CHAPTER 3 TWO-DIMENSIONAL STEADY STATE CONDUCTION 3.1 The Heat Conduction Equation • Assume: Steady state, isotropic, stationary, k = = constant (3.1)
• Special case: Stationary material, no energy generation (3.2) • Cylindrical coordinates: (3.3) Eq. (3.1), (3.2) and (3.3) are special cases of
3.2 Method of Solution and Limitations (3.4) Eq. (3.4) is homogenous, second order PDE with variable coefficients 3.2 Method of Solution and Limitations • Method: Separation of variables • Basic approach: Replace PDE with sets of ODE
• The number of sets depends on the number of independent variables • Limitations: (1) Linear equations • Examples: Eqs. (3.1)-(3.4) are linear (2) The geometry is described by an orthogonal coordinate system. Examples: Rectangles, cylinders, hemispheres, etc.
3.3 Homogeneous Differential Equations and Boundary Conditions • Example: Eq. (3.1): Replace T by cT and divide through by c NH (a)
NH Example: Boundary condition (b) Replace T by cT (c) • Simplest 2-D problem: HDE with 3 HBC and 1 NHBC • Separation of variables method applies to NHDE and NHBC
3.4 Sturm-Liouville Boundary Value Problem: Orthogonality • PDE is split into sets of ODE. One such sets is known as Sturm-Liouville equation if it is of the form (3.5a) Rewrite as (3.5b)
• w(x) = weighting function, plays a special role where (3.6) NOTE: • w(x) = weighting function, plays a special role • Equation (3.5) represents a set of n equations corresponding to n values of • The solutions are known as the characteristic functions
• Important property of the Sturm-Liouville problem: orthogonality • Two functions, and are said to be orthogonal in the interval a and b with respect to a weighting function w(x), if (3.7) • The characteristic functions of the Sturm-Liouville problem are orthogonal if:
(1) p(x) , q(x) and w(x) are real, and (2) BC at x = a and x = b are homogeneous of the form (3.8a) (3.8b) (3.8c)
Special Case: If p(x) = 0 at x = a or x = b , these Special Case: If p(x) = 0 at x = a or x = b , these conditions can be extended to include (3.9a) and (3.9b) (3.9a) and (3.9b) are periodic boundary conditions. • Physical meaning of (3.8) and (3.9)
• Relationship between the Sturm-Liouville problem, • Relationship between the Sturm-Liouville problem, orthogonality, and the separation of variables method: PDE + separation of variables ® 2 ODE One of the 2 ODE = Sturm-Liouville problem Solution to this ODE = n f = orthogonal functions
3.5 Procedure for the Application of Separation of Variables Method Example 3.1: Steady state 2-D conduction in a rectangular plate Find T (x,y) 2 HBC x y 1 . 3 Fig. T = f(x) T = 0 (1) Observations • 2-D steady state • 4 BC are needed • 3 BC are homogeneous
(2) Origin and Coordinates • Origin at the intersection of the two simplest BC • Coordinates are parallel to the boundaries (3) Formulation (i) Assumptions (1) 2-D (2) Steady (3) Isotropic (4) No energy generation (5) Constant k
(ii) Governing Equations (3.2) (iii) Independent Variable with 2 HBC: x- variable (iv) Boundary Conditions T = 0 2 HBC x y 1 . 3 Fig. T = f(x) (1) H (2) H (3) H
(4) Solution (4) non-homogeneous (i) Assumed Product Solution (a) (a) into eq. (3.2) (b)
(c) (d) where is known as the separation constant
• The separation constant is squared NOTE: • The separation constant is squared • The constant is positive or negative • The subscript n. Many values: are known as eigenvalues or characteristic values • Special case: constant = zero must be considered • Equation (d) represents two sets of equations (e)
terms (f) • The functions and or characteristic values are known as eigenfunctions (ii) Selecting the sign of the terms Select the plus sign in the equation representing the variable with 2 HBC. Second equation takes the minus sign
equations (g) and (h) become • Important case: equations (g) and (h) become (i)
(iii) Solutions to the ODE (j) (iii) Solutions to the ODE (k) (l) (m) (n)
NOTE: Each product is a solution The complete solution becomes (q) (iv) Applying Boundary Conditions
NOTE: Each product solution must satisfy the BC BC (1) Therefore (r) (s) Applying (r) to solution (k)
Therefore Similarly, (r) and (m) give B.C. (2) Therefore
Applying (k) Therefore (t) Thus (u) Similarly, BC 2 and (m) give
Therefore With the solution corresponding to vanishes. BC (3)
Which gives Results so far: therefore and Temperature solution: (3.10)
Remaining constant B.C. (4) (3.11) To determine from (3.11) we apply orthogonality. (v) Orthogonality • Use eq. (7) to determine
• The function in eq. (3.11) is the solution to (g) • If (g) is a Sturm-Liouville equation with 2 HBC, orthogonality can be applied to eq. (3.11) Return to (g) (g) Compare with eq. (3.5a) (3.5a)
and Eq. (3.6) gives and • BC (1) and (2) are homogeneous of type (3.8a). • The characteristic functions are orthogonal with respect to to Multiply eq. (3.11) by integrate from
(3.12) Interchange the integration and summation, and use Introduce orthogonality (3.7) (3.7)
Apply (3.7) to (3.12) Solve for (3.13) (5) Checking
The solution is (3.10) Dimensional check Limiting check Boundary conditions Differential equations (6) Comments Role of the NHBC
3.6 Cartesian Coordinates: Examples Example 3.3: Moving Plate with Surface Convection y x U ¥ T h , L 3 . Fig. HBC 2 Semi-infinite plate leaves a furnace at Outside furnace the plate exchanges heat by convection. Determine the temperature distribution in the plate.
Solution (1) Observations • Symmetry • At plate is at • NHBC (2) Origin and Coordinates (3) Formulation (i) Assumptions (1) 2-D (2) Steady state
(3) Constant and (4) Uniform velocity (5) Uniform and (ii) Governing Equations Define (a) (b)
(iii) Independent variable with 2 HBC: y- variable (iv) Boundary Conditions y x U ¥ T h , L Fig. 3.3 HBC 2 o (1) (2) (3) (4)
(4) Solution terms (i) Assumed Product Solution (c) (d) (ii) Selecting the sign of the terms
(e) (f) For (g) (h)
(iii) Solutions to the ODE (j) (k) And (l)
Complete solution: (m) (iv) Application of Boundary Conditions BC (1) BC (2) (n)
BC (3) With (3.17) BC (4) (o)
(v) Orthogonality Characteristic Functions: are solutions to equation (f). Comparing (f) with eq. (3.5a) shows that it is a Sturm- and Liouville equation with Thus eq. (3.6) gives and (p) 2 HBC at and are orthogonal
to Multiply both sides of (o) by integrate from and apply orthogonality (q)
(5) Checking Dimensional check Limiting check: If (6) Comments For a stationary plate,
3.7 Cylindrical Coordinates: Examples Example 3.5: Radial and Axial Conduction in a Cylinder o r z ¥ T h , L 2 HBC w Fig. 3.5 Two solid cylinders are pressed co- axially with a force F and rotated in opposite directions. Coefficient of friction is Convection at the outer surfaces.
Solution Find the interface temperature. (1) Observations • Symmetry z ¥ T h , L 2 HBC w Fig. 3.5 • Symmetry • Interface frictional heat = tangential force x velocity • Transformation of B.C., (2) Origin and Coordinates
(3) Formulation (i) Assumptions (1) Steady (2) 2-D (3) Constant and (4) Uniform interface pressure (5) Uniform and (6) Radius of rod holding cylinders is small compared to cylinder radius (ii) Governing Equations
(3.17) (iii) Independent variable with 2 HBC: r- variable (iv) Boundary conditions (1) or finite
(4) Solution (2) (3) (4) (i) Assumed Product Solution r z T h , L ¥ T h , L 2 HBC w Fig. 3.5 (3) (4) (4) Solution (i) Assumed Product Solution
(a) (b) (c) (ii) Selecting the sign of the r-variable has 2 HBC
(d) (e) For (f) (g)
(iii) Solutions to the ODE (d) is a Bessel equation: Since and is real (h) Solutions to (e), (f) and (g): (i)
(iv) Application of Boundary Conditions (j) (k) Complete solution (l) (iv) Application of Boundary Conditions BC (1) and Note:
BC (2) (m) (n) BC (2)
Since BC (3) (3.20)
(v) Orthogonality BC (4) (o) is a solution to (d) with 2HBC in and Comparing (d) with eq. (3.5a) shows that it is a Sturm- Liouville equation with Eq. (3.6): and (p)
and are orthogonal with respect to Applying orthogonality, eq. (3.7), gives (q) Interface temperature: set in eq. (3.20)
(r) (5) Checking Dimensional check: Units of Limiting check: If or then (6) Comments is not always equal to unity.
3.8 Integrals of Bessel Functions (3.21)
Table 3.1 Normalizing integrals for solid cylinders [3] Boundary Condition at (3.8a): (3.8b): (3.8c):
3.9 Non-homogeneous Differential Equations Example 3.6: Cylinder with Energy Generation L a T o r q ¢ z Fig. 3.6 Solid cylinder generates heat at a rate One end is at while the other is insulated. Cylindrical surface is at Find the steady state temperature distribution.
Solution (1) Observations • Energy generation leads to NHDE • Define q ¢ z Fig. 3.6 • Energy generation leads to NHDE • Define to make BC at surface homogeneous • Use cylindrical coordinates (2) Origin and Coordinates
(3) Formulation (i) Assumptions (1) Steady state (2) 2-D (3) Constant and (ii) Governing Equations Define
(iii) Independent variable with 2 HBC (3.22) (iii) Independent variable with 2 HBC (iv) Boundary conditions L a T o r q ¢ z Fig. 3.6 (1) finite (2) (3) (4)
(4) Solution Let: (a) Substitute into eq. (3.22) (b) Split (b): Let (c)
NOTE: • (c) is a HPDE for Therefore (d) • (d) is a NHODE for the • Guideline for splitting PDE and BC: should be governed by HPDE and three HBC. Let take care of the NH terms in PDE and BC
BC (1) finite (c-1) BC (2) (c-2) BC (3) Let (c-3)
Thus (d-1) BC (4) Let (c-4) Thus (d-2) Solution to (d)
(e) (i) Assumed Product Solution (f) (f) into (c), separating variables (g) (h)
(ii) Selecting the sign of the terms (i) (j) For (k)
(l) (iii) Solutions to the ODE (m) (j) is a Bessel equation with (n)
Solution to (k) and (l) (o) (p) Complete Solution: (q) (iv) Application of Boundary Conditions BC (c-1) to (n) and (p)
BC (c-4) to (m) and (o) BC (c-3) to (m) and (o) Equation for or (r) With The solutions to become
(s) BC (d-1) and (d-2) and (t) BC (c-2) (u)
(v) Orthogonality are solutions to equation (i). and Comparing (i) with eq. (3.5a) shows that it is a Sturm- Liouville equation with Eq. (3.6) gives and and 2 HBC at are orthogonal with respect to
Evaluating the integrals and solving for
Solution to (w) (5) Checking Dimensional check: Units of and of in solution (s) are in °C Limiting check: and
3.10 Non-homogeneous Boundary Conditions Method of Superposition: • Decompose problem Example: Problem with 4 NHBC is decomposed into 4 problems each having one NHBC ) ( y g x f L W o q ¢ ¥ T h , 3.7 Fig. DE
= + Solution: 3 2 4 ) ( y g x f L W q ¢ T h , x y L W T h , ) ( f g ¥ = x y L W ¥ T h , 4 + ) ( f 3 g 2