Laminar and
Turbulent Flow in Pipe
3/18/2024
Usama Alameedy, Ph.D.
Petroleum Engineering Dept.
University of Baghdad
2nd Class
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LAMINAR FLOW
So far, in the preceding lectures, primarily the flow of an ideal fluid has been discussed. In the case of Newtonian
fluid, the flows can be classified as (i) laminar (or viscous), and (ii) turbulent, depending on characteristic
Reynolds number
Characteristics of laminar flow:
(1) ‘No slip’ at the boundary.
(2) Due to viscosity, there is a shear between fluid layers, which is given by
⋅ for flow in X-direction.
(3) The flow is rotational.
(4) Due to viscous shear, there is continuous dissipation of energy and for maintaining the flow energy must be
supplied externally.
(5) Loss of energy is proportional to first power of velocity and first power of viscosity.
(6) No mixing between different fluid layers (except by molecular motion, which is very small).
(7) The flow remains laminar as long as is
less than critical value of Reynolds number.
Reynolds experiment
Osborne Reynolds in 1883, with the help of a simple experiment, demonstrated the existence of the following
two types of flows:
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LAMINAR FLOW
Reynolds experiment
Procedure followed:
The water was made to flow from the tank through the glass tube
into the atmosphere and the velocity of flow was varied by adjusting
valve. The liquid dye was introduced into the flow at the bell mouth
through a small tube.
Observations made:
1. When the velocity of flow was low, the dye remained in the form of a straight and stable
filament passing through the glass tube so steadily that it scarcely seemed to be in motion.
This was a case of laminar flow.
2. With the increase of velocity a critical state was reached at which the dye filament
showed irregularities and began to waver. This was a transitional state.
3. With further increase in velocity of flow the fluctuations in the filament of dye became
more intense and ultimately the dye diffused over the entire cross-section of the tube,
due to the intermingling of the particles of the flowing fluid. This was the case of a
turbulent flow.
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LAMINAR FLOW
Reynolds experiment On the basis of his experiment Reynolds discovered that:
• In case of laminar flow: The loss of pressure head ∝ velocity.
• In case of turbulent flow: The loss of head is approximately ∝ 𝑉
[More exactly the loss of head ∝ 𝑉 where n varies from 1.75 to 2.0]
Figure shows the apparatus used by Reynolds for estimating the loss of head in a pipe by measuring the pressure difference
over a known length of the pipe.
(1) The velocity of water in the pipe was determined by measuring the volume of water (Q) collected in the tank over a
known period of time (Q= VA , where A is the area of cross-section of the pipe.)
(2) The velocity of flow (V) was changed and corresponding values of hf (loss of head) were obtained.
(3) A graph was plotted between V (velocity of flow) and hf (loss of head). It may be seen from
the graph that:
(a) At low velocities the curve is a straight line, indicating that the hf (loss of head) is directly
proportional to velocity -->the flow is laminar (or viscous),
(b) At higher velocities the curve is parabolic; in this range hf α 𝑉 , where the value of n lies
between 1.75 to 2.0 --> the flow is turbulent.
(c) In the intermediate region, there is a transition zone. This is shown by dotted line.
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LAMINAR FLOW
Reynolds experiment
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It is defined as the ratio of the inertia force to the viscous force.
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FLOW OF VISCOUS FLUID IN CIRCULAR PIPES-HAGEN POISEUILLE LAW
Hagen-Poiseuille theory is based on the following assumptions:
1. The fluid follows Newton’s law of viscosity.
2. There is no slip of fluid particles at the boundary (i.e. the fluid particles adjacent to the pipe will have zero velocity).
Figure shows a horizontal circular pipe of radius R, having laminar flow
of fluid through it. Consider a small concentric cylinder (fluid element)
of radius r and length dx as a free body.
If τ is the shear stress, the shear force F is given by:
The forces acting on the fluid element are:
1. The shear force, τ × 2πr × dx on the surface of fluid element.
2. The pressure force, 𝑝 × π 𝑟 on the left end. 3. The pressure force,
on the right end.
For steady flow, the net force on the cylinder must be zero.
Eq. (1)
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Eqn. (1) shows that flow will occur only if pressure gradient exists in the direction of flow.
The negative sign shows that pressure decreases in the direction of flow.
FLOW OF VISCOUS FLUID IN CIRCULAR PIPES-HAGEN POISEUILLE LAW
• Eqn. (2) indicates that the shear stress varies linearly across the section. Its value is zero at the
center of pipe (r = 0) and maximum at the pipe wall given by:
Eq. (2)
From Newton’s law of viscosity,
In this equation, the distance y is measured
from the boundary. The radial distance r is
related to distance y by the relation:
Eq. (3)
Comparing two values of τ from eqns. 1 and 3, we have:
Integrating the above equation w.r.t. ‘r’, we get:
Eq. (4)
Eqn. (4) shows that the velocity distribution curve is a parabola (see Figure). The maximum velocity occurs at the center and is given by,
Eq. (5)
Eq. (6)
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Substitute eqn. (4) into (5), we have:
Eqn. (6) is the most commonly used equation for the
velocity distribution for laminar flow through pipes.
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FLOW OF VISCOUS FLUID IN CIRCULAR PIPES-HAGEN POISEUILLE LAW
Equation 6 can be used to calculate the discharge as follows: The discharge through an elementary ring of thickness dr at radial
distances r is given by:
Average velocity of flow,
Eq. (7)
Eqn. (7) shows that the average velocity is one-half the maximum velocity. Substituting the value of umax from eqn. (5), we have:
The pressure difference between two sections 1 and 2 at distance x1 and x2 , Is given
by
Eq. (8)
where, D is the diameter of the pipe, and L is the length. Eqn. (8) is known as the Hagen-Poiseuille equation.
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Example. An oil of viscosity 9 poise and specific gravity 0.9 is flowing through a horizontal pipe of 60 mm diameter. If the
pressure drop in 100 m length of the pipe is 1800 kN/m2, determine: (i) The rate of flow of oil; (ii) The Centre-line velocity; (iii)
The total frictional drag over 100 m length; (iv) The power required to maintain the flow; (v) The velocity gradient at the pipe
wall; (vi) The velocity and shear stress at 8 mm from the wall.
Solution
Wall shear stress,
Frictional drag for 100 m length,
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H.W
1. Oil of absolute viscosity 1.5 poise and density 848.3 kg/𝑚 flows through a 30 cm I.D. pipe. If the head loss in 3000 m length of
pipe is 20 m, assuming a laminar flow, determine (i) the velocity, (ii) Reynolds number and (iii) friction factor (Fanning’s)
2. A crude oil of viscosity 0.9 poise and relative density 0.9 is flowing through a horizontal circular pipe of diameter 120 mm and
length 12 m. Calculate the difference of pressure at the two ends of the pipe, if 785 N of the oil is collected in a tank in 25
seconds
3. A liquid with a specific gravity 2.8 and a viscosity 0.8 poise flows through a smooth pipe of unknown diameter, resulting in a
pressure drop of 800 N/ 𝑚 in 2 km length of the pipe. What is the pipe diameter if the mass flow rate is 2500 kg/h.
4. A fluid of viscosity 8 poise and specific gravity 1.2 is flowing through a circular pipe of diameter 100 mm. The maximum shear
stress at the pipe wall is 210 N/ 𝑚 . Find: (i) The pressure gradient, (ii) The average velocity, and (iii) Reynolds number of flow.
5. A fluid of density 1200 kg/ 𝑚 and viscosity 0.5 poise is flowing at a rate of 5 𝑚 /min in a circular pipe of cross-section of 1 𝑚
. Is the flow laminar or turbulent? Can you predict the maximum velocity of the fluid in the pipe ?
6. A lubricating oil of viscosity 1 poise and specific gravity 0.9 is pumped through a 30 mm diameter pipe. If the pressure drop per
metre length of pipe is 20 kN/ 𝑚 , determine: (i) The mass flow rate in kg/min, (ii) The shear stress at the pipe wall, (iii) The
Reynolds number of flow, and (iv) The power required per 50 m length of the pipe to maintain the flow
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H.W
7. Crude oil of µ = 1.5 poise and relative density 0.9 flows through a 20 mm diameter vertical pipe.
The pressure gauges fixed 20 m apart read 600 kN/ 𝑚 and 200 kN/ 𝑚 , as shown in Fig. 10.10.
Find the direction and rate of flow through the pipe.
8. A pipe 60 mm diameter and 450 m long slopes upwards at 1 in 50. An oil of viscosity 0.9 Ns/ 𝑚 and specific gravity 0.9 is
required to be pumped at the rate of 5 litres/sec. (i) Is the flow laminar? (ii) What pressure difference is required to attain
this condition? (iii) What is the power of the pump required assuming an overall efficient of 65%? (iv) What is the centre-line
velocity and the velocity gradient at pipe wall?
9. It is required to pump glycerine at the rate of 25 litres/sec. from a sump and deliver it freely at a point 120 m away and 8 m
above the level of sump through a 150 mm pipe, Fig. 10.13. (i) What is the power of the pump required assuming an overall
efficiency of 65% ? (ii) What would be the rate of rise of temperature due to viscous dissipation if the pipe is completely
insulated ? Sp. gr. of glycerine = 1.26; viscosity = 15 poise, specific heat = 248 J/N °C; K.E. correction factor, α = 2
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