Thermal Engineering

Adiabatic Boundary – Thermal Symmetry

Adiabatic Boundary – Thermal Symmetry

Special Case – Adiabatic Boundary – Perfectly Insulated Boundary

A special case of this condition corresponds to the perfectly insulated surface for which (∂T/∂x = 0). Heat transfer through a properly insulated surface can be taken to be zero since adequate insulation reduces heat transfer through a surface to negligible levels. Mathematically, this boundary condition can be expressed as:

Special Case – Thermal Symmetry

Another very important case, that can be used for solving heat transfer problems involving fuel rods, is the thermal symmetry. For example, the two surfaces of a large hot plate of thickness L suspended vertically in air will be subjected to the same thermal conditions, and thus the temperature distribution will be symmetrical (i.e. in one half of the plate will be the same temperature profile as that in the other half). As a result, there must be a maximum in in the center line of the plate and the center line can be viewed as an insulated surface (∂T/∂x = 0). The thermal condition at this plane of symmetry can be expressed as:

See also: Neumann Boundary Condition

See also:

Heat Equation

Heat Equation in Cylindrical and Spherical Coordinates

In engineering, there are plenty of problems, that cannot be solved in cartesian coordinates. Cylindrical and spherical systems are very common in thermal and especially in power engineering. The heat equation may also be expressed in cylindrical and spherical coordinates. The general heat conduction equation in cylindrical coordinates can be obtained from an energy balance on a volume element in cylindrical coordinates and using the Laplace operator, Δ, in the cylindrical and spherical form.

Cylindrical coordinates:

Spherical coordinates:

Obtaining analytical solutions to these differential equations requires a knowledge of the solution techniques of partial differential equations, which is beyond the scope of this text. On the other hand, there are many simplifications and assumptions, that can be applied to these equations and that lead to very important results. In the next section we limit our consideration to one-dimensional steady-state cases with constant thermal conductivity, since they result in ordinary differential equations.

See also:

Thermal Conduction

Conduction with Heat Generation

Conduction with Heat Generation

In the preceding section we considered thermal conduction problems without internal heat sources. For these problems the temperature distribution in a medium was determined solely by conditions at the boundaries of the medium. But in engineering we can often meet a problem, in which internal heat sources are significant and determines the temperature distribution together with boundary conditions.

In nuclear engineering, these problems are of the highest importance, since most the heat generated in nuclear fuel is released inside the fuel pellets and the temperature distribution is determined primarily by heat generation distribution. Note that, as can be seen from the description of the individual components of the total energy energy released during the fission reaction, there is significant amount of energy generated outside the nuclear fuel (outside fuel rods). Especially the kinetic energy of prompt neutrons is largely generated in the coolant (moderator). This phenomena needs to be included in the nuclear calculations.

For LWR, it is generally accepted that about 2.5% of total energy is recovered in the moderator. This fraction of energy depends on the materials, their arrangement within the reactor, and thus on the reactor type.

Note that, heat generation is a volumetric phenomenon. That is, it occurs throughout the body of a medium. Therefore, the rate of heat generation in a medium is usually specified per unit volume and is denoted by g_V [W/m³].

The temperature distribution and accordingly the heat flux is primarily determined by:

• Geometry and boundary conditions. Different geometry leads to completely different temperature field.
• Heat generation rate. The temperature drop through the body will increase with increased heat generation.
• Thermal conductivity of the medium. Higher thermal conductivity will lead to lower temperature drop.

See also:

Heat Equation

Example of Heat Equation – Problem with Solution

Heat Conduction in a Large Plane Wall

Example of Heat Equation – Problem with Solution

Consider the plane wall of thickness 2L, in which there is uniform and constant heat generation per unit volume, q_V [W/m³]. The centre plane is taken as the origin for x and the slab extends to + L on the right and – L on the left. For constant thermal conductivity k, the appropriate form of the heat equation, is:

The general solution of this equation is:

where C₁ and C₂ are the constants of integration.

1)

Calculate the temperature distribution, T(x), through this thick plane wall, if:

• the temperatures at both surfaces are 15.0°C
• the thickness this wall is 2L = 10 mm.
• the materials conductivity is k = 2.8 W/m.K (corresponds to uranium dioxide at 1000°C)
• the volumetric heat rate is q_V = 10⁶ W/m³

In this case, the surfaces are maintained at given temperatures T_s,1 and T_s,2. This corresponds to the Dirichlet boundary condition. Moreover, this problem is thermally symmetric and therefore we may use also thermal symmetry boundary condition. The constants may be evaluated using substitution into the general solution and are of the form:

The resulting temperature distribution and the centerline (x = 0) temperature (maximum) in this plane wall at these specific boundary conditions will be:

The heat flux at any point, q_x [W.m^-2], in the wall may, of course, be determined by using the temperature distribution and with the Fourier’s law. Note that, with heat generation the heat flux is no longer independent of x, the therefore:

Example of Heat Equation – Problem with Solution

Heat Conduction in a Fuel Rod

Most of PWRs use the uranium fuel, which is in the form of uranium dioxide. Uranium dioxide is a black semiconducting solid with very low thermal conductivity. On the other hand the uranium dioxide has very high melting point and has well known behavior. The UO₂ is pressed into cylindrical pellets, these pellets are then sintered into the solid.

These cylindrical pellets are then loaded and encapsulated within a fuel rod (or fuel pin), which is made of zirconium alloys due to its very low absorption cross-section (unlike the stainless steel). The surface of the tube, which covers the pellets, is called fuel cladding.

See also: Thermal Conduction of Uranium Dioxide

Thermal and mechanical behavior of fuel pellets and fuel rods constitute one of three key core design disciplines. Nuclear fuel is operated under very inhospitable conditions (thermal, radiation, mechanical) and must withstand more than normal conditions operation. For example temperatures in the centre of fuel pellets reach more than 1000°C (1832°F) accompanied by fission-gas releases. Therefore detailed knowledge of temperature distribution within a single fuel rod is essential for safe operation of nuclear fuel. In this section we will study heat conduction equation in cylindrical coordinates using Dirichlet boundary condition with given surface temperature (i.e. using Dirichlet boundary condition). Comprehensive analysis of fuel rod temperature profile will be studied in separate section.

Temperature in the centerline of a fuel pellet

Consider the fuel pellet of radius r_U = 0.40 cm, in which there is uniform and constant heat generation per unit volume, q_V [W/m³]. Instead of volumetric heat rate q_V [W/m³], engineers often use the linear heat rate, q_L [W/m], which represents the heat rate of one meter of fuel rod. The linear heat rate can be calculated from the volumetric heat rate by:

The centreline is taken as the origin for r-coordinate. Due to symmetry in z-direction and in azimuthal direction, we can separate of variables and simplify this problem to one-dimensional problem. Thus, we will solve for the temperature as function of radius, T(r), only. For constant thermal conductivity, k, the appropriate form of the cylindrical heat equation, is:

The general solution of this equation is:

where C₁ and C₂ are the constants of integration.

Calculate the temperature distribution, T(r), in this fuel pellet, if:

• the temperatures at the surface of the fuel pellet is T_U = 420°C
• the fuel pellet radius r_U = 4 mm.
• the averaged material’s conductivity is k = 2.8 W/m.K (corresponds to uranium dioxide at 1000°C)
• the linear heat rate is q_L = 300 W/cm and thus the volumetric heat rate is q_V = 597 x 10⁶ W/m³

In this case, the surface is maintained at given temperatures T_U. This corresponds to the Dirichlet boundary condition. Moreover, this problem is thermally symmetric and therefore we may use also thermal symmetry boundary condition. The constants may be evaluated using substitution into the general solution and are of the form:

The resulting temperature distribution and the centerline (r = 0) temperature (maximum) in this cylindrical fuel pellet at these specific boundary conditions will be:

The radial heat flux at any radius, q_r [W.m^-1], in the cylinder may, of course, be determined by using the temperature distribution and with the Fourier’s law. Note that, with heat generation the heat flux is no longer independent of r.

The following figure shows the temperature distribution in the fuel pellet at various power levels.

______

The temperature in an operating reactor varies from point to point within the system. As a consequence, there is always one fuel rod and one local volume, that are hotter than all the rest. In order to limit these hot places the peak power limits must be introduced. The peak power limits are associated with a boiling crisis and with the conditions which could cause fuel pellet melt. However, metallurgical considerations place an upper limits on the temperature of the fuel cladding and the fuel pellet. Above these temperatures there is a danger that the fuel may be damaged. One of the major objectives in the design of a nuclear reactors is to provide for the removal of the heat produced at the desired power level, while assuring that the maximum fuel temperature and the maximum cladding temperature are always below these predetermined values.

Temperature distribution in Fuel Cladding

Cladding is the outer layer of the fuel rods, standing between the reactor coolant and the nuclear fuel (i.e. fuel pellets). It is made of a corrosion-resistant material with low absorption cross section for thermal neutrons, usually zirconium alloy. Cladding prevents radioactive fission products from escaping the fuel matrix into the reactor coolant and contaminating it. Cladding constitute one of barriers in ‘defence-in-depth‘ approach.

Consider the fuel cladding of inner radius r_Zr,2 = 0.408 cm and outer radius r_Zr,1 = 0.465 cm. In comparison to fuel pellet, there is almost no heat generation in the fuel cladding (cladding is slightly heated by radiation). All heat generated in the fuel must be transferred via conduction through the cladding and therefore the inner surface is hotter than the outer surface.

To find the temperature distribution through the cladding we must solve the heat conduction equation. Due to symmetry in z-direction and in azimuthal direction, we can separate of variables and simplify this problem to one-dimensional problem. Thus, we will solve for the temperature as function of radius, T(r), only. In this example, we will assume that there is strictly no heat generation within the cladding. For constant thermal conductivity, k, the appropriate form of the cylindrical heat equation, is:

The general solution of this equation is:

where C₁ and C₂ are the constants of integration.

1)

Calculate the temperature distribution, T(r), in this fuel cladding, if:

• the temperature at the inner surface of the cladding is T_Zr,2 = 360°C
• the temperature of reactor coolant at this axial coordinate is T_bulk = 300°C
• the heat transfer coefficient (convection; turbulent flow) is h = 41 kW/m².K.
• the averaged material’s conductivity is k = 18 W/m.K
• the linear heat rate of the fuel is q_L = 300 W/cm and thus the volumetric heat rate is q_V = 597 x 10⁶ W/m³

From the basic relationship for heat transfer by convection we can calculate the outer surface of the cladding as:

As can be seen, also in this case we have given surface temperatures T_Zr,1 and T_Zr,2. This corresponds to the Dirichlet boundary condition. The constants may be evaluated using substitution into the general solution and are of the form:

Solving for C₁ and C₂ and substituting into the general solution, we then obtain:

∆T – cladding surface – coolant

Detailed knowledge of geometry, outer radius of cladding, linear heat rate, convective heat transfer coefficient and the coolant temperature determines ∆T between the coolant (T_bulk) and the cladding surface (T_Zr,1). Therefore we can calculate the cladding surface temperature (T_Zr,1) simply using the Newton’s Law:

∆T in fuel cladding

Detailed knowledge of geometry, outer and inner radius of cladding, linear heat rate, and the cladding surface temperature (T_Zr,1) determines ∆T between outer and inner surfaces of cladding. Therefore we can calculate the inner cladding surface temperature (T_Zr,2) simply using the Fourier’s Law:

See also:

Heat Equation

Heat Conduction in a Fuel Rod

Heat Conduction in a Fuel Rod

Most of PWRs use the uranium fuel, which is in the form of uranium dioxide. Uranium dioxide is a black semiconducting solid with very low thermal conductivity. On the other hand the uranium dioxide has very high melting point and has well known behavior. The UO₂ is pressed into cylindrical pellets, these pellets are then sintered into the solid.

These cylindrical pellets are then loaded and encapsulated within a fuel rod (or fuel pin), which is made of zirconium alloys due to its very low absorption cross-section (unlike the stainless steel). The surface of the tube, which covers the pellets, is called fuel cladding.

See also: Thermal Conduction of Uranium Dioxide

Thermal and mechanical behavior of fuel pellets and fuel rods constitute one of three key core design disciplines. Nuclear fuel is operated under very inhospitable conditions (thermal, radiation, mechanical) and must withstand more than normal conditions operation. For example temperatures in the centre of fuel pellets reach more than 1000°C (1832°F) accompanied by fission-gas releases. Therefore detailed knowledge of temperature distribution within a single fuel rod is essential for safe operation of nuclear fuel. In this section we will study heat conduction equation in cylindrical coordinates using Dirichlet boundary condition with given surface temperature (i.e. using Dirichlet boundary condition). Comprehensive analysis of fuel rod temperature profile will be studied in separate section.

Temperature in the centerline of a fuel pellet

Consider the fuel pellet of radius r_U = 0.40 cm, in which there is uniform and constant heat generation per unit volume, q_V [W/m³]. Instead of volumetric heat rate q_V [W/m³], engineers often use the linear heat rate, q_L [W/m], which represents the heat rate of one meter of fuel rod. The linear heat rate can be calculated from the volumetric heat rate by:

The centreline is taken as the origin for r-coordinate. Due to symmetry in z-direction and in azimuthal direction, we can separate of variables and simplify this problem to one-dimensional problem. Thus, we will solve for the temperature as function of radius, T(r), only. For constant thermal conductivity, k, the appropriate form of the cylindrical heat equation, is:

The general solution of this equation is:

where C₁ and C₂ are the constants of integration.

Calculate the temperature distribution, T(r), in this fuel pellet, if:

• the temperatures at the surface of the fuel pellet is T_U = 420°C
• the fuel pellet radius r_U = 4 mm.
• the averaged material’s conductivity is k = 2.8 W/m.K (corresponds to uranium dioxide at 1000°C)
• the linear heat rate is q_L = 300 W/cm and thus the volumetric heat rate is q_V = 597 x 10⁶ W/m³

In this case, the surface is maintained at given temperatures T_U. This corresponds to the Dirichlet boundary condition. Moreover, this problem is thermally symmetric and therefore we may use also thermal symmetry boundary condition. The constants may be evaluated using substitution into the general solution and are of the form:

The resulting temperature distribution and the centerline (r = 0) temperature (maximum) in this cylindrical fuel pellet at these specific boundary conditions will be:

The radial heat flux at any radius, q_r [W.m^-1], in the cylinder may, of course, be determined by using the temperature distribution and with the Fourier’s law. Note that, with heat generation the heat flux is no longer independent of r.

The following figure shows the temperature distribution in the fuel pellet at various power levels.

______

The temperature in an operating reactor varies from point to point within the system. As a consequence, there is always one fuel rod and one local volume, that are hotter than all the rest. In order to limit these hot places the peak power limits must be introduced. The peak power limits are associated with a boiling crisis and with the conditions which could cause fuel pellet melt. However, metallurgical considerations place an upper limits on the temperature of the fuel cladding and the fuel pellet. Above these temperatures there is a danger that the fuel may be damaged. One of the major objectives in the design of a nuclear reactors is to provide for the removal of the heat produced at the desired power level, while assuring that the maximum fuel temperature and the maximum cladding temperature are always below these predetermined values.

See also:

Heat Equation

Heat Conduction in a Large Plane Wall

Heat Conduction in a Large Plane Wall

Consider the plane wall of thickness 2L, in which there is uniform and constant heat generation per unit volume, q_V [W/m³]. The centre plane is taken as the origin for x and the slab extends to + L on the right and – L on the left. For constant thermal conductivity k, the appropriate form of the heat equation, is:

The general solution of this equation is:

where C₁ and C₂ are the constants of integration.

1)

Calculate the temperature distribution, T(x), through this thick plane wall, if:

• the temperatures at both surfaces are 15.0°C
• the thickness this wall is 2L = 10 mm.
• the materials conductivity is k = 2.8 W/m.K (corresponds to uranium dioxide at 1000°C)
• the volumetric heat rate is q_V = 10⁶ W/m³

In this case, the surfaces are maintained at given temperatures T_s,1 and T_s,2. This corresponds to the Dirichlet boundary condition. Moreover, this problem is thermally symmetric and therefore we may use also thermal symmetry boundary condition. The constants may be evaluated using substitution into the general solution and are of the form:

The resulting temperature distribution and the centerline (x = 0) temperature (maximum) in this plane wall at these specific boundary conditions will be:

The heat flux at any point, q_x [W.m^-2], in the wall may, of course, be determined by using the temperature distribution and with the Fourier’s law. Note that, with heat generation the heat flux is no longer independent of x, the therefore:

Example of Heat Equation – Problem with Solution

See also:

Heat Equation

Temperature distribution in Fuel Cladding

Cladding is the outer layer of the fuel rods, standing between the reactor coolant and the nuclear fuel (i.e. fuel pellets). It is made of a corrosion-resistant material with low absorption cross section for thermal neutrons, usually zirconium alloy. Cladding prevents radioactive fission products from escaping the fuel matrix into the reactor coolant and contaminating it. Cladding constitute one of barriers in ‘defence-in-depth‘ approach.

Consider the fuel cladding of inner radius r_Zr,2 = 0.408 cm and outer radius r_Zr,1 = 0.465 cm. In comparison to fuel pellet, there is almost no heat generation in the fuel cladding (cladding is slightly heated by radiation). All heat generated in the fuel must be transferred via conduction through the cladding and therefore the inner surface is hotter than the outer surface.

To find the temperature distribution through the cladding we must solve the heat conduction equation. Due to symmetry in z-direction and in azimuthal direction, we can separate of variables and simplify this problem to one-dimensional problem. Thus, we will solve for the temperature as function of radius, T(r), only. In this example, we will assume that there is strictly no heat generation within the cladding. For constant thermal conductivity, k, the appropriate form of the cylindrical heat equation, is:

The general solution of this equation is:

where C₁ and C₂ are the constants of integration.

1)

Calculate the temperature distribution, T(r), in this fuel cladding, if:

• the temperature at the inner surface of the cladding is T_Zr,2 = 360°C
• the temperature of reactor coolant at this axial coordinate is T_bulk = 300°C
• the heat transfer coefficient (convection; turbulent flow) is h = 41 kW/m².K.
• the averaged material’s conductivity is k = 18 W/m.K
• the linear heat rate of the fuel is q_L = 300 W/cm and thus the volumetric heat rate is q_V = 597 x 10⁶ W/m³

From the basic relationship for heat transfer by convection we can calculate the outer surface of the cladding as:

As can be seen, also in this case we have given surface temperatures T_Zr,1 and T_Zr,2. This corresponds to the Dirichlet boundary condition. The constants may be evaluated using substitution into the general solution and are of the form:

Solving for C₁ and C₂ and substituting into the general solution, we then obtain:

∆T – cladding surface – coolant

Detailed knowledge of geometry, outer radius of cladding, linear heat rate, convective heat transfer coefficient and the coolant temperature determines ∆T between the coolant (T_bulk) and the cladding surface (T_Zr,1). Therefore we can calculate the cladding surface temperature (T_Zr,1) simply using the Newton’s Law:

∆T in fuel cladding

Detailed knowledge of geometry, outer and inner radius of cladding, linear heat rate, and the cladding surface temperature (T_Zr,1) determines ∆T between outer and inner surfaces of cladding. Therefore we can calculate the inner cladding surface temperature (T_Zr,2) simply using the Fourier’s Law:

See also:

Heat Equation

Critical Thickness of Insulation – Critical Radius

In a plane wall the area perpendicular to the direction of heat flow adding more insulation to a wall always decreases heat transfer. The thicker the insulation, the lower the heat transfer rate. This is due to the fact the outer surface have always the same area.

But in cylindrical and spherical coordinates, the addition of insulation also increases the outer surface, which decreases the convection resistance at the outer surface. Moreover, in some cases, a decrease in the convection resistance due to the increase in surface area can be more important than an increase in conduction resistance due to thicker insulation. As a result the total resistance may actually decrease resulting in increased heat flow.

The thickness upto which heat flow increases and after which heat flow decreases is termed as critical thickness. In the case of cylinders and spheres it is called critical radius. It can be derived the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h.

As can be seen, if r₁ < r_cr, as it is in this case, the total resistance decreases and the heat rate therefore increases with the addition of insulation. This trend continues until the outer radius of the insulation corresponds to the critical radius, where the heat rate reaches its maximum. This trend is desirable for cooling of an electrical wire, since the addition of electrical insulation would aid in transferring heat dissipated in the wire to the surroundings. On the other hand, any further addition of material (beyond r_cr) would increase the total resistance and therefore decrease the heat loss. This behavior would be desirable for insulation of pipes, where insulation is added to reduce heat loss to the surroundings.

Example – Critical Thickness of Insulation

Assume a steel pipe of r₁ = 10 mm, which is exposed to natural convection at h = 50 W/m².K. This pipe is insulated by material of thermal conductivity k = 0.5 W/m.K. Determine the critical thickness of this combination:

Hence r_cr > r₁ and heat transfer will increase with the addition of insulation up to a thickness of r_cr – r₁ = (0.010 – 0.005)m = 0.005 m

Critical Thickness of Insulation – Spherical Coordinates

It can be shown in a similar manner that the critical radius of insulation for a spherical shell is:

See also:

Heat Equation

Navier-Stokes Equations

In fluid dynamics, the Navier-Stokes equations are equations, that describe the three-dimensional motion of viscous fluid substances. These equations are named after Claude-Louis Navier (1785-1836) and George Gabriel Stokes (1819-1903). In situations in which there are no strong temperature gradients in the fluid, these equations provide a very good approximation of reality.

The Navier-Stokes equations consists of a time-dependent continuity equation for conservation of mass, three time-dependent conservation of momentum equations and a time-dependent conservation of energy equation. There are four independent variables in the problem, the x, y, and z spatial coordinates of some domain, and the time t.

As can be seen, the Navier-Stokes equations are second-order nonlinear partial differential equations, their solutions have been found to a variety of interesting viscous flow problems. They may be used to model the weather, ocean currents, air flow around an airfoil and water flow in a pipe or in a reactor. The Navier–Stokes equations in their full and simplified forms help with the design of aircraft and cars, the study of blood flow, the design of nuclear reactors and many other things.

Solution of Navier-Stokes Equations

Even though the Navier-Stokes equations have only a limited number of known analytical solutions, they are amenable to fine-gridded computer modeling. The main tool available for their analysis is CFD analysis. CFD is a branch of fluid mechanics that uses numerical analysis and algorithms to solve and analyze problems that involve turbulent fluid flows. It is widely accepted that the Navier–Stokes equations (or simplified Reynolds-averaged Navier–Stokes equations) are capable of exhibiting turbulent solutions, and these equations are the basis for essentially all CFD codes. It is possible now to achieve approximate, but realistic, CFD results for a wide variety of complex two- and three-dimensional viscous flows.

The Navier–Stokes equations are also of great interest in a purely mathematical sense. Unfortunately, the highly intermittent and irregular character of turbulent flow complicates all analyses. It has not yet been proven that in three dimensions solutions always exist, or that if they do exist, then they are smooth. In fact, general solution of the Navier-Stokes equations with turbulences is often said to be the “last unsolved problem in classical mathemetical physics.”

Characteristics of Turbulent Flow

• Turbulent flow tends to occur at higher velocities, low viscosity and at higher characteristic linear dimensions.
• If the Reynolds number is greater than Re > 3500, the flow is turbulent.
• Irregularity: The flow is characterized by the irregular movement of particles of the fluid. The movement of fluid particles is chaotic. For this reason, turbulent flow is normally treated statistically rather than deterministically.
• Diffusivity: In turbulent flow, a fairly flat velocity distribution exists across the section of pipe, with the result that the entire fluid flows at a given single value and drops rapidly extremely close to the walls. The characteristic which is responsible for the enhanced mixing and increased rates of mass, momentum and energy transports in a flow is called “diffusivity”.
• Rotationality: Turbulent flow is characterized by a strong three-dimensional vortex generation mechanism. This mechanism is known as vortex stretching.
• Dissipation: A dissipative process is a process in which the kinetic energy of turbulent flow is transformed into internal energy by viscous shear stress.

Kolmogorov Microscales

In the view of Kolmogorov (Andrey Nikolaevich Kolmogorov was a Russian mathematician who made significant contributions to the mathematics of probability theory and turbulence), turbulent motions involve a wide range of scales. From a macroscale at which the energy is supplied, to a microscale at which energy is dissipated by viscosity.

For example, consider a cumulus cloud. The macroscale of the cloud can be of the order of kilometers and may grow or persist over long periods of time. Within the cloud, eddies may occur over scales of the order of millimeters. For smaller flows such as in pipes, the microscales may be much smaller. Most of the kinetic energy of the turbulent flow is contained in the macroscale structures. The energy “cascades” from these macroscale structures to microscale structures by an inertial mechanism. This process is known as the turbulent energy cascade.

The smallest scales in turbulent flow are known as the Kolmogorov microscales. These are small enough that molecular diffusion becomes important and viscous dissipation of energy takes place and the turbulent kinetic energy is dissipated into heat.

The smallest scales in turbulent flow, i.e. the Kolmogorov microscales are:

where ε is the average rate of dissipation rate of turbulence kinetic energy per unit mass and has dimensions (m²/s³). ν is the kinematic viscosity of the fluid and has dimensions (m²/s).

The size of the smallest eddy in the flow is determined by viscosity. The Kolmogorov length scale decreases as viscosity decreases. For very high Reynolds number flows, the viscous forces are smaller with respect to inertial forces. Smaller scale motions are then necessarily generated until the effects of viscosity become important and energy is dissipated. The ratio of largest to smallest length scales in the turbulent flow are proportional to the Reynolds number (raises with the three-quarters power).

This causes direct numerical simulations of turbulent flow to be practically impossible. For example, consider a flow with a Reynolds number of 10⁶. In this case the ratio L/l is proportional to 10^18/4. Since we have to analyze three-dimensional problem, we need to compute a grid that consisted of at least 10¹⁴ grid points. This far exceeds the capacity and  possibilities of existing computers.

See also:

Fluid Dynamics

Solution of Navier-Stokes Equations

Even though the Navier-Stokes equations have only a limited number of known analytical solutions, they are amenable to fine-gridded computer modeling. The main tool available for their analysis is CFD analysis. CFD is a branch of fluid mechanics that uses numerical analysis and algorithms to solve and analyze problems that involve turbulent fluid flows. It is widely accepted that the Navier–Stokes equations (or simplified Reynolds-averaged Navier–Stokes equations) are capable of exhibiting turbulent solutions, and these equations are the basis for essentially all CFD codes. It is possible now to achieve approximate, but realistic, CFD results for a wide variety of complex two- and three-dimensional viscous flows.

The Navier–Stokes equations are also of great interest in a purely mathematical sense. Unfortunately, the highly intermittent and irregular character of turbulent flow complicates all analyses. It has not yet been proven that in three dimensions solutions always exist, or that if they do exist, then they are smooth. In fact, general solution of the Navier-Stokes equations with turbulences is often said to be the “last unsolved problem in classical mathemetical physics.”

See also:

Navier-Stokes Equations