BB101 Engineering Science Chapter 5 Solid and Fluid 1

SOLID AND FLUID BB101- ENGINEERING SCIENCE

UNIT SAINS JMSK PUO/DIS 2012 Page 42

5.0 SOLID AND FLUID 5.1 State Characteristics of solid , liquid and gas

SOLID LIQUID GAS

Microscopic view

Particle Arrangement

Tightly packed, usually in a regular pattern.

Close together with no regular arrangement.

Well separated with no regular arrangement.

Shape Fixed shape Follow container Undefined Shape

Volume Fixed volume Fixed volume Undefined volume

Kinetic Energy Content

Low Average High

Compressibility Hard to compress Compressible Easily to compress

Flow Ability Does not flow Flows easily Flows easily

Movement Vibrate (jiggle) but

generally do not move from place to place.

Vibrate, move about, and slide past each

other.

Vibrate and move freely at high speeds.

5.2 Define the Density and Pressure

DEFINITION

The density of a material is defined as its mass per unit volume.

SYMBOL

The symbol of density is (rho).

FORMULA

Where:

(rho) is the density,

m is the mass in kg,

V is the volume in m3.

SI UNIT

DENSITY

V

m



Different materials usually have different densities.

Example 1:

Calculate the density and relative of wooden block which has dimensions and a mass of .

Solution:

3

222

3

4000

105102101

1040

m

kg

v

m

DEFINITION

Relative density is the ratio of the density (mass of a unit volume) of a substance to the density of a given reference material.

Specific gravity usually means relative density with respect to water.

SYMBOL

The symbol of relative density is RD.

FORMULA:

SI UNIT:

No Unit

Rel

ativ

e D

ensi

ty

y of waterThe densit

ialy of materThe densitRD



Example 2:

A solid metal cylinder has radius cm and length . Its mass is . Find the density of the metal and its relative density.

Solution:

36

22

2

1093.3

105.0

m

lrVolume

3

6

3

4.16539

1093.3

1065

m

kg

v

m

53916

1000

416539

RDDensity, Relative

.

.

waterdensity of

materialdensity of

DEFINITION:

The pressure, P, is defined as the ratio of force to area

SYMBOL:

The symbol of Pressure is P.

FORMULA:

Where:

P is the Pressure,

F is the Force in newton,

A is the Area im m2.

SI UNIT:

Application of pressure: cutting tools, injection needle and tip of thumbtack.

PRESSURE

Paal Pasc or m

N2

A

FP



Example 3:

A hammer supplies a force of . The hammer head has an area of . What is the pressure?

Solution:

Pa109.86or m

N109.86

100.7

700

5

2

5

4

A

FP

5.3 Variation Of Pressure With Depth Relating Pressure in a liquid to the Depth and Density of the liquid: Consider a cylindrical container oh height, and cross-sectional area, which is filled with a liquid of density

Volume of liquid in the container : AhV

Mass of liquid in the container : AhVm

Force on point X :

gAh

mg

F

liquid theofweight

Pressure on point X :

gh

A

gAh

A

FP

Therefore :

ghP

h liquid

ofDensity

AArea X



Example 4: What will be the: (a) the gauge pressure and (b) the absolute pressure of water at depth below the surface? (Given that:

, and ).

Solution:

a)

2

2

72.117

117720

1281.91000

m

kN

m

N

ghP watergauge

b)

272.218

10172.117

Pr

m

kN

PPessureAbsolute atmospheregauge

5.4 Pascals Principle

Pascals principle states that pressure exerted on an enclosed fluid (liquid) is transmitted equally to every part of the fluid (liquid).

Hydraulic systems can be used to obtain a large force by the application of a much smaller force.

We can turn this phenomenon to our advantage if we alter the areas exposed to equal pressures, as in an hydraulic lift:

Since the pressure must be the same everywhere:

Pressure is;

A

FP

So;

1

212

2

2

1

1

21

A

AFF

A

F

A

F

PP at at

2

1



This says that the Force at the outlet (at 2) is augmented by the size of the area of the outlet. So if we make the area 1000 times larger, we can lift 1000 times the force we apply at F1.

By applying Pascals principle on a simple hydraulic system,

Applications of Pascals principle include the hydraulic jack, hydraulic lift and hydraulic brakes.

Example of Applications of Pascals Principle (Hydraulic Lift).

2

2

1

1

A

F

A

F

2211xAxA

Where: F1 = force at 1 F2 = force at 2 A1 = cross sectional Area at 1 A2 = cross sectional Area at 2 x1 = distance moved at 1 x2= distance moved at 2



Example 5: A hydraulic car lift has a pump piston with radius . The resultant piston has a radius of . The total weight of the car and plunger is . If the bottom ends of the piston and plunger are at the same height, what input force is required to stabilize the car and output plunger?

Solution:

We need to use the area for circular objects, for both the piston and plunger. Apply Pascal's Principle:

N

r

rF

A

AFF

B

AB

B

ABA

20.131

150.0

0120.020500

2

2

2

2

5.5 Archimedes' Principle

Archimedes principle states that an object which is partially or wholly immersed in a fluid (liquid or gas) is acted upon by an upward buoyant force equal to the weight of the fluid it displaces.

An object weighs less in water than it does in the air.

This loss of weight is due to the upthrust of the water acting upon it and is equal to the weight of the liquid displaced.

If the weight of the water displaced is less than the weight of the object, the object will sink.

Otherwise the object will float, with the weight of the water displaced equal to the weight of the object.

Archimedes Principle:

The buoyant force is equal to the

weight of the displaced water.



Floatation: The principle of floatation states that a floating body displaces its own weight of the liquid in which it floats. According to Archimedes Principle: Therefore;

Figure below show four situations of object in a liquid:

Situ

atio

n 1

FWo

forceBuoyant object ofWeight

However;

displaced liquid ofWeight forceBuoyant

lloo

lloo

lo

lo

VV

gVgV

gmgm

WW

For totally submerged object;

lo VV

Situ

atio

n 2

FWo


However;


lloo

lloo

lo

lo

VV

gVgV

gmgm

WW


lo VV

Buoyant force

Weight

Rising

<

Buoyant force

Weight

Rising

>

Buoyant force = Weight of liquid displaced

Vg

mgF

displaced liquid of Weight = forceBuoyant



Situ

atio

n 3

FWo


However;


lloo

lloo

lo

lo

VV

gVgV

gmgm

WW


lo VV

Situ

atio

n 4

FWo


However;


lloo

lloo

lo

lo

VV

gVgV

gmgm

WW


lo VV

NOTE : placedliquid disobject, lo

Archimedes' Principle explains why steel ships float.

Applications of Archimedes principle can be found in ships, submarines, hot-air ballons and the hydrometer.

Buoyant force

Weight

Floating

=

=

Buoyant force

Weight

Floating

Displaced water weight < ball weight Sink

Displaced water weight = hull weight Float



Example 6:

The buoyant force acting on the object will decrease when the:

weight of the object decrease Example 7:

A concrete slab weight is , when it is fully submerged under the sea, its apparent weight is . Calculate the density of the sea water if the volume of the sea water displaced by the concrete slab is .

Solution:

N48

102150

ightApprent we weightActualforceBuoyant

According to Archimedes Principle:

3

6

6

1019

81.9104800

48

81.910480048

displaced water sea ofweight forceBuoyant

m

kg

Vg

mgFB

Object

Liquid

Buoyant Force



Example 8: Figure below shows a boat loaded with some goods floating on the sea. The density of the sea is

.

(a) Calculate the weight of the boat.

(b) Figure below shows the situation of maximum loading of the boat.

Calculate the additional weight of goods that has to be added to the boat to reach this situation.

Solution:

(a)

N

Vg

14994

8.95.11020

watersea ofWeight boat theofWeight

(b) For maximum loading;

N

Vg

44982

8.95.41020

displaced water sea ofWeight weightAdditionalboat ofWeight

Therefore;

N29988

1499444982added be togoods of weight Additional



Tutorial 5a (Density and Pressure)

1. An object has a mass of and a volume of . What is the density of the object?

2. A substance having a density of . What is the volume of the substance if the mass is

?

3. A room with a dimension of is filled with of air. What is the mass of

the air?

4. A measuring cylinder is filled up with a liquid having a mass of . What is the density

of the liquid?

5. A liquid having a density of . If 1cm3 of the liquid turn into vapors, find the

density of the vapor.

6. Liquid and having a density of and respectively. Without any changes

of volume, a liquid is added to liquid . Calculate the additional density of the liquid.

7. A pressure is exerted on the floor. Calculate the force acting on the floor.

8. A wood block with a dimension of height having a mass of

. Determine the pressure exerted by the wood block.

9. A blade with a dimension of having a force of to cut a meat. What is the pressure exerted by the blade?

10. A tank with a dimension of is filled with paraffin. ( ) Calculate :

a) The pressure exerted on the base of the tank b) The force that acted on the base of the tank

11. A density of seawater is . What is the pressure exerted by the seawater at a vertical depth of . ( Given: )

Tutorial 5b (Pascal Principle) 1. Figure shows a simple hydraulic system. Piston A & B has a cross sectional area of

and respectively.

a) What is the pressure at piston A, when load is placed on it? b) What is the pressure at piston B. c) If load W placed at piston B, determine the load W that can be pushed by piston B.

2. Figure shows a pipe system that is filled with oil. If the piston A is pushed with force:

(a) What is the pressure exerted by the oil (b) What is the force acted on piston B



3. Figure shows a simple hydraulic system.

Cross sectional area of , and Cross sectional area of

a) What is the pressure at piston , if force acting on a piston ? b) If load placed at piston , determine the load that can be pushed by piston . c) If the distance moved by the piston A is 1.5m, what is the distance moved by piston B?

d) If load is being replaced with 210kg load, what is the force acted on piston in order to support the load at piston ?

Tutorial 5c (Archimedes Principle)

1. An empty boat having a weight of is floating statically . (

) a) What is the buoyant force? b) What is the volume of the displaced water?

2. A cube of metal having a volume of is completely submerged in . a) water (

)

b) oil (

)

c) Oxygen ( )

According to the following, what is the buoyant force? 3. A metal block having a weight of is completely being submerged in to the water. The weight

of the block when it completely submerged is a) What is the volume of the block

b) What is the density of the block ( ).

4. Figure (a) shows an object is weighed in air and found to have a weight of 2.0N. While Figure (b)

shows the object is completely submerged into the water. a) What is the mass of the object? b) What is the buoyant force? c) What is the mass of the displaced water? d) What is the volume of the water?



Answer 5a: 1) 1500 kg/m3 2) 0.0003 m3 3) 97.5 kg 4) 800 kg/m3

5) 2 kg/m3 6) 840kg/m3 7) 20 N 8) 159.6 Pa 9) 8.3 x 106 Pa 10a) 15696 Pa 10b) 188352 N 11) 303129 Pa Answer 5b: 1a) 200kPa b) 200kPa c)100 kg 2a) 66.67 kPa b) 13.33N 3a) 500 Pa b) 600N c) 0.1 m d) 140N Answer 5c: 1a) 2000 N 1b) 0.2 m3 2a) 7.848 N 2b) 6.28 N 2c) 0.12 N 3a) 2 x 10-5 m3 3b) 5000 kg/m3 4a) 0.2 kg 4b) 0.2 N 4c) 0.2 kg 4d) 2 x 10-4 m3

Minimum requirement assessment task for this topic: 1 Theory Test & 1 End-of-Chapter Specification of Theory Test: CLO1- C1 & CLO3-C2, A1 Specification of Labwork: CLO2- C2, P1 **************************************************************************************************** COURSE LEARNING OUTCOME (CLO) Upon completion of this topic, students should be able to:

1. Identify the basic concept of solid and fluid (C1) 2. Apply concept of solid and fluid to prove related physics principles. (C2,P1) 3. Apply the concept of solid and fluid in real basic engineering problems. (C2, A1)

**************************************************************************************************** Compliance to PLO

PLO 1, LD1 (Knowledge)-Test 2 PLO 2, LD2 (Practical Skills)- Experiment 3

PLO 3, LD4 (Critical Thinking and Problem Solving Skills)- Test 2

BB101 Engineering Science Chapter 5 Solid and Fluid 1

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