Module 4 - Multi Dimensional Steady State Heat Conduction PDF

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10/09/2020

Module 4 Mul Dimensional Steady State Heat Conduction

Multidimensional Heat Transfer Heat Diffusion Equation

c p

2 T 2 T 2T T 2 + + = k( ) + q = k  T + q y 2 x 2 t z 2

❑ This equation governs the Cartesian, temperature distribution for a three-dimensional unsteady, heat transfer problem involving heat generation. ❑ For steady state / t = 0 ❑ No generation q = 0 ❑ To solve for the full equation, it requires a total of six boundary conditions: two for each direction. Only one initial condition is needed to account for the transient behavior.

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Two-Dimensional, Steady State Case For a 2 - D, steady state situation, the heat equation is simplified to 2 2  T T = 0, it needs two boundary conditions in each direction. + x 2 y 2

There are three approaches to solve this equation: ❑ Numerical Method: Finite difference or finite element schemes, usually will be solved using computers. ❑ Graphical Method: Limited use. However, the conduction shape factor concept derived under this concept can be useful for specific configurations. (see Table 4.1 for selected configurations) ❑ Analytical Method: The mathematical equation can be solved using techniques like the method of separation of variables. (refer to handout) Pradip Dutta

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Conduction Shape Factor ❑This approach applied to 2-D conduction involving two

isothermal surfaces, with all other surfaces being adiabatic. The heat transfer from one surface (at a temperature T1) to the other surface (at T2) can be expressed as: q=Sk(T1-T2) where k is the thermal conductivity of the solid and S is the conduction shape factor. ❑ The shape factor can be related to the thermal resistance: q=Sk(T1-T2)=(T1-T2)/(1/kS)= (T1-T2)/Rt where Rt = 1/(kS) ❑ 1-D heat transfer can use shape factor also. Ex: Heat transfer inside a plane wall of thickness L is q=kA(DT/L), S=A/L Pradip Dutta

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Example ❑ An Alaska oil pipe line is buried in the earth at a depth of 1 m. The horizontal pipe is a thin-walled of outside diameter of 50 cm. The pipe is very long and the averaged temperature of the oil is 100C and the ground soil temperature is at -20 C (ksoil=0.5W/m.K), estimate the heat loss per unit length of pipe. T2 z=1 m

From Table 8.7, case 1. L>>D, z>3D/2 2L 2 (1) = = 3. 02 ln(4z / D ) ln( 4 / 0.5) q = kS (T1 − T2 ) = (0 .5 )(3.02 )(100 + 20) = 181. 2(W ) heat loss for every meter of pipe

S=

T1 Pradip Dutta

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Example (contd...) If the mass flow rate of the oil is 2 kg/s and the specific heat of the oil is 2 kJ/kg.K, determine the temperature change in 1 m of pipe length. q 181.2  P DT , DT = q = mC = = 0. 045( C )  P 2000 * 2 mC

Therefore, the total temperature variation can be significant if the pipe is very long. For example, 45C for every 1 km of pipe length. ❑ Heating might be needed to prevent the oil from freezing up. ❑ The heat transfer can not be considered constant for a long pipe

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Example (contd...)

Ground at -20C

Heat transfer to the ground (q)

mC  P (T + dT )

mC  PT Length dx

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Example (contd...) Heat Transfer at section with a temperature T(x) 2k(dx) (T + 20 ) = 1.51(T + 20 )(dx ) q= ln(4z / D)  P T − q = mC  P ( T + dT ) Energy balance: mC dT dT + 1.51( T + 20) = 0, = − 0.000378 dx, integrate T + 20 dx T (x ) = −20 + Ce −0 .000378 x , at inlet x = 0, T(0) = 100 C, C = 120  P mC

T(x) = -20 + 120 e−0 .000378 x

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Example (contd...)

100

50 T( x ) 0

50

0

1000

2000 3000

400 05000

❑ Temperature drops exponentially from the initial temp. of 100C ❑ It reaches 0C at x=4740 m, therefore, reheating is required every 4.7 km.

x

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Numerical Methods ❑ Due to the increasing complexities encountered in the development of modern technology, analytical solutions usually are not available.

For these problems, numerical solutions obtained

using high-speed computer are very useful, especially when the geometry of the object of interest is irregular, or the boundary

conditions are nonlinear.

In numerical analysis, two different

approaches are commonly used: The finite difference and the finite element methods. In heat transfer problems, the finite difference method is used more often and will be discussed here. Pradip Dutta

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Numerical Methods (contd…) ❑ The finite difference method involves: ⬧ Establish nodal networks ⬧ Derive finite difference approximations for the governing equation at both interior and exterior nodal points ⬧ Develop a system of simultaneous algebraic nodal equations ⬧ Solve the system of equations using numerical schemes

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The Nodal Networks ❑ The basic idea is to subdivide the area of interest into subvolumes with the distance between adjacent nodes by Dx and Dy as shown. If the distance between points is small enough, the differential equation can be approximated locally by a set of finite

difference equations. Each node now represents a small region where the nodal temperature is a measure of the average temperature of the region.

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The Nodal Networks (contd…) Example Dx

m,n+1

m-1,n

m,n

m+1, n

Dy

m,n-1 m+½,n m-½,n intermediate points

x=mDx, y=nDy Pradip Dutta

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Finite Difference Approximation Heat Diffusion Equation: 2 T +

q  = , k  t

k is the thermal diffusivity  C PV  = 0,  2T = 0 No generation and steady state: q t First, approximated the first order differentiation where  =

at intermediate points (m+1/2,n) & (m-1/2,n) − Tm, n T T DT  = m +1,n x ( m+ 1/ 2, n) Dx ( m+ 1/ 2, n) Dx T DT  x ( m− 1/ 2, n) Dx Pradip Dutta

= ( m− 1/ 2, n)

Tm ,n − Tm −1,n Dx

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Finite Difference Approximation (contd...) Next, approximate the second order differentiation at m,n 2 T x 2 2 T x 2



 T /  x m +1/ 2,n −  T /  x m−1/ 2,n Dx

m, n

 m, n

T m+ 1, n + T m− 1, n − 2T m, n (Dx )2

Similarly, the approximation can be applied to the other dimension y 2 T y 2



T m, n+1 + T m, n−1 − 2T m, n

m, n

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Finite Difference Approximation (contd...) T m+1, n + T m−1, n − 2T m, n T m, n+1 + T m, n−1 − 2T m, n   2T  2T  +  x 2 + y 2   ( Dx )2 ( Dy )2   m, n To model the steady state, no generation heat equation: 2T = 0 This approximation can be simplified by specify D x=D y and the nodal equation can be obtained as Tm+1, n + T m−1, n + Tm, n+ 1 + Tm, n− 1 − 4Tm, n = 0 This equation approximates the nodal temperature distribution based on the heat equation. This approximation is improved when the distance between the adjacent nodal points is decreased: Since lim( Dx → 0) Pradip Dutta

DT T DT T , lim( Dy → 0) = = Dx x Dy y HMT/M4/V1/May 05

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A System of Algebraic Equations ❑ The nodal equations derived previously are valid for all interior points satisfying the steady state, no generation heat equation. For each node, there is one such equation. For example: for nodal point m=3, n=4, the equation is T2,4 + T4,4 + T3,3 + T3,5 - 4T3,4 =0 T3,4=(1/4)(T2,4 + T4,4 + T3,3 + T3,5) ❑ Nodal relation table for exterior nodes (boundary conditions) can be found in standard heat transfer textbooks (Table 4.2 of our textbook). ❑ Derive one equation for each nodal point (including both interior and exterior points) in the system of interest. The result is a system of N algebraic equations for a total of N nodal points. Pradip Dutta

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Matrix Form The system of equations: = C1 a 11T1 + a 12T 2 + + a 21T1 + a 22T 2 +

+

a N1T1 + a N 2T2 +

+

= C2

N

N

= CN

A total of N algebraic equations for the N nodal points and the system can be expressed as a matrix formulation: [A][T]=[C]

 a11 a12 a a22 21 where A=     a N1 a N 2 Pradip Dutta



 T1   C1  T C   , T =  2 , C =  2           T N  C N HMT/M4/V1/May 05

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Numerical Solutions ❑Matrix form: [A][T]=[C]. From linear algebra: [A]-1[A][T]=[A]-1[C], [T]=[A]-1[C] where [A]-1 is the inverse of matrix [A]. [T] is the solution vector. ❑ Matrix inversion requires cumbersome numerical computations and is not efficient if the order of the matrix is high (>10)

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Numerical Solutions (contd…) ❑ Gauss elimination method and other matrix solvers are usually available in many numerical solution package. For example, “Numerical Recipes” by Cambridge University Press or their web source at www.nr.com. ❑ For high order matrix, iterative methods are usually more efficient. The famous Jacobi & Gauss-Seidel iteration methods will be introduced in the following.

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Iteration General algebraic equation for nodal point: i −1

 a ijT j + a iiTi + j= 1

N

a T ij

j

Replace (k) by (k-1) for the Jacobi iteration

= Ci ,

j = i+ 1

(Example : a31T1 + a32T2 + a33T3 +

+

= C1, i =3)

Rewrite the equation of the form: N a C i −1 a Ti( k ) = i −  ij T j( k ) −  ij T j( k −1) a ii j= 1 aii j= i +1 aii

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Iteration (contd…) ❑ (k) - specify the level of the iteration, (k-1) means the present level and (k) represents the new level. ❑ An initial guess (k=0) is needed to start the iteration. ❑ By substituting iterated values at (k-1) into the equation, the new values at iteration (k) can be estimated ❑ The iteration will be stopped when maxTi(k)-Ti(k-1), where  specifies a predetermined value of acceptable error

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Example Solve the following system of equations using (a) the Jacobi methods, (b) the Gauss Seidel iteration method.  4 2 1  X   11  −1 2 0   Y  =  3        2 1 4  Z  16

Reorganize into new form: 1 11 1 - Y- Z 4 4 2 3 1 Y = + X+0*Z 2 2 1 1 Z = 4- X- Y 4 2 X=

4 X + 2 Y + Z = 11, − X + 2Y + 0 * Z = 3, 2 X + Y + 4 Z = 16

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Example (contd…) ❑ (a) Jacobi method: use initial guess X0=Y0=Z0=1, ❑ stop when maxXk-Xk-1,Yk-Yk-1,Zk-Zk-1  0.1 ❑ First iteration: ❑ X1= (11/4) - (1/2)Y0 - (1/4)Z0 = 2 ❑ Y1= (3/2) + (1/2)X0 = 2 ❑ Z1= 4 - (1/2) X0 - (1/4)Y0 = 13/4

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Example (contd...) Second iteration: use the iterated values X 1=2, Y1=2, Z1=13/4 X2 = (11/4) - (1/2)Y1 - (1/4)Z1 = 15/16 Y2 = (3/2) + (1/2)X1 = 5/2 Z2 = 4 - (1/2) X1 - (1/4)Y1 = 5/2 Converging Process: 13 15 5 5 7 63 93 133 31 393  [1,1,1],  2,2,  ,  , ,  ,  , ,  ,  , , 4 16 2 2 8 32 32 128 16 128          519 517 767   512 , 256 , 256  . Stop the iteration when max X 5 − X 4, Y 5 − Y 4, Z 5 − Z 4  0.1

Final solution [1.014, 2.02, 2.996] Exact solution [1, 2, 3] Pradip Dutta

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Example (contd...) (b) Gauss-Seidel iteration: Substitute the iterated values into the iterative process immediately after they are computed.

Use initial guess X 0 = Y 0 = Z 0 = 1

11 1 1 3 1 1 1 − Y − Z, Y = + X , Z = 4 − X − Y 4 2 4 2 2 2 4 11 1 1 1 0 0 Immediate substitution First iteration: X = − (Y ) − ( Z ) = 2 4 2 4 3 1 3 1 5 Y 1 = + X 1 = + (2) = 2 2 2 2 2 1 1 1 1 5 19 Z1 = 4 − X 1 − Y 1 = 4 − (2) −   = 2 4 2 42 8 X=

5 19 29 125 783  1033 4095 24541  Converging process: [1,1,1], 2, ,  ,  , , , , , 2 8    32 64 256 1024 2048 8192  The iterated solution [1.009, 1.9995, 2.996] and it converges faster Pradip Dutta

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Numerical Method (Special Cases) ❑ For all the special cases discussed in the following, the derivation will be based on the standard nodal point configuration as shown to the right.

A

m,n+1 q1

m-1,n q2

 Symmetric case: symmetrical relative to the A-A axis. In this case, Tm-1,n=Tm+1,n Therefore the standard nodal equation can be written as

m,n

m+1, n

q4 q3

m,n-1 A

axis A-A

Tm+1, n + Tm−1, n + Tm, n+1 + Tm, n−1 − 4 Tm, n = 2 Tm +1,n + Tm ,n +1 + Tm ,n −1 − 4 Tm, n = 0 Pradip Dutta

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Special cases (contd...)  Insulated surface case: If the axis A-A is an insulated wall, therefore there is no heat transfer across A-A. Also, the surface area for q1 and q3 is only half of their original value. Write the energy balance equation (q2=0): Insulated surface

A

m,n+1 q1

m,n

m+1, n

q4 q3

m,n-1 A

q1 + q3 + q4 = 0

− Tm, n − Tm, n −T T Dx T  Dx  T + k   m, n−1 + k Dy m+1, n m, n = 0 k   m, n+1 Dy Dy Dx 2  2  2Tm +1,n + Tm ,n +1 + Tm ,n −1 − 4Tm ,n = 0 This equation is identical to the symmetrical case discussed previously Pradip Dutta

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Special cases (contd...)  With internal generation G=gV where g is the power generated per unit volume (W/m3). Based on the energy balance concept:

q1 + q 2 + q 3 + q 4 + G q1 + q 2 + q 3 + q 4 + g (Dx )(Dy )(1) = 0 Use 1 to represent the dimension along the z-direction. k (T m+1,n + T m− 1,n + T m ,n+ 1 + T m,n−1 − 4T m, n ) + g (Dx ) = 0 2

Tm+1, n + Tm−1, n + Tm, n+1 + Tm, n−1 − 4T m, n +

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Special cases (contd...)  Radiation heat exchange with respect to the surrounding (assume no convection, no generation to simplify the derivation). Given surface emissivity , surrounding temperature Tsurr. qrad

Tsurr

From energy balance concept: q2+q3+q4=qrad

m,n m-1,n q2

q3

q4

m+1,n

+

+

=

m,n-1

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Special cases (contd...) − Tm ,n −T − Tm ,n T Dy T  Dy  T + k ( Dx ) m ,n−1 m ,n + k   m+ 1,n =  ( Dx ) Tm ,n 4 − Tsurr 4 k   m −1,n Dx Dy Dx  2   2 

(

)

4 k (Tm −1,n + Tm +1,n + 2Tm, n −1 − 4Tm, n ) = 2 ( Dx ) (T m4, n − T surr )

Tm −1, n + Tm +1, n + 2Tm, n −1 − 4Tm, n −

2 (Dx ) 4  (Dx ) 4 T m, n = − 2 T surr k k

Non-linear term, can solve using the iteration method

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