Chapter 8: Internal Incompressible Viscous Flow

Flows: Laminar (some have analytic solutions) Turbulent (no analytic solutions)

Depends of Reynolds number, Re = Inertial Force/Viscous Force

Re = Lu/ cartoon approach

Reynolds Number ~ ratio of inertial to viscous for(Cartoon approach)ces -- hand waving argument --

fluidelement

Inertial Force = (m) x (a) ( l3) x (U/t)U/t U/(L/U) U2/L

Inertial Force ( l3 ) x (U2/L)Inertial Force ( L3 ) x (U2/L) = L2U2

REYNOLDS NUMBER = I.F./V.F. (cartoon approach)

Reynolds Number ~ ratio of inertial to viscous for(Cartoon approach)ces -- hand waving argument --

fluidelement

Viscous Force = () x (Area) (dU/dy) x l2 dU/dy U/l

ViscousForce (U/l) x l2 ul Ul

Reynolds Number ~ ratio of inertial to viscous for(Cartoon approach)ces -- hand waving argument --Inertial Force L2U2

ViscousForce UL

Define Re as LU/ where L & U are some

characteristic length scales

Re = Lu/ from N. S. E.

REYNOLDS NUMBER a la N.S.E.

a = Du/Dt = F

Eq. 5.27a:

(u/t + uu/x + vu/y + wu/z) = - p/x + (2u/x2 + 2u/y2 + 2u/z2)

x-component incompressible, constant , Newtonian,

ignore gravity, e-m, … forces,

(u/t + uu/x + vu/y + wu/z) = - p/x + (2u/x2 + 2u/y2 + 2u/z2)

Let u’ = u/U, v’ = v/U, w’ = w/U; x’ = x/L, y’ = y/L, z’ = z/L; t’ = t/T and p’ = p/ (U2)

L and U are characteristic lengths and velocities and T=L/U

(Uu’)/(Tt’) + Uu’(Uu’)/(Lx’) + Uv’(Uu’)/(Ly’) + Uw’(Uu’)/(Lz’)

= - (1/) p’U2 /Lx’ + (2(Uu’)/(Lx’)2 + 2(Uu’)/(Ly’)2 + 2(Uu’/(Lz’)2

(Uu’)/(Tt’) + Uu’(Uu’)/(Lx’) + Uv’(Uu’)/(Ly’) + Uw’(Uu’)/(Lz’)

= - (1/) p’U2 /Lx’ + (2(Uu’)/(Lx’)2 + 2(Uu’)/(Ly’)2 + 2(Uu’/(Lz’)2

U/T = U/(L/U) = U2/L

{U2/L}[u’/t’+u’u’/x’+v’u’/y’+w’u’/z’] = - {U2/L}p’/x’ +

{U/L2}(2u’/x’2 + 2u’/y’2 + 2u’/z’2)

{U2/L}[u’/t’+u’u’/x’+v’u’/y’+w’u’/z’]

= - {U2/L}p’/x’ +{U/L2}(2u’/x’2 + 2u’/y’2 + 2u’/z’2)

[u’/t’ + u’u’/x’ + v’u’/y’ + w’u’/z’] = - p’/x’ +

{1/[UL]}(2u’/x’2 + 2u’/y’2 + 2u’/z’2)

[u’/t’ + u’u’/x’ + v’u’/y’ + w’u’/z’] = - p’/x’ +

{1/ReL}(2u’/x’2 + 2u’/y’2 + 2u’/z’2)

High Re # in some ways independent of viscosity.However, near wall viscosity always important!

(why?)

[u’/t’ + u’u’/x’ + v’u’/y’ + w’u’/z’] = - p’/x’ +

{1/ReL}(2u’/x’2 + 2u’/y’2 + 2u’/z’2)

High Re # in some ways independent of viscosity.However, near wall viscosity always important!

(because velocity gradients large)

[u’/t’ + u’u’/x’ + v’u’/y’ + w’u’/z’] = - p’/x’ +

{1/ReL}(2u’/x’2 + 2u’/y’2 + 2u’/z’2)

Two flows with the same geometry, same ReL and satisfying the above equation (i.e. no body forces,

incompressible, constant visosity, Newtonian) will have similar flow fields (dynamic similarity). Hence drag forces

measured in the lab can be extrapolated to full scale!

Drag coefficient is same for dynamically similar flows

Drag coefficient = Drag Force / (U2L2)

Lift coefficient = Lift Force / (U2L2)

The principle of dynamic similarity makes it possible to predict the performance of full-scale aircraft from wind tunnel tests.

Reynolds Experiment (1883)

Reynolds conducted many experiments using glass tubes of 7,9, 15 and 27 mm diameter and water temperatures from 4o to 44oC.

He discovered that transition from laminar to turbulent flow occurred for a critical value of uD/ (or uD/), regardless of individual values of or u or D or .

~ Nakayama & Boucher

REYNOLDS NUMBER - EMPIRACLE

Sommerfeld in a 1908 paper first referred to uD/ as the

Reynolds number

Reynolds found that the quality of the pipe inlet affected transition – with asmoother, bell-mouthed inlet transition was delayed to higher Reynolds numbers.

Laminar pipe flow is stable to infinitesimal disturbances.

REYNOLDS NUMBER - EMPIRACLE

From Reynolds’ 1883 paper

Reynolds found transition to occur around Re = 13,000,when experiment repeated a hundred years later (left) transition was found to be much less –

Note: In pipe flow turbulence does not suddenly at Retr appear throughout the pipe. It forms

turbulent slugs near the pipe entrance and grows as it is passed through the pipe.

Pipe centerline: (a) fluctuating velocity; (b) mean velocity

Re = 2550 ----- from Triton

u fluctuation

u mean

QUESTION: refer to data above, head not changing,roughness not changing, viscosity not changing,

pipe diameter not changing – so why is flow rate?

Flow rate is reduced with appearance of turbulent “slug”. Flow slows down cause increased wall friction due to turbulence. If near Recr then new Re can be < Recr so no turbulent slugs nearentrance. After turbulent slug passes, flow speeds up and Recr reoccurs and pattern repeats.

breath

Fully Developed Flow

“viscous forces are dominant” - MYO

Internal Flows can be: developing flows - velocity profile changing fully developed - velocity profile not changing

D = 27 mmVavg = 6 cm/secReD = 1600

V=0 AT WALL

Laminar Pipe FlowEntrance Length for

L/D = 0.06 Re

{L/D = 0.03 Re, Smits} {L/D = 0.06 Re, White}

{L/D = 0.13 Re, Boussinesq 1891}

As “inviscid” core accelerates, pressure must drop

Pressure gradientbalances wall

shear stressNo acceleration

Pressure gradientbalances both

wall shear stressand acceleration

Le = 140D, Re = 2300

shear stressNo acceleration

{same trends for turbulent flow}

Turbulent Pipe FlowEntrance Length for

shear stressNo average acceleration

25-40 pipe diameter - Fox…Le/D = 4.4 Re1/6 - MYO

20D < Le < 30D104 < Re < 105 = White

Entrance lengthmuch shorter nowin turbulent flow

PIPEReD = 1600

DUCT FLOW: H = 0.2 cm, Uavg = 3.2 cm, ReH = 64

LAMINAR Pipe Flow Re< 2300 (2100 for MYO)

LAMINAR Duct Flow Re<1500 (2000 for SMITS)

LAMINAR Pipe Flow Re< 2300 (2100 for MYO)LAMINAR Duct Flow Re<1500 (2000 for SMITS)

Uo = V = Q/A

OUTSIDE BLUNDARY LAYER TREAT AS INVISCID, CAN USE B.E.

Fully Developed Laminar Pipe/Duct Flow

breath

Incompressible

•Compressibility requires work, may produce heat andchange temperature (note temperature changes due to viscous dissipation usually not important)

•For water usually considered constant •For gas usually considered constant for M < 0.3 (~100m/s or 230 mph; / ~ 4%)

•Pressure drop in pipes “usually” not large enough to make compressibility an issue (water hammer in an

exception).

Time OutStatic / Dynamic / Stagnation Pressures

What are static, dynamic and stagnation pressures?

The thermodynamic pressure, p, used throughout this book refers to the static pressure. This is the pressure experienced by a fluid particle as it moves with the fluid.

static pressure

What are static, stagnation, and dynamic pressures?

The stagnation pressure is obtained when the fluid is decelerated to zero speed through an isentropic process (no heat transfer, no friction).

For incompressible flow: po = p + ½ V2

What are static, dynamic and stagnation pressures?

The dynamic pressure is defined as ½ V2.

For incompressible flow: ½ V2 = po - p

breath

Laminar Flow – TheoryFully Developed Flow

FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES

Want to know “stuff” like:What’s pressure drop for specified flow & length?What’s shear stress on bottom & top plates? Suppose plate moving?What’s leakage flow rate of hydraulic oil between piston and cylinder…

need to know what u(y) is

If gap between piston and cylinder is 0.005 mm or less than this flow can be modeled as

flow between infinite parallel plates.

(high pressure hydraulic system

like break system of car)

Assumptions: (1) steady, incompressible, (2) fully developed flow (3) no body forces, (4) no changes in z variables, (5) u = 0 at y = 0, y = a

= 0(2)= 0(3) = 0(1)

FSx + FBx = /t (cvudVol )+ csuVdAEq. (4.17)

FSx = surface forces

= pressure and shear forcesin x-direction

Could use NSE directly, instead will derive velocity profileusing a differential control volume.

u = [a2/2](dp/dx)[(y/a)2 – (y/a)]

++ = 0

FSx = 0y

(Want to know what the velocity profile is.)

- p/x + dxy/dy = 0

p/x = dp/dx = dxy/dy = constant

Left side is f(x) only [p(x)] = Right side f(y) only [u(y)]Can only be true for all x and y if both sides equal a constant

no changes in z variables, w = 0 ~ 2-Dimensional, symmetry arguments

v = 0 du/dx + dv/dy = 0 via Continuity, 2-Dim. du/dx = 0 everywhere since fully developed,therefore dv/dy = 0 everywhere,

but since v = 0 at boundary, then v = 0 everywhere!

N.S.E. for incompressible flow with and constant viscosity. v-component

Proof that p/y = 0*

(v/t + uv/x + vv/y + wv/z) = gy - p/y + (2v/x2 + 2v/y2 + 2v/z2)

Eq 5.27b, pg 215

v = 0 everywhere and always, gy ~ 0 so left with:

p/y = 0; p = f(x) only!!!

Important distinction because book integrates p/x with

respect to y and pulls p/x out of integral (pg 314), can only do that if dp/dx, which is not a

function of y.

integrate

(Want to know what the velocity profile is.)

For Newtonian fluid*

substitute

integrate

USE 2 BOUNDARY CONDITIONS TO SOLVE FOR c1 AND c2

p/x = dp/dx = dxy/dy

yx = (dp/dx)y + c1

u = 0 at y = a:

u = 0 at y = 0: c2 = 0

c1 = -1/2 (dp/dx)a

u = [1/(2)][dp/dx]y2 - [1/(2)] [dp/dx]ay = [a2/(2)][dp/dx]{(y/a)2 – y/a}

u = {(y/1)^2 -(y/1)}; channel height=1m

-0.3 -0.25 -0.2 -0.15 -0.1 -0.05 0

u (m/s)

a = 1; dp/dx = 2

Why velocity negative?

u = [a2/(2)][dp/dx]{(y/a)2 – y/a}

u(y) for fully developed laminar flow between two infinite plates

negative

BREATH

AsideDo you believe in the no slip condition?

“Because of the no slip condition at the wall we know that the velocity at the wall must be zero along the entire length of the pipe.”

Pg 311

NO SLIP CONDITION

“The fluid in direct contact with the solid boundary has the same value as the boundaryitself; there is no slip at the boundary. This isan experimental fact based on numerousobservations of fluid behavior.”

NO SLIP CONDITION

Hydrogen Bubble Flow Visualization

Parallel Plates - Re = UD/ = 140Water Velocity = 0.5 m/s

Circular Pipe – Re = UD/ = 195Water Velocity = 2.4 m/s

Experimentally found

(usually)

Upper plate moving at 2 mm/sec Re = 0.03 (glycerin, h = 20 mm)

Duct flow, umax = 2 mm/secRe = 0.05(glycerin, h = 40 mm)

No slip condition explains:

Why large particles are easy to remove by blowing but small particles are not.

Why there is dust on a fan blade.

Why it is difficult to all the soap from a dish, just by running water.

Low Reynoldsnumber

High Reynolds number

VELOCITY = 0 AT WALL

NO SLIP CONDITION

Stokes (1851) ~

“On the Effect of the Internal Friction of Fluids on the Motion of Pendulums” - showed that no-slip condition led to remarkable agreement with a wide range of experiments, including the capillary tube experiments of Poiseuille (1940) and Hagen (1939).

Neuman / Hagenbach (1858-1860) ~

Correct analytical solution to laminar pipe flow

What happens to fluid particles next to no-slip layer?

So you think you are comfortable with the no-slip condition …

“It has been argued that the no-slip condition, applicable when a viscous fluid flows over a solid surface,

may be an inevitable consequence of the fact that all such surfaces are, in practice, rough on a microscopic scale: the energy lost through viscous dissipation as a

fluid passes over and around these irregularities is sufficient to ensure that it is effectively brought to rest.”

- On the No-Slip Boundary Condition, S.Richardson Journal of Fluid Mechanics (1973), vol. 59, part 4, pp.707-719

Surface Roughness

No Slip Condition: u = 0 at y = 0

VELOCITY = 0 AT WALL NO SLIP CONDITION

Each air molecule at the table top makes about 1010 collisions per second.Equilibrium achieved after about 10 collisions or 10-9 second, during which molecule has traveled less than 1 micron (10-4 cm).~ Laminar Boundary Layers - Rosenhead

Bioluminescence on treated (lower) and untreated (upper) surface

Bioluminescence on treated (upper) and untreated (lower) surface

Flashlight

“It turns out – although it is not at all self evident – that in all circumstances where it has been experimentally checked, the velocity of a fluid is exactly zero at thesurface [with zero velocity]of a solid.”

The Feynman Lectures on Physics – 1964, Vol. II, 41-1

Slip Boundary conditions for water flows in

hydrophobic nanoscale geometries

J. H. Walther , R. L. Jaffe , T. Werder , and P. KoumoutsakosSwiss Federal Institute of Technology, CH

Keywords: nanofluidics, slip condition , hydrophobic surfaces, Abstract:

In a collaboration with experimental groups at NASA and ETH Zurich we conduct computational studies towards the development of biosensors in aqueous

environments. Examples include arrays of carbon nanotubes that may operate as artificial stereocillia (Noca et al., 2000) or as molecular sieves. Here we

present novel results assesing the validity of the no-slip boundary condition in nanofluidics for prototypical geometries such as flow past a carbon nanotube and flow between two graphite plates. The role of the geometry on the slip

length is investigated. The results show significant slip lengths (in disagreement with the macroscale notion of no-slip at wall-fluid interfaces) and are consistent

with relevent experimental works of water flows over other hydrophobic surfaces. First we report results from large scale non-equilibrium molecular

dynamics (NEMD) simulations of water flow past graphite surfaces in a setting equivalent to a nanoscale planar Couette flow (Figure 1). A graphite surface is

known to be hydrophobic(Adamson:1997), and to exhibit physiochemical similarity with carbon nanotubes in aqueous environments (Balavoine:1999). The validity of the no-slip condition employed in macroscale Navier-Stokes

modeling has been questioned by experiments of water in hydrophobic capillaries (Churaev:1984, Baudry:2001). In these experiments, the water is found to exhibit a finite fluid velocity at the fluid-solid interface, with a slip

length of 28--30nm.The present NEMD simulations use the SPC/E water model and the graphite-water interaction is modeled using a Lennard-Jones potential

calibrated to match the experimentally measured macroscopic contact angle of water on graphite, cf. (Werder:2002). The average density profile in the channel displays the well-known peaks in the vicinity of the interface and bulk properties at the center of the channel cf. Figure 2a. Setting the upper walll in motion with

speeds of 50 to 100m/s drives the water and a linear velocity profiles is established after 1-2ns as shown in Figure 2. The velocity profiles indicate a slip

length approximately Ls = 30nm in good agreement with the relevant experimental values(Churaev:1984, Baudry:2001). In order to examine

geometry effects on the no-slip condition we conduct also simulations of flows past carbon nanotubes (with diameters of 1 to 2 nm) whose axis is placed

perpedincular to the mean flow (Figure 1). In this case a slip length of 1nm is observed. A systematic study is conducted where the effects of geometry and driving mean velocity are assesed and a boundary condition for macroscale

simulations of water flows past hydrophobic surfaces is proposed.

Laminar Flow – TheoryFound u(y) / now yx(y)

(next want to determine shear stress profile,yx)

yx = (du/dy)

For dp/dx = negativeyx in top ½ is negative & shear force is in the – x directionyx in bottom ½ is positive & shear force is in the – x direction

Shear force

+ sheardirection

dp/dx = negative+

xy = a(dp/dx){y/a - 1/2}

negative

Shear force direction

Flow direction

Positive stress is defined in the + x-direction as normal to surface is in the + z-direction

Sign conventionfor stresses

Positive stress is defined in the – (x-direction) as normal to surface is in the – (z-direction)

tau = [(y/1)-1/2]; a=1, dp/dx=1

-0.6 -0.4 -0.2 0 0.2 0.4 0.6

Flow direction

xy = a(dp/dx){y/a - 1/2}

Shear force direction

positive

BREATH

CHANGE OF VARIABLES

y’ = y – a/2; y = y’ + a/2

(y’2 + ay’ + a2/4 –y’a – a2/2)/ a2 = (y’/a)2 – 1/4

y’=0

y’ = a/2

y’ = -a/2

y = a l

Laminar Flow – TheoryFound u(y), yx(y); now Q, uavg, umax

(volume flow rate, Q)

[y3/3 – ay2/2]oa = a3/3 – a3/2 = -a3/6

If dp/dx = const

( average velocity)

= Uavg

A = la

(maximum velocity)

(a2/4)/a2 – (a/2)/a = -1/4

BREATH

Laminar Flow – TheoryUpper Plate Moving

UPPER PLATE MOVING WITH CONSTANT SPEED U

Journal bearing (crankshaft inside car engine)

Velocity distribution

Pressure drivenBoundary driven

Shear stress distribution

= Uy2/(2a) + (1/(2))(dp/dx)[(y3/3) – ay2/2]; y = a= Ua/2 + (1/(2))(dp/dx)[(2a3 – 3a3)/6]= Ua/2 + (1/(12))(dp/dx)[– a3]

Volume Flow Rate

Average Velocity

Area = al

Maximum Velocity

umax = a/2y = 0

very large shear stresses at start-up

NOTE THAT STEADY FLOW FIELD IS NOT ESTABLISHED INSTANTANEOUSLY

BREATH

Laminar Flow – TheoryExample

EXAMPLE:

FSx + FBx = /t (cvudVol )+ csuVdAEq. (4.17)

Assume: (1) surface forces due to shear alone, no pressure forces (patm on either side along boundary)(2) steady flow and (3) fully developed

Fsx + FBx = 0FBx = - gdxdydzFs1 – Fs2 - gdxdydz = 0 Fs1 = [yx + (dyx/dy)(dy/2)]dxdzFs2 = [yx - (dyx/dy)(dy/2)]dxdzdyx/dy = g

d yx/dy = gyx = du/dy = gy + c1

du/dy = gy/ + c1/u = gy2/(2) + yc1/ + c2

u = gy2/(2) + yc1/ + c2 u = gy2/(2) - ghy/ +U0

At y=h, u = gh2/(2) - gh2/ + U0 u = -gh2/(2) + U0

BREATH

Laminar Flow – TheoryFully Developed Pipe Flow

V p2p1

Fully Developed Pipe Flow

021 DlApApF wx

ppAw 4

A = D2/4

V p2p1

Fully Developed Pipe Flow

A = r2/4

or = (r/2)(dp/dx)

Eq 8.13a

(r) = {r2/4}{p1-p2}/{2rl}

r ww 2

True for laminar and turbulent flow!!!

(r) on control volume

V p2p1

r ww 2

u/umax = 1 – (r/R)2

FULLY DEVELOPED LAMINAR PIPE FLOW

u/umax

r(dp/dx)/2

LAMINAR ANDTURBULENT

LAMINAR

du rdrl

u = 0, at y = R

22 rRl

onlylaminar

22 rRl

V p2p1

Eq. 8.12

22 rRl

Eq. 8.12

……

Eq. 8.12

Q = A V • dA

Q = 0R u2rdr

Q = 0R -[ R2 - r2] (dp/dx)/(4) 2rdr

Q = [(dp/dx)/(4)] (2)[ r4/4 - R2r2/2 ]0R

Q = (-R4dp/dx)/(8) Eq. 8.13b

Eq. 8.12

umax = - (R2/(4)) (dp/dx)

V = uavg = Q/R2 = (-R4dp/dx)/(8R2)

V =uavg = -(R2/(8)) (dp/dx)Eq. 8.13d

uavg = ½ umax

Eq. 8.13e

BREATH

Laminar Flow f = {(p/L)D}/{1/2uavg

2} = ?

uavg = -(R2/(8)) (dp/dx) Eq. 8.13d

uavg = (R2/(8)) (p/L); p/L = uavg8/R2

f = [uavg8/R2] D/{1/2uavg2}

f = {64/D}/{uavg} = 64/{uavgD}

f = 64/ReD

Laminar Flow f = {(p/L)D}/{1/2uavg

2} = ?

THE END

Laminar Flow – TheoryFully Developed Pipe Flow

Fox et al.’s development

APPROACH JUST LIKE FOR DUCT FLOW

dFL = p2rdr dFR = -(p + [dp/dx]dx) 2rdrdFI = -rx2rdxdFO = (rx + [d rx/dr]dr) 2(r + dr) dx

rFULLY DEVELOPED LAMINAR PIPE FLOW

dFL = p2rdr

dFR = -(p + [dp/dx]dx)2rdr

dFL + dFR = -[dp/dx]dx2rdr

dFL dFR

rFULLY DEVELOPED LAMINAR PIPE FLOW

dFI = -rx2rdxdFO = (rx + [d rx/dr]dr) 2(r + dr) dx

dFO+ dFI = -rx 2rdx + rx 2rdx + rx 2drdx + [drx/dr)]dr2rdx + [drx/dr]dr 2dr dx

dFO + dFI = rx 2drdx + [drx/dr]dr2rdx

dFL dFR

dFL + dFR + dFI + dFO = 0-[dp/dx]dx2rdr+rx 2drdx(r/r)+(drx/dr)dr2rdx = 0

[dp/dx] = rx/r + drx/dr = (1/r)d(rxr)/dr

dFL dFR

dp/dx = dxy/dy

dp/dx = (1/r)(d[rrx]/dr)because of spherical coordinates, more complicated than for duct.

dp/dx = (1/r)(d[rrx]/dr)

p is uniform at each section by symmetry.

rx is at most a function of r, because fully developed, rx f(x),symmetry, rx f().

dp/dx = constant = (1/r)(d[rrx]/dr)

dp/dx = constant = (1/r)(d[rrx]/dr)d[rrx]/dr = rdp/dx

integrating…..rrx = r2(dp/dx)/2 + c1

rx = du/drrx = du/dr = r(dp/dx)/2 + c1/r

What we you say about c1?

rx = du/dr = r(dp/dx)/2 + c1/r c1 = 0 or else rx =

rx = du/dr = r(dp/dx)/2

For dp/dx negative, get negative shear stress on CV

Shear forces on CV

dp/dx is negative

rx = du/dr = r(dp/dx)/2 + c1/r c1 = 0 or else rx =

rx = du/dr = r(dp/dx)/2Shear forces on CV

SHEAR STRESS PROFILE

FULLY DEVELOPED DUCT FLOW

FULLY DEVELOPED PIPE FLOW

= direction of shear force on CV

- for flow to right

SHEAR STRESS PROFILE

rx = r(dp/dx)/2 TRUE FOR LAMINAR AND TURBULENT FLOW

du/dr = r(dp/dx)/2TRUE ONLY FOR LAMINAR FLOW

du/dr = r(dp/dx)/2u = r2(dp/dx)/(4) + c2

u=0 at r=R, so c2=-R2(dp/dx)/(4)u = r2(dp/dx)/(4) - R2(dp/dx)/(4)u = [ r2 - R2] (dp/dx)/(4)u = -R2(dp/dx)/(4)[ 1 – (r/R)2]

BREATH

VOLUME FLOW RATE – PIPE FLOW

Q = A V • dA = 0

R u2rdr = 0

R [ r2 - R2] (dp/dx)/(4) 2rdr

Q = [(dp/dx)/(4)][ r4/4 - R2r2/2 ]0R (2)

= (-R4dp/dx)/(8)

VOLUME FLOW RATE

VOLUME FLOW RATE – a function of p/L

p/x = constant = (p2-p1)/L = -p/L

p2 = p + p

Q = (-R4dp/dx)/(8) = R4p/(8L) = D4(p/(128L)

AVERAGE FLOW RATE

uAVG = Q/A = Q/(R2) = R4p/(R28L)= R2p/(8L) = -(R2/(8)) (dp/dx)

Q = R4p/(8L)

AVERAGE FLOW RATE

uAVG = V = Q/A = Q/(R2) = R4p/(R28L)uAVG = R2p/(8L) = -(R2/(8)) (dp/dx)

MAXIMUM FLOW RATE

du/dr = (r/[2])p/x

At umax, du/dr = 0; which occurs at r = 0

umax = R2(p/x)/(4)

MAXIMUM FLOW RATE

Chapter 8: Internal Incompressible Viscous Flow

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