Sunday 4 May 2014
Conclusion
Laminar and Turbulent Flow Summary
·
Laminar
Flow Layers of water flow over
one another at different speeds with
virtually
nomixing between layers.
The flow velocity profile for laminar flow in circular pipes is parabolic in shape, with a
maximum flow in the centre of the pipe and a minimum flow at the pipe walls.
The average
flow velocity is approximately one half of the maximum velocity.
·
Turbulent Flow
The flow is
characterized by the irregular movement of particles of the fluid.
The flow velocity
profile for turbulent flow is fairly flat across the centre section of a pipe
and drops rapidly extremely close to the walls.
The average flow
velocity is approximately equal to the velocity at the centre of the pipe.
·
Viscosity is the fluid
property that measures the resistance of the fluid to deforming due to
a shear force. For most fluids, temperature
and viscosity are inversely proportional.
·
An
ideal fluid is one that is incompressible and has no viscosity.
·
An increasing
Reynolds number indicates an increasing turbulence of flow.
Example 6
Example 6 – Determining the diameter of an air duct
Heated air at 1 atm and 35°C is to be transported in a 150-m-long circular plastic duct at a rate of 0.35 m^3/s ( Fig, 8-33 ). If the head loss in the pipe is not exceed 20 m, determine the minimum diameter of the duct.
SOLUTION :
The flow rate and the head loss in an air duct are given. The diameter of the duct is to be determined.
Assumptions :
|1. The flow is steady and incompressible.
2. The entrance effects are neglible, and thus the flow is fully developed,
3. The duct involve no components such as bends, valves, and connectors.
4. Air is an ideal gas.
5. The duct is smooth since it is made of plastic 6.The flow is turbulent ( to be verified )
Properties :
The density, dynamic viscosity and kinematic viscosity of air at 35°C are ρ = 1.145 kg/m^3 , μ = 1.895 ×〖10〗^(-5) kg/m.s, and v = 1.655 × 〖10〗^(-5) m^2/s
Analysis :
The average velocity, the Reynolds number,the friction factor, and the head loss relations are expressed as ( D is in m, V is in m/s, and Re and f are dimensionless )
V= ὐ/A_c = ὐ/(〖πD〗^2/4) = (〖0.35m〗^3/s)/(〖πD〗^2/4)
Re = VD/v = VD/(1.655 ×(〖10〗^(-5) m^2)/s)
1/√f = -2.0 log ((ε/D)/3.7 + 2.51/(Re√f)) = - 2.0 log (2.51/(Re√f))
h_L=f L/D V^2/2g →20m=f 150m/D V^2/2(9.81 m/s^2 )
D= 0.267 m f = 0.0180 V = 6.24 m/s and Re = 100,800
The diameter of the duct should be >26.7 cm if head loss is not to exceed 20m.
Re > 4000 , the turbulent flow assumption is verified.
The diameter can also be determined directly from the third Swamee-Jain formula
D = 0.66 [ ε^(1.25 )( (LV^2)/(gh_L )) + vὐ^9.4 (L/(gh_L ) )^5.2 ]^0.04
= 0.66 [ 0 + (1.655× 〖10〗^(-5 ) m^2/s)(0.35 m^3/s)^9.4(150m/((9.81 m/s^2 )(20m)) )^5.2 ]^0.04
= 0.271 m
Example 5
Types of Fluid Flow Problems
Example 5- Determining the Head
Loss in a Water pipe
Water at 15°C ( ρ = 999 kg/m^3 and μ
= 1.138 × 〖10〗^(-3 )kg/m.s ) is flowing steadily in a 5-cm-diameter
horizontal pipe made of stainless steel at a rate of 6 L/s ( Fig. 8-32 ) .
Determine the pressure drop, the head loss, and the required pumping power
input for flow over a 60-m-long section of the pipe.
SOLUTION :
The flow rate through a specified
water pipe is given. The pressure drop, the head loss, and the pumping power
requirements are to be determined.
Assumptions :
1. The flow is steady and
incompressible.
2. The entrance effects are
negligible, and thus the flow is fully developed.
3. The pipe involves no components
such as bends, valves, and connectors.
4. The piping section involves no
work devices such as a pump or a turbine
Properties :
The density and dynamic
viscosity of water are given to be ρ = 999 kg/m^3 and μ = 1.138 × 〖10〗^(-3 )kg/m.s, respectively.
Analysis :
We recognize this as a problem of
the first type, since flow rate,pipe length, and pipe diameter are known. First
we calculate the average velocity and the Reynolds number to determine the flow
regime :
V = ὐ/A_C = ὐ/(〖πD〗^2/4) = 〖0.006 m〗^3/(〖π(0.05m)〗^2/4) = 3.06 m/s
Re = ρVD/μ = (〖999 kg/m〗^3 )(3.06 m/s)( 0.05m)/(1.138 ×〖10〗^(-3) kg/(m.s))=134,300
Re >4000 , the flow is turbulent,
The relative roughness of the pipe is estimated using table 8-2
ε/D = (0.002 mm)/(50 mm) = 0.000040
determine f from the
Colebrook equation on which the Moody chart is based :
1/√f = -2.0 log
((ε/D)/3.7+2.51/(Re√f)) → 1/√f = -2.0 log (0.000040/3.7 + 2.51/(134,300
√f))
f = 0.0172 , the pressure drop (
equivalent to pressure loss in this case), head loss, and the required power
input become
∆P= ∆P_(L )=f L/D 〖ρV〗^2/2=0.0172 60m/0.05m (( 999 kg/m^3 )(3.06 m/s^2
))/2((1 N)/(1 kg.m/s^2 ))
= 96,540 N/m^2 = 96.5 kPa
h_L= (∆P_L)/ρg = f L/D V^2/2g
= 0.0172 60m/0.05m (( 3.06 m/s^(2 ) ))/(2( 9.81 m/s^2 )) = 9.85 m
w_(pump= ὐ∆P)
= ( 0.006 m^3/s)(96,540 N/m2)((1 W)/(1 N.m/s)) = 579 W
Example 4
Example 4
Determining The Flow Rate of Air in a Duct
Reconsider Example 8-4. Now the duct length is doubled while its diameter is maintained constant. If the total head loss is to remain constant, determine the drop in the flow rate through the duct.
Solution
The diameter and the head loss in an air duct are given. The drop in the flow rate is to be determined.
The average velocity, Reynolds number, friction factor, and the head loss relation are expressed as ( D is in m, V is in m/s, and Re and f are dimensionless ).
V = = → V =
Re = → Re =
hL = → 20 m =
This is set of four equations in four unknowns and solving them with an equation solver gives
V̇ = 0.24 m3/s , f = 0.0195 , V = 4.23 m/s , and Re = 68,300
Then, drop in the flow rate becomes
V̇drop = V̇old ̶ V̇new = 0.35 ̶ 0.24 = 0.11 m3/s ( a drop of 31 percent )
Therefore, for a specified head loss ( or available head or fan pumping power ), the flow rate drops by about 31percent from 0.35 to 0.24 m3/s when the duct length doubles.
Alternative Solution
The equation for V as a function of f is
Average velocity through the pipe : V =
Once V is calculated, the Reynolds number can be calculated, from which a corrected friction factor is obtained from Moody Chart. We guess f = 0.04 for illustration :
Iteration
|
f ( guess )
|
V, m/s
|
Re
|
Corrected f
|
1
|
0.04
|
2.955
|
4.724 × 104
|
0.0212
|
2
|
0.0212
|
4.059
|
6.489 ×104
|
0.01973
|
3
|
0.01973
|
4.207
|
6.727 ×104
|
0.01957
|
4
|
0.01957
|
4.224
|
6.754 ×104
|
0.01956
|
5
|
0.01956
|
4.225
|
6.756 ×104
|
0.01956
|
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