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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
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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)
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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
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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
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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
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What is the direction and magnitude (lbf/ft2) of shear stress on pipe wall ???
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(P2-P1 = -P)
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Pipe wall exerts a negative shear on the fluid. Consequently the fluid exerts a positive shear
on the wall.
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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]
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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
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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???
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xy
xx
xzx
y
z
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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
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+
+
Shear force
Shear forceShear sign convention
xy
yPlot: xy(y)
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1507 by Leonardo in connection with a hydraulic project in Milan
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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
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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
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Q= f(p,a) ?Q pQ a3
l
Q l
Fully Developed Laminar Flow Between Infinite Parallel Plates
l = 2R = D
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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
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1500
1500
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1500
1500
Re = Va/
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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
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What is velocity profile, u(y),for a plate moving verticallyat Uo through a liquid bath?
dy
Uo
x
y
Shear Forces on Fluid Element
-(zx+[dzx/dy][dy/2]dxdz+(zx-[dzx/dy][dy/2]dxdzgdxdydz = 0
-dzx/dy = - g
zx = gy + c1
ASSUME FULLY DEVELOPED:
mg
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What is velocity profile, u(y),for a plate moving verticallyat Uo through a liquid bath?
dy
Uo
x
y
Shear Forces on Fluid Element
zx = gy + c1
zx = du/dyu(y) = gy2/[2] + c1y/ + c2
u(y=0) = U0
So c2 = Uo
ASSUME FULLY DEVELOPED:
mg
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What is velocity profile, u(y),for a plate moving verticallyat Uo through a liquid bath?
dy
Uo
x
y
Shear Forces on Fluid Element
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
ASSUME FULLY DEVELOPED:
mg
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What is velocity profile, u(y),for a plate moving verticallyat Uo through a liquid bath?
dy
Uo
x
y
Shear Forces on Fluid Element
zx = gy + c1
zx = du/dyu(y) = gy2/[2] + c1y/ + c2
ASSUME FULLY DEVELOPED:
mg
c2 = Uo
c1 = -gh
u(y) = gy2/[2] + -ghy/ + Uo
u(y) = g/{y2/2 –hy} + Uo
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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
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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
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What is the maximum diameter of a vertical pipe so that water running down it remains laminar?D
mgg(D2/4)(dx)
dx
rz
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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
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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
For fully developed, laminar, horizontal,pressure driven, Newtonian pipe flow:u(r) = -{(dp/dx) /4}{R2 – r2}
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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
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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 Re = uacgD/ = VD/D = Re /uavg
Re = 2300= 1.0 x 10-6 m2/s at 20oC
D = 1.96 mm for water
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uavg/umax = (y/R)1/n
Not accurate aty=R and y near 0
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u / Umax = (y/R)1/n
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from Hinze –Turbulence, McGraw Hill, 1975
n = 1.85 log10ReUmax –1.96
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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
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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
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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
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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
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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?
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Eq. 8.14
Eq. 8.13e
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3
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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?
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= f(n)
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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
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u(y)/Umax
y = (R-r)
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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
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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)
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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)
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[2-3]
[3-4]
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Assumptions: Incompressible 2uavg2
2 = 3uavg32
z2 = z3
energy/mass
energy/weight
PUMP HEAD [ 2 – 3 ]
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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
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From a History of Aerodynamics by John Anderson
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REMEMBER ~
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Fully Developed turbulent Flow4000 < Re < 105; f = 0.316/Re1/4
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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
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Leonardoda Vinci
1452-1519
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= 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
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From a History of Aerodynamics by John Anderson
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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
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D1 =50mmD2 = 25mmp1-p2=3.4kPa Q = ?
AR = ¼
K = 0.4
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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 = ?
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K = 0.78; Table 8.23ft
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z1 = 3 ft
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Could be greatly improved by rounding entrance and
applying a diffuser.
(about 30% increase in Q)
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= ?
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Eq. 8.43
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N/R1 = 0.45/(.15/2) = 6
Cp 0.62
AR 2.7
Pressure drop fixed, want to max Cp to get max V2
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The end
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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
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(a)
Shear stress on plate = 1.8 N/m2?
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(a)
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(b)
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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
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8.63
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8.63
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8.63
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8.63OLD Consider fully developed laminar flow betweeninfinite plates.
Evaluate the kinetic energy coefficient, .
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8.63
1/2
1/2
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8.63
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