Documents
Resources
Learning Center
Upload
Plans & pricing Sign in
Sign Out

Default Normal Template plunger

VIEWS: 814 PAGES: 18

									ESM 1224, SEMESTER 3, 2006/2007                                                    CHAPTER 2



                                     CHAPTER 2 - FLUIDS

2.1       Mass Density
   Fluids are materials that can flow which include both gases and liquids.
     Example:
        i) gas: air
        ii) liquid: water
   Mass density of liquid or gas is important factor that determine its behaviour
    as a fluid.
     Example:
        Air (gas) has smallest densities compared to water (liquid) because gas
        molecules are relatively far apart and contains large fraction of empty
        space. While for liquids, the molecules are much more tightly packed that
        leads to larger densities.
   Equation of mass density, 
                    = mass = m
                                          SI unit of  = kg / m3
                      volume v

   A convenient way to compare densities is to use the concept of specific gravity.
      specific gravity of substance = density of substance
                                      density of water at 4C (1000 kg/m3)

EXERCISE 1
What is the approximate mass of air in a living room 4.8m x 3.8m x 2.8m?

SOLUTION

The mass is found from the density of air and the volume of air.
                              
               m  V  1.29 kg m 3       4.8 m 3.8 m 2.8 m     66 kg

EXERCISE 2
Accomplished silver workers in India can pound silver into incredibly thin sheets
as thin as 3.00 x 10-7 m. Find the area of such sheet that can be formed from
1.00 kg of silver.

SOLUTION
      m          1.00 kg
V                          9.52 105 m3
             10 500 kg/m 3



                                                    V 9.52 10–5 m3
The area of the silver, is, therefore,         A             7
                                                                     317 m2
                                                    d 3.00 10 m
EXERCISE 3
A bottle has a mass of 35.00 g when empty and 98.44 g when filled with water.
When filled with another fluid, the mass is 88.78 g. What is the specific gravity of
this other fluid?

SOLUTION
To find the specific gravity of the fluid, take the ratio of the density of the fluid to
that of water, noting that the same volume is used for both liquids.
                            fluid m V fluid mfluid 88.78 g  35.00 g
               SJ fluid                                              0.8477
                            water m V water mwater 98.44 g  35.00 g

DEPARTMENT OF ENGINEERING                                                             23
ESM 1224, SEMESTER 3, 2006/2007                                                       CHAPTER 2



2.2     Pressure
   The pressure P exerted by a fluid is the magnitude F of the force acting
    perpendicular to a surface in the fluid divided by the area A over which the
    force acts:
                                  P = F /A

    SI Unit : N/m2 or Pascals ( Pa ) where: 1 Pa = 1 Nm 2
    Other units : atmosphere (atm) where 1 atm = 1.013 x 105 Pa

EXERCISE 4
High heeled shoes can cause tremendous pressure to be applied to a floor.
Suppose the radius of a heel is 6.00x10-3m. Find the pressure that is applied to
the floor under the heel because of the weight of a 50 kg woman.

SOLUTION
      F mg (50.0 kg)(9.80 m/s 2 )
P                               4.33 106 Pa
      A  r2     6.00 10 m 
                           3   2


EXERCISE 5
A person who weighs 625 N is riding a 98 N mountain bike. Suppose the entire
weight of the rider and bike is supported equally by the two tires. If the gauge
pressure in each tire is 7.6 x 105 pa, what is the area of contact between each
tire and the ground?

SOLUTION
Since the weight is distributed uniformly, each tire exerts one-half of the weight
of the rider and bike on the ground. According to the definition of pressure,
Equation 11.3, the force that each tire exerts on the ground is equal to the
pressure P inside the tire times the area A of contact between the tire and the
ground. From this relation, the area of contact can be found.

The area of contact that each tire makes with the ground is

  F
A 
          1
          2   Wperson  Wbike   12  625 N  98 N   4.76 104 m2
  P                  P              7.60 105 Pa

2.3    Pressure and Depth in a Static Fluid

                                                How the pressure            in   a    fluid
                                                 varies with depth?

                                                  Figure 1 shows a container of fluid
                                                   (assumed to be liquid) in which one
                                                   column of the liquid is outlined.

                                                  The liquid is at rest. (If a fluid is at
                                                   rest in a container, all portions of
                                                   the fluid     must     be in      static
                                                   equilibrium).

                                                  This column has a cross sectional
                                                   area, A and extends to a depth, h.


                Figure 1
DEPARTMENT OF ENGINEERING                                                                24
ESM 1224, SEMESTER 3, 2006/2007                                               CHAPTER 2



   3 external forces act on this volume of liquid:
       i) Force of gravity or weight, mg
       ii) Downward force, P1A
       iii) Upward force, P2A exerted by the liquid below it.

   Since the column is in equilibrium, the sum of vertical forces equal to zero.
        Fy = 0
       P2A P1A – mg = 0 OR
       P2A = P1A + mg

   But mass, m =  V where V = A x h
    therefore, P2A = P1A + Ahg    OR
                           where P1 = pressure at higher level
     P2 = P1 +hg          P2 = pressure at deeper level
                           h = vertical distance between 2 points

   according to the equation, pressure P2 at a depth h is greater than P1

 Conclusion
  Pressure in a fluid increases with depth due to the weight of the fluid above
  the point of interest.

EXERCISE 6
At a given instant the blood pressure in the heart is 1.6x10 4 Pa. If an artery in
the brain is 0.45m above the heart, what is the pressure in the artery?

SOLUTION
Pressure P1 in the brain (the higher point),

                                                    
P  P2   gh  1.6 104 Pa  1060 kg/m3 9.80 m/s 2  0.45 m   1.110 4 Pa
 1


EXERCISE 7
Measured along the surface of the water, a rectangular swimming pool has a
length of 15m. Along this length, the flat bottom of the pool slopes downward at
an angle of 11 below the horizontal, from one end to the other. By how much
does the pressure at the bottom of the deep end exceed the pressure at the
bottom of the shallow end?

SOLUTION
As the depth h increases, the pressure increases according to Equation 11.4 (P2
= P1 + gh). In this equation, P1 is the pressure at the shallow end, P2 is the
                                                                  3    3
pressure at the deep end, and  is the density of water (1.00  10 kg/m , see
Table 11.1). We seek a value for the pressure at the deep end minus the
pressure at the shallow end.
PDeep  PShallow  gh   or   PDeep  PShallow  gh
The drawing at the right shows that a value for h can be obtained from the 15-m
length of the pool by using the tangent of the 11 angle:

                                             15 m

                                                           h
                                    11




DEPARTMENT OF ENGINEERING                                                           25
ESM 1224, SEMESTER 3, 2006/2007                                                  CHAPTER 2



                                    h
                       tan11            or        h  15 m  tan11
                                  15 m
     PDeep  PShallow   g 15 m  tan11


                                                        
                      1.00  103 kg/m3 9.80 m/s 2 15 m  tan11  2.9  10 4 Pa

2.4      Pressure Gauges

    It is possible to measure pressure by relating pressure differences to the
         variation in the height of a column of liquid.
    Two basic types of pressure       gauges (used to measure pressure) are the
         mercury barometer and the open tube manometer.

1.        Mercury barometer-used for measuring atmospheric pressure, Patm.

                                     How is the Patm can be measured using
                                      this device?

                                        Pressure P2 at point A (at the bottom of
                                         mercury tube) is same as at point B, which is
                                         called Patm.

                                        So, P2 = Patm and P1=0 Pa (because the closed
                                         end of the barometer is vacuum)

                                        Using the equation P2 = P1 +gh
                                            Patm = 0 + gh
                                                       Patm = gh
                                              So,


    Usually weather forecasters report the pressure in terms of h, expresses it in
     millimetres (mm)
      Example:
         Using Patm = 1.013 x 105 and  = 13.6 x 103 kg /m3 (density of mercury),
         we find that h = 760 mm

2.        Open tube manometer - to determine the pressure of container

     Open tube refers to the fact that one end of a U shaped tube is open to
      atmosphere and the other end is connected to a system of unknown P.
     How is the P in the container can be measured using this device?

                                        P1=Patm while P2 (pressure at point B) is equal
                                            with pressure at point A.
                                        PB = PA
                                               P2 = Patm +  gh
                                                  P2 – Patm = Pg =    gh

                                        P2absolute pressure: the actual value for P.
                                        Pg = P2 – Patmgauge pressure: is the
                                         amount by which a pressure P differs from
                                         atmospheric pressure.


DEPARTMENT OF ENGINEERING                                                            26
ESM 1224, SEMESTER 3, 2006/2007                                               CHAPTER 2



    Gauge pressure, Pg, is defined to be how much the given pressure is above
     atmospheric pressure. That is, if the pressure at some point is P, then the
     gauge pressure at that point is given by Pg = P2 – Patm. It is important,
     however, to keep in mind that, to find the force exerted on an object due to
     a given pressure, the absolute pressure (that is, the pressure P) must be
     used, and not just the gauge pressure, Pg. That is, F = PA, not F = PgA

EXERCISE 8
A mercury barometer reads 747.0 mm on the roof of a building and 760.0 mm on
the ground. Determine the height of the building .Density of air is 1.29kg/m 3.

SOLUTION
P = gh
                    133 Pa 
     13.0 mm Hg           
h                  1 mm Hg     137 m    *Note that :   133 Pa = 1 mm Hg
     1.29 kg/m3 9.80 m/s2 
EXERCISE 9
An open tube mercury manometer is used to measure the pressure in an oxygen
tank, as shown in figure 2. What is the pressure in the tank if the height of the
mercury in the right-hand column is 28 cm lower than the mercury in the left-
hand column?
Patn = 1.013  105 Pa, mercury = 13.6  103 kg/m3

SOLUTION
P2 = P oxygen in the tank   P1 = Patm
              P2 = Patm –h
                 = (1.013  105)-[(28x10-2)( 13.6  103 kg/m3)(9.8)]
                 = 6.39  104 Pa

EXERCISE 10
A water tower is a familiar sight in many towns. The purpose of such tower is to
provide storage capacity and to provide sufficient pressure in the pipes that
deliver the water to customers. Figure below shows a spherical reservoir that
contains 5.25 x 105 kg of water when full. The reservoir is vented to the
atmosphere at the top.
For a full reservoir, find the gauge pressure that the water has at the faucet in (a)
house A and (b) house B. Ignore the diameter of the delivery pipes.




DEPARTMENT OF ENGINEERING                                                         27
ESM 1224, SEMESTER 3, 2006/2007                                                            CHAPTER 2



SOLUTION
The pressure at the level of house A is given by equation           P  P   gh .
                                                                         atm
Since the tank is spherical, its full mass is given by              M  V  [(4 / 3) r 3 ] .
Therefore,
                                         1/ 3                             1/3
             3M                   3M             3(5.25 105 kg) 
         r 
          3
                        or    r                                  3 
                                                                                 5.00 m
             4                  4            4 (1.000 10 kg/m ) 
                                                                 3

Therefore, the diameter of the tank is 10.0 m, and the height h is given by
                 h  10.0 m + 15.0 m = 25.0 m
(a) the gauge pressure in house A is, therefore,
      Pg= P  Patm     gh  (1.000  103 kg/m3 )(9.80 m/s 2 )(25.0 m) = 2.45 105 Pa
(b)    The pressure at house B is   P  P   gh , where
                                          atm
       h  15.0 m  10.0 m  7.30 m  17.7 m
       According to Equation 11.4, the gauge pressure in house B is
  Pg =     P  Patm   gh  (1.000  103 kg/m3 )(9.80 m/s2 )(17.7 m) = 1.73  105 Pa

EXERCISE 11
Mercury is poured into a tall glass. Ethyl alcohol is then poured on top of the
mercury until the height of the ethyl alcohol itself is 110 cm. The two fluids do
not mix and the air pressure at the top of the ethyl alcohol is one atm. What is
the absolute pressure at a point that is 7.1 cm below the ethyl alcohol-mercury
interface?

SOLUTION




DEPARTMENT OF ENGINEERING                                                                     28
ESM 1224, SEMESTER 3, 2006/2007                                                     CHAPTER 2



2.5     Pascal’s Principle

 PASCAL’S PRINCIPLE
States that any change in the pressure applied to a completely enclosed
fluid is transmitted undiminished to all parts of the fluid and the
enclosing walls.

     Other words: an external pressure applied to an enclosed fluid is transmitted
      unchanged to every point within the fluid.

     Application : Hydraulic lift

 Refer to figure (a):



                                          (a)




     A downward force, F1 is applied to a small piston of area A 1.
     The pressure is transmitted through a fluid to a larger piston of area A2. The
      pressure is the same on both sides (because same depth).

     Because pressure is same on both sides, the equation for pressure will be,

                   P  P2
                    1
                        F1 F2
                   P               OR
                        A1 A2

                   F1 A2  F2 A1     (applies only when the points 1 and 2 lie at
                                   the same depth in the fluid)

     Since A2 > A1, then F2 > F1 (The input force is greatly multiplied)

     Note that the greater the area of piston 2 the greater will be the upward
      force. As a result, it is a good machine for lifting objects.

 Refer to Figure (b)

                                                   The bottom surface of the plunger at
             (b)                                    point B is at the same level as point A,
                                                    which is at a depth h, beneath the
                                                    input piston.


                                                   PA  PB        P  gh  P2
                                                                     1
                                                    F1        F
                                                        gh  2
                                                    A1        A2

DEPARTMENT OF ENGINEERING                                                               29
ESM 1224, SEMESTER 3, 2006/2007                                            CHAPTER 2



 Conclusion:
  Compare figure (a) and (b): F1 in part (b) less than in part (a) because the
  weight of the h column of hydraulic oil provides some of the input forces to
  support the car.

EXERCISE 12
In a hydraulic car lift the input piston has an area of 1.2x10-2m2 and output
plunger has an area of 0.15m2.The combined weight of car and output plunger is
20500N.The hydraulic oil of density 800 kg/m 3 is used in this lift. Calculate the
input force needed to support the car and the output plunger if the faces of the
input piston and the output plunger area
(a) At the same level
(b) At different level h=1.10 m as shown in figure.




SOLUTION

           F1   F2
(a)           =
           A1 A 2

                     A1           (0.012m 2 )
          F1 = F2      =(20500N)             = 1640 N
                     A2            (0.15m 2 )

(b)    P2 = P1 + ρ gh

       F2   F1
         =    + ρ gh
       A 2 A1

        F1   F2
          =      - ρ gh
       A1 A 2
                 F2
       F1 = A1 (    - ρ gh) =
                 A2
                                  (20500)
                      (0.012) [           - (800)(9.8)(1.1)]
                                   (0.15)
                     = 1537 N




DEPARTMENT OF ENGINEERING                                                      30
ESM 1224, SEMESTER 3, 2006/2007                                                         CHAPTER 2



EXERCISE 13
In the Hydraulic press used in the trash comparator, the radii of the input and
output piston are 6.4x10-3m and 5.1x10-2m respectively. The height difference
between input and output piston can be neglected. What force is applied to the
trash if input force is 330N?

SOLUTION
      F  F
Using 1  2           F2 = (A2/A1)F1
      A1 A2
                                 2
         2
        R2   5.1  102 m 
             6.4  103 m   330 N   2.1  10 N
F2  2 F1                                      4
                           
    R1                    

EXERCISE 14
The hydraulic oil in a lift has a density of 8.3 x 10 2 kg/m3. The weight of the input
piston is negligible. The radii of the input piston and the output plunger are 7.7 x
10-3m and 0.125m respectively. What input force F is needed to support the
24500 N combined weight of a car and the output plunger when
(a) the bottom surfaces of the piston and plunger are at the same level?
(b) the bottom surfaces of the output plunger is 1.3 m above the input piston?

SOLUTION
                 2
(a) Using A =  r for the circular areas of the piston and plunger, the input force
required to support the 24 500-N weight is
                           
                                                      
                                       3                  2
         A2                7.70 10 m                   
F2  F1     24 500 N                                    93.0 N
        A 
                              0.125 m 
                                           2
         1                                                
                                                           


(b)  The pressure P2 at the input piston is related to the pressure P1 at the
bottom of the output plunger by equation, P2 = P1 + gh, where h is the

difference in heights. Setting                                                   
                                          P2  F2 / A2  F2 /  r22 , P  F1 /  r12 , and
                                                                       1
solving for F2, we have
          r2 
F2  F1  22    gh  r22
        r                
         1 
                     
                                             
                                                  2
                                 3
                       7.70 10 m                
        24 500 N                               
                        0.125 m 
                                     2
                                                  
                                                   

                                                                     
                                                                              2
                8.30 102 kg/m3 9.80 m/s 2 1.30 m   7.70 10 3 m              94.9 N




DEPARTMENT OF ENGINEERING                                                                    31
ESM 1224, SEMESTER 3, 2006/2007                                                   CHAPTER 2



2.6    Archimedes’ Principle

 Why things float???
 An object that is either partially or completely submerged in a fluid will
  experience an upward buoyant force exerted by the fluid.
 The buoyant force is the upward force that a fluid applies to an object that is
  partially or completely immersed in it.

 ARCHIMEDES’S PRINCIPLE
States that an object wholly or partially immersed in a fluid is buoyed up
by a force equal in magnitude to the weight of the fluid displaced by the
object.


                  FB  Wf  m f g   f gV f
        The subscript refer to,       B = buoyant force and f = fluid displaced

 How to relate buoyant force, FB and Weight of object, WO ?
 1. If FB > Wo  Object will float
 2. If FB < Wo  Object will sink
 3. If FB = Wo  Object will be in equilibrium at any submerged depth in a
    fluid.

   Usually for the object that is totally submerged Vf = Vo

    The buoyant force is = FB = Wf =          f V f g   f Vo g                            Vo




   Usually for the object that is partially submerged Vf = Vsub

    The buoyant force is = FB = Wf =          f V f g   f Vsubg                           Vsub
    (Vsub is the volume of the object submerged beneath the
    surface of the water)


   An object will float in a liquid which is more dense than that object. The
    greater the density of the liquid the more volume of the object will be seen
    above the surface of the liquid. (For example, very little volume of an iceberg
    is seen above the surface of the water since the density of the water is not
    much bigger than the density of the ice.)

EXERCISE 15
What percent volume of an iceberg floats beneath the
surface of the water?

SOLUTION
Using Archimedes’s Principle :
The buoyant force is = FB = Wf =        f Vf g
Since the object partially submerged Vf = Vsub

 FB = Wf =    f V f g   f Vsubg     …(1)


DEPARTMENT OF ENGINEERING                                                            32
ESM 1224, SEMESTER 3, 2006/2007                                                            CHAPTER 2



Applying Newton’s 2nd law to the y-direction, we get that

 Fy  may  0

FB  mg  0 FB = mo g                         …(2)

Combine (1) and (2)

 f Vsub g   f Vobjectg   Replace fluid seawater and object iceberg

seawaterVsubmerged  iceVice

sea water = 1024 kg/m3                 and   ice = 917 kg/m3 .

The percent volume submerged beneath the surface of the water.




This answers the famous question of how much of the iceberg is unseen beneath
the water’s surface! (Poor Titanic didn’t have a chance!)

EXERCISE 16
A scuba diver and her gear displace a volume of 65.0 L and have a total mass of
68.0 kg.
(a) What is the buoyant force on the diver in sea water?
(b) Will the diver sink or float?

SOLUTION
(a) The buoyant force is the weight of the water displaced, using the density of
     sea water.
               Fbuoyant  mwater g  waterVdisplaced g
                                 displaced

                                                            1  10 3 m 3 
                                               
                             1.025  10 3 kg m 3 65.0 L 
                                                                 1L
                                                                                  
                                                                            9.80 m s  653 N
                                                                           
                                                                                     2




                                                           
(b) The weight of the diver is mdiver g  68.0 kg  9.80 m s
                                                                      2
                                                                           666 N. Since the
buoyant force is not as large as her weight, she will sink , although it will be very
gradual since the two forces are almost the same.

EXERCISE 17
A flat-bottomed rectangular boat is 4.0 m long and 1.5 m wide. If the load is
2000kg including the mass of the boat, how much height of the boat will be
submerged when it floats in a lake? (density of lake water = 1000 kg/m 3)

SOLUTION
     FB = mg = ρ fluid gVfluid = WFluids displaced by the boat
                 mg = ρ fluid g( LxWxH)
                          m         2000
                H=           =                = 0.33m
                        ρxLxW   (1000)(4)(1.5)




DEPARTMENT OF ENGINEERING                                                                       33
ESM 1224, SEMESTER 3, 2006/2007                                                                                  CHAPTER 2



EXERCISE 18
A spherical balloon has a radius of 7.35 m and is filled with helium. How large a
cargo can it lift, assuming that the skin and structure of the balloon have a mass
of 930 kg? Neglect the buoyant force on the cargo volume itself.

SOLUTION

The buoyant force of the balloon must equal the weight of the balloon plus the
weight of the helium in the balloon plus the weight of the load. For calculating the
weight of the helium, we assume it is at 0 oC and 1 atm pressure. The buoyant
force is the weight of the air displaced by the volume of the balloon.

Fbuoyant  airVballoon g  mHe g  mballoon g  mcargo g      
    mcargo  airVballoon  mHe  mballoon  airVballoon  HeVballoon  mballoon  air  He Vballloon  mballoon
             
            1.29 kg m 3  0.179 kg m 3            7.35 m   930 kg 
                                                   4
                                                   3
                                                                   3
                                                                                   920 kg  9.0  10 3N

EXERCISE 19
A geologist finds that a Moon rock whose mass is 9.28 kg has an apparent mass
of 6.18 kg when submerged in water. What is the density of the rock?

SOLUTION
The difference in the actual mass and the apparent mass is the mass of the water
displaced by the rock. The mass of the water displaced is the volume of the rock
times the density of water, and the volume of the rock is the mass of the rock
divided by its density. Combining these relationships yields an expression for the
density of the rock.

                                                           mrock
mactual  mapparent  m  waterVrock  water                    
                                                           rock
                   mrock                           9.28 kg
rock  water
                   m
                             
                          1.00  10 3 kg m 3          
                                              9.28 kg  6.18 kg
                                                                 2.99  10 3 kg m 3


2.7         Fluids in Motion

      Basic types of fluid flow:

1. Fluid flow can be steady or unsteady:
  A fluid is said to display steady flow (or laminar flow) if, at any point in the
     liquid, the velocity of the flow is always the same.
  Unsteady flow (or turbulent flow) exists whenever the velocity at a point in
     the fluid changes as time passes.

2. Fluid flow can be Compressible or Incompressible:
  Most liquids are nearly incompressible that is the density of a liquid remains
     almost constant as the pressure changes.
  To a good approximation, liquids flow in an incompressible manner
  In contrast, gases are highly compressible.

3. Fluid flow can be viscous on nonviscous
  Viscosity – degree of internal friction in the fluid, associated with the
     resistance between 2 adjacent layers.
  Honey does not flow readily due to higher viscosity.
  Water is less viscous and flows more readily (no internal friction force
     between adjacent fluid layers)

DEPARTMENT OF ENGINEERING                                                                                                34
ESM 1224, SEMESTER 3, 2006/2007                                                      CHAPTER 2



 Conclusion:
 An incompressible, non viscous and steady flow of fluid is called as an
   ideal fluid.

   Fluids which have the ideal properties obey two important equations.
        1. The Equation in Continuity
        2. Bernoulli’s Equation

2.8     The Equation in Continuity

   The mass flow rate of a fluid with a density         ,   flowing with a speed      in a
    pipe of cross-sectional area A, is the mass per second (kg/s) flowing past a
    point and is given by

                               Mass flow rate =         A
      where : A is the cross-sectional area of the pipe at that point
      SI unit : m3/s

 Idea: If the fluid is incompressible, the flow rate must be the same
  everywhere along the pipe (Conservation of fluid particles: what goes in must
  come out). This leads directly to the Continuity Equation:

   Equation of continuity express the fact that mass is conserved: what flows
    into one end of a pipe flows out the other end, assuming there are no
    additional entry or exit points in between. Expressed in terms of the mass
    flow rate, the equation of continuity is

                                    1 A11  2 A22
    where the subscripts 1 and 2 denote two points along the pipe.

   If a fluid is incompressible, the density at any points is the same,
                                    1   2  A11  A2 2

   The product A  is known as the volume flow rate Q:
                             Q = volume flow rate = A 
    SI unit : kg/s

   The equation shows that where the tube’s cross-sectional area is large, the
    fluid speed is small, and conversely, where the tube’s cross-sectional area is
    small, the speed is large. Refer to the figure




                                V smaller                                                V higher




 Conclusion
 This equation tells us that the product of the area and the speed or (Av) is
  constant. The only way this can be true is if, when the area decreases, the
  speed increases.
 Example:



DEPARTMENT OF ENGINEERING                                                                 35
ESM 1224, SEMESTER 3, 2006/2007                                             CHAPTER 2



 if you are holding a garden hose without a nozzle on the end, and the water is
  flowing out of the end of the hose, you can see that the flow is a rather slow
  but steady flow (mass of water per unit time).

 However, if you place your thumb over the end of the hose where the water is
  flowing out and press your thumb down on the opening, the water suddenly
  sprays out very quickly from the small opening that is left. – That is, a smaller
  area through which the water can flow means the faster the speed of the fluid
  flow. Refer to figure




   Another example is water flow in a river. If the river is very wide, then the
    flow will be relatively slow. However, if the river narrows appreciably, then
    the water speed will greatly increase, resulting in what is known as “rapids”.
    Again – less area for the fluid to flow through means a faster resulting speed
    of fluid flow

EXERCISE 20




Water flows from left to right through the five section A,B,C,D and E of a pipe. In
which sections does the water speed

    (a) Increase?
    (b) Decrease?
    (c) Remain constant?

SOLUTION
  (a) Increase?             [Ans: B]
  (b) Decrease?             [Ans: D]
  (c) Remain constant?      [Ans: A,C,E]




DEPARTMENT OF ENGINEERING                                                       36
ESM 1224, SEMESTER 3, 2006/2007                                                                   CHAPTER 2



EXERCISE 21
A water hose 1.00 cm in radius is used to fill a 20 liter bucket. If it takes 1.0
minute to fill the bucket, what is the speed at which the water leaves the hose?
[1 liter = 103cm3]

SOLUTION
                                                                  20 x10 3 cm 3
Volume flow rate = Q = Av = 20 liter/minute =
                                                                     1x60 s

                                         20 x10 3 cm 3    1
                                  v=                   x
                                            1x60 s       r 2
                                      20 x10 3 cm 3       1
                                    =               x
                                         1x60 s        (1cm) 2
                                    = 106 cm/s

EXERCISE 22
A 15-cm-radius air duct is used to replenish the air of a room 9.2 m  5.0 m  4.5 m
every 16 min. How fast does air flow in the duct?

SOLUTION
We apply the equation of continuity at constant density, Eq. 10-4b.
Flow rate out of duct  Flow rate into room
                           V               Vroom
Aduct vduct   r 2 vduct  room  vduct  2            
                                                             9.2 m 5.0 m 4.5 m   3.1 m s
                           t to fill       r t to fill                         60 s 
                                                           0.15 m  16 min 
                                                                     2
                             room               room
                                                                                1 min 
                                                                                       

EXERCISE 23
A nozzle with a radius of 0.250 cm is attached to a garden hose with a radius of
0.90cm. The flow rate through hose and nozzle is 0.5 liters per second.
Calculate the velocity of water : a) in the hose and
                                  b) in the nozzle
SOLUTION
       L 10 3 m 3
Q  0.5 x                     = 5 x 10-4 m3/s
       s   1L
                      5 x104
Q  A1v1         v1                   = 1.96 m/s velocity in the hose
                        r 2
From equation of continuity

                              A1
A1v1=A2v2 =>          v2        xv1    = 25.46 m/s velocity in nozzle
                              A2




DEPARTMENT OF ENGINEERING                                                                            37
ESM 1224, SEMESTER 3, 2006/2007                                                    CHAPTER 2



2.9    Bernoulli’s Equation and it’s applications




   Bernoulli’s equation is the direct result of the application of conservation
    energy to fluid. It is written as :

                              p +½ ρv2 + ρgh = constant

 Or in other words:
    The sum of the pressure P, KE per unit volume ( 
                                                             2
                                                                 / 2 ), and the PE per unit
    volume ( gh ) has a constant value at all points along a streamline.

   For any two points:
                                2
                    P1 +½ ρv1       + ρgh1= P2 +½ ρv22 + ρgh2

   You should recognize the second term as a kinetic energy term, and the third
    term as a potential energy term!)

 Note:
 If the flow is at a constant height, h = 0 (horizontal
  pipe, as shown in figure))

    Then,         P + ½ ρv2 = constant

   If there is no height difference, Bernoulli's equation
    relates the pressure at two points along the flow to
    the fluid velocities:

               P1 - P2 = ½ ρ(v22 - v12)

EXERCISE 24
a) An ideal fluid flows at 4.0 m/s in a horizontal circular pipe. If the pipe narrows
to half of its original radius, what is the flow speed in narrow section?
b) If the fluid is water and the pressure at the narrow section is 1.8 x 10 5Pa, what
is the pressure at the wide section?

SOLUTION
a) From the equation of continuity, A1v1=A2v2

            A v r v
                          2
        v2  1 1  1 21  16 m s
             A2    r2

DEPARTMENT OF ENGINEERING                                                               38
ESM 1224, SEMESTER 3, 2006/2007                                                                              CHAPTER 2



b)        Since the fluid is flowing horizontally, h is constant, so Bernoulli’s Equation
          can be written as                       : p1 +½ ρv12 = p2 +½ ρv2               2




          So,                         p1 = p2 + ½ (v22-v12)
                                      = 1.8 x105 + ½(1000)(162-42)
                                      = 3.0 x 105 Pa.
EXERCISE 25
A 6.0-cm-diameter pipe gradually narrows to 4.0 cm. When water flows through
this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa
and 24.0 kPa, respectively. What is the volume rate of flow?

SOLUTION
Use the equation of continuity (Eq. 10-4) to relate the volume flow of water at
the two locations, and use Bernoulli’s equation (Eq. 10-5) to relate the pressure
conditions at the two locations. We assume that the two locations are at the same
height. Express the pressures as atmospheric pressure plus gauge pressure. We
use subscript “1” for the larger diameter, and subscript “2” for the smaller
diameter.

                                    A1      r2     r2
A1v1  A2 v2             v2  v1       v1 12  v1 12
                                    A2      r2     r2
P0  P1  1 v12   gy1  P0  P2  1 v2   gy2
          2                          2
                                         2
                                                            
                                            r14                   2 P1  P2 
P1  1 v12  P2  1 v2  P2  1 v12
                       2
                                                        v1                     
     2             2            2
                                            r24                     r14    
                                                                    r 4  1
                                                                    2      

                2 P1  P2                                      
                                                                2 32.0  10 3 Pa  24.0  10 3 Pa     
                                                   
                                                     2
A1v1   r12                     3.0  10 2 m
                                                                                                
                 r   4
                                                                                3.0  10 m 2   4
                                                                                                         
                   1                                 1.0  10
                     1

                 r   4
                       
                                                                     3
                                                                         kg m 3                       1
                                                                                  
                                                                                 2.0  10 2 m         
                                                                                                  4
                     2
                                                                                                        
        5.6  10 3 m 3 s


EXERCISE 26
A small crack occurs at the base of a 15m high dam. The effective crack area
through which water leaves is 1.3 x 10-3 m2.
(a) Ignoring viscous losses, what is the speed of water flowing through the crack?
(b) How many cubic meters of water per second leave the dam?


SOLUTION                                                                              P1 = 1 atmosphere
(a) The drawing shows two points,                                         1
                                                                                      v1 = 0 m/s
labeled 1 and 2, in the fluid. Point 1 is
at the top of the water, and point 2 is
where it flows out of the dam at the
bottom. Bernoulli’s equation can be
used to determine the speed v2 of the                                    y1
water exiting the dam.                                                                                       2 P2 = 1 atmosphere
According to Bernoulli’s equation, we
have                                                                                                           v2
                                                                                                              y2
     P
     1
          1
          2
               v1
                 2
                       gy1  P2         1 v2
                                           2   2      gy2

DEPARTMENT OF ENGINEERING                                                                                          39
ESM 1224, SEMESTER 3, 2006/2007                                            CHAPTER 2



Setting P1 = P2, v1 = 0 m/s, and solving for v2, we obtain


                                 
v2  2 g  y1  y2   2 9.80 m/s 2 15.0 m   17.1 m/s

(b) The number of cubic meters per second of water that leaves the dam is the
volume flow rate Q. According to Equation 11.10, the volume flow rate is the
product of the cross-sectional area A2 of the crack and the speed v2 of the water;
Q = A2v2.
The volume flow rate of the water leaving the dam is

                           
Q  A2v2  1.30 103 m2 17.1 m/s   2.22 102 m3/s




DEPARTMENT OF ENGINEERING                                                      40

								
To top