chapter 13 t. norah ali al moneef 1. motion is steady: velocity, density, pressure at a given point...
TRANSCRIPT
2
•Motion is steady: velocity, density, pressure at a given point don’t change
•Moves without turbulence
T . Norah Ali Al moneef
3
Steady/Unsteady flow• In steady flow the velocity of particles is constant with time• Unsteady flow occurs when the velocity at a point changes with
time. Turbulence is extreme unsteady flow where the velocity vector at a point changes quickly with time e.g. water rapids or waterfall.
• When the flow is steady, streamlines are used to represent the direction of flow.
• Steady flow is sometimes called streamline flow
T . Norah Ali Al moneef
• Streamlines never cross. A set of streamlines can define a tube of flow, the borders of which the fluid does not cross
Turbulent Flow Turbulent flow is an extreme kind of unsteady flow and occurs when there are sharp obstacles or bends in the path of a fast-moving fluid.
In turbulent flow, the velocity at a point changes erratically from moment to moment, both in magnitude and direction.
4T . Norah Ali Al moneef
Viscous/Non viscous flow
• A viscous fluid such as honey does not flow readily, it has a large viscosity.• Water has a low viscosity and flows easily.• A viscous flow requires energy dissipation.• Zero viscosity – requires no energy. with no dissipation of energy. Some
liquids can be taken to have zero viscosity e.g. water.• An incompressible, non viscous fluid is said to be an ideal fluid
7T . Norah Ali Al moneef
The volume flow rate, Q is defined as the volume of fluid that flows past an imaginary (or real) interface.
A v represents the volume of fluid per second that passes through the tube and is called the volume flow rate Q
8T . Norah Ali Al moneef
• If the fluid is incompressible, the density remains constant throughout
• Av represents the volume of fluid per second that passes through the tube and is called the volume flow rate Q
2211 vAvA AvQ
This just means that the amount of fluid moving in any “section of pipe” must remain constant.
•If the area is reduced the fluid must speed up!
tAvxAV
• Speed is high where the pipe is narrow and speed is low where the pipe has a large diameter
9T . Norah Ali Al moneef
The product of the area and the fluid speed at all points along a pipe is constant for an incompressible fluid.
10T . Norah Ali Al moneef
Q: Have you ever used your thumb to control the water flowing from the end of a hose?
A: When the end of a hose is partially closed off, thus reducing its cross-sectional area, the fluid velocity increases.
This kind of fluid behavior is described by the equation of continuity.
Example: Oil is flowing at a speed of 1.22 m/s through a pipeline with a radius of 0.305 m. How much oil flows in 1 day?
11T . Norah Ali Al moneef
• A river flows in a channel that is 40. m wide and 2.2 m deep with a speed of 4.5 m/s.
• The river enters a gorge that is 3.7 m wide with a speed of 6.0 m/s. • How deep is the water in the gorge?
The area is width times depth. A1 = w1d1
Use the continuity equation. v1A1 = v2A2
v1w1d1 = v2w2d2
Solve for the unknown d2.
d2 = v1w1d1 / v2w2 =(4.5 m/s)(40. m)(2.2 m) / (3.7m)(6.0 m/s) = 18 m
example
example
T . Norah Ali Al moneef 13
If pipe 1 diameter = 50mm, mean velocity 2m/s, pipe 2 diameter 40mm takes 30% of total discharge andpipe 3 diameter 60mm. What are the values of discharge and mean velocity in each pipe?
example
T . Norah Ali Al moneef 16
Example:The volume rate of flow in an artery supplying the brain is 3.6x10-6 m3/s. If the radius of the artery is 5.2 mm, A) determine the average blood speed.
B) Find the average blood speed if a constriction reduces the radius of the artery by a factor of 3 (without reducing the flow rate).
18T . Norah Ali Al moneef
if the cross-section area, A, is 1.2 x 10-3m2 and the discharge, Q is 24 l / s , then the mean velocity,
if the area A 1 = 10 x10-3 m2 and A 2 = 3 x10-3 m 2 and the upstream mean velocity, 1 = 2.1 m / s , then the downstream mean velocity
Example
Example
20T . Norah Ali Al moneef
• The figure shows 3 straight pipes through which water flows. The figure gives the speed of the water in each pipe.
• Rank them according to the volume of water that passes through the cross-sectional area per minute, greatest first. 6
Example:
sameExample:
• Water flows smoothly through the pipe shown in the figure, descending in the process.
Rank the four numbered sections of pipe according to
• (a) the volume flow rate Q through them
• (b) the flow speed v through them• (c) the water pressure p within them,
greatest first.a)Same b) 1,2,3,4 c ) 4,3,2,1
22
Pressure
T . Norah Ali Al moneef
Pressure P is the magnitude of the force acting perpendicular to a surface divided by the area A over which the force acts.
The greater the force, the greater the pressure is. The greater the area, the smaller the force is.
T . Norah Ali Al moneef 24T . Norah Ali Al moneef
ExampleThe mattress of a water bed is 2.00m long by 2.00m wide and 30.0cm deep. a) Find the weight of the water in the mattress.
The volume density of water at the normal condition (0oC and 1 atm) is 1000kg/m3. So the total mass of the water in the mattress is
Since the surface area of the mattress is 4.00 m2, the pressure exerted on the floor is
m
P
Therefore the weight of the water in the mattress is Wb) Find the pressure exerted by the water on the floor when the bed rests in its normal position, assuming the entire lower surface of the mattress makes contact with the floor.
MWV kg31020.1300.000.200.21000
mg N43 1018.18.91020.1
A
F
A
mg 3
4
1095.200.4
1018.1
T . Norah Ali Al moneef 25T . Norah Ali Al moneef
13.3 Bernoulli’s Equation
• Relates pressure to fluid speed and elevation• Bernoulli’s equation is a consequence of Work
Energy Relation applied to an ideal fluid• Assumes the fluid is incompressible and
nonviscous, and flows in a nonturbulent, steady-state manner
27T . Norah Ali Al moneef
This work is negative because the force on the segment of fluid is to the left and the displacement is to the right. Thus, the net work done on the segment by these forces in the time interval
29T . Norah Ali Al moneef
• States that the sum of the pressure, kinetic energy per unit volume, and the potential energy per unit volume has the same value at all points along a streamline
T . Norah Ali Al moneef 30T . Norah Ali Al moneef
Bernoulli’s Principle
• In steady flow, the speed, pressure and elevation of an ideal fluid are related. Two situations.
• (1) Horizontal pipe, changing Area A.– Pressure drop in thin pipe why?– Fluid speeds – acceleration – force required
• (2) Elevation change but same area A– Lower fluid is under more pressure
T . Norah Ali Al moneef 31T . Norah Ali Al moneef
Atmospheric Pressure
• The atmosphere exerts a pressure on the surface of the Earth and all objects at the surface
• Atmospheric pressure is generally taken to be 1.00 atm = 1.013 x 105 Pa = Patm
T . Norah Ali Al moneef 32
Fluid At Rest In a Container :•Pressure in a continuously distributed uniform static fluid varies only withvertical distance and is independent of the shape of the container.• The pressure is the same at all points on a given horizontal plane in a fluid.• For liquids, which are incompressible, we have: pb + ρgh1 = patm + ρgh2
pb = patm + ρg (h2 - h1) = patm + ρgd
13.4 static consequences of Bernoulli's equation
T . Norah Ali Al moneef
h1
h2
d
Y a = y b = y c = yd
y A = y B = y C = y D
33
Absolute Pressure and Gauge PressureThe pressure P is called the absolute pressure
Remember, P = Patm + ρ g hP – Patm = ρ g h ( so ρ g h is the gauge pressure)
T . Norah Ali Al moneef
As a fluid moves along if it changes speed or elevation then the pressure changes and vice versa.Bernoulli’s equation is really just conservation of energy for the fluid
• Bernoulli’s equation.• P + ½ ρv2 + ρg h = constant• Bernoulli’s equation shows that as the velocity of a fluid speeds up
it’s pressure goes down…..this idea used in airplane wings and frisbees (difference in pressure leads to upward force we call Lift)
Gauge pressure: Pressure expressed as the difference between the pressure of the fluid and that of the surrounding atmosphereUsual pressure gauges record gauge pressure. To calculate absolute pressure:
P = P atm + P gauge
, because it is the actual value of the system’s pressure.
34T . Norah Ali Al moneef
PA – Patm = ρ g h
PA=PC
PA>PB
• P atm is atmospheric pressure =1.013 x 105 Pa• The pressure does not depend upon the shape of the
container
ghPP atm PA = PB = PC = PD
35T . Norah Ali Al moneef
Variation of Pressure with Depth• If a fluid is at rest in a container, allportions of the fluid must be in static equilibrium• All points at the same depth must be at the same pressure– Otherwise, the fluid would not be inequilibrium
• Pressure changes with elevation
•The pressure gradient in the vertical direction is
negative
• The pressure decreases as we move upward
in a fluid at rest
• Pressure in a liquid does not change due to
the shape of the container• The fluid would flow from the higher pressure
region to the lower pressure region•the pressure at a point in a fluid depends only on density, gravity and depth
37T . Norah Ali Al moneef
the Manometer is a device to measure pressures. A common simple manometer consists of a U shaped tube of glass filled with some liquid. Typically the liquid is mercury because of its high density
both ends of the tube are open to the atmosphere. Thus both points A and B are at atmospheric pressure. The two points also have the same vertical height.
the top of the tube on the left has been closed. We imagine that there is a sample of gas in the closed end of the tube.The right side of the tube remains open to the atmosphere. The point A, then, is at atmospheric pressure.The point C is at the pressure of the gas in the closed end of the tube.The point B has a pressure greater than atmospheric pressure due to the weight of the column of liquid of height h.The point C is at the same height as B, so it has the same pressure as B. And we have already seen that this is equal to the pressure of the gas in the closed end of the tube.Thus, in this case the pressure of the gas that is trapped in the closed end of the tube is greater than atmospheric pressure by the amount of pressure exerted by the column of liquid of height h.
Case 2
Case 1
38T . Norah Ali Al moneef
Perhaps the closed end of the tube contains a sample of gas as before, or perhaps it contains a vacuum.The point A is at atmospheric pressure.The point C is at whatever pressure the gas in the closed end of the tube has, or if the closed end contains a vacuum the pressure is zero.Since the point B is at the same height as point A, it must be at atmospheric pressure too. But the pressure at B is also the sum of the pressure at C plus the pressure exerted by the weight of the column of liquid of height h in the tube.We conclude that pressure at C, then, is less than atmospheric pressure by the amount of pressure exerted by the column of liquid of height h.If the closed end of the tube contains a vacuum, then the pressure at point C is zero, and atmospheric pressure is equal to the pressure exerted by the weight of the column of liquid of height h. In this case, the manometer can be used as a barometer to measure atmospheric pressure.
Case 3
39T . Norah Ali Al moneef
The manometer
ghPP
yygPP
atm
atm
)( 12
gygyPP
gyPgyP
gyPP
gyPPPP
atmA
atmA
atmB
AB
A
12
21
2
1
Pressure Measurements:
• Manometers are devices in which one or more columns of a liquid are used
to determine the pressure difference between two points.
• U-tube manometer• One end of the U-shaped tube is open to the atmosphere• The other end is connected to the pressure to be measured
T . Norah Ali Al moneef 40T . Norah Ali Al moneef
• Pressure at B is P=Patm+ρgh
Pressure at A = Pressure at B P A = P B = Patm+ ρgh
Pressure in a continuous static fluid is the same at any horizontal level so,
Pressure at A > Patm
T . Norah Ali Al moneef
41T . Norah Ali Al moneef
Pressure at c =pressure at D
PC = PD
PC = PA + ρsgh
PB = Patm + ρ gh
PA + ρsgh = Patm + ρ gh
PA = Patm + ρ g h – ρsgh
Pblood =PA = Patm + ρ g h – ρsgh
Blood Pressure measurements by cannulation
T . Norah Ali Al moneef 42
Blood Pressure
• Blood pressure is measured with a special type of manometer called a sphygmomano-meter
• Pressure is measured in mm of mercury
Sphygmometer
43T . Norah Ali Al moneef
Pressure with depthEstimate the amount by which blood pressure changes in an actuary in the foot P2 and in the aorta P1 when the person is lying down and standing up
Take density of blood = 1060kg/m3
PaPP
dings
PaghPP
downLying
412
12
104.135.18.91060
tan
0
T . Norah Ali Al moneef 45
A fluid of constant density ρ = 960 kg / m3 is flowing steadily through the above tube. The diameters at the sections are d 1 =100 mm and d 2 = 80 mm . The gauge pressure at 1 is p1 = 200 k N/ m2 ,and the velocity here is u 1 = 5 m /s . We want to know the gauge pressure at section 2.
The tube is horizontal, with y1 = y2 so Bernoulli gives us the following equation for pressure at section 2:
P2 = 2 x10 5 + 960 { 52 – ( 7.8 )2} /2 = 182796.8 pa
example
46T . Norah Ali Al moneef
An example of the U-Tube manometer Using a u-tube manometer to measure gauge pressure of fluid density ρ = 700 kg/m3, and the manometric fluid is mercury, with a relative density of 13.6. What is the gauge pressure if: h1 = 0.4m and h2 = 0.9m?
pB = pC pB = pA + ρgh1 pB = pAtmospheric + ρman gh2 We are measuring gauge pressure so patmospheric = 0 pA = ρman gh2 - ρgh1
a) pA = 13.6 x 103 x 9.81 x 0.9 - 700 x 9.81 x 0.4 = 117 327 N, b) pA = 13.6 x 103 x 9.81 x (-0.1) - 700 x 9.81 x 0.4 = -16 088.4 N, -The negative sign indicates that the pressure is below atmospheric
T . Norah Ali Al moneef 47
A MANOMETER (U-TUBE) is a variation of the Barometer in which the pressure of two gases may be compared. Here the difference in the pressures of the gases on the arms of the manometer is equal to the pressure 'exerted' by the column of fluid, ph = ρ g h.
example
T . Norah Ali Al moneef 48
Example: The horizontal constricted pipe illustrated in the Figure, known as a Venturi tube, can be used to measure the flow speed of an incompressible fluid. Determine the flow speed at point 2 if the pressure difference P1 & P2 is known.
T . Norah Ali Al moneef 49T . Norah Ali Al moneef
P = ρ g hh = P / ρ g = 50 X 103 / 10 3 X 9.8 = 5.1 m
example
Calculate the column height of water corresponding to a pressure of 50 kPa
50T . Norah Ali Al moneef
Example: Fluid is flowing from left to right through the pipe. Points A and B are at the same height, but the cross-sectional areas of the pipe differ. Points B and C are at different heights, but the cross-sectional areas are the same.Rank the pressures at the three points, from highest to lowest.
A) A and B (a tie), CB) C, A and B (a tie)C) B, C, AD) C, B, AE) A, B, C
E) PA > PB > PC
T . Norah Ali Al moneef 51
Example: A stream of water of water d = 0.10 m flows steadily from a tank of diameter D = 1.0 m as shown in the figure. What flow-rate is needed from the inlet to maintain a constant water volume in the header tank depth? The depth of water at the outlet is 2.0 m . Can regard outlet as a free jet (note water level at(1) is not going down).
T . Norah Ali Al moneef 52
Pressure Measurements:
• A long closed tube is filled with mercury and inverted in a dish of mercury
• Measures atmospheric pressure as ρ g h
Vacuum pressure = 0
T . Norah Ali Al moneef
For mercury, h = 760 mm. How high will water rise?
No more than h = patm/g = 10.3 m
h patm
p=0
atmatm
PP gh h
g
Vacuum!•Since the closed end is at vacuum, it does not exert any force.
53T . Norah Ali Al moneef
We can set (assume the hole is on the ground or is where we measure height from). We also have 1 atm. So we have
v2h
example
T . Norah Ali Al moneef 56
P2 =p1 + ρ (v12 –v2
2 ) / 2 = 180X103 + 103X(22 – 182) = 20X 103 pa
example
T . Norah Ali Al moneef 59
If a fluid is at rest in a container, all portions of the fluid must be in static equilibrium
• If the height doesn’t change much, Bernoulli becomes: y1 = y2
• Where speed is higher, pressure is lower.
• Speed is higher on the long surface of the wing – creating a net force of lift.
222
12
212
11 vPvP FL
T . Norah Ali Al moneef 62
1.5m
Water flows through the pipe as shown at a rate of .015 m3/s. If water enters the lower end of the pipe at 3.0 m/s, what is the pressure difference between the two ends?
A2 = 20 cm2
5.18.9
2
12
12
1
2
1
/sm 5.720
015.0
1035.710
pp
pp
10
3223
12
2
1
2
221
2
2
221
2
11
42
P
g
gg
A
Q
AVQ
yyVV
yVyV
V
T . Norah Ali Al moneef 64
ExampleEstimate the force exerted on your eardrum due to the water above when you are swimming at the bottom of the pool with a depth 5.0 m.
We first need to find out the pressure difference that is being exerted on the eardrum. Then estimate the area of the eardrum to find out the force exerted on the eardrum.
F
Since the outward pressure in the middle of the eardrum is the same as normal air pressure
Estimating the surface area of the eardrum at 1.0cm2=1.0x10-4 m2, we obtain
ghW Pa4109.40.58.91000
APP 0 N9.4100.1109.4 44
PP atm
T . Norah Ali Al moneef 65
Absolute and Relative PressureHow can one measure pressure?
One can measure pressure using an open-tube manometer, where one end is connected to the system with unknown pressure P and the other open to air with pressure P0.
This is called the absolute pressure, because it is the actual value of the system’s pressure.
In many cases we measure pressure difference with respect to atmospheric pressure due to changes in P0 depending on the environment. This is called gauge or relative pressure.
P
The common barometer which consists of a mercury column with one end closed at vacuum and the other open to the atmosphere was invented by Evangelista Torricelli.
Since the closed end is at vacuum, it does not exert any force. 1 atm is
0P
The measured pressure of the system is
h
P P0
0PP
ghP 0
gh
gh )7600.0)(/80665.9)(/10595.13( 233 msmmkg
atmPa 110013.1 5
T . Norah Ali Al moneef 66
Q: Is it possible to stand on the roof of a five-story (20 m) building and drink, using a straw, from a glass on the ground?
20m
Even if a person could completely remove all of the air from the straw, the height to which the outside air pressure pushes the water up the straw would not be high enough for him/her to drink the water, no matter how hard he/she sucks
T . Norah Ali Al moneef 67
Applications of Bernoulli's Equation
The tarpaulin that covers the cargo is flat when the truck is stationary but bulges outward when the truck is moving.
T . Norah Ali Al moneef 68
•Air speeds up in the constricted space between the car & truck creating a low-pressure area. Higher pressure on the other outside pushes them together.
•Hold two sheets of paper together, as shown here, and blow between them. No matter how hard you blow, you cannot push them more than a little bit apart!