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## STEADY-STATE, MULTI-DIMENSIONAL CONDUCTION

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**STEADY-STATE, MULTI-DIMENSIONAL**CONDUCTION • Analytical Method: • Separation of Variables • Conduction Factor and Dimensionless Conduction Heat Rate • Numerical Method: • Finite Difference Method • Finite Volume Method**Eigenvalue problem in the direction of homogeneous boundary**condition Analytical Method Separation of Variables + Boundary conditions 1) Equation : linear and homogeneous 2) Boundary condition : homogeneous at least in one direction**over**or boundary conditions : Sturm-Liouville Theory Sturm-Liouville System : Hermitian operator (self-adjoint operator) w(x) : weighting function Corresponding eigenvalue problem**boundary terms**= boundary terms thus Definition of inner product : formally self-adjoint operator**i)**ii) when or and can be finite. iii) when either at a or b boundary terms homogeneous boundary conditions**When**Ex) self-adjoint operator**: at least piecewise continuous function**: eigenfunction Properties of Hermitian operator • eigenvalues : all real • eigenfunctions : orthogonal • eigenfunctions : a complete set orthogonality: : least square convergence**Example : Laplace equation**y x assume then Ex) 1) in (x,y) coordinate system (Cartesian) with homogeneous boundary conditions in the x direction This is a special case of S-L system with eigenfunctions : trigonometric functions, sinx or cosx**r**z assume , then or or This is a special case of S-L system with Remark : transform with Bessel equation : Ex) 2) in (r,z) coordinate system (cylindrical) with homogeneous boundary conditions in the r direction eigenfunctions : Bessel functions**r**q with homogeneous boundary conditions in the q direction assume equation for or with or This is a special case of S-L system with Legendre equation : Remark 3) in (r,q) coordinate system (spherical) eigenfunctions : Legendre functions**T2**T1 T1 T1 Let Two dimensional conduction in a thin rectangular plate or a long rectangular rod with no heat generation y W T(x,y) b.c. x 0 L b.c.**y**1 Let W 0 q(x,y) 0 x 0 L 0 or Dividing both sides by XY, = constant**y**1 W 0 q(x,y) 0 x 0 L 0 boundary conditions**For X(x):**b.c. To be non-trivial : eigenvalue : eigenfunction**For Y(y):**b.c. particular solution:**Multiply both sides by**and integrate fromx = 0 toL**Solution:**where**Td**0 0 0 Td Ta Tb Ta 0 0 Tb 0 0 0 = + + + 0 0 0 Tc 0 Tc -T3(x) T0 T0 T0 0 adiabat 0 0 0 0 0 = 0 0 0 0 = 0 0 + 0 + + T1 0 0 0 adiabat 0 0 0 0 -T3(x) Method of Superposition 1) inhomogeneous boundary condition T = T1 + T2 + T3 + T4 2) inhomogeneous equation T = T1 + T2 = T1+ T3 + T4**T(ro,z) = 0**T(r,l) = Tl T(r,0) = 0 or Cylindrical Rod r z ro l boundary conditions r: finite, z:**T(ro,z) = 0**r T(r,l) = Tl T(r,0) = 0 z ro l = finite Boundary conditions = finite**= finite,**Let then or thus as = finite: such that For R(r): b.c.**eigenfunction:**l1ro l2ro l3ro l4ro**For Z(z):**b.c.**Solution:**where**Conduction Shape Factor and**Dimensionless Heat Transfer Rate Conduction Shape Factor S: shape factor S : determined analytically two-dimensional conduction resistance**A**T1 T2 L T2 T1 Ex) Plane wall Cylindrical wall**Conduction shape factors and dimensionless conduction heat**rates for selected systems**or**Dimensionless Heat Transfer Rate Objects at isothermal temperature (T1) embedded with an infinite medium of uniform temperature (T2) Characteristic length Dimensionless conduction heat rate**(m,n+1)**cell (m-1,n) (m+1,n) (m,n) Dy (m,n-1) Dx Numerical Method Finite Difference Method grid discretization node (nodal point)**or**Similarly,**If we neglect (truncation**error) and when**(m,n+1)**(m-1,n) (m+1,n) (m,n) Dy (m,n-1) Dx Finite Volume Method (Energy Balance Method) Assumption of linear temperature profile**Then,**Similarly, If**Convection Boundary Conditions**1) Side (m,n+1) h, T∞ (m,n) (m-1,n) Dy (m,n-1) Dx If or**2) Corner**(m,n) (m-1,n) (m,n-1) Dy Dx If**3) Internal corner**(m,n+1) (m-1,n) (m,n) (m+1,n) Dy (m,n-1) Dx If or**1) Analytical method (matrix inversion) → Cramer’s rule**Ifn = 10, 3×106operations. Ifn = 25, IBM360 1017years 2) Direct (elimination) method (n < 40) Gauss-Jordan elimination method Augmented matrix of operations**3) Gauss-Seidel iteration method (n > 100)**diagonal dominance → sufficient condition for convergence convergence criterion**Dx =**0.25 m Dy = 0.25 m 1 2 1 3 4 3 5 6 5 7 8 7 Example 4.3 Ts = 500 K Fire clay brick 1 m ×1 m k = 1 W/m.K Ts = 500 K Ts = 500 K Find: Temperature distribution and heat rate per unit length T∞ = 300 K h = 10 W/m2.K Air Nodes at the plane surface with convection: When Dx = Dy,**inner nodes:**convection nodes: Node 1: Node 3: Node 5: Node 2: Node 4: Node 6: Node 7: Node 8: Ts = 500 K 1 2 1 3 4 3 Ts = 500 K Ts = 500 K 5 6 5 7 8 7 T∞ = 300 K h = 10 W/m2.K**Using a standard matrix inversion routine, it is a simple**matter to find the inverse of , giving