Wasit University- College of Engineering- Civil dept ... · Calculating the force on pipe bends is important to design the support system. In pipe bend the inlet and outlet velocities

Wasit University- College of Engineering- Civil dept.

Fluid Mechanic

1

Momentum Equation

The analysis of forces on vanes and pipe bends, the thrust produced by a rocket

or turbojet, and torque produced by a hydraulic turbine are all examples of the

application of the momentum equation. Application of this equation allows the

engineer to analyze forces and moments produced by flowing fluids.

When forces act on a particle, the particle accelerates according to Newton’s

second law of motion:

By definition, the mass (m) is constant, so the equation may be written using

momentum:

…….. 4.11

Depending of equation 4.11, the momentum force can be defined as the net

force acting on the fluid which equal to the change in momentum of flow per

unit time.

∑ 𝐹 = 𝜌. 𝜏∆𝑉

∆𝑡= 𝜌

∀

∆𝑡∆𝑉 = 𝜌. 𝑄. ∆𝑉

So, ∑ 𝐹 = 𝜌. 𝑄. (𝑉2 − 𝑉1) ………….4.12

1) Pipe reducer and nozzle

In pipe reducer and nozzle, the inlet and outlet velocities are in the same

direction as shown in Fig. 4.8 which represent reducer fitting in pipe line,

equation 4.12 written as:


Fluid Mechanic

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𝑃1𝐴1 − 𝑃2𝐴2 − 𝑅𝑥 = 𝜌. 𝑄. (𝑉2 − 𝑉1) , then Rx can be found.

Fig. 4.8

2) Pipe Bends

Calculating the force on pipe bends is important to design the support system. In

pipe bend the inlet and outlet velocities are in different directions. There are two

cases of pipe bend can be illustrated as below:

Case 1: pipe bend in horizontal plan

According Fig. 4.9, equation 4.12 written as:

Fig. 4.9

∑ 𝐹𝑥 = 𝑃1𝐴1 − 𝑃2𝐴2 cos 𝜃 − 𝑅𝑥 = 𝜌. 𝑄. (𝑉2 cos 𝜃 − 𝑉1) , to find Rx


Fluid Mechanic

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∑ 𝐹𝑦 = 0 − 𝑃2𝐴2 sin 𝜃 − 𝑅𝑦 = 𝜌. 𝑄. (𝑉2 sin 𝜃 − 0) , to find Ry

The resultant can be get by: 𝑅 = √𝑅𝑥2 + 𝑅𝑦

2

The resultant inclined with horizontal with angle of: ∅ = tan−1 𝑅𝑥

𝑅𝑦

Case 2: pipe bend in perpendicular plan

According Fig. 4.10, the summation of forces in x-direction is same as in case

of horizontal plan. The summations of forces in y-direction include the effect of

fluid weight in pipe bend Wf.

Fig. 4.10

∑ 𝐹𝑦 = 0 − 𝑃2𝐴2 cos 𝜃 − 𝑊𝑓 − 𝑅𝑦 = 𝜌. 𝑄. (𝑉2 sin 𝜃 − 0) , to find Ry

The weight of fluid can be founded as below:

𝑊𝑓 = 𝛾∀

the volume of fluid is a cross section area times

the length of center for the pipe bend. The length

of center line can be calculated as below:

Fig. 4.11


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For pipe bend of θ ≤ 90ᵒas shown in Fig. 4.11,

the radius for center line of bend given by:

𝐻 = 𝑅 − 𝑅 cos 𝜃 = 𝑅(1 − cos 𝜃)

∴ 𝑅 =𝐻

1 − cos 𝜃

For pipe bend of θ > 90ᵒas shown in Fig. 4.12, the radius for center line of bend

given by:

𝑦 = 𝑅 sin 𝜃

𝑅 = 𝐻 − 𝑦 = 𝐻 − 𝑅 sin 𝛼

𝐻 = 𝑅(1 + sin 𝛼)

𝑅 =𝐻

1 + 𝑅 sin 𝛼

Then, length of pipe bend centerline

is given by;

𝐿 = 𝜋𝑅𝜃

180

Volume of fluid in pipe bend given by;

∀=𝜋

3𝐿(𝑟1

2 + 𝑟1𝑟2 + 𝑟22)

Example 4.26: 300 l/s of water flow through a horizontal 300mm by 200mm

pipe bend having θ=60ᵒ. Calculate the force by the fluid on the bend and its

direction if the pressure gage reading at point 1 is 70kPa?

Fig. 4.12


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Example 4.27: 300 l/s of water flow through a perpendicular 300mm by

200mm pipe bend having θ=60ᵒ. Calculate the force by the fluid on the bend

and its direction if the pressure gage reading at point 1 is 70kPa?


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Example 4.28: 300 l/s of water flow through a vertical 300mm by 200mm pipe

bend having θ=150ᵒ. Calculate the force by the fluid on the bend and its

direction if the pressure gage reading at point 1 is 150kPa?


Fluid Mechanic

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3) Momentum force of flow through diversion

For flow through diversion shown in Fig.13, the momentum equation become as

follow:

Fig. 4.13

∑ 𝐹𝑥 = (𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑢𝑡)𝑥 − (𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑖𝑛)𝑥

∑ 𝐹𝑦 = (𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑢𝑡)𝑦 − (𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑖𝑛)𝑦

The main pipe supply each branching pipe according to its diameter as follow:

𝑄1 =𝐷1

2

𝐷12+𝐷2

2+𝐷32 × 𝑄0 , 𝑄2 =

𝐷22

𝐷12+𝐷2

2+𝐷32 × 𝑄0 , 𝑄3 =

𝐷32

𝐷12+𝐷2

2+𝐷32 × 𝑄0, ….etc.

∴ ∑ 𝐹𝑥 = ∑ 𝜌𝑄𝑉𝑥𝑜𝑢𝑡− 𝜌𝑄𝑉𝑥𝑖𝑛

𝐹𝑜 − 𝐹1 cos 𝜃1 − 𝐹2 cos 𝜃2 − 𝐹3 cos 𝜃3 − 𝑅𝑥 = 𝜌[𝑄1𝑉1 cos 𝜃1 + 𝑄2𝑉2 𝑐𝑜𝑠 𝜃2 +

𝑄3𝑉3 𝑐𝑜𝑠 𝜃3] − 𝜌[𝑄𝑜𝑉𝑜] , then Rx can be founded.

In the same way:

∴ ∑ 𝐹𝑦 = ∑ 𝜌𝑄𝑉𝑦𝑜𝑢𝑡− 𝜌𝑄𝑉𝑦𝑖𝑛


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0 − 𝐹1 sin 𝜃1 − 𝐹2 sin 𝜃2 + 𝐹3 sin 𝜃3 + 𝑅𝑦 = 𝜌[𝑄1𝑉1 sin 𝜃1 + 𝑄2𝑉2 𝑠𝑖𝑛 𝜃2 −

𝑄3𝑉3 𝑠𝑖𝑛 𝜃3] − 𝜌[0] , then Ry can be founded.

The resultant can be get by: 𝑅 = √𝑅𝑥2 + 𝑅𝑦

2

The resultant inclined with horizontal with angle of: ∅ = tan−1 𝑅𝑥

𝑅𝑦

Example 29: 500 l/s of water flowing through the main pipe of 500mm

diameter as shown in Fig. 4.13. The water is diverted into the three horizontal

pipes having diameters of D1=300mm, D2=250mm and D3=200mm. The

branching pipes inclined with horizontal by θ1= 60ᵒ, θ2=30ᵒ and θ3=45ᵒ. The

pressure in the main pipe is 300kPa. Calculate the magnitude and direction of

resultant force on the branch pipe?


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4) Hydraulic Jump

A hydraulic jump is an abrupt change from a shallow high speed flow to a deep

low speed flow of lower energy. It occurs when either the bed slope and

changes to water depth are insufficient to compensate for the high frictional

losses associated with rapid flow, or when a height differential is imposed by

upstream and downstream conditions (sluice gates, weirs, ...). Rapid flow may

be created by, for example, a steep spillway or sluice gate. The formation of a

hydraulic jump at the base of a spillway may be desirable to remove surplus

energy and reduce downstream erosion. Across a hydraulic jump:

the continuity equation for unit width of flow section;

𝑦1𝑉1 = 𝑦2𝑉2 → 𝑉2 =𝑦1

𝑦2𝑉1

the momentum for unit width of flow section;

𝐹1 − 𝐹2 = 𝜌. 𝑞. (𝑉2 − 𝑉1)

𝐹1 = 𝑃1𝐴1 =(𝛾𝑦1)𝑦1

2=

𝛾𝑦12

2 & 𝐹2 = 𝑃2𝐴2 =

(𝛾𝑦2)𝑦2

2=

𝛾𝑦22

2

𝛾

2(𝑦1

2 − 𝑦22) =

𝛾

𝑔(

𝑞2

𝑦2−

𝑞2

𝑦1) =

𝛾

𝑔(

𝑉22𝑦2

2

𝑦2−

𝑉12𝑦1

2

𝑦1) =

𝛾

𝑔(𝑉2

2𝑦2 − 𝑉12𝑦1)


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𝑔

2(𝑦1

2 − 𝑦22) = 𝑉1

2 [𝑦1

2

𝑦22 𝑦2 − 𝑦1]

𝑔

2(𝑦1 − 𝑦2)(𝑦1 + 𝑦2) = 𝑉1

2𝑦1

𝑦2

(𝑦1 − 𝑦2)

𝑉12 =

𝑔

2

𝑦2

𝑦1

(𝑦1 + 𝑦2)

The Fraud number is 𝐹𝑟 =𝑉1

𝑔𝑦1

The energy losses due the dissipation;

𝑦1 +𝑉1

2

2𝑔= 𝑦2 +

𝑣22

2𝑔+ 𝑦2 + ℎ𝑓

Example 30: water flow in a horizontal open channel at a depth of 6.0m, the

flowrate per 1m of width is 3.7m3/s. Determine is the hydraulic jump occurred

downstream or note. Then calculate the depth of water downstream and the

power dissipated in it.

4


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Solved Problems

Problem 4.10: For the Fig. shown, calculate the force on the bolt. The

discharge of water is 60 l/s.

Problem 4.11: Water flows through this nozzle at a rate of 0.43m3/s and

discharges into the atmosphere. D1= 300mm, and D2=100mm. Determine the

force required at the flange to hold the nozzle in place. Neglect gravitational

forces (Wf =0).

Problem 4.12: The gage pressure at point 1 for the vertical 90° pipe bend is 35

kPa. If the pipe diameter is 0.9m and the water flow rate is 2.83 m3/s, what

magnitude and direction of the resultant force of water on the pipe bend?


Fluid Mechanic

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Problem 4.13: For this wye fitting, which lies in a horizontal plane, the cross-

sectional areas at sections 1, 2, and 3 are 0.1m2, 0.1m

2, and 0.025m

2,

respectively. At these same respective sections the pressures are 1.5kPa,

1.35kPa, and 0kPa, and the water discharges are 0.6 m3/s to the right, 0.36 m

3/s

to the right, and exits to atmosphere at 0.24 m3/s. What x-component of force

would have to be applied to the wye to hold it in place?


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Unit 5

Real Fluid Flow

This unit describes how to classify flow in a conduit by considering (a) whether

the flow is laminar or turbulent, and (b) whether the flow is developing or fully

developed. Classifying flow is essential for selecting the proper equation for

calculating head loss.

Laminar Flow and Turbulent Flow

The basic deference between the ideal and real fluid flow is that the ideal fluid

flow is not viscous while a real fluid flow is viscous. Viscosity introduces

resistance to motion by developing shear (friction) stress between the fluid

layers and between the fluid layers and boundary. The fluid layer directly in

contact with boundary has not relative motion with boundary because the

shearing resistance in real fluid flow causes the fluid to adhere with boundary.

Reynold was first to distinguish between the kinds of flow by introduction of

dimensionless term, Re, later called Reynold’s number. Reynold number is the

ratio of inertia force to the viscous force.

𝑅𝑒 = 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒

𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒=

𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐.

𝑎𝑟𝑒𝑎 × 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠=

𝜌𝐿3𝑉𝑑𝑣𝑑𝑦

𝐿2𝜇𝑑𝑣𝑑𝑦

𝑅𝑒 =𝜌𝑉𝐿

𝜇=

𝑉𝐿

𝜈 …………….. 5.1


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Fig. 5.1: Reynolds experiments

The flow of any fluid in circular pipe will be laminar if Re < 2100, transition

flow if 2100< Re < 4000, and turbulent if Re > 4000.

Example 5.1: water of kinematic viscosity 𝑣=1.15x10-6

m2/s flow in circular

pipe of 30mm diameter. Calculate the largest flowrate for which laminar flow

can be expected.

Solution:

𝑅𝑒 =𝑉𝐷

𝜈 ∴ 2100 =

𝑉 × 0.03

1.15 × 10−6 → 𝑉 =

0.085𝑚

𝑠

Q=VA=5.7x10-5

m3/s

Example 5.2: An oil (SG=0.86) and viscosity μ=7.2x10-3

Pa.s is flowing

through a pipe of 300mm. If the flowrate change to 1.0, 6 and 15 l/s

respectively. Classify the type of flow according to Reynold’s no.


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Instability of laminar flow:

In velocity profile of pipe flow, the velocity of fluid increase in direction of

center line. So, the inertia force will increase in that direction. If the inertia

forces are large, the laminar flow becomes unstable. Rouse found that the

instability of laminar flow depend on increase in velocity gradient, increase the

mass density of fluid, increase distance from boundary and decrease the

viscosity. Depending on previous four effects and by dimensional analysis, he

found the following non dimensional parameter called 𝜒 (𝑐ℎ𝑖):

𝜒 =𝜌𝑦2𝑑𝑣

𝑑𝑦

𝜇=

𝑦2𝑑𝑣

𝑑𝑦

𝜈 …………….. 5.2

𝜒 = 0 at y=0 and at dv/dy =0, at some intermediate point 𝜒 reach the maximum

value. When 𝜒𝑚𝑎𝑥 > 500, the instability occurs at this point.

Example 5.3: Oil of ρ=850kg/m3

and μ=0.8x10-2

Pa.s. is flowing in pipe of

60mm diameter and the velocity distribution 𝑉 = 0.2 (1 −𝑟2

𝑅2). Is the flow is

laminar or turbulent according Reynold’s no.? Where is the point which

represents the maximum unstable flow in the pipe according to 𝜒 value?


Fluid Mechanic

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Turbulent Flow

Turbulent shear relations:

Turbulent cause transfers of fluid particles from layer to another. The fluid

molecular flow in heterogeneous and cause eddies. So, the velocity distribution

becomes more and more uniform. The instantaneous turbulent velocity u, v, w

composed of temporal mean velocity u, v, w and three flocculation (pulsation)

components u , v

, w

in x, y, z directions which are measure by hot wire

anemometer method. The x-component of velocity for example is given by:

u= u+ u

Fig. 5.2 turbulent flocculation in direction of flow.


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The root mean squer rms value of pulsation velocities is measure the intensity

of turbulence.

𝑟𝑚𝑠 = (𝑢 2)1/2

𝒐𝒓 𝑟𝑚𝑠 = (𝑣 2)1/2

𝒐𝒓 𝑟𝑚𝑠 = (𝑤 2)1/2

𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡𝑢𝑟𝑏𝑢𝑙𝑎𝑛𝑐𝑒 = (𝑢 2)

1/2 𝒐𝒓 (𝑣 2)

1/2 𝒐𝒓 (𝑤 2)

1/2

𝑉

In turbulent flow the shear stresses should be the sum of eddy viscosity and

dynamic viscosity.

𝜏 = (𝜇 + 𝜂)𝑑𝑣

𝑑𝑦 ……………. 5.3

Where 𝜂 in analogy called dynamic eddy viscosity. It has the same dimension

of dynamic viscosity. Kinematic eddy viscosity 𝜖 is defined as:

𝜖 =𝜂

𝜌 ………………….. 5.4

Kinematic eddy viscosity 𝜖 represents the characteristic of turbulent flow. It is

proportion with size of eddies which represent by:

Scale of turbulence l

Velocity of eddy rotation which represent by the intensity of turbulence

which measured by root mean rms value of pulsation velocities.

𝜖 = 𝑙𝑣 𝑤ℎ𝑒𝑟𝑒 𝑣 = √𝑣 2

Reynolds confirmed that the flocculation velocities caused an effective mean

shearing stress in turbulent flow and could be written as:

𝜏 = 𝜌𝑣 𝑢 …………….. 5.5

Where 𝑣 𝑢 is the mean value of product 𝑣 and 𝑢 .


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Prandtl found that:

Which mean the correlation between 𝑢 and 𝑣 depending on continuity equation

( Fig. 5.3b). 𝑙 is a distance called mixing length which varied with 𝑣 𝑎𝑛𝑑 𝑢 .

Fig. 5.3: notation of mixing length theory

So, 𝑢 = 𝑙𝑑𝑣

𝑑𝑦 𝑎𝑛𝑑 𝑣 = 𝑙

𝑑𝑣

𝑑𝑦 ……………….. 5.6

Substitute eq. 5.6 in eq. 5.5 produce:

𝜏 = 𝜌𝑙2 (𝑑𝑢

𝑑𝑦)

2 ……………….. 5.7

∵ 𝜏 = 𝜂𝑑𝑢

𝑑𝑦 then, 𝜂 = 𝜌𝑙2 𝑑𝑢

𝑑𝑦 𝑎𝑛𝑑 𝜖 = 𝑙2 𝑑𝑢

𝑑𝑦

Then Von Karman give that the mixing length depend on distribution of mean

point velocity in turbulent flow 𝑙 = 𝜅𝑦 where 𝜅 (kappa) is dimensionless

constant of turbulence having value of 0.4.


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∴ 𝜏 = 𝜌𝜅2𝑦2 (𝑑𝑢

𝑑𝑦)

2 𝑃𝑟𝑎𝑛𝑑𝑡𝑙 − 𝑉𝑜𝑛 𝐾𝑎𝑟𝑚𝑎𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ………5.7

Example 5.4: A turbulent flow of fluid (SG=0.9) occur in a pipe of 600mm

diameter. The velocity profile is measured approximately 𝑣 = 3.56𝑦1/7 where v

in (m/s) and y in (m) from the wall. The shearing stress in the fluid at y=

150mm is 12.44 Pa. Calculate the value of 𝜂, mixing length and turbulence

constant 𝜅.

Example 5.5: A turbulent flow of water occur in a pipe of 2m diameter. The

velocity profile is measured approximately 𝑣 = 10 + 0.8𝑦 . Calculate the value

of 𝜂, mixing length and turbulence constant 𝜅.


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Fig. 5.4


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Boundary-Layer Thickness and Shear-Stress Correlations

Unlike the laminar boundary layer, there is no analytically derived equation for

the thickness of the turbulent boundary layer. the resistance force expressed in

terms of the average shear-stress coefficient, Cf. The variation of Cf with

Reynolds number is shown by the solid line in Fig. 5.4. This curve corresponds

to a boundary layer that begins as a laminar boundary layer and then changes to

a turbulent boundary layer after the transition Reynolds number. This is the

normal condition for a flat-plate boundary layer.


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Fig. 5.4

Table 5.1 summarizes the equations for boundary layer- thickness, and for local

shear-stress and average shear-stress coefficients for the boundary layer on a flat

plate.


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Unit 6

Pipes Real Fluid Flow

This unit presents the Darcy-Weisbach equation, which is used for calculating

head loss in a straight run of pipe. Also, this unit present solution of practical

pipe flow problems results from application of energy principle, continuity

equation and the principles and equations of fluid resistance. Resistance to flow

in pipes is offered not only by long reaches of pipe but also by pipe pipe

fittings, such as bends and valves which is dissipate energy by producing

relatively large scale turbulence.

Pipe Head Loss

Summing of forces in the stream wise direction shown in Fig. 6.1 gives:

Fig. 6.1


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27

Next, apply the energy equation . Recognize that V1 = V2, and α1=α2.Thus, the

energy equation reduces to:

Combine Eqs. (6.2) and (6.3) and replace ∆𝐿 by L. Also, introduce a new

symbol to hf represent head loss in pipe.

Rearrange the right side of Eq.6.4

Friction factor f that gives the ratio of wall shear stress to kinetic pressure:

Combining eqs. 6.5and 6.6 gives the Darcey-Weisbach equation:


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28

Darcy-Weisbach equation is used for either laminar flow or turbulent flow and

for either round pipes or non-round conduits such as a rectangular duct.

Stress Distributions in Pipe Flow

Summing of forces in the stream wise direction shown in Fig. 6.2 gives:

Fig. 6.2


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29

Laminar Flow in a Round Tube

To derive an equation for the velocity profile in laminar flow, use Eq. (2.6)

where y is the distance from the pipe wall. Change variables by

letting where ro is pipe radius and r is the radial coordinate.

Substitute Eq. (6.11) into Eq. (6.10).

In Eq. (6.12), the left side of the equation is a function of radius r, and the right

side is a function of axial location s. This can be true if and only if each side of

Eq. (6.12) is equal to a constant. Thus,

Where ∆ℎ is the change in piezometric head over a length of ∆𝐿conduit.

Combine Eqs.(6.12) and (6.13):

Integrate Eq. (6.14):


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30

Fig. 6.3

To evaluate the constant of integration C in Eq. (6.15), apply the no-slip

condition, which states that the velocity of the fluid at the wall is zero. Thus,

Solve for C and substitute the result into Eq. (6.15):

The maximum velocity occurs at r=0:

Combine Eqs. (6.16) and (6.17):

Discharge and Mean Velocity V:To derive an equation for discharge Q,

introduce the velocity profile from Eq. (6.16) into the flow rate equation :


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31

To derive an equation for mean velocity, apply Q=AV :

Comparing Eqs. (6.20) and (6.17) reveals that:

Substitute D/2 for ro in eq. 6.20:

Head Loss and Friction Factor f: To derive an equation for head loss in a round

tube, ∆ℎ is the change in piezometric head over a length of ∆𝐿conduit between

two point along pipe.

To derive an equation for the friction factor f, combine Eq. (6.21) with the

Darcy-Weisbach equation:


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32

Equation (6.22) shows that the friction factor for laminar flow depends only on

Reynolds number.


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Turbulent Flow

The characteristics of turbulent flow, presents equations for calculating the

friction factor f, and presents a famous graph called the Moody diagram.

Turbulent flow occurs when Re > 4000. turbulent flow modeled by using an

empirical approach. This is because the complex nature of turbulent flow which

prevent from establishing a general mathematical solution. There are many

equations for shear stress and head loss in turbulent pipe flow. the logarithmic

velocity distribution is the most significant equation which is same of turbulent

boundary-layer equations (eq. 5.11).

To derive an equation for f in turbulent flow, substitute Eq. 6.23 into the

definition of mean velocity in continuity equation. After integration, algebra,

and substitute the constants to better fit experimental data, the result is:

Moody Diagram

Colebrook work by acquiring data for commercial pipes and then developing an

empirical equation, called the Colebrook-White formula, for the friction factor.

Moody used the Colebrook-White formula to generate a design chart similar to

that shown in Fig. 6.4. This chart is now known as the Moody diagram for

commercial pipes.


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34

Fig. 6.4


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35

In the Moody diagram, the variable e denote the relative roughness. In the

Moody diagram, the abscissa is the Reynolds number Re, and the ordinate is the

resistance coefficient f. To find f, given Re and e/D one goes to the right to find

the correct elative roughness curve. Then one looks at the bottom of the chart to

find the given value of Re and, with this value of Re, moves vertically upward

until the given curve is reached. Finally, from this point one move horizontally

to the left scale to read the value of f. If the curve for the given value of is not

plotted, then one simply finds the proper position on the graph by interpolation

between the curves that bracket the given e/D.

By using the Colebrook-White formula, Swamee and Jain developed an explicit

equation for friction factor:

Turbulent flow on smooth pipe

For turbulent flow on smooth Blasius has shown that when 300 < Re < 100000

may be closely approximated by line whose equation:

𝑓 =0.316

𝑅𝑒0.25

Solving Turbulent Flow Problems

The problems that involve turbulent flow in a pipe can be classify into three

cases based on the goal of the problem and based on what information is

known.


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36

Case 1 is when the goal is to find the head loss, given the pipe length, pipe

diameter, and flow rate. This problem is straightforward because it can be

solved using algebra. The solution is straightforward.

1. Calculate the mean velocity using the flow rate equation.

2. Calculate the Reynolds number.

3. Calculate the relative roughness and then look up f on the Moody

diagram.

4. Find head loss by applying the Darcy-Weisbach equation.

Case 2 is when the goal is to find the flow rate, given the head loss (or pressure

drop), the pipe length, and the pipe diameter. This problem can be solved in two

approaches:

a. direct (i.e., noniterative) solution when head loss is specified. The

strategy will be to use the horizontal scale on the top of the Moody

diagram.

1. Calculate the parameter on the top of the Moody diagram.

2. Using the Moody diagram, find the friction factor f.

3. Calculate mean velocity using the Darcy-Weisbach Equation.

4. Find discharge using the flow rate equation

b. iterative solution when V is unknown, so there is no direct way to use the

Moody diagram.

1. Apply the energy equation from section 1 to section 2.

2. First trial. Guess a value of f and then solve for V.


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37

3. Second trial. Using V from the first trial, calculate a new value of f.

4. Convergence. If the value of f is constant within a few percent

between trials, then stop. Otherwise, continue with more iterations.

5. Calculate flow rate using the flow rate equation.

Case 3 is when the goal is to find the pipe diameter, given the flow rate, length

of pipe, and head loss (or pressure drop). This problem usually requires an

iterative approach.

1. Derive an equation for pipe diameter by using the Darcy-Weisbach

equation.

2. For iteration 1, guess f, solve for pipe diameter, and then recalculate f.

3. To complete the problem, build a table in a spreadsheet program.


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41

Solved Problems

Problem 1: Oil (SG=0.9) flows in a 200mm diameter directed in horizontal

plane. It is found that the velocity profile 𝑣 = 1 −𝑟2

𝑅2 where r in m and v in m/s.

The fluid dynamic viscosity = 0.12 Pa.s. Calculate 1) the flowrate, the mean

velocity, and velocity at r =50mm. 2) the shearing stress on the pipe walland in

the oil at distance of y= 30mm. 3) the head loss for each 1000m of this pipe


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42

line. 4) the power required to move the liquid 1000m. 5) if the discharge will be

doubled, what will effect on the the above requirements.


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43

Problem 2: in the syringe in Fig. The fluid has ρ= 900kg/m3 and μ= 0.002Pa.s.

What steady force F required to produce flow of 0.4 l/s through the needle.

Problem 3: fluid flow from tank shown in Fig. at Q= 45 ft3/h. Find the

kinematic viscosity. Is the flow is laminar?


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44

Problem 4: 200 l/s of oil (SG= 0.9) flowing in a pipe of 200mm diameter. If

fluid dynamic viscosity = 1.0 Pa.s. Calculate 1) the type of flow laminar or

turbulent. 2) the head loss for each 10m of this pipe line.. 3) the shearing stress

on the pipe wall and in the oil at distance of y= 30mm 4) the velocity at a

distance of y= 30mm. 5) the maximum velocity. 6) the mean velocity.


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Problem 5: water flow through a pipe of e=0.00085ft and length of 1200ft and

diameter of 0.5ft. Find the head loss due to friction. The kinematic viscosity is

1.05 x 10-5

ft2/s and velocity 9.7 ft/s.

Problem 6: a 96in diameter pipe of e= 0.00085ft and 1000ft length carry fluid

of = 1.21x10-5

ft2/s. the head loss due to friction is 1.5 ft. what the discharge

capacity of the pipe?


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Problem 7: water is being discharged from an open from an open tank through

pipe of 24in diameter, 130ft length and e= 0.00085ft as shown in Fig. find the

flowrate when 𝑣= 1.05x10-6ft2/s.


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Problem 8: Liquid of μ= 2.92x10-4

Pa.s. and ρ=719kg/m3 is being flow through

pipe of e=0.5mm. the pressure at point 1 is 2.5 kPa as shown in Fig. find the

pipe diameter needed to discharge liquid at rate of 0.1m3/s.

ℎ𝑓 = 16.36𝑚

Trial 1: Assume f=0.02

D=0.25m

V=Q/A=2.037m/s

Re=1.25*106

e/D= 0.002

from Moody diagram f=0.0235

Trial 2: Assume f=0.0235

D=0.2582m

V=Q/A=1.909m/s

Re=1.21*106

e/D= 0.00194

from Moody diagram f=0.0235

The solution is correct. Use

D=0.252m


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Problem 9: water flow at 100mm diameter of pipe having e= 0.00026m with

velocity of 5m/s. the length of pipe is 100m. Determine the pressure drop and

the power lost due to friction when 𝑣= 1.02x10-6

m2/s.

Problem 10: Determine the discharge capacity of 150mm diameter of pipe

having e = 0.000046 to carry water (𝑣=1.02x10-6m2/s) if the pressure loss due

to friction is 35kPa. The pipe length is 100m.


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49

The Minor Losses

When fluid flows through a component such as a partially open value or a bend

in a pipe, viscous effects cause the flowing fluid to lose mechanical energy. To

characterize component head loss, engineers use the minor loss coefficient K.

where ∆ℎ is drop in piezometric head that is caused by a component, and V is

mean velocity. Most values of K are found by experiment. To find K, flow rate

is measured and mean velocity is calculated using Pressure and elevation

measurements are used to calculate the change in piezometric head.

Pipe inlet: The loss coefficient for the abrupt pipe inlet is approximately 0.5.

Other values of head loss are summarized in Fig.6.5.

Fig. 6.5

Sudden expansion: head loss coefficient for sudden expansion in pipe line

given as below:


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Sudden contraction: head loss coefficient for sudden contraction in pipe line

given as below:

Other components. The loss coefficients for a number of other fittings and

flow transitions are given in Table 6.1. The K was found by experiment.


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51

Combined Head Loss Equation

The total head loss is given by Eq. (6.27), which is:

To develop an equation for the combined head loss, substitute Eqs. 6.7and 6.26

in Eq. 6.27:

Solution:


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Example 6.6 can be solve by equivalent length method as below:


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53

Pipes in series

When pipes of different diameters or material are connected end to end to form

a pipe line, they are said to be in series. The total loss of energy (or head) will

be the sum of the losses in each pipe plus local losses at connections (minor

losses).

HL=∑ ℎ𝑓 𝑄1 = 𝑄2 = ⋯

Example 6.7: Pipes in Series Example

Consider the two reservoirs shown in figure, connected by a single pipe that

changes diameter over its length. The surfaces of the two reservoirs have a

difference in level of 9m. The pipe has a diameter of 200mm for the first 15m

(from A to C) then a diameter of 250mm for the remaining 45m (from C to B).

find Q.

For the entrance use Ke = 0.5 and the exit KE = 1.0. The join at C is sudden. For

both pipes use f = 0.01.

Solution:

Total head loss for the system HL = height difference of reservoirs

hf1 = head loss for 200mm diameter section of pipe


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54

hf2 = head loss for 250mm diameter section of pipe

hme = head loss at entry point

hmj = head loss at join of the two pipes

hmE = head loss at exit point

So

HL = hf1 + hf2 + hme + hmj + hmE = 9m …….. (1)

All losses are, in terms of Q:

Substitute these into eq.1

and solve for Q, to give Q = 0.062 m3/s


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59

Pipes in parallel

When two or more pipes in parallel connect two reservoirs, as shown in Figure,

for example, then the fluid may flow down any of the available pipes at possible

different rates. But the head difference over each pipe will always be the same.

The total volume flow rate will be the sum of the flow in each pipe. The

analysis can be carried out by simply treating each pipe individually and

summing flow rates at the end.

Q=∑ 𝑄𝑖 𝑤ℎ𝑒𝑟𝑒 𝑖 = 1,2,3, 4, … … . . ℎ𝑓1= ℎ𝑓2

= ⋯

There are two types of problems:

1. When the elevation of HGL at A and B known. The discharge is required.

In this case, the discharges added to find the Q because hf (drop of HGL)

is known.

2. Q is known as well as pipe and fluid characteristics. The distribution of

flow and head loss need to be calculated.

For this type of problem the solution will be as follow:

a. Assume discharge 𝑄1′ through pipe 1.

b. Solve for ℎ𝑓1′ using assumed discharge 𝑄1

′ .

c. Using ℎ𝑓1′ find 𝑄2

′ , 𝑄3′ , ……

d. With these discharges (𝑄1,′ 𝑄2

′ , 𝑄3′ …….) for common head loss, now ,

assume that given Q is split up among the pipes in the same proportion

as 𝑄1,′ 𝑄2

′ , 𝑄3′ …… thus;

𝑄1 =𝑄1

′

∑ 𝑄𝑖′ , 𝑄2 =

𝑄2′

∑ 𝑄𝑖′ , 𝑄3 =

𝑄3′

∑ 𝑄𝑖′ ……..

e. Check the correctness of these discharges by computing ℎ𝑓1′ , ℎ𝑓2

′ , ℎ𝑓3′

for the computed Q1, Q2, Q3, …..


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60

Example 6.13: Pipes in Parallel Example

Two pipes connect two reservoirs (A and B) which have a height difference of

10m. Pipe 1 has diameter 50mm and length 100m. Pipe 2 has diameter 100mm

and length 100m. Both have entry loss KL = 0.5 and exit loss KL=1.0 and Darcy

f of 0.032. Calculate:

a) rate of flow for each pipe

b) the diameter D of a pipe 100m long that could replace the two pipes and

provide the sameflow.

Solution

a)Apply Bernoulli to each pipe separately. For pipe 1:


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66

Branched pipes

If pipes connect more than two reservoirs, as shown in Fig. 6.5, then the

problem becomes more complex. One of the problems is that it is sometimes

difficult to decide which direction fluid will flow. For these problems it is best

to use the Darcy equation expressed in terms of discharge.

Fig. 6.5

When three or more pipes meet at a junction then the following basic principles

apply:

1. The continuity equation must be satisfy i.e. total flow into the junction must

equal total flow out of the junction;𝑄1 = 𝑄2 + 𝑄3 𝑜𝑟 𝑄3 = 𝑄1 + 𝑄2

2. at any one point there can only be one value of head, and

3. Darcy’s equation must be satisfied for each pipe.

In the figure above if the elevation H.G.L. at the junction J above z2 then flow

will be into it. While if the elevation H.G.L. at the junction J above z2 then flow

will be out of it.


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67

The algebraic solution is rather tedious so a trial and error method is usually

recommended. For example this procedure usually converges to a solution

quickly:

1. estimate a value of the H.G.L. at the junction, J

2. compute Q for each pipe.

3. check to see if continuity is (or is not) satisfied from the fourth equation

4. if the flow into the junction is too high choose a larger H.G.L. at J and vice

versa.

5. return to step 2

Example 6.19: In Fig. shown, find Q for the water with following data: use

f=0.015

Pipe Length (m) Diameter (m) e/D

1 3000 1.0 0.0002

2 600 0.45 0.002

3 1000 0.6 0.001


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68

Solution

Trial 1: Assume 𝐻𝐺𝐿𝐽 = 𝑧𝐽 +𝑃𝐽

𝛾= 23𝑚

ℎ𝑓𝑖= 𝐻𝐺𝐿𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟 − 𝐻𝐺𝐿𝐽

Pipe


1 7 1.38

2 -5 -0.278

3 -14 -0.811

∑ 𝑄 0.291

Then, the flow into the junction is high choose a larger H.G.L. at J

Trial 2: Assume 𝐻𝐺𝐿𝐽 = 𝑧𝐽 +𝑃𝐽

𝛾= 24.6𝑚

Pipe


1 5.4 1.205

2 -6.6 -0.32

3 -15.6 -0.856

∑ 𝑄 0.029


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69

Draw the linear relation between 𝑧𝐽 +𝑃𝐽

𝛾 and ∑ 𝑄

By extrapolating linearly 𝑧𝐽 +𝑃𝐽

𝛾= 24.8𝑚 and the exact solution will be:

Pipe


1 5.6 1.183

2 -6.8 -0.325

3 -15.8 -0.862

∑ 𝑄 -0.004


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70

Pipe Networks

The pipe networks, like the water distribution systems for municipalities, have

one or more sources and numerous loads: one for each household and

commercial establishment. The solution of a problem for a given layout and a

given set of sources and loads requires that two conditions be satisfied:

1. The continuity equation must be satisfied. That is, the flow into a junction of

the network must equal the flow out of the junction. This must be satisfied for

all junctions.𝑄𝑖𝑛 = 𝑄𝑜𝑢𝑡 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛

2. The algebraic sum of head losses around a given loop must be equal to zero.

The Hardy Cross method is as follows.

1. the flow distribute throughout the network so that loads at various nodes

are satisfied. In the process of distributing the flow through the pipes of

the network, the engineer must be certain that continuity is satisfied at all

junctions .

2. The first guess at the flow distribution obviously will not satisfy

requirement 2 regarding head loss; therefore, corrections are applied. For

each loop of the network, a discharge correction is applied to yield a zero

net head loss around the loop (∑ ℎ𝑓 = 0).

The Darcy Welsbach equation can be written as:

for loop in network, let Qo to be the assumed value of flowrate and Q the actual

value of flowrate. The correction to be applied is:


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71

∆𝑄 = (𝑄 − 𝑄𝑜) → 𝑄 = 𝑄𝑜 + ∆𝑄

∑ ℎ𝑓 = ∑ 𝑟𝑄2 = ∑ 𝑟(𝑄𝑜 + ∆𝑄)2 = 0

∑ ℎ𝑓 = ∑ 𝑟𝑄𝑜2 + 2∆𝑄 ∑ 𝑟𝑄𝑜 − ∑ 𝑟∆𝑄2 = 0

∆𝑄 =∑ 𝑟𝑄2

2 ∑ 𝑟𝑄=

∑ ℎ𝑓

2 ∑ℎ𝑓

𝑄

Example 6.20: solve the network shown in Fig.

Assume the distribution of flow as shown in Fig. below and find hf=rQ2 for

each pipe. Then if hf ≠0 find ∆𝑄 =∑ ℎ𝑓

2 ∑ℎ𝑓

𝑄

Loop I

line Q hf hf/Q ∆𝑄

DA 70 29400 420

∆𝑄 =∑ ℎ𝑓

2 ∑ℎ𝑓

𝑄

= 28575

2 × 675= 21.17

AC 35 3675 105

DC -30 -4500 150

sum 28575 675


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Loop II


AB 15 225 15

∆𝑄 =∑ ℎ𝑓

2 ∑ℎ𝑓

𝑄

= 2799

2 × 126.5= 11.06

AC -13.83 -574 41.5

CB -35 -2450 70

sum 2799 126.5

Repeat the solution again

Loop I


DA 48.83 14306.21 292.98

∆𝑄 =∑ ℎ𝑓

2 ∑ℎ𝑓

𝑄

= 1237.43

2 × 557.14= 1.11

AC 2.77 23.019 8.31

DC -51.17 -13091.8 255.85

sum 1237.43 557.14

Loop II


AB 26.06 679.12 26.06

∆𝑄 =∑ ℎ𝑓

2 ∑ℎ𝑓

𝑄

= 483.48

2 × 78.88= 3.06

AC -1.66 -8.27 4.98

CB -23.92 -1144.33 47.84

sum -483.48 78.88


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Unit 7

Dimensional Analysis and Similitude

Fluid mechanics is more heavily involved with experimental testing than other

disciplines because the analytical tools currently available to solve the

momentum and energy equations are not capable of providing accurate results.

For analyzing model studies and for correlating the results of experimental

research, it is essential that researchers employ dimensionless groups. Because

of size and expense it is not always feasible to carry out tests on a full-scale

prototype. Thus engineers will test a subscale model and measure the pressure

drop across the model.

Dimensions

Dimensions and Units

A dimension is the type of physical quantity. A unit is a means of assigning a

numerical value to that quantity.

Primary Dimensions

In fluid mechanics the primary or fundamental dimensions, together with their

SI units are: mass M (kilogram, kg) , length L (metre, m) and time T (second, s)

Dimensions of Derived Quantities

Dimensions of common derived mechanical quantities are given in the table 7.1.


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76

Table 7.1

Dimensional Analysis

Dimensional analysis is a means of simplifying a physical problem by

appealing to dimensional homogeneity to reduce the number of relevant

variables. Example 7.2 shows how to use the step-by-step Buckingham’s 𝝅

Theorem method.


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79

Solution


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80

Common 𝝅 Groups

Variables that have significance in a general flow field are the velocity V, the

density 𝜌, the viscosity 𝜇, and the acceleration due to gravity g. In addition, if

fluid compressibility were likely, then the bulk modulus of elasticity, K, should

be included. If there is a liquid gas interface, the surface tension effects may

also be significant. The 𝝅 groups, their symbols, and their names are

summarized in Table 7.2.


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81

Similitude

Similitude is the theory and art of predicting prototype performance from model

observations which cannot be obtained by analytical means alone. The theory of

similitude involves the application of 𝜋groups, such as the Reynolds number or

the Froude number, to predict prototype performance from model tests. In civil

engineering, model tests are always used to predict flow conditions for the

spillways of large dams. In addition, river models assist the engineer in the

design of flood-control structures as well as in the analysis of sediment

movement in the river.

Geometric Similitude means that the model is an exact geometric replica of the

prototype. Consequently, if a 1:10 scale model is specified, all linear

dimensions of the model must be 1:10 of those of the prototype.


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82

It follows that the ratio of corresponding areas between model and prototype

will be the square of the length ratio:

The ratio of corresponding volumes will be given by:

Dynamic Similitude: means that the forces that act on corresponding masses in

the model and prototype are in the same ratio(Fm/Fp) throughout the entire flow

field. For example, the ratio of the kinetic to viscous forces must be the same

for the model and the prototype. The dynamic similarity will yield similarity of

flow patterns. Consequently, the flow patterns for the model and the prototype

will be the same if geometric similitude is satisfied and if the relative forces

acting on the fluid are the same in the model as in the prototype.

The force ratios can be illustrated by considering the flow over the spillway

shown in Fig. 7.1. Here corresponding masses of fluid in the model and

prototype are acted on by corresponding forces. These forces are the force of

gravity Fg, the pressure force Fp , and the viscous resistance force Fv. These

forces add vectorially as shown in Fig. 7.1 to yield a resultant force FR, which

will in turn produce an acceleration of the volume of fluid in accordance with

Newton’s second law of motion. Hence, because the force polygons in the

prototype and model are similar, the magnitudes of the forces in the prototype

and model will be in the same ratio as the magnitude of the vectors representing

mass times acceleration:


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83

Thus the Froude number for the model must be equal to the Froude number for

the prototype to have the same ratio of forces on the model and the prototype.

Equating the ratio of the forces producing acceleration to the ratio of viscous

forces, leads to:

The same analysis can be carried out for the Mach number and the Weber

number. To summarize, if the independent 𝜋groups for the model and prototype

are equal, then the condition for dynamic similitude is satisfied.

To have complete similitude between the model and the prototype, it is

necessary to have both geometric and dynamic similitude.

Example 7.4: A cylinder 0.16m in diameter is to be mounted in a stream of

water in order to estimate the force on a tall chimney of 1m diameter which is

subject to wind of 33m/s. Calculate (A) the speed of the stream necessary to

give dynamic similarity between the model and chimney, (b) the ratio of forces.


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86

Example7.5: Explain briefly the use of the Reynolds number in the

interpretation of tests on the flow of liquid in pipes. Water flows through a 2cm

diameter pipe at 1.6m/s. Calculate the Reynolds number and find also the

velocity required to give the same Reynolds number when the pipe is

transporting air. Obtain the ratio of pressure drops in the same length of pipe for

both cases. For the water the kinematic viscosity was 1.3x10-6

m2/s and the

density was 1000 kg/m3. For air those quantities were 15.1x10-6

m2/s and

1.19kg/m3.


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Unit 8

Open Channels Flow

Open channel flow involves the flows of a liquid in a channel or conduit that is

not completely filled. There exists a free surface between the flowing fluid

(usually water) and fluid above it (usually the atmosphere). The main deriving

force is the fluid weight-gravity forces the fluid to flow downhill. Under steady,

fully developed flow conditions, the component if the weight force in the

direction of flow is balanced by the equal and opposite shear force between the

fluid and the channel surface. The main topics of this unit can be concluded as:

1. General Characteristics of Open-Channel Flow

2. Energy Considerations

3. Uniform Depth Channel Flow

Open Channel Flow vs. Pipe Flow

1. There can be no pressure force driving the fluid through the channel or

conduit.

2. For steady, fully developed channel flow, the pressure distribution within

the fluid is merely hydrostatic.

There are many examples of Open Channel Flow such as:

1. The natural drainage of water through the numerous creek and river

systems.

2. The flow in canals, drainage ditches, sewers, and gutters along roads.

3. The flow of small rivulets, and sheets of water across fields or parking

lots.


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90

The variables in Open-Channel Flow represented by:

1. Cross-sectional shape.

2. Bends.

3. Bottom slope variation.

4. Character of its bounding surface.

Classification of Open-Channel Flow

For open-channel flow, the existence of a free surface allows additional

types of flow. The extra freedom that allows the fluid to select its free-

surface location and configuration allows important phenomena in open

channel flow that cannot occur in pipe flow. So, the fluid depth, y, varies

with time, t, and distance along the channel, x, are used to classify open-

channel flow:

Classification - Type I


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91

1. Uniform flow (UF): The depth of flow does not vary along the channel

(dy/dx=0).

2. Nonuniform flows:

a. Rapidly varying flows (RVF): The flow depth changes considerably over

a relatively short distance dy/dx~1.

b. Gradually varying flows (GVF): The flow depth changes slowly with

distance dy/dx <<1.

Classification - Type II

1. Laminar flow: Re < 500.

2. Transitional flow:

3. Turbulent flow: Re > 2000.

R = ρVRh / μ

Where: V is the average velocity of the fluid. Rh is the hydraulic radius of

the channel.

Most open-channel flows involve water (which has a fairly small viscosity)

and have relatively large characteristic lengths, it is uncommon to have

laminar open-channel flows.

Classification - Type III

1. Critical Flow: Froude number Fr =1.

2. Subcritical Flow: Froude number Fr <1.

3. Supercritical Flow: Froude number Fr >1.


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92

If V=0, the wave speeds equally in all directions. If V<c, the wave can move

upstream. Such flow conditions, V<c, or Fr<1, are termed subcritical. If

V>c, Such conditions, V>c, or Fr>1, are termed supercritical. If V=c or

Fr=1, the upstream propagating wave remains stationary and the flow is

termed critical.

Energy Considerations

With the assumption of a uniform velocity profile across any section of the

channel, the one-dimensional energy equation become

hL is the head loss due to viscous effects between sections (1) and (2).


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93

x and y are taken as the distance along the channel bottom and the depth

normal to the bottom.

The slope of the channel bottom or bottom slope is constant over the

segment. Very small slop for most open-channel flows.

Specific Energy

Specific energy, E, can be defined as:

So, the energy equation can be written as:

When the head losses are negligible, Sf=0, the sum of the specific energy

and the elevation of the channel bottom remains constant.


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If the cross-sectional shape is a rectangular of width b;

q=Q/b=Vyb/b=Vy Where q is the flowrate per unit width. Then,

For a given channel, b= constant, q = constant. Then, E = E (y) is specific

energy diagram.

For a given q and E, equation (8.1) is a cubic equation with three solutions ,

ysup, ysub, and yneg.

If E >Emin, two solutions are positive and yneg is negative (has no physical

meaning and can be ignored). These two depths are term alternative depths.

Approach y=E, Very deep and very slowly.


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Approach E > Emin, two possible depths of flow, one subcritical and the

other supercritical and ysup < ysub.

Approach y=0, Very shallow and very high speed.

To determine the value of Emin

Substitute (8.2) into (8.1)

1. The critical conditions (Fr=1) occur at the location of Emin.

2. Flows for the upper part of the specific energy diagram are subcritical

(Fr<1)

3. Flows for the lower part of the specific energy diagram are supercritical

(Fr>1)


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Example 8.1 Specific Energy diagram -Qualitative

Water flows under the sluice gate in a constant width rectangular channel as

shown in Fig.a. Describe this flow in terms of the specific energy diagram.

Assume inviscid flow (Sf=0).

Inviscid flow Sf=0

Channel bottom is horizontal z1=z2 (or So=0)

E1=E2 and q1=q2

The specific energy diagram for this flow is as shown in Fig. b.

The flowrate can remain the same for this channel even if the upstream depth

is increased. This is indicated by depths y1’ and y2’ in Fig c.

To remain same flowrate, the distance between the bottom of the gate and

the channel bottom must be decreased to give a smaller flow area (y2’ < y2 ),

and the upstream depth must be increased to give a bigger head (y 1’ > y1 ).

On the other hand, if the gate remains fixed so that the downstream depth

remain fixed (y2’’ = y2 ), the flowrate will increase as the upstream depth

increases to y 1’’ > y1. Then q”>q0


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Example 8.2 Specific Energy diagram – Quantitative

Water flows up a 0.5-ft-tall ramp in a constant width rectangular channel at a

rate of q = 5.75 ft3/s/ft . If the upstream depth is 2.3 ft, determine the

elevation of the water surface downstream of the ramp, y2 + z2. Neglect

viscous effects.

Solution:

With hL=0,

The corresponding elevations of the free surface are either

Which of these flows is to be expected?


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98

This can be answered by use of the specific energy diagram obtained from

Eq.(8.2), which for this problem is

The diagram is shown in Fig. (b).

The upstream condition corresponds to subcritical flow; the downstream

condition is either subcritical or supercritical, corresponding to points 2 or

2’.

Note that since E1=E2+(z2-z1)=E2+0.5 ft, it follows that the downstream

conditions are located to 0.5 ft to the left of the upstream conditions on the

diagram.

The surface elevation is y2 + z2 = 1.72ft + 0.50ft = 2.22ft


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Uniform Depth Channel Flow

Many channels are designed to carry fluid at a uniform depth all along their

length such as irrigation canals, rivers and creeks. Uniform depth flow

(dy/dx=0) can be accomplished by adjusting the bottom slope, S0, so that it

precisely equal the slope of the energy line, Sf. A balance between the

potential energy lost by the fluid as it coasts downhill and the energy that is

dissipated by viscous effects (head loss) associated with shear stress

throughout the fluid.


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The Chezy Equation

Under the assumption of steady uniform flow, the x component of the

momentum equation

where F1 and F2 are the hydrostatic pressure forces across either end of the

control volume. P is wetted perimeter.


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101

𝜏𝑜 = 𝜆𝛾𝑉2

2𝑔

𝜆𝑉2

2𝑔= 𝑅ℎ𝑆𝑜 → 𝑉2 =

2𝑔

𝜆 𝑅ℎ𝑆𝑜 𝑤ℎ𝑒𝑟𝑒 𝜆 =

𝑓

4

𝑉 = √8𝑔

𝑓√𝑅ℎ𝑆𝑜 = 𝐶√𝑅ℎ𝑆𝑜 𝐶ℎ𝑒𝑧𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛

The Manning Equation

In 1889, R. manning (1816-1897), developed the following somewhat

modified equation for open-channel flow to more accurately describe the Rh

dependence:

Where c =1 if SI units are used, c=1.49 if USCS units are used.

n; is the Manning resistance coefficient. Its value is dependent on the surface

material of the channel’s wetted perimeter and is obtained from experiments.

It has the units of s/m1/3

or s/ft1/3

.


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The best hydraulic cross section

The best hydraulic cross section is defined as the section of minimum area

for a given flowrate Q, slope, So, and the roughness coefficient, n.

∴ 𝐴 = 𝐾 𝑃2

5⁄


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Example 8.3 Uniform Flow, Determine Flow Rate

Water flows in the canal of trapezoidal cross section shown in Fig. The

bottom drops 1.4 ft per 1000 ft of length. Determine the flowrate if the canal

is lined with new smooth concrete. Determine the Froude number for this

flow.

Solution:


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Example 8.4 Uniform Flow, Determine Flow Depth

Water flows in the channel shown in Fig. E10.3 at a rate o Q = 10.0 m3/s. If

the canal lining is weedy, determine the depth of the flow.

Solution:

Example 10.5 Uniform Flow, Maximum Flow Rate

Water flows in a round pipe of diameter D at a depth of 0 ≤ y ≤ D, as shown

in Fig. a. The pipe is laid on a constant slope of S0, and the Manning

coefficient is n. At what depth does the maximum flowrate occur? Show that

for certain flowrate there are two depths possible with the same flowrate.

Explain this behavior.


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105

A graph of flowrate versus flow depth, Q = Q(y), has the characteristic

indicated in Fig. (b).

The maximum flowrate occurs when y=0.938D, or θ=303º

Q = Qmax when y = 0.938D

Example 8.6 Uniform Flow, Effect of Bottom Slope

Water flows in a rectangular channel of width b = 10 m that has a Manning

coefficient of n = 0.025. Plot a graph of flowrate, Q, as a function of slope

S0, indicating lines of constant depth and lines of constant Froude number.


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For given value of Fr, we pick various value of y, determine the

corresponding value of So from above Eq , and then calculate Q=VA, with V

from either Manning or V=(gy)1/2

Fr. The results are indicated in Fig.

below.

Example 8.7 Uniform Flow, Variable Roughness

Water flows along the drainage canal having the properties shown in Fig. If

the bottom slope is S0 = 1 ft/500 ft=0.002, estimate the flowrate when the

depth is y = 0.8 ft + 0.6 ft = 1.4 ft.


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Example 8.8 Uniform Flow, Best Hydraulic Cross Section

Water flows uniformly in a rectangular channel of width b and depth y.

Determine the aspect ratio, b/y, for the best hydraulic cross section.


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