Lecture Week 4 1

8/12/2019 Lecture Week 4 1

1/21

ME 3345 Heat Transfer

Objective:

Introduction To Steady State 2-D Heat Transfer

Numerical Solutions to Multidimensional Heat

Transfer


2/21

2

2

1T T

tx

2-D and Transient Heat Conduction Equations

How can we solve this equation, given 4 B.C.s?

How can we solve this equation, given

2 B.C.s and the initial condition?

( , )T x y

( , )T x t

Partial differential equations

Parabolic equation

LaPlace Equation (special form of Poisson Equations)

Chapter 4

Chapter 5

2 2

2 2 0

T T

x y

T2


3/21

1. Exact solutions or analytical solutions

- can only be obtained for limited cases with regular B.C.s

- complicated functions and series are involved

- serve as a tool to verify numerical solutions

- excellent sources of analytical solutions in many found in

many texts: (e.g., Carslaw and Jaeger, Conduction of

Heat in Solids, 1959)

2. Graphical methods- physically intuitive and tedious

- limited use for simple geometries only

- Not good for high accuracy calculations

3. Numerical methods

- powerful

- easy to get wrong results: be careful

- 1 & 2 are often used to ensure that the numerical

solutions are correct


4/21

Main points: Isotherms and Heat Flow Lines

FLUX PLOT


5/21

Main points: Symmetry lines

Lines of symmetry are adiabatic.

Symmetry lines are also heat flow lines.

We need only to study 1/8 of the whole

structure to save time.

Long square duct


6/21

2 2

2 2

1

1

1

2

0

(0, )

( , )

( , 0)

( , )

T T

x y

T y T

T L y T

T x T

T x W T T1

T2

T1 T1

L

W

x

y

2 2

2 2 0

(0, ) 0

( , ) 0

( , 0) 0

( , ) 1

x y

y

L y

x

x W

1

2 1

LetT T

T T

2-D steady-state:


7/21

Separation of Variables

Assume ( , ) ( ) * ( )x y f x g y

Hope to obtain equations for ( ) and ( ) separately.f x g y

Find all solutions and the one that meets all the B.C.s.

The Fourier series is often involved.

Let ( , ) ( )* ( )x y X x Y y


8/21

1

( , ) sin sinhnn

n x n yx y C

L L

You can verify that it satisfies the partial different equation

as well as three of the B.C.s. For it to be the solution, wemust have

Solving for Cn, we obtain:

1

( , ) 1 sin sinhnn

n x n W x W C

L L

The solution to the differential equation for this problem is

in the form of:

1

2 1 cos( , ) sin sinh sinh

n

n n x n y n W x y

L Ln L


9/21


10/21

Introduction to Finite Difference Method

Several different numerical methods have been developed.

Finite difference method is more intuitive and easy to apply.

Other methods include

Finite element method

Finite volume method

Boundary element method, etc.


11/21

Numerical method -- continued

Discretization: a nodal network, called mesh orgrid. We canrefine the mesh but we cannot obtain continuous solutions.

Nodal points or simply nodes, nodal property such as nodal

temperature - it is the temperature of the node but it also

represents the temperature around the node (average).

Basic principle: Through approximation, convert the

differential equations to a set of algebraic equations that can be

solved numerically.

A coarse mesh A fine mesh


12/21

,m nT = average temperature around nodeP(m, n).


13/21

Discretization of a Continuous Function

1

1/ 2

m m

m

T TdT

dx x

1

1/ 2

m m

m

T TdTdx x

21/ 2 1/ 2 1 1

2 2

2

( )

m m m m m

m

T T

x x T T Td T

xdx x

T

x

m

m+1

m+1/2m 1m 1/2

xx


14/21

2 2

2 2 0

T T

x y

21, 1, ,

2 22

( )

m n m n m n

m

T T TTx x

2, 1 , 1 ,

2 2

2

( )

m n m n m n

m

T T TT

y y

2-D Steady State

If , then we obtainx y

1, 1, , 1 , 1 ,4 0m n m n m n m n m nT T T T T

Finite difference equation for (m, n)

UseNlinear algebraic equations to solveNunknowns.


15/21

The Energy Balance Method

m,n

m,n 1

m,n+1

m+1,nm 1, n

x

y

y

x

m,n m+1,n

x

y

m,n m+1,n

qin

1yk

xqin = (Tm+1,nTm,n)


16/21

m,n

m,n 1

m,n+1

m+1,nm 1, n

x

y

y

x

2-D Steady State with Heat Generation

0in gE E

1, , 1, ,

, 1 , , 1 ,

( ) ( )

( ) ( )

0

m n m n m n m n

m n m n m n m n

y y

k T T k T T x x

x xk T T k T T

y y

q x y

m,n

m+1,n

1ykx

1xk

y

m,n+1

m 1, n

m,n 1


17/21

21, 1, , 1 , 1 ,

( )4 0m n m n m n m n m nq xT T T T T

k

If ,x y

Corner Nodes and others -- Read pp. 200 - 205


18/21

What do we end up with is a set of linear algebraic equations.

Assume there areNunknown nodal temperatures.

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

.........

.........

........

.........

N N

N N

N N NN N N

a T a T a T C

a T a T a T C

a T a T a T C

1

, 1, 2,...,N ij j ij

a T C i N

1 1[ ] [ ] [ ]N N N NA T C

11 12 1 1 1

21 22 2 2 2

1 2

... ...

... ...

. .... ... ... . .

... ...

N

N

N N NN N N

a a a T C

a a a T C

a a a T C

1[ ] [ ] [ ]T A C

Matrix inversion solution:


19/21

1. Energy balance

2. Grid-Independent Study

3. Compare with exact solution if available


20/21


21/21

Documents

Lecture Week 4 1