BME II/II, FM: Viscous effects

FLUID MECHANICS
Chapter Six: Viscous effects
For: By:
BME II/II Raj Kumar Chaulagain
Department of Automobile and Mechanical Engineering Lec., Mechanical Engineering
Thapathali Campus, IOE, TU Department of Automobile and Mechanical Engineering
Thapathali Campus, IOE, TU

June, 2019
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 BME II/II, FM: Viscous effects

CHAPTER OVER VIEW

1. Reynold’s experiment
2. Shear stress and velocity gradient
3. Frictional loss in pipe flow
4. Laminar flow between parallel plates
5. Laminar flow in circular pipe
6. Boundary layer flow over flat plate
7. Force on immerged body
8. Drag force on flat plate due to B.L.
9. Head losses in pipe flow
a. Major loss and Moody diagram
b. Minor loss

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 BME II/II, FM: Viscous effects

1. REYNOLD’S EXPERIMENT

• Apparatus for Reynold’s experiment:

• Tank containing water at constant head
• Small tank containing liquid dye (sp. Wt. as water)
• A long and uniform glass tube with bell mouth entrance and valve at exit.
• When velocity of flow was varied by regulating the valve, following result were
obtained.
• Head loss due to friction  flow velocity
• hL v n (n varies 1.75-2)

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 BME II/II, FM: Viscous effects

1. REYNOLD’S EXPERIMENT

Laminar Flow:
• Fluid particles move along the smooth path in layers, with one layer
sliding smoothly over an adjacent layer.
Turbulent Flow:
• The fluid particles move in very irregular paths, causing an exchange of
momentum from one portion of the fluid to other portion
To determine weather the flow is laminar or turbulent Reynold gave Reynold’s no.
• Re = Inertia force / viscous force
• Re = Fi /Fv =  A v2 / (  v/D A) =  v D / 

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 BME II/II, FM: Viscous effects

1. REYNOLD’S EXPERIMENT

• Re<2,000- flow is laminar

• Re>4,000- flow is turbulent, in between is transition flow.
• In engineering practice, Re high Re medium Re low
• most flow conditions are turbulent. Conversing Straight Diverging
• The pipe diameter is much larger than 25 mm. Flow Flow Flow
• Laminar flow is found only in viscous flow. Rough Flow Smooth Flow
• Laminar cases are studied because to derive theories simple and accurate relation.

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 BME II/II, FM: Viscous effects

2. SHEAR STRESS AND VELOCITY GRADIENT

• Newton’s law of viscosity is given for one-dimensional flow by

du
 yx =
dy
• Velocity gradient is max. close to the boundary that results max. shear stress near
boundary and vice versa.

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 BME II/II, FM: Viscous effects

3. FRICTIONAL LOSS IN PIPE FLOW

• For a horizontal pipe flow condition, applying , Bernoulli’s equation:

• Frictional force = F. Resistance per unit area per unit velocity * wetted area* V2

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3. FRICTIONAL LOSS IN PIPE FLOW

f’= f/2, where f is coeff. Of friction, then

This is Darchy Weisbach equation for major loss in pipe flow.

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 BME II/II, FM: Viscous effects

4.VISCOUS FLOW BETWEEN PARALLEL PLATES

• Forces acting on fluid element

dFAB = p  y 1
p
dFCD = −( p + x)y 1
x
dFBC = −  x 1
d
dFAD = ( +  y )  x 1
dy
• Combining all forces and dividing by xy following equation is obtained.
p d
=
x dy
• This is the pressure gradient in the direction of motion of the fluid is equal to the shear gradient normal to the direction of motion.

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 BME II/II, FM: Viscous effects

4.VISCOUS FLOW BETWEEN PARALLEL PLATES

A. Velocity distribution
p d d du
= = ( ) • Applying boundary conditions
x dy dy dy y=0 u=0 and y=t u=0
1 p d 2u 1 p
== 2 c1 = − 2 ( x )t
 x dy
du 1  p  c =0
=   y + c1 2
dy   x 

u=
1  p  2
  y + c1 y + c2
u=−
1 p
( ) ty − y 2
2 x
 
2  x 

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4.VISCOUS FLOW BETWEEN PARALLEL PLATES

B. Ratio of maximum to average velocity

  t 1  p 
Flow, Q =  V  dA =    ( y 2
− ty )dy 1
0 2  x
A
 
1  p  3
Q=−  t Q 1  p  t 3 1  p  2
12  x  Average , u = = −   = −  t
A 12   x  1 t 12   x 
1 p 2
max .velocity − is − at , y = t / 2  umax =− ( )t
8 x

u max
= 3u / 2
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4.VISCOUS FLOW BETWEEN PARALLEL PLATES

C. Drop in pressure head for a given length

p p2 − p1 − p
= =
x L L
Q t 3p
=
L 12 L
12  uL
p =
t2

D. Shear stress distribution

1 p
 = − ( )t − 2 y 
2 x
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4.VISCOUS FLOW BETWEEN PARALLEL PLATES

Example 1: Determine pressure gradient, the shear stress at the two horizontal
parallel plates and the discharge per meter width for the laminar flow of oil viscosity
2.4525Ns/m2 with a maximum velocity of 2m/s between two horizontal parallel
fixed plates which are 100mm apart.

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5. VISCOUS FLOW IN CIRCULAR PIPE

▪ In laminar flow the paths of individual particles of fluid do not cross, so the flow
may be considered as a series of concentric cylinders sliding over each other –
rather like the cylinders of a collapsible pocket telescope.
▪ Lets consider a cylinder of fluid with a length L, radius r, flowing steadily in the
center of pipe.

Figure Cylindrical of fluid flowing steadily in a pipe

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5. VISCOUS FLOW IN CIRCULAR PIPE

• The fluid is in equilibrium, shearing forces equal the pressure forces.
• Shearing force = Pressure force
2rL = PA = Pr 2
P r (linear nature) (1)
=
L 2
• Taking the direction of measurement r (measured from the center of pipe), rather than the use of
y (measured from the pipe wall), the above equation can be written as;
(2)
• Equating (1) with (2) will give: = − du r
dr

P r du
= −
L 2 dr
du P r
=−
dr L 2
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 BME II/II, FM: Viscous effects

5. VISCOUS FLOW IN CIRCULAR PIPE

• In an integral form this gives an expression for velocity, with the values of r = 0 (at the pipe
center) to r = R (at the pipe wall)
P 1 r = R
u=−  rdr
L 2 r = 0

ur =−
(R 2 − r 2 ) P
(parabolic nature) (3)
4 L

where P = change in pressure

L = length of pipe
R = pipe radius
r = distance measured from the center of pipe

• The maximum velocity is at the center of the pipe, i.e. when r = 0.

R 2 P
u max = −
4 L
• It can be shown that the mean velocity is half the maximum velocity, i.e. V=Umax/2
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 BME II/II, FM: Viscous effects

5. VISCOUS FLOW IN CIRCULAR PIPE

Figure: Shear stress and velocity distribution in pipe for laminar flow
• The discharge may be found using the Hagen-Poiseuille equation, which is given by the
following; P D 4
Q=
L 128 (4)
• The Hagen-Poiseuille expresses the discharge Q in terms of the pressure gradient ,
diameter of pipe, and viscosity of the fluid.
 dP P 
 = 
 dx L 
• Pressure drop throughout the length of pipe can then be calculated by
8LQ
P = (5)
R 4
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6. BOUNDARY LAYER

• When real fluid past a solid boundary, a layer of fluid, which comes in contact
with the boundary surface, follows to it on account of viscosity.
• This layer of fluid cannot slip away from the boundary surface it attains the
same velocity as that of the boundary.
• The velocity of the flowing fluid increases gradually from zero at the boundary
surface to the velocity of the main stream. This region is known as boundary
layer.
• The large variation of velocity in a relatively small distance, exists large
velocity gradient (dv/dy) normal to the boundary surface.

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6. BOUNDARY LAYER

• Away from the boundary layer this retardation due to the presence of viscosity is
negligible and the velocity there will be equal to that of the main stream.
• The resistance due to viscosity is confined only in the boundary layer. The fluid
outside the boundary layer may be considered as ideal.

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6. BOUNDARY LAYER

• Near the leading edge of the edge of a plate, the BL is wholly laminar. For a BL
the velocity distribution is parabolic.

• Thickness of BL increases with the distance from the leading edge as more and
more fluid is slow down by the viscous boundary, become unstable and breaks
into turbulent boundary layer over a transition region.

• Thickness of boundary layer = δ (at the distance x from the edge) is the distance
from the boundary in which velocity reaches 99% velocity of the free stream.

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6. BOUNDARY LAYER

Boundary layer separation

• When pressure decreases in the direction of flow, the flow is accelerated then the pressure
gradient (dp/dx 0) both inertia and pressure force tend to reduce the effect of boundary layer
in the direction of flow.
• When pressure increases in the direction of flow (dp/dx >0), the pressure force acts opposite
to the direction of the flow and further increases the retarding effect of the viscous force.
• That increases in the thickness of the B.L. in the direction of flow.
• If these forces act over a long stretch, the boundary layer gets separated from the surface and
moves into the main stream. This phenomenon is called separation.
• The point at which the boundary layer is separate from the surface is called the point of
separation.

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 BME II/II, FM: Viscous effects

6. BOUNDARY LAYER

• A represent a point in the region of accelerated flow, with a normal velocity

distribution in the boundary layer (either laminar or turbulent)
• B is the point where the velocity outside the boundary layer reaches a maximum.
• C, D, and E are points downstream where the velocity outside the boundary layer
decreases, resulting in an increase in pressure in accordance with ideal-flow
theory.
• Thus the velocity finally brought to a stop at D.

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6. BOUNDARY LAYER

• Now the increasing pressure calls for further retardation; but this is
impossible, and so the boundary layer actually separates from the wall.
• Large turbulent eddies are formed down stream of the point of separation
the disturbed region in which eddies are formed is called Turbulent Wake.
• It is loss of energy.
• The flow separation depends upon:
• curvature of the surface
• the Reynolds' Number of flow
• the roughness of the surface

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 BME II/II, FM: Viscous effects

6. BOUNDARY LAYER

• Although the laminar and turbulent boundary layers behave in essentially the same
manner at a point of separation, the location of the separation point on a given
curved surface will be very different for the two cases.
• Fig. Turbulent wake behind a flat plate held normal to the flow

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6. BOUNDARY LAYER

Prevention of separation of Boundary layer

• Stream lining the body shape.
• Increasing the surface roughness i.e. tripping the boundary layer from laminar to turbulent.
• Injecting high velocity fluid in the boundary layer.
• Providing slots near the leading edge.
• Guidance of flow in a confine passage.
• Providing a rotation cylinder near the leading edge.
• Energizing the flow by introducing optimum amount of swirl in the incoming flow.
• Sucking the retarded flow.

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6. BOUNDARY LAYER

Effect of pressure gradient on boundary layer development

• So far, we have assumed zero pressure gradient in the flow direction across the
flat surfaces considered.
• In the curved surface, there is the presence of pressure gradient dp/dx which
effectively means a du/dx term, i.e., the flow stream velocity is seen to vary as
show in the figure.
• If pressure is decreased in downstream direction, boundary layer thickness tends
to reduce.
• If pressure is increased in downstream direction, boundary layer thickness
rapidly increase.

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7. FORCE ON IMMERSED BODY

• In engineering fields there are various problems which involve the fluid around
the submersed bodies.
• In such problems either a field may be flowing around submerged stationary body
or body may be flowing through a large mass of stationary fluid.
• Examples:
• Motion of very small objects such as sand particles in air or water.
• Large bodies such as airplane, submarines, automobiles, ships etc moving
through air or water
• Structure such as buildings and bridges etc which are submerged in air or
water.

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7. FORCE ON IMMERSED BODY

• A body wholly immersed in a real fluid may be subjected to two kinds of forces.
• Drag force : The component of force in the direction of flow on a submerged
bodies is called drag force (FD).
• Lift force : The component of force in the perpendicular to the flow is called the
lift force (FL).
• Symmetrical body moving through an ideal fluid at a uniform velocity, the
pressure distribution around a body is symmetrical and resultant (FD and FL)
force acting on the body is zero.
• Symmetrical body moving through an real fluid at a uniform velocity (FD, no
FL)

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7. FORCE ON IMMERSED BODY

• The production of lift force requires asymmetry of flow, while drag force exists
always.
• It is possible to create drag without lift but impossible to create lift without drag.
• The fluid viscosity affects the flow around the body causes the force on the body
accordingly;
• At low Reynolds' Number the fluid is deformed in very wide zone around the
body causing pressure force & friction force.
• As Reynolds' Number increases, viscous effects are confined to the boundary
layer causes predominant the friction force on the boundary.

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7. FORCE ON IMMERSED BODY

Expressions for Drag & Lift

Pressure and friction forces on in elementary surface of an immersed body

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7. FORCE ON IMMERSED BODY

• Consider a body held stationary in a stream of real fluid moving at a uniform

velocity U.
Let θ = inclination of the tangent to the small element dA with the direction of
flow.
• Then the force acting on dA of the surface of the body can be considered to have
two components, tangential component dA (called shear force).
• And normal component pdA (called pressure force) acting along the directions
tangential and normal to the surface respectively.

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7. FORCE ON IMMERSED BODY

• The summation of component of the forces acting over the entire surface of the
body in the direction of fluid flow is drag force, FD and perpendicular to fluid
flow is lift force, FL.
• FD =  pdA sin  +   dA cos 
A A
0

• FL =   dAsin  −  pdAcos
A
0
A

• Here,  pdA sin 

A
is called the pressure drag and
  dA cos 
A
0 is called the friction drag or skin / shear drag.
• The contribution of shear stresses to the lift may be neglected since shear
stresses are small as compared.

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7. FORCE ON IMMERSED BODY

Calculation of the drag and the lift forces:

• For the body moving through a fluid density  at a uniform velocity U, the mathematical
expression for the calculation of the drag and the lift forces are given by
• FD = (CD ρAV2)/2. and FL = (CL ρAV2)/2
Where, CD = coefficient of drag
CL = coefficient of lift
A = characteristics area
= area projected on a plane perpendicular to the relative motion of the
fluid, in the case of calculating FD.
= area projected on a plane perpendicular to the direction lift force, in the case
of calculating FL.

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7. FORCE ON IMMERSED BODY

Pressure Drag and Friction Drag:

• The contribution of pressure drag and friction drag to the total drag depends on
the flowing parameters.
• characteristics of fluid
• shape of body
• orientation of the body immersed in the fluid.

• When the plate is held at an angle with the direction of flow, the total drag will
be equal to the sum of pressure drag and friction drag.

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7. FORCE ON IMMERSED BODY

1. When a thin plate is placed parallel to the direction of flow the pressure drag will
be zero & the total drag is entirely due to shear stress, thus equal to friction drag.

2. When the same plate is held with its axis normal to flow direction the friction
drag will be zero. In this case the total drag is due to the pressure force only.

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7. FORCE ON IMMERSED BODY

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7. FORCE ON IMMERSED BODY

• Example 2: Experiments were conducted in a wind tunnel with a wind speed of 50km/hr on flat
plate of size 2m long and 1m wide. The density of air is 1.15kg/m3, the co-efficient of lift and drag
are 0.75 and 0.15 respectively. Determine:
a. The lift force
b. The drag force
c. Magnitude and direction of resultant force
d. Power exerted by air on the plate
Here, A=2x1= 2m2
U= 50km/hr = 13.89m/s
ρ = 1.15kg/m3
CD = 0.75 and CL = 0.15

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7. FORCE ON IMMERSED BODY

a. FL = (CL ρAV^2)/2 = 166.4 N

b. FD = (CD ρAV^2)/2 = 33.28 N
c. 𝐹𝑅 = √(𝐹 2 𝐷 + 𝐹 2 𝐿 ) = 170 N
𝜃 = 𝑡𝑎𝑛−1 (𝐹𝐿 /𝐹𝐷 ) = 78.7º
d. Power exerted by air on the plate,
P = FD . U = 462.3w (where, N m/s = w)

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8. DRAG FORCE ON FLAT PLATE DUE TO B.L.

• The drag force on the plate can be known if the velocity profile near the plate is
known.
• Shear stress is given by,
• Drag force or shear force on a small distance ∆x is given by
∆FD= 𝜏0 .∆x.b (where b is the width of plate)

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8. DRAG FORCE ON FLAT PLATE DUE TO B.L.

• The drag force must be equal to rate of change of momentum over the distance ∆x.
• For this, consider ABCD as a control volume of fluid over ∆x.
• Now mass entering through AD, =
𝛿
‫׬‬0 𝜌 ∗ 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑤𝑖𝑡ℎ𝑖𝑛 𝐵𝐿 ∗ 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑡𝑟𝑖𝑝 𝑜𝑓 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑑𝑦
𝛿
=‫׬‬0 𝜌. 𝑢. 𝑏. 𝑑𝑦
• Again mass leaving through BC,
𝜕
= Mass through AD + (mass through AD). ∆x
𝜕𝑥
𝛿 𝜕 𝛿
=‫׬‬0 𝜌. 𝑢. 𝑏. 𝑑𝑦 + (‫׬‬0 𝜌. 𝑢. 𝑏. 𝑑𝑦). ∆x
𝜕𝑥

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8. DRAG FORCE ON FLAT PLATE DUE TO B.L.

• From continuity equation, 𝑚ሶ 𝐴𝐷 + 𝑚ሶ 𝐷𝐶 = 𝑚ሶ 𝐵𝐶
• Or, mass rate entering DC, 𝑚ሶ 𝐷𝐶 = 𝑚ሶ 𝐵𝐶 − 𝑚ሶ 𝐴𝐷
𝛿 𝜕 𝛿 𝛿
=‫׬‬0 𝜌. 𝑢. 𝑏. 𝑑𝑦 + (‫׬‬0 𝜌. 𝑢. 𝑏. 𝑑𝑦). ∆x - ‫׬‬0 𝜌. 𝑢. 𝑏. 𝑑𝑦
𝜕𝑥
𝜕 𝛿
= (‫׬‬0 𝜌. 𝑢. 𝑏. 𝑑𝑦). ∆x
𝜕𝑥
Now calculating momentum flux through Control Vol.
• Momentum flux entering through AD,
δ
=‫׬‬0 momentum flux through strip of thickness dy
δ
=‫׬‬0 mass rate through strip x velocity
δ δ
=‫׬‬0 ρ. u. b. dy. u = ‫׬‬0 ρ. u2 . b. dy

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8. DRAG FORCE ON FLAT PLATE DUE TO B.L.

• Momentum flux leaving side BC,

𝛿 𝜕 𝛿
= ‫׬‬0 𝜌. 𝑢2 . 𝑏. 𝑑𝑦 +
𝜕𝑥
(‫׬‬0 𝜌. 𝑢2 . 𝑏. 𝑑𝑦). ∆x
• Momentum flux entering side DC,
= mass rate through DC x velocity
𝜕 𝛿 𝜕 𝛿
= (‫׬‬0 𝜌. 𝑢. 𝑏. 𝑑𝑦). ∆x. U = (‫׬‬0 𝜌. 𝑢. 𝑈. 𝑏. 𝑑𝑦). ∆x
𝜕𝑥 𝜕𝑥
• Rate of change of momentum of Control Vol., = ∅𝐵𝐶 − ∅𝐴𝐷 − ∅𝐷𝐶
𝛿 𝜕 𝛿 𝛿 𝜕 𝛿
= ‫׬‬0 𝜌. 𝑢2 . 𝑏. 𝑑𝑦 + (‫׬‬0 𝜌. 𝑢2 . 𝑏. 𝑑𝑦). ∆x - ‫׬‬0 𝜌. 𝑢2 . 𝑏. 𝑑𝑦 - (‫׬‬0 𝜌. 𝑢. 𝑈. 𝑏. 𝑑𝑦). ∆x
𝜕𝑥 𝜕𝑥
𝜕 𝛿
= 𝜌𝑏 {‫׬‬0 ( 𝑢2 − 𝑢𝑈)𝑑𝑦}. ∆x
𝜕𝑥

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8. DRAG FORCE ON FLAT PLATE DUE TO B.L.

• The total external force in the direction of rate of change of momentum,

= - 𝜏0 .∆x.b
• According to the momentum principle,
𝜕 𝛿
• - 𝜏0 .∆x.b = 𝜌𝑏 {‫׬‬0 ( 𝑢2 − 𝑢𝑈)𝑑𝑦}. ∆x
𝜕𝑥
𝜕 𝛿 𝜕 𝛿𝑢 𝑢
• 𝜏0 = 𝜌 {‫׬‬0 ( 𝑢𝑈 − 𝑢2 )𝑑𝑦} = 𝜌𝑈 2 {‫׬‬0 (1- ) dy}
𝜕𝑥 𝜕𝑥 𝑈 𝑈
𝜏0 𝜕 𝛿𝑢 𝑢
• = { ‫׬‬ (1- ) dy}
𝜌𝑈 2 𝜕𝑥 0 𝑈 𝑈
𝜏0 𝜕θ 𝛿𝑢 𝑢
• = [𝑎𝑠 θ = ‫׬‬0 (1- ) dy] = momentum thickness
𝜌𝑈 2 𝜕𝑥 𝑈 𝑈

• This is the equation known as Von Karman momentum integral equation for
Boundary Layer.

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8. DRAG FORCE ON FLAT PLATE DUE TO B.L.

• Example 3: A plate of 600mm length and 400mm wide is immersed in fluid of sp.
Gr. 0.9 and kinematic viscosity of 10−4 m2/s. the fluid is moving with a velocity
of 6m/s. Determine:
a. Thickness of boundary layer
b. Shear stress at the end of the plate
c. Drag force on one side of the plate
𝑈𝐿 𝜌𝑈 2 4.91𝑥
Take 𝑅𝑒 = , 𝜏0 = 0.332 . , 𝛿= , CD = 0.007
𝜇 √𝑅𝑒 √𝑅𝑒

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8. DRAG FORCE ON FLAT PLATE DUE TO B.L.

𝑈𝐿
• 𝑅𝑒 = = 3.6x10−4
𝜇
a. Thickness of boundary layer,
4.91𝑥
𝛿=
√𝑅𝑒
= 15.5mm

b. Shear stress at the end of the plate

𝜌𝑈 2
𝜏0 = 0.332 . = 56.6 N/m2
√𝑅𝑒
c. Drag force on one side of the plate,
FD = (CD ρAV2)/2 = 26.8N. (A = Lxb = 0.6x 0.4)

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9. HEAD LOSSES IN PIPES

• In steady incompressible flow in a pipe the irreversibility are

expressed in terms of a head loss or drop in EGL.
• Head loss (hl):
• It is mainly due to the major loss and minor loss.
1. Major head loss (hf):
• It is due to f, L, v, D.
2. Minor head loss :
• due to pipe joints and obstructions.

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9. HEAD LOSSES IN PIPES

A. MAJOR HEAD LOSS

Darcy Weisbach Equation:
• Darcy Weisbach determine experimentally a relation, called DWE

• hf = f L v 2 / 2 g D
• v = average velocity
• above equation is determined experimentally
• hf is determined by a manometer

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9. HEAD LOSSES IN PIPES

Friction factor
• f = f (v, D, , , , ’, m)
• f = f ({vD/}, /D, ’/D, m) all are dimensionless.
• /D = relative roughness
•  = size of the roughness projection = absolute value of roughness
• ’ = Arrangement of the roughness projection, dimension of length
• m = form factor = It depends upon the roughness of the shape of the individual
roughness element = dimensionless

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9. HEAD LOSSES IN PIPES

Smooth and rough pipe

• 1 = viscous sub layer
• Hydraulically smooth pipe if 16
• Hydraulically rough pipe if 1 0.3
• Neither completely rough nor smooth pipe if 0.3 1 6
• If irregularities (roughness) on any actual surface is such that the effect of
projection do not pierce through the various sub layer, the surface is hydraulic
smooth.
• f = f(Re), for smooth flow
• f = f (e/D, Re), for neither rough nor smooth flow
• f = f(e/D), for rough flow
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9. HEAD LOSSES IN PIPES

Moody diagram
• Moody prepared a chart that has been plotted with the aid of the help of preceding
equations. The chart obtained gives the value of the friction factor for typical pipe.
• This chart for friction factor is also called as the moody chart.

• Friction factor chart is the graphical representation of the numerical value of

friction factor for closed conduits plotted which the help of the empirical
relationships obtained for the experimental results.

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9. HEAD LOSSES IN PIPES

The chart shows that there are four Zones
• Laminar flow zone : (f = 64/Re)

• Critical range: (Where values are uncertain because the flow might be either
laminar or turbulent (f = 0.316/Re0.25)

• Transition Zone : Where f is function of both Re and e/D

• Complete turbulence zone : where f is independent of Re and depends solely upon
the relative roughness e/D. (f = F (e/D)

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9. HEAD LOSSES IN PIPES

Steps to use friction factor Chart

• Find e/D (relative roughness) from the chart (suggested by Moody) for different
materials & manufacturing processes.

• Determine Reynold’s number, Re

• Take the point of e/D line with the corresponding value of Re.

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9. HEAD LOSSES IN PIPES

Moody chart
Material Roughne
ss,
k (mm)

Glass smooth
Brass 0.002
Concrete
Smooth 0.04
Rough 2.0
CI 0.26
GI 0.15
Wrought 0.046

Steel
Commerc
ial 0.046
Riveted 3
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9. HEAD LOSSES IN PIPES

f
• In practical cases, the value of may in error by + 5% for smooth pipes and by
+10% for rough pipes.
• For old pipes values of e may be much higher, but there is much variations in the
degree with which pipe roughness increases with age.
• Equation of moody diagram for 10-6 = e/d = 10-2 and 5,000 = Re = 108

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9. HEAD LOSSES IN PIPES

Example 4: Find volume flow rate Q of water flowing on pipe of diameter 20-cm of
material with relative roughness 0.0006 when head loss found per kilometer is 12.2
m. Material Roughness,
k (mm)
Glass smooth
Brass 0.002
Concrete
Smooth 0.04
Rough 2.0
CI 0.26
GI 0.15
Wrought 0.046
Steel
Commercial 0.046
Riveted 3
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 BME II/II, FM: Viscous effects

Relative roughness same as previous problem

x

f = 0.019
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 BME II/II, FM: Viscous effects

Now use the Darcy-Weisbach equation to get V,

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9. HEAD LOSSES IN PIPES

B. MINOR HEAD LOSSES

❖Minor losses are the losses due to the local disturbances of the flow in conduits
such as changes in cross-section, projecting gaskets, elbows, valves and other
similar obstructions.

❖In case of short pipe or channel the minor losses are taken as major quantity such
as in suction line of pump with strainer and foot valves.

❖Usually these losses are insignificant in comparison with the fluid friction in the
length considered.

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9. HEAD LOSSES IN PIPES

• Minor loss in diverging flow is much larger than that in converging flow.
• Minor loss generally increases with an increase in the geometric distortion
of the flow.
• Though minor losses are usually confined to a very short length of path,
the effects may not disappear for a considerable distance downstream.
• Minor loss in laminar flow are insignificant.
• The losses may be represented either of the two ways.
• In terms of minor loss factor K.
• In terms of equivalent certain length of straight pipe (or in terms of the number
of pipe diameters)

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9. HEAD LOSSES IN PIPES

1. Loss of Head at Entrance

• The loss of head as entrance is given by he = Ke v2/2g
• Where, v is the mean velocity in the pipe and Ke is the loss coefficient.

• The entrance loss is caused primarily by the turbulence created by the enlargement
of the stream after it passes through the section of vena contracta, which is formed
immediately after the edge of the entering mouth.

• The value of Ke is much more depended on the conditions at the entrance to the
pipe.

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9. HEAD LOSSES IN PIPES

2. Contraction loss
2.1 Loss due to Sudden contraction
• The loss of head for a sudden contraction
• hc = Kc v2/2g
• suffix 1 : for upstream side, 2 for downstream.
D2 / D1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Kc 0.50 0.45 0.42 0.39 0.36 0.33 0.28 0.22 0.15 0.06 0.00

2
1
𝑘𝑐 = −1
𝐶𝑐

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9. HEAD LOSSES IN PIPES

2.2 Loss due to Gradual contraction

• hc = Kc v2/2g
• With a smoothly curved transition a loss co-efficient Kc as small as 0.05 is
possible.

• For conical reducers a minimum kc of about 0.10 is obtained with a total cone
angle of 20 to 40 degrees.

• Larger total cone angles result in higher values of kc.

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9. HEAD LOSSES IN PIPES

3.Expansion loss
3.1 Loss Due to Sudden Expansion
• The loss due to sudden expansion is greater than the loss due to a corresponding
contraction.
• In an expansion where the diverging paths of the flow encourages the formation
of eddies within the flow. And the separation of the flow from the wall of the
conduit induces pockets of eddying turbulence outside the flow region. 2 (v −v )
h = 1 2
e
2g
• But, in converging flow there is a dampening effect on eddy formation and the
conversion from pressure energy to kinetic energy is quite efficient.

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9. HEAD LOSSES IN PIPES

D1/D2 0.0 0.2 0.4 0.6 0.8
K 1.00 0.87 0.70 0.41 0.15

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9. HEAD LOSSES IN PIPES

3.2 Loss Due to Gradual Expansion:

• To reduce the expansion loss accompanying a reduction in velocity a diffuser may
be used.

• The loss of head will be some function of the angle of divergence and also of the
ratio of the two areas, the length of the diffuser being determined by these two
variables.

• The loss due to a gradual enlargement is

• hx= k’ (V1-V2)2/2g
• k’ is the function of cone angle of the diffuser.

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9. HEAD LOSSES IN PIPES

4. Losses in Pipe Fittings

• The loss of head in pipe fittings may be expressed in terms of
• h = k v2/2g
• where, v is the velocity in a pipe of the nominal size of the fitting.
• Table below shows, values of loss factors for different pipe fittings
Type K
Exit (pipe to tank) 1.0
Entrance (tank to pipe) 0.5
90 elbow 0.9
45 elbow 0.4
T-junction 1.8
6/23/2019 GateBy:valve 0.25
Lec., R. K. Chaulagain, Thapathali - 25IOE, TU
Campus, 67
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9. HEAD LOSSES IN PIPES

• In the expression hb = Kb v2/2g

• Kb Depends on
• The radius of curvature, r,
• The diameter of the pipe, D
• Angle of bend
• Kb varies according to the value of r/D ratio for smooth pipe. For rough pipes Kb
depends on both the r/D ratio and e/D ratio.
• Sharp bends result in separation downstream of the bend.
• The turbulence in the separation zone causes flow resistance.
• Greater radius of bend reduces flow resistance

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9. HEAD LOSSES IN PIPES

5. Loss in Bends and Elbows
• The centrifugal force on the particles near the centre of the pipe, where the
velocities are high, is larger than the centrifugal force on the particles near the
walls of the pipe, where the velocities are law because of this unbalanced
condition a secondary flow develops.
• This combines with the axial velocity to form a double spiral flow which persists
for some distance.
• Bend loss is not directly proportional to the angle of the bend.

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9. HEAD LOSSES IN PIPES

6. Losses due to obstruction

• Loss is due to reduction in cross section area
A = area of pipe
a = area of obstruction
Cc = contraction co-efficient
2 2
𝑉 𝐴
ℎ𝑜𝑏𝑠 = −1
2𝑔 𝐶𝑐 𝐴 − 𝑎
7. Losses at exit
𝑉2
• Loss is due to velocity at exit, ℎ𝑜 =
2𝑔

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9. HEAD LOSSES IN PIPES

8. Loss in pipeline with pump
• In case of pump, the friction head is equivalent to same added lift, so that the effect is
the same as if the pump lifted the fluid a height .
• The power delivered by the pump is given by
• preq = gQ{Δz +ΣhL}
• p (req to run)= pdel+ p
• In case of the pump with nozzle,
the received kinetic energy of the
fluid is also accounted ie
• preq = gQ{Δz +ΣhL+v2/2g}
• p (req to run)= pdel+ p
• A pump gives an abrupt rise in EGL and HGL.
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9. HEAD LOSSES IN PIPES

9. Loss in pipeline with turbine
• In case of the turbine, the power delivered to it is decreased by the friction loss in
the pipeline (penstock). And which is given by
• Pdel= gQ{Δz -ΣhL}
• p (Net del)= pdel - t

• A turbine gives an abrupt

Drop in HGL and EGL.

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9. HEAD LOSSES IN PIPES

Example 5. When a suddenly contraction is introduced in a horizontal pipe line from 50cm to 25cm,
the pressure changes from 10,500kg/m2 to 6900kg/m2. Calculate the rate of flow. Assume co-
efficient of contraction of jet to be 0.65. Following this, if there is a sudden enlargement from 25cm
to 50cm and if the pressure at the 25cm section is as previous, what is the pressure at the 50cm
enlarged section?

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9. HEAD LOSSES IN PIPES

• Solution:

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9. HEAD LOSSES IN PIPES

• Bernoulli’s equation
Including contraction loss

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9. HEAD LOSSES IN PIPES

• P4=78.5kPa

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 BME II/II, FM: Viscous effects

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