Fluid Statics & the Hydrostatic Equation

J. Gordon Leishman

doi:https://doi.org/10.15394/eaglepub.2022.1066.9

10 Fluid Statics & the Hydrostatic Equation

Introduction

In aerodynamic problem solving, a primary concern is describing and understanding the motion or dynamics of the air. First, remember that air is just one type of fluid, for which there are many types in the form of gases and liquids. However, suppose the fluid is not moving or stationary, commonly called a stagnant fluid. In that case, its characteristics or properties (e.g., its pressure, density, etc.) can be described more easily using the physical principles of fluid statics.

A fluid in static equilibrium can be defined as a state where every fluid particle is either at rest or has no relative motion with respect to the other particles in the fluid. There are no viscous forces on stagnant fluids because there is no relative motion between the fluid elements. Under these conditions, two types of forces on the fluid must be considered:

Body forces, e.g., weight, inertial accelerations.
Surface forces, e.g., the effects of an external pressure acting over an area.

The term hydrostatics often refers to the field of fluid statics in general. Hydrostatic principles apply to all types of fluids, both gases, and liquids. However, hydrostatic principles are frequently used to analyze liquids and other predominantly incompressible fluids, i.e., fluids with constant density. The term pneumatics is used when the principles of fluid statics are applied to gases such as air. Examples of engineering problems that could be analyzed by using hydrostatics and pneumatics include the following:

The transmission of forces and power using pressurized liquids.
The measurement of pressure using various types of manometry.
The buoyancy and stability of floating objects, such as ships in the water, and lighter-than-air aircraft, such as balloons and blimps.
Design of water retaining structures such as dams and reservoirs.
The pressure and other fluid properties inside containers or vessels, including under the action of body forces or acceleration fields.
The properties of the atmosphere on Earth or other planets.

Learning Objectives

Understand the meaning of a stagnant fluid and the underlying principles of fluid statics.
Be able to derive, understand, and use the general equation for a hydrostatic pressure field.
Understand how to develop and apply the hydrostatic equation.
Use the principles of hydrostatics and the hydrostatic equation to solve some fundamental engineering problems.

Pascal’s Law

The pressure at a point in a fluid at rest is equal in all directions, so the pressure is referred to as an isotropic scalar quantity that is formally embodied in Pascal’s Law. The proof of Pascal’s Law proceeds by considering an infinitesimally small right-angled, three-dimensional fluid wedge at rest. However, the outcome is most easily obtained if the wedge is examined in just one view (or one plane), as shown in the figure below, which appears as a right-angled triangle with horizontal side length $dx$ , vertical side length $dy$ , and diagonal length $ds$ , with the angle $\theta$ .

The pressures acting on the respective faces are $p_{x}$ in the $x$ -direction, $p_{y}$ in the $y$ -direction, and $p_{s}$ along the diagonal. Therefore, for horizontal pressure force equilibrium on the fluid per unit length (into the page), then

(1) $\begin{equation*} p_x dy - p_s ds \cos \theta = 0 \end{equation*}$

For the vertical pressure force equilibrium per unit length, then

(2) $\begin{equation*} p_y dx - p_s ds \sin \theta - dW = 0 \end{equation*}$

where $dW$ is the weight of the fluid per unit length, i.e.,

(3) $\begin{equation*} dW = \varrho g \, (dx \, dy/2) \end{equation*}$

From the geometry of the triangle in the figure, then $dx = ds \sin \theta$ and $dy = ds \cos \theta$ . Also, as the triangle shrinks to point B in the fluid, then the product $dx \, dy$ becomes negligible, i.e., the effects of the weight of fluid inside can be neglected. So, it becomes apparent that

(4) $\begin{equation*} p_x dx - p_s dx = 0 \end{equation*}$

and

(5) $\begin{equation*} p_y dy - p_s dy = 0 \end{equation*}$

Therefore, from these two latter equations, the conclusion reached is that as the triangle shrinks to the point, then

(6) $\begin{equation*} p_x = p_y = p_s = \mbox{~constant} = p \end{equation*}$

as shown in the figure below. This latter proof is easily extended to three dimensions to give the result that $p_x = p_y = p_z = \mbox{constant} = p$ , which states in words that that pressure at any point in a fluid at rest is the same in all directions; this is called Pascal’s Law.

Pascal’s Law states that the pressure at a point is constant and acts equally in all directions.

Archimedes’ Principle

Archimedes’ principle is commonly used in solving problems in hydrostatics. This principle states that the upward or buoyancy force exerted on a body immersed in a fluid equals the weight of the fluid that the body displaces. This outcome is true whether the body is wholly or partially submerged in the fluid and is independent of the body’s shape. The resulting force acts in the upward direction at the center of mass of the displaced fluid. The principle is commonly used in problems involving floatation or “floating bodies,” such as boats, ships, balloons, and airships.

For example, as shown in the figure below, consider a balloon floating in the air filled with some volume of another (lighter) gas, e.g., warmer air and so of a lower density, in the case of a hot-air balloon. Using Archimedes’ principle, the balloon’s buoyancy force will equal the weight of displaced air. The balloon’s net weight will include the weight of its structure (including anything attached) plus the weight of gas contained within it. Therefore, the net force acting on the balloon will equal the difference between the balloon’s net weight and the weight of the displaced air (i.e., the “upthrust”). In the case of an airship or blimp, it will be filled with helium gas rather than hot air. However, high-altitude balloons, such as weather balloons, will be filled with helium gas.

The net force acting on a lighter-than-air aircraft is equal to the difference between the downward acting weight of the vehicle and the upward force (upthrust), which is equal to the weight of the displaced air.

In either case, if the upthrust is greater than the weight, then the balloon or airship will rise, and if the upthrust is less than the weight, then the balloon or airship will descend. Equilibrium, or neutral buoyancy, will be achieved when these two forces are equal and are in exact vertical balance.

Take the airship as an example. Using Archimedes’ principle, the upforce (lift), $L$ , on the airship will be equal to the net weight of the air displaced less the weight of the helium inside the envelope, i.e.,

(7) $\begin{equation*} L = \left( \varrho_{\rm air} - \varrho_{\rm He} \right) g {\cal{V}} \end{equation*}$

where ${\cal{V}}$ is the volume of the gas envelope, $\varrho_{\rm air}$ is the density of air, and $\varrho_{\rm He}$ is the density of helium gas. If the weight of the airship is $W$ , then for vertical force equilibrium and neutral buoyancy, then

(8) $\begin{equation*} W = \left( \varrho_{\rm air} - \varrho_{\rm He} \right) g {\cal{V}} \end{equation*}$

Therefore, the volume of the helium can be determined to give a certain buoyancy force to overcome the airship’s weight. This is the fundamental principle for all lighter-than-air flight vehicles.

Hydrostatic Pressure Field

Remember that pressure is a point quantity and can have a different value from one part to another in the fluid. With the concept of pressure already being discussed, the formal derivation of the hydrostatic variation of pressure in a stagnant fluid can proceed with the development of a particular case of this result, called the hydrostatic equation.

Consider in the differential sense an infinitesimally small volume of fluid in a stagnant flow, i.e., a fluid element of volume $dx \, dy \, dz$ , as shown in the figure below. As the volume becomes infinitesimally small in the limit, it shrinks to point B.

Pressure forces acting on an infinitesimally small stationary volume of fluid with pressure gradients in three directions.

Acting on this small fluid volume are:

Pressure forces from the surrounding fluid, which vary from point to point, i.e., $p = p(x, y, z)$ .
Body forces, in general, which are usually expressed as per unit mass, i.e., $\vec{f_b} = f_x \vec{i} + f_y \vec{j} + f_z \vec{k}$ .
A specific body force in the form of gravity, which manifests as a vertically downward force from the weight of fluid inside the element, which can be written as $\vec{f_b} = -g \vec{k}$ per unit mass, i.e., the negative sign indicating that the force of gravity acts downward.

Notice the units of a body force per unit mass has dimensions of acceleration, i.e., length/time². Therefore, body forces can also arise from any inertial accelerations of the fluid. While fluid forces from electromagnetic or electrostatic effects are not a consideration for air, certain types of fluids have properties that respond to such fields and may be explicitly selected for that purpose.

Consider the pressure forces in the $x$ direction, i.e., on the $y$ – $z$ face of area $dy \, dz$ . Let $\partial p/\partial x$ be the change of pressure $p$ with respect to $x$ . Notice that the partial derivative must be used because the pressure could vary in all three spatial directions. The net pressure force in the $x$ direction is

(9) $\begin{equation*} d F_x = p \left( dy \, dz \right) - \left( p + \frac{\partial p}{\partial x} dx \right) dy \, dz = -\frac{\partial p}{\partial x} \left(dx \, dy \, dz \right) \end{equation*}$

Similarly, for the $y$ direction, the net pressure force is

(10) $\begin{equation*} d F_y = p \left( dx \, dz \right) - \left( p + \frac{\partial p}{\partial y} dy \right) dx \, dz = -\frac{\partial p}{\partial y} \left(dx \, dy \, dz \right) \end{equation*}$

and in the $z$ direction the net pressure force is

(11) $\begin{equation*} d F_z = p \left( dx \, dy \right) - \left( p + \frac{\partial p}{\partial z} dz \right) dx \, dy = -\frac{\partial p}{\partial z} \left(dx \, dy \, dz \right) \end{equation*}$

Therefore, the net resultant pressure force on the fluid element is

(12) $\begin{eqnarray*} d\vec{F} & = & dF_x \vec{i} + dF_y \vec{j} + dF_z \vec{k} \\[6pt] & = & -\left( \frac{\partial p}{\partial x} \vec{i} + \frac{\partial p}{\partial y} \vec{j} + \frac{\partial p}{\partial z} \vec{k} \right) dx \, dy \, dz \nonumber \\[6pt] & =& -\nabla p \, (dx \, dy \, dz ) \end{eqnarray*}$

where the gradient vector operator $\nabla$ has now been introduced.

If $\varrho$ is the mean density of the fluid element, then the total mass of the fluid element, $dm$ is $\varrho \left(dx \, dy \, dz \right)$ , as shown in the figure below. Any coordinate system can be assumed, but usually, $z$ is defined as positive in the vertical direction, in which case the gravitational acceleration is $-g\vec{k}$ . As such, gravity manifests as a vertically downward force from the weight of fluid inside the element, i.e., the net gravitational force is $-\varrho \left(dx \, dy \, dz \right)g\vec{k}$ .

Body forces per unit mass acting on a fluid can include inertial accelerations and acceleration under gravity, the latter manifesting as weight.

Furthermore, any additional body forces per unit mass acting on the element can be described as

(13) $\begin{equation*} \vec{f_b} = f_x \vec{i} + f_y \vec{j} + f_z \vec{k} \end{equation*}$

Now, if a fluid element is at rest and in equilibrium, then the sum of the pressure forces and the gravitational force and the body force must be zero, i.e.,

(14) $\begin{equation*} -\nabla p (dx \, dy \, dz ) + \varrho (dx \, dy \, dz) \vec{f_b} - \varrho \left(dx \, dy \, dz \right)g\vec{k} = 0 \end{equation*}$

or

(15) $\begin{equation*} \nabla p = \varrho \vec{f_b} -\varrho g\vec{k} \end{equation*}$

In scalar form the foregoing equation can be written as

(16) $\begin{eqnarray*} \frac{\partial p}{\partial x} & = & \varrho f_x \nonumber \\[6pt] \frac{\partial p}{\partial y} & = & \varrho f_y \nonumber \\[6pt] \frac{\partial p}{\partial z} & = & \varrho f_z -\varrho g \nonumber \end{eqnarray*}$

Therefore, these three partial differential equations describe the pressure variations within a stagnant fluid. Notice that the center of gravity is at the centroid of the fluid element, so there is no gravitational moments acting on the element.

Effects of Gravity Alone

If the only body force is the effects of gravitational acceleration or gravity force (the weight) acting on the element, then

(17) $\begin{equation*} d\vec{W} = -\varrho (dx \, dy \, dz) g \vec{k} \end{equation*}$

where $g$ is acceleration under gravity and because it manifests as a downward force then there is a minus sign in the $z$ direction, i.e., it is downward because $z$ is measured positive upward. Therefore, in this case

(18) $\begin{equation*} \nabla p = -\varrho g \vec{k} \end{equation*}$

or in scalar form

(19) $\begin{eqnarray*} \frac{\partial p}{\partial x} & = & 0 \nonumber \\[6pt] \frac{\partial p}{\partial y} & = & 0 \nonumber \\[6pt] \frac{\partial p}{\partial z} & = & -\varrho g \nonumber \\ \end{eqnarray*}$

Because the pressure does not depend on $x$ or $y$ , the above equation can now be written as
the ordinary differential equation

(20) $\begin{equation*} \frac{dp}{dz} = -\varrho g \end{equation*}$

This latter ordinary differential equation is called the hydrostatic equation. The hydrostatic equation relates the change in pressure $dp$ in a fluid to a change in vertical height $dz$ . This equation has many uses in manometry, hydrodynamics, and atmospheric physics, including the derivation of the properties of the International Standard Atmosphere (ISA), which is considered later.

Special Solutions to the Hydrostatic Equation

There are several special cases of interest that lead to convenient solutions to the hydrostatic equation, namely the pressure changes for a:

Constant density fluid.
Constant temperature gas.
Linear temperature gradient in a gas.

In the case of gases, once the pressure is known then the corresponding density and/or temperature of the gas can also be solved for using the equation of state. Remember that the equation of state can be used to relate pressure, density, and temperature of a gas. If one of these quantities is unknown, then the values of two others can be used to determine the unknown quantity. In the case of liquids, there is no equivalent equation of state because ${\cal{V}} =$ constant.

Constant Density Fluid

Suppose the density of a fluid is assumed to be a constant then Eq. 20 can be easily integrated from one height to another to find the corresponding change in pressure. Separating the variables gives

(21) $\begin{equation*} \int_{p_1}^{p_2} dp = -\varrho g \int_{z_1}^{z_2} dz \end{equation*}$

and so

(22) $\begin{equation*} p_2 - p_1 = -\varrho g \left( z_2 - z_1 \right) \end{equation*}$

or

(23) $\begin{equation*} \Delta p = -\varrho \Delta z \end{equation*}$

or even

(24) $\begin{equation*} p_1 + \varrho g z_1 = p_2 + \varrho g z_2 \end{equation*}$

or perhaps in a more familiar form as

(25) $\begin{equation*} p + \varrho g z = \mbox{constant} \end{equation*}$

where $z$ is “height.” This simple equation has many uses in various forms of manometry.

Constant Temperature Gas

If the temperature of a gas is assumed to be constant, i.e., what is known as an isothermal condition, then $T= T_0 =$ constant. In this case, then

(26) $\begin{eqnarray*} \frac{dp}{dz} & = & -\varrho g \\[8pt] T & = & T_0 \\[8pt] p & = & \varrho R T \end{eqnarray*}$

Substituting for $\varrho$ and assuming isothermal conditions then

(27) $\begin{equation*} \frac{dp}{p}= - \frac{g}{R \, T_0} dz \end{equation*}$

Integrating from $z_1$ to $z_2$ where the pressures are $p_1$ and $p_2$ , respectively, gives

(28) $\begin{equation*} \int_{p_1}^{p_2} \frac{dp}{p}= -\frac{g}{R \, T_0} \int_{z_1}^{z_2} dz \end{equation*}$

and performing the integration gives

(29) $\begin{equation*} \ln p_2 - \ln p_1= \ln \left( \frac{p_2}{p_1} \right) = -\frac{g}{R \, T_0} \left( z_2 - z_1 \right) \end{equation*}$

Therefore,

(30) $\begin{equation*} \frac{p_2}{p_1} = \exp \left( -\frac{g (z_2 - z_1)}{R \, T_0} \right) \end{equation*}$

Notice also that for isothermal conditions

(31) $\begin{equation*} \frac{\varrho_2}{\varrho_1} = \frac{p_2}{p_1} \end{equation*}$

Linear Temperature Gradient in a Gas

If the temperature of a gas decreases linearly with height, i.e., $T = T_0 - \alpha z$ where $\alpha$ is the thermal gradient or “lapse rate,” then in this case

(32) $\begin{eqnarray*} \frac{dp}{dz} & = & -\varrho g \\[6pt] T & = & T_0 - \alpha z \\[6pt] p & = & \varrho R T \end{eqnarray*}$

Substituting the other two equations into the differential equation gives

(33) $\begin{equation*} \frac{dp}{dz} = -\varrho g = - \left( \frac{p}{R T} \right) g = -\frac{p g}{R (T_0 - \alpha z)} \end{equation*}$

which is the governing equation that now needs to be solved. By means of separation of variables and integrating from $z_1$ to $z_2$ where the pressures are $p_1$ and $p_2$ , respectively, gives

(34) $\begin{equation*} \int_{p_1}^{p_2} \frac{dp}{p} = - \frac{g}{R} \int_{z_1}^{z_2}\frac{dz}{T_0 - \alpha z} \end{equation*}$

Performing the integration gives

(35) $\begin{equation*} \frac{p_2}{p_1} = \bigg( 1 - \frac{\alpha}{T_0} \left(z_2 - z_1 \right) \bigg)^{\displaystyle{\frac{g}{R\alpha}}} \end{equation*}$

The corresponding density ratio is

(36) $\begin{equation*} \frac{\varrho_2}{\varrho_1}= \bigg( 1 - \frac{\alpha}{T_0} \left( z_2 - z_1 \right) \bigg)^{\left(\displaystyle{\frac{g}{R\alpha}} - 1 \right)} \end{equation*}$

“Rules” of Manometry

Several basic principles or “rules” of manometry directly follow in using the constant density solution to the hydrostatic equation, i.e., p + $\varrho g z = \mbox{constant}$ . Learning how to use these rules properly is obtained by studying exemplar problems in hydrostatics. These rules are:

1. The pressure at two points 1 and 2 in the same fluid at the same height is the same if a continuous line can be drawn through the fluid from point 1 to point 2, i.e., $p_1$ = $p_2$ if $z_1$ = $z_2$ .
2. Any free surface of a fluid open to the atmosphere has atmospheric pressure, $p_a$ , acting upon the surface.
3. In most practical hydrostatic problems with air, the atmospheric pressure can be assumed to be constant at all heights unless the change in vertical height is significant, e.g., more than a few feet or meters.
4. For a liquid in a container, then the shape of a container does not matter in hydrostatics, i.e., the pressure at a point in the liquid depends only on the vertical height of the fluid above it. This result is called the hydrostatic paradox.
5. The pressure is constant across any flat fluid-to-fluid interface.

Hydrostatic Paradox

Consider the situation shown in the figure below, which shows three different-shaped containers that are filled with the same liquid. What are the pressure values on the container’s bottom in each case? The answer is that the pressures are all the same because the hydrostatic equation shows the pressure at a point in a liquid depends only upon the height of the liquid directly above it, $H$ , and not at all upon the container’s shape.

According to the principles of hydrostatics, then the pressure on the bottom of each container is the same, regardless of the shape of the container and volume (or weight) of liquid above it.

The pressure at the bottom of each container is $p =\varrho \, g \, H$ , where $\varrho$ is the density of the liquid. If each container has the same area $A$ at the bottom, then the pressure force, which is $F = p A$ on each container’s base, is the same. This outcome is independent of the volume of liquid, and the weight of the liquid in each container is different, and is referred to as the hydrostatic paradox.

Proof of Archimedes’ Principle

Having introduced the concept of hydrostatic pressure and the hydrostatic equation, the proof of Archimedes’ principle is now relatively straightforward because the source of buoyancy is a consequence of a pressure difference acting on the body. To this end, consider a solid rectangular body of dimensions $\Delta x$ , $\Delta y$ , and $\Delta z$ completely immersed in a fluid of constant density, as shown in the figure below.

A solid body immersed in a fluid will produce an upforce (or upthrust) equal to the weight of the fluid displaced, which is referred to as Archimedes’ principle.

The hydrostatic pressure force on the top face is lower than that on the bottom face, the difference being the source of the upthrust (buoyancy) on the body. If the upper face is at a distance $z_1$ below the surface of the fluid (i.e., below the hydrostatic pressure interface), then the pressure there on this face will be $p_1 = \varrho g z_1$ . The lower face is at a distance $z_2$ below the surface, so the pressure acting on this face will be $p_2 = \varrho g z_2$ . Notice that the net pressure forces in the $x$ and $y$ direction will balance out. The net pressure force in the upward $z$ direction, which is the buoyancy force, $F_B$ , will be

(37) $\begin{eqnarray*} F_B & = & \varrho g z_2 (\Delta x \, \Delta y) - \varrho g z_1 (\Delta x \, \Delta y) = \varrho g (z_2 - z_1) (\Delta x \, \Delta y) \\[6pt] \nonumber & = & \varrho g ( \Delta z) ( \Delta x \, \Delta y) = \varrho g ( \Delta x \, \Delta y \, \Delta z) = \varrho g \, \Delta \cal{V} \end{eqnarray*}$

It can be seen that $\Delta x \, \Delta y \, \Delta z$ is simply the volume of the body, $\Delta \cal{V}$ , and so $\varrho \, \Delta\cal{V}$ is the mass of the fluid that is displaced by the body. The weight of the of the fluid, $W$ , displaced by the body is then $W = \varrho g \, \Delta\cal{V}$ .

Therefore, the proof concludes by stating that the upthrust on the body is equal to the weight of the fluid displaced by the body, i.e., Archimedes’ principle, which is a consequence of the net difference in hydrostatic pressure force acting on the body. This latter result can be generalized to an immersed body of any shape, including one partially immersed in a fluid.

U-tube Manometer

A U-tube manometer is the oldest and most basic type of pressure measuring instrument, and it is still used today as a reference instrument because of its inherent accuracy. A U-tube manometer is simply a glass tube in the form of a “U” partially filled with a liquid, as shown in the left figure below. This liquid is often water, alcohol, oil, mercury (rarely used today), or some other liquid of known density. Light liquids, such as water, are used to measure small pressure differences, and heavier liquids are used for larger pressure differences. This type of manometer has no moving parts and requires no formal calibration other than the use of an accurate length scale. The scale may be a simple ruler or something more sophisticated with a vernier scale.

The operating principle of liquid manometers is that they measure differences in pressure by balancing the weight of a liquid contained between the two applied pressures, as shown in the figure below. The height difference between the liquid levels in the two legs of the U-tube is proportional to the pressure difference, and so the measurement of height (length) can be related to the value of differential pressure.

The relative levels of the fluid in U-tube manometer respond to the differences in pressure applied to each leg.

Using one leg of the manometer as a reference, say $p_1$ , which could continue to be open to the atmosphere or connected to another (known) reference pressure, and connecting the other leg to an unknown applied pressure, say $p_2$ m then the difference in column heights $\Delta h$ indicates the difference in the two pressures, i.e.,

(38) $\begin{equation*} p_2 - p_1 = \varrho_{\rm liq} \, g \Delta H \end{equation*}$

where $\varrho_{\rm liq}$ is the density of the liquid used in the U-tube manometer. Notice that pressure that is measured with respect to atmospheric pressure is called gauge pressure. If the reference pressure is a vacuum, then the pressure is referred to as absolute pressure.

The figure below summarizes what happens with a U-tube manometer depending on whether the applied pressure $p_2$ is greater than, equal to, or less than the reference pressure $p_{\rm ref}$ . For example, if the opening of each leg of a U-tube manometer is exposed to the same pressure, then the height of the liquid columns will be equal.

If one leg of a U-tube manometer is exposed to lower or higher pressures relative to atmospheric pressure then the columns of liquid will rise and fall accordingly. If the opening of each leg of a U-tube manometer is exposed to the same pressure, then the height of the liquid columns will be equal.

A liquid manometer is used to measure pressure in terms of a difference in column heights, which is often expressed in units of inches or centimeters at a specific temperature. The height measurements are easily changed to standard units of pressure using appropriate conversion factors, e.g., 1 inch of water is equivalent to 5.1971 lb/ft $^2$ or 0.0361 lb/in $^2$ .

Several variations of the basic U-tube manometer concept include inclined and reservoir or well-type manometers. All manometers respond to changes in pressure by vertical changes in fluid height, so with an inclined manometer tube, as shown below, a vertical change in the fluid levels is stretched over a longer length. Consequently, an inclined-tube manometer has better sensitivity (and resolution) for measuring lower pressures.

An alternative type of liquid manometer, in this case an inclined manometer. By inclining the tube there is an increased movement of liquid for a given difference in pressure thereby making it more sensitive.

For a well-type or reservoir-type manometer, as shown in the image below, if pressure is applied to the well, then this level will fall very slightly compared to the more significant rise in the column. The differential pressure can be obtained by measuring the change in the fluid level in the well relative to the vertical height of the column.

Another type of liquid manometer, in this case a well manometer. Well type manometers are often used for reference pressure measurements or calibrations of other pressure measuring devices.

Worked Example #1

A U-tube manometer is connected to the inlet and outlet of a water pump, as shown below, the left side is connected to the inlet at pressure $p_{\rm in}$ and the right side to the outlet at pressure $p_{\rm out}$ . Assuming that the inlet and outlet conditions are at the same elevation, determine the pump’s pressure increase. Note: Remember that specific gravity SG = $\varrho/\varrho_{H_2 0}$ . The density of water can be assumed to be 1.93 slugs/ft $^3$ .

The pressure at point “a” will be equal to the pressure produced by each arm of the U-tube manometer. For the left arm, then

$p_a = \varrho_{\rm Hg} \, g h_1 + \varrho_{\rm Hg} \, g h_2 + \varrho_{\rm H_{2}O} \, g h_3 + p_{\rm in}$

Similarly, for the right arm, then

$p_a = \varrho_{\rm Hg} \, g h_1 + \varrho_{\rm H_{2}O} \, g h_2 + \varrho_{\rm H_{2}O} \, g h_3 + p_{\rm out}$

Therefore, the difference in pressure becomes

$\Delta p = p_{\rm out} - p_{\rm in} = \varrho_{\rm Hg} \, g h_2 - \varrho_{\rm H_{2}O} \, g h_2 = \left( \varrho_{\rm Hg} - \varrho_{\rm H_{2}O} \right) g h_2$

and substituting the known numerical values gives

$\Delta p = \left( 13.6 - 1.0 \right) \times 1.93 \times 32.17 \times (6.0/12) = 391.16~\mbox{lb/ft$^2$}$

Worked Example #2

A fuel tank that is vented to the atmosphere is partly filled with fuel. The tank is subject to external vertical and lateral accelerations, $a_z$ and $a_x$ , respectively. Notice from the figure below that the fluid level tilts under the action of accelerations. Find an expression for the slope, $\alpha$ , of the free surface of the fuel in terms of the accelerations.

The governing hydrostatic equation for the pressure in the fluid subject to a uniform external acceleration $\vec{a} = a_x \vec{i} + a_y \vec{j} + a_z \vec{k}$ is

$\nabla p = \varrho \vec{a} - \varrho g \vec{k}$

So for a uniformly accelerating fluid in static equilibrium in the $x$ – $z$ plane then

$\begin{eqnarray*} \frac{\partial p}{\partial x} & = & \varrho (-a_x) \nonumber \\[6pt] \frac{\partial p}{\partial y} & = & \varrho (-a_y) \nonumber \\[6pt] \frac{\partial p}{\partial z} & = & \varrho ((-a_z) - g) \nonumber \end{eqnarray*}$

the minus sign on the acceleration terms appearing because these are inertial terms, i.e., in the opposite direction to the accelerations. The relevant subset of equations in this particular two-dimensional case are

$\begin{eqnarray*} \frac{\partial p}{\partial x} & = & -\varrho a_x \nonumber \\[6pt] \frac{\partial p}{\partial z} & = & -\varrho (a_z + g) \nonumber \end{eqnarray*}$

Integrating these equations and solving for the pressure $p$ leads to

$p = -\varrho a_x x - \varrho (a_z + g ) z + \mbox{constant}$

The shape of the free surface is obtained by letting $p$ = constant. Therefore, the shape of the free surface becomes a linear function of $x$ , i.e.,

$a_x x + (a_z + g) z = \mbox{constant}$

and the angle that the free surface makes with the horizontal will be

$\alpha = \tan^{-1} \left( \frac{-a_x}{a_z + g} \right)$

Notice from the foregoing equation that the fluid level will be unaffected by the vertical acceleration, $a_z$ , and only the horizontal acceleration, $a_x$ , affects the level.

Baffling Balloon Behavior!

A helium-filled party balloon tied to a string attached to the seat is floating inside a car. In which direction will the balloon move if the car accelerates forward? To see a demonstration use this link: Baffling Balloon Behavior. Can you explain the physics of this behavior using the principles of hydrostatics?

The answer is obtained by using the Archimedes’ principle. The net up-force from buoyancy, $L$ , on the balloon will be equal to the net weight of the air displaced, less the weight of the helium inside the balloon, as shown in the figure below, i.e.,

$L = \varrho_{\rm air} g {\cal{V}} - \varrho_{\rm He} g {\cal{V}} = \left( \varrho_{\rm air} - \varrho_{\rm He} \right) g {\cal{V}}$

where ${\cal{V}}$ is the volume of the balloon, $\varrho_{\rm air}$ is the density of air, and $\varrho_{\rm He}$ is the density of the helium gas inside the balloon. We are told to neglect the weight of the balloon itself. Therefore, the tension, $T$ , on the string is

$T = L = \left( \varrho_{\rm air} - \varrho_{\rm He} \right) g {\cal{V}}$

If the car accelerates forward with a constant acceleration, $a$ , in the $x$ direction, then the balance of forces on the balloon will change, as shown in the figure. Because the air mass inside the car is stagnant, the acceleration to the left (positive $x$ direction) manifests as an inertial body force on the stagnant air directed to the right (negative $x$ direction). This acceleration will create a horizontal pressure gradient (the higher pressure biased to the rear of the car) that will lead to a net horizontal buoyancy force directed toward the left, i.e.,

$F_x = \left( \varrho_{\rm air} - \varrho_{\rm He} \right) a {\cal{V}}$

In other words, because of the lighter helium gas in the balloon, the horizontal buoyancy force is greater than its inertial force. Therefore, it can be concluded the helium balloon will lean forward relative to the car when it is accelerating.

To find the angle of the string, then for horizontal and vertical force equilibrium

$T \cos \theta = F_x$

and

$T \sin \theta = L$

Therefore,

$\tan \theta = \frac{F_x}{L} = \frac{ \left( \varrho_{\rm air} - \varrho_{\rm He} \right) a {\cal{V}}} {\left( \varrho_{\rm air} - \varrho_{\rm He} \right) g {\cal{V}}} = \frac{a}{g}$

So, finally, the angle of the string is

$\theta = \tan^{-1} \left( \frac{a}{g} \right)$

Digital Manometers & Pressure Transducers

Liquid manometers are useful applications, have limitations in that they are large and bulky and are more suited to a laboratory than routine use in the field. They also come is large banks suitable for use in the wind tunnel, as shown in the photograph below. One side of the manometer bank is referenced to static pressure and the each individual manometer is connected to a separate pressure tap, say on a wing.

A liquid-filled manometer bank for use in the wind tunnel, which solid-state pressure transducers have essentially now replaced.

Such types of manometers, however, cannot be interfaced with a computer and they take a very long time to read the scales. However, digital pressure manometers are available in convenient, portable, battery-powered sizes for ease of use, with single or multiple inputs and outputs for controlling measurements and transferring data. In addition, calibrations or correction factors can be incorporated into the software used to control the digital manometer.

A pressure sensor or pressure transducer is a solid-state device used for pressure measurement. A pressure sensor generates a voltage or other analog signal that is a function of the pressure imposed upon it. The sensor must be calibrated by applying known pressures and recording the output voltage, the resulting relationship being known as the calibration function. In most cases, this relationship is linear, so the calibration is represented by a single calibration constant. The sensor’s signal may also be conditioned by amplifying and/or filtering, followed by converting the signal to digital form.

Today, pressure sensors are used in thousands of industrial applications and come in all sorts of sizes, shapes, and pressure ranges. For example, miniature pressure transducers are often used in aerospace applications because of their low weight and ability to be placed closely together, say over the surface of a wing. In wind tunnel applications, pressure transducer modules, such as the one shown below, are often used. Pressure tubes can be connected to each of the ports, and the module has self-contained electronics that allow digital values of the pressures to be read directly by a computer and stored for analysis.

A miniature solid-state pressure transducer block, in this case one produced by the Scanivalve company and used at the ERAU wind tunnel. The module allows 64 pressure values to be measured digitally at high-speed and with excellent accuracy.

Summary & Closure

The principles of hydrostatics apply to stationary or stagnant fluids, i.e., those with no relative motion between the fluid elements, so these fluid problems are easier to understand and predict. The types of problems that can be analyzed using hydrostatics include: buoyancy, finding pressures on submerged objects or inside fluid-filled containers, or any other type of fluid problem where the fluid is stagnant. Besides pressure, other fluid properties of interest are temperature and density. The essential equations to remember are those for the hydrostatic pressure field and the hydrostatic equation. The latter is a particular case of the former when gravity is the only acting body force. The equation of state is also helpful in analyzing hydrostatic problems, which can be used to relate pressures, densities, and temperatures. Another use of the hydrostatic equation is to help analyze the properties of the Earth’s atmosphere.

5-Question Self-Assessment Quickquiz

For Further Thought or Discussion

A balloon filled with helium is released at sea level. What happens to the balloon?
A container filled with water is rotated. What happens to the shape of the water surface?
In a density stratified container of water the density varies with depth according to $\varrho (z)$ = 1000 + 1.1 $z$ for 0 < $z$ < 100 m, where $z$ = 0 is the surface. Show how to determine the pressure field.
Sometimes, in making pressure measurements, an inclined manometer is used. Explain why.
Explain the principle of a hydraulic lift for obtaining a large force from a small force.
A cruise ship weighs 100,000 tons. How much volume of seawater does it displace?
Precision liquid manometers are still used today, which may appear inconvenient if not old-fashioned. Explain why.

Additional Online Resources

To improve your understanding of hydrostatics, take some time to navigate to some of these online resources:

Wikipedia has a good site on hydrostatics.
Video on the concept hydrostatic pressure from the University of Colorado.
Video on visualizing the idea of hydrostatic pressure.
For an interactive demonstration, click launch on How do you control a hot air balloon?
For more on Archimedes and his principle, check out Eureka! The Archimedes Principle.

License

Icon for the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License

Introduction to Aerospace Flight Vehicles by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

Digital Object Identifier (DOI)

https://doi.org/https://doi.org/10.15394/eaglepub.2022.1066.9