Fluid statics
Contents
2. Fluid statics#
In the previous section, we have considered many of the basics principles and concepts within fluid dynamics. In general, the motion of a fluid can be unordered and complex, which makes it difficult to analyse. However, many fluid problems do not involve motion of the fluid itself, but instead focus on the pressure distribution in a static fluid. Such problems include the forces that fluids impose on submerged surfaces, such as a sluice gate or a dam, as well as floating bodies and buoyancy forces.
In this section we will therefore investigate such problems by deriving and applying the hydrostatic condition: the pressure variation in the fluid owing only to its weight. We will then apply this to several problems in the observation and measurement of pressure within the atmosphere, instrumentation (such as manometry), submerged surfaces and buoyancy.
2.1. Pressure and pressure gradient#
In the previous section, we contemplated the pressure force, which we said imposed equal force on the fluid volume at each face of our cube. In order toe quantify that, let’s now extend this to fluid in the shape of a wedge: we consider a wedge because we are also interested to know whether there is any reliance on the horizontal orientation of a fluid volume.
Consider the small wedge of fluid of size \(\Delta x\) by \(\Delta y\) by \(\Delta s\) shown below. We visualise this in the \(x\)-\(z\) cross-section, since \(z\) will generally be considered to be the coordinate corresponding to depth. The \(y\) coordinate therefore extends into the page, and we assume that the volume has a width \(b\) into the page.
Fig. 2.1 A fluid volume shaped as a wedge, undergoing pressure forces. Note the \(y\)-direction extends into the page.#
The weight of the fluid volume is readily calculated as \(W = \rho g \cdot(\frac{1}{2}b\Delta x\Delta z)\), where the latter component arises from the formula for the area of a triangle. Then, since the element is ‘small’ (meaning that we can use our continuum assumption), we presume that the pressure on each face is constant. Moreover, since the fluid is at rest, there are no shear forces. Since the summation of forces must equal zero, we can observe the force balance in each of the \(x\) and \(z\) directions:
However, by trigonometry we know that
which gives
This illustrates two important concepts of hydrostatic flows:
There is no pressure change in the horizontal direction.
There is a vertical pressure change in pressure, which is proportional to the density, gravity and, most significantly, the depth change.
In the limit, as our fluid wedge now shrinks to a point, we have that \(\Delta z\to 0\). Reading the equations above clearly indicates therefore that
Validating our assumption that the pressure \(p\) is a point property that is independent of orientation.
2.2. The hydrostatic equation#
Now let us consider the balance of forces owing to pressure on a small cylinder of fluid volume lying at rest, as shown in the following figure.
Fig. 2.2 A cylindrical fluid volume of length \(\delta z\) and the forces acting upon it when the fluid is at rest.#
Assuming the top and bottom surfaces have an area of \(\delta A\), and a pressure \(p(z)\) at the bottom surface and \(p(z+\delta z)\) at the top surface, we can now work to compute the forces that act on this volume whilst the fluid is at rest:
Weight: This is perhaps the easiest to calculate: we know that the cylinder’s volume is given by \(\delta A\delta z\), so its weight must therefore be \(W = \rho g \delta A \delta z\).
Pressure at the bottom surface: Our standard equation for pressure applies here. We know that pressure is force per unit area, so therefore \(F_{\text{bottom}} = p(z) \delta A\).
Pressure at the top surface: This one is a little more tricky. Since we know that our volume height \(\delta z\) is small, we can approximate the pressure at the top from the pressure at the bottom using the Taylor series representation.
Note
Recall that for a function \(f\), we have that the Taylor series around a point \(x + \delta x\) is given as
This simple linear approximation is sufficent for our purposes, but of course this can be written as more terms in higher-order derivatives.
In our case then, we have that
and so we may compute that
Now, as in our previous investigation, we can perform the standard force balance in the \(z\) direction, taking care to apply signs correctly so that each force is applied in a consistent way to our chosen coordinate direction. This gives the balance equation:
Cancelling the \(\delta A\) term, we see that this equation simplifies then to
This is an ordinary differential equation which we call the hydrostatic equation.
If we are interested to calculate the pressure \(p_1 = p(z_1)\) and \(p_2 = p(z_2)\) between two points \(z_1\) and \(z_2\), then we can integrate this equation to give us the integral relationship
but to proceed further we need to know more about the density function \(\rho\).
Warning
From the equation above it is clear that in general the hydrostatic equation relies on the density \(\rho\). For a general fluid, this may well vary as a function of \(z\).
2.2.1. Incompressible fluids#
We will very frequently be considering liquids such as water, which are essentially incompressible. For our purposes then, we may assume that \(\rho(z)\) is constant, in which case we obtain the relationship
Therefore, if we know the pressure at one point and are given a height difference to the second point, we can calculate its pressure with the relationship above. In particular, since \(z\) points upwards, this means that as \(z\) decreases the pressure increases. This validates our intuition: for example if we are diving in a pool, we notice the increased pressure from the water on our ears as we swim deeper.
Example: Diving underwater
Let us consider a diver who descends to a depth of \(20\,\mathrm{m}\) below sea level in water of density \(\rho = 1025\,\mathrm{kg}\,\mathrm{m^{-3}}\). Relative to the surface, what pressure does the diver experience?
This calculation is straightforward from our formula above. Let’s consider that \(z=0\) defines the sea level, and we therefore set \(z_2 = 0\); similarly, \(z_1 = -20\,\mathrm{m}\) denotes the position of the diver. We select them this way around, since \(z\) is increasing and therefore \(z_1 < z_2\). We are interested to calculate the pressure difference between the top and the bottom surface, i.e. \(p_1 - p_2\). From the hydrostatic equation we have:
Therefore the diver experiences a relative pressure \(p_1 - p_2 = -(p_2 - p_1) = 201\,\mathrm{kPa}\).
Since we frequently work with problems such as the above, where we are calculating pressure difference with respect to some free surface, the direction with respect to \(z\) can often be a little confusing. To make things a bit simpler, we can instead define the height \(h = z_0-z\) from the free surface at location \(z_0\). This then gives the simpler form:
In the above equation you can validate that this does indeed give the same answer as using the original hydrostatic equation. Typically, we would define the free surface at \(z_0 = 0\); this then gives the simpler expression
2.2.2. Gauge, atmospheric and absolute pressures#
In the above, we have inadvertently introduced a concept that is important in the study of hydrostatics and fluid dynamics generally. In particular, since we tend to perform our experiments on Earth vs. other celestial bodies (!) most experiments, including the problem of the diver above, are subject also the atmospheric pressure \(p_{\text{atm}}\). This is the pressure that is imposed on us by the weight of air above us. For most problems, we assume that atmospheric pressure is taken at roughly mean sea-level pressure, equivalent to \(101325\,\mathrm{Pa}\).
The diver above therefore experienced a pressure not of that owing to the water above him, but the air above this also. The total pressure \(p_{\text{total}}\) experienced by the diver would be given as instead:
where \(p_{\text{total}}\) is referred to as the absolute pressure. The relative pressure difference \(p = p - p_{\text{atm}}\), which is what we have been looking atbove, is consequently known as the gauge pressure: so-called because almost all pressure gauges will read a value of \(0\,\mathrm{Pa}\) when open to the atmosphere.
2.2.3. Pressure head#
Another important concept that is very commonly found in hydrostatics fluids problems is the pressure head, which is given by a rearrangement of the formula above for \(h\):
The concept of gauge pressure is a basic reformulation of the pressure to that of the depth, or ‘head’ instead: that is, it is the height of a column of liquid corresponding to a particular pressure that is exerted by the liquid on the base of its container.
Warning
In referring to pressure as ‘head’ of a given fluid, it is critical that the density of the fluid is stated. For example, a pressure of \(100\,\mathrm{kPa}\) corresponds to a head of \(h = 10.1\,\mathrm{m}\) of water of mass density \(\rho=1000\,\mathrm{kg}\mathrm{m}^{-3}\). However if instead we consider the same pressure, but under a volume of mercury at 20°C with density \(\rho = 13.6\,\mathrm{g}\,\mathrm{cm}^{-3}\) at 20°C, the head would now be \(h = 10^5 / (13600 \cdot 9.81) = 0.75\mathrm{m}\).
2.2.4. Specific weight and specific gravity#
In doing so, we might also formulate the pressure head in terms of the specific weight of the fluid, denoted as \(\gamma = \rho g\), which is the weight per unit volume of a material. This gives a slightly simpler form of the pressure head as
Warning
It is very easy to confuse specific weight with specific gravity. These two are not the same. Specific gravity (denoted as \(\mathrm{SG}\)) is a measure of the relative density of a material to a chosen reference material, i.e. \(\text{SG} = \rho_{\text{material}} / \rho_{\text{reference}}\). The selection of the reference material depends on the type of fluid under consideration:
for liquids, we select a reference material of water of density \(\rho=1000\,\mathrm{kg}\,\mathrm{m}^{-3}\);
for gases, we select a reference material of air at 20°C and sea level pressure, with density \(\rho = 1.205\,\mathrm{kg}\,\mathrm{m}^{-3}\).
2.3. Manometry#
One of the primary applications of fluid statics and the hydrostatic formula is in measuring pressure differences using a piece of equipment known as a manometer. A simple manometer might be visualised as in the figure below:
Fig. 2.3 A simple U-tube manometer.#
The underlying principle of a manometer is to use two fluids of different densities which are immiscible: that is, they do not mix. Immiscible fluids are those such as oil and water, or air and water. In this figure, the water is at the bottom, represented by the blue coloured fluid. Air is not explicitly shown.
The manometer works by measuring the height difference between the two fluids in each end of the tubes. To visualise this, imagine you are blowing into the end of one end of the U-tube: doing so will impart a force on the air, which in turn imparts a force on the water. Intuitively you will probably realise that the level of water in the other end of the U-tube will increase. By using the hydrostatic equation, we can equate this to a pressure difference between the two ends of the tube.
Example: calculating pressure from manometer readings
First, let’s choose a reference level so that below this level, only the same fluid is found in the manometer. Clearly at point A, we will find such a level. Point A is at a depth \(h = l\) from point B. Then, from our equation above, the pressure at \(A\) is calculated as
where \(\rho_a\) is the density of the fluid above point A (in this case, air). Now let’s look to the other side of the U-tube at point C. We can compute the pressure here as
where \(\rho_b\) is the density of the blue fluid (in this case, water). Furthermore, we know that pressure does not vary in the horizontal direction: therefore, the pressure at point C is equal to that at point A, and thus
Therefore the difference in pressure can be expressed as
At this point, we may be told that, for example \(p_1\) is at atmospheric pressure: this would then allow us to find \(p_2\) concretely. There will be some examples of this in the tutorial sessions.
Warning
In the derivation above, we have used a critical point: Any two points at the same elevation in a continuous mass of the same static fluid will be at the same pressure.
2.4. Forces on submerged bodies#
Submerged bodies are commonly found within engineering problems. For example, a containment structure such as a dam will require the computation of hydrostatic forces on the surfaces that hold the fluid back, in order to e.g. determine how thick its walls should be. In this section we will apply our hydrostatic equation to calculate these forces when such bodies are found in static equilibrium.
This becomes more challenging than e.g. point mass systems, because we know that the pressure that a submerged body will experience is a function of the density of the fluid as well as its depth. Additionally, the wall may be flat and/or lie at some angle to the surface, or indeed in general it may be curved, as shown below:
Fig. 2.4 An example of pressure force distribution on surfaces that are flat vs. angled.#
However, we can use many of the concepts from our mechanics knowledge to date to bring to bear on this problem. If we can compute the resultant force of the pressure against our surface: including its magnitude, point of application and direction, then this becomes a simpler problem of balancing forces and moments, as we have already seen in 4CCE1MEC. Firstly we investigate the resultant force: then we discuss moments.
2.4.1. Resultant force on a flat plate#
First, let us consider a submerged flat plate at an angle \(\theta\), as shown in the figure below. We will assume that the fluid the plate is submerged in is of uniform density.
Fig. 2.5 A submerged body of an arbitrary shape, oriented at fixed angle \(\theta\).#
The figure shows both the side view, in which we can observe the profile of the plate, and the plan view which shows its shape. In this figure we assume the depth is given by \(h\).
Warning
The \(x\) and \(y\) coordinates are not located on the side view, but rather on the plan view in the same plane as the plate. They are chosen so that their origin lies on the centroid of the shape.
Firstly, we aim to replace the distributed pressure force with a single resultant force which is defined by \(F\) in the figure above. First, let us consider the magnitude of \(F\). To do this, we know that on a small area \(\delta A\) of the surface, the force owing to pressure is given by
The total pressure force across the whole surface can be found by summing all contributions, and then adopting the limit as \(\delta A\to 0\): in other words, by calculating the surface integral of the pressure force across \(A\) as
You will have encountered such integrals in the 5CCE2ENM Engineering Mathematics module. This force describes the magnitude of the force acting on the area: for a flat plate, the force will clearly be perpendicular to the plate, and the position in which will act is called the centre of pressure, which in the figure is denoted by the point \(CP\). Our task is therefore to determine where this centre of pressure lies.
The first step is to understand a bit more about the nature of \(F\). We know that the pressure \(p\) can be expressed as
If we define a coordinate \(\xi = h/\sin\theta\), which descends from the surface in the line of the plate as shown in the figure, then we have that \(h = \xi \sin\theta\) and so
where we note that, since \(\theta\), \(\rho\) and \(g\) are all constant, we can remove them from the integral. By definition, the integral on the right hand side is very close to the definition of the plate, which we denote by the position \(CG\) in the figure. In the \(\xi\) coordinate system, we would define this as
Substituting this into the formula above we therefore obtain the expression
Now, from our figure we can see that \(\xi_{CG}\sin\theta = h_{CG}\), where \(h_{CG}\) is the depth of the plate at the centre of gravity. Thus our formula now becomes:
where \(p_{CG} = \rho g h_{CG}\) is the pressure on the plate as the plate’s centroid. In other words, the force on one side of the plate is the pressure at the plate’s centroid times the plate area, and this is independent of the shape of the plate or the angle \(\theta\) we’re considering. This therefore immediately:
validates the concept that we can now start to think of this problem as a simple force-moment balance problem;
gives us an easy way to compute the magnitude of the resultant force, given the location of the centroid of the shape.
At this stage it is important to reiterate the different locations under consideration:
the resultant force is going to act at the centre of pressure \(CP\);
this will be related to the centroid of the plate, which we define as the point \(CG\).
In particular note that the centroid is only related to the shape of the plate and has no particular dependency on the fluid dynamics at hand: doubtless you will have computed the centroid of a shape many times in previous courses and modules.
2.4.2. Moment balance on a flat plate#
To compute the centre of pressure for the submerged plate, we will compute moments about the centroid of the plate. Let us consider that in the coordinates of the plate, the centre of pressure can be found at the position \((x_{CP}, y_{CP})\). Resolving moments first in the \(y\)-direction, we see that
Then if we let \(\xi = \xi_{\text{CG}} - y\), we have that
where \(I_{xx}\) is the area moment of inertia of the plate area around its centroidal \(x\) axis. Note that since the coordinate system is defined with the origin at the centroid of the plate, we have that \(\int_A y dA = 0\) by symmetry. Thus, we can finally rearrange for \(y_{CP}\) to obtain
There are a few points to unpick here:
The negative sign in the equation above indicates that \(y_{CP}\) will lie somewhere below the centroid, owing to the pressure forces being located at a deeper level than the centre of gravity.
Moreover, the location now depends on the angle \(\theta\), as we might expect.
Finally, we note that if we were to increase the depth of the plate, then \(y_{CP}\) will move closer to the centroid, since every term in the equation above remains constant expect for \(p_{CG}\), which increases.
The calculation for \(x_C\) is done in precisely the same manner, and yields
where \(I_{xy}\) is the product of inertia of the plane.
Example: Area and product moments of inertia
As a reminder and quick cheat sheet, here are some of the common mass and product moments of inertia you might find for some simple shapes. Be sure to consult your notes for 4CCE1MEC for a reminder on how to calculate some of these!
Fig. 2.6 Common moments of inertia. Adopted from White, Fluid Mechanics, 2011.#
2.4.3. Pressure on a gate#
To motivate what has been quite a heavy theory section, let’s now try and apply this in practice.
Example: pressure on a gate
Consider the problem in the figure below, where a gate \(AB\) rests against two solid walls. The gate is hinged at the point \(B\), and is in static equilibrium. The gate is \(5\,\mathrm{m}\) wide (into the page). We would like to compute:
the force on the gate due to seawater pressure;
the horizontal force \(P\) exerted by the wall at point \(A\);
the reactions at the hinge \(B\).
Fig. 2.7 A gate AB rests up against a wall at an angle \(\theta\). The gate is \(5\,\mathrm{m}\) wide.#
By geometry the gate is \(10\,\mathrm{m}\) long from \(A\) to \(B\), and its centroid is halfway between at an elevation \(3\,\mathrm{m}\) above point \(B\). The depth \(h_{CG}\) is threfore is thus \(15 - 3 = 12\,\mathrm{m}\). Now, since the gate has a depth of \(5\,\mathrm{m}\), its are area is \(5\times 10 = 50\,\mathrm{m}^2\). Since the atmospheric pressure acts on both the gate and the top of the tank we can neglect this aspect. Then, from our equation above, we can calculate the resultant force magnitude as
Now let’s try to simplify things to a statics-type problem by considering a free body diagram of the gate.
Here we can see the application of all of the forces involved in this problem, where the centre of pressure and centre of gravity are shown explicitly. The distance \(\ell\) we can compute from our formulae; we know that the gate is rectangular, so
and so
To find the horizonal force \(P\), we can now consider moments about \(B\). The distance from \(B\) to the resultant force \(F\) is (5-\ell), and the horizontal component of \(P\) is given by \(P\sin\theta\). Therefore taking moments counterclockwise around \(B\), we have that
Finally, since we now know \(P\) and \(F\), we can calculate reactions by summing the forces on the gate in the horizontal and vertical directions respectively:
2.4.4. Forces on curved surfaces#
Finally, we conclude this section by discussing a route towards calculating the forces imposed on curved submerged surfaces. In principle this is the same process as the above, but integration on a generally-shaped surface is difficult and often impractical to perform by hand.
Instead therefore we consider an alternative method. Many problems involve curved surfaces that are ‘easy’ for us to decompose into simpler shapes. By considering the force balances on each of these shapes, we can work out forces for more complex submerged surfaces. The easiest way to demonstrate this is through an example, so let’s consider one!
Example: forces on a circular surface
Consider the problem in the figure below, where we have a dam filled with water with a quarter-circle shape, which has a width \(w = 3\,\mathrm{m}\) into the page. We would like to calculate the resultant force on the curved surface at the bottom of the dam.
Fig. 2.8 A curved dam holds back a volume of water.#
Let’s first break the problem down into bits we do know how to solve:
looking at the dashed, vertical line which is imaginary, we know from our previous sections that if this were the wall, we’d easily be able to calculate the resultant force and the position of the force.
moreover we can also calculate the weight owing to the water above the circular section (and also maybe the atmosphere on top, if we so desire).
If we therefore break down the problem into small sections indicated by the dashed lines, we can create a free body diagram for the circular cross-section and resolve our forces to work out the induced force on the wall.
Fig. 2.9 View of the fluid block with the curved wall, where forces act on the left-hand and top ‘imaginary’ surfaces.#
In the diagram we have several things going on:
\(F_1\) corresponds to the weight of water (and the atmosphere above the water) that will act on this body of water through the top surface.
\(F_2\) is the force owing from the column of water to the left, which we can calculate from our previous sections.
\(W\) is the weight of the water in the circular cross-section, which acts from the centroid of the shape.
Finally, since we know the resultant force \(\vec{F}_R\) will impose a force on the wall, there is an equal and opposite force of the wall on the water, which we denote by \(\vec{F}_N = -\vec{F}_R\). Since everything is in static equilibrium and there is no acceleration, we can balance the forces in each direction to determine \(\vec{F}_N\) and thus \(\vec{F}_R\).
In the \(x\)-direction, we have from our previous sections that \(F_2\) is given by the force acting on the ‘imaginary’ plate on the left hand side. The centre of gravity for this plate lies directly in the middle of the line, which is submerged at a depth of \(h_{CG} = (5-2) + 1 = 4\,\mathrm{m}\) and its area is \(A = 2w = 6\,\mathrm{m}^2\). Thus,
In the \(y\)-direction, we can compute the forces using the density and volume of each of the regions under consideration:
The top volume above the curved section is \(V = 2\cdot 3 \cdot w = 18\,\mathrm{m}^3\);
The volume of the curved section is given by \(V = \pi R^2/4 \cdot w = 3\pi \,\mathrm{m}^3\)
We therefore have that
Our resultant force therefore has a magnitude of \(\sqrt{F_{N,x}^2 + F_{N,y}^2} = 357.5\,\mathrm{kN}\).
The other question that we might have now is: how do we compute the location of where the force is applied? For this, we can take moments about the bottom left-hand corner of the quarter-circle:
the force on the left-hand surface is applied at the ‘centre of pressure’ for the imaginary surface, \(y_{CP}\).
the force on the top surface is applied in the middle of imaginary line.
the weight of the fluid acts at the centroid of the quarter-circle.
The calculation itself is left as an exercise to the reader!
2.5. Buoyancy#
Finally, we conclude this section by a brief discussion of buoyancy. We are probably intuitively very familiar with this concept; bodies that are immersed in a fluid experience an upwards force as dictated by Archimedes’ principle: the magnitude of the force is equal to the weight of the displaced fluid. This force is the resultant force of the pressure forces surrounding the body. We can use our knowledge of hydrostatics to validate Archimedes’ observation!
Consider a cylindrical object that is fully submerged in a fluid of density \(\rho\)
Fig. 2.10 Calculating the buoyancy force for a simple cylindrical object.#
In this example, the horizontal forces that act on the side walls of the cylinder due to pressure all cancel: any any given cross-section, there is an equal and opposite pressure on the opposing side of the cylinder. However, we know that pressure forces will vary with depth. If we consider the forces acting only on the top (\(F_1\)) and bottom (\(F_2\)) of the cylinder, we can readily calculate that
The resultant force in the upward direction is therefore
This is the buoyancy force, which we can plainly see is the same as Archimedes’ principle: the force is given by the weight of the displaced fluid.
Calculation of buoyancy forces is highly important when we consider e.g. the stability of a ship. The buoyancy force will act on the centre of gravity of the displaced fluid; however this does not necessarily coincide with the centre of gravity of the immersed body. Depending on where these two forces are applied, a moment can occur that causes the ship to overturn and capsize.
Fig. 2.11 Buoyancy and its impact on stability. \(B\) denotes the application point of the buoyancy force, and \(G\) the centre of mass of the object. The body is tilted by a small angle \(\Delta\theta\) and the so-called metacentre \(M\) is calculated from the intersection of the line of symmetry and the horizontal location of the new buoyancy position \(B'\). If \(M\) lies below \(G\), as in (c), this causes instability. Adopted from White, Fluid Mechanics, 2011.#
We will not consider this precise effect in this course, but you can refer to the textbook by White for further details.
2.6. Summary & further reading#
In this section we have outlined the principles behind hydrostatic forces and fluid statics problems. In particular we have seen that:
pressure varies with depth, but not with horizontal position;
pressure is subject to the hydrostatic equation;
this principle can be applied to manometers in order to measure pressure;
forces on submerged flat plates can be derived, using the depth of submersion and moments of inertia;
forces on submerged curved plates can also be derived using force balances;
submerged objects are subject to buoyancy forces, according to Archimedes’ principle.
In the next section, we will move on to discuss fluid dynamics: the calculation of properties of fluids that are in motion.
Further reading
For further reading, consult chapter 2 of White (2011). In particular:
sections 2.1-2.3 for discussion of the hydrostatic equation;
section 2.4 for the application to manometry;
section 2.5-2.6 for forces on plane and curved surfaces; and
section 2.8 on buoyancy.