3. Fluid dynamics

In the previous section, we started to consider basic fluid static problems, where the fluid is at rest and we can consider the hydrostatic forces that are imposed on objects submerged within the fluid. However it is clear that in the vast majority of problems, the fluid will be in motion, and thus we need to contemplate how forces affect fluid motion.

To do this, we will need to formulate the concepts we know about the conservation of basic physical properties:

Note

We should also consider conservation of angular momentum; however in this course we will not consider this aspect.

• conservation of mass;

• conservation of momentum;

• conservation of energy.

However, we must instead look at these concepts in the context of fluid mechanics. This can be done in one of two ways:

• We can consider ‘small-scale’ analysis, where a small element of the body is examined. This leads to an analytical approach that, whilst giving us knowledge of all of the flow’s properties, including the velocity and pressure at all points in the body. However, it leads to differential equations that are challenging to solve in general.

• We can alternatively consider a ‘larger-scale’ analysis, wherein we define a body of a known size within the fluid. By considering how fluid flows in and out of these so-called control volumes, and applying basic laws of conservation, we can observe some of the properties of the flow: for example, the force imposed by the fluid, or the flowrate of fluid in and out of the control volume. However, we will not be able to recover all knowledge of the flow, such as the velocity at any given point within the control volume.

In this section, we will start by considering the latter option; in the coming sections we will outline the differential approach and examine the resulting equations. Let us first begin by contemplating the definition of a control volume, and then see how we can apply conservation properties to understand something about fluid dynamics problems.

3.1. The control volume approach

All the laws of mechanics are written for a system. This is an arbitrary quantity of mass within some body, where everything external to the system are its surroundings and the separation between the system and surroundings are its boundary. Conceptually, once we have chosen a system, then its mass must remain constant.

Fig. 3.1 A system of mass \(m\) with its boundary.

When we are considering a problem involving particles or rigid bodies, this approach is standard because the definition of a system is usually fairly straightforward: we encounter bodies that have boundaries which are well-defined.

However, in a fluid, any given body will continuously deform over time. Therefore, if we select a system approach in which the mass in the system remains constant, we have to observe how its boundary evolves in time. In the majority of fluid problems, this will be exceptionally difficult as the flow can evolve in strange and potentially unpredictable ways, as shown in the figure below.

Fig. 3.2 A fluid system of mass \(m\) evolves over time. In the system approach its mass remains fixed and thus we need to track its boundary.

The easy way to get around this is to not choose a body of fixed mass, but instead a body of fixed boundary. This is the control volume (or CV) approach. Unlike a system, fluid can flow in and out of a CV and we can examine the conservation of properties as it does so.

With that said, we need to start with the system approach, as this is the one in which we understand that e.g. mass, momentum and energy are all conserved. In working with control volumes, we therefore must work out how to translate between the CV and system approaches. This is called the Reynolds transport theorem, named after one of the most influential historical figures in fluid dynamics, Osborne Reynolds (1842-1912).

In the following sections we will proceed in steps to work this out:

• defining the conservation equations for a system;

• examining mass and volume flowrate in and out of CVs

• defining the Reynolds transport equation.

3.2. Conservation equations for a system

Let us start by considering the different quantities that are preserved in a system, and how they are all represented mathematically.

3.2.1. Conservation of mass

This is simply stated that the mass of a system \(m_{\mathrm{syst}}\) must remain constant, i.e. that

\[ m_{\mathrm{syst}} = \mathrm{const} \quad\mathrm{or}\quad \frac{dm_{\mathrm{syst}}}{dt} = 0. \]

Of course for most solid mechanics problems, this is such an obvious statement that we often forget to even mention it. However in the context of fluids, this is a particularly important statement.

3.2.2. Conservation of momentum

If the surroundings exert a total force \(\vec{F}\) on the system, then Newton’s second law implies that the mass will begin to accelerate, i.e. that

\[ \vec{F} = m\vec{a} = m\frac{d\vec{v}}{dt} = \frac{d}{dt}(m\vec{v}) \]

so that if the total momentum of the system \(\vec{M}_{\mathrm{syst}} = m\vec{v}\) then we have that

\[ \frac{d\vec{M}_{\mathrm{syst}}}{dt} = \vec{F}. \]

It is worth emphasising that this is a vector equation, so we will often consider the components of the equation individually.

3.2.3. Conservation of energy

Finally, we may consider that, particularly for problems such as we will cover in the second half of this course, heat \(\delta Q\) may be added to the system, or work \(\delta W\) may be done by the system over a time period \(\delta t\). Then system energy \(dE\) must therefore change according to the equation

\[ \delta Q - \delta W = dE \Rightarrow \frac{dE_{\mathrm{syst}}}{dt} = \dot{Q} - \dot{W} \]

where \(\dot{Q}\) and \(\dot{W}\) are the rate of heat transfer and work done, respectively. \(E_{\mathrm{syst}}\) is the total energy of the system; i.e. its sum of internal energy, kinetic energy and potential energy. This is the first law of thermodynamics, which we will revisit in the second half of the course.

All of these are straightforward to formulate for a solid body, but difficult for a fluid since we do not know the evolution of the boundary in a meaningful manner. To overcome this, we will therefore need to understand how fluid flows in and out of a control volume, which we cover in the next section.

3.3. Mass and volume flowrate

3.3.1. Uniform flow through a surface

Consider the flow of a fluid of density \(\rho\) through a surface of area \(A\). Let us suppose that the water is travelling with a uniform velocity \(u\) and is completely normal to the surface.

Across a short time \(\delta t\), we would observe a small cylinder of fluid that is ‘extruded’ through the surface of \(A\), of length \(\delta x\). The volume of the cylinder is therefore \(A\delta x\), and from this we can compute its mass \(\delta m = \rho\delta V = A\rho\delta x\). This looks something like the figure below.

Fig. 3.3 Fluid flows through a tube of area \(A\) at uniform velocity \(u\) across a time \(\delta t\).

The mass flowrate \(\dot{m}\) through \(A\) is therefore given by

\[ \dot{m} = \frac{dm}{dt} = \lim_{\delta t\to 0} \frac{\rho A \delta x}{\delta t} = \rho A \frac{dx}{dt} = \rho A u. \]

More succinctly this can be stated as

\[ \boxed {\dot{m} = \rho u A} \]

The flowrate per unit area is called the mass flux, which is defined as

\[ \dot{f} = \rho u \]

Alternatively, we might consider the volume of fluid flowing through \(A\). In this case, a very similar analysis (this time neglecting density \(\rho\) will yield the volumetric flow rate \(Q\)

\[ \boxed{Q = uA \Rightarrow \dot{m} = \rho Q} \]

3.3.2. Flow through an arbitrary surface

The above is only valid for flow through a flat surface, where the velocity is entirely normal to the surface. In general, we might be considering any surface \(S\), where fluid could flow at any angle to the surface.

In this setting however we can consider much the same process. Take a small area \(\delta A\) on which we can consider the velocity \(\vec{v}\) constant, and assume that \(\vec{v}\) passes through this area at an angle \(\theta\) from the surface normal \(\vec{n}\), as shown in the figure below.

Fig. 3.4 Fluid flows through a curved surface at a non-normal angle. We break the surface up into small surface areas, as shown on the right.

We then have that the volume extruded through \(\delta A\) is given by

\[ \delta V = \| \vec{u}\| \delta t \delta A\cos\theta = (\vec{u}\cdot\vec{n}) \delta A\delta t \]

Rearranging this a bit gives that

\[ \frac{\delta V}{\delta t} = \vec{u}\cdot\vec{n} \delta A \]

Taking limits as all of these quantities go to zero and noting that the integral of \(dV/dt\) is simply the total volumetric flow rate \(Q\) through \(S\), we have therefore that

\[ \boxed{Q = \int_{S} \vec{u}\cdot\vec{n}\, dA} \]

Alternatively we can compute the mass flow rate by multiplying by the density in this derivation so that

\[ \dot{m} = \int_S \rho(\vec{u}\cdot\vec{n})\, dA \]

Example: validating our previous result

Let’s consider the ‘simple’ example from the section above, where we have a fluid of uniform density flowing at a constant velocity \(u\) at a normal direction to a surface \(S\) of area \(A\).

Since the flow is normal to the surface, we have that \(\vec{u}\cdot\vec{n} = u\). The mass flow rate can then be computed as

\[ \dot{m} = \int_S \rho (\vec{u}\cdot{n})\,dA = \int_S \rho u\, dA = \rho u\int_S dA \]

where the last equality follows from the fact that \(u\) is constant, so it can be removed from the constant. Since the last integral is simply the surface area \(A\), we have that \(\dot{m} = \rho u A\) as before.

3.4. The Reynolds transport theorem

We have talked at some length about the need to reformulate basic laws in terms of control volumes. Mathematically however, what does the system form look like as our starting point? Let’s consider the conservation of mass. Imagine we choose a system with volume \(V(0)\) at time \(t=0\). Then at time \(t\), we would expect this to deform into a volume \(V(t)\). At any time \(t\) we can write the mass of the volume as the volume integral

\[ m_{\mathrm{syst}} = \int_{V(t)} \rho\, dV. \]

The conservation of mass can therefore be stated as

\[ \frac{dm_{\mathrm{syst}}}{dt} = 0 \Rightarrow \frac{d}{dt}\int_{V(t)} \rho\, dV = 0. \]

What we would like get to is a form similar to this for a control volume CV, i.e.

\[ \frac{d}{dt}\int_{CV}\rho \, dV = \cdots \]

The Reynolds transport theorem is what will allow us to get there!

3.4.1. Analysis of a control volume

Let’s consider a control volume as shown in the figure below. We assume that at the start point at time \(t\), the control volume and the system lie at the same position. Then, the flow will deform the system volume across a time \(\delta t\), whilst our CV remains fixed in space.

Fig. 3.5 Fluid deforms the system from the circle on the left hand side to the shape on the right hand side. Dashed lines represent the system, solid lines represent the control volume. Arrows on the right denote the velocity field direction that deforms the system.

Furthermore, since we are not just interested in mass, but also other quantities such as momentum, let us generalise a bit at this point. Let \(B\) represent any property of the fluid within a system or CV, and let \(\beta\) represent the same quantity but as an intensive value, i.e. the amount of \(B\) per unit mass. \(B\) would then be known as the corresponding extensive property. We have in general then that

\[ B = \int_{V(t)} \beta\rho \, dV \]

For example, in the case of mass, \(B = m\) and \(\beta = 1\), which recovers the volume integral that we observed above. Visually, we can see in the figure that the velocity field causes the following three changes to the system:

• a change of the amount of the property within the control volume;

• outflow of \(\beta\) through the top and bottom;

• inflow of \(\beta\) through the left and right hand sides.

How do we formulate this mathematically? First, let:

• \(B_{\mathrm{syst},t}\) denote the amount of \(B\) in the system at time \(t\); and

• \(B_{\mathrm{CV},t}\) denote the amount of \(B\) in the CV at time \(t\)

Then, there will be:

• some rate of change of \(B\) into the CV, which we will call \(\dot{B}_{\mathrm{in}}\);

• and a rate of change of \(B\) out of the CV, \(\dot{B}_{\mathrm{out}}\).

Across a small time \(\delta t\) we can therefore calculate that an amount of \(\dot{B}_{\mathrm{in}}\delta t\) flows into the volume, whilst \(\dot{B}_{\mathrm{out}}\delta t\) flows out of the volume. This allows us to establish the relationship that \(B_{\mathrm{syst},t+\delta t} = B_{{\mathrm{CV},t+\delta t}} - \dot{B}_{\mathrm{in}}\delta t + \dot{B}_{\mathrm{out}}\delta t\).

We have, therefore, that the change in \(B\) in the system can be written as

\[\begin{split} \begin{align*} B_{\mathrm{syst},t+\delta t} - B_{\mathrm{syst}, t} &= (B_{{\mathrm{CV},t+\delta t}} - \dot{B}_{\mathrm{in}}\delta t + \dot{B}_{\mathrm{out}}\delta t) - B_{\mathrm{syst}, t} \\ &= (B_{{\mathrm{CV},t+\delta t}} - \dot{B}_{\mathrm{in}}\delta t + \dot{B}_{\mathrm{out}}\delta t) - B_{\mathrm{CV}, t} \end{align*} \end{split}\]

since at time \(t\) the CV and system coincide so that \(B_{\mathrm{syst},t} = B_{\mathrm{CV},t}\). Dividing by \(\delta t\) and rearranging, we have that

\[ \frac{B_{\mathrm{syst},t+\delta t} - B_{\mathrm{syst}, t}}{\delta t} = \frac{B_{{\mathrm{CV},t+\delta t}} - B_{{\mathrm{CV},t}}}{\delta t} - \dot{B}_{\mathrm{in}} + \dot{B}_{\mathrm{out}} \Rightarrow \frac{dB_{\mathrm{syst}}}{dt} = \frac{dB_{\mathrm{CV}}}{dt} - \dot{B}_{\mathrm{in}} + \dot{B}_{\mathrm{out}} \]

when \(\delta t\to 0\).

This is the Reynolds transport theorem: restated in words, it says that the rate of change of a property of the system \(dB_{\mathrm{syst}}/dt\) is related the rate of change of a property within the control volume \(dB_{\mathrm{CV}}/dt\), plus the net flux of the property out of the control volume, \(\dot{B}_{\mathrm{out}} - \dot{B}_{\mathrm{in}}\).

3.4.2. Reformulating in integral form

Although this is a correct form of the theorem, it is not terribly easy to use for our purposes. Instead, we can reformulate the above in terms of surface and volume integrals. Firstly, we know that within the control volume, we may write \(B_{\mathrm{CV}}\) in terms of the volume integral

\[ B_{\mathrm{CV}} = \int_{\mathrm{CV}} \beta\rho \, dV \]

Additionally, the flux of \(B\) in and out of the control volume’s surface, which we refer to as the control surface CS, can be reformulated using our knowledge of the flowrate from previous sections as:

\[ \dot{B}_{\mathrm{out}} - \dot{B}_{\mathrm{in}} = \int_{\mathrm{CS}} \beta \rho (\vec{u}\cdot\vec{n})\, dA \]

We therefore have that

\[ \boxed{ \frac{dB_{\mathrm{syst}}}{dt} = \frac{d}{dt}\left(\int_{\mathrm{CV}} \beta\rho \, dV\right) + \int_{\mathrm{CS}} \beta \rho (\vec{u}\cdot\vec{n})\, dA} \]

This is a far more useful form of the theorem that we can use for investigation of the conservation of our properties.

3.5. Conservation of mass

Let us start by considering the conservation of mass. In this case, using the Reynolds transport theorem and noting that \(dm_{\mathrm{syst}}/dt = 0\), we can select \(B=m\) and \(\beta=1\) to obtain the relationship

\[ 0 = \frac{dm_{\mathrm{syst}}}{dt} = \frac{d}{dt}\left(\int_{\mathrm{CV}} \rho \, dV\right) + \int_{\mathrm{CS}} \rho (\vec{u}\cdot\vec{n})\, dA \]

Moving CVs

Actually it is not required for the CV to remain in the same location at all times: the CV can move with constant velocity and the derivation below remains exactly the same. However if the CV accelerates then it is called non-inertial and thus we need to consider additional forces.

This is referred to as the continuity equation. Since the control volume lies at a fixed location, we can move the derivative into the integral to obtain

\[ \int_{\mathrm{CV}} \frac{d\rho}{dt} \, dV = - \int_{\mathrm{CS}} \rho (\vec{u}\cdot\vec{n})\, dA \]

If we further assume that the fluid is steady, then by definition its density must remain constant in time and this further reduces to

\[ \int_{\mathrm{CS}} \rho (\vec{u}\cdot\vec{n})\, dA = 0. \]

Finally, if the control volume has \(N\) one-dimensional inlets and \(M\) outlets, and we can assume that the velocity at those inlets and outlets is uniform, then we can write

\[ \sum_{i=1}^N (\rho_i A_i V_i)_{\mathrm{in}} = \sum_{j=1}^M (\rho_i A_i V_i)_{\mathrm{out}} \]

where we have obtained this sum by breaking down the integral into separate integrals for each inlet and outlet. \(\rho_i\) is the density of fluid entering/leaving the inlet/outlet, and \(A_i\) is the cross-sectional area of the inlet/outlet.

This is a simple approximation that is widely used in engineering, and essentially says that the mass flow into the control volume must balance the mass flow out of the volume. Furthermore, if the fluid is incompressible so that \(\rho\) is constant, we can simplify this equation even further, since the equation above yields

\[ \int_{\mathrm{CS}} (\vec{u}\cdot\vec{n})\, dA = 0. \]

after density is removed from the integral. This then gives

\[ \sum_{i=1}^N (\rho_i A_i V_i)_{\mathrm{in}} = \sum_{j=1}^M (\rho_i A_i V_i)_{\mathrm{out}} \Rightarrow \sum_{i=1}^N Q_{i,\mathrm{in}} = \sum_{j=1}^M Q_{i,\mathrm{out}} \]

where \(Q\) denotes volumetric flow rates through each inlet/outlet. If you recall from 4CCE1SUS, this should be familiar!

Example: flow through a streamtube

For a steady incompressible flow, consider the control volume given by a streamtube: a tube consisting of a collection of streamlines, as shown in the figure below.

Fig. 3.6 A streamtube consisting of streamlines, which are all perpendicular to the velocity field.

By considering the conservation of mass on this streamtube, let us relate the flow velocity at the inflow and outflow, given the cross-sectional areas \(A_{\mathrm{in}}\) and \(A_{\mathrm{out}}\). In this example, we are simply going to select the streamtube as the control volume, since the flow is steady.

The conservation relation above gives us that

\[ Q_{\mathrm{in}} = Q_{\mathrm{out}} \]

We therefore have that

\[ V_{\mathrm{in}} A_{\mathrm{in}} = V_{\mathrm{out}} A_{\mathrm{out}} \Rightarrow V_{\mathrm{out}} = V_{\mathrm{in}}\cdot\frac{A_{\mathrm{in}}}{A_{\mathrm{out}}} \]

Moreover, the volumetric flow is constant across every section of the tube.

3.6. Conservation of momentum

Recall that we wish to preserve the momentum in the system sense, so that

\[ \frac{d\vec{M}_{\mathrm{syst}}}{dt} = \vec{F} \]

where \(\vec{M}_{\mathrm{syst}}\) is the momentum and \(\vec{F}\) is the net force on the system. We can more conveniently express the left hand side of this equation by the volume integral

\[ \frac{d}{dt} \int_{V(t)} \rho \vec{u}\, dV = \vec{F} \]

To apply the Reynolds transport theorem, then after selecting a control volume we can thus replace the variable \(B\) by the momentum \(m\vec{u}\), and the variable \(\beta = \vec{u}\), so that the conservation of momentum statement reads that

\[ \vec{F} = \frac{d}{dt} \int_{\mathrm{CV}} \vec{u}\rho\, dV + \int_{\mathrm{CS}} \vec{u}\rho (\vec{u}\cdot\vec{n})\, dS \]

It is worth noting that this is a vector equation, which makes the notation above quite compact but perhaps difficult to use. You will probably find it more convenient to express the equation above in terms of its \(x\) and \(y\) components separately, i.e.

\[\begin{split} \begin{align*} F_x &= \frac{d}{dt} \int_{\mathrm{CV}} u\rho\, dV + \int_{\mathrm{CS}} u\rho (\vec{u}\cdot\vec{n})\, dS\\ F_y &= \frac{d}{dt} \int_{\mathrm{CV}} v\rho\, dV + \int_{\mathrm{CS}} v\rho (\vec{u}\cdot\vec{n})\, dS \end{align*} \end{split}\]

where \(F_x\) and \(F_y\) are the components of the vector \(\vec{F} = (F_x, F_y)\).

Still, this form of the equations remain quite unwieldy! Let’s proceed to simplify them for the cases we will typically be interested in here: steady flows. In this case, we know that \(\vec{u}\) and \(\rho\) do not depend on time, so that its derivative with respect to time is zero. Therefore, the first integral of the equations above is zero, giving the simplified equations

\[ \vec{F} = \int_{\mathrm{CS}} \vec{u}\rho (\vec{u}\cdot\vec{n})\, dS \]

Or in other words, the net forces are balanced by the total mass flow through the control surface. If we consider that many engineering problems involve a fixed number of \(N\) inlets and \(M\) outlets, as in the mass conservation example above, then we can further reduce this to

\[ \boxed{\vec{F} = \sum_{i=1}^N (\dot{m}_i V_i)_{\mathrm{out}} - \sum_{j=1}^M (\dot{m}_j V_j)_{\mathrm{in}}} \]

where we have assumed that:

• the velocity \(V_i\) at each outlet is uniform;

• the sign of \(\dot{m}_i\) is always positive: i.e. we have incorporated direction into the mass flow via the subtraction of outlet mass flows.

This is a commonly used approach in the approximation of many engineering problems, and can be used to calculate e.g. the forces imposed by a jet of fluid on a wall. To motivate this, let’s use an example!

Example: force imposed by a streamtube

A fixed control volume of a streamtube in steady flow has a uniform inlet flow \((u_1, A_1, \rho_1)\) and a uniform exit flow \((u_2, A_2, \rho_2)\) as shown in the figure below. Find an expression for the net force on the control volume.

Fig. 3.7 A streamtube.

In this problem we can clearly just apply our change of momentum formula directly, so that the net of the forces \(\sum\vec{F}\) will balance the change in momentum of the fluid as it bends in the streamtube. We have one inlet at section 1, and one outlet at section 2, so

\[ \sum\vec{F} = \dot{m}_2\vec{V}_2 - \dot{m}_1\vec{V}_1 = (\rho_2 A_2 V_2)\vec{V}_2 - (\rho_1 A_1 V_1)\vec{V}_1. \]

Warning

Remember that the net force, such as we calculate above, is that which acts on the control volume in order to balance the rate of change of momentum. In several problems we will be more interested in the force imposed by the fluid on, for example, a wall. This is easy to compute from the force of the wall on the fluid, since it must be equal and opposite to the force of the fluid on the wall, but it is necessary to pay careful attention to the correct orientation of the desired force!

Example: vane deflecting a jet

Consider the the figure below, where a vane at rest deflects a jet of water through an angle \(\theta\) without changing its velocity magnitude. We assume that the friction on the vane is negligible. Let us compute the force \(F\) that the vane imposes on the fluid, as well as the angle \(\phi\) between the horizontal and \(\vec{F}\).

Fig. 3.8 A jet of water is deflected by a vane through an angle \(\theta\). The vane imposes a force \(\vec{F}\) on the fluid, which acts at an angle of \(\phi\) to the horizontal.

To achieve this we first need to select an appropriate control volume. Importantly it needs to contain all of the features we wish to resolve. In this case it is fairly straightforward: we select a CV that contains the jet’s inflow, outflow and the vane itself. The CV therefore has one inlet and one outlet.

In terms of forces, we’re told the friction on the vane is negligible, so we can neglect this. For simplicity we will also neglect weight of the fluid and, since atmospheric pressure acts equally on the CV, this can be neglected also.

Conservation of momentum tells us that the force imposed by the vane on the fluid must balance the change in momentum through the inlet and outlet, i.e. that

\[ \vec{F} = \dot{m}_{\mathrm{out}} \vec{u}_{\mathrm{out}} - \dot{m}_{\mathrm{in}} \vec{u}_{\mathrm{in}} \]

Since we know that the jet’s velocity is unchanged, we thus know that \(\dot{m}_{\mathrm{out}} = \dot{m}_{\mathrm{in}} = \dot{m} = \rho A V\). Computing forces in the \(x\) and \(y\) directions separately, we have first in the \(x\)-direction that

\[ F_x = \dot{m}V\cos\theta - \dot{m}V = \dot{m}V(\cos\theta - 1) \]

Notice that since \(\cos\theta\leq 1\) and \(\dot{m}\) and \(V\) are both positive, this implies that \(F_x\leq 0\), so that it will always point to the left, as we see in the force diagram above.

In the \(y\)-direction, there is no contribution to the velocity from the inflow, since the incoming velocity is horizontal. Therefore, \(F_y = \dot{m}V\sin\theta\). The magnitude of \(\vec{F}\) is thus given by

\[\begin{split} \begin{align*} \|\vec{F}\| = \sqrt{F_x^2 + F_y^2} = \dot{m}V\sqrt{(\cos\theta - 1)^2 + \sin^2\theta} &= \dot{m}V\sqrt{2-2\cos\theta} \\ &= 2\dot{m}V\sqrt{\frac{1-\cos\theta}{2}} \\ &= 2\dot{m}V\sin\left(\frac{\theta}{2}\right). \end{align*} \end{split}\]

We can also compute the angle of action of \(\vec{F}\) which we call \(\phi\) between the horizontal and \(\vec{F}\). Simple trigonometry yields

\[ \phi = 180^\circ - \arctan\frac{F_y}{F_x} = 90^\circ + \frac{\theta}{2}. \]

3.6.1. Moving control volumes

We briefly noted earlier that control volumes are permitted to move so long as they do not accelerate. This then enables us to use this analysis in the study of problems involving, for example, moving plates. We just need to account for the relative velocity between any fluid and the CV. This is sometimes a useful thing to be able to consider, so let’s examine this further in the example below.

Example: a moving cart

A water jet of velocity \(V_j\) impinges normal to a flat plate which is moving to the right at velocity \(V_c\), as shown in the figure below. Find the force required to keep the plate moving at a constant velocity if the jet density \(\rho = 1000\,\mathrm{kg}\,\mathrm{m}^{-3}\), the jet area \(A = 3\,\mathrm{cm}^2\), \(V_j = 20\,\mathrm{m}\,\mathrm{s}^{-1}\) and \(V_c = 15\,\mathrm{m}\,\mathrm{s}^{-1}\).

We start by making some assumptions: the flow is steady with respect to the plate, the weight of the jet and plate can be neglected, and the jet splits into equal upward and downward half-jet.

Then we take a control volume as shown in part (b) of the figure. This cuts through the plate support and allows us to examine the forces \(R_x\) and \(R_y\). The CV moves at a speed of \(V_c\), and is therefore fixed relative to the cart. There are two outlets and one inlet.

First let us calculate the jet velocity at the outlets. The flow at the inlet travels at speed \(V_j\) with respect to a fixed observer: however in the frame of reference of the CV, it travels with a relative speed of \(V_j-V_c\). Mass conservation tells us that

\[ \dot{m}_{\text{in}} = \dot{m}_{\text{out}} \Rightarrow \rho A_j V_j = \rho A_1 V_1 + \rho A_2 V_2 \]

Since moreover \(A_1 = A_2 = \frac{1}{2} A_j\), we have that

\[ V_1+V_2 = 2(V_j - V_c) \]

and so if we assume that the problem is symmetric and there are no gravity effects, then

\[ V_1 = V_2 = V_j - V_c = 5\,\mathrm{m}\,\mathrm{s}^{-1}. \]

Now we can compute the force required to prevent acceleration (and thus travel at a constant velocity) by comparing the two components of momentum conservation. In the \(x\)-direction we have that

\[ R_x = \dot{m}_1 u_1 + \dot{m}_2 u_2 - \dot{m}_j u_j \]

and we know that \(\dot{m}_1 = \dot{m}_2 = \frac{1}{2}\dot{m}_j\), so \(\dot{m}_1 = \dot{m}_2 = \frac{1}{2}\rho A_j(V_j-V_c)\). In the \(x\)-direction, \(u_1 = u_2 = 0\), so

\[ R_x = -\dot{m}_j u_j = -[\rho A_j(V_j-V_c)](V_j-V_c) = -7.5\,\mathrm{N} \]

after we substitute values. The negative sign indicates that the force acts to the left: i.e. we have to impose a force to keep the plate from accelerating to the right after it is hit with the jet. This should match our intuition for this problem.

The vertical force is given by conservation of momentum in the \(y\)-direction as

\[ R_y = \dot{m}_1 v_1 + \dot{m}_2 v_2 - \dot{m}_j v_j \]

Since in this direction \(v_1 = V_1\), \(v_2 = -V_2\) and \(v_j = 0\), we have that

\[ R_y = \dot{m}_1 V_1 - \dot{m}_2 V_2 = \frac{1}{2}\dot{m}_j (V_1 - V_2) \]

However we earlier surmised that \(V_1 = V_2\), so this means that \(R_y = 0\). This also makes sense: the symmetry of the jet deflection that we assumed means that forces from the deflection upwards balance those from the deflection downwards.

3.6.2. Tips for using conservation of momentum

The use of conservation of momentum is very handy; the assumptions that we make in the problems above are obviously not reflective of reality, but do allow us to get rough estimates for forces involved in otherwise challenging fluid dynamics problems.

To successfully apply the conservation of momentum approach, consider the following aspects:

• Consider the choice of CV carefully. A careful choice of CV is required to simplify the problem, and needs to include all the necessary forces, inlets and outlets of the problem.

• Pay attention to coordinates and direction. Drawing a diagram where you explicitly outline your choice of coordinate system and the forces can help in avoiding issues with accidental negation of direction.

• What are the forces involved? For example, are you considering the appropriate reaction force of a jet on a wall or the wall on a jet? Does atmospheric pressure come into account somewhere?

• Does the frame of reference move? If so, make sure it does not accelerate (i.e. it is non-inertial) and that the movement is accounted for in the relative velocity of any jets or moving bodies.

3.7. Bernoulli’s equation

Bernoulli’s equation is perhaps one of the most ubiquitous equations in fluid dynamics, and is extremely useful in a number of settings. Broadly speaking, it allows us to connect the pressure, velocity and potential energy along a streamline. However to do so, it makes certain assumptions about the flow, so it is important to keep in mind these restrictions when trying to apply it. In particular, we are going to assume that:

• the flow is steady in time;

• there are no forces due to viscosity, i.e. the fluid is inviscid;

• the fluid is incompressible, so that the density is constant.

Warning

Although many problems are steady and incompressible, in reality virtually no fluid is inviscid (although there are some examples, such as superfluid helium-4). It is therefore extremely important that Bernoulli’s equation is applied with great care in regions of the flow where viscosity is negligible.

In particular, applying it near a wall, where we have already seen that viscosity plays an important role in imposing no-slip conditions, can yield results that are far removed from reality!

In this section we will briefly derive Bernoulli’s equation, using our knowledge of conservation of mass and momentum. We won’t use the Reynolds transport theorem, instead using a first-principles approach.

Let us consider the figure below, where we have a fluid element on a streamline.

Fig. 3.9 A fluid element lies on a streamline. The cross-sectional area on the left- and right-hand sides are \(\delta A\).

We need to select a coordinate system, and in this case, it makes sense to consider the problem in a coordinate system that aligns with the streamline, so that we have a tangential vector at each point along the streamline, and \(s(t)\) denotes the distance that the element moves along the streamline at a time \(t\). The magnitude of the fluid velocity is then represented by the quantity \(u(s)\).

Let us then consider the acceleration of the fluid element. This is given by:

\[ a = \frac{du}{dt}(s) = \frac{du}{ds}\frac{ds}{dt} = u\frac{du}{ds} = \frac{1}{2} \frac{d}{ds}(u^2)\]

where we have used the chain rule and the product rule to simplify to the right hand side. Newton’s second law therefore gives us that

\[ F = ma = \frac{1}{2}\rho\delta A\delta s \frac{d}{dt}(u^2) \]

where \(m = \rho\delta A\delta s\) and \(\delta A\) is the cross-sectional area of the fluid element. Now, what are the forces that are involved?

• Since the fluid is inviscid, there are no viscous forces to look at.

• We do however have a pressure force.

• We also need to consider the weight of the fluid volume \(\delta W\).

The pressure forces are relatively straightforward: they act on the surface of area \(\delta A\), so we have that

\[ F_p = -p(s+\delta s) \delta A + p(s)\delta A \approx -\left(p(s) + \delta s\frac{dp}{ds}\right) \delta A + p(s)\delta A = -\delta A\delta s\frac{dp}{ds}\]

where we have applied Taylor’s theorem on the right hand side. For gravity, the component acts downwards at an angle \(\theta\) as

\[ F_g = -\rho g \delta A\delta s\cos\theta = -\rho g\delta A\delta s \frac{dz}{ds} \]

In the above, \(z\) refers to an elevation level and \(\theta\) is the angle between the velocity and the \(z\) direction, as shown in the triangle in the figure. This also tells us immediately that \(\cos\theta = dz/ds\). Putting this all together then, we have that

\[ F_p + F_g = -\delta s\delta A\frac{dp}{ds} - \rho g\delta A\delta s \frac{dz}{ds} = \frac{1}{2}\rho\delta A\delta s \frac{d}{dt}(u^2) \]

Dividing by \(\delta s\delta A\) further gives

\[ \frac{1}{2}\rho\frac{d}{ds}(u^2) = -\frac{dp}{ds} - \rho g\frac{dz}{ds}. \]

Finally, since \(\rho\) is constant we can place it inside the derivatives, and this gives

\[ \frac{d}{ds}\left( \frac{p}{\rho} + \frac{1}{2}u^2 + gz \right) = 0 \Rightarrow \frac{p}{\rho} + \frac{1}{2}u^2 + gz = c \]

where \(c\) is a constant. This is Bernoulli’s equation, and states that the quantity \(\frac{p}{\rho} + \frac{1}{2}u^2 + gz\) is preserved along a streamline.

It’s more common to write this in a slightly different way: if we are given two points along a streamline, they will have pressures \(p_1\) and \(p_2\), velocities \(u_1\) and \(u_2\), and elevations \(z_1\) and \(z_2\). We can then rewrite the above as:

\[ \boxed{\frac{p_1}{\rho} + \frac{1}{2}u_1^2 + gz_1 = \frac{p_2}{\rho} + \frac{1}{2}u_2^2 + gz_2} \]

3.7.1. Application

To motivate the use of Bernoulli’s equation, let’s consider an application.

Example: draining a tank

Consider the tank below, where fluid flows out of a container through a hole at height \(h\) below the surface of the fluid. What is the velocity of the jet?

Fig. 3.10 Fluid flows out of a tank of water of height \(h\).

We assume the flow is steady - this might sound counterintuitive, but we could quite safely make the assumption that the tank is very large with a constantly topped-up level, or that the hole is small.

Considering a streamline connecting the surface with the jet, we can apply Bernoulli quite happily, noting that \(z_1 - z_2 = h\), so that

\[ \frac{p_1}{\rho} + \frac{1}{2}u_1^2 + gh = \frac{p_2}{\rho} + \frac{1}{2}u_2^2 \]

Now how do we relate the velocities and pressures?

• At the surface (point 1), we can essentially assume that the velocity is zero so that \(u_1 = 0\).

• Similarly at both points 1 and 2, we can assume that we experience atmospheric pressure, so \(p_1 = p_2 = p_{\text{atm}}\).

Atmospheric pressure and density cancel from both sides to yield

\[ gh = \frac{1}{2}u_2^2 \Rightarrow u_2 = \sqrt{2gh} \]

This is called Torricelli’s law after the Italian scientist Evangelista Torricelli who noted this back in 1643!

3.7.2. Stagnation, static and dynamic pressures

It’s often the case in many of our investigations that the relative change in elevation is negligible. Then, Bernoulli’s equation reduces to

\[ p_1 + \frac{1}{2}\rho u_1^2 = p_2 + \frac{1}{2}\rho u_2^2 = p_o = \text{constant} \]

The quantity \(p_o\) is the pressure at any point in the frictionless flow where the velocity is zero. This is called the stagnation pressure, and will be the highest pressure possible in the flow if we neglect elevation changes. The place where a velocity of zero occurs is called a stagnation point. An example of this is on an aeroplane, where the front nose and the leading edges of aerofoils are the points of highest pressure.

The pressures \(p_1\) and \(p_2\) are examples of static pressures in the moving fluid. Finally, the terms \((1/2)\rho u_i^2\) both have dimensions of pressure, and these are called the dynamic pressure.

Warning

Note that the zero-velocity condition of no-slip flow along a fixed wall does not result in stagnation pressure. The no-slip condition is a frictional effect, and the Bernoulli equation does not apply in this regime.

A useful visual interpretation of Bernoulli’s equation is to sketch two grade lines of a flow.

• The energy grade line (EGL) shows the height of the total Bernoulli constant \(h_o = z + p/\gamma + v^2/2g\). In frictionless flow with no work or heat transfer, the EGL has constant height.

• The hydraulic grade line (HGL) shows the height corresponding to elevation and pressure head \(z + p/\gamma\), i.e. the EGL without the velocity head \(V^2/2g\). The HGL is the height to which liquid would rise in a tube attached to the flow.

Fig. 3.11 Hydraulic and energy grade lines for frictionless flow in a duct.

The figure above demonstrates the EGL and HGL through two sections of a duct. The tubes measure the static pressure head and thus outline the HGL. The stagnation velocity tubes measure the total head \(z + p/\gamma + V^2/2g\), i.e. including the velocity head. In this case the EGL is constant and the HGL rises since the velocity drops.

You will explore these properties in the first lab, where you will see that in general the EGL will drop due to friction losses and obstructions in the geometry.

3.8. Summary

We have covered a lot of ground in this section:

• formulating the control volume approach;

• conservation equations

• mass and volume flowrates;

• conservation of mass and momentum, within the control volume via the Reynolds transport theorem;

• and Bernoulli’s equation.

In the next section, we will consider internal flows, i.e. the flow of fluid through an internal body such as a pipe or channel.

Further reading

For further reading, consult chapter 3 of White. In particular:

• section 3.1 for an overview of conservation laws;

• section 3.2 for the Reynolds transport theorem;

• section 3.3 for conservation of mass;

• section 3.4 for conservation of momentum;

• and section 3.5 for the Bernoulli equation.