given : incompressible flow in a circular channel and re = 1800, where d = 10 mm

Given: Incompressible flow in a circular channel and Re = 1800, where D = 10 mm.

Find: (a) Re = f (Q, D,) (b) Re = f(dm/dt, D,) (c) Re for same Q and D = 6 mm

Incompressible flow in a circular channel. Re = 1800, where D = 10mm

Find: (a) Re = f (Q, D, ); (b) Re = f(dm/dt, D, );

Equations: Re = DUavg/ = DUavg/ Q = AUavg dm/dt = AUavg A =

D2/4

(a) Re = DUavg/ = DQ/(A) = 4DQ/(D2) = 4Q/(D)

(b)Re = DUavg/ = (dm/dt)D/(A) = (dm/dt)D4/(D2) = 4(dm/dt)/(D)

Incompressible flow in a circular channel. Re = 1800, where D = 10mm

Find: (c) Re for same flow rate and D = 6 mm

(a) Re = DUavg/ = DQ/(A) = 4DQ/(D2) = 4Q/(D)

Q = Re (D)/4

(c) Q1 = Q2

Re1D1/4 = Re2D2/4 Re2 = Re1(D1/D2)

Re2 = 1800 (10 mm / 6 mm) = 3000

For most engineering pipe flow systems turbulence occurs around Re = 2300. On a log-log plot of volume flow rate, Q, versus tube diameter, plot lines that cor-respond to Re = 2300 for standard air and water at 15o.

Q = ReD/4 Q = 2300 D/4

Air: = / = 1.46 x 10-5 m2/s at 15oC Table A-10Water: = / = 1.14 x 10-6 m2/s at 15oC Table A-8

Re = 2300

Re = Q/(A) = 4DQ/(D2) = 4Q/(D)Q = ReD/4

Why gethigher Q’s

for same Reand D in airthan water?

air ~ 13 water

What is the direction and magnitude (lbf/ft2) of shear stress on pipe wall ???

(P2-P1 = -P)

Pipe wall exerts a negative shear on the fluid. Consequently the fluid exerts a positive shear

on the wall.

Given: Fully developed flow between two parallel plates, separated by h. Flow is from left to right.

y = h/2

y = 0

y = h/2

Plot: xy(y)

Net Pressure Force

u(y) = [h2/(2)][dp/dx][(y/h)2 – ¼]

u(y) = [a2/(2)][dp/dx][(y/a)2 – ¼] Eq. 8.7

(y=0 at centerline & a = h)u(y) = [h2/(2)][dp/dx][(y/h)2 – ¼]

xy = du/dy

xy = [h2/(2)][dp/dx][2y/h2] = [dp/dx][y]

xy = [dp/dx][y] (for y=0 at centerline)

Net Pressure Force

xy = [dp/dx]{y–[a/2]}(for y=0 at bottom plate)

0

h/2

-h/2

xy(y) = [dp/dx][y] dp/dx < 0

So xy is < 0 for y > 0

And xy is > 0 for y < 0

|Maximum xy| = y(+h/2) and y(-h/2)

u

Shear stress forces

xy

y??? Sign of shearstress and directionof shear stressforces “seem”contradictory???

xy

xx

xzx

y

z

sign convention for stress (pg 26): A stress component is positive when the direction of the stress component and the normal to the plane at which it acts are both positive or both negative.

Stresses shown in figure are all positive

+

+

Shear force

Shear forceShear sign convention

xy

yPlot: xy(y)

1507 by Leonardo in connection with a hydraulic project in Milan

D = 6mmL = 25 mmP = 1.5MPa (gage)M = ?, SAE 30 oil at 20o

Q = f(p,a), f =?Velocity of M = 1 mm/mina = ?

Q

D = 6mmL = 25 mmP = 1.5MPa (gage)M = ?SAE 30 oil at 20o

Velocity of M = 1 mm/mina = ?

M = ?

Fully DevelopedLaminar Flow

Between InfiniteParallel Plates

Q= f(p,a) ?Q pQ a3

l

Q l

Fully Developed Laminar Flow Between Infinite Parallel Plates

l = 2R = D

D = 6mmL = 25 mmP = 1.5MPa (gage)M = ?SAE 30 oil at 20o

Velocity of M = 1 mm/mina = ?

If V = 1 mm/min, what is a?

v

Q = a3pl/(12L) Q = VA =UavgDa

1500

1500

1500

1500

Re = Va/

What is velocity profile, u(y),for a plate moving verticallyat Uo through a liquid bath?

Uo

x

y

Shear Forces on Fluid Element

Shear stress

B.C.zx

y


dy

Uo

x

y


-(zx+[dzx/dy][dy/2]dxdz+(zx-[dzx/dy][dy/2]dxdzgdxdydz = 0

-dzx/dy = - g

zx = gy + c1

ASSUME FULLY DEVELOPED:

mg


dy

Uo

x

y


zx = gy + c1

zx = du/dyu(y) = gy2/[2] + c1y/ + c2

u(y=0) = U0

So c2 = Uo


mg


dy

Uo

x

y


zx = gy + c1

zx = du/dyu(y) = gy2/[2] + c1y/ + Uo

du/dy (y=h) = 0du/dy = zx / = 0 at y=h0 = gh/ + c1/c1 = -gh


mg


dy

Uo

x

y


zx = gy + c1

zx = du/dyu(y) = gy2/[2] + c1y/ + c2


mg

c2 = Uo

c1 = -gh

u(y) = gy2/[2] + -ghy/ + Uo

u(y) = g/{y2/2 –hy} + Uo

velocity profile for water film on vertically moving plate

-2000

-1500

-1000

-500

0

500

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

distance from plate to water surface

velo

city

At y = 0velocity at edge of film 0

but du/dy = 0

What is the maximum diameter of a vertical pipe

so that water running down it remains laminar?

D

mgg(D2/4)(dx)

dx

rz

ReD = VD/ = uavgD/

D = ? = 2300()/V

What is the maximum diameter of a vertical pipe so that water running down it remains laminar?D

mgg(D2/4)(dx)

dx

rz

What is the maximum diameter of a vertical pipe so that water running down it remains laminar? Assume: Fully Developed

D

g(D2/4)(dx)+ rz2rdx=0

dx


D


dx

For fully developed, laminar, horizontal,pressure driven, Newtonian pipe flow:u(r) = -{(dp/dx) /4}{R2 – r2}


D


dx


D


dx Re = uacgD/ = VD/D = Re /uavg

Re = 2300= 1.0 x 10-6 m2/s at 20oC

D = 1.96 mm for water

uavg/umax = (y/R)1/n

Not accurate aty=R and y near 0

u / Umax = (y/R)1/n

from Hinze –Turbulence, McGraw Hill, 1975

n = 1.85 log10ReUmax –1.96

Turbulent Flow

0.720.740.760.780.80.820.840.860.88

1.00E+03 1.00E+04 1.00E+05 1.00E+06 1.00E+07

Re=UmaxD/nu n uavg/umax4000 4.703811 0.74544120000 5.996905 0.79111950000 6.733095 0.810498100000 7.29 0.82293110000 7.366576 0.824514200000 7.846905 0.833835500000 8.583095 0.8463461000000 9.14 0.8546291100000 9.216576 0.8556982000000 9.696905 0.8620653200000 10.07453 0.866689

n = 1.85 log10(ReUmax) –1.96

Uavg/Umax = 2n2/((n+1)(2n+1))

ReUmax

Uavg/Umax

For F.D. laminar flow uavg = ½ umax

Turbulent Flow

0.720.740.760.780.80.820.840.860.88

1.00E+03 1.00E+04 1.00E+05 1.00E+06 1.00E+07

Uavg/Umax

ReUmax

Uavg/Umax = (1 – r/R)1/n = (y/R)1/2; n = f(Re)

Power Law Velocity Profiles for Fully-Developed Flow in a Smooth Pipeu/U = Uavg/Umax

close to wall

Eq.1: u/u* = yu*/; u*=(wall/)1/2;

note: wall = du/dy ~ (u/y) = u*2 u/u* = yu*/

Eq.5: u/u* = 2.5ln(yu*/) + 5.5 or u/u* = 5.75l0g(yu*/) + 5.5

Consider fully developed laminar pipe flow.

Evaluate the kinetic energy coefficient, .A(u2/2)udA = A (u2/2)VdA

= (dm/dt) V2/2u = u(r) = V = V(r); uavg = V

= A u3dA / ((dm/dt) V2)

Question: What is for inviscid flow?

Eq. 8.14

Eq. 8.13e

Fully developed turbulent pipe flow.

Evaluate the kinetic energy coefficient, .A(u2/2)udA = A (u2/2)VdA

= (dm/dt) V2/2u = u(r) = V = V(r); uavg = V

= A u3dA / ((dm/dt) V2)

Question: What is for inviscid flow?

= f(n)

1.02

1.04

1.06

1.08

1.1

1.12

1.14

1.00E+03 1.00E+04 1.00E+05 1.00E+06 1.00E+07

Re = UmaxD/

Velocity profile getting flatterIf no viscosity completely flat

and = 1

u(y)/Umax

y = (R-r)

0.72

0.74

0.76

0.78

0.8

0.82

0.84

0.86

0.88

0.00E+00 1.00E+06 2.00E+06 3.00E+06 4.00E+06 5.00E+06 6.00E+06

Re = UmaxD/

Uavg

/Umax

Given: Smooth pipe, fully developed turbulent flow, Avg velocity = 1.5 m/s, diameter = 50mm, Re = 75,000,p1=590 kPa (gage), z1 = 0, p2 = atmosphere, z2 = 25m

Find: Head loss between 1 and 2

25 m(1) (2)

Dimensions of L2/t2 (energy per unit mass)

Given: Average velocity = 1.5 m/s, Diameter = 50mm, Re = 75,000, p1=590 kPa (gage), z1 = 0, p2 = atmosphere, z2 = 25m

25 m(1) (2)

[2-3]

[3-4]

Assumptions: Incompressible 2uavg2

2 = 3uavg32

z2 = z3

energy/mass

energy/weight

PUMP HEAD [ 2 – 3 ]

Assumptions: Incompressible 3uavg3

2 = 4uavg42

p4 = atm; z3 = 0

PIPE LOSSES [ 3 – 4 ]

(p3/ + 3Vavg2/2 + gz3) - (p4/ + 4Vavg

2/2 + gz4) = hLT hLT = hl + hlm = (fL/D + K)Vavg

2/2

(p3/ + 3Vavg2 + gz3) - (p4/ + 4Vavg

2 + gz4) = hLT

hLT = p3/ –gz4 = 50 (lbf/in2)(ft3/1.94 slug)(144 in2/ft) – 32.2(ft/sec2)90(ft)(lbf-sec2/slug-ft)

hLT = 813 lbf-ft/slug

0 0

or H = hLT/g = 25.2 ft

From a History of Aerodynamics by John Anderson

REMEMBER ~

Fully Developed turbulent Flow4000 < Re < 105; f = 0.316/Re1/4

Boundary Layer Theory – Schlicting, 1979

This = our fDarcy

= 64/ReD

= 0.3164/ReD1/4

1/1/2 = 2.0 log(ReD 1/1/2) – 0.8

Leonardoda Vinci

1452-1519

= 16/Re

Re = UavgR /

Be careful that you know if using Darcy or Fanning friction factor and if Re is bases on D or R

From a History of Aerodynamics by John Anderson

D1 =50mmD2 = 25mmp1-p2=3.4kPa

Q = ?

(p1/ + Vavg12/2 + gz1) = (p2/ + Vavg2

2/2 + gz2) + hLT

hLT = hl + KVavg22/2

Vavg1 = Vavg2(A2/A1) = Vavg2AR

p1/ + Vavg22AR2 /2 = p2/ + Vavg2

2/2 + KVavg22/2

0

D1 =50mmD2 = 25mmp1-p2=3.4kPa Q = ?

AR = ¼

K = 0.4

p1/ + Vavg22AR2 /2 = p2/ + Vavg2

2/2 + KVavg22/2

(p1 – p2) / = (Vavg22 /2)(1 – 0.0625 + 0.4)

D1 =50mm; D2 = 25mm; p1-p2=3.4kPa; = 999 kg/m3

Q = ?

K = 0.78; Table 8.23ft

z1 = 3 ft

Could be greatly improved by rounding entrance and

applying a diffuser.

(about 30% increase in Q)

Eq. 8.43

N/R1 = 0.45/(.15/2) = 6

Cp 0.62

AR 2.7

Pressure drop fixed, want to max Cp to get max V2

The end

Given: Laminar, fully developed flow between parallel plates

= 0.5 N-sec/m2; dp/dx = -1200 N/m3

Distance between the plates, h = 3mm

Find: (a) the shear stress, yx, on the upper plate(b) Volume flow rate, Q, per unit width, l.

(a) yx = du/dy

(b) Q = u(y)dA = u(y)ldy

y = 0 at centerline

(a)

Shear stress on plate = 1.8 N/m2?

8.63OLD Consider fully developed laminar flow of water between infinite plates. The maximum flow speed, plate spacing, and width are 6 m/s, 0.2 mm, and 30mm respectively. Evaluate the kinetic energy coefficient, .

= 999 kg/m3

= 1 x 10-6 m2/s

8.63OLD Consider fully developed laminar flow betweeninfinite plates.

Evaluate the kinetic energy coefficient, .

8.63

1/2

1/2

given : incompressible flow in a circular channel and re = 1800, where d = 10 mm

Documents