E Velocity Disn in laminar shell momentum bal PDF

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Chemical engineering topics intended for undergrad level...

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MOMENTUM TRANSFER

E. VELOCITY DISTRIBUTION IN THE LAMINAR FLOW It could be said that the current state of our understanding of and technology for fluid transport stems from our triumph in the theoretical modeling of the laminar fluid flow. A most basic and pivotal accomplishment of engineers and scientists is the derivation of the equation for the velocity distribution of laminar flow in various flow geometries. Once the velocity profile is known, other flow parameters such as average velocity, volumetric or mass flowrate, etc. become quite easy to calculate. We are even able to extend such basic understanding of the laminar flow to the more challenging turbulent flow.

Fig 1. https://waterfallnow.com/laminar-flow-fountain-videos/

Learning Outcomes • Explain why we need to know the complete velocity distribution profile in a fluid continuum • Explain what is meant by steady-state in fluid flow system • Carry out the application of shell momentum balance over a thin “shell” of a falling fluid film and in fluid flow within a circular tube, and in other analogous systems (annulus flow, creeping flow around a solid sphere) • Identify the most used boundary conditions involved in shell momentum balance calculation • Derive, from the velocity distribution profile (or momentum flux), the commonly required quantities in fluid system applications (vmax, vave, volume rate of flow, film thickness, force of the fluid on the surface) • Derive the Hagen-Poiseuille and Stoke’s law equations • Calculate the Reynold’s number NRe • Identify the ranges of NRe that corresponds to laminar, transition and turbulent regions • Solve some application problems of the Hagen-Poiseuille and Stoke’s equation

Momentum transfer in fluids basically happens in two ways: (1) by MOLECULAR TRANSPORT (  yx ) which we already discussed and should review:

Eqn. (1)

F V = A Y

For above, F is the constant force required to maintain the motion of the lower plate, A is the area of the lower plate in contact with the fluid, V is the constant velocity and Y is the distance between plates. The constant of proportionality μ is called the viscosity of the fluid. F/A is clear a shearing stress quantity and is now replaced by the notation 

yx

which is the

force in the x direction on a unit area perpendicular to the y direction. Furthermore, we replace V/Y by -dvx/dy (remember that we can vary these values “differentially”). Then, in terms of these symbols, Eqn (1) becomes:

 yx = − 

dvx dy

Eqn. (2)

Hence, the Newton’s law of viscosity could be stated as: “The shearing force per unit area of fluid flow is proportional to the negative of the velocity gradient.”

Momentum interpretation for  yx : The shear stress  yx in Eqn (2) can also be interpreted as a flux of x-momentum in the y direction, which is the rate of flow of momentum per unit area. The shear stress can be written as:

 yx =

kg  m / s mass  velocity momentum = = m2  s area  time m2 s

Eqn. (3)

This was shown by considering the interaction between two adjacent layers of a fluid which have different velocities, and hence difference momentum in the x direction. The random motion of the molecules in the faster moving layer send some of the molecules into the slower moving layer, where they collide with the slower moving molecules and tend to speed them up or increase their momentum in the x direction. Also, in the same fashion, molecules in the slower layer tend to retard those in the faster layer. This exchange of molecules between layers produces a transfer or flux of x-directed momentum from high-velocity to low-velocity layers.

x

Figure 3. Molecular illustration the momentum transfer in laminar flow that is due to molecular diffusion. The negative sign indicates that the momentum is transferred within the gradient from high to low velocity regions.

(2) the CONVECTIVE TRANSPORT which happen due to the bulk flow of the fluid.

How do we describe CONVECTIVE MOMENTUM TRANSPORT? Imagine a fluid flowing across a unit area:

Figure 2. Diagram to illustrate convective transport.

In the diagram above, vx is the velocity vector at x-direction. By virtue of its velocity, this fluid carries with it momentum expressed as ρvx per unit volume. We can verify the units as:

v x =

Eqn. (1)

kg m mass mv momentum =  velocity = 3 = 3  m3 m s volume m

To convert this momentum density quantity to momentum flux (remember flux is amount per time per area), we need to multiply the quantity again with vx:

v x v x = velocity 

m kg m mv momentum mass  velocity =  3  = 2 = = v x2 s m s m  s area  time volume

Eqn. (2)

SHELL MOMENTUM BALANCE AND BOUNDARY CONDITIONS One important approach in deriving important fluid flow equations is by setting up momentum balances over a thin “shell” or “strip” of the fluid continuum. We will first begin by analyzing steady-state flow. In steady state flow, any flow property at a particular point in the fluid continuum, for example velocity, is not changing with respect to time. A general form of the momentum balance that we will use is the following: Rate of momentum in by convective /molecular transport

–

rate of momentum out by convective /molecular transport

+

sum of forces acting on the system

Eqn. (3) = 0

General procedure of setting up and solving viscous flow problems by shell momentum balance: 1. Identify the non-vanishing velocity component and the spatial variable on which it depends 2. Write the momentum balance over a thin shell perpendicular to the relevant spatial variable

3. Let thickness of shell approach zero (apply limit theorem) and use the definition of first derivative to obtain the differential equation for momentum flux. 4. Integrate to get the momentum flux distribution applying the appropriate boundary conditions (see next section for discussion about it). 5. Insert Newton’s law of viscosity to obtain the differential equation for velocity 6. Integrate this equation to get the velocity distribution, applying the appropriate boundary condition 7. Use the derived velocity distribution to get other flow quantities e.g., v max, vave, force exerted by fluid on the solid surfaces, etc.

Boundary conditions: these are statements about the velocity or stress at the boundaries of the system (a) At solid-fluid interfaces: fluid velocity is equal to the velocity of the solid surface. Also known as the “no-slip condition”, “skin velocity” (b) At liquid-liquid interfacial plane: there is continuous behavior of momentum, velocity components (c) At liquid-gas interface: momentum is taken to be zero since gas viscosity is much less than liquid viscosity

1. Flow of Falling Film This is quite a commonly-encountered physical system in chemical engineering design. Such films are studied in connection with wetted wall towers, evaporation and gas absorption, condensers, etc.

Figure 3. Falling film setup (from BSL)

velocity distribution is yet arbitrary, but we know it must be decreasing to zero at the ramp surface

Figure 4. The section L magnified; shaded region is the shell or strip (From BSL)

Assumptions: (a) Viscosity and density of fluid is constant (b) Laminar flow applies, rectilinear motion (streamlines of flow are straight, parallel to ramp) (c) Entrance and exit disturbances are absent Step 1: Postulates: relevant spatial variable = x, v z = v z (x) ; vz varies with x Maximum vz is at x = 0, vz = 0 at ramp surface (or when x = δ ); v x = v y = 0 Step 2: Momentum balance over the shell:

τ in

ρv 2 out

τ in at ∆X L face

ρv 2 in at ∆x ∆y or ∆x W face

τ out

Figure 5. The shell or strip magnified, momentum quantities indicated (from BSL)

Convective momentum in at ∆X W face =

Convective momentum out at ∆X W face = molecular momentum in at W L face =

(note that unit is

z=0

xWv z vz

WL xz

molecular momentum out at W L face = weight force acting on the shell =

xWvz vz

m v p = ) s s

z =L

x

WL xz x +x

Cos β

xWLg cos 

β

by Eqn 3: Rate of momentum in by convective /molecular transport

 xW vz

2 z =0

rate of momentum out by convective /molecular transport

–

−  xW vz

2 z =L

+

sum of forces acting on the system

+ WL xz x − WL xz

x+  x

= 0

Eqn. (4)

+  xWLg cos  = 0

Since velocity along z is constant, the first two terms cancel out. It is very easy for you to provide the details of the mathematical work to show that the equation is reduced to:

( xz

x + x

−  xz x )

x

= g cos 

Step 3: Using the definition of derivative:

 ( xz

Lim 

 x⎯ ⎯→ 0

x + x



−  xz x )

x

 = g cos  

d xz = g cos  dx Step 4: Solving the differential equation:

 d

xz

=  (g cos )dx

 xz = g cosx + c1 0 = g cos (0) + c 1   xz = g cos  x

BC 1 : xz = 0 at x = 0  c1 = 0

This is the equation for momentum flux. It is found to be liner with respect to x.

Step 5: since the flow is laminar, we can use the Newton’s law of viscosity:

 xz = − 

dvz dx

Eqn. (5)

dvz = g cos x dx   g cos   dvz = −   x dx   

−

Step 6: Integrate

 g cos   = −   xdx     g cos   x 2  + c 2 vz = −   2 

 dv

z

  g cos    0 = −  + c2   2 

BC 2 : v z = 0 at

 g cos    c2 =     2 

2

 g cos  2 v z = −   x  2 

x = 2



 g cos  2 +    2 

Again, you should be able to supply the rest of the mathematical work to show that finally, the equation is:

g cos  2 vz = 2

  x 2 1 −        

Eqn. (6)

The derived equations for  xz (linear w.r.t. x) and vz (parabolic w.r.t. x) are plotted below:

6

Step 7: Calculation of other important fluid flow quantities: (a) Maximum velocity vmax. In the falling film, this is obviously at x = 0

v z,max

g cos  2 = 2

  0  2  g cos  2 1 −    = 2     

Eqn. (7)

(b) Average velocity or vz,ave over a cross-section of flow (in our falling film, the crosssection is assumed a uniform rectangle:

vz, ave =

vz ,ave =

1 v z dA A  A

 v z dA A

 dxdy

A =  dxdy

where w

w

=

  vz dxdy 0 0 w



=

0 0

g cos  2   x  1 −   2    

2

  dxdy 

w

  dxdy

  dxdy

0 0

0 0

The denominator is clearly the cross-sectional area of the film Wδ and so:

1  g cos  2 vz ,ave = 2 W

  x 2 0 0 1 −    dxdy   w

Integrating the inner integral (dx term) and apply limits, we get:

1 g cos  2 vz ,ave = 2 W

2 1 g cos  2  2W  2  g cos  2  dy =  =   0 3 W 2 2  3  3  Eqn (8) w

2 vz ,ave = v z ,max 3 (c) Mass rate of flow = average velocity x area of flow x fluid density

m =

2   g cos 2   2 g cos 3W W   ( ) =   3 2 3 

Eqn. (9)

Note that average velocity x area of flow is equal to volume rate of flow Q (other books would denote Q as VRF).

 3 v z ,ave (d) Thickness of film,  =   g cos 

  

1 3

  3m =  2     gW cos  

1 3

Eqn. (10)

2. FLOW THROUGH A CIRCULAR TUBE The flow of fluids in circular tubes is encountered frequently in engineering such as in the design of piping systems. Again, we will be limiting our analysis to the laminar flow and use cylindrical coordinate system which is but natural to describe positions in a circular pipe. We consider a steady laminar flow of a fluid of constant density in a very long tube of length L and radius R. “very long” here is to specify that no entrance and exit effects on the flow is considered; we assume that flow direction is parallel everywhere (straight streamlines) to the tube surface.

Convective momentum in at z=0 3

∆r Pressure p0

Molecular momentum out at r + ∆r

1 Molecular momentum in at r

Convective momentum out at z = L 4

2

Pressure pL

Figure 7. Cylindrical shell of fluid over which momentum balance is made to get the velocity profile and the Hagen-Poiseuille formula.

Listing the components of the momentum balance: 1

Rate of momentum in across cylindrical surface at r

(2rL rz ) | r

2

Rate of momentum out across cylindrical surface at r + ∆r

(2rL  rz ) | r + r

3

Rate of momentum in across annular surface at z = 0

(2rrv z )( v z ) | z= 0

4

Rate of momentum in across annular surface at z = L

( 2rrv z )( v z ) | z =L

5.a Gravity force acting on cylindrical shell 5.b Pressure force acting on annular surface at z = 0

( 2r rL )g

(2rr ) p0

5.b Pressure force acting on annular surface at z = L

(2rr ) pL

Adding the components of the momentum balance:

(2rL rz ) |r −(2rL rz ) | r+  r +(2r rv z )( v z ) | z=0 −(2r rv z )( v z ) | z= L + (2 r rL) g + 2  r( p 0 − pL ) = 0

Eqn. (11)

Because the fluid is assumed to be incompressible, v z is the same at z = 0 and at z = L, hence the 3rd and 4th terms cancel out. Dividing Eqn. (11) by 2ΠL∆r and take the limit as ∆r goes to zero:

 (r  rz ) r+  r − ( r rz ) r   p0 − pL  + g  r =  Lim r L     r⎯ ⎯→ 0   The left side is the first derivative while the right side can be simplified by using the complete definition of fluid pressure:

p = P+  gz and since

p0 = P0 +  gz0

z 0 = 0 and

pL = PL +  gzL

zL = L

 P + g(0) − ( PL + gL)  d +  g r ( rrz ) =  0 dr L   d  P − PL ( rrz ) =  0 dr  L

 r 

By integration (you must show the details!):

P0 − PL  c r + 1 r  2L 

  rz = 

There is no applicable boundary condition that we can use here but by practical analysis, the momentum flux must not be infinite at the tube center r = 0. This will force us to decide that c1 must be zero to eliminate such impracticality in our equation. Hence, the final equation for momentum flux is:

 P0 − PL  r  2L 

 rz = 

Eqn. (12)

To arrive at the velocity distribution, we will use the Newton’s law of viscosity (in cylindrical notation):

 rz = −

dvz ; dr

dv z  P − PL = − 0 dr  2 L

and hence:

 r 

Which upon integration yields:

 P −P v z = −  0 L  4 L

 2  r + c2 

where

 P −P c2 =  0 L  4 L

 2 R 

By using the boundary condition at fluid-solid interface (vz = 0 at r = R). Finally:

(P −P )R 2   r  vz = 0 L 1−   4 L   R 

2

  

Eqn. (13)

Figure 8. Schematic diagram of the velocity and momentum flux distributions in fluid flow in circular tube.

Calculation of other flow parameters: (a) Maximum velocity vmax occurs at r = 0

v z. max =

( P0 − PL ) R 2 4 L

Eqn. (14)

(b) Average velocity or vz,ave over a cross-section of flow:

v z ,ave =

1 v z dA A  A

where

A =  dA = rdrd

Review: differential area (dA) of a circular sector subtended by the angle dθ

dr RECALL:

rdθ

dθ

The shaded region is the differential area dA which, if dΘ and dr are small enough, the shaded region could be taken as a rectangle of area dr x rdΘ or dA = rdrdΘ

r θ Figure 9. Illustration for the derivation of dA in the cylindrical coordinate system.

2 R

v dA   v rdrd = =   rdrd    rdrd z

z

v z ,ave

A

0 0 2 R

=

( P0 − PL ) R 2 (P0 − PL )D 2 = 8L 32 L

Eqn. (15)

0 0

where D = 2R The mathematical details in deriving Eqn (15) is not shown but you should be able to do this. (c) Volume rate of flow Q = average velocity x area of flow

( P0 − PL ) R2  ( P0 − PL ) R 4 2 Q=  R = 8 L 8 L

Eqn.(16)

This equation for volume flow rate in circular tube (Eqn. 16), or alternatively Eqn. 15 (according to Geankoplis) is called the HAGEN-POISEUILLE LAW.

ASSIGNMENT: 1. Work out the details of the derivation on Eqn. (15) 2. Present the derivation of the velocity distribution for laminar fluid flow in circular tube that is oriented horizontally. Set the horizontal direction as the z-axis. 3. Based on the quantities present in the Hagen-Poiseuille law/equation, define it according to your own understanding (operational definition). 4. Derive the relationship between maximum velocity and average velocity for laminar fluid flow in circular tube. 5. Answer Problems a. 2.6-3, b. 2.6-4 and c. 2.10-1 of Geankoplis book, provide detailed solutions.

3. ANNULAR FLOW Annular comes from the word annulus: the area/region that is in-between a smaller pipe inserted inside a bigger pipe. This type of flow is also important in chemical engineering as we usually see it in double pipe heat exchangers, annular reactors, etc.

Figure 10. The annular region in a double pipe set-up. You can read on how the shell momentum balance is applied to determine the velocity distribution equation for annular flow (see BSL book). We will revisit this but through another technique based on the equations of change in fluid flow.

4. CREEPING FLOW AROUND A SOLID SPHERE Creeping flow refers to very slow flow (NRE < about 0.1). There is no eddying downstream from the obstruction of flow (sphere is generally used as a model shape).

Sphere of radius R

v∞ Figure 11. Streamlines in creeping flow around a sphere that is settling in a stationary fluid. Fluid approaches from below with velocity v  and viscosity μ

Analysis for this system is quite complex because the flow streamlines are not rectilinear (see Fig. 11), hence we can only provide the total force acting on the sphere directly. The total force F by the fluid acting on the sphere is given as:

Total force = buoyant force + form drag + friction drag

F
...