Fluid Ch.1

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Fluids and their

Properties

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Fluids and their PropertiesFluid Mechanics

Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha

Fluids (Liquidsor gases)which a substance deforms continuously, or flows,

when subjected to shearing forces.

If a fluid is at rest, there are no shearing forces acting.

In fluids we usually deal with continuous streamsof fluid without

beginning or end.

Shear Stress in Moving ParticlesIf fluid is in motion, shear stress are developed this occur if the fluidparticles move relative to each other with different velocities.

However, if the fluid velocity is the same at every point (fluid particles are at

rest relative to each other), no shear stress will be produced.

The following figure exhibit the velocity profilein a circular pipe:

Note that fluid next to the pipe wall has zerovelocity (fluid sticksto wall),

But if the fluid moved away from the wall, velocity increases to maximum.Change in velocity (v) with distance(y) is (velocity gradient):

Velocity gradient = dvdyThis also called (rate of shear strain)

v

v=f(y)

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Newtons Law of Viscosity:

= dvdy (Pa=N/m)

= shear stress.

=dynamic viscosity (will be discussed later).dvdy = velocity gradientNewtonian & Non-Newtonian FluidNewtonian Fluids:

Fluids obey ( ) Newtons law of viscosity are Newtonian fluids.

For this type of fluids, there is a linear relationshipbetween shear stress

and the velocity gradient.

Dynamic viscosity ()is the slope of the line. Dynamic viscosity ()is constant for a fluid at the same temperature. As temperature increase ()decreases slope decreases. Most common fluids are Newtonian, for example: Air, Water, Oil, etc

The following graph explains the linear variation of shear stress with rate of

shear strain (velocity gradient) for common fluids:

Slope =

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Properties of Fluids

Density:There are many ways of expressing density:

1. Mass Density (): Mass per unit volume = MV (Kg/m)Where, M = mass of fluid (Kg), V = volume of fluid (m)Typical values:

Water= 1000 Kg/m , Mercury=13546 Kg/m , Air = 1.23Kg/m2. Specific Weight(): also called (unit weight): is the weight per unit

volume.

= g = WV (N/m)Where, W = weight of fluid(N) = M g , V = volume of fluid (m)Typical values:

Water=9814 N/m , Mercury=132943 N/m , Air = 12.07Kg/m3. Relative Density(): also called (specific gravity SG):

= S G =

(Unitless)

Typical values:

Water =1 , Mercury=1.34. Specific Volume(): is the reciprocal of mass density. = 1 (m/Kg)

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Viscosity:its the ability of fluid to resist shear deformation due to

cohesion and interaction between molecules of fluid.

There are two types of viscosity:

1.

Coefficient of Dynamic Viscosity(): = dvdy = dvdy= ForceAreaVelocityDistance

= ForceAreaDistanceDistanceTime= ForceTimeArea

But Force = = = Units:

= kg ms

Another Unit: N = K g = = = . S = P a . SNote that ()is often expressed in Poise (P)where 10P=1Pa.S2. Coefficient of Kinematic Viscosity(): =

Units: = = Note that ()is often expressed in Stokes (St)where 10S t = 1 Surface Tension():is a force per unit length (N/m)

p

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R is the radius of the droplet, is the surface tension, is the pressuredifference between the inside and outside pressure.

The force developed around the edge due to surface tension along the line:

= =

The force produced from the pressure difference p: = = (Inside direction, because inside pressure largere thanoutside pressure)Now, these two forces are in equilibrium so we equating them:

2 R = p R p = 2R = p pCapillarity

Capillary effect:is the rise or fall of a liquid in a small-diameter tubecaused by surface tension.

h: is the height of water rises.

R: is the radius of the tube (d=2R).:is the angle of contactbetweenliquid and solid

The weight of the fluid is balanced

with the vertical force caused bysurface tension.W = m g = V gdcos() = 4 d h g h = 4cos()gd (only for tube)

if any thing else you should derive the equation with the same concept (see p. 12)Notes: If < 90 water is rise, If > 90 water is fall, If = 90 no rise or fall (h = 0.0). If d is too small is too small and may be neglected. For clean glass in contact with Water = 0 . 0 (rise). For clean glass in contact with Mercury = 1 3 0 (fall).

2R = d

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Compressibility & Bulk ModulusAll materials are compressible under the application of an external force.

The compressibility of a fluid is expressed by its bulk modulus (K)such

that: K = Change in Pressure = p = P PVolumetric Strain = V VV = VV K = pV/VThe ve sign refer to the volume decrease as the pressure increase.

K can be expressed by another form as following:

K = pLarge values of bulk modulus refer to incompressibility which indicates the

large pressures are needed to compress volume slightly.

Famous Unit Conversion 1m=0.3048 ft

1ft=12 inch 1inch=2.54 cm

1Ib = 4.45 N

=9810 . (.) = 62.4 Ib/ft g = 9 . 8 1 . =32.2ft/S To transform from to multiply by

v =r where,=velocity in , r = radius of rotation (m)

1slug = 14.59 Kg.

Volume of cone (V) = R h ( ( 1m = 10Liter , 1Liter=10 ml , 1m = 10ml. 1m = 10cm.

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Problems

1.

If 9 m3 of oil weighs 70.5 KN, Calculate its:

A.

Specific Weight: = WeightVolume = 70.59 =7.833 KN/m =7.83310N/m.B.Density:

(Newtons ) = g =g = 7.83310

9.81 =798.47 Kg/m.C.Specific Volume:

=1 =

1798.47 = 1 .2 5 1 0

m

/Kg

.2.A cylindrical tube of volume 335 ml is filled with soda. If the Mass of tube

when its filled by soda is 0.37 Kg and the weight of tube when its empty is

0.155 N. Determinethe specific weight, density, and specific gravity for

soda.

Solution

= WV V = V = 3 3 5 ml = 3 3 5 1 0mM = 0 .3 7 KgW = M g = 0.37 9.81 = 3.63 NW = 0 . 1 5 3 N W = W W = 3.63 0.153 = 3.47 N = 3.4733510 = 10358.2 N/m.

= g =g = 10358.29.81 = 1055.88 Kg/m.S . G = @ =

1055.881000 =1.0558.

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3.A quantity of soda of mass Mis filled in a container having volume V. The

density of soda in this container is 1005Kg/m. If the soda is distributedfrom this container to three similar containers having the same volumeV.

Calculate specific gravityand specific weightof the soda in each one of

three containers after distribution occurs.

Solution

The mass of soda (M) is the samebefore and after distribution

M = M(Solution Key) = MV M = V

V = V ( = 1 0 0 5 , V = V , =? ? , V =3V)

1 0 0 5 V = 3 V 1 0 0 5 = 3 =335 Kg/m = 335 Kg/m (For soda in each container)S . G = @ =

3351000 =0.335(For soda in each container). = g = 335 9.81 = 3286.35 N/m.

4.

A 10-kg block slides down a smooth inclined surface as shown in Figurebelow. Determine the terminal velocity of the block if the 0.1-mm gap

between the block and the surface contains oil having dynamic viscosity of

0.38 Pa.s. Assume the velocity distribution in the gap is linear, and the area

of the block in contact with the oil is 0.1 m2.

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Solution

Important Notes:

In all problems, the direction of shear force is in reverse directionofmovement, because its the resistanceof fluid against moving object. So

the shear force resists the movement of the object exactly like friction

force (if the object moves above rough surface).

The summation of forcesequal zero in all directions, because the

velocity is constant (we assume it constant) and so the acceleration is

zero.

Return to the above problem, the free body diagramof the block is:

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F = 0 .0 Wsi n(20) = A = Wsin(20)A W = Mg = 10 9.81 = 98.1 N , A = 0.1 m(given)

= 98.1sin(20)0.1 = 335.5 Pa. = vy v = y = 335.50.110

0.38 =0.0883 m/s.

5.

A cylinder of 40 cm length and 10 cm diameter rotates about a vertical axis

inside a fixed cylindrical tube of 105 mm diameter, and 0.4 m length. If the

space between the tube and the cylinder is filled with liquid of dynamic

viscosity of 0.2 N.s/m2.Determinethe external torque that led the cylinder

to rotate by speed of 700 rev. /min.

Solution

Torque = Shear force radius of rotating cylinderShear force (F) = Side Area of rotating cylinder (A) = du

dy

=0.2N.s/m du = velocity of rotating cylinder in (m/s).=700rev/min=700 260 = 73.3 rad/s.v = radius of rotating cylinder = 73.3 0.12

v = 3.66 m/s.dy = ( always is: distance perpendicular to the direction of shear force)dy = 1051002 =2.5mm=0.0025m

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= 0 . 2 3.660.0025 = 292.8 N/mA = dL = 0.1 0.4 = 0.1256 m Shear force (F) = 292.8 0.1256 = 36.7 N. Torque = 36.7 0.05 = 1.838 N. m .

6.

A cylindrical body of 70 mm diameter and 150 mm length falls freely in a80 mm diameter circular tube as shown in the figure below. If the space

between the cylindrical body and the tube is filled with oil of viscosity 0.9

poise. Determine the weight of the body when it falls at a speed of 1.5

m/s.

105 mm

100 mm

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Solution

The weight of cylindrical body (downward) equalsthe resist shear force

(upward) due to the equilibrium.

F = 0 . 0 W = AThe shear stress caused by cylindrical tube is:

= dudy = 0.9 pois, but 10pois = 1pa. s 0.9poise =. = 0.09 pa. s.

du = velocity of freely falling = 1.5m/s.

d y = 8 0 7 02 =5mm=0.005m = 0 . 0 9 1.50.005 = 27 pa.Shear force = side area of cylindrical body F = d L = 3 . 1 4 0 . 0 7 0 . 1 5 2 7

F = 0.89 NThe weight of cylindrical body=F = 0.89 N.7.A movable plate of 0.5m

2area (for each face) is located between two large

fixed plates as shown in the figure below. Two differentNewtonian fluids

having the viscosities indicated are contained between the plates.

Determine the magnitude of the force acting on the movable plate when

it moves at a speed of 4.0 m/s.

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SolutionThe free body diagram of the plate is:

F = 0 . 0 F = A + A F = A ( + )u = u =4m/s. = uy = 0 .0 1

40.003 = 13.33 Pa. = u

y=0.02 4

0.006= 13.33 Pa.

F = 0 . 5 (13.33+13.33) = 13.33 N.

F A A

A

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8.Fluid flow through a circular pipe is one-dimensional, and the velocity

profile forlaminar flow is given by u(r) = u 1 where Ris theradius of the pipe, ris the radial distance from the center of the pipe, and

uis the maximum flow velocity, which occurs at the center. Determine:(a).a relation for the drag force applied by the fluid on a section of the pipeof length L.

(b). the value of the drag force for water flow at 20C with R = 0 . 0 8 m , L =15m,u =3m/s,=0.001kg/m.s.

Solution

a)

The shear stress at the surface of the pipe (at r = R)is given by:

= dudr (Negative sign is because u decresed with r increased)

= ddru 1 rR (But,uis constant) = u ddr 1 r

R = u 0 2rR (because R is constant)

= 2ruR (at the surface of the pipe r = R) = 2uR .

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Note: Always we calculate the shear stress at the fixed surface(like pipe

surface in the above problem) because it gives maximum shear stress.

The drag force that causes shear stress on the pipe surface is:

F = 0 . 0 F = A

(A = side area of tube = 2RL) F = 2uR 2RL F =4Lu.b)F =4Lu F = 4 15 0.001 3 = 0.565 N.9.

A layer of water flows down an inclined fixed surface with the velocityprofile shown in Figure below. Determine the magnitudeof shearing stress

that the water exerts on the fixed surfacefor U= 2m/s and h=0.1m =1.1210pa. s

Solution

The shear stress at the fixedsurface (at y = 0.0)is given by:

= + dudy (Positive sign is because u incresed with y increased)uU = 2 yh y

h u = U 2 yh y

h (But,U=2m/s and h=0.1m)

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u = 2 2y0.1 y

0.1 u = 4 0 y 2 0 0 y

dudy =40400y (But, at the fixed surface y = 0.0)

dudy = 40

= 1 . 1 2 1 0 4 0 = 4 4 . 8 1 0N/m.10.When a 2-mm-diameter tube is inserted into a liquid in an open tank, the

liquid is observed to rise 10 mm above the free surface of the liquid. The

contact angle between the liquid and the tube is zero, and the specific weight

of the liquid is 1.2 x 104N/m

3. Determinethe value of the surface tension

for this liquid.

Solution

For tube the capillary rise can be expressed by the relation tha we previously

derived:

h = 4cos()gd = hgd4cos()h = 0.01m , g = = 1.2 10N/m, d = 0.002m , = 0

= 0.011.210

0.0024cos(0) = 0.06 N/m.11.Derive an expression for the capillary height change hfor a fluid of surface

tension and contact angle between twovertical parallelplatesa distanceWapart, as in figure.

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Solution

Assume the side length of each plate is (b).

F = 0 . 0 F = Weight of water column

F = 2 b c o s() (2 because there are two plates)W e i g h t = M g = V g ( V = W h b ) W e i g h t = W h b g 2 b c o s() = W h b g h = 2cos()gW .12.

Assume that the surface tension of 7.34x10-2

N/m act at an angle relativeto the water surface as shown in Figure below.

a.

If the mass of the double-edge blade is 0.64 x 10-3

Kg, and the totallength

of its sidesis 206 mm. Determine the value of required to maintainequilibrium between the blade weight and the resultant surface tension force.

b.

If the mass of the single-edge blade is 2.61 x 10-3

Kg, and the total

length of its sides is 154 mm. Explain why this blade sinks. Supportyour answer with the necessary calculations.

Solution

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a.

Note: The surface tension exists about all sides of the blade so to calculatethe surface tension force Fwe should multiply the value of by the totallength of blade sides as following:F = total length = 7.34 10 0.206 = 0.0152 N F = 0.0152 N with inclination of with horizontal The weight of the blade is:W = M g = 0 . 6 4 1 0 9.81=6.27810N (downward)By Equilibrium F =0.0

0.0152sin() =6.27810

= 2 4 . 4

.

b.

F = total length = 7.34 10 0.154 = 0.0113 NF, =0.0113sin()W = M g = 2 . 6 1 1 0 9.81=0.0256N

Now, the maximum value of sin() = 1so, the maximum valueofF, = 0.0113 N.Note that the value of W=0.0256N (downward) is greater than the valueofF, =0.0113 Nso, the blade will sink downward.13.A rigid cylinder (15 mm inside diameter) contains column of water 500mm

length. If the bulk modulus of water is Kwater= 2.05 x 109N/m

2. What will

be the column length if a 2 KN force is applied its end by frictionless

plunger? Assume no leakage.

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Solution

K = pV/VK = 2 . 0 5 1 0Pa. p = P P (P =0.0 because no applied forces in first case)P = ForceArea = 2 1 04 0.015 = 11.317 10Pa.p= 1 1 .3 1 7 1 0 0 = 11.317 10Pa.

VV = V VV = L A 0 . 5 A0 . 5 A = L 0 . 50.5

K = 2.0510 =

... L = 0.497 m

.

Fluid Ch.1

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