16. pipe flow iv (6.5) - lth · vvr 120 fluid mechanics 16. pipe flow iv (6.5) • smooth turbulent...

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16. Pipe flow IV (6.5)

• Smooth turbulent flow• Rough turbulent flow• Friction coefficient, Moody´s diagram• Non-circular pipesExercises: D8


Nikuradse’s experiments

• A major contribution on determining the friction factor as a functionof Reynolds number, Re, and pipe roughness

• A series of experiments where friction factor and velocity distribution were determined for various Reynolds numbers

• In the experiments, pipes were artificially roughened by stickinguniform sand grains to smooth pipes


a) δL >> kS: f = f(y0), y0 = y0(δL), δL = δL(Re) ⇒ f = f(Re)b) δL ≈ kS: f = f(y0), y0 = y0(δL, kS), ⇒ f = f(Re, kS/D)c) δL << kS: f = f(y0), y0 = y0(kS) ⇒ f = f(kS)


Thickness of laminar sublayer

Laminar sublayer thickness, δL (from measurements):

Friction velocity, v*, can be determined from:

R = radiusτ = τ0 r/R r = distance from pipe wall

τ0 = shear stress at pipe wall

*

4vLνδ =

824 0

*

20 fVv

gV

DLf

gDLhf ==⇒==

ρτ

ρτ

where ν = kinematic viscosity


Example – laminar sublayer

Water is flowing in a 100 mm pipe with an average velocityof 1 m/s. Pipe friction factor is 0.02 and kinematic viscosity,ν = 1·10-6 m2/s.

mv

smfVv

L 08.04

/05.08

*

*

==

==

νδ


ff


FORMULAS USED TO DETERMINE THE FRICTION FACTOR

Prandtl’s and von Karman’s semi-empirical laws

• For smooth pipes

• For rough pipes

Colebrook-White transition formula:

This formula is applicable to the whole turbulent region forcommercial pipes (iterative solution)

)/7.3log(21Dkf S

=

)Re

51.27.3

log(21fD

kf

S +−=

)51.2

Relog(21 ff= Eq. 6.16

Eq. 6.17

Eq. 6.18


Example – turbulent pipe flow.

200 l/s of water is pumped through a straight 300 mm pipe. The pipe has a roughness value of ks=0.3 mm and the water temperature is 20°C. Calculate the head loss for 1000 m of pipe and required pump power assuming that the end points of the pipe are at the same level.


D8 A horizontal rough pipe of 150 mm diameter carries water at 20°C. It is observed that the fall of pressure along this pipe is 184 kPa per 100 m when the flowrate is 60 l/s. What size of smooth pipe would produce the same pressure drop for the same flowrate?


NON-CIRCULAR PIPES – THE HYDRAULIC RADIUS

Although the majority of pipes have circular cross-sections, there aresome cases where one has to consider flow in rectangular or othertypes of non-circular pipes.

Head loss calculations for non-circular pipe sections are done using thehydraulic radius concept

The hydraulic radius, Rh, is defined as the area, A, of the pipe sectiondivided by the “wetted perimeter” (circumference), P.

Rh = A/P


NON-CIRCULAR PIPES – THE HYDRAULIC RADIUS (cont.)

For a circular pipe: Rh = area/wet perimeter = (πD2/4)/(πD) = D/4; or D = 4Rh

This value for the diameter may be substituted into the Darcy-Weisbachequation, Reynold’s number, and relative roughness:

Using these equations, the head loss for non-circular cross-sections may bedetermined, using an equivalent diameter for a non-circular pipe.

The hydraulic radius approach works well for turbulent flow, but not for laminarflow.

h

sh

Hf R

kDskRV

gV

RLfh

4,

4(Re,

2

2

4)

===μ

ρ


Example – non-circular pipe.

Calculate the loss of head and the pressure drop when air at an absolute pressure of 101.3 kPa flows through 600 m of a 450 mm by 300 mm smooth rectangular horizontal duct (pipe) with a mean velocity of 3 m/s. ρair= 1.225 kg/m3, dynamic viscocity μair = 1.789⋅10-5 Pa⋅s.

16. pipe flow iv (6.5) - lth · vvr 120 fluid mechanics 16. pipe flow iv (6.5) • smooth turbulent...

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