Fluid Mechanics Chapter 1-5

C H A P T E R I:

Basic Properties of Fluids _____________________________________________________________________________________

Fluid Mechanics Is the science of the mechanics of fluids (liquids and gases) and is based on the

same fundamental principles that are employed in the mechanics of solids. It parti-

cularly deals with the actions of fluids at rest or in motion and with the applications

of devices in engineering using fluids.

*Mechanics is the study of the behavior of a physical system under the action of forces.

Three (3) branches of Fluid Mechanics:

1. Fluid Statics -- Is the study of fluids at rest.

2. Fluid Dynamics -- Is the study of fluids in motion and deals with the velocities (or Kinematics) and streamlines without considering the forces that causes them to move.

3. Hydrodynamics -- deals with the forces exerted upon liquids in motion including

the relations between velocities and accelerations involved in

such fluid that is in motion.

3.1 Hydraulics -- deals with the application of fluid mechanics to engineering devices involving liquids usually water or oil. It deals with such problems as the flow of fluids through pipes, or in open channels, the designs of storage dams, pumps, and water turbines. Or with any other devices for the control or use of liquids such as nozzles, valves, jets, and flow meters.

*** Fluids are substances which owing to the nature of their internal structure offer

comparatively little resistance to a change in form and are divided into liquids and gases.

FLUID MECHANICS/ ENGR. G.S. ROBLES

Differences between liquids and gases: For liquids: 1. 2. 3. For gases: 1. 2. 3. ** Incompressibility ** Compressibility

Distinction between a solid and a liquid: 1. 2. “ The distinction is that any fluid no matter how viscous , would yield in time to the slightest

stress. But a solid no matter how plastic, requires a certain magnitude of stress to be exerted before it will flow”….


GENERAL PROPERTIES OF A LIQUID:

1. DENSITY (MASS DENSITY) defined as the ratio between mass and volume.

2. SP. WEIGHT (UNIT WEIGHT defined as the ratio between weight and volume. or SP. WEIGHT)

Mass Density of Water (H20) at STP: 1000 kgm/ cu.m. = 62.427 lbm/ cu.ft. = 1 gmm/ cu.cm.

where: F or Wt -- force or weight m -- mass go -- observed gravitational acceleration gc -- gravitational constant = 32.174 lbm-ft/ lbf-s

2 = 9.806 kgm-m/ kgf-s

2 = 1 kgf -m/ N -s2

gs -- std. gravitational acceleration = 9.806 m/ s2 = 32.174 ft/ s2

3. Specific Volume is the volume occupied by a unit mass of fluid.


4. Sp. Gravity or Relative Density is the ratio of the mass density of fluid in question to the mass density of an equal volume of water.

5. Bulk Modulus of Elasticity is the incremental change in volume when the pressure is changed by an incremental amount,

6. Viscosity is considered as the property of a fluid which determines its resistance to shearing stress. Oftentimes, it is also called as coefficient of viscosity, absolute viscosity, or dynamic viscosity.

= shear stress/ rate of shear strain Pa – s

7. Kinematic Viscosity is the ratio of the dynamic viscosity to its corresponding density.

= dynamic viscosity / mass density m2/ s; ft2/ s

LIQUID PRESSURE the liquid pressure at any point is equal to the product of the weight density of the liquid and the depth of the point in question. That it increases as the depth also increases..

** Equivalent Pressure Head two pressure heads are said to be equivalent if they cause the same intensity of pressure. From the liquid pressure formula and by the said definition….

Sp. Gravity A ( Ht A ) = Sp. Gravity B ( Ht B )


Absolute pressures, Gage pressures, vacuum pressures:

P abs = P atm + P gage

P abs = P atm - P vacuum

STD. REFERENCE FOR ATMOSPHERIC OR BAROMETRIC PRESSURE (BASED ON SEA LEVEL):

The standard atmospheric pressure Patm changes from 101.325 kPa at sea level to the following elevations:

89.88 kPa at 1000 m 26.50 kPa at 10,000 m 79.50 kPa at 2000 m 5.530 kPa at 20,000 m 54.05 kPa at 5000 m

Common devices used for measuring pressures (gage pressure): Conversion values or equivalent identities of units: 1 newton = 1 dyne =

1 poise = 1 dyne-s/ cm2 1 lb = 444,800 dynes

1 bar = 100 KPa 1 poise = 0.10 Pa-s Basic Constants for volume: 1 cu. ft. = 7.482 gal 1 drum = 55 gal (petroleum, unrefined)

1 gal = 3.7854 Li 1 barrel = 42 gal ( refined petroleum products and other liquids)

1 cu.m. = 1000 Li 1 stoke = 1 cm2/ s


PROBLEM SOLVING:

1. What is the mass of a liter of saltwater expressed in pounds (lbsm)?.

2. What is the weight of a 45 kg boulder if it is brought to a place where the acceleration due to gravity is 395 meter per seconds per minute?

3. A cylindrical tank 80 cm in diameter and 90 cm high is filled with a liquid. The tank and the

liquid weighed 420 kgs. The weight of the empty tank is 40 kgs. What is the unit weight of the liquid?

4. A lead cube has a total mass of 80 kgs. What is the length of its side if the relative density

of lead is 11.3?

5. If the viscosity of water at 70 0C is 0.42 centipoise and its relative density is 0.978, determine its absolute viscosity in Pa-s and its kinematic viscosity in m2/ s and in stokes.

6. A 10m diameter cylindrical tank has a height of 5m and is full of water at 20 0C. (Unit weight = 9.879 kN/ cu.m.) If the water is heated to a temperature of 50 0C (Unit weight = 9.689 kN/ cu.m.), solve for:

a. The weight of the water at its initial temperature b. The volume of the water when heated to its final temperature. c. The volume of water that will spill over the edge of the tank.

7. Water has a dynamic viscosity of 1 centipoise. Compute its dynamic viscosity in terms of

lbs-s/ sq. ft…

8. At what height in meters would a vertical column of water be supported by standard atmospheric pressure? When mercury was used instead?

9. A cubic meter of air at barometric pressure weighs 12 Newtons. What is its specific volume?

10. If the pressure 3m below the free surface of a liquid is 14 KPa, what would be its relative density?

11. Find the Bulk Modulus of Elasticity of a liquid in ksi, if a pressure of 150 psi applied to 10 cu. ft. of the liquid causes a volume reduction of 0.02 cu. Ft.

12. A submarine is cruising 600 ft below the ocean’s surface. Determine the absolute pressure on the submarine’s surface. Assume acceleration due to gravity to be constant even at that depth.

13. A beer barrel has a mass of 20 lbs and a volume of 5 gallons. Assuming the beer’s density is like that of water, what would be the total mass and weight of the beer barrel when it is filled with beer?

14. A city of 6,000 population has an average total consumption per person per day of 100 gallons. Compute the daily total consumption of the city in cu.m. per seconds.

15. A lunar excursion module (LEM) weighs 1500 kgf on earth where go = 9.75 m/ s2. What would be its weight in the moon’s surface where go = 1.70 m/ s2?


16. The mass of a given airplane at sea level (go = 32.1 fps2) is 10 tons. Find its mass in lbsm, slugs, and gravitational weight when it is travelling at a 50,000 ft elevation. The acceleration of gravity, go decreases by 3.33 x 10-6 for each foot of elevation.

17. A fluid moves in a steady flow manner between two (2) sections in a flowline. At section 1: A1 = 10 ft2 ѵ1 = 100 fpm v1 = 4 ft3/ lbm

At section 2:` A2 = 2 ft2 ρ2 = 0.20 lbm/ ft3

Calculate the mass flow rate and the speed at section 2….. 18. If a pump discharges 75 gpm of water whose specific weight is 61.5 lbf/ ft3 (go = 31.95 fps2),

find (a) the mass flowrate in lbm/min, and (b) the total time required to fill a vertical cylindrical tank 10 ft in diameter and 12 ft high.

19. A cargo ship has tanks for carrying fuel oil. The tank dimensions are 1m x 5m x 15m. How

many barrels could be filled by the ship’s tank? In how many gallons.


C H A P T E R I I:

Principles of Hydrostatic Pressure _____________________________________________________________________________________

LIQUID PROPERTIES/ BEHAVIOUR WITHOUT ANY EXTERNAL FORCE EFFECTS (ONLY ATMOSPHERIC

PRESSURE):


Increases with depth

Does not depend upon the surface

area/ shape of its confining vessel

Acts in all directions

Pressure taken at any point in the

same horizontal level is constant

Pressure of a gas above the free surface

of a confined liquid is transferred

undiminished in every directions.

Holes for jets Spouting can

Weakest jet

Strongest jet

table

water

PRESSURE INCREASES WITH DEPTH:

Suppose that a spouting can or

simply a can is poured with water. Drilling

three (3) holes at the side at different

elevations causes the formation of water jet

or streamlines of water to leak at the holes.

With the gravitational force exerted

upon the water, the water’s natural

tendendency would be to “leak” or

“discharge” water into the said holes.

Taking a closer look, the uppermost

hole would provide the weakest streamline/

jet of water as there is less pressure at the

top.

The lowermost hole would provide the greatest streamline/ jet of water which is due to the

fact that pressure located beneath is of the greatest value of all.

FBD of a certain volume of liquid

showing the action of forces…


PRESSURE DOES NOT DEPEND UPON THE

SURFACE AREA/ SHAPE OF ITS CONFINING

VESSEL:

Irregardless of the shape to which

the liquid is stored/ confined, pressure

brought about by the body of liquid would

not depend upon the shape/ contour of

confining vessel.

Liquid pressure is always a function

of the liquids height/ depth. That is liquid

pressure is directly proportional with the

depth or height of the liquid.

A spring is a device that can be used to store energy. If such were to be submerged in a body of liquid at a particular depth and held in equilibrium, the spring will be compressed. This is brought about by the pressure exerted by the volume of liquid which varies with depth.

Irregardless of its positioning, the spring will still be deflected…

LIQUID PRESSURE ACTS IN ALL DIRECTIONS

The funnel containing liquid shown is bent at different directions/ angles. With the presence of a liquid and a certain depth/ height, liquid pressure could be seen acting in all directions or parts of the funnel containing the liquid only. However, the pressure acting on the given funnel is of different values at different locations/ elevations in particular. Suppose that a point was taken at one

particular portion of the funnel. The pressure

that will be acting on that particular point will be

coming in at all directions or at all portions of the

circumference of the point. And with the point

having a negligible size, pressure in this case

would be the same at all points/ directions

P3

P1

P2

Exploded view of a point taken in a liquid

A cylinder with a lid spring mounted on the inside…

Consider Gas Law:

P1V1/ T1 = P2V2/ T2 = P3V3/ T3 with change in condition(P, V, and T)

but without a change in mass, m…..

from General Gas Law Equation: P V = m R T

LIQUID PRESSURE AT THE SAME HORIZONTAL PLANE IS CONSTANT:


PRESSURE OF A GAS ABOVE THE FREE

SURFACE OF A CONFINED LIQUID IS

TRANSFERRED UNDIMINISHED IN EVERY

DIRECTIONS.

If the pressure of a confined liquid varies

with “depth or height” , the same could not be

said of a confined gas that is hovering above the

free surface of the confined liquid.

As for gases, its pressure does not vary

with the height or depth to which it is confined. It

has the same pressure all throughout its

confining vessel.

Irregardless of the liquid’s confining vessel to which it would be contained, pressure taken along the same horizontal plane is said to be constant. This even if one space is smaller or larger than the other.

From the configuration, pressure at the

handle portion is the same with the pressure at the right wall of the water container.

But that the pressure at the container’s

lip is of the largest value as pressure varies directly with depth…

Pwall Phandle

Such principle is used

so as to indicate the level of

liquid inside liquid storage

tanks…

Photo insert shows

one of the many shapes of a

level gauge…

LIQUID PROPERTIES/ BEHAVIOUR UNDER THE ACTION OF EXTERNAL FORCES:


Liquids can transmit motion and

force

Liquids can increase/ decrease force A confined liquid under pressure/ action

of force has its pressure transferred

undiminished in every direction

(PASCAL’S LAW)

A CONFINED LIQUID UNDER PRESSURE

HAS ITS PRESSURE TRANSFERRED

UNDIMINISHED IN EVERY DIRECTION

When pressure is exerted on a

confined liquid, it is transmitted

undiminished. Force on the piston has

created a pressure of 50 psi (pounds per

square inch), upon the liquid in the

pressure cylinder. Notice that all gauges

read the same throughout the system.

Pressure is transmitted undiminished to all

parts of the system. If gauge A reads 50 psi,

gauges B, C, D, and E will also read 50 psi.

FORCE

LIQUID CAN TRANSMIT MOTION AND

FORCE:

In the Figure, you will see that any

movement of piston A will cause piston B to

move an equal amount. This is a

transmission of motion though a liquid. If a

200 lb force is placed on piston A, piston B

will support 200 lbs. Both pistons are the

same size.

Applying an external force besides

the given weight would add force over the

given cross sectional area thus increasing its

pressure at A and then will cause piston B to

move upwards.


LIQUID CAN INCREASE/ DECREASE

FORCE:

When a force is applied to piston A, it can be increased if it is transmitted to a larger piston B. If piston A has a surface area of 1 sq. inch, the 200 lb force on piston A represents a pressure of 200 pounds per square inch (psi). According to Pascal's Law, this 200 psi force will be transmitted undiminished.

If piston B has a surface area of 20 sq. inch, piston A will exert a 200 lb force on each square inch of piston B. This would produce a mechanical advantage (MA) of twenty, and the original 200 lb force would be increased to 4000 lbs. The force may be further increased by either making piston A smaller or piston B larger. Reversing the configuration’s process decreases the force that is conveyed.

THE HYDRAULIC JACK

The hydraulic jack which is one

tool that every motorists should have is a

powerful yet simple tool. Figure shows

how a fluid can be used to produce a

powerful lifting force using the principles

of hydraulics. When the jack handle raises

piston A, piston A will form a vacuum. This

will draw check valve 1 open and close

check valve 2. When the handle is

depressed with a force that exerts 200 lbs

pressure (or any force) on piston A, check

valve 1 will close, check valve 2 will open

and 200 psi will be transmitted to piston B.

If piston B has a surface area one hundred

times greater than the 1 sq. inch area of A,

piston B will raise a weight of 200,000 lbs.

HYDRAULIC PRINCIPLES IN VEHICLE BRAKE SYSTEMS:

When a driver depresses the brake pedal, force is transmitted undiminished to each caliper or

wheel cylinder. The caliper pistons or wheel cylinders transfer this force (increased or decreased,

depending on piston area) to the friction linings.

When the master cylinder piston moves, the caliper pistons or wheel cylinders will move until

the brake friction components are engaged. This causes the vehicle to stop gradually.


SCHEMATIC DIAGRAM OF A TYPICAL BRAKE SYSTEM (DISC TYPE):

Maintains even viscosity throughout a wide temperature variation. Does not freeze at the coldest possible temperature (hygroscopic has the ability to absorb and

retain moisture) that the vehicle may encounter. Boiling point is above the highest operating temperature of the brake system parts. Does not corrode the brake system’s metal parts. It acts as a lubricant for pistons, seals, and cups

to reduce internal wear and friction. It does not deteriorate (swell) the brake system's plastic and rubber parts.

Under no circumstances put anything but brake fluid into the brake system. Any mineral or petroleum-based oils such as motor oil, transmission fluid, power steering fluid, kerosene, or gasoline in even the smallest amounts will swell and destroy the rubber cups and seals in the system. **Warning: Brake fluid is poison. Keep it away from skin and eyes. Do not allow brake fluid to splash on painted surfaces.

(a) Pressure applied to a car’s brake pedal is transmitted by the brake fluid to the car’s wheels.

(b) The same force per unit area is supported by different sized pistons that are at the same height and are in contact with a static fluid, because the fluid pressure on each piston is the same. Thus a small force applied to a small piston balances a large force on a large piston.

(c) The force-multiplying effect shown in (b) is applied in this hydraulic lift.

CROSS SECTION OF A DISC BRAKE:


VARIATION OF PRESSURE WITH DEPTH IN A FLUID:

Consider a prism with a given cross-sectional area A and a length L that is submerged at

an inclination in a body of water. With the prism considered at rest, equilibrium conditions

apply.

Forces along the length of the prism:

SAMPLE PROBLEMS:

1. The hydraulic press shown contains a confined liquid with a relative density of 0.65. If a force of

190 kgs is applied on a circular lid A with a diameter of 90cm, what maximum load at B maybe

placed on a 1.2m x 6m rectangular platform lid? The difference in elevation between point

A and point B is 0.20 m..

2. A 0.3m diameter pipe is connected to a 0.02m diameter pipe and both are rigidly held in place. Both pipes are horizontal with pistons at each end. If the space between the pistons is filled with water, what force will have to be applied to the larger piston to balance a force of 90N applied to the smaller piston?


3. The bottom of a river is 12m below the water surface. Underneath which is a silt has a specific

gravity of 1.75 and a thickness “t”. If the pressure at the bottom of the silt is 0.450 MPa, what

would be the silt’s thickness?

4. A cylindrical tank having a diameter of 1.5m and height of 4m is open at one end and closed at

the other end. It is placed below the water surface with its open end down. How deep below

the water surface should the tank be placed if the depth of water inside the tank is 1.8m?

5. The hydraulic press shown is filled with oil of 0.82 specific gravity. Neglecting the weight of the

two pistons, what force “F” on the handle is required to support the 10 kN weight?

6. For the configuration shown, what

force directed at piston D is

necessary to cause the system to be

in its equilibrium state??? If gas was

used instead of water, would the

necessary force increase or

decrease???

7. Assuming normal barometric pressure, how deep in the

ocean is the point where an air bubble, upon reaching

the surface, has six times its volume that it had at the

bottom?

8. From the composite layers of liquid

stored in a vessel, at which liquid will

a pressure of 0.7 MPa absolute will

be first achieved?


9. The basic elements of a hydraulic press is shown

in the figure below. The plunger has an area of

1 in2, and a force, F1, can be applied to the plunger through a lever mechanism having a

mechanical advantage of 8 to 1. If the large piston has an area of 150 in2, What pressure is

exerted at the top if the smaller plunger is to be exerted with a 30 lbs of force.

ACTIVITY QUESTIONS TO FURTHER IMPROVE YOUR

UNDERSTANDING OF PRESSURE:

1. A small amount of water is boiled in a 1 gallon metal can. The can is removed from the heat and

the lid put on. Shortly thereafter the can collapses… Explain…

2. A person’s ability to do work is greatly affected when working at very high altitudes/ mountain

ranges. Does the same hold true in the case of automobiles???

Explain how the tube known as a

“SIPHON”, can transfer liquid from one

container to a lower one even though the

liquid must flow uphill for part of its

journey. (Note: The tube must be filled

with liquid to start with).

3. If you dangle two pieces of paper vertically a

few inches apart, and blow between them, how

do you think the papers will move?? Try it and

see… Explain…

4. The two open tanks have the same bottom area,

A but have different shapes. When the depth, h,

of a liquid in the two tanks is the same, the

pressure on the bottom of the two tanks will be

the same. However, the weight of the liquid in

each of the tanks is different. How do you

account for this apparent paradox???


C H A P T E R I I I:

Manometry; Manometers

Manometers are devices or apparatus’ that measure a change in pressure. It is usually made-up of a clear glass tube (normally bent in the form of a letter “U”) with one or more substances of known specific gravity that moves proportionally to the force that is generated by the pressure it is measuring.

TYPES OF MANOMETERS:

1. Open type manometer: Is a type of manometer whose one leg is open to the atmosphere and

is capable of measuring gage pressure.

1.a. Piezometers: Is the simplest form of manometer which is tapped into

the wall of a container or conduit and in which the liquid can freely rise without overflowing. The height of the liquid in the tube gives the pressure head directly.

Piezometers are limited to measuring small amounts

of pressure due to the impracticability of providing a long Tube to accommodate higher pressures. Moreover, it can only be used for liquid pressure measurement since a gas does not form any free surface. In measuring pressures of fluids in motion, precautions

should be taken in making connections. The hole must be drilled exactly normal to the inner surface of the container or conduit wall and the piezometer tube must not project beyond this surface. All burrs and roughness on the inner surface must be removed for this will have an effect on the piezometric head.

To reduce the error due to capillarity, the tube diameter

should be at least 1.25 cm. The disadvantages of a piezometer could be overcomed by

using a more complex form of manometer. This makes use of a bent tube (or loop) that’s makes use of more than type fluid. As such fluids used should be immiscible so as to form a meniscus between them.


A differential manometer tapped into a conduit:

2. Differential manometers: Is a type of manometer whose ends are both closed to the atmosphere and thus is not subjected to atmospheric pressure. Its purpose is to determine the pressure difference between pipes, vessels, or tanks with known internal pressures.

2.a. Micromanometer: Is a special type of differential manometer that is used

for measuring difference in gas pressures. One example is illustrated wherein this type provides a higher precision and is used when the pressure difference is too small a value that cannot be measured by a typical differential manometer. Obviously, the primary purpose of this gage or device is to magnify the reading to permit a greater accuracy.


PROCEDURE OF COMPUTATION IN SOLVING MANOMETRY PROBLEMS:

1. Draw a sketch of the manometer approximately to scale. 2. Starting with one end of the Manometer (left or right maybe taken), label the contact points of fluids

of different specific gravity. 3. After the labeling process, add the pressure heads of the liquids as the elevation decreases and or

subtract the pressure heads of the liquids as the elevation increases. 4. Using the liquid pressure formula, you may now obtain unknown values from the derived equation.

SAMPLE PROBLEMS:


1. For the differential manometer shown, what would be the new mercury deflection If the pressure at point A is increased by 40 kPa. Initial mercury deflection is 250 mm.

2. A U-tube manometer of 10 mm diameter is to contain mercury thereby occupying it inside. Supposedly, 12 mL of water is poured unto the right-hand leg, what would be the new heights at both legs?

3. Three different liquids with properties as indicated fill the tank and manometer tubes as shown. Determine the specific gravity of fluid #3…


4. Determine the elevation difference, ΔH between the water levels in the two open tanks shown..

5. Determine the angle, Ø of the inclined tube shown if the pressure at pt A is 1 psi greater than pt. B…

6. The closed tank is filled with water and is 5 ft. long. The pressure gage on the tank reads 7 psi. Find (a) the height, h in the open water column; (b) the pressure acting on the bottom tank surface AB.


7. A mercury manometer is used to measure the pressure difference in the in the two pipelines. Fuel oil has a unit weight of 53 lbs/ ft3 and is flowing in A and SAE 30 lube oil, unit weight of 57 lbs/ ft3 is flowing in B. An air pocket has become entrapped in the lube oil as indicated. Determine the pressure in pipe B, if the pressure in pipe A is 15.3 psi.

8. A closed cylindrical tank filled with water has a hemispherical dome and is connected to an inverted piping system. The liquid in the top part of the piping system has a specific gravity of 0.8 and the remaining parts of the system are filled with water. If the pressure gage reading at pt. A is 60 kPa, determine: (a) the pressure in pipe B, and (b) the pressure head in mm of mercury at the top of the dome (point C)..

9. Water, oil, and an unknown fluid are contained in the vertical tubes as shown. Determine the density of the unknown fluid.

10. Compartments A and B of the tank are closed and filled with air and a liquid with a specific gravity equal to 0.6. Determine the manometer reading, h, if the barometric pressure is of standard value, and that the gage pressure reads 0.5 psi…

CHAPTER IV

Buoyancy and Archimedes Principle

INTRODUCTION:

It is a common experience that an object feels lighter and weigh less in a liquid that it does in air. This could be demonstrated by weighing a heavy object in water by a spring scale. Also, objects made of wood or other light materials float on water but at times there are instances wherein very heavy objects also tend to float on water (consider cases of ships and submarines).

These are observations suggesting that fluids (liquids in particular) exert an upward component of a force on a body immersed in it. The force that tends to lift the “body” is called the “BUOYANT FORCE”. This buoyant force is caused by the increase of pressure in a fluid (liquid) with respect to depth.

Referring to the configuration shown, It would be noted

that liquid pressure increases with depth and since force is directly proportional with pressure, thus F2 > F1. The buoyant force, BF provided by the liquid is merely a summation of all acting forces (vertical)….

BF = F2 - F1

Consider P = F/A and P = γ h, thus:

= P2A2 - P1A1 = γ2 h2 A2 - γ1 h1 A1

where: Δh = h2 - h1; A1 = A2 = A; γ1 = γ2 = γ Δh A = V, volume of object immersed in the fluid (liquid) = V, volume of fluid (liquid) displaced by the object

BF = γ (Δh) A ** Experimentally, the buoyant force that is given-off by a liquid could be determined. That is by weighing the object to which is to be subjected to the said buoyant force initially in air. And then weighing the same object when it is immersed in the body of liquid that will provide the buoyant force…

ARCHIMEDES’ PRINCIPLE:

“ The buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced** by that object”…

** By the term “fluid displaced”, we mean a volume of fluid equal to the volume of the submerged object, or that part of the object submerged if it floats or is only partially submerged (the fluid that used to be where the object is)….


BF = γliquid Vdisplaced

Weight of displaced liquid is the

BUOYANT FORCE

HOW SUCH PRINCIPLE WAS ACTUALLY DISCOVERED: Archimedes’ discovery of such law was accidental, he was given an order by his king in which he had to devise a way of determining whether his king’s blacksmith was actually lulling the king into believing that the crown he had made was actually made-up of pure gold. As he was thinking of ways to solve such mystery, he was actually taking a bath in a bathtub. This he noticed that he was actually displacing some water as some parts of his body lay submerged in water. As he finally was able to found some evidence, he even run naked out of the tub shouting “EUREKA!” EUREKA! (meaning he finally found it)..

Archimedes’ devised an experiment in which he weighed the king’s supposedly gold crown and a actual pure gold having the same weight in air. Then weighed both objects in water. At that time it is already known that materials such as solid have this property called “density”. Gold has a relative density of 19.3 higher than most metals. As such if one was the pure gold (consider weight and volume), it would produce a greater or heavier “density”. In the end, it was proven that the king was actually being fooled, made into believing that the crown was actually a “pure gold” where it was not.

The catch here was that although both the

said materials had the same weight, they had different volumes as suggested by the displaced water.

BUOYANT FORCE IN ACTION:

Supposedly a solid body was dropped into a body of liquid, it will sink, float, or remain at rest at any point in the fluid depending on the materials density.

A. POSITIVELY BUOYED

B. NEUTRALLY BUOYED

C. NEGATIVELY BUOYED

Other contributing factors:

1.

2.

ACTIVITY QUESTIONS:

1. Buoyancy may also be considered as the ability of objects to float along a liquid, does sinking at the very

bottom of a liquid indicate that there is no buoyant force acting along the object?

2. At where would you weigh more? In air? Or in water?

3. Supposed that you had a ball of cork and a lead cube having the same amount of volume, are they

subjected to the same buoyant force when totally immersed in a same body of liquid? At what instance

are they the same? At what instance are they not the same?


A

B

C

4. If a pound of coin was accidentally dropped in ditch full of mercury, would you still be able to see that

pound of coin?

5. Do heavier objects sink all the time? If not how and why?

6. Submarines employ the principles of buoyancy to travel up or down beneath the ocean’s water.

How is this made possible?

7. As in the case of fish or other marine mammals, how are they able to float effortlessly in water?

8. Does buoyancy only apply to objects immersed/ submerged in a body of liquid?

9. A piece of grape dropped in a glass of water would settle at the bottom. Can you make that piece of grape

to float with using any mechanical devices?

SAMPLE PROBLEMS:

1. A scuba diver and her gear displaces a volume of 65 Li and have a total mass of 68 kgs. Is the diver

sinking? Or floating?

2. A 0.48 kg piece of wood floats in water but is found to sink in alcohol (specific gravity = 0.79), in which it

has an apparent mass of 0.047 kg. What is the relative density of wood?

3. An iceberg having a relative density of 0.92 is

floating on saltwater of 1.03 relative density.

If the volume of the iceberg above the seawater

surface is 1000 cu.m., what is the total volume

of the iceberg?

4. A wooden spherical ball with a specific gravity of 0.42

and a diameter of 300mm is dropped from a height of

4.3m above the surface of water in a pool of unknown

depth. This ball barely touched the bottom of the pool

before it began to float. Determine the depth of the

pool.

5. A block of wood requires a force of 40 N to keep it immersed in water and a force of 100 N to keep it

immersed in glycerine (sp. Gr. = 1.3), what would be the relative density of wood?


6. If a 5 kg steel plate is attached to one end of a 0.1 m

x 0.3m x 1.2m wooden pole, what is the length of

pole above water? Consider specific gravities of

0wood and steel to be 0.5 and 7.85 respectively

7. A ship travels from the sea and onto a lake.

As it went from salt water into fresh water,

it sinks by 7.62 cm. To compensate for the

“sagging” it uses the 72, 730 kg coal

onboard rising to a height 15.24 cm.

a. Find the original draft in salt water.

b. Find the original draft in fresh water

c. Find the weight of displaced seawater.

8. An open cylindrical tank 350 mm in diameter and 1.8

m high is inserted with its open end down into a

body of water. For stabilization, a 1,300 N block of

concrete is tied at the bottom having a specific

gravity of 2.4. To what depth will the open end be

submerged in water?

9. A concrete cube having 0.5 m sides is to be held

in equilibrium underwater by attaching a light

foam buoy to it. If the unit weights of concrete

and foam are 23.58 kN/ m3

and 0.79 kN/ m3

respectively, What is their total weight?

10. A composite cubical block having an edge equal to

3m has its upper half of sp. gravity = 0.80 and its

lower half of sp. gravity 1.4. It is resting on two

layers fluid with an upper sp. gravity of 0.90 and a

lower sp. gravity of 1.2. Find the height of the top

of the cube above the interface of the two layer

fluid.


1.5m

Sg = 0.90

Sg = 1.20

Sg = 0.0

Sg = 1.40

C H A P T E R V:

Hydrostatic Force on Surfaces

INTRODUCTION: Any object exposed to a liquid, say a gate valve mechanism in a piping system, the wall of a liquid storage tank, or even the hull of a ship at rest, are all subjected to fluid pressure distributed over its surface. That when there is pressure acting, it will also mean that forces are acting in the form of “hydrostatic forces”.

The designs of many engineering systems such as water dams, and liquid storage tanks are mostly

based on the determination of such forces acting on various surfaces (plane & curved surfaces) that requires the understanding of fluid statics (that deals with problems associated with fluids at rest; or where there is no relative motion between adjacent fluid layers.


Water flowing in long stretches of

piping are pressurized by pumps

(prime mover). As such, the internal

side is subject to pressure with the

portion restricting/ stopping the flow

of water having the highest pressure.

Photo inset of a gate mechanism/

stopper inside a gate valve being

subjected to hydrostatic /

pressureforce while the water is not

moving.

Water tanks or any tanks used to hold other types

of fluids are examples of pressure vessels in that

they are used to store fluids before they are

diverted for their respective uses.

Photo inset shows a water tank with a certain

cross-section being shown, with liquid pressure

being a function of the liquid’s height/ depth, the

depth of liquid on such types of pressure vessels

can no longer be neglected. The length of arrows

indicating where the largest pressure/ force can be

located,.

Liquid pressure could be seen acting on

the ship’s hull (as well as buoyant force) in

two (2) different cases.

On plane surfaces they act perpendicular,

while on curved surfaces, they act at the

point of tangency….

A. HYDROSTATIC FORCE ON PLANE SURFACES:

“The hydrostatic force on any plane surface of Area A, submerged in a body of fluid of Unit

Weight δ, is equal to the product of the area and intensity of pressure at its centroid

(center of gravity)”

F -- hydrostatic force A -- submerged area δfluid -- unit weight of fluid ђ -- vertical distance of the centroid with respect to the fluid surface

ÿ -- inclined distance of the centroid with respect to the fluid surface hp -- vertical distance of the center of pressure with respect to the fluid surface yp -- inclined distance of the center of pressure with respect to the fluid surface c.g. -- centroid(center of gravity) -- portion where concentration of weight is to be located c.p. -- center of pressure -- portion where the hydrostatic force is to be located e -- eccentricity(distance between centroid and center of pressure), m

Location of center of pressure: …from water surface “o” where:

…from center of gravity or centroid “g”


F = A P centroid

= A δfluid ђ

yp = Io / A ў

Io = Ig + A ў2

e = Ig / A ў

Water/ liquid surface

where: A -- area of submerged object or body

ў -- distance from the water surface to centroid along the y-axis e -- eccentricity Ig -- moment of inertia of the submerged surface about the centroidal axis

“ for any vertically submerged object ђ = ў ”

TABLE A: Moment of Inertia, Areas, location of centers of

gravity of plane figures.


SAMPLE PROBLEMS: 1. A vertical rectangular sluice gate at the

bottom of the dam is 0.6m wide and 1.8m high and is exposed to water pressure on one side corresponding to a head of 15m above its center. Assuming the gate on stem to weigh 2.23 kN and the coefficient of friction of gate on runners to be 0.25, find the hydrostatic force acting on the depth, the frictional force between the gate and runners, and the force necessary to raise the gate.

2. A gate with a circular cross-section is held

closed by a lever 1m in length attached to a buoyant cylinder. The cylinder is 25 cm in diameter and weighs 0.2 kN. The gate is attached to a horizontal shaft so it can pivot about the center. The liquid is water. The chain and the lever attached to the gate have negligible weight. The depth of water above the gate hinge is 10m. Compute the length of the chain such that the gate is just on the verge of opening.

3. In the figure shown, the gate is 1.2m square and

weighs 6.69 kN. Neglecting the thickness of the gate and the weight of the chain, compute the minimum force required to open the gate provided that the water exerts a force of 12, 714 N on the gate.


4. From the figure shown, the gate is 1m wide and

is hinged at the bottom of the gate. Determine the minimum volume of concrete with unit weight of 23.6 kN/ cu.m. needed to keep the gate in a closed position.

5. Gate AB is connected by a vertical rod to which a spherical ball is attached by a cable. The gate is 4m wide and the spherical ball has a relative density of 2.40. Compute the radius of the spherical ball needed to maintain the gate AB in its current position.

6. Compute the height of water for which the gate AB will start to fail if it has a length of 16 ft and width of 8 ft. The gate supports a load of 11 kips. Neglect the weight of the gate which is inclined at 600 with the horizontal and hinged at B and pulley friction.


7. The gate in the figure shown is 1.5 m wide, and

is hinged at point A, and rests against a smooth wall at B. Suppose that the minimum stress on the pin used at point A is at 20 MPa, would a pin of 10 cm diameter be possible to be used?

8. From the composite layers of fluids stored on a

vessel shown,, how far along the rectangular gate’s surface is the hydrostatic force acting?

Fluid Mechanics Chapter 1-5

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Transcript of Fluid Mechanics Chapter 1-5