## Chapter 8 : Laminar Flow¶

### Example 8.1 Page no 286¶

In [1]:
# Determine maximum velocity and shear stress

# Given

from math import *

P1 = 200                 # Pressure at inlet in kPa

P2 = 260                # Pressure at outlet in kPa

d = 0.004                      # diameter in m

L = 8                          # length of pipe in meters

z = 6                          # height of the pipe from the ground

g = 9.81                       # acceleration due to gravity in m/s**2

# properties of kerosene

mu = 19.1*10**-4               # viscosity of kerosene at 20 deg C

S = 0.81                       # specific gravity of kerosene

rho = 1000                     # density in kg/m**3

# Solution

# calculating direction of flow

p1 = (P1+g*z*S)*1000                # point 1

p2 = (P2)*1000                     # point 2

# direction of flow is from point 1-2

# shear stress

Sp = -((p1-p2)/sqrt(L**2+z**2))

r = d/2

Tau_max = r*Sp/2

print "(a) Maximum shear stress =",round(Tau_max,3),"N/m**2"

# maximum velocity

Vmax = r**2*Sp/(4*mu)

print "(b) Maximum velocity =",round(Vmax,3),"m/s"

# discharge

Q = pi*r**4*Sp/(8*mu)

print "(c) Discharge = ",round(Q,7),"m**3/s"

# calculate reynolds number

V = Vmax/2

R = rho*V*d*S/mu

print "Reynolds number =",round(R,0),"is less than 2000, the flow is laminar and the calculations are valid"

(a) Maximum shear stress = 1.232 N/m**2
(b) Maximum velocity = 0.645 m/s
(c) Discharge =  4.1e-06 m**3/s
Reynolds number = 547.0 is less than 2000, the flow is laminar and the calculations are valid


### Example no 8.2 Page no 289¶

In [2]:
# Determine the head loss

# Given

d = 0.02                  # diameter of the pipe in m

l = 30                    # length of the pipe in m

v = 0.1                   # velocity in m/s

g = 9.81                  # acceleration due to gravity in m/s**2

# for water at 5 deg C

nu = 1.54*10**-6          # kinematic viscosity of water in m**2/s

# Solution

R = v*d/nu

print "R = ",round(R,0),"is lesss than 2000 , the flow is laminar"

f = 64/R                 # friction factor

Hf = f*l*v**2/(2*g*d)    # head loss due to friction

H=Hf*100

print "Head loss = ",round(H,2),"cm of water"

R =  1299.0 is lesss than 2000 , the flow is laminar
Head loss =  3.77 cm of water


### Example 8.3 Page no 290¶

In [3]:
# Horsepower required to pump 50 tons of oil

# Given

from math import *

# oil properties

S = 0.92             # specific gravity

gma = S*62.4         # density in lbs/ft**3

nu=0.0205            # viscosity in ft**2/s

W = 50         # weight of oil

d = 9                # diameter of the pipe in inches

g = 32.2             # acceleration due to gravity in ft/s**2

# Solution

Q = W*2000/(gma*3600)       # discharge in ft**3/s

A = pi*d**2/(4*144)         # area of pipe

V = Q*1000/(A)                     # velocity in ft/s

R = V*0.75/(nu*1000)                  # Reynolds number

print "R =",round(R,2),"is less than 2000 and hence flow is laminar"

f = 64/R               # friction factor

Hf = (f*5280*(V/1000)**2)/(2*g*0.75)

Hp = gma*Q*Hf/(550)

print "Horse power required to pump the oil = ",round(Hp,1)

R = 40.07 is less than 2000 and hence flow is laminar
Horse power required to pump the oil =  10.6


### Example 8.4 Page no 291¶

In [4]:
# Viscosity of the liquid in poise

# Given

from math import *

V = 50                      # Volume in m**3

d = 5                       # diameter in m

d1 = 0.1                    # diameter of bore

l = 10                      # length of the tube

t = 20*60                   # time in seconds

rho = 0.88                  # density in g/cm**3

H1 = 5                    # height from the base in m

A = pi*d**2/4

a = pi*d1**2/4

# Solution

# From derivation we obtain a equation for T

H2 = H1-(V/A)

mu = t*rho*a*(0.1)*98.1/(32*A*10*log(H1/H2))

print "Viscosity of the liquid =",round(mu,4),"poise"

Viscosity of the liquid = 0.0182 poise


### Example 8.5 Page no 297¶

In [5]:
# Velocity distribution; Discharge ; shear on the upper plate

# Given

from math import *

# properties of kerosene oil at 20 deg C

S = 0.81                # specific gravity of oil

mu = 4*10**-5           # viscosity of oil in lb.s/ft**2

gma = 62.4*S            # density in lbs/ft**3

p1 = 6.51               # pressure at point 1 in psia

p2 = 8                  # pressure at point 2 in psia

h = 0.006               # distance between the plate in ft

l = 4                   # length of the plate in ft

theta = pi/6            # angle of inclination

# Solution

# point 1

P1 = p1*144 + gma*l*sin(theta)

# point 2

P2 = p2*144

# flow is taking from poont 2-1

Sp = (P2-P1)/4

# equation for u = 2154.75*y-359125*y**2

y = h

# discharge per ft width

q = (2154.75*y**2/2) - (359125*y**3/3)

print "Discharge q = ",round(q,3),"per unit ft of the plate"

# to find shear at the top of the plate take du/dy = 0

dV = 2154.75 - 718250*h

# shear stress

T = -mu*dV

print "Shear stress on the plate = ",round(T,3),"lbs/ft**2 and resisting the motion of the plate"

Discharge q =  0.013 per unit ft of the plate
Shear stress on the plate =  0.086 lbs/ft**2 and resisting the motion of the plate

In [ ]: