Properties Of Matter Sheet 3

Physics For Premedical Students Sheet # 3

Properties of Matter

Dr. Maan A. Ibrahim

2008 - 2009

Fluid Pressure Pressure The pressure is : the force that is applied over a given area.

SI = Pascal (Pa) Basic unit = N/m² = kg/m.s² • With same force exerted on two different areas  bigger pressure with smaller

area p ressu re =

fo r c e (SI unit: pascal (Pa)=N/m2; 1KPa=1000Pa) area

Block cannot pop the balloon but the pin can pop up the balloon. b/c pin has a very small area  big pressure

•ex) A weight of 4000N box of cube shape.

Each side of the square bottom is

2m.. What pressure does the box exert on the ground? 4000N ÷(2m×2m)=4000N÷4m2=1000N/m2=1000Pa=1KPa Fluid pressure

2

• Static pressure in fluids: if fluid is at rest (not flowing or moving) pressure is

equal in all directions  Pascal’s Principle •water pressure increases when depth increases. Pascal's Principle - pressure in fluid (liquid/gas) gets evenly distributed

hydraulic press (ex car lift in garage) P = F1/A1 = F2/A2 so, by increasing A2, you can increase F2

Pressure of a static (=not moving) fluid (hydrostatic pressure) Liquid Pressure

= density × gravitational acceleration ×

height(=depth) P is the hydrostatic pressure (in pascals); ρ is the water density (in kilograms per cubic meter); 1g/cm3 = 10-3kg/106

m3=1000kg/m3

g is gravitational acceleration (in meters per second squared); h is the height of fluid above (in meters). Pressure on Dams

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Liquid Pressure = density x gravitational acceleration × depth P1 = 1000 kg/m3 × 9.8m/s2 x 3m

P2= 1000 kg/m3 × 9.8N/kg x 6m

= 1000 kg/m3 × 9.8N/kg x 3m

= 58,800N/m2= 58,800Pa = 58.8Kpa

= 29400N/m2=29000.4Pa =29.4 KPa • Pascal vase – several shape vases all connected at the bottom • When water put in, water flows until all vases have same height (=depth) of water in all vases • Pressure depends on height, not on amount of fluid

When a body is submerged in a fluid, the fluid exerts a force perpendicular to the surface of the body at each point on the surface. However, for a fluid, the force does not necessarily need to be a constant over the whole surface. This leads us to a more refined definition of pressure. We define the pressure p at a point in a fluid as the ratio of the normal force dF on a small area dA around that point, to the area p=

dF dA

From this definition we can immediately see that pressure also has units of Pascals. The pressure due to a fluid tends to compress a body. The ratio of the pressure to the fractional decrease in volume is called the bulk modulus

4

B=−

p ∆V

V

The inverse of the bulk modulus is called the compressibility, k. Air Pressure and the Atmosphere •air pressure depends on the depth/height of air on top of us. (=altitude) • altitude increases (when you go high)  pressure decreases (b/c height of air

decreases) • Normal atmospheric pressure (normal air pressure at sea level)

1 atmosphere = 760 mmHg= 29.92 in Hg = 14.7 lb/in2 = 101,300Pa =101.3 KPa = 1,013.0 millibar

(barometer) Fluid Mechanics Let us now look at the properties inherent in a fluid. A fluid differs from a solid in that it cannot support a shear stress. In this sense, both liquids and gases can be described as fluids (in fact, under certain extreme conditions, even solids exhibit fluid-like properties). The density of a fluid is defined as its mass per unit volume

5

ρ=

m V

Density is considered to be inherent to the fluid and can be used to characterize it. The specific gravity of a fluid is the ratio of its density to that of water. It is a dimensionless number. Since the density of water is defined to be 1 gm/cm3, the specific gravity of a fluid is the density without the units, if the density was also given in gm/cm3. Properties of Matters 1)

Density

Mass density is the mass per unit volume. The volumes of different matters are equal, but they aren’t equal in their masses. Density

ρ=

m v

The density’s unit in SI unit and in the basic unit are same = kg/m³ The density depend on: 1-

The kind of matter.

2-

Temperature.

3-

Pressure.

The density of solids is higher than the density of liquids and the density of liquids is higher than the density of gases. The densities of solids and liquids are approximately independent of pressure. Effect of temperature on the Density: Most substances expand when their temperature increase , but the densities of substances decrease , the relation is inversely proportion Accept the water , is exempted between temperatures 0 and 4 °C.

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Specific Gravity of substance: SG (Specific Gravity) =

ρ substance ρ water

kg/m³

Atmospheric Pressure: The weight of the air in the upper of the Earth’s atmosphere exerts pressure on the layer of the air below . Mercury barometer 1 atm = 1.013 x

=760 mmHg

Pressure in a fluid Pressure is defined as the perpendicular force on a surface per unit surface area.

P=

F A

P0 A

[N/m2 = Pascal = Pa]

Mg

h PA

Mass density is mass per unit volume. ρ=

M V

[kg/m3, g/cm3, …]

Water has a density of about 1 g/cm3 (= 103 kg/m3). Due to gravity, pressure increases with depth in a fluid as P = P0 + ρgh ,

where P0 is the pressure at the top of the fluid and P is the pressure at a depth h. As seen in the diagram to the right, additional pressure below a section of a fluid is

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required to keep this section from sinking.

Since this section of fluid is in

equilibrium, PA − P0 A − Mg = 0

But M = ρ V = ρ Ah, so

PA − P0 A − ρAhg = 0 P = P0 + ρgh If the fluid container is open to the atmosphere, then P0 is atmospheric pressure. At sea level P0 = 1.013 x 105 Pa ( = 14.7 lb/in2 = 1 atm) Example: What is the downward force exerted by the atmosphere on top of a 2 m x 1 m desk top. F = P0A = (1.013 x 105 Pa)(2 m2) = 1.026 x 105 N The reason why this enormous force doesn’t crush the desk is because of a nearly equal upward force on the bottom of the table. Example: At what depth in water is the pressure 2 atm?

P = 2 P0 = P0 + ρ g h ρ g h = P0 P0 1.013 x 10 5 Pa h= = = 10.3 m ρ g ( 10 3 kg / m 3 )( 9.8 m / s 2 ) Example: A mercury manometer consists of an inverted tube of mercury as shown to the right. The top end is closed and the void at the top is essentially a vacuum. The bottom end is open and is in an open container of mercury. What is the height of the column of mercury in the tube? The specific gravity of mercury is 13.6. 8

P = P0 + ρgh P = Patm ( bottom )

P~0

P0 = 0 ( top ) h=

Patm

ρg

h

1.013 x 10 5 N / m 2

=

3

3

Patm

2

( 13.6 x 10 kg / m )( 9.8 m / s ) h = 0.76 m = 76 cm

Hydraulic press A hydraulic press uses a fluid to magnify an applied force. A force F1 applied to the small piston of area A1 increases the pressure in the fluid by P1 = F1/A1. This pressure increase is transmitted uniformly throughout the fluid (Pascal’s principle). This additional pressure results in a lift on the large piston. P2 = P1 F2 F1 = A2 A1

F1

A F2 = F1 2 A1

F2 fluid

Example:

In a hydraulic press, the diameter of the small piston is 2.5 cm and the diameter of the large piston is 10 cm. If the force applied to the small piston is 500 N, what is the force applied to the large piston?

F2 = F 1

A2 A1

= ( 500 N )

π ( 0.05 m ) 2 π ( 0.0125 m ) 2

= 2000 N

Archimede’s Principle Archimede’s principle states that an object submerged in

B

fluid is buoyed up by a force equal to the weight of the displaced by the object.

fluid W

B = W f = m f g = ρ f Vobj g

9

a

The buoyant force, B, is just a consequence of the fact that the pressure below the object is greater than above it. To understand Archimede’s principle, we envision replacing the object with fluid of the same size and shape. This fluid must be in equilibrium and have the same buoyant force as the object. Thus, its weight and the buoyant force must be the same. Example: A cubical block of aluminum 10 cm on edge is suspended in water by a cord. What is the tension in the cord? The density of Al is 2.7 x 103 kg/m3. Since the block is in equilibrium, T B

mg

T + B − m Al g = 0 T = m Al g − B = m Al g − m water g

= ρ Al Vg − ρ w Vg = ( ρ Al − ρ w ) gV = ( 2.7 x 10 3 kg / m 3 − 1 x 10 3 kg / m 3 )( 9.8 m / s 2 )( 0.1m ) 3 = 16.7 N

Example: An ice cube floats in a glass of water. What fraction of its volume is below water?

The specific gravity of ice is

0.917.

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W ice = B m ice g = m w g

ρiceVg = ρwVbelow g Vbelow ρ = ice = 0.917 V ρw Thus, 91.7% of the volume of the ice is below the water. Fluid Dynamics Equation of continuity The rate at which fluid mass flows through two different parts of the same pipe must be the same. Thus,

v1

∆M 1 = ∆M 2

A1

ρ 1 ∆V 1 = ρ 1 ∆V 2 ρ 1 A 1 ∆x 1 = ρ 1 A 2 ∆x 2

v2 A2

ρ 1 A 1 v 1 ∆t = ρ 1 A 2 v 2 ∆t

If the fluid is incompressible, i.e., its density is nearly the same throughout the pipe, then we have

A 1v 1 = A 2 v 2

(Eq. of continuity)

Example: A hose of diameter 3 cm has a nozzle of diameter 1 cm. If the water flows at 2 m/s in the hose, what is the water speed as it goes through the nozzle? A1 π r1 2 ( 3 )2 v 2 = v1 = v1 = ( 2m / s ) = 18 m / s A2 ( 1 )2 π r2 2

Bernoulli’s equation: Bernoulli’s equation gives a relationship in a flowing fluid between the fluid’s pressure, flow speed, and elevation. It is based on conservation of energy and 11

holds for an ‘ideal’ fluid.

The ideal fluid would be (1) non-viscous, (2)

incompressible, (3) steady in its flow, and (4) non-turbulent.

P + 1 ρ v 2 + ρ g y = constant Bernoulli’s equation 2

This means that if you pick any two points in a flowing fluid, P 1 + 12 ρ v 1 2 + ρ g y 1 = P 2 + 12 ρ v 2 2 + ρ g y 2

where p is the absolute pressure. This is called Bernoulli's Equation. If the fluid is at rest (v1 = v2 = 0), then Bernoulli’s equation is the same as the earlier equation giving P as a function of depth in a fluid – P1 = P 2 + ρ g ( y 2 − y1 )

Qualitatively, Bernoulli’s equation says that the pressure is lower in a region of a fluid where its speed is greater. Example: An airplane wing has curvature and angle of attach such that the air speed above the wing is greater than below. If v(below) = 100 m/s and v(above) = 103 m/s and the area of the wing is 10 m2, what is the lift on the wing? The density of air is about 1.3 kg/m3. The difference in elevation below and above the wing is nearly the same, so y1 ~ y2 and ∆P = P 1 − P 2 = 21 ρ ( v 2 2 − v 1 2 ) = 21 ( 1.3 kg / m 3 )(( 105 m / s ) 2 − ( 100 m / s ) 2 ) = 666 N / m 2 Lift = F = ∆PA = ( 666 N / m 2 )( 10 m 2 ) = 6 ,666 N

Example: Water enters a house through a pipe 5 cm in inside diameter at absolute pressure of 1 x 105 Pa. The pipe leading inside the house to the second floor 2.5 m above has an inside diameter of 3 cm. When the flow velocity at the inlet pipe is 1.5 m/s, what is the flow velocity and pressure at the upstairs bathroom?

12

h

v

Solution : A1 v 1 = A2 v 2 A v 2 = 1 v1 A2 =

( 5 cm ) 2 (1.5 m ) s ( 3 cm ) 2

= 4.2 ms

From Bernoulli's equation, the pressure is

(

)

p2 = p1 − 1 ρ v 22 − v12 − ρg ( y2 − y1 ) 2

(

)

(

) (

2 kg  = 1 × 10 5 Pa − 12  1 × 10 3 3   4.2 ms − 1.5 ms m  

)2  −  1 × 10 3 mkg  9.8 sm ( 2.5 ms ) 3

2

= 6.8 × 10 4 Pa

Properties of a System Density, ρ, is defined as the mass per unit volume

ρ=

m ass V

alternatively, we may define the specific volume as the volume per unit mass:

v =

V m ass

• Pressure: An intensive property of a fluid, defined as the average force exerted by the fluid per unit area:

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P = Force/Area Atmospheric Pressure: The atmosphere surrounds Earth for a depth of about 70 kilometers. The weight of this air exerts a force at sea level of about: Patm = 1 atm. = 14.7 lbf/in2 (psi) = 101 kN/m2 (kPa) = 1.01 bars The atmospheric pressure is reduced as we climb to higher elevations, in relation to the decrease in the weight of air above the surface. Absolute Pressure: If we increase the force exerted on a fluid, the resultant pressure is increased. The pressure on the gas, measured in this fashion, is said to be the absolute pressure. Most scientific measurements are based on absolute pressure measurements. Gauge Pressure: In many common applications, it may not be convenient to determine the atmospheric pressure.

In such applications gauge pressure is

often used:

Pgauge = W/A Patm W

Pabs = (W+PatmA)/A Pabs = (W+PatmA)/A

A Patm W

Pabs = (W+PatmA)/A Pabs = (W+PatmA)/A

A

Pgauge = Pabs - Patm Pvaccuum = Patm - Pabs 14

• Manometers Manometers are simple, inexpensive and accurate devices used to measure fluid pressure.

A Gas

PA = 1 atm z

B

PB = PA + m·g/A = PA + ρ [(A·h)] g/A = PA + ρ·g·z fluid

Because of the low density of the gas, the gas pressure will be virtually uniform within the container. We see then that the differential height of the fluid in the column is a direct measurement of the gauge pressure of the gas : Pgauge = ρ·g·z Example : Assume that the manometer fluid is an oil with a specific gravity of 0.87 and that we can read elevational differences of 0.001 m (1 mm). Solution : Pgauge = (0.87 x 1000 kg/m3)(9.81 m/s2)(0.001m) = 8.53 Pa = 8.42x10-5 atm = 1.24 x10-3 psi 15

FLUID PROPERTIES Basic Units  Every fluid has certain characteristics by which its physical condition may be

described. These characteristics are called the properties of the fluid.  Properties are expressed in terms of a number of basic dimensions (length, mass or force, time, temperature) which are quantified by basic units.  There are two systems of units: the SI system of units and the traditional units. SI System of Units 

The basic units of mass, length, time, and temperature are the kilogram (kg), meter (m), second (s) and Kelvin (K).

 The dimensions and units of other quantities are derived from those of the basic dimensions and units, e.g.  Specific Gravity, S 

The ratio of the specific weight of a given fluid to the specific weight of water at 4oC is defined as the Specific Gravity.

S = 

γ fluid γ

= o

water @ 4 C

ρ fluid ρ

water @ 4 o C

The specific gravity is dimensionless. Why ?

 Ideal Gas Law 

The equation of state for an ideal gas is :

P = ρ RT where P = absolute pressure; T = absolute temperature

ρ = density; R = gas

constant. The gas constant is related to the universal gas constant Ru by 16

Ru R= M where M = molar mass. (Note: Ru = 8.314 KJ/Kgmol . K) Viscosity The term viscosity (also referred to as the body of paint) describe the liquid paint’s resistance to flow. Viscosity results from the attractive forces that occur between the particles that make up paint. Attractive forces not only prevent particle from separating from one another but also restrict the movement of the body of liquid itself. Because attractive forces decrease with increasing temperature, temperature has a strong effect on paint viscosity. viscosity - the resistance of a liquid to flow; the higher the viscosity, the more sluggish to flow Viscosity arises from the forces between molecules; strong intermolecular forces hold molecules together and do not let them move past one another. Viscosity usually decreases as the temperature rises. Molecules have more energy at high temperatures and can wriggle past their neighbors more readily. The viscosity of water at 100 oC is only one-sixth of its value at 0 oC. Viscosity usually decreases with increasing temperature.  In a viscous fluid, the shear stress is proportional to the time rate of strain. Mathematically it is expressed as τ =µ

dV dy

where τ is the shear stress, V is the fluid velocity and y is the distance measured from the wall. dV/dy is the rate of strain (which is also the velocity gradient normal to the wall). The proportionality factor, µ , is the dynamic, or absolute, viscosity. It is a property of the fluid.

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 The SI unit of viscosity is N.s/m2.

 Consider the velocity distribution next to a boundary. The following observations can be made:  The velocity gradient becomes smaller with distance from the boundary. Therefore, the maximum shear stress is at the boundary. 

The fluid velocity is zero at the stationary boundary. Viscosity causes the fluid to adhere to the surface. This is known as the no-slip condition.

 Kinematic viscosity is defined as

ν = µ ρ  The viscosity of a gas increases with the temperature of the gas. The variation of gas viscosity with absolute temperature can be estimated with Sutherland’s equation, µ  T  = µ o  To 

3 2

To + S T+S

where µ o is the viscosity at temperature To and S is Sutherland’s constant. The viscosity of a liquid decreases as temperature increases. An equation for the variation of liquid viscosity with absolute temperature is : µ = C eb T

where C and b are constants. Surface Tension  The molecules at the surface of a liquid have a greater attraction for each other than they do for molecules below the surface. The surface behaves as if it were a “skin” or “membrane” stretched over the fluid mass.

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 Because of the membrane effect, each portion of the surface exerts “tension” on adjacent portions of the surface or on objects that are in contact with the liquid surface.  The tension acts in the plane of the surface. The magnitude of the tension per

unit length is defined as the surface tension, The concept of surface tension has been used to explain several commonly observed phenomena. Examples of these are:  A steel needle will float on water if placed gently on the surface because the surface tension force supports the needle. 

Capillary action in a small-diameter tube .

 The excess pressure created inside droplets and bubbles. Vapor Pressure  The pressure at which a liquid will boil is called its vapor pressure.

 The vapor pressure increases as the temperature increases.  Illustration: At 100oC, the vapor pressure of water is 101.3 kPa, whereas at 10oC

the vapor pressure is 1.23 kPa. This means that water at 101.3 kPa will boil at 100oC. But if the pressure of water is lowered to 1.23 kPa, boiling will take place at 10oC.  Application: In flowing fluids it is possible to have very low pressure due to the

fluid motion. The liquid will boil if the pressure is lowered to the vapor pressure. This may occur in flow through the irregular, narrowed passages of a valve or the suction side of a pump. The vapor bubbles formed will collapse in regions of higher pressure downstream. This phenomenon (i.e. the formation and subsequent collapse of vapor bubbles) is called Cavitation. It can lead to structural damage and, therefore, should be avoided.

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The weight of the atmosphere exerts a pressure of 14.7 pounds per square inch of force at sea level. This is to say that a 1 inch column of air as tall as the atmosphere, would weigh 14.7 pounds. We commonly call this . 1 Atmosphere of pressure, or 1 Atom. (=101.3 kPa the SI unit) Water exerts much more pressure. It only takes a 1 inch column of sea water 33 feet tall to weigh 14.7 pounds. This means that at a depth of 33 feet deep in the ocean, there is a total pressure of 29.4 pounds per square inch (psi). This would be 2 ATMs of pressure. One ATM from the water, + one ATM from the air. We call this the ambient pressure or absolute pressure. Absolute pressure differs from gauge pressure. Gauge pressure would be the pressure that would show up on a gauge at this depth. A pressure gauge would start at 0 at the surface and show 14.7 psi at our depth of 33 feet. Gauge pressure would ignore the 14.7 psi of atmospheric pressure. Continuity Equation Consider the pipe section shown above. The pipe cross sectional area changes from A1 to A2. The question here is what happens to the fluid velocity.

If the pipe is not leaky and the fluid does not compress, then :

Fluid in = Fluid out 20

we can calculate each of these sides. Also since the fluid does not compress, the density is constant throughout the fluid. Thus: in a time ∆t, we have:

ρ A1 ( v1 ( ∆ t ) ) = ρ A 2 ( v 2 ( ∆ t ) ) This is easily solved to give: A1 v 2 =A v1 2

This is known as the “Continuity Equation” for fluids. Next, you may have wondered just how a fluid flow works (yes... they are pretty clear things right up to the point that you start to siphon a fluid). undiminished through the fluid. Let’s see how this all works: If you place a weight say m1 on plate A1, then the pressure pulse is given by:

∆P1 =

m1g A1

According to Pascal’s principle, this overpressure must be the same at plate 2. Thus, at plate 2, ∆P2 = ∆P1

This overpressure will lift a weight m2. How much will it lift? m1g A1

=

m2g A2

⇒ m 2 = m1

( ) A2 A1

The essence is that the smaller A1 is, the bigger the mass m2 that can be lifted. This is basically how a hydraulic lift or a hydraulic press works. Solids and Fluids This chapter deals with properties of solids and fluids (liquid and gases). One of the properties of a material is its elasticity, which is a measure of the extent to which a force can deform the material. In particular, the elastic modulus, is defined as 21

elastic modulus =

stress strain

The elastic modulus is a measure of the stiffness of a material. Stress is the applied force per unit area (N/m2 = Pascal) and strain is the fractional change in a dimension due to the stress. There are three main types of elastic moduli. ∆L

Young’s modulus – elasticity of length

L0

F

Y=

F

tensile stress F/A = tensile strain ∆L / L0

A

Up to a certain limit, the stress is typically proportional to the strain; i.e., Y is a constant. Beyond a certain stress called the elastic limit, this proportionality ceases to exist and the material will be irreversibly strained. Additional stress will eventually break the material. A

Shear modulus – elasticity of shape F S=

∆x

F

shear stress F / A = shear strain ∆x / h

h F

Bulk modulus – elasticity of volume

B=

volume stress ∆F / A ∆P =− =− volume strain ∆V / V ∆V / V

∆P is the change in the pressure (force per unit area normal to the surface). The minus sign in the above equation is because an increase in pressure gives a decrease in volume and B is positive. Example:

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A copper wire has a length of 1 m and a diameter of 2 mm. How much force is required to stretch the wire by 1 mm? Y(copper) = 11 x 1010 Pa. Y=

F/A FL = ∆L / L A ∆L

YA∆L ( 11 x 10 10 Pa ) π ( 1 x 10 − 3 m ) 2 ( 1 x 10 − 3 m ) F= = = 346 N L 1m

---------------------------------------------------------------------------------------------------Problems and Answer 1-

What is the mass of the air in a living room has dimension of 4.0m wide, 6.0m length, and 2.5m height ?

Solution: (Mass) m= ? ρ=

→ m= ρ.V

(volume ) V = wide x length x height = 4.0m x 6.0m x 2.5m = 60 m³ ρ of the air = 1.29 kg/m³ m= ρ.V = 1.29 kg/m³ x 60 m³ = 77.4 kg 2-

If you have a cube of gold its length 6 cm and its mass 4168 g. Find density of gold?

ρ= m= 4168 g = 4168 x V= ρ= 3-

kg = 4.168 kg

= 216 cm³ = =

m³ = 2.16 x

= 19296.3 kg/m³

What volume of water has the same mass as 2.0 m³ of lead ?

23

m³

the

Solution: V of lead = 2 m³ ρ of lead = 11.3 x

kg/m³ = 11.3 ×

m of lead = ρ.V = (11.3 x mass of water = mass of lead m of water = 11.3 × ρ of water= 1.00 ×

kg/m³

V of water =

11.3 m³

4 – the nucleus of uranium has a diameter of 1.5 x kg. What is the density of the nucleus ? Solution: ρ= m= 4 x

kg

Sphere volume = V=

r³ = 0.75 ×

Radius (r) =

m

(Prefixes form) = 0.75 x = 0.75 × (10 x

) = 15-

24

and a mass of 4 x

= 7.5 ×

= 7.5 fm

(0.75 ×

V=

³ = 1.767 ×

ρ= 5-The dimension of rectangular of wood is 10m x 8m x 6m. what is the mass of the rectangular if its density is 0.6 g/cm³ ? Solution: V= 10 × 8 × 6= 480 m³

M= ? ρ=

=

=6x

m= ρ.V = 480 × (6 x

= 288 ×

6- How much is the pressure at a depth of 11.0 km in the Pacific Ocean . Assume the density of sea water is 1.025 g/cm³ and the atmospheric pressure is 1.01 x Pa ? Solution: h = 11.0 km = 11 x

m

ρ = 1.025 g/cm³ = 1.025 x P = ρ h g = (11 x = 1.106 x

kg/m³ (1.025 x

Pa

25

) x 9.81

7- Two fish swim in the fresh water lake, one ‫ ا‬at a depth of 14 m and the other at a depth of 98.0 m. What is the different in pressure on the fish? Select the reasonable values for P and ρ? Solution: ρ = 1.025 x

kg/m³ freshfirst P1 = ρ h1 g

h1=14m h2=98m

= 1.025 x = 1.41 x

x 9,81 Pa

Second fresh P2 = ρ h2 g = 1.025 x = 9.85 x P =

1.01

P total = P + P2 = ( 1.01 x = 1.42 x

Pa

Pt = P + P2 = ( 1.01 x = 9.956 x

) + (1.41 x

) + (9.85 x Pa

8- Suppose a plunger has a cross sectional area of 0.1 m 2 while a second plunger has an area of 0.01m2. The system is connected together via yucky orange fluid. 26

How much weight will the large plunger be able to pick up if 10 kg is placed on the small plunger? Solution: let 1 be the smaller plunger and let 2 be the larger plunger. Then, according to Pascal’s principle:

m1g A1

m2g A2

=

⇒ m 2 = m1

( ) A2 A1

The ratio of areas is a factor of 10. Thus,

m 2 = 10 ( 101 ) = 100kg 9- Suppose you eat lots of bad food that eventually blocks the cross sectional area of an artery by a factor of ½ . How much lower is the blood pressure in this region of the artery than an unblocked region ? Solution: The continuity equation tells us how much faster the fluid flows in the blocked region. Let 1 be unblocked and 2 be blocked. Then :

v 2 = v1

( )⇒v A1 A2

2

= v1

( ) = 2v A1 1A 2 1

1

10- Write the values in the prefixes form ? Pressure

in Pressure in prefixes form

Pa 2 x

2 x

4 x

4 x

6 x

6 x

1.01 x

1.01 x

2.8 x

2.8 x

1 x

= 1 p Pa

= 20 P Pa 400 x

400 G Pa

= 60 x K Pa

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Problems 1- The prefixes used to indicate multiplication of units are : G (giga) = ......

c (centi) = ......

M (mega) = .......

m (milli) = .......

k (kilo) = ........

µ (micro) =........

2- Atmosphere of pressure, or 1 Atom ( = ............ kPa

the SI unit)

The SI unit of γ is .......... Absolute pressure is the sum of ............................. plus .......................... ...................... is the resistance to flow of a liquid . The elastic modulus is defined as ......................... 3- The dynamic viscosity of air at 15oC is 1.78 × 10-5 N.s/m2. Using Sutherland’s equation, find the viscosity at 200oC . 4- What happens if the viscosity is too low ? 5- What happens if the viscosity is too high? 6- Explain why a water drop has surface tension. 7- The atmospheric pressure at the top and the bottom of a building are read by a barometer to be 96 kPa and 98 kPa. If the density of air is 1.0 kg/m3 . What is the height of the building?

[204 m]

8- Determine the atmospheric pressure at a location where the barometric reading is 750 mm Hg. Take the density of mercury to be 13600 kg/m3.

[99.96 kPa]

9- The absolute pressure in water at a depth of 5 m is read to be 145 kPa. Determine (a)the local atmospheric pressure , and (b) the absolute pressure at a depth of 5 m in liquid whose specific gravity is 0.85 at the same location. [96 kPa, 137.65 kPa] .

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