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 =  DU avg /  = DU avg / Q = AU avg dm/dt =  AU avg A =  D 2 /4 (a)Re = DU avg / = DQ/(A ) = 4DQ/(  D 2 ) = 4Q/(  D ) (b)Re =  DU avg /  = (dm/dt)D/(A  ) = (dm/dt)D4/(  D 2  ) = 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 = DU avg / = DQ/(A ) = 4DQ/(  D 2 ) = 4Q/(  D ) Q = Re (  D )/4 (c) Q 1 = Q 2 Re 1  D 1 /4 = Re 2  D 2 /4 Re 2 = Re 1 (D 1 /D 2 ) Re 2 = 1800 (10 mm / 6 mm) = 3000

For most engineering pipe flow systems turbulence occurs around Re = 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 15 o. Q = Re  D /4 Q = 2300  D /4 Air: =  /  = 1.46 x m 2 /s at 15 o C Table A-10 Water: =  /  = 1.14 x m 2 /s at 15 o C Table A-8

Re = 2300 Re = Q/(A ) = 4DQ/(  D 2 ) = 4Q/(  D ) Q = Re  D /4 Why get higher Q’s for same Re and D in air than water? air ~ 13 water

What is the direction and magnitude (lbf/ft 2 ) of shear stress on pipe wall ???

(P 2 -P 1 = -  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) = [h 2 /(2  )][dp/dx][(y/h) 2 – ¼]

u(y) = [a 2 /(2  )][dp/dx][(y/a) 2 – ¼] Eq. 8.7 (y=0 at centerline & a = h) u(y) = [h 2 /(2  )][dp/dx][(y/h) 2 – ¼]  xy =  du/dy  xy =  [h 2 /(2  )][dp/dx][2y/h 2 ] = [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 And  xy is > 0 for y < 0 |Maximum  xy | = y(+h/2) and y(-h/2) u Shear stress forces  xy y ??? Sign of shear stress and direction of shear stress forces “seem” contradictory???

 xy  xx  xz x 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 sign convention  xy y Plot:  xy (y)

1507 by Leonardo in connection with a hydraulic project in Milan

D = 6mm L = 25 mm P = 1.5MPa (gage) M = ?, SAE 30 oil at 20 o Q = f(  p,a), f =? Velocity of M = 1 mm/min a = ? Q

D = 6mm L = 25 mm P = 1.5MPa (gage) M = ? SAE 30 oil at 20 o Velocity of M = 1 mm/min a = ? M = ? Fully Developed Laminar Flow Between Infinite Parallel Plates

Q= f(  p,a) ? Q   p Q  a 3 l Q l Fully Developed Laminar Flow Between Infinite Parallel Plates l = 2  R =  D

D = 6mm L = 25 mm P = 1.5MPa (gage) M = ? SAE 30 oil at 20 o Velocity of M = 1 mm/min a = ? If V = 1 mm/min, what is a? v Q = a 3  pl/(12  L) Q = VA =U avg  Da

1500

Re = Va/

What is velocity profile, u(y), for a plate moving vertically at U o through a liquid bath? UoUo x y Shear Forces on Fluid Element Shear stress B.C. z x y

What is velocity profile, u(y), for a plate moving vertically at U o through a liquid bath? dy UoUo x y Shear Forces on Fluid Element -(  zx +[d  zx /dy][dy/2]dxdz +(  zx -[d  zx /dy][dy/2]dxdz -  gdxdydz = 0 -d  zx /dy = -  g  zx =  gy + c 1 ASSUME FULLY DEVELOPED: mg

What is velocity profile, u(y), for a plate moving vertically at U o through a liquid bath? dy UoUo x y Shear Forces on Fluid Element  zx =  gy + c 1  zx =  du/dy u(y) =  gy 2 /[2  ] + c 1 y/  + c 2 u(y=0) = U 0 So c 2 = U o ASSUME FULLY DEVELOPED: mg

What is velocity profile, u(y), for a plate moving vertically at U o through a liquid bath? dy UoUo x y Shear Forces on Fluid Element  zx =  gy + c 1  zx =  du/dy u(y) =  gy 2 /[2  ] + c 1 y/  + U o du/dy (y=h) = 0 du/dy =  zx /  = 0 at y=h 0 =  gh/  + c 1 /  c 1 = -  gh ASSUME FULLY DEVELOPED: mg

What is velocity profile, u(y), for a plate moving vertically at U o through a liquid bath? dy UoUo x y Shear Forces on Fluid Element  zx =  gy + c 1  zx =  du/dy u(y) =  gy 2 /[2  ] + c 1 y/  + c 2 ASSUME FULLY DEVELOPED: mg c 2 = U o c 1 = -  gh u(y) =  gy 2 /[2  ] + -  gh y/  + U o u(y) =  g/  {y 2 /2 –h y} + U o

At y = 0 velocity 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 mg  g  (D 2 /4)(dx) dx  rz Re D = VD/ = u avg D/ D = ? = 2300( )/V

What is the maximum diameter of a vertical pipe so that water running down it remains laminar? D mg  g  (D 2 /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  (D 2 /4)(dx) +  rz 2  rdx =0 dx

What is the maximum diameter of a vertical pipe so that water running down it remains laminar? Assume: Fully Developed D  g  (D 2 /4)(dx) +  rz 2  rdx =0 dx For fully developed, laminar, horizontal, pressure driven, Newtonian pipe flow: u(r) = -{(dp/dx) /4  }{R 2 – r 2 }

What is the maximum diameter of a vertical pipe so that water running down it remains laminar? Assume: Fully Developed D  g  (D 2 /4)(dx) +  rz 2  rdx =0 dx 

What is the maximum diameter of a vertical pipe so that water running down it remains laminar? Assume: Fully Developed D  g  (D 2 /4)(dx) +  rz 2  rdx =0 dx Re = u acg D/ = VD/ D = Re /u avg Re = 2300 = 1.0 x m 2 /s at 20 o C D = 1.96 mm for water

u avg /u max = (y/R) 1/n Not accurate at y=R and y near 0

u / U max = (y/R) 1/n

from Hinze –Turbulence, McGraw Hill, 1975 n = 1.85 log 10 Re Umax –1.96

n = 1.85 log 10 (Re Umax ) –1.96 U avg /U max = 2n 2 /((n+1)(2n+1)) Re Umax U avg /U max For F.D. laminar flow u avg = ½ u max

U avg /U max Re Umax

U avg /U max = (1 – r/R) 1/n = (y/R) 1/2 ; n = f(Re) Power Law Velocity Profiles for Fully-Developed Flow in a Smooth Pipe u/U = U avg /U max 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 * / ) or u/u * = 5.75l0g(yu * / ) + 5.5

Consider fully developed laminar pipe flow. Evaluate the kinetic energy coefficient, .  A (u 2 /2)  udA =  A (u 2 /2)  VdA =  (dm/dt) V 2 /2 u = u(r) = V = V(r); u avg = V  =  A  u 3 dA / ((dm/dt) V 2 ) Question: What is  for inviscid flow?

Eq Eq. 8.13e

3

Fully developed turbulent pipe flow. Evaluate the kinetic energy coefficient, .  A (u 2 /2)  udA =  A (u 2 /2)  VdA =  (dm/dt) V 2 /2 u = u(r) = V = V(r); u avg = V  =  A  u 3 dA / ((dm/dt) V 2 ) Question: What is  for inviscid flow?

 = f(n)

Re = U max D/  Velocity profile getting flatter If no viscosity completely flat and  = 1

u(y)/U max y = (R-r)

Re = U max D/ U avg / U max

Given: Smooth pipe, fully developed turbulent flow, Avg velocity = 1.5 m/s, diameter = 50mm, Re = 75,000, p 1 =590 kPa (gage), z 1 = 0, p 2 = atmosphere, z 2 = 25m Find: Head loss between 1 and 2 25 m(1) (2) Dimensions of L 2 /t 2 (energy per unit mass)

Given: Average velocity = 1.5 m/s, Diameter = 50mm, Re = 75,000, p 1 =590 kPa (gage), z 1 = 0, p 2 = atmosphere, z 2 = 25m 25 m(1) (2)

[2-3] [3-4]

Assumptions: Incompressible  2 u avg2 2 =  3 u avg3 2 z 2 = z 3 energy/mass energy/weight PUMP HEAD [ 2 – 3 ]

Assumptions: Incompressible  3 u avg3 2 =  4 u avg4 2 p 4 = atm; z 3 = 0 PIPE LOSSES [ 3 – 4 ] (p 3 /  +  3 V avg 2 /2 + gz 3 ) - (p 4 /  +  4 V avg 2 /2 + gz 4 ) = h LT h LT = h l + h lm = (fL/D + K)V avg 2 /2 (p 3 /  +  3 V avg 2 + gz 3 ) - (p 4 /  +  4 V avg 2 + gz 4 ) = h LT h LT = p 3 /  –gz 4 = 50 (lbf/in 2 )(ft 3 /1.94 slug)(144 in 2 /ft) – 32.2(ft/sec 2 )90(ft)(lbf-sec 2 /slug-ft) h LT = 813 lbf-ft/slug 0 0 or  H = h LT /g = 25.2 ft

From a History of Aerodynamics by John Anderson

REMEMBER ~

Fully Developed turbulent Flow 4000 < Re < 10 5 ; f = 0.316/Re 1/4

Boundary Layer Theory – Schlicting, 1979 This = our f Darcy = 64/Re D = /Re D 1/4 1/ 1/2 = 2.0 log(Re D 1/ 1/2 ) – 0.8

Leonardo da Vinci

= 16/Re Re = U avg R / 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

D 1 =50mm D 2 = 25mm p 1 -p 2 =3.4kPa Q = ? (p 1 /  + V avg1 2 /2 + gz 1 ) = (p 2 /  + V avg2 2 /2 + gz 2 ) + h LT h LT = h l + KV avg2 2 /2 V avg1 = V avg2 (A 2 /A 1 ) = V avg2 AR p 1 /  + V avg2 2 AR 2 /2 = p 2 /  + V avg2 2 /2 + KV avg2 2 /2 0

D 1 =50mm D 2 = 25mm p 1 -p 2 =3.4kPa Q = ? AR = ¼ K = 0.4

p 1 /  + V avg2 2 AR 2 /2 = p 2 /  + V avg2 2 /2 + KV avg2 2 /2 (p 1 – p 2 ) /  = (V avg2 2 /2)(1 – ) D 1 =50mm; D 2 = 25mm; p 1 -p 2 =3.4kPa;  = 999 kg/m 3 Q = ?

K = 0.78; Table 8.2 3ft

z 1 = 3 ft

Could be greatly improved by rounding entrance and applying a diffuser. (about 30% increase in Q)

= ?

Eq. 8.43

N/R 1 = 0.45/(.15/2) = 6 C p  0.62 AR  2.7 Pressure drop fixed, want to max C p to get max V 2

The end

Given: Laminar, fully developed flow between parallel plates  = 0.5 N-sec/m 2 ; dp/dx = N/m 3 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/m 2 ?

(a)

(b)

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/m 3 = 1 x m 2 /s

8.63

8.63OLD Consider fully developed laminar flow between infinite plates. Evaluate the kinetic energy coefficient, .

8.63 1/2

8.63