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Week 8: Literature review - RANS derivation and analysis

Aim: To apply Reynold's decomposition to the Navier Stoke's equations to obtain the expression for Reynold's stress. Also understanding the terms Reynold's stress and the difference between turbulent and molecular viscosity. Theory: Most of the engineering problems deal with turbulent flows. A characteristic feature of…

Sridhara Kumudavalli
updated on 28 Jul 2020

Aim: To apply Reynold's decomposition to the Navier Stoke's equations to obtain the expression for Reynold's stress. Also understanding the terms Reynold's stress and the difference between turbulent and molecular viscosity.

Theory: Most of the engineering problems deal with turbulent flows. A characteristic feature of a turbulent flow is a small scale, high-frequency random fluctuations imposed on the main flow which has a definable direction. In fluid dynamics, a turbulent flow is characterized by irregular chaotic changes in pressure and flow velocity. A turbulent regime refers to irregular flows in which eddies, swirls, and flow instabilities occur. It is in contrast to a laminar flow, which occurs when a fluid flows in parallel layers, with no disruption between those layers. As mentioned before the turbulent regime is extremely common in natural phenomenon and engineering applications. Few examples are smoke coming out of a chimney, waterfalls, external flow over all kinds of vehicles such as cars, airplanes, ships, etc.

Need for Turbulence Models:

In order to capture the turbulence effect, we need to solve NS and a few more equations. The problem is turbulence has a very small time scale and the computation mesh required to solve the system would be very small. For example, in an IC engine approximately we would require a mesh size of 1e-5m. This method requires a huge computation power to solve the equations at these scales and this method is known as Direct Numerical Simulation (DNS).To run such a large number of cells it would nearly be impossible to solve using normal computers and rather time-consuming. Hence we use turbulence models in order to capture the turbulence effect in coarser mesh size. Basically, there are 2 approaches used to model turbulence.

Reynolds-averaged Navier Strokes (RANS)
Large Eddy Simulation (LES).

We will be discussing in detail the Reynolds-averaged Navier Stokes (RANS) model.

RANS:

It is believed that using NS equations we can model any kind of flow including turbulent flows. The major drawback is that when there are higher values of Reynold's number the resolution of NS equation is challenging and not stable. Thus, a small perturbation in the parameters, boundary conditions, or initial conditions may lead to a different solution. This issue can be partially resolved by the use of Reynolds-averaged Navier Stokes equations (RANS). The Reynolds-averaged Navier Stokes equations (RANS) are time-averaged equations of motion for fluid flows. The ideology behind these equations is Reynold's decomposition, where an instantaneous quantity is decomposed into its time-averaged (mean values) and fluctuating quantities.

The variation of velocity in a turbulent flow field is shown in the above graph. Velocity at any instant can be divided into a mean component and fluctuating component. The instantaneous velocity component along x, y, and z directions are as follows:

$u_{i} (x, y, z, t) = u_{m} (x, y, z) + u_{f} . (x, y, z, t)$ $u_i(x,y,z,t) = u_m(x,y,z)+u_f.(x,y,z,t)$

$v_{i} (x, y, z, t) = v_{m} (x, y, z) + v_{f} . (x, y, z, t)$ $v_i(x,y,z,t) = v_m(x,y,z)+v_f.(x,y,z,t)$

$w_{i} (x, y, z, t) = w_{m} (x, y, z) + w_{f} . (x, y, z, t)$ $w_i(x,y,z,t) = w_m(x,y,z)+w_f.(x,y,z,t)$

This process is known as Reynold's Decomposition. The subscript i represents instantaneous component, m represents mean component and f represents fluctuating component.

The mean component can be given as:

$u_{m} (x, y, z) = \frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} u (x, y, z, t) d t$ $u_m(x,y,z)=1/t_(Int)int_0^(t_(Int)) u(x,y,z,t)dt$

The above equation is the time average of the instantaneous velocity over the turbulence time step.

The fluctuating component can be given as:

$\frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} u_{f} . (x, y, z, t) d t = 0$ $1/t_(Int) int_0^(t_(Int))u_f.(x,y,z,t)dt = 0$

Now we will apply Reynolds decomposition to Navier stokes equation as follows.

Continuity Equation

$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}$ $(delu)/(delx) + (delv)/(dely)$

Applying Reynold's decomposition to continuity equation.

$\frac{\partial (u_{m} + u_{f})}{\partial x} + \frac{\partial (v_{m} + v_{f})}{\partial y} = 0$ $(del(u_m+u_f))/(delx) + (del(v_m+v_f))/(dely) = 0$

Integrating over the time interval.

$\frac{1}{t_{I n t}} [\int_{0}^{t_{I n t}} \frac{\partial (u_{m} + u_{f})}{\partial x} d t + \int_{0}^{t_{I n t}} \frac{\partial (v_{m} + v_{f})}{\partial y} d t] = 0$ $1/t_(Int)[int_0^(t_(Int))(del(u_m+u_f))/(delx) dt + int_0^(t_(Int))(del(v_m+v_f))/(dely) dt ]=0$

Splitting the integrals.

$\frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} \frac{\partial (u_{m})}{\partial x} d t + \frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} \frac{\partial (u_{f})}{\partial x} d t + \frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} \frac{\partial (v_{m})}{\partial y} d t + \frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} \frac{\partial (v_{f})}{\partial y} d t = 0$ $1/t_(Int) int_0^(t_(Int)) (del(u_m))/(delx) dt + 1/t_(Int) int_0^(t_(Int)) (del(u_f))/(delx) dt + 1/t_(Int) int_0^(t_(Int)) (del(v_m))/(dely) dt + 1/t_(Int) int_0^(t_(Int)) (del(v_f))/(dely) dt = 0$

We know that the integral of the fluctuating component is zero. Hence, we can write above equations as

$\frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} \frac{\partial (u_{m})}{\partial x} d t + \frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} \frac{\partial (v_{m})}{\partial y} d t = 0$ $1/t_(Int) int_0^(t_(Int)) (del(u_m))/(delx) dt + 1/t_(Int) int_0^(t_(Int)) (del(v_m))/(dely) dt = 0$

The above equation can be written as

$\frac{\partial (u_{m})}{\partial x} \frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} d t + \frac{\partial (v_{m})}{\partial y} \frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} d t = 0$ $(del(u_m))/(delx) 1/(t_(Int)) int_0^(t_(Int))dt + (del(v_m))/(dely) 1/(t_(Int)) int_0^(t_(Int))dt = 0$

Thus the continuity equation can be simplified as,

$\frac{\partial (u_{m})}{\partial x} + \frac{\partial (v_{m})}{\partial y} = 0$ $(del(u_m))/(delx) + (del(v_m))/(dely) = 0$

Momentum Equation

$ρ (\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}) = - \frac{\partial p}{\partial x} + μ \frac{\partial^{2} u}{\partial x^{2}}$ $rho((delu)/(delt)+u(delu)/(delx)+v(delu)/(dely))=-(delp)/(delx)+mu(del^2u)/(delx^2)$

Taking density term into another side we obtain

$\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} = - \frac{1}{ρ} \frac{\partial p}{\partial x} + ν \frac{\partial^{2} u}{\partial x^{2}}$ $(delu)/(delt) + u(delu)/(delx) + v(delu)/(dely) = -1/rho (delp)/(delx) + nu (del^2u)/(delx^2)$

$\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} + u (\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}) = - \frac{1}{ρ} \frac{\partial p}{\partial x} + ν \frac{\partial^{2} u}{\partial x^{2}}$ $(delu)/(delt) + u(delu)/(delx) + v(delu)/(dely) + u((delu)/(delx) + (delv)/(dely)) = -1/rho (delp)/(delx) + nu (del^2u)/(delx^2)$

$\frac{\partial u}{\partial t} + 2 u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} + u \frac{\partial v}{\partial y} = - \frac{1}{ρ} \frac{\partial p}{\partial x} + ν \frac{\partial^{2} u}{\partial x^{2}}$ $(delu)/(delt) + 2u(delu)/(delx) + v(delu)/(dely) + u(delv)/(dely) = -1/rho (delp)/(delx) + nu (del^2u)/(delx^2)$

We know that,

$\frac{\partial (u v)}{\partial y} = u \frac{\partial v}{\partial y} + v \frac{\partial u}{\partial y}$ $(del(uv))/(dely) = u(delv)/(dely) + v(delu)/(dely)$

$\frac{\partial u}{\partial t} + 2 u \frac{\partial u}{\partial x} + \frac{\partial (u v)}{\partial y} = - \frac{1}{ρ} \frac{\partial p}{\partial x} + ν \frac{\partial^{2} u}{\partial x^{2}}$ $(delu)/(delt) + 2u(delu)/(delx) + (del(uv))/(dely)= -1/rho (delp)/(delx) + nu (del^2u)/(delx^2)$

Simplifying the above equations we obtain,

$\frac{\partial u}{\partial t} + \frac{\partial u^{2}}{\partial x} + \frac{\partial (u v)}{\partial y} = - \frac{1}{ρ} \frac{\partial p}{\partial x} + ν \frac{\partial^{2} u}{\partial x^{2}}$ $(delu)/(delt) + (delu^2)/(delx) + (del(uv))/(dely) = -1/rho (delp)/(delx) + nu (del^2u)/(delx^2)$

Now, applying Reynold's decomposition into the above equation and integrating it over a time interval.

$\frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} \frac{\partial (u_{m} + u_{f})}{\partial t} d t + \frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} \frac{\partial {(u_{m} + u_{f})}_{m}^{2}}{\partial x} d t + \frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} \frac{\partial (u_{m} + u_{f}) (v_{m} + v_{f})}{\partial y} d t = - \frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} \frac{1}{ρ} \frac{\partial (p_{m} + p_{f})}{\partial x} d t + ν \frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} \frac{\partial^{2} (u_{m} + u_{f})}{\partial x^{2}} d t$ $1/t_(Int) int_0^(t_(Int)) (del(u_m+u_f))/(delt) dt + 1/t_(Int) int_0^(t_(Int)) (del(u_m+u_f)^2)/(delx) dt + 1/t_(Int) int_0^(t_(Int)) (del(u_m+u_f)(v_m+v_f))/(dely) dt = - 1/t_(Int) int_0^(t_(Int)) 1/rho (del(p_m+p_f))/(delx) dt + nu 1/t_(Int) int_0^(t_(Int)) (del^2(u_m+u_f))/(delx^2) dt$

We will be neglecting the integrals of fluctuating terms and by time integrating the rest of the terms, we get the below equation.

$\frac{\partial u_{m}}{\partial t} + \frac{\partial {(u^{2})}_{m}^{2}}{\partial x} + \frac{\partial (u_{m} v_{m})}{\partial y} = - \frac{\partial p_{m}}{ρ \partial x} + ν \frac{\partial^{2} u_{m}}{\partial y^{2}} - \frac{1}{t_{I n t}} \int_{0}^{t_{I n t}} ⎛ ⎝ \frac{\partial {(u^{2})}_{f}^{2}}{\partial x} + \frac{\partial (u_{f} v_{f})}{\partial y} d t ⎞ ⎠$ $(delu_m)/(delt) + (del(u^2)_m)/(delx) + (del(u_mv_m))/(dely) = - (delp_m)/(rhodelx) + nu (del^2u_m)/(dely^2) - 1/t_(Int) int_0^(t_(Int)) ((del(u^2)_f)/(delx) + (del(u_fv_f))/(dely) dt)$

When we consider a boundary layer the change in square of velocity across the x-direction can be taken as zero.i.e the velocity gradient is not as large when compared to the y-axis and hence the square of the x-velocity can be considered to be zero. Using this assumption and simplifying the above equation we obtain.

$\frac{\partial u_{m}}{\partial t} + \frac{\partial {(u^{2})}_{m}^{2}}{\partial x} + \frac{\partial (u_{m} v_{m})}{\partial y} = - \frac{\partial p_{m}}{ρ \partial x} + \frac{1}{ρ} \frac{\partial}{\partial y} (μ \frac{\partial u_{m}}{\partial y} + \frac{ρ}{t_{I n t}} \int_{0}^{t_{I n t}} (u_{f} v_{f}) d t)$ $(delu_m)/(delt) + (del(u^2)_m)/(delx) + (del(u_mv_m))/(dely) = - (delp_m)/(rhodelx) + 1/rho (del)/(dely)(mu (delu_m)/(dely) + rho/t_(Int) int_0^(t_(Int))(u_fv_f)dt)$

In the above equation, the momentum diffusion terms consist of 2 components, They are:

Molecular Viscosity $(μ \frac{\partial u_{m}}{\partial y})$ $(mu (delu_m)/(dely))$ . This is also known as viscous stress.
Momentum diffusivity due to turbulence $(\frac{ρ}{t_{I n t}} \int_{0}^{t_{I n t}} (u_{f} v_{f}) d t)$ $(rho/(t_(Int)) int_0^(t_(Int))(u_fv_f)dt)$ . This is also known as Reynold's Stress.

Reynold's Stress:

In fluid dynamics, Reynold's stress is the component of the total stress tensor in a fluid obtained from the averaging operation over the Navier Stokes equations to account for turbulent fluctuations in fluid momentum. Turbulence modeling is all about calculating this stress and to solve this Reynold's stress there is 2 turbulence model available which are the k-epsilon and k-omega model.

Molecular Viscosity and Turbulent Viscosity

Molecular Viscosity: Molecular viscosity is the same as viscosity. The coefficient of molecular viscosity is the same value as dynamic viscosity. Molecular viscosity is the transport of mass motion momentum solely by the random motion of individual molecules not moving together in coherent groups. Molecular viscosity is analogous in laminar flow to eddy viscosity in a turbulent flow.

Turbulent Viscosity: Turbulent viscosity is also known as Eddy Viscosity which occurs in turbulence. The turbulent transfer of momentum by eddies gives rise to internal fluid friction, in a manner analogous to the action of molecular viscosity in a laminar flow, but occurring on a larger scale. Eddy Viscosity is a function of flow, not of the fluid.