Fluid mechanics, turbulent flow and turbulence modeling. Lars Davidson. Division of Fluid Dynamics. Department of Applied Mechanics. Fluid mechanics, turbulent flow and turbulence - Lars homeranking.info - Ebook download as PDF File .pdf), Text File .txt) or read book online. PDF | Cambridge Core - Fluid Dynamics and Solid Mechanics - A Voyage Through Turbulence - edited by Peter A. Davidson.
|Language:||English, Spanish, Dutch|
|ePub File Size:||25.79 MB|
|PDF File Size:||11.33 MB|
|Distribution:||Free* [*Regsitration Required]|
An Introduction to Turbulence. Models. Lars Davidson, homeranking.info /˜lada. Department of Thermo and Fluid Dynamics. CHALMERS UNIVERSITY. Lars Davidson. Division files/solids-and-fluids turbulent-flow turbulence- homeranking.info . Transport equations for turbulent kinetic energy. What are the fundamental principles of plasma turbulence? Also: P. A. Davidson. Turbulence .. Intermittency & non-Gaussian pdf's.
The pressure must decrease in the streamwise direction so that the pressure gradient term. The vorticity transport equation in three dimensions 44 4. These methods are still in use in wind engineering and for helicopters. The flow is shown in Fig. The kinetic energy is the sum of the kinetic energy of the three fluctuating velocity components. The left side of Eq.
University Press Scholarship Online. Sign in. Not registered? Sign up. Publications Pages Publications Pages. Search my Subject Specializations: Select your specializations: Classical Archaeology: Classical, Early, and Medieval Plays and Playwrights: Classical, Early, and Medieval Poetry and Poets: Classical, Early, and Medieval Prose and Writers: Classical, Early, and Medieval World History: Civil War American History: Users without a subscription are not able to see the full content.
More This book presents the subject of turbulence. L x1 Figure 3. The left hand side is zero since the flow is fully developed. Now we apply Eq. Two-dimensional boundary layer flow over flat plate 38 is zero because we assume that the flow is fully developed. The second term is zero on all four surfaces and the first term is zero on the lower and upper surfaces see Exercises below. The unit normal vector is denoted by nj which points out from the volume.
The last term corresponds to the viscous dissipation term. We replace the pressure P with p using Eq. Now we want to replace x1 and x2 with a new dimensionless variable. At the same time we define a new dimensionless streamfunction.
In Section 3. Exercise 23 From the dissipation. Exercise 19 For the fully developed flow. The flow is analyzed in Appendix C. Exercise 21 Show that the second term in Eq. Is it the same as that obtained from the force balance if not.
The numerical solution is given in Table 3. Exercise 22 Using the exact solution. Exercise 20 Show that the first and second terms in Eq. Blasius numerical solution of laminar flow along a flat plate. The v1 momentum equation requires that the horizontal viscous stresses balance the pressure difference.
As shown in Fig. The fluid particle is located in the lower half of fully developed channel flow. The v1 velocity is given by Eq. The v1 velocity field is indicated by dashed vectors.
Note that the v2 momentum equation see Eqs. Let us have a look at the momentum equations in order to show that the viscous terms indeed can be formulated with the vorticity vector. Figure 4. Surface forces acting on a fluid particle. Looking at Fig. In incompressible flow the viscous terms read see Eqs. The vorticity is always created at boundaries. Inviscid flow is often called irrotational flow i. Exercise 25 Write out Eq. An imbalance in shear stresses left side of Eq. The main points that we have learnt in this section are: The viscous terms are responsible for creating vorticity.
Exercise 24 Prove the first equality of Eq. The exception is when vorticity is transported into an inviscid region. In vector notation. The trick we have achieved is to split the convective term into one term without rotation first term on the right side of Eq.
We have learnt that physically it means rotation of a fluid particle and that it is only the viscous terms that can cause rotation of a fluid particle.
Equation 4. As usual we start with the Navier-Stokes equation. In this section we will derive the transport equation for vorticity in incompressible flow. The vorticity transport equation in three dimensions 44 4. There is a small difference between the three terms because there may be vorticity in inviscid flow that is convected into the flow at the inlet s. The terms inviscid no friction.
From Eq. Since the vorticity vector in Eq. The last term on line 2 is zero because the gravitation vector. That makes sense. The vorticity transport equation in three dimensions 45 These equations states that the gradient of stagnation pressure.
Imagine a slender. This means that vorticity in the x2 direction. Dashed lines denote fluid element before bending or tilting. The volume of the fluid element must stay constant during the stretching the incompressible continuity equation. The off-diagonal terms in Eq. The diagonal terms in this matrix represent vortex stretching. Vortex tilting. Vortex stretching. Assume is has a vorticity. Dashed lines denote fluid element before stretching.
Vortex stretching Vortex tilting. This means that vorticity diffuses in the same way as temperature does. In fully developed channel flow. Vortex stretching and tilting are physical phenomena which act in three dimensions: We have the same situation for the vorticity. The vorticity transport equation in two dimensions 47 creates vorticity in the x1 direction. Vorticity is useful when explaining why turbulence must be threedimensional.
Its gradient. The diffusion time. As shown above. This equation is exactly the same as the transport equation for temperature in incompressible flow. It is only when the wall-normal temperature derivative at the walls are non-zero that a temperature field is created in the domain. The boundary layer thickness. This means that. In a boundary layer. Boundary layer. The vorticity in a developing boundary layer is created at the leading edge of the plate note that in channel flow, vorticity is indeed created along the walls because in this case the streamwise pressure gradient is not zero.
The vorticity generated at the leading edge is transported along the wall by convection and at the same time it is transported by diffusion away from the wall. Below we will estimate the boundary layer thickness using the expression derived for the Rayleigh problem. Hence, if we can estimate how far from the wall the vorticity diffuses, this gives us an estimation of the boundary layer thickness. Consider the boundary layer in Fig. Exercise 26 Note that the estimate above is not quite accurate because in the Rayleigh problem we assumed that the convective terms are zero, but in a developing boundary layer, as in Fig.
The proper way to solve the problem is to use Blasius solution, see Section 3. Blasius solution gives see Eq. Exercise 27 Assume that we have a developing flow in a pipe radius R or between two flat plates separation distance h. We want to find out how long distance it takes for the the boundary layers to merge.
Make a comparison with this and Eq. The vorticity is then zero by definition since the curl of the divergence is zero. This is easily seen by inserting Eq. This is of great important since many analytical methods exist for the Laplace equation. The velocity field in potential flow is thus given by the continuity equation, Eq. Do we have any use of the Navier-Stokes equation? The answer is yes: We use the Navier-Stokes equation Eq.
The gravity force is conservative because when integrating this force, the work i. We start this section by repeating some basics of complex analysis.
We can approach the point z0 both in the real coordinate direction, x, and in the imaginary coordinate direction, y. The complex derivative is defined only if the value of the derivative is independent of how we approach the point z0. Equations 4. Another way to derive Eq. It can also be expressed in polar coordinates see Fig. Now we return to fluid mechanics and potential flow.
This is easily seen by taking the divergence of the streamfunction, Eq. The complex plane in polar coordinates. Real and imaginary axes correspond to the horizontal and vertical axes, respectively. Hence the complex potential, f , also satisfies the Laplace equation. Furthermore, f also satisfies the Cauchy-Riemann equations, Eq.
Thus we can conclude that f defined as in Eq. Taking the first and the second derivatives of Eq. Parallel flow. When we divide the fourth line with r2 and add it to the second line we find that the Laplace equation Eq. The flow is shown in Fig.
The solution in form of a vector plot and contour plot of the streamfunction is given in Fig. The polar velocity components are obtained as see. The streamfunction. Note that this flow is the same as we looked at in Section 1. Potential flow. Potential flow 54 0.
Stagnation flow. Line source. The angle. The streamfunction is zero along the lower boundary. Writing Eq. Potential flow 56 First. It is defined as a closed line integral along line C. The streamfunction corresponds to the imaginary part of f and we get see Eq. It was introduced in Section 1. The circulation. When origo is approached. This is easily seen by integrating vr Eq. The reason is that the inviscid assumption zero viscosity is not valid in this region.
Vortex line. Flow around a cylinder of radius r0. Pressure coefficients. Re50 Re Re Re 1 0. Integration of surface pressure. The markers show the time-averaged location of separation. We assume in Eq. To find the lift force. The surface pressure is obtained from Bernoulli equation see Eq. The velocity field at the cylinder surface. It should be stressed that although Eqs. The lift coefficient is obtained as see Fig. This requirement is not valid neither in the boundary layers nor in the wake.
How do we find the lift and drag force? The only force per unit area that acts on the cylinder surface is the pressure in viscous flow there would also be a viscous stress. Usually the lift force is expressed as a lift coefficient. Here we will introduce the use of additional circulation which alters the locations of the stagnation points and creates lift. Flow around a cylinder of radius r0 with maximal additional circulation.
The lift force on the lower surface side cancels the force on the upper side. Flow around a cylinder of radius r0 with additional circulation. Same argument for the drag force: This approach is used in potential methods for predicting flow around airfoils in aeronautics mainly helicopters and windpower engineering. We add the complex potential of a vortex line see Eq. We found in Eq. The last term cannot give any contribution to the lift because it is constant on the entire.
The pressure is obtained from Bernoulli equation as see Eq. For a limiting value of the circulation. What about the present case? Potential flow 61 On polar form it reads see Eqs. The larger the circulation. The two positions are indicated with a and b in Fig. The velocity at the surface. Side view. Recall that the relative velocity of the air is in the negative x1 direction.
The reason to the sign of the lift force can easily be seen from Fig. Instead of adding a circulation. Here the object is often vice versa. You want the ball to go as far as possible. The ball experiences a force. The loop uses the Magnus effect. The stagnation points. Table tennis. One way to improve the chance that this will happen is to make a loop.
This means that you hit the ball slightly on the top. This has interesting application in sports. Hence we only need to include the third term in Eq. The lift force is downwards because the stagnation points are located on the upper surface. In Eq. In the drag integral see Eq.
Another example where the Magnus effect is important is golf. A rotating cylinder produces lift. The drag is. In table tennis. The rotation causes a lift. Hence you hit it with a. The opponents erects a wall of players between the goal and the location of the free-kick.
If you are interested in football you may be pleased to learn that by use of fluid dynamics it is now scientifically proven that it was much harder to make a good freekick in worldcup than in .
Here the lift is used sideways. Top view. Because the ball was rotated 45 degrees before the second freekick see.
The ship moves with speed Vship. This makes the ball rotate clockwise. The freekicks are made 25m from the goal. As an experiment. A free-kick uses the Magnus effect.
Flettner rotor in blue on a ship. The reason that the ball turns to the right first after the wall and not before is that the forward momentum created by F the player is much larger than FL. The Magnus effect helps to achieve this. Figure 7b in that paper is particularly interesting. The result is a lift force in the positive x2 direction which makes the ball go further. A final sports example is football. The player who makes the free-kick wants to make the ball go on the left side of the wall.
The result of the two freekicks is that the two footballs reach the goal three meters from each other in the vertical direction. The ship is moving to the right with speed Vship. The diameter of this rotor can be a couple of meter and have a length i. Note that if the wind comes from the right instead of from the left.
Finally we give an engineering example of the use of the Magnus effect. It has recently gained new interest as the cost of fuel is rising. The boundary layers. A Flettner rotor is a rotating cylinder or many on a ship.
The Magnus effect creates a force in the orthogonal direction to the relative windspeed. The first Flettner rotors on ships were produced in Rear stagnation point at the upper surface suction side.
Streamlines from potential flow. The Division of Fluid Dynamics recently took part in an EU project where we studied the flow around rotating cylinders in relation to Flettner rotors .
The Flettner rotor rotates in the clockwise direction. When this flow is computed using potential methods. This is called the Kutta condition. In aeronautics. Streamlines from potential flow with added circulation.
The flow on the pressure lower side cannot be expected to make a o turn at the trailing edge and then go in the negative x1 direction towards the stagnation point located on the suction side. For low angles of attack which is the case for. The boundary layers and the wake are illustrated in red. Outside these regions the flow is essentially inviscid.
The added circulation is negative clockwise. These methods are still in use in wind engineering and for helicopters. The boundary layer is thinner on the pressure lower side than on the suction upper side. At the Division of Fluid Dynamics we have an on-going PhD project where we use potential methods for computing the aerodynamic loads for windturbine rotorblades .
The magnitude of the circulation is determined by the requirement that the stagnation point should be located at the trailing edge. Rear stagnation point at the trailing edge. The solution is to move the stagnation points in the same way as we did for the cylinder flow in Section 4. This is achieved by adding a circulation in the clockwise direction. The flow around airfoils is a good example where the flow can be treated as inviscid in large part of the flow.
The lift of a two-dimensional airfoil or a two-dimensional section of a three-dimensional airfoil is then computed as see Eq. We want to move the rear stagnation point towards the trailing edge.
The region covered by a large eddy may well enclose also smaller eddies. The flow consists of a spectrum of different scales eddy sizes. Large Reynolds Numbers. Turbulent flow is always three-dimensional and unsteady. The increased diffusivity also increases the resistance wall friction and heat transfer in internal flows such as in channels and pipes.
The turbulence increases the exchange of momentum in e. Even though we have small turbulent scales in the flow they are much larger than the molecular scale and we can treat the flow as a continuum. Also the flow and combustion in engines.
The small eddies receive the kinetic energy from slightly larger eddies. The largest eddies are of the order of the flow geometry i. The boundary layers and the wakes around and after bluff bodies such as cars. Turbulent flow occurs at high Reynolds number. Turbulent flow is irregular and chaotic they may seem random. The largest eddies extract their energy from the mean flow. At the other end of the spectra we have the smallest eddies which are dissipated by viscous forces stresses into thermal energy resulting in a temperature increase.
Even though turbulence is chaotic it is deterministic and is described by the Navier-Stokes equations. Turbulent flow is dissipative. It has a characteristic velocity and length called a velocity and length scale.
We do not have any exact definition of an turbulent eddy. A turbulent eddy cascade process. Air movements in rooms are turbulent. This process of transferring energy from the largest turbulent scales eddies to the smallest is called the cascade process. There is no definition on turbulent flow. The slightly larger eddies receive their energy from even larger eddies and so on. Typical examples are flow around as well as in cars.
Turbulence 66 5 Turbulence 5. In turbulent flow the diffusivity increases. The dissipation is proportional to the kinematic viscosity. We assume that these scales are determined by viscosity. Cascade process with a spectrum of eddies. These scales extract kinetic energy from the mean flow which has a time scale comparable to the large scales. Through the cascade process.
At the smallest scales the frictional forces viscous stresses become large and the kinetic energy is transformed dissipated into thermal energy. The kinetic energy transferred per unit time from eddy-to-eddy from an eddy to a slightly smaller eddy is the same for each eddy size. The energy-containing eddies are denoted by v0. In reality a small fraction is dissipated at all scales. The friction forces exist of course at all scales. The argument is as follows.
It turns out that it is often convenient to use Fourier series to analyze turbulence. We get two equations. The dimensions of the left and the right side must be the same. The more energy that is to be transformed from kinetic energy to thermal energy. Energy spectrum 68 viscosity: Since the kinetic energy is destroyed by viscous forces it is natural to assume that viscosity plays a part in determining these scales. We can think of them as eddies.
Having assumed that the dissipative scales are determined by viscosity and dissipation. In this case the left side of Eq. The wavenumber. Intuitively we assume that large eddies have large fluctuating velocities which implies large kinetic energy. Range for small. Range for the large. The turbulent scales are distributed over a range of scales which extends from the largest scales which interact with the mean flow to the smallest scales where dissipation occurs.
The total turbulent kinetic energy is obtained by integrating over the whole wavenumber. An example of a Fourier series and spectra are given in Appendix E. Let us think about how the kinetic energy of the eddies varies with eddy size. The dimension of wavenumber is one over length. It is convenient to study the kinetic energy of each eddy size in wavenumber space. Spectrum for turbulent kinetic energy. The energy spectrum. In that case. For readers not familiar to Fourier series. Let g be a fluctuating velocity component.
This energy transfer takes places via the production term. The eddies in this region are independent of both the large. Energy per time unit. The turbulence is also in this region isotropic. Dissipation range. Inertial subrange. The scales of the eddies are described by the Kolmogorov scales see Eq.
Energy spectrum 70 space. The eddies in this region represent the mid-region. Part of the energy extracted per unit time by the largest eddies is transferred per unit time to slightly smaller scales. II and III which are discussed below. In this region we have the large eddies which carry most of the energy.
The kinetic energy of all eddies of all size is computed by Eq. The eddies are small and isotropic and it is here that the dissipation occurs. We find region I. These eddies interact with the mean flow and extract energy from the mean flow. The existence of this region requires that the Reynolds number is high fully turbulent flow. The kinetic energy is the sum of the kinetic energy of the three fluctuating velocity components. Energy spectrum 71 We get two equations. Note that is not true instantaneously.
Using our knowledge in tensor notation. In the inertial subrange. As discussed on p. Inserted in Eq. We isotropic turbulence.
Applying Eq. An eddy loses part of its kinetic energy during one revolution. This means that — in average — the eddies have no preferred direction. Disturbances are amplified by interaction between the vorticity vector and the velocity gradients. With a computational grid we must resolve all eddies. Equations 5. This means that the eddy range wavenumber range of the intermediate region region II. These eddies interact with the mean flow and extract energy from the mean flow.
The existence of this region requires that the Reynolds number is high fully turbulent flow. The kinetic energy is the sum of the kinetic energy of the three fluctuating velocity components. Energy spectrum 71 We get two equations. Note that is not true instantaneously. Using our knowledge in tensor notation.
In the inertial subrange. As discussed on p. Inserted in Eq. We isotropic turbulence. Applying Eq. An eddy loses part of its kinetic energy during one revolution. This means that — in average — the eddies have no preferred direction. Disturbances are amplified by interaction between the vorticity vector and the velocity gradients. With a computational grid we must resolve all eddies.
Equations 5. This means that the eddy range wavenumber range of the intermediate region region II. Number of eddies at each generation with their axis aligned in the x1. The cascade process created by vorticity generation 1st 2nd 3rd 4th 5th 6th 7th x1 1 0 2 2 6 10 22 72 x2 0 1 1 3 5 11 21 x3 0 1 1 3 5 11 21 Table 5. Two idealized phenomena in this interaction process can be identified: We find that the ratio of the velocity.
This is the very reason why it is so expensive in terms of computer power to solve the NavierStokes equations. Note that also the circulation. Equation 5. The large original eddy. At the same time. Family tree of turbulent eddies see also Table 5. As was mentioned above. Five generations. We have neglected the viscosity since viscous diffusion at high Reynolds number is much smaller than the turbulent one and since viscous dissipation occurs at small scales see p.
For each generation the eddies become more and more isotropic as they get smaller. The large original eddy 1st generation is aligned in the x1 direction. Figure 5. The radius of the red fluid element decreases. The rotation rate of the fluid element black circles in Fig.
At each stage. The small eddies have no preferred direction. Here a generation is related to a wavenumber in the energy spectrum Fig. In other words. A fluid element is stretched by line to solid line. It creates eddies in the x2 and x3 direction 2nd generation. The creation of multiple eddies by vortex stretching from one original eddies is illustrated in Fig. The smaller the eddies.
They are isotropic. The vorticity and velocity field becomes chaotic and three-dimensional: Vortex stretching and vortex tilting qualitatively explain how interaction between vorticity and velocity gradient create vorticity in all three coordinate directions from a disturbance which initially was well defined in one coordinate direction.
From the discussion above we can now understand why turbulence always must be three-dimensional Item IV on p. The turbulence is also maintained by these processes. Once this process has started it continues. As a result. This shows how vorticity in one direction is transferred to the other two directions through vortex tilting.
Term I: Another reason is that when we want to solve the Navier-Stokes equation numerically it would require a very fine grid to resolve all turbulent scales and it would also require a fine resolution in time turbulent flow is always unsteady. When we time average Eq. Turbulent mean flow 76 6 Turbulent mean flow 6. It represents correlations between fluctuating velocities. This is called the closure problem: It is an additional stress term due to turbulence fluctuating velocities and it is unknown.
This equation is the time-averaged Navier-Stokes equation and it is often called the Reynolds equation. The continuity equation applies both for the instantaneous velocity. The width i. In addition to the viscous shear stress. Here the velocity gradient is largest as the velocity drops down to zero at the wall over a very short distance. First we re-write the left side of Eq. One important quantity is the wall shear stress which is defined as.
Flow between two infinite parallel plates. Note that the two terms on the left side are of the same order.
In fully developed wall friction velocity. The streamwise momentum equation. The first term on the left side of Eq. The wall region adapted from Ch. The wall region can be divided into one outer and one inner region. The inner region includes the viscous region dominated by the viscous diffusion and the logarithmic region dominated by turbulent diffusion. The buffer region acts as a transition region between these two regions where viscous diffusion of streamwise momentum is.
Note that the total shear stress is constant only close to the wall Fig. At the wall. In the logarithmic layer the viscous stress is negligible compared to the Reynolds stress. Wall region in fully developed channel flow 1 0. In the inner region.
DNS data [ Reynolds shear stress. Hence the friction velocity. In this region the flow is assumed to be independent of viscosity. The Reynolds shear stress. Velocity profiles in fully developed channel flow. DNS What about the length scale? Near the wall. Hence it seems reasonable to take the wall distance as the characteristic length scale. Another way of deriving the expression in Eq.
Symmetry plane of channel flow. The turbulent part shear stress. In the outer region of the boundary layer. Since nothing changes in the x3 direction. The constant. As can be seen in Fig. Figure 6. This is the force opposing the movement of the fluid. Looking at Eq. This opposing force has its origin at the walls due to the viscous wall force viscous shear stress multiplied by area. On the other hand. Start by looking at Fig. With the same argument.
The pressure must decrease in the streamwise direction so that the pressure gradient term. This agrees — fortunately — with our intuition. As can be seen. The gradient of the shear stress. The pressure gradient is constant and equal to one: The reason is that although the time-averaged flow is two-dimensional i. We can imagine that the fluid air. It is equal to one near the wall and decreases rapidly for increasing wall distance.
Another way to describe the behaviour of the pressure is to say that there is a pressure drop. Here the forces are two orders of magnitude larger than in Fig. The force that balances the pressure gradient is the gradient of the Reynolds shear stress.
Fully developed channel flow. What is the difference between that flow and a boundary layer flow? In turbulent flow the velocity profile in the center region is much flatter than in laminar flow cf. We can of course make a force balance for a section of the channel. In this region the Reynolds shear stress gradient term is driving the flow and the opposing force is the viscous force.
This makes the velocity gradient near the wall and the wall shear stress. The former is largest because the mean flow is in this direction. The turbulent kinetic energy. Forces in a boundary layer. Velocity profiles in a boundary layer along a flat plate. DNS convective terms are not zero or negligible.
Normal Reynolds stresses and turbulent kinetic energy. The flow in a boundary layer is described by Eq. The flow in a boundary layer is continuously developing. Consider the probability density functions of the fluctuations. Probability density functions 20 1.
With a probability density. The second moment corresponds to the variance of the fluctuations or the square of the RMS. The mean velocity can of course also be computed by integrating over time. In order to extract more information. From the velocity signals we can compute the probability densities sometimes called histograms. Figure 7. The skewness. This is confirmed in the PDF distribution.
The PDF function in Fig. Probability density functions 87 3 0. Point 2. Point 3. The time history shows that the positive and the negative values have the same magnitude. Probability density functions of time histories in Fig. A probability density function is symmetric if positive values are as frequent and large as the negative values. At this point the time history Fig. Let us analyze the data at the three points. The fluctuations at this point are much smaller and the positive values are as large the negative values.
The resulting probability densities functions are shown in Fig. Point 1. The time history of the velocity fluctuation Fig. The flatness gives 4 this information. There is yet another statistical quantity which sometimes is used for describing turbulent fluctuations. The variance the square of RMS tells us how large the fluctuations are in average. Probability density functions 88 Note that f must be normalized see Eq. I Trick 1: Trick 2: Transport equations for kinetic energy 89 8 Transport equations for kinetic energy N this section and Section 9 we will derive various transport equations.
There are two tricks which often will be used. Both tricks simply use the product rule for derivative backwards. The Exact k Equation 90 2 8. We subtract Eq. Equations 8. The second term on the right side of Eq. Now we can assemble the transport equation for the fluctuation. I Convection. For the first term in Eq. III The two first terms represent turbulent diffusion by pressure-velocity fluctuations. The term excluding the minus sign is always positive it consists of velocity gradients squared.
It is largest for large wavenumbers. This term originates from the convection term the first term on the right side of Eq. The Reynolds stresses which appear in P k are larger for large eddies than for small eddies. It can be written as. There are two arguments for this: In the fully turbulent region of a boundary layer. This requirement is best satisfied by the large scales. It is largest for the energy-containing eddies. When the production term is negative.
The large turbulent scales extract energy from the mean flow. The velocity-fluctuation term originates from the convection term the last term on the right side of Eq. Figure 8. The dissipation term stems from the viscous term see Eq. IV Dissipation. This term including the minus sign is almost always positive.
In order to extract energy from the mean flow. This term is responsible for transformation of kinetic energy at small scales to thermal energy. When the viscous stress vector is in the opposite direction to the fluctuating velocity. It can be noted that the production term is an acceleration term. When the acceleration term and the fluctuating velocity are in opposite directions i.
The last term is viscous diffusion. The magnitude of the velocity gradient is given by Eq. A force multiplied with a velocity corresponds to work per unit time. The largest eddies created by the velocity profile A are much smaller than those created by the velocity profile B.
The size of the largest eddies dashed lines for different velocity profiles. Now we know that the energy spectrum see Eqs. How do we know that the majority of the dissipation takes place at the smallest scales? The energy that is transferred in spectral space i. Insertng Eq.
The Exact k Equation 94 In the cascade process see Section 5. The cascade process assumes that this energy transfer per unit time is the same for each eddy size. Recall that the viscous dissipation. The energy transferred from eddy-to-eddy per unit time in spectral space can also be used for estimating the velocity gradient of an eddy. Consider the inertial subrange. This is called local equilibrium.
The Exact k Equation: Now consider Fig. Closer to the wall Fig. Note that the production and the dissipation terms close to the wall are two orders of magnitude larger than in the logarithmic region Fig.
The dissipation term and the viscous local equilibrium. At the wall the turbulent fluctuations are zero which means that the production term is zero. This is because the dissipation term is at its largest for small. Since the region near the wall is dominated by viscosity the turbulent diffusion terms due to pressure and velocity are also small.
In the outer region Fig. The left side is — since the flow is fully developed — zero. The production term term II in Eq. In homogeneous turbulence the spatial transport terms i. If the channel walls are very long and wide compared to the distance between the walls.
Some flows may have one or two homogeneous directions. The velocity gradient is largest at the wall see Fig. How are the spectral transfer and the spatial transport related?
The reason that we in Section 5. The turbulence kinetic energy is produced by its main source term. Note that the derivatives of the instantaneous turbulent fluctuations are non-zero even in homogeneous turbulence. In Section 8. Since P k is largest here so is also k. The dissipation term term IV in Eq. V kdV. It may first be transported in physical space towards the center.
Consider a large. Integration of the convection term over the entire volume. Large eddies which extract their energy from the mean flow may not give their energy to the slightly smaller eddies as assumed in Figs. In Townsend . Inside the domain. In summary. This should be kept in mind when thinking in terms of the cascade process.
The reason is that the transfer of turbulent kinetic energy. This statement simply means that production is equal to dissipation. This shows that the net effect of the convection term occurs only at the boundaries. The overall effect of the transport terms 97 In non-homogeneous turbulence.
In the inertial range Region II. The turbulence in the buffer layer and the logarithmic layer of a boundary layer see Fig. We say that the turbulence at these scales is in local equilibrium. Mathematically these terms are called divergence terms. Multiply the time-averaged Navier-Stokes equations. Similarly for the diffusion term. Comparison of mean and fluctuating dissipation terms. This corresponds to the dissipation term in Eq. Note that the first term in Eq.
On the right side we find: The cascade process assumes that the term in red is negligible 2v see also Figs. This is seen in Fig. The gradient of the time-averaged velocity field. In the viscous-dominated wall region. The energy flow from the mean flow to internal energy is illustrated in Fig. Transfer of energy between mean kinetic energy K. The major part of the energy flow goes from K to k and then to dissipation.
Two-equations models are commonly used in these simplified models. This is important: Transport equations for Reynolds stresses 9 Transport equations for Reynolds stresses N this section we will derive the transport equation for the Reynolds stress tensor. This is an unknown in the time-averaged Navier-Stokes equations.
Adding Eqs. This approach is very similar to that we used when deriving the k equation in Section 8. In Section 8 we derived transport equations for kinetic turbulent energy. Now write Eq. Subtract Eq. Note that it is permitted to use any other index than k in some terms but you must not use i and j. Re-writing also the third terms in the two groups in Eq. Indices i and j appear once in each term not more. Transport equations for Reynolds stresses terms in its transport equation must also be symmetric.
You could. Compare Eq. It is called the Reynolds stress equations. IV Note that the manipulation in Eq. It is an equation for a second-order tensor which consists of nine equations. Put the first term in Eq. After a derivation. We can now put everything together. Transport equations for Reynolds stresses The product rule is used backwards to merge the two first terms so that the third line in Eq.
An alternative — and maybe easier — way to derive Eq. The physical meaning of this term is to redistribute energy between the normal stress components if we transform Eq. For 2D boundary layer flow.
Transport equations for Reynolds stresses 0. Note that the trace of the pressure-strain term is zero. The terms in the wall-normal stress equation. In the middle there is a heat source. A simple example is the one-dimensional unsteady heat conduction equation Eq.
A positive source term in a transport equation. Source terms Q wall wall x1 Figure 9. The pressure-strain term. As we saw for the k equation. If Q is positive. As mentioned previously. The production term in Eq. Figure 9. Here we find — as expected — that the pressure-strain term. T will increase and vice-versa.
One-dimensional unsteady heat conduction. Another difference is that the pressure diffusion term. Indeed it should be. Reynolds shear stress vs. The production term — which should be a source term — is here negative. That implies there is no correlation between two fluctuating velocity components. It can be summarized as: This is because dissipation takes place at the smallest scales and they are isotropic. For channel flow. If you want to learn more how to derive transport equations of turbulent quantities.
Compare Fig. This means that when the. A fluid particle is moving downwards particle drawn with solid line from x2. At its new location the v1 velocity is in average smaller than at its old. P k is almost always positive. Consider the flow in a boundary layer. B to x2. A where the streamwise velocity is v1 x2.
Thus the streamwise fluctuation is positive. A x2 x1 Figure 9. B which has streamwise velocity v1 x2. If we look at the other particle dashed line in Fig. B comes down to x2. A it has an excess of streamwise velocity compared to its new environment at x2.
In cases where the shear stress and the velocity gradient have the same sign for example. The normalized two-point correlation reads norm A B11 x1. Pick two points along the x1 axis. Consider a flow where the largest eddies have a length scale of Lint. It is obvious that if we move point A and C closer to each other. B11 increases. Correlations B11 Figure For flows with small eddies. We expect that the two point correlation. Two-point correlation. Schematic relation between the two-point correlation.
B11 t. Assume that the turbulent fluctuation at point A is transported by the mean integral time scale. Auto correlation If the flow is homogeneous see p.