Chaptwer 3. Fluid Mechanics 1 (Section-I) Fluid Kinematics. P.E

All rights reserved. 2003 A. Jaafar

Chapter 3: Fluid kinematics

The Bernoulli Equation• Newton’s Second Law•F=ma along a streamline•F=ma normal to a streamline•Physical interpretations•Static, Stagnation, Dynamic and Total Pressure•Examples of use of the Bernoulli Equation•The energy line and the hydraulic grade line•Restrictions of use of the Bernoulli Equation

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Newton’s 2nd Law

The net force acting on the fluid particle must equal its mass times its acceleration

F=maFor inviscid fluid, we are assuming that the fluid motion is governed by pressure and gravity forces only

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Newton’s second law of motionAs a fluid particle moves from one location to another, it usually experiences an acceleration or deceleration. consider the motion of inviscid fluids. That is, the fluid is assumed to have zero viscosity.

StreamlinesFor steady flows each particle slides along its path, and its velocity vector is tangent to the path. The lines that are tangent to the velocity vectors throughout the flow field are called streamlines.

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Bernoulli equation states that the sum of pressure head,Velocity head, andElevation head is constant along a streamline

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Physical InterpretationAn equivalent form of the Bernoulli Equation

Hzg

Vp

2

2

Constant along a streamline

Elevation head-related to potential energy of the particle

Velocity head-vertical distance needed for the fluid to fall freely (neglecting friction) if it is to reach V from restPressure head

-height of the column of fluid that is needed to produce the pressure p

Total head

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Example 1 Some animals have learned to take advantage of the Bernoulli

effect. For example, a typical prairie dog burrow contains two entrances – a flat front door and a mounded back door. When the wind blows with velocity Vo across the front door, the average velocity across the back door is greater than Vo because of the mound. Assume the air velocity across the back door is 1.07Vo. For a wind velocity of 6 m/s, what pressure difference, p1-p2, is generated to provide a fresh air flow within the burrow.

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Static, stagnation, Dynamic and Total

Pressure TpzVp 2

2

1Constant along a streamline

Dynamic pressure

Hydrostatic pressure

Total pressure

Static pressureActual thermodynamic pressure

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Static, stagnation, Dynamic pressure


Static, stagnation, DynamicThe way to measure the static pressure would be to drill a hole in a flat surface and fasten a piezometer tube as indicated by the location of point 3 in Figure. P3=ρgh4-3

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Point 2 is a stagnation point.

ρ V2/2 is called dynamic pressure


Static, stagnation, Dynamic and Total

Pressure (cont.)Then, p2 is called the stagnation pressure

The pressure at stagnation point, p2, is greater than the static pressure, p1

There is a stagnation point on any stationary body that is placed into a flowing fluid

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Example 2Natural gas (methane) flow from a 7.6cm diameter gas main, through a 2.5cm diameter pipe and in to a burner of a furnace at a rate of 2.8m3/h. determine the pressure in the gas main if the pressure in the 2.5cm pipe is to be 15.2cm of water (take density of methane =0.667kg/m3)

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Examples of use of the Bernoulli Equation

Free Jets

Assumptionsz1=h, z2=0

Reservoir is large, V1=0

Reservoir is open to atmosphere, p1=0 gage

Fluid leaves as a free jet, p2=0 gageOnce outside nozzle, the stream continues as a free jet, p5=0 gage

Figure 3 : Vertical flow

from a tank

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Example 3Determine the flow rate from the tank

as shown in figure

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(Ans.=3.7liters/s)


Q = A2V2=0.0037m3/s


Example 4 If viscous effects are neglected and the tank is large, determine the flow rate

from the tank shown in Fig.

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Z1 = 0.7, V1 = 0,

Z2 = 0 and P2 = 0

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Datum line


Example 4Water flows through the pipe contraction shown in Figure for the given 0.2-m difference in the manometer level, determine the flow rate as a function of the diameter of the small pipe, D. (Answer=0.0156m3/s)

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Solution in the class

Q=0.1555 m3/sOpen


Water flows through the pipe contraction shown in Figure For the given 0.2-m difference in the manometer level, determine the flow rate as a function of the diameter of the small pipe, D.

Example 5

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Confined FlowsIn many cases, fluid is confined and its pressure cannot be prescribed a priori – need to use the concept of conservation of mass

Figure 5 : Steady flow into and out of a tank

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Confined Flows (cont.)

In such case, mass is conserved, i.e. inflow rate must equal to the outflow rate

In general, following Bernoulli, an increase in velocity (could be due to reduction of flow area) is accompanied by a decrease in pressureFor flows of liquids, this may result in cavitation, a potentially dangerous situation that results when liquid pressure is reduced to vapor pressure and the liquid “boils”.

ible)incompress(ifor 211222111 VAVAVAVA

AVmAVQ ,

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Example 5Water is siphoned from a large tank and discharges into the atmosphere through a 5.08 cm diameter tube as shown in Figure. The end of the tube is 0.9 m below the tank bottom, and viscous effects are negligible. (a) Determine the volume flow rate from the tank. (b) Determine the maximum height, H, over which the water can be siphoned without cavitation occurring. Atmospheric pressure is 101.3 Kpa and the water vapor pressure Is 1.8 Kpa

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Free Jets (cont.) – If exit of tank is not smooth, well contoured nozzle, the diameter of the jet will be less than the diameter of the hole – vena contracta effect– Contraction coef., Cc=Aj/Ah

holejet

Figure 3 : Typical flow patterns and contraction coef. for various round exit configurations

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Flowrate measurement Assumptions – steady, inviscid and incompressible

Figure 6 : Typical devices for measuring flowrate in pipes

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Flowrate measurement (cont.)Between points (1) and (2)

212

212

2211

222

12

212

11

1

)(2

hence

and

AA

ppAQ

VAVAQ

VpVp

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Flowrate measurement (cont.)The actual measured flowrate, Qactual will be smaller than this theoretical results because of the assumptions made in deriving the Bernoulli Equation

Other flowmeters based on Bernoulli equation are used to measure flowrates in open channels such as flumes and irrigation ditches.

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Determine the flow rate through the Venturi meter shown in figure if ideal conditions exist. (6 liters/s)

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All rights reserved. 2003 A. Jaafar= 6.1 Liters/s

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Water flows steadily from large tank and exits through a vertical, constant diameter pipe as shown in the following figure. The air in the tank is pressurized to 50 kN/m2. Assume inviscid and incompressible flow for the water. The specific weight of water is 9.8 kN/m3. Determine 1.Determine the height, h to which the water rises. 2.Calculate the water velocity in the pipe. 3.Find the pressure in the horizontal part of the pipe. 4.Propose TWO(2) changes we can make in order to decrease the height of the fountain to half of the initial height. Justify your answers


ρg = 1000x9.81=9810

SinceP1= 50,000 Pa, P2=0V1=0, V2=0Z1=2, Z2=h=???


SinceP1= 50,000 Pa, P2=0V1=0, V2=?Z1=2, Z2=4

ρg = 1000x9.81=9810


SinceP1= 50,000 Pa, P2=????V1=0, V2=7.79Z1=2, Z2=0

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Chaptwer 3. Fluid Mechanics 1 (Section-I) Fluid Kinematics. P.E

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