INTERNAL INCOMPRESSIBLE VISCOUS FLOW. Nazaruddin Sinaga Laboratorium Efisiensi dan Konservasi Energi Universitas Diponegoro. Outline. Flow Measurements. Osborne Reynolds Experiment. Laminar and Turbulent Flow. Entrance Length. Developing Flow. 2. 1. 5. Types of Flow.
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Laboratorium Efisiensi dan Konservasi Energi
where = density
= dynamic viscosity
= kinematic viscosity ( = /)
V = mean velocity
D = pipe diameter
In general, flow in commercial pipes have been found to conform to the
Laminar Flow: Re <2000
Transitional Flow : 2000 < Re <4000
Turbulent Flow : Re >4000
- In reality, because fluids are viscous, energy is lost by flowing fluids due to friction which must be taken into account.
A typical velocity distribution in a pipe flow
Cylindrical of fluid flowing steadily in a pipe
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;
Equatting (6.5) with (6.6) will give:
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.
It can be shown that the mean velocity is half the maximum velocity, i.e. V=umax/2
The discharge may be found using the Hagen-Poiseuille equation, which is given by the following;
The Hagen-Poiseuille expresses the discharge Q in terms of the pressure gradient , diameter of pipe, and viscosity of the fluid.
Pressure drop throughout the length of pipe can then be calculated by
Shear stress and velocity distribution in pipe for laminar flow
where = dynamic viscosity
= eddy viscosity which is not a fluid property but depends upon turbulence condition of flow.
where Uy is the velocity at a distance y from the wall, UCL is the velocity at the centerline of pipe, and R is the radius of pipe. This equation is known as the Prandtl one-seventh law.
Figure below shows the velocity profile for turbulent flow in a pipe. The shape of the profile is said to be logarithmic.
Velocity profile for turbulent flow
In the above equations, U represents the velocity at a distance y
from the pipe wall, U* is the shear velocity =
y is the distance form the pipe wall, k is the surface roughness and
is the kinematic viscosity of the fluid.
The momentum balance in the flow direction is thus given by
Blausius OK for smooth pipe
Abrupt inlet, K ~ 0.5
V = Q/A and A=R2
A = (0.15/2)2 = 0.01767 m2
V = Q/A =0.03/.0.01767 =1.7 m/s
Re = (1000x1.7x0.15)/(1.30x10-3) = 1.96x105 > 2000 turbulent flow
To find , use Moody Diagram with Re and relative roughness (k/D).
k/D = 0.06x10-3/0.15 = 4x10-4
From Moody diagram, 0.018
The head loss may be computed using the Darcy-Weisbach equation.
The pressure drop along the pipe can be calculated using the relationship:
ΔP=ghf = 1000 x 9.81 x 8.84
ΔP = 8.67 x 104 Pa
Da/Db = 40/100 = 0.4
From Table 6.3: K = 0.70
Thus, the head loss is
Applying Bernoulli equation between section 1 and 2
P1 = P2 = Patm = 0 (atm) and V1=V2 0 Thus equation (1) reduces to:
HL1-2 = hf + hentrance + hbend + hexit
Thus, the head added by the pump: Hp = 39.3 m
Pin = 130.117 Watt ≈ 130 kW.
Pipelines in series
Referring to Figure 6.11, the Bernoulli equation can be written between points 1 and 2 as follows;
Or we can say that the different of reservoir water level is equivalent to the total head losses in the system.
The total head losses are a combination of the all the friction losses and the sum of the individual minor losses.
HL1-2 = hfa + hfb + hentrance + hvalve + hexpansion + hexit.
Since the same discharge passes through all the pipes, the continuity equation can be written as; Q1 = Q2
Figure 6.12 Pipelines in parallel
Q1 = Qa + Qb = Q2(6.19)
HL1-2=hLa = hLb(6.20)
Continuity: Q = Q1 + Q2
Pipes in parallel: hf1 = hf2
Substitute (2) into (1)
0.8165V2 + 1.778 V2 = 7.074
V2 = 2.73 m/s
V1 = 0.8165 V2 = 0.8165x2.73 = 2.23 m/s
Q1 = 0.0063 m3/s
Recheck the answer:
Q1+ Q2 = Q
0.0063 + 0.0137 = 0.020
(same as given Q OK!)
Since HL > 0.01 m, then correction has to be applied.
Since HL ≈ 0.01 m, then it is OK.
Thus, the discharge in each pipe is as follows (to the nearest integer).
Given: Kerosene (S=0.94, m=0.048 N-s/m2). Horizontal 5-cm pipe. Q=2x10-3 m3/s.
Find: Pressure drop per 10 m of pipe.
Given: Glycerin@ 20oC flows commercial steel pipe.
Find: Estimate the elevation required in the upper reservoir to produce a water discharge of 10 cfs in the system. What is the minimum pressure in the pipeline and what is the pressure there?
Given: Commercial steel pipe to carry 300 cfs of water at 60oF with a head loss of 1 ft per 1000 ft of pipe. Assume pipe sizes are available in even sizes when the diameters are expressed in inches (i.e., 10 in, 12 in, etc.).
Assume f = 0.015
Get better estimate of f
Use a 90 in pipe
Assume f = 0.020
Given: The pressure at a water main is 300 kPa gage. What size pipe is needed to carry water from the main at a rate of 0.025 m3/s to a factory that is 140 m from the main? Assume galvanized-steel pipe is to be used and that the pressure required at the factory is 60 kPa gage at a point 10 m above the main connection.
Find: Size of pipe.
Use 12 cm pipe
Given: The 10-cm galvanized-steel pipe is 1000 m long and discharges water into the atmosphere. The pipeline has an open globe valve and 4 threaded elbows; h1=3 m and h2 = 15 m.
Find: What is the discharge, and what is the pressure at A, the midpoint of the line?
D = 10-cm and assume f = 0.025
Near cavitation pressure, not good!
So f = 0.025
Given: If the deluge through the system shown is 2 cfs, what horsepower is the pump supplying to the water? The 4 bends have a radius of 12 in and the 6-in pipe is smooth.
So f = 0.0135