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**Navier-Stokes Equation**

Newtonian Fluid Constant Density, Viscosity Cartesian, Cylindrical, spherical coordinates The Navier Stokes equations may be written in terms of shear stresses or simplified for Newtonian fluids. In many textbooks, one encounters NS equation for newtonian fluids which have a constant viscosity (i.e. temperature is a constant for the system)

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**Cartesian Coordinates**

The equation in cartesian co ordinates

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**Cylindrical Coordinates**

Centrifugal force Coriolis force The equation in cylindrical co ordinates and spherical co ordinates are given for your reference. Note that the vector form is compact and ‘hides’ the details while still being conducive to physical interpretation The ‘red circle’s indicate the places where the text book gives incorrect formula

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**Spherical Coordinates**

Similar formulas for spherical coordinates

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**Spherical Coordinates**

And the correct formula for the 3rd coordinate BSL has gq here instead of gf

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**Spherical Coordinates (3W)**

The same equation expressed differently in the text book, along with the correction 3W &R have the formula in terms of However, the expression for is incorrect in the book

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Continuity Equation of continuity in different coordinate systems

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**Newton’s law of viscosity**

Constant Density, zero dilatational viscosity Cylindrical coordinates Like wise Newton’s law of viscosity in different coordinate systems

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**Newton’s law of viscosity**

Spherical coordinates: Constant Density, zero dilatational viscosity Newton’s law of viscosity, in spherical co ordinates

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**N-S Equation: Examples**

ODE vs PDE Spherical and cylindrical coordinates Eqn for pipe flow (Hagen Poiseulle) Flow between rotating cylinders (not solved in class) Thin film flow with temp variation (not solved in class, steps were discussed briefly. BSL ‘worked out’ example) Radial flow between circular plates (BSL 3B.10) We will look in to some examples where NS equations are used to solve the problems. Normally we will use cylindrical co ordinates for pipe flow. We will use cartesian coordinates for flow in a rectangular channel or flow between infinite flat plates etc. So, the geometry dictates which co ordinates should be used, so that equations are relatively simple. Spherical co ordinates are normally not necessary, but we will also see an example where it is useful. Some of the simpler problems can be handled easily by shell balance; but we will use the NS eqn, just so that you can begin with ‘easy’ problems and then move on to more difficult ones. Again, from a mathematics point of view, we will first consider problems that need solving an ODE. (i.e. NS equation will simplify to an ODE). Then we will move on to those that will need solving a PDE. (Of course, NS equation is a PDE to begin with!).

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**Example problems 1. Pressure driven steady state flow of fluid**

between two infinite parallel plates inside a circular tube 2. Steady state Couvette flow of a fluid between two infinite parallel plates with top plate moving at a known velocity between two circular plates of finite radius, with the top plate rotating at a known angular velocity between two circular cylinders with outer cylinder rotating at a known angular velocity (end effects are negligible) between a cone and plate (stationary plate and cone is rotating at a known angular velocity). Angle of cone is very small (almost a parallel plate with almost zero gap) 3. Coutte Poisseuille flow between two parallel plates I haven’t ‘keyed in’ the example problems into the powerpoint and hence they are not available online now. Hopefully, sometime later…

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**N-S Equation: Examples**

PDE Please refer to the book “Applied Mathematical Methods for Chemical Engineers” by Norman W Loney (CRC press), pages 330 to 342 for “worked out” examples for Momentum Transfer problems involving PDE. Either multi dimensional or time dependent (however multidimensional and time dependent cases are not discussed in detail) Steady state in Rectangular channel: pressure driven , coutte flow Plan suddenly moving with constant velocity (or stress) from time t=0 (Stokes problem) Sudden pressure gradient in a cylindrical tube (unsteady flow , converging to Hagen-Poisseuille’s flow (Bessel functions) Flow between two (non rotating) cylinders, caused by boundary movement (coutte flow). Unsteady vs steady (not discussed in class or covered in tutorial) We will cover some examples in this. The rest is left as homework! If you have a system which involves solving unsteady state problems or multi dimensions, the NS eqn cannot be reduced to ODE (normally). We will take flow in a rectangular channel or sudden application of pressure in a fluid in a pipe. We will obtain what is called ‘exact’ solution. When we get very complex equations, we may throw away some terms saying that they are not ‘significant’. We have to do that with care. We will look into ‘creeping flow approximation’ and ‘boundary layer approximation’. The solutions of those equations (if solved analytically) or sometimes referred to as ‘exact solutions’ and if they are solved numerically, they are approximate solutions. So, when you read literature, beware of the meaning of ‘exact solution’. 1. Similar equations are encountered in Heat, Mass and Momentum transfers. So it is worthwhile to learn to solve these eqns in this course. 2. While you are solving the problem, the boundary and initial conditions help you in ‘guiding’ which way to go, when you encounter more than one way to go about. 3. We will try two different approaches; the separation of variables and combination of variables. 4. You may get solution in the form of Bessel functions, if the problem is in cylindrical co ordinates; in Legendre polynomials, if the problem is in spherical co ordinates and in cosine and sine series (Fourier Series), for cartesian co ordinates (rectangular). 5. There may be other methods to solve the same problem, for example, using Laplace Transforms. 6. If you have an equation that is 2nd order in ‘x’ and first order in ‘t’, then you need one condition in ‘t’ and two conditions in ‘x’. We will take it as the requirement, in a crude way. The definition of well posed problem, existence and uniqueness of a solution, etc are beyond the scope of this course. The above list is a summary and if you do not understand all of it right now, do not worry. Things will become clearer as you ‘work out’ the problems.

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**Guidelines for solving PDE in Momentum Transfer**

Method: If the problem involves finite scales, “separation of variable” method should be tried If the problem involves infinite (or semi-infinite) distances, “combination of variables” method should be tried You can consider these as ‘rules of thumb’. In the next couple of pages, we will go over few more points; however, you are likely appreciate them when you actually try to solve few problems.

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**Guidelines for solving PDE in Momentum Transfer**

Solution forms, for finite scales: Applying the separation of variables directly may not always give proper results If the equation is non-homogenous For time dependent problems, first try to get steady state solution (and try that as the ‘particular solution’ for the equation). Unsteady state solution may be the ‘general solution’ for the corresponding homogenous equation For multi dimensional problems, first try to get solution for ‘one dimensional’ problem and try that as particular solution. The ‘correction term’ may be the ‘general solution’ for corresponding homogenous equation. Even if the equation is homogenous, you can try the above methods of obtaining ‘steady state’ or ‘one dimensional’ solution. The ‘complete solution’ will be the sum of ‘steady state + transient’ solution OR ‘one dimensional solution + correction for presence of plates’ (for example). Always make sure that the ‘correction term’ goes to zero in the appropriate limit (eg time --> infinity, or the ‘width of the channel --> infinity) If the scales are finite, one shouldn’t jump into separating variables right away. You should see if the eqn is homogenous or not. Even if it is homogenous, you should try simpler solutions for simpler equation and ‘correction terms’ for the additional complexity. Keeping the physical nature of the problem in mind will help you identify the method.

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**Guidelines for solving PDE in Momentum Transfer**

Other relevant Information: Problems in Cartesian coordinates tend to give Cosine/ Sine series solution. In cylindrical coordinates, Bessel functions. In spherical coordinates, Legendre functions When you attempt a ‘complete solution’ as ‘steady state+ transient’ (OR ‘one dimensional + correction’), make sure that you also translate the boundary conditions correctly While solving for the ‘transient’ or ‘correction’ terms, you may encounter a situation where you have to choose an arbitrary constant (either positive or negative or zero). Usually the constant will not be zero. Choose the constant as positive or negative, depending on the boundary conditions (otherwise, you will proceed only to realize that it will not work). The last point, about choosing the constant correctly will become clear when you understand how the solution evolves. In case of 2D steady state problems, try solving the problem by first taking ‘x’ direction as infinite plate and ‘y’ direction as ‘correction’. You will get a solution. Next try by taking ‘y’ as infinite plate. Note which choice of constant gave correct answer in each case. You will understand how to choose the constant. If you choose the constant incorrectly, you will find out it doesn’t work; one or more of the boundary conditions will not be satisfied.

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**Stoke’s first problem (Please refer to BSL for solution)**

This is fairly straight forward and the solution is given in detail in BSL (and perhaps in many other books). We will go over other problems. Note that the concept of ‘combination of variables’ is applied here to solve the PDE. This is a useful technique particularly when one of the BC involved is at ‘infinity’.

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**N-S Equation: Example: Steady state flow in Rectangular channel**

Steady state in Rectangular channel: pressure driven flow, incompressible fluid Vy = Vz =0 Vx is function of y and z gravity has no component in x direction When you look at pressure driven flow in a rectangular channel, first think of what may be ‘intuitive’ solutions. Can we use hydraulic diameter and then use the Hagen-Poiseulle’s equation? That is one method worth trying since it is easy. We understand that this may give pressure drop vs flow rate, but we don’t expect it to satisfy NS eqn. How about a bit more sophistication? Can we use the result of one dimensional flow (between plates) and ‘intelligently’ combine with other direction? We find that this also doesn’t satisfy NS equation (although it will satisfy the Boundary Conditions). We will try actually solving NS eqn. Since none of the boundary conditions are at ‘infinity’, we should try using separation of variables method. 1 Method employed: Find a particular solution satisfying above equation; then find a general solution satisfying following differential eqn 3 2

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**N-S Equation: Example; Rectangular channel**

Hint: To obtain a physically meaningful format, we can take particular solution to resemble one dimensional flow (when b goes to infinity) 4 Note: Check that the above solution is a valid particular solution Before trying to get general solution, write down the boundary conditions for the over all solution Vx Notice: In all these problems, always write down the boundary conditions first. 5

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**N-S Equation: Examples; rectangular channel**

Translate that to get the boundary conditions for Vx-general We know 3 Hence, from equation 5, 6 Use separation of variables method And remember to translate the boundary conditions whenever you change the solution format ‘artificially’. Eqn 7a 2 implies 7b

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**N-S Equation: Examples; rectangular channel**

implies Since LHS is only a function of y and RHS is fn of z, both must be equal to a constant We say 8 Note: Why do we say , why not ? What will happen if you try that? Or if we say ? In any case, the chosen constant leads to The rest is boring mathematics (or interesting mathematics; depends on your point of view) 9 From 6 implies and implies

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**N-S Equation: Examples; rectangular channel**

Hence, substituting in 9 7a Now, from 6 implies Using superposition principle Yes, you are supposed to remember what the superposition principle is. Again, from 6 implies

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**N-S Equation: Examples; rectangular channel**

Using Fourier cosine expansion for an even function we can find Kn Equating the co-efficients, we get And how to express periodic functions in Fourier series. Remember even function cosine series and odd function sine series ? Hence, general solution part is

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**N-S Equation: Examples; rectangular channel**

“Complete” solution for the original problem is given by Note: When “b” goes to infinity, the ‘correction’ part goes to zero Finally verify that the ‘correction’ part will go to zero at the appropriate limit

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**N-S Equation: Other examples**

To determine the velocity profile in a rectangular channel, where the top plate is moving at a constant velocity of V-zero, under steady state conditions Try out a solution of the form “V-parallel-plate + V-correction” Use separation of variable techniques, to determine V-correction What happens if you try separation of variable in the first place? To determine the unsteady state solution for a flow in a cylindrical pipe, caused by sudden application of pressure Try a solution of the form ‘Steady state + Transient’, just like the one we saw for flow between parallel plates You will get Bessel Equations. Just like we represented functions in rectangular coordinates by sine and cosine functions, we can represent functions in cylindrical co ordinates by Bessel functions, because they are orthogonal. Outline for solutions for some other ideal cases are given here. In all cases, verify that the ‘correction’ or transient term will go to zero at the appropriate limit

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