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8/11/2019 Fluid Ch.1
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Chapter (1)
Fluids and their
Properties
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Page (1)
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
= 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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
Dr.Khalil M. Al-astal Eng. Ahmed S. Al-Agha
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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Fluids and their PropertiesFluid Mechanics
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
.