝑃
(𝑟𝑜 2 − 𝑟𝑖 2) 𝑟
2
2
𝑣𝑧 = ( )[𝑟 − 𝑟𝑜 − ln(𝑟 Τ𝑟 ) ln 𝑟 ]
4𝜇 𝜕𝑧
𝑜
𝑜 𝑖
(𝑟𝑜 2 − 𝑟𝑖 2 ) 1
𝑑𝑣𝑧
1 𝜕𝑃
1
= 0.0 = ( )[2𝑟 −
( )]
𝑑𝑟
4𝜇 𝜕𝑧
ln(𝑟𝑜 Τ𝑟𝑖 ) (𝑟 Τ𝑟𝑜 ) 𝑟𝑜
(𝑟𝑜 2 − 𝑟𝑖 2 ) 1
2𝑟 =
( )
Τ
ln(𝑟𝑜 𝑟𝑖 ) 𝑟
2 − 𝑟 2)
(𝑟
𝑜
𝑖
𝑟2 =
2 ln(𝑟𝑜 Τ𝑟𝑖 )
24
Steady, Axial, Laminar Flow in an Annulus
➢ The position of maximum velocity (𝑟𝑚𝑎𝑥 ):
➢ The maximum velocity occurs at the radius, (𝑟 = 𝑟𝑚𝑎𝑥 ), where,
(𝑑𝑣𝑧 Τ𝑑𝑟) = 0.0, Thus,
2 − 𝑟 2)
(𝑟
𝑜
𝑖
2
𝑟𝑚𝑎𝑥 =
2 ln(𝑟𝑜 Τ𝑟𝑖 )
(𝑟𝑜 2 − 𝑟𝑖 2 )
𝑟𝑚𝑎𝑥 =
2 𝑙𝑛(𝑟𝑜 Τ𝑟𝑖 )
❑ The hydraulic diameter (𝐷ℎ ):
➢ A criterion based on the conventional Reynolds number (which is
defined in terms of the tube diameter) cannot be directly applied to
the annulus, since there are really “two” diameters involved. For
tube cross sections other than simple circular tubes it is common
practice to use an “effective” diameter, termed the hydraulic
diameter, 𝐷ℎ , which is defined as:
4 𝑥 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
𝐷ℎ =
𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 25
Steady, Axial, Laminar Flow in an Annulus
4 𝑥 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
𝐷ℎ =
𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟
➢ The wetted perimeter is the perimeter in contact with the fluid. For
an annulus equal to → 2𝜋(𝑟𝑜 + 𝑟𝑖 )
𝐷ℎ =
4𝜋(𝑟𝑜 2 − 𝑟𝑖 2 )
2𝜋(𝑟𝑜 +𝑟𝑖 )
=
2[(𝑟𝑜 −𝑟𝑖 )(𝑟𝑜 +𝑟𝑖 )]
(𝑟𝑜 +𝑟𝑖 )
𝐷ℎ = 2 (𝑟𝑜 − 𝑟𝑖 )
❑ The Reynolds number (𝑅𝑒):
➢ In terms of the hydraulic diameter, the Reynolds number is 𝑅𝑒 =
(𝜌V𝐷ℎ Τ𝜇) where V = (𝑄 ∕cross-sectional area), and it is
commonly assumed that if this Reynolds number remains below
2100 the flow will be laminar.
𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒
𝑅𝑒 =
𝑉𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒
𝜌𝑉𝑚𝑒𝑎𝑛 𝐷ℎ
𝑅𝑒 =
𝜇
26
CLOSING NOTES AND ANNOUNCEMENTS
➢ Procedure for solving the incompressible continuity and Navier–
Stokes equations.
✓ Step 1: Set up the problem and geometry (sketches are helpful),
identifying all relevant dimensions and parameters.
✓ Step 2: List all appropriate assumptions, approximations,
simplifications, and boundary conditions.
✓ Step 3: Simplify the differential equations of motion (continuity
and Navier–Stokes) as much as possible.
✓ Step 4: Integrate the equations, leading to one or more constants
of integration.
✓ Step 5: Apply boundary conditions to solve for the constants of
integration.
✓ Step 6: Verify your results.
27
Summary
✓ Some Simple Solutions for Viscous incompressible Fluids
“Introduction”
✓ Steady, Laminar Flow Between Two Fixed Parallel Plates
“Channel Flow”
✓ Flow Between Two Parallel Plates Bottom-Fixed, Upper-Moving
with Constant Velocity 𝑢 “Coutte Flow”
✓ Steady, Laminar Flow in Circular Tubes “Hagen-Poiseuille Flow”
✓ Steady, Axial, Laminar Flow in an Annulus “Annular flow”
28
References
❑ Munson, R. et. al. “Fundamentals of Fluid Mechanics”, US: John
Wiley & Sons, Inc., 2002 (Main)
✓ Chapter (6): Introduction (page 352)
✓ Chapter (6): Channel Flow (pages 352~354)
✓ Chapter (6): Couette Flow (pages 355~357)
✓ Chapter (6): Hagen-Poiseuille Flow (pages 357~359)
✓ Chapter (6): Flow in an Annulus (pages 360~362)
❑ Cengel, A. and Cimbala, M. “Fluid Mechanics: Fundamentals
and Applications”, US: McGraw-Hill Series, 2006
✓ Chapter (9): Differential Analysis of Fluid Flow (pages 437~449)
29
THANK YOU
30
0
