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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics1
Fluid Statics[3-1]
Dr. Mohammad N. Almasri
Fluid Mechanics
[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Fluid Pressure
Fluid pressure is the normal force exerted by the fluid per unit area at some location within the fluid
Fluid pressure has the units:
N/m2
Pascal → 1 Pa = 1 N/m2
psi → pound-force per square inch = 6,894.757 Pa
Bar → 1 bar = 105 Pa = 100 Kpa
Atmosphere → 1 atm = 101,325 Pa = 1.01325 bars
1 kgf/cm2 = 9.807 × 104 N/m2
2
psia (pounds-force per square inch absolute) — gauge pressure plus local atmospheric pressure
[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Absolute Pressure
Absolute pressure is the actual pressure at a given position
However, the absolute pressure is measured relative to absolute zero pressure
This is why most pressure-measuring devices (called gages) read zero in the atmosphere
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Gage and Vacuum Pressures
When the pressure gages read a pressure value this value would be the difference between the absolute pressure and the local atmospheric pressure
This difference is called gage pressure (Pgage)
Pgage = Pabs – Patm
If Pgage is negative then it is called vacuum pressure
Pvac = Patm - Pabs
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Gage and Vacuum Pressures
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Gage Pressure
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The gage used to measure the air pressure in an automobile tire reads the gage pressure
Therefore, the common reading of 32 psi indicates a pressure of 32 psi above the atmospheric pressure
At a location where the atmospheric pressure is 14.3 psi, for example, the absolute pressure in the tire is 32 + 14.3 = 46.3 psi
[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
A vacuum gage connected to a chamber reads 5.8 psi at a location where the atmospheric pressure is 14.5 psi
Determine the absolute pressure in the chamber
We know that Pabs = Patm – Pvac = 14.5 – 5.8 = 8.7 psi
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Pressure at a Point
Consider a small wedge-shaped fluid element of unit length (into the page) in equilibrium
The mean pressures at the three surfaces are Pn, Px, and Pz, and the force acting on a surface is the product of mean pressure and the surface area
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Pressure at a Point
By writing the equation of equilibrium for the x direction, we obtain:
For the z direction, we obtain:
Now, when we divide this equation by the product ΔlΔycosα and the fluid element shrinks to a point (Δl→0)m the last term disappears. Thus we have:
Ultimately, this implies that :
Since α is arbitrary and Pn is independent of α, we conclude that the pressure at a point in a static fluid acts with the same magnitude in all directions:
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Variation of Pressure with Depth
To obtain a relation for the variation of pressure with depth, consider the element shown in the figure with a length of Δl, ΔA in cross-sectional area, and inclined at angle αwith the horizontal
A force balance in the vertical z-direction gives:
Upon simplification:
However, if we let the length approach zero, then Δp/Δl=dp/dl. Note also that sin α = dz/dl. Therefore:
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Variation of Pressure with Depth
This is the basic equation for hydrostatic pressure variation with elevation
It implies that pressure changes with the vertical distance but remains constant in the other directions
The pressure changes inversely with elevation. If one travels upward in the fluid (positive z direction), the pressure decreases; and if one goes downward (negative z), the pressure increases
Of course, a pressure increase is exactly what a diver experiences when descending in a lake or pool
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Variation of Pressure with Depth
If the specific weight of the fluid is uniform then it is constant. Thus the previous equation (dp/dz = -γ) becomes after integration:
The sum of the terms p/γ and z is called the piezometric head. This is constant throughout an incompressible static fluid
One can easily relate the piezometric head at one point to that of another point in the fluid as in the following:
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Variation of Pressure with Depth
Thus, the pressure difference between two points in a constant density fluid is proportional to the vertical distance Δz between the points and the density ρ of the fluid
In other words, pressure in a fluid increases linearly with depth as we go down and vice versa
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Variation of Pressure with Depth
A pressure difference of 10 psi can be specified in terms of pressure head as 23.1 ft of water (γ = 62.4 lb/ft3), or 518 mm of Hg (γ = 133 kN/m3)
As illustrated by the figure, a 23.1-ft-tall column of water with a cross-sectional area of 1 in.2 weighs 10 lb
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Variation of Pressure with Depth
If we take point 1 to be at the free surface of a liquid open to the atmosphere where the pressure is the atmospheric pressure (Patm) then the pressure at a depth h from the free surface becomes:
P = Patm + ρgh
or
Pgage = ρgh
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
Determine the pressure in psi at a depth of 20 ft below the free surface of a body of water if γ = 62.4 lb/ft3
P = γ h = 62.4 × 20 = 1,248 lb/ft2
P = 1,248/144 = 8.67 psi
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
Determine the pressure at a depth of 9 m in oil of specific gravity 0.75
P = γ h = 9.79 × 0.75 × 9 = 66.1 kPa
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
What depth of oil of specific gravity 0.75 will produce a pressure of 40 psi? What depth of water?
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
Find the pressure at the bottom of a tank containing glycerin under pressure
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
What is the water pressure at a depth of 35 ft in the tank shown?
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
Oil with a specific gravity of 0.80 forms a layer 0.90 m deep in an open tank that is otherwise filled with water. The total depth of water and oil is 3 m. What is the gage pressure at the bottom of the tank?
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
Because of a leak in a buried gasoline storage tank, water has seeped in to the depth shown in the figure. The specific gravity of the gasoline is 0.68
Determine the pressure at the gasoline—water interface and at the bottom of the tank
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Pressure Transmission
In a closed system, a pressure change produced at one point in the system will be transmitted throughout the entire system
This phenomenon of pressure transmission, along with the ease with which fluids can be moved, has led to the widespread development of hydraulic controls for operating equipment such as hydraulic presses
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Pressure Transmission
In the figure the air pressure from a compressor establishes the pressure in the oil system, which in turn acts against the piston in the lift
It can be seen that if a pressure of 600 kN/m2, for example, acts on the 25-cm diameter piston, then a force equal to p×A, or 29.5 kN, will be exerted on the piston
To handle larger or smaller loads it is necessary only to increase or decrease the pressure
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
A hydraulic jack has the dimensions shown. If one exerts a force F of 100 N on the handle of the jack, what load, F2, can the jack support? Neglect lifter weight
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
In the figure the areas of the plunger A and cylinder B are 6.00 and 600 in2, respectively, and the weight of B is 9000 lb. The vessel and the connecting passages are filled with oil of specific gravity 0.750. What force F is required for equilibrium, neglecting the weight of A?
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Pressure Transmission
The pressure is the same at all points on a horizontal plane in a given fluid
Note that the pressures at points A, B, C, D, F. F, and C are the same since they are at the same depth, and they are interconnected by the same static fluid
However, the pressures at points H and I are not the same since these two points cannot be interconnected by the same fluid although they are at the same depth
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Pressure Transmission
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Measurement of Pressure – Barometer
Absolute atmospheric pressure is measured by a device called barometer and thus the atmospheric pressure is often referred to as the barometric pressure
If we immerse the open end of a tube in a liquid that is open to the atmosphere (atmospheric pressure), and if we exhaust air from the tube, the liquid will rise in the tube
If the tube is long enough and if we have removed all the air, the only pressure on the surface of the liquid in the tube will be that of its own vapor pressure, and the liquid will have reached its maximum possible height
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Measurement of Pressure – Barometer
From the concepts developed earlier, we see that the pressure at 0 within the tube and at a on the surface of the liquid outside the tube must be the same; that is:
P0 = Pa = Patm
In addition, we have:
P0 = γy + Pvapor or
Patm = γy + Pvapor
If we neglect the vapor pressure, then we would have:
Patm = γy
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Measurement of Pressure – Barometer
The liquid used in barometers is usually mercury because:
its density is sufficiently great to enable a reasonably short tube to be used
Its vapor pressure is negligibly small at ordinary temperatures
Standard sea-level atmospheric pressure in different ways are:
29.92 in Hg = 760 mm Hg
33.91 ft of water = 10.34 m of water
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
Determine the atmospheric pressure at a location where the barometric reading is 740 mm Hg and the gravitational acceleration is g = 9.81 m/s2. assume that the temperature of mercury to be 10°C at which the density is 13,570 kg/m3
We know that:
Patm = γy + Pvapor
Neglecting the vapor pressure we have:
Patm = γy = ρgh = 13,570×9.81×0.74 = 98.5×103 Pa
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
What would be the reading on a barometer containing carbon tetrachloride at 68°F at a time when the atmospheric pressure was equivalent to 30.26 in Hg?
For carbon tetrachloride at 68°F we have:
ρ = 3.08 slugs/ft3 and Pvapor = 1.9 psia
Convert Patm to the units of psia
Patm = 30.26 × 14.696 / 29.92 = 14.86 psia
Y = (Patm – Pvapor)/ρg = (14.86 – 1.9)/(3.08×32.2) = 18.82 ft of carbon tetrachloride
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Measurement of Pressure – Piezometer
Measuring pressure involves the use of devices that use liquid columns. These measuring devices are called manometers
The simplest type of manometer, called a piezometer tube, consists of a vertical tube, open at the top, and attached to the container in which the pressure is desired
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Measurement of Pressure – Piezometer
Because manometers involve columns of fluids at rest, the fundamental equation describing their use is
P = γh + P0 which gives the pressure at any elevation in terms of the vertical distance h between P and P0
Application of this equation to the piezometer tube indicates that the pressure PA can be determined by a measurement of h1 through the relationship
PA = γ1h1
where γ1 is the specific weight of the liquid in the container
Since the tube is open at the top, the pressure P0 can be set equal to zero (we are now using gage pressure for PA)
Because point (1) and point A within the container are at the same elevation then PA = P1
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Measurement of Pressure – U-Tube Manometer
Another type of manometer is used and consists of a tube formed into the shape of a U
To find the pressure PA in terms of the various column heights, we start at one end of the system and work our way around to the other end, simply utilizing equation:
P1 = γh + P2
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Measurement of Pressure – U-Tube Manometer
Figure 1. The liquid is at the same height in each leg and pressure is equal at both ends
Figure 2. When positive pressure is applied to one leg, the liquid is forced down in that leg and up in the other. The difference in height, "h," which is the sum of the readings above and below zero, indicates the pressure
Figure 3. When a vacuum is applied to one leg, the liquid rises in that leg and falls in the other. The difference in height, "h," indicates the amount of vacuum
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Measurement of Pressure – U-Tube Manometer
Thus, we will start at point A and work around to the open end
The pressure at points A and (1) are the same
As we move from point (1) to (2) the pressure will increase by y1h1
The pressure at point (2) is equal to the pressure at point (3), since the pressures at equal elevations in a continuous mass of fluid at rest must be the same
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Measurement of Pressure – U-Tube Manometer
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With the pressure at point (3) specified we now move to the open end where the pressure is zero
As we move vertically upward the pressure decreases by an amount y2h2
As such we can write
PA + γ1h1 – γ2h2 = 0
And the pressure PA can be written as column heights as:
PA = γ2h2 – γ1h1
[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Measurement of Pressure – U-Tube Manometer
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
A closed tank contains compressed air and oil (SGoil = 0.90). A U-tube manometer using mercury (SGHg = 13.6) is connected to the tank. The column heights are h1 = 36 in., h2 = 6 in., and h3 = 9 in
Determine the pressure reading (in psi) of the gage
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
Following the general procedure of starting at one end of the manometer system and working around to the other, we will start at the air—oil interface in the tank and proceed to the open end where the pressure is zero. The pressure at level (1) is:
This pressure is equal to the pressure at level (2), as these two points are at the same elevation in a homogeneous fluid at rest. As we move from level (2) to the open end, the pressure must decrease by YHgh3, and at the open end the pressure is zero. Thus, the manometer equation can be expressed as:
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
Assume that the gage pressure remains at 3.06 psi, but the manometer is altered so that it contains only oil. That is, the mercury is replaced by oil. A simple calculation shows that in this case the vertical oil-filled tube would need to be h3 = 11.3 ft tall, rather than the original h3 = 9 in. There is an obvious advantage of using a heavy fluid such as mercury in manometers
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Measurement of Pressure – Differential Manometers
The U-tube manometer is also widely used to measure the difference in pressure between two containers or two points in a given system
Consider a manometer connected between containers A and B as is shown in the figure. The difference in pressure between A and B can be found by again starting at one end and working around to the other end
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Measurement of Pressure – Differential Manometers
For example, at A the pressure is PA, which is equal to P1, and as we move to point (2) the pressure increases by γ1h1
The pressure at P2 is equal to P3, and as we move upward to point (4) the pressure decreases by γ2h2
Similarly, as we continue to move upward from point (4) to (5) the pressure decreases by γ3h3
Finally, P5 = PB, as they are at equal elevations. Thus,
PA + γ1h1 – γ2h2 – γ2h2 = PB
PA – PB =γ2h2 + γ2h2 – γ1h1 49
The pressure difference
[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
The nozzle creates a pressure drop, PA – PB, along the pipe. The pressure drop is measured with a differential U-tube manometer of the type illustrated
(a) Determine an equation for PA – PB in terms of the specific weight of the flowing fluid, γ1, the specific weight of the gage fluid, γ2 and the various heights indicated
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
If we start at point A and move vertically upward to level (1), the pressure will decrease by γ1h1 and will be equal to the pressure at (2) and at (3).
We can now move from (3) to (4) where the pressure has been further reduced by γ2h2
The pressures at levels (4) and (5) are equal, and as we move from (5) to B the pressure will increase by γ1 (h1 + h2). Thus, in equation form
PA – γ1h1 – γ2h2 + γ1 (h1 + h2) = PB
PA – PB =h2(γ2 – γ1)
51
It is to be noted that the only column height of importance is
the differential reading, h2
[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Measurement of Pressure – Inclined Manometers
To measure small pressure changes, a manometer of the type shown in the figure is frequently used
One leg of the manometer is inclined at an angle θ and the differential reading l2 is measured along the inclined tube. The difference in pressure PA – PB can be expressed as:
PA + γ1h1 – γ2l2 sin θ – γ3h3 = PB
PA – PB = γ2l2 sin θ + γ3h3 – γ1h1
where it is to be noted the pressure difference between points (1) and (2) is due to the vertical distance between the points, which can be expressed as l2 sin θ
Thus, for relatively small angles the differential reading along the inclined tube can be made large even for small pressure differences
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
For the open tank, with piezometers attached on the side, containing two different immiscible liquids as shown in the figure, find:
(a) The elevation of the liquid surface in piezometer A
(b)The elevation of the liquid surface in piezometer B
(c) The total pressure at the bottom of the tank
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
(a) Liquid A will simply rise in piezometer A to the same elevation as liquid A in the tank (i.e., to elevation 2 m)
(b) Liquid B will rise in piezometer B to elevation 0.3 m (as a result of the pressure exerted by liquid B) plus an additional amount, hA, as a result of the overlying pressure of liquid A, PA
PA = γh = (0.72 × 9.79)(1.7) = 11.98 kPa
hA = p/ γ = 11.98/(2.36 × 9.79) = 0.519 m
Liquid B will rise in piezometer B to elevation 0.3 + 0.5 19 = 0.819 m
(c) Pressure at bottom = (0.72 × 9.79)(1.7) + (2.36 × 9.79)(0.3) = 18.9 kPa
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
Determine the pressure at A in psi gage due to the deflection of the mercury, SG = 13.57 in the U-tube gage shown in the figure
Pressure head at B = pressure head at C
PA/γ + 2.00 ft water = 2.65 × 13.57 ft water
Solving, PA/ γ = 34.0 ft water, and PA = (62.4 × 34.0)/144 = 14.7 psi
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
A manometer is attached to a tank containing three different fluids, as shown in the figure. Find the difference in elevation of the mercury column in the manometer (for instance find the value of y in the figure)
pressure at A = pressure at B
30 + (0.82 × 9.79)(3) + (9.79)(3.00) = (13.6 ×9.79)(y)
Y = 0.627 m56
[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
Oil of specific gravity 0.750 flows through the nozzle shown in the figure and deflects the mercury in the U-tube gage. Determine the value of h if the pressure at A is 20.0 psi
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
59
For a gage pressure at A of -10.89 kPa, find the specific gravity of the gage liquid B in the figure
[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
For a gage reading at A of –2.50 psi, determine:
(a) the elevations of the liquids in the open piezometer columns E, F, and G
(b) the deflection of the mercury in the U-tube gage i
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
Since the unit weight of the air (about 0.08 lb/ft3) is very small compared with that of the liquids, the pressure at elevation 49 may be considered to be -2.50 psi without introducing significant error in the calculations
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
The loss through a device X is to be measured by a differential gage using oil of specific gravity 0.750 as the gage fluid. The flowing liquid has SG 1.50. Find the change in pressure head between A and B for the deflection of the oil shown in the figure
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
A differential mercury manometer is connected to two pressure taps in an inclined pipe. Water at 50°F is flowing through the pipe. The deflection of mercury in the manometer is 1 inch. Find the change in piezometric pressure and piezometric head between the two points
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
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[3] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
Example
69