Chapter (four)
fluid hydrostatic
 Fluid is a material that has the ability to flow.
 The fluids include liquids, gases and Plasma.
 Gases are compressible and take the space of any container (have no definite
volume or shape).
 Liquids are incompressible and have definite volume (take shape of container
but no definite shape).
 In this chapter we will take about fluid properties and laws in case of static.
Density and Relative density
Density (ρ): it’s the mass per unit volume of a substance.
𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝜌 =
𝑚
𝑘𝑔
𝑔
𝑔
, 𝒖𝒏𝒊𝒕𝒔 3 𝑜𝑟
𝑜𝑟
, 𝑫𝒊𝒎𝒔𝒔𝒊𝒐𝒏 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 𝑀. 𝐿−3 .
3
𝑉
𝑚
𝑐𝑚
𝑙𝑖𝑡
Density depend on the type of the material (atomic weight, molecular spacing)
and temperature only.
Relative Density or specific density or specific weight or specific gravity: it’s the
ratio between the density of the material to the density of the water at same
temperature.
𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝑖𝑠
𝑜𝑟
𝑚𝑠
𝑠𝑎𝑚𝑒 𝑣𝑜𝑙𝑢𝑚𝑒, 𝑠𝑎𝑚𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒
𝑚𝑤
𝑉𝑤
𝜌𝑠
𝑠𝑎𝑚𝑒 𝑚𝑎𝑠𝑠, 𝑠𝑎𝑚𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑟
𝑠𝑎𝑚𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑡𝑎𝑢𝑟𝑒
𝑉𝑠
𝜌𝑤
Relative has no unit or dimensional formal because it’s a ratio between same
physical quantities; (Unit less – dimensionless).
The density of water = 1000
𝑘𝑔
𝑚3
=1
𝑔
𝑐𝑚3
. so the numerical value of relative density
of substance = the numerical value of its density in unit of
𝑔
𝑐𝑚3
.
Application of density:
1
Measuring the density of electrolytic solution of the car’s battery:
During discharge, the sulphuric acid (concentrate H2So4) during its reaction with
lead plates forming lead Sulphate, the density of the electrolytic solution
decreases (Dilute H2So4).
On recharging, the sulphate radical is released from the lead plates back to the
solution increase its density (conc. H2So4)
So we can identify the charging level of the battery by measuring the density of its
electrolytic solution.
Measuring density of the blood and urine in clinical medicine:
The blood density in normal person ranges from 1040
If the blood density below 1040
𝑘𝑔
𝑚3
𝑘𝑔
𝑘𝑔
𝑚
𝑚3
3 to 1060
.
, it refers to Anemia disease.
The normaldensity of urine is 1020
𝑘𝑔
𝑚3
. in some diseases this value may increase
due t the increasing if the secretion of salt.
Pressure
The pressure at a point (p):- it’s the average normal force acts on a unit surface
area around the point.
𝐹
𝑊
𝐴
𝑉
𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝑝 = Where F is force (N), A area (𝑚2 ). Or 𝑝 =
volume𝑫𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏𝒂𝒍 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝑀. 𝐿−1 . 𝑇 −2 , 𝒖𝒏𝒊𝒕𝒔
. W work, V
𝑁
𝐽
𝑚
𝑚3
2 , 𝑃𝑎,
… … . . 𝑒𝑡𝑐.
Pressure at point inside a liquid (fluid):
Suppose that a horizontal plate (X) of Area (A) in 𝑚2 is placed at a depth (h) in m
below the surface of a liquid of density (ρ) in
𝑘𝑔
𝑚3
.
The force (F) exerts by the liquid column on the plate equal its weight.
2
𝑊 =𝑚∗𝑔 =𝜌∗𝑉∗𝑔 =𝜌∗𝐴∗ℎ∗𝑔
𝑃=
𝑊
𝐴
=
𝜌∗𝑔∗𝐴∗ℎ
𝐴
= 𝜌 ∗ 𝑔 ∗ ℎ(the liquid pressure)
𝑃 = 𝑃𝑎 + 𝜌 ∗ 𝑔 ∗ ℎ(absolute pressure)
Liquid pressure: it’s the pressure exerts by a liquid at a point inside it. It is equal to
the weight of the liquid column whose cross section area is 1 𝑚2 at this point and
its height is the vertical distance between that point and the free liquid surface.
Items
Graph
Equation
Slope
Int.Y-axis
Conclusion:
Liquid Pressure
Absolute pressure(total pressure)
𝑃 =𝜌∗𝑔∗ℎ
𝜌∗𝑔
Zero
𝑃 = 𝑃𝑎 + 𝜌 ∗ 𝑔 ∗ ℎ
𝜌∗𝑔
Pa
The pressure is the same at same
horizontal level inside same fluid.
The liquid pressure doesn’t depend on
the area.
Notes:
The liquid pressure acting on the base of the tank doesn’t depend on its area.
Because the pressure is the normal force acting on the unit area, and Area is not a
factor the pressure depends on according to 𝑃 = 𝜌 ∗ 𝑔 ∗ ℎ.
The pressure at any point in a liquid acts in all direction with same magnitude
Becausethe pressure is the normal force acting on the unit area.
3
The pressure acting on one of the walls of a liquid container =
𝑷𝟏+𝑷𝟐
𝟐
.
Where P1, P2 the pressure of the start and end point.
P, h relationships
Items
Dir.h
From the surface
From the bottom
Graph
Relation
Pa
Slope
Pa=0
Direct
Int. Y axis
𝜌∗𝑔
Inverse
The end point (max h)
𝜌∗𝑔
The atmospheric pressure
The atmospheric pressure: it’s the weight of the air column over unit area of the
earth’s surface; Or the pressure occurring due to the weight of mercury column of
height 76 cm at S.T.P. Or the weight of mercury column with height =76 cm, cross
section area of 1𝑚2 at 0̊C at sea level (S.T.P). (The density of mercury at 0 ⁰C is
13595
𝑘𝑔
𝑚3
𝑚
and gravity of 9.8 2 .).
𝑠
S.T.P: Standard temperature and pressure (0̊C, Sea level)
Measuring the atmospheric pressure: Mercury Barometer
4
It consists of a glass tube of length one meter with a unit cross section and
completely filled with mercury
When the tube is inverted in a dish filled with mercury the surface of the mercury
in the tube falls to a certain level.
Torricelli vacuum: it’s the zone above mercury level in
barometer tube its pressure is nearly zero.
The atmospheric pressure acting to up the mercury and
the weight of mercury acting to down it.
When two forces are equals the mercury will stop
moving and the its height represents the atmospheric
pressure (cm.Hg)
𝑃𝑎 = 𝜌𝑔ℎ + 𝑍𝑒𝑟𝑜 = 0.76 ∗ 9.81 ∗ 13595
𝑁
1 𝑎𝑡𝑚 = 1.013 ∗ 105 2
𝑚
Notes:
The mercury surface in the barometric tube keeps
having the same height above the free mercury
surface in the dish whatever the inclination of that
tube.
As the inclination of the barometric tube increase
the Torricelli vacuum decrease till it disappeared
when the vertical height of the closed end ≤Pa (76
cm.Hg) above the free surface.
5
Pa decrease with the increase height from the
earth’s surface so the reading of the barometer will
decrease too, the height of the building or
mountain can be measured from the following:
𝜌𝑎𝑖𝑟 ∗ 𝑔 ∗ ℎ = 𝜌ℎ𝑔 ∗ 𝑔 ∗ ℎ2 − 𝜌ℎ𝑔 ∗ 𝑔 ∗ ℎ1
𝜌ℎ𝑔
ℎ=
∗ (ℎ2 − ℎ1)
𝜌𝑎𝑖𝑟
We use mercury instead of water due to the high
density as the height is inversely proportional with
the density to make the process of measurement
easier.
Application on Pressure:
Measuring the blood pressure:
There are two values have to be measured,
The first one is Systolic pressure, max pressure; the contraction of the heart, the
blood is pushed from the left ventricle to the Aorta then to arteries.
The second one is Diastolic pressure, min pressure, the relaxation of the heart,
blood in arteries.
Normal people pressure values are 120 Torr-80 Torr
For the sick person there is a disturbance in the rate of the blood flow combined
by flutter (noisy sound) heard by the doctor stethoscope.
Adjusting the car tyres pressure:
If the pressure of the tyres is high enough, well inflated, the area of the tyres, in
contact with road, will decrease decreasing the friction force between tyres and
road.
But if the pressure is not enough, underinflated, the area will increase increasing
the friction force. Force = pressure * Area.
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U-Shaped tube
U-shaped tube used to calculate the Density& Specific density for several
materials using the liquid pressure equation.
Operation steps:(immiscible liquid)
1- Pour enough amount of water in the U-tube fixed in a vertical position;water
rises in the two arms at same height.
2- Pour our liquid (oil) in one arm (side); the water will rise in the other arm
(moves down in the oil arm volume is constant) due to the difference in the
pressure.
3- The pressure is the same in same horizontal level (bottom of the oil is ours
reference separating surface between oil and water in the oil arm).
4- Measure h1, h2
5- By knowing the density of water we can calculate the density of the other
liquid (oil) by substitute in thisequation ℎ1 ∗ 𝜌1 = ℎ2 ∗ 𝜌2.
Another way (miscible liquid):
1- Pour enough amount of mercury into the U tube
2- Pour Alcohol in one arm and water in the other arm till the surface of
mercury in both arms is at the same level (Horizontal).
3- Measure h1, h2 staring from the separating surface between the mercury
and the other liquid (water & Alcohol).
4- Knowing the density of water; apply our pressure rules ℎ1 ∗ 𝜌1 = ℎ2 ∗ 𝜌2.
We can also calculate the Relative Density from the above steps using same
equation.𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒅𝒆𝒏𝒔𝒊𝒕𝒚 =
𝝆𝟏
𝝆𝟐
=
𝒉𝟐
𝒉𝟏
.
Note: the volume is the same even the area in changing and the pressure doesn’t
depend on the area only height.
1- 𝐯𝐨𝐥𝐮𝐦𝐞: A ∗ x = B ∗ y, 𝑆𝑜 𝒚 =
𝑨
𝑩
∗ 𝒙.Where A, B is the cross section area of
the U tube arms, x is the downward step, y is the upward step; so 𝒉 = 𝒙 + 𝒚.
7
2- Pressure: doesn’t depend on area 𝜌1 ∗ ℎ1 = 𝜌2 ∗ ℎ2.
Note: height measure from the separating surface between the two liquids.
The manometer
Manometer: it’s a device used to measure the gauge pressure or absolute
pressure of a gas container.
Construction
1- It consists of U glass tube filled with enough amount of manometric liquid
(mercury or water)
2- The gas container (tank or reservoir) is connected to one end through a tap
(clip) to control the measurement process. While the other end in free (Atm).
Operation
1- When the clip (tap) opened the pressure of the gas and the atm pressure will
pressurized the liquid surface.
2- The difference in the manometric height represents the gauge pressure of
the gas container. 𝜟𝒑 = 𝝆 ∗ 𝒈 ∗ 𝒉 Respect the sign (-ve or +ve).
3- If the atm pressure is greater than the gas tank the liquid will rise in the gas
arm and the absolute pressure of gas reservoir𝑷𝒈 = 𝑷𝒂 − 𝜟𝒑.
4- If the gas tank pressure is greater than the atm pressure the liquid will rise in
the free arm and the absolute pressure of gas tank𝑷𝒈 = 𝑷𝒂 + 𝜟𝑷.
Where:
ΔP: gauge pressure, difference between gas pressure and atm pressure.
Pg: absolute pressure, total gas tank pressure.
Pa: atmospheric pressure.
h: here has many expression, the difference height, Liquid pressure, manometer
reading, gauge pressure, pressure difference.
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Pascal principle
Pascal’s principle: the pressure applied to a liquid enclosed inside a container is
transmitted equally to every portion of the liquid and to the walls of the container.
Application (The hydraulic press):
Construction
Two pistons with different area connected by incompressible fluid in closed
system.
Operation
When we apply a force in a small piston a pressure will transfer to the larger one
through the incompressible fluid according to Pascal’s principle.
Ideal Case
The mechanical advantage of Hydraulic press η (eta)
It’s the ratio between the Area of large piston and the area of the small one.
Pressure, volume and time are constant, force is directly proportional to the Area,
power and energy is constant according to the conservation law of energy.
𝑃1 = 𝑃2,
𝐹1 𝐹2
=
= 𝑐𝑜𝑛𝑠𝑡, 𝑉1 = 𝑉2 = 𝐿1 ∗ 𝐴1 = 𝐿2 ∗ 𝐴2 = 𝑐𝑜𝑛𝑠𝑡,
𝐴1 𝐴2
𝑊1 = 𝑊2 = 𝐹1 ∗ 𝐿1 = 𝐹2 ∗ 𝐿2, 𝑃𝑜𝑤𝑒𝑟1 = 𝑃𝑜𝑤𝑒𝑟2 = 𝐹1𝑣1 = 𝑓2 ∗ 𝑣2
𝑭𝟐 𝑨𝟐 𝒓𝟐𝟐 𝒅𝟐𝟐 𝑭 𝑨 𝑳𝟏 𝒚𝟏 𝒗𝟏
𝜼=
=
=
=
= = =
=
=
𝑭𝟏 𝑨𝟏 𝒓𝟏𝟐 𝒅𝟏𝟐 𝒇 𝒂 𝑳𝟐 𝒚𝟐 𝒗𝟐
Real Case
In real case there is losses due to the friction, bubbles (compressibility) so the
output energy (work) or power is less than the input one.
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𝑬𝒇𝒇 =
𝑾𝒐𝒖𝒕 𝑶𝑷 𝑭𝟐 ∗ 𝑳𝟐 𝑭 ∗ 𝒚𝟐 𝑭𝟐 ∗ 𝒗𝟐 𝑭 ∗ 𝒗𝟐
=
=
=
=
=
𝑾𝒊𝒏
𝑰𝑷 𝑭𝟏 ∗ 𝑳𝟏 𝒇 ∗ 𝒚𝟏 𝑭𝟏 ∗ 𝒗𝟏 𝒇 ∗ 𝒗𝟏
Where:
F2 or F is the force created in the second or larger piston (N).
F1 or f is the acting force in the first or smaller piston (N).
A2 or A is the cross section area of the larger piston (𝑚2 ).
A1 or a is the cross section area of the smaller piston (𝑚2 ).
r2 is the radius of the larger piston (m), d2 is its diameter (m).
r1 is the radius of the smaller piston (m), d1 is its diameter (m).
L2 or y2 is the vertical distance travelled by the larger piston (m).
L1 or y1 is the vertical distance travelled by the smaller piston (m).
v2 is the velocity of the larger piston (m/s).
v1 is the velocity of the smaller piston (m/s).
Wout is output work from the larger piston (N.m or J).
Win is input work to the small piston (N.m or J).
OP is the output power from the large piston (W).
IP is the input power to the small piston (W).
Archimedes’ principle
Archimedes’s principle: when a body is immersed wholly or partially in a fluid, it
experiences to an up thrust force in the vertical direction equal to the weight of
the volume of the fluid displaced by the body wholly or partially.
(There is an upwards force acting on a wholly or partially immersed body in a fluid;
this force equal to the fluid weight displaced by this body.)
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Proving Archimedes’ principle
1- Consider an imaginary cylinder of a volume (Vd) of a liquid of density (ρl)
enclosed in a container.
2- There are two balanced forces acting on it making it suspend in the liquid:
a- The horizontal forces: cancel each other (equal in magnitude opposite in
direction having the same line of action.)
b- The vertical forces:
 Weight of the liquid downwards (Fg)L=Vol*ρ*g
 Up thrust force (buoyant force) (Fb) acting upwards results from the
difference in pressure between lower and upper surface.
3- Calculating the force:
The upper base is affected by a downward force F1=P1*A; the lower base is
affected by an upward force F2=P2*A
Where the up thrust force Fb is the difference between these two forces and
acting upwards.
𝛥𝐹 = 𝐹𝑏 = 𝐹2 − 𝐹1 = 𝑃2 ∗ 𝐴 − 𝑃1 ∗ 𝐴 = 𝐴 ∗ (𝑃2 − 𝑃1) =
𝐹𝑏 = 𝐴(ℎ2 ∗ 𝜌𝑙 ∗ 𝑔 − ℎ1 ∗ 𝜌𝑙 ∗ 𝑔) = 𝐴 ∗ ℎ ∗ 𝜌𝑙 ∗ 𝑔 = 𝑉𝑑 ∗ 𝜌𝑙 ∗ 𝑔
𝑤ℎ𝑒𝑟𝑒 ℎ = (ℎ2 − ℎ1), 𝑉𝑑 = 𝐴 ∗ ℎ
So the up thrust force = the weight of the displaced fluid by the body.
𝑭𝒃 = 𝝆𝒍 ∗ 𝒈 ∗ 𝑽𝒅
Where:
Fb is the buoyancy Force (Floating force or up thrust force) (N).
ρl is the density of the fluid (𝐾𝑔/𝑚3 ).
Vd is the fluid displaced volume by the body= the volume of the body if totally
immersed, if not take the volume of the body part inside (immersed in) the fluid
(𝑚3 ).
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g is the gravity (𝑚/𝑠 2 ).
Note:
- Specific gravity showing the Archimedes principle in water only; if S.G > 1 the
body will sink, if S.G ˂ 1 the body will float, if S.G = 1 the body will hang in the
water; this principle used in submarine to move it up and down changing its
average specific density by filling or empty an air tanks.
- The resultant force acting on the body immersed in the fluid equal the
difference between its weight and the buoyant force acting on it in the
direction of the greater one.
- 𝐹𝑛𝑒𝑡 = |𝐹𝑤 − 𝐹𝑏| = |𝜌𝑠 ∗ 𝑔 ∗ 𝑉𝑠 − 𝜌𝑙 ∗ 𝑔 ∗ 𝑉𝑑| = 𝑔 ∗ |𝑉𝑠 ∗ 𝜌𝑠 − 𝑉𝑑 ∗ 𝜌𝑙|
- 𝑭𝒏𝒆𝒕 = 𝒈 ∗ |𝑽𝒔 ∗ 𝝆𝒔 − 𝑽𝒅 ∗ 𝝆𝒍|; In the direction of the big one.
- For fully immersed body Vs = Vd = V. So 𝑭𝒏𝒆𝒕 = 𝒈 ∗ 𝑽 ∗ (𝝆𝒔 − 𝝆𝒍).
- So the value of total density of the body determines its direction in the fluid;
if its value is greater than the fluid density it will sink, if less it will float, if
equal it will hang.
Floating Body
Sinking Body
Suspended (Hanging) Body
- Generally at rest position or uniform velocity motion the net force = Zero
𝐹𝑛𝑒𝑡 = 𝐹𝑤 − 𝐹𝑏 = 𝑍𝑒𝑟𝑜 𝑆𝑜 𝑭𝒘 = 𝑭𝒃
Floating principle: the weight of floating or suspended body is equal to the weight
of the liquid displaced by the immersed part of that body.
What’s meant by the apparent weight of a body 60 N
The difference between weight of that body in air and the up trust (Buoyant) Force
of that fluid on it is 60 N. Or the weight of that body in the new fluid is 60 N.
Application on floating:
1- The hydrotherapy technique:
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Some patient suffer from their joints and muscles (unable to lift their limbs);
in this case natural therapy is very important, the patient’s body is immersed
in water his weight is nearly zero (weightless) so it’s easy to move his joints.
2- Weightlessness experiment:
Weightlessness experiments takes place in a containers completely filled
with a liquid solution whose concentration can be adjusted such that Fb=Fw
3- Submarine technique:
The submarine floats when its tanks inside it are filled with air, while it sinks
(submerged) when tanks filled with water. Also fish and whales fill air sacs
with air to enable them to float and empty from air them when go down.
4- The diver's equipment:
On diving to small (shallow) depth, the diver breaths compressed air (air
under pressure greater than Atm pressure) to equalize the pressure on his
lungs with the outer pressure.
On diving to a great depth the diver have to change the pressure of his diving
suite to control the buoyancy force Fb.
13
Chapter (five)
hydrodynamic
This chapter deals with the fluid in case of motion (dynamic).
We can classify fluid flow:
Laminar flow: the velocity is relatively small and the flow is so smooth (Steady
flow, streamline flow).
Turbulent flow: the velocity is high and the flow is aggressive.
Additional Information
Steady: means no change with time.
Uniform: means no change with position.
1-D, 2-D, 3-D: the studying directions x, y, z.
Our course deals with 1-D steady flow
Streamline: it's an imaginary smooth path representing the fluid particle motion.
The streamline properties:
1- They are imaginary lines show the kind of the flow.
2- They never intersect with each other.
3- The tangent to the streamline represents the instant velocity.
Density of the streamlines: it's the number of streamlines passing normally
through a unit area at a point (expresses the velocity at this point).
The conditions necessary for obtaining Laminar (streamline) flow:
1- The fluid is incompressible, has constant density (Q́=constant).
2- The velocity is constant with time (Steady flow).
3- The flow is irrotational (no eddy or vortex).
Note: viscous and non-viscous (friction forces between the layers).
At high speed the flow is turbulent there is vortex (eddy)
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Additional information
There is a number used to determine the type of flow called Reynolds No (Re)
𝑅𝑒 =
𝜌∗𝑣∗𝑑
𝜇
Where: ρ fluid density, v velocity, d pipe diameter, μ fluid viscosity.
If Re ≤ 2000 the flow is laminar, Re ≥ 3000 the flow is turbulent
We will study the laminar flow in a full fluid tube
The rate of the flow and continuity equation:
Volume rate of flow (Qv): it's the volume of liquid flowing normally through the
cross section of the tube in one second (𝑄𝑣 =
𝑉
𝑡
=
𝐴∗𝐿
𝑡
= 𝐴 ∗ 𝑣 = 𝑐𝑜𝑛𝑠𝑡)(𝑚3 /𝑠)
Mass rate of flow (Mass conservation Law): it's the mass of liquid flowing
normally through the cross section of the tube in one second (𝑄𝑚 = 𝜌 ∗ 𝑄𝑣 = 𝜌 ∗
𝐴 ∗ 𝑣 = 𝑐𝑜𝑛𝑠𝑡)(𝐾𝑔/𝑠)
Where: ρ fluid density, v velocity, A cross section area.
The equation of continuity: the velocity of the fluid in a tube at a point is inversely
proportional the cross section area of the tube at this point 𝑣 𝛼
1
𝐴
𝑄𝑚1 = 𝑄𝑚2, 𝜌 ∗ 𝐴1 ∗ 𝑣1 = 𝜌 ∗ 𝐴2 ∗ 𝑣2 𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 𝜌 𝑎𝑠 𝑖𝑡 ′ 𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡;
𝑄𝑣1 = 𝑄𝑣2 = 𝐴1 ∗ 𝑣1 = 𝐴2 ∗ 𝑣2 𝑆𝑜
𝑣1 𝐴2
=
𝑣2 𝐴1
Viscosity: it's the resistance of the fluid to flow. Or it's the property responsible for
resistance of the fluid to flow or to the body motion through it due to the frictional
forces between its adjacent layers and the attraction forces between them when
they slid over each other under the effect of an external force. Or it's the property
responsible for resisting the relative motion of liquid layers.
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Note: viscosity doesn't related to the density of the fluid; example water and oil
the density of water is greater than the density of oil but the viscosity of oil is
greater than the one for water.
The non-turbulent viscous laminar flow (viscous steady flow): A kind of flow in
which the velocity of flow between the adjacent layers changes without forming
vortexes.
The coefficient of viscosity (ηvs): it's the tangential force acting on unit area to
produce a unit change in velocity between two layers at unit distance apart from
each other; M.E (ηvs =
𝐹∗𝑑
𝐴∗𝑉
); Unit (𝑁.
𝑠
𝑚2
,
𝐾𝑔
𝑚∗𝑠
, 𝑃𝑎 ∗ 𝑠); D.F (𝑀 ∗ 𝐿−1 ∗ 𝑇 −1 ).
Application of viscosity:
Lubrication: it's used in mechanical parts in contact to prevent heat and metal
wear due to friction force by using high viscous fluid (Oil for closed part and grease
for open one).
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Experiments
Viscosity:
Ep.1:
1- Put two empty beakers below two similar funnels.
2- Pour same volume of alcohol in one funnel and glycerin in other one at the
same time.
3- See that the velocity of the alcohol is higher than the glycerin.
Ep.2
1- Two similar beakers; one contains volume of water and the other contains
the same volume honey.
2-
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Take away
Define
Density, relative Density, Liquid pressure, Atmospheric pressure, Pascal’s principle,
mechanical advantage of hydraulic press, the mechanical efficiency of hydraulic
press, Archimedes’ principle, floatation principle, apparent weight in a fluid .
What’s meant by:
1- The pressure at a point is 10
𝑁
𝑚2
.
2- The atmospheric pressure at the top of a mountain is 0.9 bar.
3- The specific density of Aluminum is 2.7.
4- The normal force acting on the unit area = 9 ∗ 105 𝑁.
5- The density of the oil is 970
𝑘𝑔
𝑚3
.
6- The pressure at point inside a liquid = 2 ∗ 105 𝑁/𝑚2 .
7- The pressure difference in a tire of a car = 5 Atm.
8- The apparent weight of a piece of iron in water = 60 N.
9- The up thrust force acting on a floating body = 50 N.
10The buoyant force acting on a wholly immersed body = 50 N.
11The mechanical advantage of the hydraulic press=95.
Give reasons:
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The relative density is unit less.
The tyres of the military cars are broad.
The needle has pointed end.
The base of the dam is made thicker than the top.
Mercury is preferred in barometer.
The atmospheric pressure is neglected in submarine pressure calculation.
The pressure exerted by a liquid on the base of the container doesn’t depend
on the area of the base.
8- The liquid used in manometer may be mercury or water.
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9- The up thrust force exerted by fresh water on a ship equals to the up thrust
force exerted by the salty water on the same ship although density of salty
water is greater than that of the fresh (pure) water.
10The apparent weight of floating or suspended body =Zero.
11The weight of a body in air is bigger than its weight when it’s immersed
in liquid.
12Swimming in sea water is easier than swimming in the Nile.
13When a tank filled to its rim with water then a cube of wood is put
above the water the weight of the water doesn’t changed.
14A nail made of steel sink in water but a ship made of the same material
floats on it.
15There is no ideal hydraulic press.
16The heavy object is put on the large piston.
Prove that:
1- The pressure exerted by a liquid at a point inside it is 𝜌 ∗ 𝑔 ∗ ℎ.
2- For ideal hydraulic press 𝜼 ==
𝒓𝟐𝟐
𝒓𝟏𝟐
=
𝒅𝟐𝟐
𝒅𝟏𝟐
=
𝑭
𝒇
=
𝑨
𝒂
=
𝒚𝟏
𝒚𝟐
=
𝒗𝟏
.
𝒗𝟐
3- Archimedes’ Principle the buoyant force = the weight of the fluid displaced
by the object. Fb=Vol*ρL*g.
What are the factors affecting the following:
1- Pressure at a point inside the liquid.
2- Specific gravity of the liquid.
3- The buoyant (up thrust) Force by a fluid.
Complete
1- 3 atm= …… 𝑁/𝑚2 = ……… cm.Hg = ………. Torr= ……… Bar.
2- 700 mm.Hg = ………. millibar.
3- Force (F) acts on an area (A) makes an angle (θ) with the normal to the
surface, the pressure equation is …………
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Problems:
1- A swimming pool has a length of 30 m, width 10 m, height of water is 3 m
given that Pa = 1.013 ∗
105 𝑁
𝑚2
, 𝜌𝑤 = 1000
𝑘𝑔
10𝑚
𝑚
𝑠2
3,𝑔 =
. 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒.
a- Water pressure acting on the base.
b- The weight of the water in the pool.
c- The pressure at a point 125 cm above its bottom.
d- The water pressure acting on its walls.
2- Some fashions accessories are electroplated with gold coating 0.85 ∗
10−4 𝑐𝑚 thick. Calculate the area that can be coated by 1 Kg of gold knowing
that ρg= 19.3
𝑔
𝑐𝑚3
.
3- A container of mass 0.03 Kg when empty, 55gm when it’s full f water and 370
gm when it’s full of mercury. Calculate the relative density of mercury.
4- A submarine is located in sea water at depth 100m, knowing that the density
𝑚
of sea water is 1030𝐾𝑔/𝑚3 , g=9.8 2 , the interior pressure is Pa; Calculate the
𝑠
force acting on a circular windows with radius 21 cm.
If the max design pressure for the submarine body is 12 atm, is it safe at 100
m and why?
5- .A layer of water of thickness 50 cm rests on a layer of mercury of thickness
20 cm. what is the difference in pressure between two points one at the
interface between the water and mercury and the other at the bottom of the
𝑚
𝑘𝑔
13600𝑘𝑔
𝑠
𝑚
𝑚3
mercury layer (given that g=10 2 , ρw=1000
3 , 𝜌ℎ𝑔 =
.)
6- A U-tube with Area = 2 𝑐𝑚2 contains water; a 20 𝑐𝑚3 of oil is poured in one
of its arms calculate the height of water above the separating surface in the
other arm (ρw=1000
𝐾𝑔
𝐾𝑔
𝑚
𝑚3
3 , 𝜌𝑜𝑖𝑙 = 800
).
7- U-tube with 1 𝑐𝑚2 narrow arm area and 2 𝑐𝑚2 large one filled partially with
water (ρw=1000
𝑘𝑔
𝑚3
); when an amount of oil (ρoil=800
𝐾𝑔
𝑚3
) is poured in the
narrow arm a water raised 2 cm above the separating surface
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8- The following table showing the relation between the pressure (P) at a point
inside water lake and its depth from the free surface (h) (neglecting the temp
effect)
h (m)
5
10
15
20
25
30
P (bar)
1.5
2
2.5
3
3.5
4
Plot the graph between the P on Y axis and h on X axis then find the value of
atmospheric pressure above the water lake and the density of water.
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