SOLID AND FLUID
BB101- ENGINEERING SCIENCE
5.0 SOLID AND FLUID
5.1 State Characteristics Characteristics of solid , liquid and and gas
SOLID
LIQUID
GAS
Tightly packed, usually in a regular pattern.
Close together with no regular arrangement.
Well separated with no regular arrangement.
Fixed shape Fixed volume
Follow container Fixed volume
Undefined Shape Undefined volume
Low
Average
High
Hard to compress Does not flow Vibrate (jiggle) but generally do not move from place to place.
Compressible Flows easily Vibrate, move about, and slide past each other.
Easily to compress Flows easily
Microscopic view Particle Arrangement Shape Volume Kinetic Energy Content Compressibility Flow Ability Movement
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 ρ (rho).
DENSITY
FORMULA Where: •ρ (rho) is the density, •m is the mass in kg, •V is the volume in m 3.
m V
SI UNIT
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SOLID AND FLUID
BB101- ENGINEERING SCIENCE
Different materials usually have different densities.
DEFINITION Relative Relative density is the ratio of the density (mass of a unit volume) of a substance to the density of a given reference reference material.
y t i s n e D e v i t a l e R
Specific gravity usually means relative density with respect to water.
SYMBOL •The symbol of relative density is RD. RD.
FORMULA:
SI UNIT:
No Unit Example 1: Calculate the density and relative of wooden block which has dimensions and a mass of . Solution:
m v
40 10 2 10 5
1
10
4000
2
2
3
10
2
kg m3
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SOLID AND FLUID
BB101- ENGINEERING SCIENCE
Example 2: A solid metal cylinder cylinder has radius radius cm and length . Its mass is . Find the density of the metal and its relative density. Solution:
Volume
Relative Density,RD
2
r l
0.5
10
6
2 2
3.93 10 m
m v
65 10
3
3.93 10 6 kg 16539.4 m3
3
density of material density of water
16539.4
1000 16.539
DEFINITION: The pressure, P, is defined as the ratio of force to area SYMBOL: •The symbol of Pressure is P. FORMULA: Where:
PRESSURE
•P is the Pressure, •F is the Force in newton, •A is the Area im m 2.
P
F A
SI UNIT UNI T:
Application of pressure: cutting tools, injection needle and tip of thumbtack.
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SOLID AND FLUID
BB101- ENGINEERING SCIENCE
Example 3: A hammer supplies supplies a force of . The hammer head has an area of . What is is the pressure? Solution:
P
F A
700 7.0 10
4
9.86 105
N m
9.86 105 Pa
or
2
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
h X
Volume of liquid in the container
:
V Ah
Mass of liquid in the container container Force on point X
: :
m V Ah F weight of the liquid
Pressure on point X
mg Ah g
:
P
F A Ah g
A h g
Therefore
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:
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SOLID AND FLUID
BB101- ENGINEERING SCIENCE
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)
P ga ug e
gh water
1000 9.81 12 117720 117.72
N
m2 k N m2
b)
Absolute Absolute Pr essure P ga ug e P atmosphere
117.72 101
218.72
kN m2
5.4 Pascal’s Principle Pascal’s principle states that pressure exerted on an enclosed fluid (liquid) is transmitted equally to every part of the f luid (liquid). (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 hydraulic lift:
1
2
Since the pressure must be the t he same everywhere: Pressure is;
P
F A
So;
P at 1 P at 2 F 1 A1
F 2 UNIT SAINS JMSK PUO/DIS 2012
F 2 A2
F 1 A2 A1 Page 46
SOLID AND FLUID
BB101- ENGINEERING SCIENCE
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 F 1. By applying Pascal’s principle on a simple hydraulic system,
F F A A 1
2
1
2
A x 1
1
A x 2
2 Where: F 1 = force at 1 F 2 force at 2 2 = A1 = cross sectional Area at 1 A2 = cross sectional Area at 2 x 1 = distance moved at 1 x 2 2= distance moved at 2
Applications Applications of Pascal’s principle principle include include the hydraulic hydraulic jack, hydraulic hydraulic lift and hydraulic hydraulic brakes. Example of Applications of Pascal’s Principle (Hydraulic Lift).
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SOLID AND FLUID
BB101- ENGINEERING SCIENCE
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:
A A A B r A 2 F B 2 r B 0.01202 20500 2 0.150 131.20 N
F A F B
5.5 Archimedes' Principle Archimedes’ principle 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 weighs less in water 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.
Archimedes Principle: The buoyant force is equal to the weight of the displaced water.
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.
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SOLID AND FLUID
BB101- ENGINEERING SCIENCE
Floatation: The principle of floatation states that a floating body displaces its own weight of the liquid in which it floats. According to Archimedes’ Archimedes’ Principle: Principle: Buoyant force = Weight W eight of liquid displaced Therefore;
Buoyant force = Weight of liquid displaced F mg Vg
Figure below show four situations of object in a liquid:
Weight of object
Buoyant force
W o F However; Buoyant force
Buoyant force Weight of liquid displaced 1 n o i t a u t i S
Rising
W o Weight
W l
mo g ml g oV o g l V l g
>
oV o
V
l l
For totally submerged object;
V o
Weight of object
V l
Buoyant force
W o F However;
2 n o i t a u t i S
Buoyant force
Buoyant force Weight of liquid displaced W o W l
Rising
mo g ml g Weight
<
oV o g l V l g oV o
l V l
For totally submerged object;
V o
UNIT SAINS JMSK PUO/DIS 2012
V l
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SOLID AND FLUID
BB101- ENGINEERING SCIENCE
Weight of object
Buoyant force
W o F Buoyant force
However;
Floating
Buoyant force Weight of liquid displaced 3 n o i t a u t i S
Weight
W o W l mo g ml g oV o g l V l g
l V l
oV o
=
For totally submerged object;
V o
Weight of object
V l
Buoyant force
W o F Buoyant force
However;
Floating
Buoyant force Weight of liquid displaced 4 n o i t a u t i S
Weight
W o W l mo g ml g oV o g l V l g oV o
=
l V l
For totally submerged object;
V o
V l
NOTE : o object, l liquid dis place placed d
Archimedes' Archimedes' Principle explains why steel ships float.
Sink
• Displaced water weight < ball weight
Float
• Displaced water weight = hull weight
ballons and the Applications of of Archimedes’ principle principle can be found foun d in ships, submarines, hot-air ballons hydrometer.
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SOLID AND FLUID
BB101- ENGINEERING SCIENCE
Example 6:
Object Liquid Buoyant Force
The buoyant force acting on the t he 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:
Buoyant force Actual weight
150 102
48 N
Apprent weight
According to Archimedes’ Archimedes’ Principle: Principle:
Buoyant force weight of sea water displaced F B
mg Vg
48 4800 10
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6
9.81
48
4800 1019
10
6
9.81
kg m3
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SOLID AND FLUID
BB101- ENGINEERING SCIENCE
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)
Weight of the boat Weight of sea water Vg
(b)
1020 1.5 9.8
14994 N
For maximum loading;
Weight of boat Additional weight
Weight of sea water displaced Vg
1020 4.5 9.8
44982 N
Therefore;
Additional weight of goods to be added
UNIT SAINS JMSK PUO/DIS 2012
44982 14994
29988 N
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SOLID AND FLUID
BB101- ENGINEERING SCIENCE
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 substanc e 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 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 Pis ton A & B has a cross sectional area of and respectively.
a) What is the pressure at piston piston A, when load is placed on it? b) What is the pressure at piston B. c) If load W placed at at piston B, determine the load W that can can be pushed by piston B. 2. Figure shows a pipe system that is filled with 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
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SOLID AND FLUID
BB101- ENGINEERING SCIENCE
3. Figure shows a simple hydraulic system.
Cross sectional area of , and Cross sectional area of a) b) c) d)
What is the pressure at piston , if force acting on a piston ? If load placed at piston , determine the load that can be pushed by piston . If the distance moved moved by the piston A is 1.5m, what is the distance distance moved by piston B? 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’ (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 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 is the volume of the the block block b) What is the density density of 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 completely submerged into the water. a) What is is the mass of the object? b) What is the buoyant force? c) What is the mass of the displaced water? d) What is is the volume of the the water? water?
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Answer 5a: 1) 1500 kg/m 3 5) 2 kg/m 3 9) 8.3 x 106 Pa
BB101- ENGINEERING SCIENCE
2) 0.0003 m3 3) 97.5 kg 4) 800 kg/m3 6) 840kg/m3 7) 20 N 8) 159.6 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 -5 3 3 3a) 2 x 10 m 3b) 5000 kg/m 4a) 0.2 kg 4b) 0.2 N 4c) 0.2 kg 4d) 2 x 10-4 m3
2c) 0.12 N
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. 2. 3.
Identify the basic basic concept of solid and fluid (C1) (C1) Apply concept concept of of solid and fluid to prove prove related related physics physics principles. principles. (C2,P1) Apply the concept of of solid and fluid in real basic engineering engineering problems. problems. (C2, A1) 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
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