abstrac1coutte flow with heat transfer: numerical studies

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    Abstract

    This project work report provides a full solution

    of simplified Navier-Stokes equations for the

    Incompressible Fully Developed Couette Flow

    with heat transfer.

    The known analytical solution to the problem is

    compared with the numerical solution. For

    discrete problem formulation implicit Crank-

    Nicolson method is used. Finally, the system of

    equation (TriDiagonal) is solved with Thomas

    algorithm TDMA (TriDiagonal matrix algorithm).

    Results are compared with the known analytical

    solution to the problem.

    1 Introduction

    Figure-1 shows the main problem. There is a

    viscous Fully Developed flow between two

    parallel plates. Upper plate is moving in x-

    direction with constant velocity(U). Lower one

    is stationary. Temperature of the lower and

    upper plates are T0and T1, respectively.

    Figure-1: Schematic representation of Fully

    Developed Couette Flow with plates at different

    temperature problem

    2 Fundamental Equations

    Continuity Equation:

    (mass conservation equation)

    1)

    X-Momentum equation:

    2)

    Y-Momentum Equation:

    3)

    Energy Equation:

    4)

    where, = viscous dissipation term

    (5)

    Assumptions:

    1. Steady fully developed flow, which meansvelocity and temperature profiles do not change

    in the flow direction.

    2. Zero pressure gradient, zero body forces

    3. Laminar, Incompressible flow

    4. Constant properties of the fluid (, cp, , k)

    with above assumptions,

    (6)

    Therefore, from equation(1), ; v= constant;Since v= constant and v=0 at bottom wall,

    v=0 everywhere

    X-momentum equation reduces to,

    (7)

    Y-momentum equation reduces to,

    Hydrostatic pressure gradient would exist if

    the body forces were present

    Energy Equation will become as,

    (8)

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    Now non-dimensionalize momentum and

    energy equations using the following

    combinations:

    ; ;

    (9)

    Non-dimesionalized momentum equation

    becomes:

    (10)

    Non-dimesionalized momentum equation

    becomes:

    (11)

    where,

    Br= Pr * Ec;

    where, Pr= Prandtl Number =

    Prandtl Number is fluid property,

    and Ec = Eckert Number =

    Boundary Conditions:

    The solution to the momentum and energy

    equation will give the following,

    (12) (13)

    When Ec Pr

    The Non dimensional temperature profile

    becomes linear. When Pr is of the order of

    unity, small Ec implies dissipation effects

    can be ignored.

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    4 Numerical Solution

    4.1 Velocity Profile

    As we obtained in the last section,

    X-momentum equation (10),

    (14)

    with boundary condition,

    Finite-Difference Representation

    In our solution of equation (14), we will use

    Crank-Nicolson Scheme, so discrete

    representation of that equation can be written

    as:

    (15)

    (16)

    with j=1,2,...,n (index 1 is at the bottom plate

    and index n is at the upper plate)

    and (points 0 and n+1 are imaginary points)

    and

    4.2 Temperature Profile

    As we obtained in the last section,

    Energy equation (11),

    (17)

    We know that , and therefore

    Again using Crank-Nicolson Scheme, discrete

    representation of equation(17) can be written

    as:

    (18)

    (19)

    with j=1,2,...,n (index 1 is at the bottom plate

    and index n is at the upper plate)

    and (points 0 and n+1 are imaginary points)

    and

    5 Solving The System of Linear Equations

    The system of equations (16) and (18) can

    either be solved by Gauss elimination method

    or Thomas method (TDMA- TriDiagonal Matrix

    Algorithm). I have used Thomas method.

    The discretization equations can be written as

    (20)

    for i=1,2,3,...,N. Thus variable is related to theneighboring variables and . To accountfor the special form of the boundary-point

    equations, let us set

    and ,

    so that the variables and will not haveany meaningful role to play.

    Also

    and

    is given so we have

    = given value of = given value of

    Now for the matter of forward substitution, we

    seek a relation

    (21)after we have just obtained

    (22)

    Substituting equation (22) into equation (20)

    leads to,

    (23)

    which can be rearranged to look like equation

    (20). In other words, the coefficients and then stand for

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    (24-a)

    (24-b)These are recurrence relations.

    Now set ,Now we are in position to start the back

    substitution via equation (21).

    Results

    Dimensionless Velocity Profile

    Dimensionless Temperature profile

    Effect of Eckert number on Dimensionless

    temperature Profile

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