Fluid Statics 23 Fluid Mechanics lecture notes by David S. Ancalle (updated 8/3/2020) Module 2 Fluid Statics 2.1 Approaching Pressure 2.1.1 Pressure at a Point Pressure at a point can be defined as an infinitesimal normal compressive force divided by an infinitesimal area over which it acts. Pressure at a point does not vary with direction. Consider a wedge-shaped element with a uniform width , sides Δ and Δ, and hypotenuse Δ. A force, results from a pressure on the hypotenuse, and forces and act on the other sides. If we draw a free body diagram and take forces in the x and z directions, we get: Then, taking a sum of forces in the x-axis: ∑ = ⟹ − sin = ΔΔ − ΔΔ sin = 1 2 ΔΔΔ ΔΔ − ΔΔ = 1 2 ΔΔΔ − = 1 2 Δ As the element shrinks to a point, Δ → 0 and the limit becomes: − =0⟹ = Repeating this process for the z-axis: ∑ = ⟹ − cos − = = ΔΔ = ΔΔ = ΔΔ = = ΔΔΔ 2 Δ = Δ cos Δ = Δ sin
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Fluid Statics 23
Fluid Mechanics lecture notes by David S. Ancalle (updated 8/3/2020)
Module 2
Fluid Statics
2.1 Approaching Pressure
2.1.1 Pressure at a Point
Pressure at a point can be defined as an infinitesimal normal compressive force divided by an infinitesimal area over
which it acts.
Pressure at a point does not vary with direction. Consider a wedge-shaped element with a uniform width 𝛥𝑦, sides
Δ𝑥 and Δ𝑧, and hypotenuse Δ𝑠. A force, 𝐹𝑠 results from a pressure 𝑃𝑠 on the hypotenuse, and forces 𝐹𝑥 and 𝐹𝑧 act on
the other sides. If we draw a free body diagram and take forces in the x and z directions, we get:
Then, taking a sum of forces in the x-axis:
∑ 𝐹𝑥 = 𝑚𝑎𝑥 ⟹ 𝐹𝑥 − 𝐹𝑠 sin 𝜃 = 𝑚𝑎𝑥
𝑃𝑥Δ𝑦Δ𝑧 − 𝑃𝑠Δ𝑠Δ𝑦 sin 𝜃 =1
2𝜌Δ𝑥Δ𝑦Δ𝑧𝑎𝑥
𝑃𝑥Δ𝑦Δ𝑧 − 𝑃𝑠Δ𝑦Δ𝑧 =1
2𝜌Δ𝑥Δ𝑦Δ𝑧𝑎𝑥
𝑃𝑥 − 𝑃𝑠 =1
2𝜌𝑎𝑥 Δ𝑥
As the element shrinks to a point, Δ𝑥 → 0 and the limit becomes:
𝑃𝑥 − 𝑃𝑠 = 0 ⟹ 𝑃𝑥 = 𝑃𝑠
Repeating this process for the z-axis:
∑ 𝐹𝑧 = 𝑚𝑎𝑧 ⟹ 𝐹𝑧 − 𝐹𝑠 cos 𝜃 − 𝐹𝑊 = 𝑚𝑎𝑧
𝐹𝑠 = 𝑃𝑠Δ𝑠Δ𝑦
𝐹𝑥 = 𝑃𝑥Δ𝑦Δ𝑧
𝐹𝑧 = 𝑃𝑧Δ𝑥Δ𝑦
𝑥
𝑧
𝐹𝑊 = 𝛾𝑉 = 𝜌𝑔Δ𝑥Δ𝑦Δ𝑧
2
𝜃
Δ𝑥 = Δ𝑠 cos 𝜃
Δ𝑧
=Δ
𝑠si
n𝜃
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Fluid Mechanics lecture notes by David S. Ancalle (updated 8/3/2020)
𝑃𝑧Δ𝑥Δ𝑦 − 𝑃𝑠Δ𝑠Δ𝑦 cos 𝜃 − 𝜌𝑔Δ𝑥Δ𝑦Δ𝑧
2=
1
2𝜌Δ𝑥Δ𝑦Δ𝑧𝑎𝑧
𝑃𝑧Δ𝑥Δ𝑦 − 𝑃𝑠ΔxΔ𝑦 − 𝜌𝑔Δ𝑥Δ𝑦Δ𝑧
2=
1
2𝜌Δ𝑥Δ𝑦Δ𝑧𝑎𝑧
𝑃𝑧 − 𝑃𝑠 − 𝜌𝑔Δ𝑧
2=
1
2𝜌Δ𝑧𝑎𝑧
𝑃𝑧 − 𝑃𝑠 =1
2𝜌(𝑎𝑧 + 𝑔)Δ𝑧
As the element shrinks to a point, Δ𝑧 → 0 and the limit becomes:
𝑃𝑧 − 𝑃𝑠 = 0 ⟹ 𝑃𝑧 = 𝑃𝑠
So, we see that, at a point, the pressure in all directions does not vary. This analysis holds true for other shapes.
2.1.2 Pressure Variation
Consider an infinitesimally small element with dimensions 𝑑𝑥, 𝑑𝑦, 𝑑𝑧, that undergoes a pressure 𝑃(𝑥, 𝑦, 𝑧), where
𝑃0 is the pressure at the center of the element. We can determine the pressure at the sides of the element using the
chain rule:
𝑑𝑃 =𝜕𝑃
𝜕𝑥𝑑𝑥 +
𝜕𝑃
𝜕𝑦𝑑𝑦 +
𝜕𝑃
𝜕𝑧𝑑𝑧
So, if we move from the center of the element to an arbitrary side 𝑠 (where 𝑠 can be any side, 𝑥, 𝑦, or 𝑧), then the
pressure at that side is:
𝑃 = 𝑃0 +𝜕𝑃
𝜕𝑠
𝑑𝑠
2
Note that the distance moved is 𝑑𝑠
2 because we are moving from the center of the element to the side, so we only
move half of the entire length of the element in that direction. We can use this equation to find the pressure on all of
the sides of the element, and we can use those pressures to determine the forces acting on the element:
On the top face: 𝑃 = 𝑃0 +
𝜕𝑃
𝜕𝑧
𝑑𝑧
2 𝐹 = − (𝑃0 +
𝜕𝑃
𝜕𝑧
𝑑𝑧
2) 𝑑𝑥 𝑑𝑦
On the bottom face: 𝑃 = 𝑃0 −
𝜕𝑃
𝜕𝑧
𝑑𝑧
2 𝐹 = (𝑃0 −
𝜕𝑃
𝜕𝑧
𝑑𝑧
2) 𝑑𝑥 𝑑𝑦
On the front face: 𝑃 = 𝑃0 +
𝜕𝑃
𝜕𝑥
𝑑𝑥
2 𝐹 = − (𝑃0 +
𝜕𝑃
𝜕𝑥
𝑑𝑥
2) 𝑑𝑦 𝑑𝑧
𝑑𝑦 𝑑𝑥
𝑑𝑧
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Fluid Mechanics lecture notes by David S. Ancalle (updated 8/3/2020)
On the back face: 𝑃 = 𝑃0 −
𝜕𝑃
𝜕𝑥
𝑑𝑥
2 𝐹 = (𝑃0 −
𝜕𝑃
𝜕𝑥
𝑑𝑥
2) 𝑑𝑦 𝑑𝑧
On the left face: 𝑃 = 𝑃0 −
𝜕𝑃
𝜕𝑦
𝑑𝑦
2 𝐹 = (𝑃0 −
𝜕𝑃
𝜕𝑦
𝑑𝑦
2) 𝑑𝑥 𝑑𝑧
On the right face: 𝑃 = 𝑃0 +
𝜕𝑃
𝜕𝑦
𝑑𝑦
2 𝐹 = − (𝑃0 +
𝜕𝑃
𝜕𝑦
𝑑𝑦
2) 𝑑𝑥 𝑑𝑧
Newton’s second law can now be applied in each direction, taking 𝑚 = 𝜌𝑉 = 𝜌 𝑑𝑥 𝑑𝑦 𝑑𝑧 and 𝐹𝑊 = 𝜌𝑔𝑉:
∑ 𝐹𝑥 = (𝑃0 −𝜕𝑃
𝜕𝑥
𝑑𝑥
2) 𝑑𝑦 𝑑𝑧 − (𝑃0 +
𝜕𝑃
𝜕𝑥
𝑑𝑥
2) 𝑑𝑦 𝑑𝑧 = 𝜌 𝑑𝑥 𝑑𝑦 𝑑𝑧 𝑎𝑥
−𝜕𝑃
𝜕𝑥𝑑𝑥 𝑑𝑦 𝑑𝑧 = 𝜌 𝑑𝑥 𝑑𝑦 𝑑𝑧 𝑎𝑥
𝜕𝑃
𝜕𝑥= −𝜌𝑎𝑥
∑ 𝐹𝑦 = (𝑃0 −𝜕𝑃
𝜕𝑦
𝑑𝑦
2) 𝑑𝑥 𝑑𝑧 − (𝑃0 +
𝜕𝑃
𝜕𝑦
𝑑𝑦
2) 𝑑𝑥 𝑑𝑧 = 𝜌 𝑑𝑥 𝑑𝑦 𝑑𝑧 𝑎𝑦
𝜕𝑃
𝜕𝑦= −𝜌𝑎𝑦
∑ 𝐹𝑧 = (𝑃0 −𝜕𝑃
𝜕𝑧
𝑑𝑧
2) 𝑑𝑥 𝑑𝑦 − (𝑃0 +
𝜕𝑃
𝜕𝑧
𝑑𝑧
2) 𝑑𝑥 𝑑𝑦 − 𝜌𝑔 𝑑𝑥 𝑑𝑦 𝑑𝑧 = 𝜌 𝑑𝑥 𝑑𝑦 𝑑𝑧 𝑎𝑧
−𝜕𝑃
𝜕𝑧𝑑𝑥 𝑑𝑦 𝑑𝑧 − 𝜌𝑔 𝑑𝑥 𝑑𝑦 𝑑𝑧 = 𝜌 𝑑𝑥 𝑑𝑦 𝑑𝑧 𝑎𝑧
𝜕𝑃
𝜕𝑧 = −𝜌𝑎𝑧 − 𝜌𝑔 = −𝜌(𝑎𝑧 + 𝑔)
We can now apply these expressions to find the pressure differential in any direction:
𝑑𝑃 =𝜕𝑃
𝜕𝑥𝑑𝑥 +
𝜕𝑃
𝜕𝑦𝑑𝑦 +
𝜕𝑃
𝜕𝑧𝑑𝑧 = −𝜌𝑎𝑥𝑑𝑥 − 𝜌𝑎𝑦𝑑𝑦 − 𝜌(𝑎𝑧 + 𝑔)𝑑𝑧
2.1.3 Pressure in Fluids at Rest
An object at rest does not undergo acceleration in any direction. Therefore, for a fluid at rest, the pressure
differential becomes:
𝑑𝑃 = −𝜌𝑎𝑥𝑑𝑥 − 𝜌𝑎𝑦𝑑𝑦 − 𝜌(𝑎𝑧 + 𝑔)𝑑𝑧 = −𝜌𝑔𝑑𝑧
which we can express as:
𝑑𝑃
𝑑𝑧= −𝜌𝑔 = −𝛾
This tells us the following: there is no pressure variation in the x and y directions, pressure only varies in the vertical
direction; pressure decreases as we move up and increases as we move down; and pressure variation for a fluid at
rest depends on the specific weight of the fluid.
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Fluid Mechanics lecture notes by David S. Ancalle (updated 8/3/2020)
If the specific weight of the fluid is constant (e.g. an incompressible liquid with no variations in temperature or
composition), then we can integrate the pressure differential:
∫ 𝑑𝑃 = ∫ −𝛾𝑑𝑧 = −𝛾 ∫ 𝑑𝑧
Δ𝑃 = −𝛾Δ𝑧
2.2 Hydrostatic Pressure
2.2.1 Pascal’s Law
Fluid statics: the study of fluids at rest.
When a system is at rest: ∑ �⃑� = 0, ∑ �⃑⃑⃑� = 0
We define pressure as 𝑃 = 𝑑𝐹/𝑑𝐴. Following this definition, we can express the average pressure over an area as
𝑃𝑎𝑣𝑔 = 𝐹/𝐴.
In 1647-1648, Blaise Pascal established that a change in pressure in an enclosed fluid at rest is equal at all points in
the fluid. This is known as Pascal’s Law.
Consider an enclosed container full of a fluid:
If we apply a force 𝐹1 on the left area 𝐴1, then the pressure applied on the fluid is 𝑃1 = 𝐹1/𝐴1. Pascal’s Law states
that the change in pressure will be equal at all points in the fluid, which means that on the right area 𝐴2, the pressure
will be increased by 𝑃2 = 𝑃1. We can then determine the resulting force on the right area: 𝐹2 = 𝑃2𝐴2.
Mathematically:
𝑃1 =𝐹1
𝐴1
= 𝑃2 =𝐹2
𝐴2
∴ 𝐹2 = 𝐹1
𝐴2
𝐴1
Notice that if 𝐴2 is larger than 𝐴1, then the force applied on the right side will be increased. This makes hydraulic
lifts possible.
𝐹1 𝐹2𝐴1
𝐴2
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Fluid Mechanics lecture notes by David S. Ancalle (updated 8/3/2020)
Ex. In the figure above, 𝐹1 = 1 𝑘𝑁, 𝐴1 = 2 𝑚2, and 𝐴2 = 4 𝑚2. Determine 𝐹2.
Ans:
𝐹2 = 𝐹1
𝐴2
𝐴1
= (1) (4
2) = 2 𝑘𝑁
Note: If the area is twice as large, the force will be twice as large as well.
2.2.2 Pressure in Incompressible Fluids
We define hydrostatic pressure as the pressure exerted by a fluid at rest on a point submerged in that fluid. While the
term literally means “pressure of water at rest,” we will use this term to refer to pressure from any liquid. Consider a
container with an incompressible fluid. We will analyze a very small cube of fluid:
We can determine the pressure acting on the top face of the fluid as follows:
𝑃 =𝐹𝑊
𝐴=
𝛾𝑉
𝑑𝑥 𝑑𝑦=
𝛾 𝑑𝑥 𝑑𝑦 ℎ
𝑑𝑥 𝑑𝑦= 𝛾ℎ
where:
• 𝐹𝑊 is the weight of the column of water above the cube
• 𝐴 is the area of the top face of the cube
• 𝛾 is the specific weight of the column of water above the cube
• 𝑉 is the volume of the column of water above the cube
Notice that the pressure exerted by the fluid on the top of the cube only depends on the depth of the cube, and not its
geometry. We also did not include atmospheric pressure (which can be conceptually defined as the pressure exerted
on the water by the column of air above it).
This equation agrees with the equation for pressure in fluids at rest that we derived in the previous section Δ𝑃 =
−𝛾Δ𝑧, where Δ𝑧 = −ℎ (since ℎ is measured from top to bottom).
ℎ
𝑑𝑦 𝑑𝑥
𝑑𝑧
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Fluid Mechanics lecture notes by David S. Ancalle (updated 8/3/2020)
2.2.3 Pressure Head
For incompressible fluids, the pressure can be expressed in dimensions of length by solving for ℎ:
𝑃 = 𝛾ℎ ⟹ ℎ =𝑃
𝛾
This quantity is called the pressure head, and is used as a unit in pressure measurements. Conceptually, this head
represents the height of a column of liquid with specific gravity 𝛾 that produces a gage pressure 𝑃. Pressure is
commonly measured as the length of a column of water (𝑆 = 1) or a column of mercury (𝑆 = 13.6). The unit
conversions are as follows:
1 𝑚𝑚𝐻2𝑂 = 9.81 𝑃𝑎
1 𝑚𝑚𝐻𝑔 = 133 𝑃𝑎
From the conversions, it can be deduced that mercury is used to measure higher pressures than water. Pressure head
can also be measured in English units.
2.2.4 Pressure in Compressible Fluids
On compressible fluids, density and specific weight vary with pressure. Therefore, for a compressible fluid (such as
an ideal gas), we will have to express density as a function of pressure. Applying the ideal gas law to the pressure
differential yields:
𝑑𝑃
𝑑𝑧= −𝜌𝑔 = −
𝑃
𝑅𝑇𝑔
If we take the temperature of the fluid to be constant, we can integrate the above expression to get:
∫1
𝑃𝑑𝑃
𝑃
𝑃0
= ∫ −𝑔
𝑅𝑇𝑑𝑧
𝑧
𝑧0
ln𝑃
𝑃0
= −𝑔
𝑅𝑇(𝑧 − 𝑧0)
𝑃 = 𝑃0 𝑒−𝑔
𝑅𝑇(𝑧−𝑧0)
where 𝑃0 is a reference pressure at an elevation 𝑧0.
2.2.5 Pressure in the Atmosphere
We have learned that the temperature in the troposphere decreases linearly by 𝑇 = 𝑇0 − 𝛼𝑧. We can now determine
the pressure variation in the troposphere. Applying the ideal gas law to the pressure differential yields
𝑑𝑃
𝑑𝑧= −𝜌𝑔 = −
𝑃
𝑅𝑇𝑔
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Fluid Mechanics lecture notes by David S. Ancalle (updated 8/3/2020)
Rearranging, expressing 𝑇 as a function of 𝑧, integrating between an elevation of 0 and z, and solving for pressure:
∫1
𝑃𝑑𝑃
𝑃
𝑃𝑎𝑡𝑚
= ∫ −𝑔
𝑅(𝑇0 − 𝛼𝑧)𝑑𝑧
𝑧
0
ln𝑃
𝑃𝑎𝑡𝑚
=𝑔
𝛼𝑅ln
𝑇0 − 𝛼𝑧
𝑇0
𝑃 = 𝑃𝑎𝑡𝑚 (𝑇0 − 𝛼𝑧
𝑇0
)
𝑔𝛼𝑅
Solving for values between 0 < 𝑧 ≤ 1000 shows that the pressure decrease is very small, and changes in static air
pressure can be neglected unless the elevation difference is relatively large.
The temperature in the stratosphere, 𝑇𝑠, is constant. So, we determine the pressure by integrating from 𝑧𝑠, the lowest
elevation in the stratosphere to an elevation 𝑧.
𝑃 = 𝑃𝑠 𝑒−
𝑔𝑅𝑇𝑠
(𝑧−𝑧𝑠)
2.3 Measuring Pressure
2.3.1 U-tube Manometers
Consider pressurized flow in a pipe. If we punch open a hole in the top wall of the pipe and insert a tube, water will
flow upward until the pressure of the water is equal to atmospheric pressure. The fluid in the tube will be static, and
so we can use hydrostatic equations to find the pressure at different points in the tube and in the pipe.
Manometers are instruments that use this principle to measure pressures. In the figure below, a cross-sectional view
of the pipe is shown, and a manometer is connected to the side of the pipe. The manometer connects the pipe to a
point with a known pressure (i.e. ○2 , where the pressure is atmospheric). By measuring the pressure difference
between points ○1 and ○2 , we can determine the pressure at the pipe. If we want to measure gage pressures, then
𝑃2 = 𝑃𝑎𝑡𝑚 = 0.
𝑃 = 𝑃𝑎𝑡𝑚
ℎ
𝑃 = 𝛾ℎ
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𝑃2 + Δ𝑃 = 𝑃1
0 + 𝛾ℎ = 𝑃1
𝑃1 = 𝛾ℎ
Remember that pressure increases with depth. Therefore, since we move “downward” from 𝑃2 to 𝑃1, the pressure
difference is positive. If we would have started measuring from the pipe to the tube, our equation would have been:
𝑃1 − 𝛾ℎ = 𝑃2 = 0
𝑃1 = 𝛾ℎ
Ex. In the U-tube manometer above, ℎ = 2𝑓𝑡 and the fluid is water. What is the pressure in the pipe?
Ans:
𝛾 = 62.4 𝑙𝑏/𝑓𝑡3
𝑃 = 𝛾ℎ = (62.4)(2) = 124.8 𝑙𝑏/𝑓𝑡2
These types of manometers are called U-tube manometers. Manometers that use a single fluid are usually only used
to measure very small pressures. To measure larger pressures, a heavier fluid can be inserted in the manometer.
Measuring the pressure differences from ○1 to ○3 yields:
𝑃1 + 𝛾1ℎ − 𝛾2𝐻 = 𝑃3 = 0
𝑃1 = 𝛾2𝐻 − 𝛾1ℎ
Notice that the pressure at ○2 and ○2’ is the same, since there is no difference in elevation.
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2.3.2 Differential Manometers
A differential manometer is used to measure pressure differences between two pipes or between two points in a
conduit. The pressure difference is computed in the same way as the U-tube manometer, but the end of the
manometer will not be open to the atmosphere.
2.3.3 Piezometer
A piezometer is a simple type of manometer that consists of a vertical tube that is inserted in a vessel with a fluid.
Piezometers are useful for measuring small pressures. The static pressure of a point in the liquid at a depth 𝑑 is:
𝑃 = 𝛾(ℎ + 𝑑)
2.3.4 Micromanometers
Another type of manometer is the micromanometer, which is used to measure very small pressure changes.