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PROBLEM 1.11 A rectangular plate of length L and height H slides down an inclined surface with a velocity U. Sliding friction results in surface heat flux q o . The front and top sides of the plate exchange heat by convection. The heat transfer coefficient is h and the ambient temperature is T . Neglect heat loss from the back side and assume that no frictional heat is conducted through the inclined surface. Write the twodimensional steady state heat equation and boundary conditions. x h y T W U h T 0 L g (1) Observations. (i) This is a two-dimensional steady state problem. (ii) Four boundary conditions are needed. (iii) Rectangular geometry. (iv) Specified flux at surface (x,0) points inwards. (v) Plate velocity has no effect other than generating frictional heat at the inclined plane. (vi) Use Section 1.6 as a guide to writing boundary conditions. (2) Origin and Coordinates. The origin and Cartesian coordinates x,y are as shown. The coordinates move with the plate. (3) Formulation. (i) Assumptions. (1) Steady state, (2) two-dimensional, (3) no energy generation, (4) all frictional heat is added to plate (inclined surface is insulated) and (5) stationary material (plate does not move relative to coordinates). (ii) Governing Equations. The heat equation in Cartesian coordinates is given by (k T ) (k T ) (k T ) q c p ( T U T V T W T ) x x y y z z t x y z (1.7) where c p is specific heat and is density. The above assumptions give U V W Eq. (1.7) simplifies to q 0 z t ( k T ) ( k T ) 0 x x y y (a) If we further assume constant conductivity, we obtain 2T 2T 0 x2 y2 (iii) Boundary Conditions. Four boundary conditions are needed. They are: (1) Specified flux at boundary (x,0) (2) (b) PROBLEM 1.11 (continued) k T ( x,0) qo y (2) Convection at surface ( x,W ). Pretending that heat flows in the positive y-direction, conservation of energy, Fourier’s law of conduction and Newton’s law of cooling give k T ( x,W ) hT ( x,W ) T y (3) Convection at surface (0,y). Pretending that heat flows in the positive x-direction, conservation of energy, Fourier’s law of conduction and Newton’s law of cooling give k T (0, y ) hT T (0, y ) x (4) Insulated boundary at surface ( L, y). Fourier’s law gives T ( L, y ) 0 x (4) Checking. Dimensional check: Each term in boundary conditions (1), (2) and (3) has units of flux. Limiting check: Boundary conditions (2) and (3) should reduce to an insulated condition if h = 0. Similarly, boundary condition (1) should describe an insulated surface if qo 0. Setting h = 0 in conditions (2) and (3) and qo 0 in (1) gives the description for an insulated surface. (5) Comments. (i) The four boundary conditions are valid for constant and variable conductivity k. (ii) Formulating boundary conditions requires an understanding of what takes place physically at the boundaries. (iii) Plate motion in this example should not be confused with convection velocity appearing in the heat equation (1.7). PROBLEM 2.17 A section of a long rotating shaft of radius ro is buried in a material of very low thermal conductivity. The length of the buried section is L. Frictional heat generated in the buried section can be modeled as surface heat flux. Along the buried surface the radial heat flux is q r and the axial heat flux is q x . The exposed surface exchanges heat by convection with the ambient. The heat transfer coefficient is h and the ambient temperature is T . Model the shaft as semi-infinite fin and assume that all frictional heat is conducted through the shaft. Determine the temperature distribution. (1) Observations. (i) This is a constant area fin. (ii) Temperature distribution can be assumed one dimensional. (iii) The shaft has two sections. Surface conditions are different for each of sections. Thus two equations are needed. (iv) Shaft rotation generates surface flux in the buried section and does not affect the fin equation. h T dx qr x L 0 qx (2) Origin and Coordinates. The origin is selected at one end of the fin and the coordinate x is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area, (4) uniform surface flux in the buried section, (5) Bi 1, (6) constant conductivity, (7) no radiation (8) no energy generation and (9) uniform h and T . (ii) Governing Equations. Since surface conditions are different for the two sections, two fin equations are needed. Let T1 ( x ) = temperature distribution in the buried section, 0 x L T2 ( x ) = temperature distribution in the exposed section, L x 2L The heat equation for the buried section must be formulated using fin approximation. Conservation of energy for the fin element dx gives dq (a) q x qr 2 ro dx q x x dx dx dq qx x dx where q x is the rate of heat conducted in the x-direction, dx given by Fourier’s law dT1 dx Substituting (b) into (a) and rearranging q x k d 2T1 dx 2 2q r 0, kro 0 xL (b) dx 2 roqr qx (c) The heat equation for constant area fin with surface convection is given by equation (2.9) PROBLEM 2.17 (continued) d 2T2 dx 2 hC (T2 T ) 0 , kAc xL (d) where C is the circumferance and Ac is the cross section area given by C 2 ro (e) A ro2 (f) Using (c) and (d) to define m m hC kAc 2h kro (g) Subatituting (e) into (d) d 2T2 2 m 2 (T2 T ) 0 dx (iii) Boundary Conditions. Four boundary conditions are needed. (h) (1) Specified flux at x = 0 k dT1 (0) q x dx (i) (2) Equality of temperature at x = L T1 ( L) T2 ( L) (j) dT1 ( L) dT2 ( L) dx dx (k) T2 () finite (l) (3) Equality of flux at x = L (4) Finite temperature at x (4) Solution. The solutions to (c) is T1 ( x) q r 2 x A1 x B1 kro (m) where A1 and B1 are constants of integration. The solution to (h) is given in Appendix A, equation (A-2b) (n) T2 ( x) T A2 e mx B2 e mx where A2 and B2 are constants of integration. Application of the four boundary conditions gives the four constants q (o) A1 x k B1 T q q r 2 q x 1 q L L 2 r L x kro k m kro k (p) PROBLEM 2.1 Radiation is used to heat a plate of thickness L and thermal conductivity k. The radiation has the effect of volumetric energy generation at a variable rate given by q ( x) qo e bx To L radiation q(x) 0 x where qo and b are constant and x is the distance along the plate. The heated surface at x = 0 is maintained at uniform temperature To while the opposite surface is insulated. Determine the temperature of the insulated surface. (1) Observations. (i) This is a steady state one-dimensional problem. (ii) Use a rectangular coordinate system. (iii) Energy generation is non-uniform. (iv) Two boundary conditions are needed. The temperature is specified at one boundary. The second boundary is insulated. (v) To determine the temperature of the insulated surface it is necessary to determine the temperature distribution in the plate. (2) Origin and Coordinates. A rectangular coordinate system is used with the origin at one of the two surfaces. The coordinate x is oriented as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) stationary and (4) constant properties. (ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives d 2 T q 0 k dx 2 (a) where k = thermal conductivity q = rate of energy generation per unit volume T = temperature x = independent variable Energy generation q varies with location according to q qo e bx (b) where qo and b are constant. Substituting (b) into (a) gives d 2T qo bx e 0 k dx 2 (c) (iii) Boundary Conditions. Since (c) is second order with a single independent variable x, two boundary conditions are needed. (1) Specified temperature at x = 0 T(0) = To (2) The surface at x = L is insulated 2-1 (d) PROBLEM 2.1 (continued) dT ( L ) 0 dx (4) Solution. Separating variables and integrating (c) twice T ( x) qo 2 b k e bx C1 x C2 (e) (f) where C1 and C2 are constants of integration. Application of boundary conditions (d) and (e) gives C1 and C2 q q C1 o e bL , C2 To 2o (g) bk b k Substituting (g) into (f) q (h) T ( x ) 2o 1 e bx bxebL To b k The temperature of the insulated surface is obtained by setting x = L in (h) T ( L) qo 1 (1 bL)e bL To 2 b k (i) Expressing (h) and (i) in dimensionless form gives T ( x) To 1 e bL( x / L ) (bL)( x / L)e bL q o (j) b2k and T ( L) To 1 (1 bL) e bL qo (k) b2k (5) Checking. Dimensional check: Each term in solution (h) has units of temperature. Boundary conditions check: Setting x = 0 in (h) gives T(0) = To . Thus boundary condition (d) is satisfied. To check condition (e) solution (h) is differentiated to obtain q dT 2o be bx be bL dx b k (m) Evaluation (m) at x = L shows that condition (e) is satisfied. Differential equation check: Direct substitution of (h) into equation (c) shows that the solution satisfies the governing equation. Global energy conservation check: The total energy generated in the plate, E g , must be equal to the energy conducted from the plate at x = 0. Energy generated is given by qo (1 e bL ) b 0 0 Energy conducted at x = 0, E (0) , is determined by substituting (m) into Fourier’s law L E g q( x )dx qo L e bx dx 2-2 (k) PROBLEM 2.1 (continued) q dT (0) E (0) k o (1 e bL ) dx b (l) The minus sign indicates that heat is conducted in the negative x-direction. Thus energy is conserved. Limiting check: For the special case of q 0, temperature distribution should be uniform given by T ( x ) To . Setting qo 0 in (h) gives the anticipated result. (6) Comments. (i) The solution to the special case of constant energy generation corresponds to b 0. However, setting b 0 in solution (h) gives 0/0. To obtain the solution to this case it is necessary to set b 0 in differential equation (c) and solve the resulting equation. (ii) The dimensionless solution is characterized by a single dimensionless parameter bL. 2-3 PROBLEM 2.2 Repeat Example 2.3 with the outer plates generating energy at a rate q and no energy is generated in the inner plate. (1) Observations. (i) The three plates form a composite wall. (ii) The geometry can be described by a rectangular coordinate system. (iii) Due to symmetry, no heat is conducted through the center plate. Thus the center plate is at a uniform temperature. (iv) Heat conduction is assumed to be in the direction normal to the three plates. (v) This problem reduces to a single heat generating plate in which one side is maintained at a specified temperature and the other side is insulated. (2) Origin and Coordinates. Since only one plate needs to be analyzed, the origin and coordinate x are selected as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state and (3) constant conductivity of the heat generating plate. (ii) Governing Equations. Let the subscript 2 refer to the heat generating plate. Based on the above assumptions, eq. (1.8) gives To d 2T2 q L2 k2 (a) 0 q dx 2 k 2 k1 x L1 where L2 qq k2 k 2 = thermal conductivity q = rate of energy generation per unit volume To 0 T = temperature x = coordinate (iii) Boundary Conditions. Since (a) is second order with a single independent variable x, two boundary conditions are needed. (1) Specified temperature at x = 0 T2 (0) To (b) dT2 ( L2 ) 0 dx (c) (2) Insulated surface at x L2 where L2 is the thickness of the heat generating plate. (4) Solution. Integrating (a) twice gives T2 ( x ) q 2 x C1 x C 2 2k 2 (d) where C1 and C2 are constants of integration. Application of boundary conditions (b) and (c) gives C1 and C2 2-4 PROBLEM 2.2 (continued) C1 q L2 , C 2 To k2 (e) Substituting (e) into (d) gives the temperature distribution in the heat generating plate q x 2 (2 L2 / x ) 1 To T2 ( x ) 2k 2 (f) To determine the temperature distribution in the center plate we apply continuity of temperature at x L2 T1 ( L2 ) T2 ( L2 ) (g) Noting that the center plate must be at a uniform temperature, evaluating (f) at x L2 and substituting into (g) gives qL22 (h) T1 ( x ) T1 ( L2 ) T2 ( L2 ) To 2k 2 Expressing (f) and (h) in dimensionless form gives T2 ( x) To 1 x 2 ( x / L2 ) 2 L2 q L22 k2 (i) and T2 ( x) To 1 2 q L22 k2 (j) (5) Checking. Dimensional check: Each term in (f) has units of temperature. Boundary conditions check: Substitution of solutions (f) into (b) and (c) shows that the two boundary conditions are satisfied. Differential equation check: Direct substitution of (f) into equation (a) shows that the solution satisfies the governing equation. Global energy conservation check: The total energy generated in the plate, E g , must be equal to the energy conducted from the plate at x = 0. Energy generated is given by E g qAL2 (k) where A is surface area. Energy conducted at x = 0, E (0) , is determined by differentiation (f) and substituting into Fourier’s law dT (0) E (0) k 2 A 2 q AL2 dx (m) The minus sign indicates that heat is conducted in the negative x-direction. Equations (k) and (m) show that energy is conserved. 2-5 PROBLEM 2.2 (continued) Limiting check: If energy generation vanishes, the composite wall should be at a uniform temperature To . Setting q 0 in (f) and (h) gives T1 ( x ) T2 ( x ) To . (6) Comments. (i) An alternate approach to solving this problem is to consider half the center plate and the heat generating plate as a composite wall, note symmetry, write two differential equations and four boundary conditions and solve for T1 ( x ) and T2 ( x). Clearly this is a longer approach to solving the problem. (ii) No parameter characterizes the solution. 2-6 PROBLEM 2.3 A plate of thickness L1 and conductivity k1 moves with a velocity U over a stationary plate of thickness L2 and conductivity k 2 . The pressure between the two plates is P and the coefficient of friction is . The surface of the stationary plate is insulated while that of the moving plate is maintained at constant temperature To . Determine the steady state temperature distribution in the two plates. To x L1 L2 0 U k1 k2 (1) Observations. (i) The two plates form a composite wall. (ii) The geometry can be described by a rectangular coordinate system. (iii) Heat conduction is assumed to be in the direction normal to the two plates. (iv) Since the stationary plate is insulated no heat is conducted through it. Thus it is at a uniform temperature. (v) All energy dissipated due to friction at the interface is conducted through the moving plate. (vi) This problem reduces to one-dimensional steady state conduction in a single plate with specified flux at one surface and a specified temperature at the opposite surface. (vii) Plate motion plays no role in the governing equation since no heat is conducted in the direction of motion. (2) Origin and Coordinates. Since only one plate needs to be analyzed, the origin is selected at the interface and the coordinate x is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant conductivity of the moving plate, (4) no energy generation and (5) uniform energy dissipation at the interface. (ii) Governing Equations. Let the subscript 1 refer to the moving plate. Based on the above assumptions, eq. (1.8) applied to the moving plate gives d 2T1 0 dx 2 (a) (iii) Boundary Conditions. Since (a) is second order with a single independent variable x, two boundary conditions are needed. (1) Specified heat flux at x = 0 . Fourier’s law gives k1 dT1 (0) PU dx (b) T1 ( L1 ) To (c) where k1 = thermal conductivity P = interface pressure U = plate velocity = coefficient of friction (2) Specified temperature at x L1 where L1 is the thickness of the moving plate. 2-7 PROBLEM 2.3 (continued) (4) Solution. Integrating (a) twice gives T1 ( x) C1 x C2 (d) where C1 and C2 are constants of integration. Application of boundary conditions (b) and (c) gives C1 and C2 PU PU L1 C1 , C 2 To (e) k1 k1 Substituting (e) into (d) gives the temperature distribution in the moving plate T1 ( x ) PUL1 k1 1 ( x / L1 ) To (f) To determine the temperature distribution in the stationary plate we apply continuity of temperature at x = 0 T2 (0) T1 (0) (g) Noting that the stationary plate must be at a uniform temperature, evaluating (f) at x = 0 and substituting into (g) gives PUL1 T2 ( x ) T2 (0) T1 (0) To (h) k1 Expressing (f) and (h) in dimensionless form gives T1 ( x) To x 1 PUL1 L1 k1 T1 ( x) To 1 PUL1 k1 (i) (j) (5) Checking. Dimensional check: Each term in (f) has units of temperature Boundary conditions check: Substitution of solutions (f) into (b) and (c) shows that the two boundary conditions are satisfied. Differential equation check: Direct substitution of (f) into equation (a) shows that the solution satisfies the governing equation. Limiting check: If no energy is added at the interface ( 0 or U = 0 or P = 0), the two plates should be at a uniform temperature To . Setting any of the quantities , U or P equal to zero in (f) and (h) gives T1 ( x ) T2 ( x ) To . (6) Comments. (i) An alternate approach to solving this problem is to consider the two plates as a composite wall, write two differential equations and four boundary conditions and solve for T1 ( x ) and T2 ( x). Clearly this is a longer approach to solving the problem. (ii) No parameter characterizes the solution. 2-8 PROBLEM 2.4 A thin electric element is wedged between two plates of conductivities k1 and k 2 . The element dissipates uniform heat flux q o . The thickness of one plate is L1 and that of the other is L2 . One plate is insulated while the other exchanges heat by convection. The ambient temperature is T and the heat transfer coefficient is h. Determine the temperature of the insulated surface for one-dimensional steady state conduction. h,T L1 k1 + L2 k2 x - 0 (1) Observations. (i) The two plates form a composite wall. (ii) The geometry can be described by a rectangular coordinate system. (iii) Heat conduction is assumed to be in the direction normal to the two plates. (iv) Since no heat is conducted through the insulated plate, its temperature is uniform equal to the interface temperature. (v) All energy dissipated by the element at the interface is conducted through the upper plate. (vi) This problem reduces to one-dimensional steady state conduction in a single plate with specified flux at one surface and convection at the opposite surface. (vii) To determine the temperature of the insulated surface it is necessary to determine the temperature distribution in the upper plate. (2) Origin and Coordinates. Since only one plate needs to be analyzed, the origin is selected at the interface and coordinate x is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant conductivity of the upper plate, (4) no energy generation and (5) uniform energy dissipation at the interface. (ii) Governing Equations. Let the subscript 1 refer to the upper plate. Based on the above assumptions, eq. (1.8) applied to the upper plate gives d 2T1 dx 2 0 (a) (iii) Boundary Conditions. Since (a) is second order with a single independent variable x, two boundary conditions are needed. (1) Specified heat flux at x = 0 dT1 (0) qo dx (2) Convection at x L1 . Fourier’s law and Newton’s law of cooling give dT ( L ) k1 1 1 h T1 ( L1 ) T dx k1 (b) (c) where L1 is the thickness of the upper plate. (4) Solution. Integrating (a) twice gives T1 ( x) C1 x C2 2-9 (d) PROBLEM 2.4 (continued) where C1 and C2 are constants of integration. Application of boundary conditions (b) and (c) gives C1 and C2 C1 q qo , C2 o (hL1 / k1 ) 1 T k1 h (e) Substituting (e) into (d) and rearranging gives the temperature distribution in the upper plate T1 ( x) qo hL1 (1 x ) 1 T h k1 L1 (f) To determine the temperature distribution in the insulated plate we apply continuity of temperature at x = 0 T2 (0) T1 (0) (g) Noting that the stationary lower plate must be at a uniform temperature, evaluating (f) at x = 0 and substituting into (g) gives q hL T2 ( x) T2 (0) T1 (0) o 1 1 T (h) h k1 Equation (h) gives the temperature of the insulated surface. Expressing (f) and (h) in dimensionless form gives T1 ( x) T x Bi (1 ) 1 qo L1 h (i) and T1 ( x) T 1 Bi q o (j) h where Bi is the Biot number defined as Bi hL1 k1 (k) (5) Checking. Dimensional check: Each term in (f) has units of temperature. Boundary conditions check: Substitution of solution (f) into (b) and (c) shows that the two boundary conditions are satisfied. Differential equation check: Direct substitution of (f) into equation (a) shows that the solution satisfies the governing equation. Limiting check: (i) If no energy is added at the interface, the two plates should be at a uniform temperature T . Setting q o 0 in (f) and (h) gives T1 ( x) T2 ( x) T . (ii) If h = 0, 2-10 PROBLEM 2.4 (continued) no heat can leave the plates and thus their temperature will be infinite. Setting h = 0 in (f) and (h) gives T1 ( x) T2 ( x) . Qualitative check: Increasing interface energy input increases the temperature level in the two plates. Equations (f) and (h) exhibit this behavior. (6) Comments. (i) An alternate approach to solving this problem is to consider the two plates as a composite wall, write two differential equations and four boundary conditions and solve for T1 ( x) and T2 ( x). Clearly this is a longer approach to solving the problem. (ii) The problem is characterized by a single dimensionless parameter hL1 / k1 which is the Biot number. This parameter appears in problems involving surface convection. 2-11 PROBLEM 2.5 A thin electric element is sandwiched between two plates of conductivity k and thickness L each. The element dissipates a flux q o . Each plate generates energy at a volumetric rate of q and exchanges heat by convection with an ambient fluid at T . The heat transfer coefficient is h. Determine the temperature of the electric element. h,T 0 +L x q _ q L h,T (1) Observations. (i) The two plates form a composite wall. (ii) The geometry can be described by a rectangular coordinate system. (iii) Heat conduction is assumed to be in the direction normal to the two plates. (iv) Symmetry requires that half the energy dissipated by the element at the interface is conducted through the upper plate and the other half is conducted through the lower plate. Thus only one plate needs to be analyzed.(v) The problem reduces to one-dimensional steady state conduction in a single plate with energy generation having specified flux at one surface and convection at the opposite surface. (vi) To determine the temperature of the element it is necessary to determine the temperature distribution in one of the two plates. (2) Origin and Coordinates. The upper plate is considered for analysis. The origin is selected at the upper surface and coordinate x is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant conductivity of the upper plate, (4) uniform energy generation and (5) uniform energy dissipation at the interface. h,T 0 +L (ii) Governing Equations. Let the subscript 1 refer to the upper plate. Based on the above assumptions, eq. (1.8) applied to the upper plate gives d 2T1 q 0 k dx 2 x q _ qo 2 (a) (iii) Boundary Conditions. Since (a) is second order with a single independent variable x, two boundary conditions are needed. (2) Specified heat flux at x = L q dT1 ( L) o dx 2 (2) Convection at x 0 . Fourier’s law and Newton’s law of cooling give dT (0) k 1 h T T1 (0) dx k (b) (c) where L is the thickness of the upper plate. (4) Solution. Integrating (a) twice gives T1 ( x) q 2 x C1 x C 2 2k 2-12 (d) PROBLEM 2.5 (continued) where C1 and C2 are constants of integration. Application of boundary conditions (b) and (c) gives C1 and C2 C1 qo q L q q L , C 2 T o 2k 2h k h (e) Substituting (e) into (d) and rearranging gives the temperature distribution in the upper plate T1 ( x) T q o q L q o q L q 2 x x 2h h k 2k 2k (f) Expressing (f) in dimensionless form gives 2 qo T1 ( x) T x Bi x 1 1 Bi L 2 L q L / h 2q L (g) where Bi is the Biot number defined as Bi hL k (h) To determine the temperature of the electric element set x = L in (g) q T1 ( L) T Bi o 1 Bi 1 q L / h 2q L 2 (i) (5) Checking. Dimensional check: Each term in (g) is dimensionless. Boundary conditions check: Substitution of solution (f) into (b) and (c) shows that the two boundary conditions are satisfied. Differential equation check: Direct substitution of (f) into equation (a) shows that the solution satisfies the governing equation. Limiting check: (i) If no energy is added at the interface and there is no energy generation the two plates should be at a uniform temperature T . Setting qo q 0 in (f) gives T1 ( x) T . (ii) If h = 0, no heat can leave the plate and thus the temperature will be infinite. Setting h = 0 in (f) gives T1 ( x) . Qualitative check: Increasing interface energy input and/or energy generation increases the temperature level in the plate. Equation (f) exhibit this behavior. (6) Comments. (i) An alternate approach to solving this problem is to consider the two plates as a composite wall, write two differential equations and four boundary conditions and solve for T1 ( x ) and T2 ( x). Clearly this is a longer approach to solving the problem. (ii) The solution is characterized by two parameters: the convection parameter Bi and the ratio of energy dissipated in the element to energy generation, q o / q L. 2-13 PROBLEM 2.6 One side of a plate is heated with uniform flux q o while the other side exchanges heat by convection and radiation. Surface emissivity is , heat transfer coefficient h, ambient temperature T , surroundings temperature Tsur , plate thickness L and the conductivity is k. Assume one-dimensional steady state conduction and use a simplified radiation model. Determine: Tsur [a] The temperature distribution in the plate. [b] The two surface temperatures for: L h = 27 W/m2 o C , k = 15 W/m o C , L = 8 cm, = 0.95, q o =19,500 W/m2 , Tsur = 18 o C , T = 22 o C . h,T x 0 k qo (1) Observations. (i) This is a one-dimensional steady state problem. (ii) Two boundary conditions are needed. (iii) Cartesian geometry. (iv) Specified flux at one surface and convection and radiation at the opposite surface. (v) Since radiation is involved in this problem, temperature units should be expressed in kelvin (2) Origin and Coordinates. The origin and coordinate x are shown. (3) Formulation. (i) Assumptions. (1) Steady state, (2) one-dimensional, (3) constant properties, (4) no energy generation, (5) small gray surface which is enclosed by a much larger surface, (6) uniform conditions at the two surfaces and (7) stationary material. (ii) Governing Equations. Based on the above assumptions, the heat Cartesian coordinates is given by d 2T 0 dx 2 equation in (a) (iii) Boundary Conditions. Two boundary conditions are needed. They are: (1) Specified flux at x 0 k d T ( 0) q o dx (b) where k thermal conductivity = 15 W/m o C qo surface flux = 19,500 W/m2 (2) Convection and radiation at x L . Conservation of energy, Fourier’s law of conduction, Newton’s law of cooling and Stefan-Boltzmann radiation law give k dT ( L) 4 h T ( L) T T 4 ( L) Tsur dx where h heat transfer coefficient = 27 W/m2 o C = 27 W/m2 K 2-14 (c) PROBLEM 2.6 (continued) L plate thickness = 8 cm T = ambient temperature = 22 o C = (22 + 273.15) K = 295.15 K Tsur = surroundings temperature = 18 o C = (18 + 273.15) K = 291.15 K emissivity = 0.95 Stefan-Boltzmann constant = 5.67 108 W/m2 K 4 (4) Solution. Integrating (a) twice gives T ( x) C1 x C2 (d) where C1 and C2 are constants of integration. Application of boundary conditions (b) and (c) gives C1 qo k q q 4 qo h o L C 2 T ( o L C 2 ) 4 Tsur k k (e) (f) Equation (f) gives the constant C2 . However, this equation must be solved for C2 by trial and error. (5) Checking. Dimensional Check: Each term in equations (b), (c) and (f) has units of flux. Differential equation check: Solution (d) satisfies the governing equation (a). (6) Computations. Equation (e) gives C1 19,500 (W/m2 ) o 15 ( W/m- C) 1300 o C K 1300 m m Equation (f) gives - 19,500 (W/m2 ) 19,500 (W/m ) 27 ( W/m - K) 0.08 (m) C 2 295.15 (K) 15 ( W/m - K) 2 (0.95)5.67 10 2 8 4 - 19,500 (W/m2 ) 4 4 ( W/m - K ) 0.08 (m) C 2 (291.15) (K ) 15 ( W/m - K) 2 4 Solving this equation for C2 by trial and error gives C 2 761.9 K Substituting C1 and C2 into (d) gives T ( x) 761.9 (K ) 1300(K/m) x Surface temperatures at x = 0 and x = L are determined using (g) 2-15 (g) PROBLEM 2.6 (continued) T (0) 761.9 K T ( L) 657.9 K (7) Comments. (i) A trial and error procedure was required to solve this problem. (ii) Care should be exercised in selecting the correct units for temperature in radiation problems. (iii) With the two temperatures determined a numerical check based on conservation of energy can be made. Energy flux entering plate = Energy flux conducted through the plate = Energy flux leaving Energy flux entering plate = qo 19,500 W/m2 Energy flux conducted through the plate = T ( 0) T ( L ) L/k q Energy flux leaving h o L C 2 T k q o 4 4 ( k L C 2 ) Tsur Numerical computations show that each term in (h) is equal to 19,500 W/m2 . 2-16 (h) PROBLEM 2.7 A hollow shaft of outer radius Rs and inner radius Ri rotates inside a sleeve of inner radius Rs and outer radius Ro . Frictional heat is generated at the interface at a flux q s . At the inner shaft surface heat is added at a flux q i. The sleeve is cooled by convection with a heat transfer coefficient h. The ambient temperature is T . Determine the steady state one-dimensional temperature distribution in the sleeve. h , T Ro Rs R i shaft q s sleeve q i (1) Observations. (i) The shaft and sleeve form a composite cylinder. (ii) Use a cylindrical coordinate system. (iii) Heat conduction is assumed to be in the radial direction only. (iv) At steady state, energy entering the shaft at Ri must be equal to energy leaving at Ro . (v) Frictional energy dissipated at the interface and energy added at Ri must be conducted through the sleeve. (vi) This problem reduces to one-dimensional steady state conduction in a hollow cylinder (sleeve) with specified flux at the inside surface and convection at the outside surface. (2) Origin and Coordinates. The origin is selected at the center and the coordinate r is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant conductivity of the sleeve, (4) no energy generation, and (5) uniform frictional energy at the interface. (ii) Governing Equations. Based on the above assumptions, eq. (1.11) applied to the sleeve gives d (r dT ) 0 (a) dr dr (iii) Boundary Conditions. Since (a) is second order with a single independent variable r, two boundary conditions are needed. (1) Specified heat flux at r Rs . Fourier’s law gives dT ( Rs ) k q sl dr (b) where qsl is the heat flux added to the sleeve at r Rs . This flux is the sum of frictional energy generated at the interface and energy added at Ri . That is qsl qs qo (c) where q o = heat flux leaving the shaft at R s q s = heat flux generated by friction at the interface R s qsl = total heat flux entering the sleeve at R s At steady state, conservation of energy applied to the shaft requires that energy entering the shaft at Ri be equal to energy leaving it at R s . Thus 2-17 PROBLEM 2.7 (continued) 2Ri qi 2Rs qo Or Ri qi Rs Substituting (c) and (d) into (b) gives the boundary condition at R s qo dT ( Rs ) R q s i qi dr Rs (2) Convection at r Ro . Fourier’s law and Newton’s law of cooling give dT ( Ro ) k h T ( Ro ) T dr k (d) (e) (f) (4) Solution. Integrating (a) twice gives T (r) C1 ln r C2 (g) where C1 and C2 are constants of integration. Application of boundary conditions (e) and (f) gives C1 and C2 C1 Rs qs ( Ri / Rs )qi , k C2 T Rs qs ( Ri / Rs )qi(k / hRo ) ln Ro k (h) Substituting (h) into (g) and rearranging gives the temperature distribution in the sleeve Rs qs ( Ri / Rs )qiln( Ro / r ) (k / hRo ) T k Expressing (i) in dimensionless form gives T (r) T (r ) T qiRi 1 ln( Ro / r ) (1 / Bi ) q sRs q Rs k (i) (j) where the Biot number Bi is defined as Bi hRo k (k) (5) Checking. Dimensional check: Each term in (i) has units of temperature. Each term in (j) is dimensionless. Boundary conditions check: Substitution of solutions (i) into (e) and (f) shows that the two boundary conditions are satisfied. Differential equation check: Direct substitution of (i) into (a) shows that the solution satisfies the governing equation. Limiting check: (i) If no energy is added to the shaft at Ri and no frictional heat is generated at the interface R s , the sleeve should be at a uniform temperature T . Setting qi qs 0 in 2-18 PROBLEM 2.7 (continued) (i) gives T (r) T . (ii) If h = 0, no heat can leave the sleeve and the temperature will be infinite. Setting h = 0 in (i) gives T ( r ) . (6) Comments. (i) An alternate approach to solving this problem is to consider the shaft and sleeve as a composite cylinder, write two differential equations and four boundary conditions and solve for the temperature distribution in both shaft and cylinder. Clearly this is a longer approach to solving the problem. (ii) Two parameters characterize the solution: The Biot number and heat ratio (qiRi / q sRs ) . 2-19 PROBLEM 2.8 A hollow cylinder of inner radius ri and outer radius ro is heated with flux q i at its inner surface. The outside surface exchanges heat with the ambient and surroundings by convection and radiation. The heat transfer coefficient is h, ambient temperature T , surroundings temperature Tsur and surface emissivity is . Assume steady state one-dimensional conduction and use a simplified radiation model. Determine: T sur h, T q i ro [a] The temperature distribution in the cylinder. [b] The inner and outer surface temperatures for: ri r q i = 35,500 W/m2 , T = 24 o C , Tsur = 14 o C , k = 3.8 W/m o C , ro = 12 cm, ri = 5.5 cm, h = 31.4 W/m2 o C , = 0.92 (1) Observations. (i) This is a one-dimensional steady state problem. (ii) Two boundary conditions are needed. (iii) Cylindrical geometry. (iv) Specified flux at the inner surface and convection and radiation at the outer surface. (v) Since radiation is involved in this problem, temperature units should be expressed in kelvin (2) Origin and Coordinates. The origin and coordinate r are shown. (3) Formulation. (i) Assumptions. (1) Steady state, (2) one-dimensional, (3) constant properties, (4) no energy generation, (5) small gray surface which is enclosed by a much larger surface, (6) uniform conditions at the two surfaces and (7) stationary material. (ii) Governing Equations. Based on the above assumptions, the heat equation in Cartesian coordinates is given by d (r dT ) 0 (a) dr dr (iii) Boundary Conditions. Two boundary conditions are needed. They are: (2) Specified flux at r ri k d T (ri ) qi dr (b) where k thermal conductivity = 3.8 W/m o C qo surface flux = 35,500 W/m2 ri inside radius = 5.5 cm (2) Convection and radiation at r ro . Conservation of energy, Fourier’s law of conduction, Newton’s law of cooling and Stefan-Boltzmann radiation law give k dT (ro ) 4 h T (ro ) T T 4 (ro ) Tsur dr 2-20 (c) PROBLEM 2.8 (continued) where h heat transfer coefficient = 31.4 W/m2 o C = 27 W/m2 K ro outside radius = 12 cm T = ambient temperature = 24 o C = (24 + 273.15) K = 297.15 K Tsur = surroundings temperature = 14 o C = (14 + 273.15) K = 287.15 K emissivity = 0.92 Stefan-Boltzmann constant = 5.67 108 W/m2 K 4 (4) Solution. Integrating (a) twice gives T (r ) C1 ln r C2 (d) where C1 and C2 are constants of integration. Application of boundary conditions (b) and (c) gives C1 qi ri k ri q qi h i ri ln ro C 2 T ro k (e) qi 4 4 ( k ri ln ro C 2 ) Tsur (f) Equation (f) gives the constant C2 . However, this equation must be solved for C2 by trial and error. (5) Checking. Dimensional Check: Each term in equations (b), (c) and (f) has units of flux. Differential equation check: Solution (d) satisfies the governing equation (a). (6) Computations. Equation (e) gives C1 35,500 (W/m2 ) o 3.8 ( W/m- C) 0.055(m) 513.82 o C K 513.82 m m Equation (f) gives - 0.055 (m) 0.055(m) 35,500 (W/m 2 ) 31.4 ( W/m 2 - K) 35,500(W/(W/m 2 ) ln 0.12(m). C 2 297.15 (K) 0.12(m) 3.8 ( W/m - K) 4 - 0.055(m) (0.92)5.67 10 8 ( W/m 2 - K 4 ) 35,500 (W/m 2 ) ln 0.12(m) C 2 (287.15) 4 (K 4 ) 3.8 ( W/m - K) Solving this equation for C2 by trial and error gives C 2 484.8 K Substituting (e) and C2 into (d) gives T (r ) 484.8 (K) 2-21 qi ri ln r k (g) PROBLEM 2.8 (continued) To determine the inside surface temperature set r ri in (g) T (ri ) 484.8 (K) 35,500( W / m 2 o C) 0.055(m) ln 0.055(m) 1005.48 K 3.8( W / m o C) Similarly, the outside surface temperature is 35,500( W / m 2 o C) T (ro ) 484.8 (K) 0.055(m) ln 0.055(m) 604.625K 3.8( W / m o C) (7) Comments. (i) A trial and error procedure was required to solve this problem. (ii) Care should be exercised in selecting the correct units for temperature in radiation problems. (iii) With the two temperatures determined a numerical check based on conservation of energy for a tube of length L can be made: Energy entering tube at ri = Energy conducted through tube = Energy leaving tube at ro Energy entering tube = q 2 ri qi W/m L Energy conducted through tube = Energy flux leaving at ro q T (ri ) T ( ro ) r 1 L ln o 2 k ri q 4 2 ro h T (ro ) T 2 ro T 4 (ro ) Tsur L Numerical computations show that each term in (h) is equal to 12267.9 W/m . 2-22 (h) PROBLEM 2.9 Radiation is used to heat a hollow sphere of inner radius ri , outer radius ro and conductivity k. Due to the thermal absorption characteristics of the material the radiation results in a variable energy generation rate given by ro r2 T o q (r ) qo 2 r ro ri where q o is constant and r is the radial coordinate. The 0 inside surface is insulated and the outside surface is maintained at temperature To . Assume steady state oneq(r ) dimensional conduction, determine the temperature distribution. radiation (1) Observations. (i) Radiation results in a non-uniform energy generation. (ii) Use a spherical coordinate system. (iii) At steady state the total energy generated in the sphere must be equal to the energy leaving the surface by convection and radiation. (iv) This is a one-dimensional steady state conduction problem in a hollow sphere with insulated inner surface and convection and radiation at the outer surface. (2) Origin and Coordinates. The origin is selected at the center and the coordinate r is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant conductivity, (4) uniform surface and ambient conditions, and (5) the sphere is modeled as a small gray surface which is enclosed by a much larger surface. (ii) Governing Equations. Based on the above assumptions, eq. (1.13) gives 1 d 2 dT q ( r ) 0 dr k r 2 dr (a) where q depends on r according to q (r ) qo r2 (b) ro2 Substituting (b) into (a) 1 d 2 dT q r 2 ( r ) 0 dr k r2 r 2 dr (c) o (iii) Boundary Conditions. Since (a) is second order with a single independent variable r, two boundary conditions are needed. (1) Insulated boundary at ri dT (ri ) 0 dr PROBLEM 2.9 (continued) 2-23 (d) (2) Specified temperature at ro T (ro ) To (e) (4) Solution. Separating variables and integrating (a) twice gives T qo 20kro2 r4 C1 C2 r (f) where C1 and C2 are constants of integration. Application of boundary condition (d) gives qo ri5 C1 5k ro2 (g) Boundary condition (2) gives C 2 To qo ro2 5k 1 ri5 5 4 ro (h) Substituting (g) and (h) into (f) 4 ri5 ri5 4 T (r ) To r 4 ro 4 r ro 20kro2 Expressing (i) in dimensionless form gives qo T (r ) To q ro2 k 5 5 1 1 ri 1r r 1 r i o 20 5 ro 5 ro r 20 ro (i) 4 (j) (5) Checking. Dimensional check: Each term in (i) has units of temperature and each term in (j) is dimensionless. Boundary conditions check: Equation (i) satisfies boundary conditions (d) and (e). Differential equation check: Direct substitution of (i) into (c) shows that the solution satisfies the governing equation. Limiting check: If no energy is generated in the sphere the temperature will be uniform equal to To . Setting qo 0 in (i) gives T (r ) To . Conservation of energy check: The total energy generated in the sphere must leave the outer surface by conduction. Total energy generate is obtained by integrating (b) over the volume of the hollow sphere. Energy conducted the outer surface is obtained by applying Fourier’s law at r ro using solution (i). Both are found to be equal to Total energy generated 4 qo 5 (ro ri5 ) 2 5 ro (7) Comments. According to (j), the solution is characterized by a single dimensionless geometric parameter ri / ro . 2-24 PROBLEM 2.10 An electric wire of radius 2 mm and conductivity 398 W/ m oC generates energy at a rate of 1.25 105 W/ m 3 . The surroundings and ambient temperatures are 78 oC and 82 oC , respectively. The heat transfer coefficient is 6.5 W/ m 2 oC and surface emissivity is 0.9. Neglecting axial conduction and assuming steady state, determine surface and centerline temperatures. (1) Observations. (i) The wire generates energy. (ii) Use a cylindrical coordinate system. (iii) Heat conduction is assumed to be in the radial direction only. (iv) At steady state the total energy generated in the wire must be equal to the energy leaving the surface by convection and radiation. (v) To determine center and surface temperatures it is necessary to solve for the temperature distribution in the wire. (vi) This is a one-dimensional steady state conduction problem in a solid cylinder with energy generation. Tsur h, T r ro 0 q Ts (2) Origin and Coordinates. The origin is selected at the center and the coordinate r is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant conductivity, (4) uniform energy generation, (5) uniform surface and ambient conditions, and (6) the wire is modeled as a small gray surface which is enclosed by a much larger surface. (ii) Governing Equations. Based on the above assumptions, eq. (1.11) gives 1 d (r dT ) q 0 r dr dr k (a) where k = thermal conductivity = 398 W/m o C q = energy generation rate = 1.25 10 5 W/m 3 (iii) Boundary Conditions. Since (a) is second order with a single independent variable r, two boundary conditions are needed. (2) Symmetry gives dT (0) 0 dr (b) (2) Convection and radiation at the surface r ro . Conservation of energy at the surface and using Fourier’s law, Newton’s law of cooling and Stefan-Boltzmann radiation law give k dT ( ro ) 4 h T ( ro ) T T 4 ( ro ) Tsur dr where h heat transfer coefficient = 6.5 W/m2 o C ro wire radius = 2 mm 0.002 m Tsur surroundings temperature = 82 o C 355 K 2-25 (c) PROBLEM 2.10 (continued) T ambient temperature = 78 o C 351 K emissivity = 0.9 Stefan-Boltzmann constant = 5.67 108 W/m2 K 4 (4) Solution. Separating variables and integrating (a) twice gives T (r) q 2 r C1 ln r C2 4k (d) where C1 and C2 are constants of integration. Application of boundary condition (b) gives C1 0 Thus (d) becomes T (r ) C2 (e) q 2 r 4k (f) Boundary condition (2) gives 4 q ro 4 h C 2 ( q ro2 / 4k ) T C 2 ( q ro2 / 4k ) Tsur 2 (g) This equation gives C2 . However, it can not be solved explicitly for C2 . Solution must be obtained by trial and error. Once C2 is determined, equation (f) gives center and surface temperatures by setting r 0 and r ro , respectively. Thus T (0) C2 and T ( ro ) C 2 q 2 ro 4k (h) (i) (5) Checking. Dimensional check: Each term in (g) has units of flux and each term in (i) has units of temperature. Boundary conditions check: Equation (f) satisfies boundary condition (b). Condition (c) can not be checked since the constant C2 is not known explicitly. Differential equation check: Direct substitution of (f) into (a) shows that the solution satisfies the governing equation. Limiting check: If no energy is generated in the wire and T Tsur , the wire temperature will be uniform equal to the ambient temperature. Setting q 0 and T Tsur in (f) gives T ( r ) C2 . Equation (g) gives C2 T . (6) Computations. Substituting numerical values in (g) and solving for C2 by trial and error gives C2 361.2 K. Equations (h) and (i) give center and surface temperatures as T (0) 361.2 K and T ( ro ) 361.2 K (7) Comments. The wire is essentially at uniform temperature. This is due to the fact that k is high and ro and h are small. 2-26 PROBLEM 2.11 The cross-sectional area of a fin is Ac (x) and its circumference is C (x ). Along its upper half surface heat is exchanged by convection with an ambient fluid at T . The heat transfer coefficient is h. Along its lower half surface heat is exchanged by radiation with a surroundings at Tsur . Using a simplified radiation model formulate the steady state heat equation for this fin. (1) Observations. (i) This is a fin problem with surface convection and radiation. (ii) To formulate the fin equation apply conservation of energy to an element of the fin using fin approximation and Fourier, Newton and Stefan-Boltzmann laws. (2) Origin and Coordinates. The origin is selected at one end of the fin and the coordinate x is directed as shown. h, T h, T dqc Tsur dqr x 0 dqc h, T Ac qx dAs dqr qx dqx dx dx dx (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4) Bi 1, and (5) the fin is modeled as a small gray surface which is enclosed by a much larger surface. (ii) Governing Equations. To formulate the heat equation for this fin we select an element dx at location x as shown. This element is enlarged showing its geometric characteristics and energy interchange at its surfaces. Conservation of energy for the element under steady state conditions requires that E in = E out (a) where E in = rate of energy added to element E out = rate of energy removed from element Energy enters the element by conduction at a rate q x and leaves at a rate q x + (d q x /dx) dx. Energy also leaves by convection at a rate dqc and by radiation at a rate dqr . We now formulate each term in (a) E in = q x (b) dq E out = q x x dx dqc dqr dx (c) dqx dx dqc dqr 0 dx (d) and Substituting (b) and (c) into (a) 2-27 PROBLEM 2.11 (continued) We introduce Fourier, Newton and Stefan-Boltzmann laws to eliminate q x , dqc and dqr , respectively. Thus, q x k Ac dT dx (e) dqc h(T T )dAs / 2 and (f) 4 dqr T 4 Tsur dAs / 2 (g) where Ac Ac (x ) is the cross-sectional area through which heat is conducted. The infinitesimal area dAs / 2 is half the surface area of the element where heat is exchanged by convection at the top half and by radiation at the bottom half. Substituting (e), (f) and (g) into (d) we obtain d dx (kAc ( x ) dT ) h (T T ) 1 dAs (T 4 Tsur4 ) 1 dAs dx 2 dx 2 dx 0 (h) The gradient of surface area dAs / dx is given by eq. (2.6a) dAs C ( x ) 1 ( dy s / dx) 2 dx 1/ 2 (i) where y s (x ) is variable which describes the fin profile and C(x) is the circumference. If k is assumed constant, (h) becomes d 2T 1 dAc dT h dAs dAs 4 4 (T T ) (T Tsur )0 2 Ac dx dx 2kAc dx 2kAc dx dx (j) (4) Checking. Dimensional check: Each term in (j) has the same units. Limiting check: If there is no radiation, (j) should reduce to eq. (2.5b). Setting 0 in (j) gives eq. (2.9). (5) Comments. (i) Radiation introduces non-linearity in the fin equation. (ii) To obtain the corresponding equation for a fin with no surface convection set h = 0 in (j) and remove the ½ factor if radiation takes place over the entire surface. 2-28 PROBLEM 2.12 A constant area fin of length 2 L and cross-sectional area Ac generates heat at a volumetric rate q . Half the fin is insulated while the other half exchanges heat by convection. The heat transfer coefficient is h and the L L ambient temperature is T . The base and h, T tip are insulated. Determine the steady x q 0 state temperature at the mid-section. At h, T what location is the temperature highest? (1) Observations. (i) This is a constant area stationary fin. (ii) Temperature distribution can be assumed one dimensional. (iii) Surface conditions are different for each half. Thus two equations are needed. (iv) Both ends are insulated. (v) The highest temperature is at the insulated end x = 0. (2) Origin and Coordinates. The origin is selected at one end of the fin and the coordinate x is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area, (4) uniform energy generation, (5) Bi 1, (6) constant conductivity, (7) no radiation and (8) uniform h and T . (ii) Governing Equations. Since half the fin is insulated while the other half exchanges heat by convection, two fin equations are needed. Let T1 ( x ) = temperature distribution in the first half, 0 x L T2 ( x ) = temperature distribution in the second half, L x 2L The fin equation for each section is obtained from eq. (2.5b) dAs q d 2T 1 dAc dT h ( T T ) 0 dx k dx 2 Ac ( x ) dx dx kAc ( x ) (2.5b) Note that Ac is constant and dAs / dx C , where C is the circumference. Eq. (2.5b) becomes d 2T hC q (T T ) 0 (a) 2 kAc k dx In the first half the fin is insulated along its surface and thus the convection term vanishes. Equation (a) gives d 2T1 q (b) 0, 0 xL k dx 2 For the second half (a) gives d 2T2 q m 2 (T2 T ) 0, 2 k dx L x 2L (c) where m2 hC k Ac 2-29 (d) PROBLEM 2.12 (continued) (iii) Boundary Conditions. Four boundary conditions are needed. (1) Insulated surface at x = 0 dT1 (0) 0 dx (e) T1 ( L) T2 ( L) (f) dT1 ( L) dT2 ( L) dx dx (g) (2) Equality of temperature at x = L (3) Equality of flux at x = L (4) Insulated surface at x = 2L dT2 ( 2 L ) 0 dx (4) Solution. Separating variables and integrating (b) twice T1 ( x ) q 2 x Ax B 2k (h) (i) where A and B are constants of integration. The solution to (c) is given in Appendix A, equation (A-2b) q (j) T2 ( x ) T C sinh mx D coshmx k m2 where C and D are constants of integration. Application of the four boundary conditions gives the four constants (k) A0 B T qL2 q qL tanh 2mL sinh mL coshmL 2k k m 2 k m tanh 2mL coshmL sinh mL (l) C q L tanh 2ml k m sinh mL tanh 2mL cosh mL (m) D q L 1 k m tanh 2mL cosh mL sinh mL (n) Substituting into (i) and (j) gives the solutions as qL2 q qL tanh 2mL sinh mL coshmL q 2 T1 ( x ) T x 2k k m 2 k m tanh 2mL coshmL sinh mL 2k and T2 ( x ) T q qL coshmx tanh 2mL sinh mx k m 2 k m tanh 2mL coshmL sinh mL 2-30 (o) (p) PROBLEM 2.12 (continued) Expressing (o) and (p) in dimensionless form gives T1 ( x) T q L2 k 1 1 1 tanh 2mL sinh mL cosh mL 1 ( x / L) 2 2 2 ( mL) (mL) tanh 2mL cosh mL sinh mL 2 (q) and T2 ( x) T 1 1 cosh(mL)( x / L) tanh 2mL sinh(mL)( x / L) (mL) tanh 2mL cosh mL sinh mL (r) q L (mL) k Examination of (o) shows that the temperature of the insulated half decreases as the distance from the origin is increased. Thus the highest temperature occurs at x = 0. 2 2 (5) Checking. Dimensional check: Each term in (o) and (p) has units of temperature and each term in (q) and (r) is dimensionless. Boundary conditions check: Substitution of solutions (o) and (p) into equations (e)-(h) shows that the four boundary conditions are satisfied. Differential equation check: Direct substitution of (o) into (b) and (p) into (c) shows that the governing equations are satisfied. Limiting check: (i) If no energy is generated, the entire fin should be at the ambient temperature. Setting q 0 in (o) and (p) gives T1 ( x) T2 ( x) T (ii) If h = 0, no heat can leave the fin and thus the temperature should be infinite. According to (d) if h = 0, then m = 0. Setting m = 0 in (o) and (p) gives T1 ( x) T2 ( x) (6) Comments. (i) When conditions along a fin are not uniform it is necessary to use more than one fin equation. (ii) An alternate approach to solving this problem is to recognize that all the energy generated in the insulated half must enter the second half at x = L. Thus one can consider the second half as a constant area fin with specified flux at the base x = L and an insulated tip at x = 2L. The solution to the temperature distribution in this half gives the temperature at x = L. The insulated section can be treated as a fin with insulated base at x = 0 and a specified temperature at the tip x = L. (iii) The solution is characterized by a single parameter mL. 2-31 PROBLEM 2.13 A constant area fin of length L and cross-sectional area Ac is maintained at To at the base and is insulated at the tip. The fin is insulated along a distance b from the base end while L b heat is exchanged by convection along its h,T remaining surface. The ambient temperature x q is T and the heat transfer coefficient is h. h,T Determine the steady state heat transfer rate To from the fin. (1) Observations. (i) This is a constant area stationary fin. (ii) Temperature distribution can be assumed one-dimensional. (iii) The fin has two sections. Surface conditions are different for each section. Thus two equations are needed. (iv) One end is maintained at a specified temperature and the other end is insulated. (v) To determine fin heat transfer rate the temperature distribution in the two-section fin must be determined. (2) Origin and Coordinates. The origin is selected at one end of the fin and the coordinate x is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area, (4) uniform energy generation, (5) Bi 1, (6) constant conductivity, (7) uniform h and T , and (8) no radiation. (ii) Governing Equations. Since one section is insulated while the other exchanges heat by convection, two fin equations are needed. Let T1 ( x ) = temperature distribution in the first section, 0 x b T2 ( x ) = temperature distribution in the second half, b x L The fin equation for each section is obtained from eq. (2.5b) dA q d 2T 1 dAc dT h (T T ) s 0 2 Ac ( x ) dx dx kAc ( x ) dx k dx (2.5b) Note that Ac is constant and dAs / dx C , where C is the circumference. Eq. (2.5b) becomes d 2T hC q (T T ) 0 2 kAc k dx (a) In the first section the fin is insulated along its surface and thus the convection term vanishes. Equation (a) gives d 2T1 q 0, (b) 0 xb k dx 2 For the second section equation (a) gives d 2T2 q m 2 (T2 T ) 0, 2 k dx where 2-32 bxL (c) PROBLEM 2.13 (continued) hC k Ac (iii) Boundary Conditions. Four boundary conditions are needed. m2 (d) T1 (0) To (e) T1 (b) T2 (b) (f) dT1 (b) dT2 (b) dx dx (g) (1) Specified temperature at x = 0 (2) Equality of temperature at x = b (3) Equality of flux at x = b (4) Insulated surface at x = L dT2 ( L ) 0 dx (4) Solution. Separating variables and integrating (b) twice T1 ( x) q 2 x A1 x B1 2k (h) (i) where A1 and B1 are constants of integration. The solution to (c) is given in Appendix A, equation (A-2b) q (j) T2 ( x) T A2 sinh mx B2 cosh mx k m2 where A2 and B2 are constants of integration. Application of the four boundary conditions give the four constants (To T q b 2 q )(m sinh mb m tanh mL cosh mb) 2 q b 2k km A1 k cosh mb tanh mL sinh mb mb sinh mb mb tanh mL cosh mb B1 To q b 2 q (To T ) tanh mL 2 2k km A2 cosh mb tanh mL sinh mb mb sinh mb mb tanh mL cosh mb q b 2 q 2k km 2 B2 cosh mb tanh mL sinh mb mb sinh mb mb tanh mL cosh mb (k) (l) (m) To T 2-33 (n) PROBLEM 2.13 (continued) Fin heat transfer rate q f is equal to the energy leaving surface area C(L-b) by convection. Using Newton’s law of cooling L q f hC (T2 T )dx (o) b where T2 ( x) is given by (j). An alternate and simpler approach is based on conservation of energy for the two-section fin. Thus (p) q f q(0) E g where E = rate of energy generated in the fin g q (0) = rate of energy conducted at the base E g is given by E g CLq Fourier’s law gives q(0) kAc Using (i) dT1 (0) dx q(0) kAc A1 (q) (r) (s) Substituting (q) and (s) into (p) gives the fin heat transfer rate q f kAc A1 CLq (t) (5) Checking. Dimensional check: Each term in (i) and (j) has units of temperature. Boundary conditions check: Substitution of solutions (i) and (j) into equations (e)-(h) shows that the four boundary conditions are satisfied. Differential equation check: Direct substitution of (i) into (b) and (j) into (c) confirms that the governing equations are satisfied. Limiting check: (i) If q 0 and To T , the temperature distribution in the fin should be uniform equal to T . For this case equations (k), (m) and (n) give A1 A2 B2 0 . Substituting into equations (i) and (j) gives T1 ( x) T2 ( x) T . (ii) If q b 0 , the problem reduces to a constant area fin with specified temperature at the base, insulated tip and convection at its surface. The solution to this case is given eq. (2.14) of Section 2.2.9. Setting q b 0 in (j) and rearranging the result gives T2 ( x ) T (To T ) coshm( L x ) coshmL This is identical to eq. (2.14). (6) Comments. When conditions along a fin are not uniform it is necessary to use more than one fin equation. 2-34 PROBLEM 2.14 Heat is removed by convection from an electronic package using a single fin of circular cross section. Of interest is increasing the heat transfer rate without increasing the weight or changing the material of the fin. It is recommended to use two identical fins of circular cross-section with a total weight equal to that of the single fin. Evaluate this recommendation using a semi-infinite fin model. (1) Observations. (i) This is a constant area semi-infinite fin problem. (ii) Determine the heat transfer rate from a circular fin and examine the effect of radius on the heat transfer rate. (iii) Assume that the base is at a specified temperature and that heat transfer from the surface is by convection. h ,T 0 T o x h ,T (2) Origin and Coordinates. The origin is selected at the base and the coordinate x is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area, (4) no energy generation, (5) Bi 1, (6) constant conductivity, (7) convection at the surface, (8) uniform heat transfer coefficient and ambient temperature, (9) specified base temperature and (10) no radiation. (ii) Governing Equations. The fin equation for this case is given by eq. (2.10) d 2 m2 0 dx 2 where T ( x) T (2.10) (a) and m2 = hC k Ac (b) For a circular rod of radius r the cross-sectional area Ac and the circumference C are given by (c) Ac r 2 and C 2 r (d) Substituting (c) and (d) into (b) 2h m2 (e) kr (iii) Boundary Conditions. The two boundary conditions are (1) Specified temperature at the base x = 0 2-35 PROBLEM 2.14 (continued) T (0) To or T (0) To (0) 0 (f) (3) Finite temperature at x T ( ) finite or ( ) finite (g) (4) Solution. The solution to eq. (2.10) is ( x) C1 exp(mx) C2 exp( mx) (h) where C1 and C2 are constants of integration. Conditions (f) and (g) give C1 and C2 C1 0 and C 2 To T (i) Substituting into (h) and using (a) gives the temperature distribution in the fin T ( x ) T (To T )e mx (j) Fourier’s law gives the heat transfer rate from the fin q f kAc dT (0) k Ac m(To T ) dx (k) Substituting (c) and (e) into (k) q f (To T ) 2 h k r 3 / 2 (l) We now compare the heat transfer rate from a single fin of radius r with that from two fins having the same weight as the single fin. Let r2 be the radius of each of the two fins. Equality of mass gives 2 r22 r 2 Thus r (m) r2 2 The total heat transfer rate from two fins of radius r2 each, q 2 f , is obtained by substituting (m) into (l) q2 f 2 (To T ) 2 h k ( r / 2 ) 3 / 2 21 / 4 (To T ) 2 h k r 3 / 2 (n) Taking the ratio of (n) and (l) q2 f qf 21 / 4 1.19 (5) Checking. Dimensional check: Each term in (j) has units of temperature. The heat transfer rate q f in equations (l) and (n) is expressed in units of watts. 2-36 PROBLEM 2.14 (continued) Boundary conditions check: Substitution of solutions (j) into equations (f) and (g) shows that the boundary conditions are satisfied. Differential equation check: Direct substitution of (j) into eq. (2.10) confirms that the governing equation is satisfied. Limiting check: (i) If To T , the temperature distribution in the fin should be uniform equal to T . Setting To T in (j) gives T ( x) T . (ii) If k = 0 or h = 0, the heat transfer from the fin should vanish. Setting k = 0 or h = 0 in equation (l) gives q f 0. (6) Comments. Two fins having the same weight as a single fin transfer 19% more heat than a single fin. 2-37 PROBLEM 2.15 A plate which generates heat at a volumetric rate q is placed inside a tube and cooled by convection. The width of the plate is w and its thickness is . The heat transfer coefficient is h and coolant temperature is T . The temperature at the interface between the tube and the plate is Tw . Because of concern that the temperature at the center may exceed design limit, you are asked to estimate the steady state center temperature using a simplified fin model. The following data are given: w q h, T Tw Tw h = 62 W/ m 2 oC , w 3 cm, = 2.5 mm, o 7 3 q = 2.5 10 W/ m , Tw 110 C , T 94 o C , k = 18 W/ m oC . (1) Observations. (i) This is a constant area fin. (ii) Temperature distribution can be assumed one dimensional. (iii) Temperature distribution is symmetrical about the center. Thus the temperature gradient at the center is zero. (iv) Consider half the plate as a fin which extends from the center to the tube surface. One end is maintained at a specified temperature and the other end is insulated. h ,T T w D x 0 w /2 h ,T insulation (2) Origin and Coordinates. The origin is selected at the center of tube and the coordinate x is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant area, (4) uniform energy generation, (5) Bi 1, (6) constant conductivity, (7) uniform h and T , (8) negligible radiation, and (9) negligible temperature variation along the tube. (ii) Governing Equations. The fin equation is obtained from eq. (2.5b) dA q d 2T 1 dAc dT h (T T ) s 0 2 Ac ( x ) dx dx kAc ( x ) dx k dx (2.5b) Note that Ac is constant and dAs / dx C , where C is the circumference. Eq. (2.5b) becomes d 2T q m 2 (T T ) 0 2 k dx (c) where hC k Ac (d) Ac D (e) m2 where Ac is the cross-sectional given by 2-38 PROBLEM 2.15 (continued) where D is the length of plate along the tube and is plate thickness. Neglecting heat loss from the end surface w / 2 , the circumference is given by C 2D (f) Substituting (e) and (f) into (d) 2h k (iii) Boundary Conditions. The two conditions are m2 (g) dT (0) 0 dx (h) T ( w / 2) Tw (i) (1) Insulated surface at x = 0 (2) Specified temperature at x = w/2 (4) Solution. The solution to (c) is given in Appendix A, equation (A-2b) T ( x ) T q C1 sinh mx C 2 coshmx k m2 (j) where C1 and C2 are constants of integration. Application of boundary conditions (h) and (i) gives the two constants (k) C1 0 and q (l) C2 (Tw T ) 1 2 k m coshmw / 2 Substituting into (j) T ( x ) T q q (Tw T ) coshmx 2 2 km k m coshmw / 2 (m) The maximum temperature may occur at x = 0. Setting x = 0 in equation (m) gives T (0) T q q 1 (Tw T ) 2 2 km k m coshmw / 2 (n) (5) Checking. Dimensional check: Each term in solution (m) has units of temperature. Boundary conditions check: Substitution of solution (m) into equations (h) and (i) shows that the boundary conditions are satisfied. Differential equation check: Direct substitution of (m) into (e) confirms that the governing equation is satisfied. Limiting check: If Tw T and q 0 , the temperature distribution in the plate should be uniform equal to T . Setting Tw T and q 0 in (m) gives T ( x) T . 2-39 PROBLEM 2.15 (continued) (6) Computations. Equation (g) is used to compute m 2(62)(W/m2 o C) m 2755.6 (1 / m 2 ) o 18( W/m C)(0.0025)(m) 2 m = 52.493 (1/m) Substituting the given data into (n) T (0) 94( o C) 2.5 107 ( W/m3 ) 18( W/m o C)2755.6(1 / m 2 ) o 2.5 107 ( W/m3 ) 1 o 110 C 94 C o 2 18( W/m C)2755.6(1 / m ) cosh52.493(1 / m)0.015( m) T(0) = 230.1 o C (7) Comments. (i) Examination of (m) shows that the maximum temperature may not always be at the center. For example, in the limiting case of q 0 the highest temperature will be at the tube surface while the center will be at the lowest temperature. However, for the data given in this problem the maximum temperature occurs at the center. (ii) The solution is expressed in dimensionless form to determine the governing parameters. Equation (m) is rewritten as cosh(mw)( x / w) T ( x) T q q 1 Tw T k m 2 (Tw T ) k m 2 (Tw T ) cosh(mw) / 2 This form shows that two parameters characterizes the solution: mw and 2-40 q k m (Tw T ) 2 (o) . PROBLEM 2.16 An electric heater with a capacity P is used to heat air in a spherical chamber. The inside radius is ri , outside radius ro and the conductivity is k. At the inside surface heat is exchanged by convection. The inside heat transfer coefficient is hi . Heat loss from the outside surface is by radiation. The surroundings temperature is Tsur and surface emissivity is . Assuming one-dimensional steady state conduction, use a simplified radiation model to determine: [a] The temperature distribution in the spherical wall. [b] The inside air temperatures for the following conditions: h = 6.5 W/m2 o C , P = 1, 500 W, = 0.81, k = 2.4 W/m o C , Tsur = 18 o C , ri = 10 cm, Tsur r k hi + P ro ri _ ro = 14 cm. (1) Observations. (i) This is a one-dimensional steady state problem. (ii) Two boundary conditions are needed. (iii) Spherical geometry. (iv) Specified heat transfer rate at the inner surface and radiation at the outer surface. (v) Since radiation is involved in this problem, temperature units should be expressed in kelvin (2) Origin and Coordinates. The origin and coordinate r are shown. (3) Formulation. (i) Assumptions. (1) Steady state, (2) one-dimensional, (3) constant properties, (4) no energy generation, (5) small gray surface which is enclosed by a much larger surface, (6) uniform conditions at the two surfaces and (7) stationary material. (ii) Governing Equations. Based on the above assumptions, the heat equation in Cartesian coordinates is given by d 2 dT (r ) 0 (a) dr dr (iii) Boundary Conditions. Two boundary conditions are needed. They are: (3) Specified heat transfer rate at r ri k 4 rl 2 d T (ri ) P dr (b) where k thermal conductivity = 2.4 W/m o C = 2.4 W/m o K P electric heater capacity = heat transfer rate = 1,500 W ri inside radius = 10 cm (2) Radiation at r ro . Conservation of energy, Fourier’s law of conduction and StefanBoltzmann radiation law give 2-41 PROBLEM 2.16 (continued) k dT (ro ) 4 T 4 (ro ) Tsur dr (c) where ro outside radius = 14 cm Tsur = surroundings temperature = 18 o C = (18 + 273.15) K = 291.15 K emissivity = 0.81 Stefan-Boltzmann constant = 5.67 108 W/m2 K 4 (4) Solution. Integrating (a) twice gives T (r ) C1 C2 r (d) where C1 and C2 are constants of integration. Application of boundary conditions (b) and (c) gives C1 and C2 P C1 (e) 4 k 1 4 4 P P C2 Tsur 4 kro 4 ro2 Substituting (e) and (f) into (d) (f) 1 4 4 P P P T (r ) Tsur 4 kr 4 kro 4 ro2 (g) Expressing (g) in dimensionless form gives 1 T (r ) 1 ro kr T 1 o sur P 4 r P kro 4 P 1 2 4 4 ro Tsur (h) To determine the inside air temperature Ti apply Newton’s law of cooling at the inside surface (i) P 4 ri2 h [Ti T (ri ) where h 6.5 (W/m2 o C) 6.5 (W/m2 o K) Solving (i) for Ti Ti T (ri ) where T (ri ) is obtained from (h) by setting r ri . 2-42 P 4 ri2 h (j) PROBLEM 2.16 (continued) (5) Checking. Dimensional Check: Each term in equations (g) and (j) has units of temperature. Each term in (h) is dimensionless. Differential equation check: Solution (g) satisfies the governing equation (a). Boundary conditions check: Solution (g) satisfies boundary conditions (b) and (c). Limiting check: If the heater is turned off (P = 0) the sphere and inside air will be at the same temperature as the surroundings. Setting P = 0 in (g) and (j) gives T (r ) Ti Tsur . (6) Computations. Equation (g) is used to determine T (ri ) Equation (f) gives T (ri ) 1500( W ) 4 2.4( W/m - K )0.1(m) o 1500( W ) 4 2.4( W/m - o K )0.14(m) 1 4 1500( W ) 108 4 4 (291.15) (K ) 753.74 K 4 (0.81)5.67( W/m 2 -K 4 )(0.14) 2 (m 2 ) Substituting into (j) Ti 753.74 (K ) 1,500( W) 2,590 K 4 (0.1) (m 2 )6.5( W/m 2 K) 2 (7) Comments. (i) The inside air temperature is high. This is due to the small inside surface area and small heat transfer coefficient. (ii) Care should be exercised in selecting the correct units for temperature in radiation problems. 2-43 PROBLEM 2.17 A section of a long rotating shaft of radius ro is buried in a material of very low thermal conductivity. The length of the buried section is L. Frictional heat generated in the buried section can be modeled as surface heat flux. Along the buried surface the radial heat flux is q r and the axial heat flux is q x . The exposed surface exchanges heat by convection with the ambient. The heat transfer coefficient is h and the ambient temperature is T . Model the shaft as semi-infinite fin and assume that all frictional heat is conducted through the shaft. Determine the temperature distribution. (1) Observations. (i) This is a constant area fin. (ii) Temperature distribution can be assumed one dimensional. (iii) The shaft has two sections. Surface conditions are different for each of sections. Thus two equations are needed. (iv) Shaft rotation generates surface flux in the buried section and does not affect the fin equation. h T dx qr x L 0 qx (2) Origin and Coordinates. The origin is selected at one end of the fin and the coordinate x is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area, (4) uniform surface flux in the buried section, (5) Bi 1, (6) constant conductivity, (7) no radiation (8) no energy generation and (9) uniform h and T . (ii) Governing Equations. Since surface conditions are different for the two sections, two fin equations are needed. Let T1 ( x ) = temperature distribution in the buried section, 0 x L T2 ( x ) = temperature distribution in the exposed section, L x 2L The heat equation for the buried section must be formulated using fin approximation. Conservation of energy for the fin element dx gives dq (a) q x qr 2 ro dx q x x dx dx dq qx x dx where q x is the rate of heat conducted in the x-direction, dx given by Fourier’s law dT1 dx Substituting (b) into (a) and rearranging q x k d 2T1 dx 2 2q r 0, kro (b) dx 2 roqr qx 0 xL (c) The heat equation for constant area fin with surface convection is given by equation (2.9) 2-44 PROBLEM 2.17 (continued) d 2T2 dx 2 hC (T2 T ) 0 , kAc xL (d) where C is the circumferance and Ac is the cross section area given by C 2 ro (e) A ro2 (f) Using (c) and (d) to define m m hC kAc 2h kro (g) Subatituting (e) into (d) d 2T2 2 m 2 (T2 T ) 0 dx (iii) Boundary Conditions. Four boundary conditions are needed. (h) (1) Specified flux at x = 0 k dT1 (0) q x dx (i) (2) Equality of temperature at x = L T1 ( L) T2 ( L) (j) dT1 ( L) dT2 ( L) dx dx (k) T2 () finite (l) (3) Equality of flux at x = L (4) Finite temperature at x (4) Solution. The solutions to (c) is T1 ( x) qr 2 x A1 x B1 kro (m) where A1 and B1 are constants of integration. The solution to (h) is given in Appendix A, equation (A-2b) (n) T2 ( x) T A2 e mx B2 e mx where A2 and B2 are constants of integration. Application of the four boundary conditions gives the four constants q (o) A1 x k B1 T q q r 2 q x 1 q L L 2 r L x kro k m kro k 2-45 (p) PROBLEM 2.17 (continued) A2 0 B2 (q) q 1 q r L x e mL 2 m kro k (r) Substituting into (m) and (n) gives the solutions as T1 ( x) T q q q q r 2 q x 1 q L L 2 r L x r x 2 x x kro k m kro k kro k (s) and T2 ( x) T q 1 q r L x e m( L x ) 2 m kro k (t) Expressing (s) and (t) in dimensionless form gives 2 q x ro T1 ( x) T x 1 L q r L q r L2 kro q x ro x 1 1 2 L mL q r L (u) and q x ro mL(1 x / L ) T2 ( x) T 1 2 e 2 mL qr L q r L kro (v) (5) Checking. Dimensional check: Each term in (s) and (t) has units of temperature and each term in (u) and (v) is dimensionless. Boundary conditions check: Substitution of solutions (s) and (t) into equations (i)-(l) shows that the four boundary conditions are satisfied. Differential equation check: Direct substitution of (s) into (c) and (t) into (d) shows that the governing equations are satisfied. Limiting check: (i) If no energy is added to the buried section, the entire fin should be at the ambient temperature T . . Setting q x q r 0 in (s) and (t) gives T1 ( x) T2 ( x) T (ii) If h = 0, no heat can leave the fin and thus the temperature should be infinite. According to (g) if h = 0, then m = 0. Setting m = 0 in (s) and (t) gives T1 ( x) T2 ( x) (6) Comments. (i) When conditions along a fin are not uniform it is necessary to use more than one fin equation. (ii) The solution is characterized by two dimensionless parameters: q r mL and x o . q r L 2-46 2-47 PROBLEM 2.18 A conical spine with a base radius R and length L exchanges heat with the ambient by convection. The heat transfer coefficient is h and the ambient temperature is T . The base is maintained at To . Using a fin model, determine the steady state heat transfer rate. T h, T o R (1) Observations. (i) Temperature y s distribution in the fin and Fourier’s 0 x law of conduction give the heat h, T transfer rate. (ii) This is a variable area fin with specified temperature at L the base and convection at the surface. (iii) Temperature distribution can be assumed one-dimensional. (iv) Eq. (2.5) should be used to formulate the heat equation for this fin. (2) Origin and Coordinates. The origin is selected at the tip and the coordinate x is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4) Bi 1, (5) constant properties, (6) uniform h and T , and (7) negligible radiation. (ii) Governing Equations. Applying Fourier’s law at x = L gives the fin heat transfer rate q f kAc ( L) dT ( L) dx (a) where Ac (L) is the cross-sectional area at x = L, given by Ac ( L) R 2 (b) Substituting (b) into (a) q f kR 2 dT ( L) dx (c) Temperature distribution in the fin is obtained from the solution to the heat equation for the fin. Based on the above assumptions, eq. (2.5b) gives dA d 2T 1 dAc dT h (T T ) s 0 2 Ac ( x ) dx dx kAc ( x ) dx dx (d) The cross-sectional area Ac (x ) for a conical fin is given by Ac y s2 (e) y s ( R / L) x (f) Ac ( x ) ( R / L) 2 x 2 (g) where y s is the local radius given by Substituting (f) into (e) 2-47 PROBLEM 2.18 (continued) Eq. (2.6a) is used to formulate dAs / dx dAs C ( x) 1 (dys / dx) 2 dx Using (f) into the above dAs 2 ( R / L) 1 ( R / L) 2 dx 1/ 2 1/ 2 (h) x Substituting (g) and (h) into (d) and rearranging d 2 d x 2x 2 x 0 2 dx dx (i) T T (j) 2hL 1 ( R / L) 2 kR (k) 2 where and 2 (iii) Boundary Conditions. The two boundary conditions are (1) Finite temperature at x = 0 T (0) finite, or (0) T (0) T finite (l) (2) Specified temperature at x = L T ( L) To , or ( L) T ( L) T To T (m) (4) Solution. Equation (i) is a second order differential equation with variable coefficient. It may be a Bessel equation. Comparing (i) with eq. (2.26) gives A 1/ 2 B0 C 1/ 2 D 2 i, p D / i 2 n 1 Since n is an integer and D is imaginary the solution to (i) is given by eq. (2.29) ( x) T ( x) T x 1/ 2 C1 I1 (2 x1/ 2 ) C2 K1 (2 x1/ 2 ) (n) where C1 and C2 are constants of integration. Since K1 (0) (Table 2.1), condition (l) requires that (o) C2 0 Boundary condition (m) gives (T T ) L1 / 2 C1 o (p) I 1 ( 2 L1 / 2 ) Substituting (o) and (p) into (n) 2-48 PROBLEM 2.18 (continued) T ( x) T I (2 x1 / 2 ) ( L / x)1 / 2 1 To T I 1 (2 L1 / 2 ) (q) With the temperature distribution known, equation (c) gives the heat transfer rate. Using eq. (2.42) to determine the derivative of I 1 ( 2 x 1 / 2 ) and substituting into (c) gives q f R 2hk 1 ( R / L) 3/ 2 2 1/ 4 (To T ) I 2 (2 L1 / 2 ) I1 (2 L1 / 2 ) (r) Expressing the above in dimensionless form gives qf 2 R 3 / 2 hk 1 ( R / L) 2 1/ 4 (To T ) I 2 (2 L1 / 2 ) I 1 (2 L1 / 2 ) (s) (5) Checking. Dimensional check: (i) The argument of the Bessel function, (2 x 1 / 2 ) is dimensionless. (ii) Equation (r) gives the correct units for q f . Boundary conditions check: Equation (q) satisfies boundary conditions (l) and (m). Limiting check: If To T , the temperature distribution in the fin should be uniform equal to T and the heat transfer should vanish. Setting To T in (q) gives T ( x) T . Similarly, setting To T in (r) gives q f 0. (6) Comments. (i) Selecting the origin at the tip is motivated by the knowledge that K n (0) . This leads to C2 0. (ii) The minus sign in equation (r) indicates that for To T heat flows in the negative x-direction. This is consistent with the second law of thermodynamics. (iii) The solution is characterized by a single dimensionless parameter L1 / 2 2-49 PROBLEM 2.19 In many applications it is desirable to reduce the weight of a fin without significantly reducing its heat transfer capacity. Compare the heat transfer rate from two straight fins of identical material. The profile of one fin is rectangular while that of the other is triangular. Both fins have the same length L and base thickness . Surface heat transfer is by convection. Base temperature is To and ambient temperature is T . The following data are given: h = 28 W/ m 2 oC , k = 186 W/ m oC , 1 cm L = 18 cm, (1) Observations. (i) The heat transfer rate from a straight rectangular fin should be compared with that of a straight triangular fin. (ii) To determine the temperature distribution and heat transfer rate for the two fins the heat equation for each geometry should be formulated and solved. Alternatively, the efficiency graph of Fig. 2.16 can be used to determine the heat transfer rate from each fin. (iii) The heat transfer rate for insulated rectangular fins is given by eq. (2.15). h, T ys 0 x h, T L x 0 h, T (2) Origin and Coordinates. The origin is selected at the tip of the triangular fin and the coordinate x is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4) Bi 1, (5) constant properties, (6) uniform h and T , (7) the tip of the rectangular fin is insulated, (8) the side surfaces are insulated and (10) no radiation. (ii) Governing Equations. The heat transfer rate for the straight rectangular fin is given by eq. (2.15) 1/ 2 (a) q fr hCr kAcr (To T ) tanh mr L where the subscript r refers to the rectangular fin and mr hCr / kAc r (b) Acr is the cross-sectional area and Cr is the circumference. Let W be the depth of each plate. Thus Acr W (c) and (d) Cr 2W Substituting (b), (c) and (d) into (a) and rearranging yields qfr (To T )W 2hk tanh 2h / k L 2-50 (e) PROBLEM 2.19 (continued) We now turn our attention to the triangular fin. Applying Fourier’s law at x = L gives the heat transfer rate dT ( L) q f t kAc t ( L) (f) dx where the subscript t refers to the straight triangular fin and Act (L) is the cross-sectional area at x = L, given by Ac t ( L ) W (g) Substituting (g) into (f) dT ( L) q f t kW (h) dx Temperature distribution in the fin is obtained from the solution to the heat equation for the fin. Based on the above assumptions, eq. (2.5b) gives dAs t d 2T 1 dAc t dT h ( T T ) 0 dx dx 2 Ac t ( x ) dx dx kAc t ( x ) (i) The cross-sectional area Ac t ( x ) for a straight triangular fin is given by Act ( x) 2W y s (j) y s ( / 2 L) x (k) Act ( x) (W / L) x (l) where y s is the half local width Substituting (k) into (j) Eq. (2.6a) is used to formulate dAs / dx dAs t dx Ct ( x) 1 (dys / dx) 2 1/ 2 (m) where the circumference Ct (x ) is given by Ct 2W Using (k) and (o) into (m) dAs t dx 2W 1 ( / 2 L) 2 (n) 1/ 2 x (o) Substituting (l) and (o) into (i) and rearranging x2 where d 2 d x 2 x 0 2 dx dx T T and 2-51 (p) (q) PROBLEM 2.19 (continued) 2 2hL 1 ( / 2 L) 2 k (r) (iii) Boundary Conditions. The two boundary conditions are (1) Finite temperature at x = 0 T (0) finite, or (0) T (0) T finite (s) (2) Specified temperature at x = L T ( L) To , or ( L) T ( L) T To T (t) (4) Solution. Equation (p) is a second order differential equation with variable coefficient. It may be a Bessel equation. Comparing (p) with eq. (2.26) gives A0 B0 C 1/ 2 D 2 i, p D / i 2 n0 Since n is zero and D is imaginary the solution to (p) is given by eq. (2.29) ( x ) T ( x ) T C1 I 0 (2 x1/ 2 ) C2 K 0 (2 x1/ 2 ) (u) where C1 and C2 are constants of integration. Since K1 (0) (Table 2.1), condition (s) requires that (v) C2 0 Boundary condition (t) gives (To T ) C1 (w) I 0 (2 L1 / 2 ) Substituting (v) and (w) into (u) I (2 x1/ 2 ) T T 0 (x) To T I 0 ( 2 L1 / 2 ) With the temperature distribution known, equation (h) gives the heat transfer rate. Using eq. (2.46) to determine the derivative of I 0 (2 x1/ 2 ) and substituting into (h) gives qft (To T )W 2hk 1 ( 1/ 2 ) 2 1 / 4 I1 (2 L / 2 L) 1/ 2 I 0 (2 L ) (y) (5) Checking. Dimensional check: (i) The argument of the Bessel function, ( 2 x 1 / 2 ), is dimensionless. (ii) Each of equation (e) and (y) is dimensionally consistent. Boundary conditions check: Equation (x) satisfies boundary conditions (s) and (t). Limiting check: If To T , the temperature distribution in the fin should be uniform equal to T and the heat transfer should vanish. Setting To T in (x) gives T ( x) T . Similarly, setting To T in (y) gives q f t 0. 2-52 PROBLEM 2.19 (continued) (6) Computations. Substituting numerical values in (e) gives the heat transfer rate for the rectangular fin q fr (To T )W W ( 2( 28) m C 2 o )186( W m C o )(0.01)(m) tanh 2( 28)(W/m 2 o C) 186W/m C)0.01(m) o 0.18( m) 7.72 W m o C To determine the heat transfer rate for the triangular fin we first compute the argument 2 L1 / 2 of the Bessel functions I o (2 L1/ 2 ) and I1 (2L1/ 2 ) 2 2 L1 / 2 2 0.01( m) 2( 28 )( W/m 2 o C)0.18( m) 1/ 2 1 0.18( m) = 1.9757 o 0 . 36 ( m ) 186 ( W/m C)0.01( m) Tables of Bessel functions give I o (1.9757) = 2.2427 I1 (1.9757) 1.556 Substituting into (y) gives the heat transfer rate for the triangular fin qft (To T )W ( 2(28) W m C 2 o )186( W m C o 556 )0.01(m) 1 (0.01(m) / 0.36(m))2 1/ 4 21..2427 7.08 W m o C The minus sign indicates that heat flows in the negative x-direction if To T . (7) Comments. (i) Selecting the origin at the tip is motivated by the knowledge that K n (0) . This leads to C2 0. (ii) The minus sign in equation (y) indicates that for To T heat flows in the negative x-direction. This is consistent with the second law of thermodynamics. (iii) By using a triangular fin the weight is reduced by 50% while the heat transfer rate is decreased by 8.3%. (iv) The efficiency graph of Fig. 2.16 can be used to solve this problem. However, this approach introduces small errors due to reading the efficiency graph. Fin efficiency is defined as qf (A) f hAs (To T ) or qf hAs f (B) (To T )W W where f is fin efficiency and As is surface area. Surface area for the two fins is given by Asr 2WL (C) As t 2W L2 ( / 2) 2 (D) Applying (B) to the two fins and using (C) and (D) 2-53 PROBLEM 2.19 (continued) q fr 2 fr hL (E) 2 f t h L2 ( / 2) 2 (F) (To T )W and qft (To T )W To determine f from Fig. 2.16 the parameter Lc 2h / k is computed Lc 2h / k 0.18( m) 2(28)(W/m2 o C) / 186( W/m o C)0.01( m) 0.99 Fig. 2.16 gives fr 0.78 and f t 0.7 Substituting into (E) and (F) gives q fr (To T )W 7.86 W m o C 7.06 W m o C and qft (To T )W This result shows that the error in using the efficiency graphs is small. 2-54 PROBLEM 2.20 The profile of a straight fin is given by y s ( x / L) 2 where L is the fin length and is half the base thickness. The fin exchanges heat with the ambient by convection. The ambient temperature is T and the heat transfer coefficient is h. The base temperature is To . Assume that ( / L) 1 , determine the steady state fin heat transfer rate. h, T ys 0 h, T L x To (1) Observations. (i) Temperature distribution in the fin and Fourier’s law of conduction give the heat transfer rate. (ii) This is a variable area straight fin with specified temperature at the base. (iii) Temperature distribution can be assumed one-dimensional. (iv) Eq. (2.5) should be used to formulate the heat equation for this fin. (2) Origin and Coordinates. The origin is selected at the tip and the coordinate x is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4) Bi 1, (5) constant properties, (6) uniform h and T , and (7) negligible radiation. (ii) Governing Equations. Applying Fourier’s law at x = L gives the fin heat transfer rate q f kAc ( L) dT ( L) dx (a) where Ac (L) is the cross-sectional area at x = L, given by Ac ( L) 2 W (b) where W is the fin depth. Substituting (b) into (a) q f 2k W dT ( L) dx (c) Temperature distribution in the fin is obtained from the solution to the heat equation for the fin. Based on the above assumptions eq. (2.5b) gives dA d 2T 1 dAc dT h (T T ) s 0 2 Ac ( x ) dx dx kAc ( x ) dx dx (d) The cross-sectional area Ac (x ) for a straight fin is Ac 2 y s2W (e) y s ( x / L) 2 (f) where y s is the local width given by 2-55 PROBLEM 2.20 (continued) Substituting (f) into (e) Ac ( x ) 2W ( x / L) 2 (g) Eq. (2.6a) is used to formulate dAs / dx dAs C ( x) 1 (dys / dx) 2 dx 1/ 2 Using (f) into the above and noting that C = 2W dAs 2W 1 (2 x / L2 ) 2 dx 1/ 2 2W (h) Substituting (g) and (h) into (d) and rearranging x2 d 2 d 2x 0 2 dx dx where T T (i) (j) and hL2 k (k) (iii) Boundary Conditions. The two boundary conditions are (1) Finite temperature at x = 0 T (0) finite, or (0) T (0) T finite (l) (2) Specified temperature at x = L T ( L) To , or ( L) T ( L) T To T (m) (4) Solution. Equation (i) is a second order differential equation with variable coefficient. Examination of the coefficients shows that it is an Euler equation. Comparing (i) with eq. (2.49) gives a 0 and a1 2 (n) The solution to (i) depends on the roots r1, 2 of eq. (2.50) r1, 2 (a1 1) (a1 1) 2 4a0 2 (o) Substituting (n) into (o) and using (k) r1, 2 (1 / 2) (1 / 4) hL2 / k (p) r1 (1 / 2) (1 / 4) hL2 / k (q) Thus the two roots are 2-56 PROBLEM 2.20 (continued) and r2 (1 / 2) (1 / 4) hL2 / k (r) The solution to (i) is given by eq. (2.51) ( x) T ( x) T C1 x r1 C2 x r2 (s) where C1 and C2 are constants of integration. Application of boundary condition (l) gives C2 0 (t) Boundary condition (m) gives r1 C1 (To T ) / L (u) Substituting (t) and (u) into (s) gives the temperature solution T ( x ) T (To T )( x / L) r1 (v) Expressing the above in dimensionless form gives T ( x) T ( x / L) r1 (To T ) (w) Fin heat transfer rate is obtained by substituting (v) into (c) and using (q) qf Wk ( / L)(To T ) 2 (1/ 4) hL / k (1/ 2) 2 (x) (5) Checking. Dimensional check: (i) Each term in (v) has units of temperature. (ii) Each term in (w) and (x) is dimensionless. Boundary conditions check: Equation (v) satisfies boundary conditions (l) and (m). Differential equation check: Direct substitution of (v) into (i) confirms that the governing equation is satisfied. Limiting check: If To T , the temperature distribution in the fin should be uniform equal to T and the heat transfer should vanish. Setting To T in (v) gives T ( x) T . Similarly, setting To T in (x) gives q f 0. (6) Comments. (i) Not every second order differential equation with variable coefficient is a Bessel equation. In this example the equation is Euler. (ii) The minus sign in equation (x) indicates that for To T heat flows in the negative x-direction. This is consistent with the second law of thermodynamics. (iii) The solution is characterized by a single dimensionless parameter hL2 / k . This parameter appears in (q) and (x). 2-57 PROBLEM 2.21 A circular disk of radius Ro is mounted on a tube of radius Ri . The profile of the disk is described by y s Ri2 / r 2 Ro Ri h, T ys r r h, T where is half the thickness at the To tube. The disk exchanges heat with the surroundings by convection. The heat transfer coefficient is h and the ambient temperature is T . The base is maintained at To and the tip is insulated. Assume that 4 / Ri 1, use a fin model to determine the steady state heat transfer rate from the disk. (1) Observations. (i) Temperature distribution in the fin and Fourier’s law of conduction give the heat transfer rate. (ii) This is a variable area annular fin with specified temperature at the base, insulated tip and convection at the surface . (iii) Temperature distribution can be assumed one-dimensional. (iv) Use eq. (2.5) to formulate the heat equation for the disk. (2) Origin and Coordinates. The origin is selected at the tube center and the coordinate r is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4) Bi 1, (5) constant properties, (6) uniform h and T , and (7) negligible radiation. (ii) Governing Equations. Applying Fourier’s law at r Ri gives the heat transfer rate dT ( Ri ) q f kAc ( Ri ) (a) dr where Ac ( Ri ) is the cross-sectional area at r Ri , given by Ac ( Ri ) 4 Ri (b) Substituting (b) into (a) q f 4 Ri k dT ( Ri ) dr (c) Temperature distribution in the fin is obtained from the solution to the heat equation for the fin. Based on the above assumptions eq. (2.5b) gives dA d 2T 1 dAc dT h (T T ) s 0 2 Ac ( r ) dr dr kAc ( r ) dr dr (d) The cross-sectional area Ac (r ) for an annular fin is Ac ( r ) 4 r y s (e) where y s is the local fin thickness given by y s Ri2 / r 2 2-58 (f) PROBLEM 2.21 (continued) Substituting (f) into (e) Ac ( r ) 4 Ri2 / r (g) Eq. (2.6a) is used to formulate dAs / dr dAs C (r ) 1 (dy s / dr) 2 dr 1/ 2 Using (f) into the above and noting that C ( r ) 4 r dAs 4r 1 (4 2 Ri4 / r 6 ) dr 1/ 2 However, since 4 / Ri 1, it follows that (4 2 Ri4 / r 6 ) 1 . Thus the above simplifies to dAs 4 r dr (h) Substituting (g) and (h) into (d) and rearranging d 2 d r r 2 r 4 0 2 dr dr (i) T T (j) h k Ri2 (k) 2 where and 2 (iii) Boundary Conditions. The two boundary conditions are (1) Specified temperature at r = Ri ( Ri ) To T (2) Insulated tip at r = Ro (l) d ( Ro ) 0 dr (m) (4) Solution. Equation (i) is a second order differential equation with variable coefficient. It may be a Bessel equation. Comparing (i) with eq. (2.26) gives A 1 B0 C2 D i / 2, p D / i / 2 n 1/ 2 Since n is not an integer and D is imaginary the solution to (i) is given by eq. (2.30) ( r ) T ( r ) T r C1 I1/ 2 ( r 2 / 2) C2 I 1/ 2 ( r 2 / 2) 2-59 (n) PROBLEM 2.21 (continued) where C1 and C2 are constants of integration. However using equations (2.35) and (2.36), the Bessel functions in (n) can be expressed in terms of hyperbolic functions as I1/ 2 ( r 2 / 2) (2 / r ) 1 / sinh( r 2 / 2) (o) I 1/ 2 ( r 2 / 2) (2 / r ) 1 / cosh( r 2 / 2) (p) and Substituting (o) and (p) into (n) T (r ) T 2C1 sinh( r 2 / 2) 2C 2 cosh ( r 2 / 2) (q) Boundary conditions (l) and (m) give C1 and C2 C1 C2 ( / 2)(To T ) tanh( Ro2 / 2) cosh( Ri2 / 2) tanh( Ro2 / 2) sinh( Ri2 / 2) ( / 2)(To T ) cosh( Ri2 / 2) tanh( Ro2 / 2) sinh( Ri2 / 2) (r) (s) Substituting equations (r) and (s) into(q) cosh( r 2 /2) tanh( R o2 /2) sinh( r 2 /2) T T To T cosh( Ri2 /2) tanh( R o2 /2) sinh( Ri2/2) (t) Differentiating (t) and substituting into (c) gives the fin heat transfer rate qf (To T ) hk Ri2 4 sinh( Ri2 /2) tanh( R o2 /2) cosh( Ri2/2) cosh( Ri2 /2) tanh( R o2 /2) sinh( Ri2/2) (u) (5) Checking. Dimensional check: (i) The arguments of the sinh and cosh are dimensionless. (ii) Each term in (t) and (u) is dimensionless. Boundary conditions check: Equation (t) satisfies boundary conditions (l) and (m). Differential equation check: Direct substitution of (t) into (i) confirms that the governing equation is satisfied. Limiting check: If To T the temperature distribution in the fin should be uniform equal to T and the heat transfer should vanish. Setting To T in (t) gives T (r) T . Similarly, setting To T in (u) gives q f 0. (6) Comments. (i) The minus sign in equation (u) indicates that for To T heat flows in the negative x-direction. (ii) Neglecting the term (4 2 Ri4 / r 6 ) reduces the fin equation to a Bessel differential equation and facilitates obtaining a solution. (iii) The problem is characterized by two dimensionless parameters: Ri2 and Ro / Ri . 2-60 PROBLEM 2.22 A very large disk of thickness is h, T h, T mounted on a shaft of radius Ri . The shaft rotates with angular velocity . r h, T h, T The pressure at the interface between Ri the shaft and the disk is P and the coefficient of friction is . The disk exchanges heat with the ambient by convection. The heat transfer coefficient is h and the ambient temperature is T . Neglecting heat transfer to the shaft at the interface, use a fin model to determine the interface temperature. (1) Observations. (i) Temperature distribution in the fin gives the temperature at the interface. (ii) This is a variable area annular fin with uniform thickness. (iii) Heat is generated at the shaft-disk interface due to friction and is conducted through the disk. Thus the heat flux is specified at the base r Ro . Disk temperature far away from the interface approaches ambient level. (iv) Heat exchange at the surface is by convection. (v) Temperature distribution can be assumed one-dimensional. (vi) Eq. (2.5) should be used to formulate the heat equation for the disk. (2) Origin and Coordinates. The origin is selected at the shaft center and the coordinate r is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4) Bi 1, (5) constant properties, (6) uniform h and T , (7) all frictional energy is conducted through the disk and (8) negligible radiation. (ii) Governing Equations. Temperature distribution in the disk is obtained from the solution to the fin heat equation. Based on the above assumptions eq. (2.5b) gives dAs d 2T 1 dAc dT h ( T T ) 0 dr dr 2 Ac ( r ) dr dr kAc ( r ) (a) The cross-sectional area Ac (r ) for an annular fin of constant thickness is Ac ( r ) 4 r (b) Eq. (2.6a) is used to formulate dAs / dr dAs C (r ) 1 (dy s / dr) 2 dr 1/ 2 (c) where C(r) is the circumference and y s is the disk half thickness given by C ( r ) 4 r ys (d) (e) dAs 4r dr (f) Substituting (d) and (e) into (c) 2-61 PROBLEM 2.22 (continued) Substituting (b) and (f) into (a) and rearranging r2 d 2 d r 2 r 2 0 2 dr dr where T T (g) (h) and 2 2h k (i) (iii) Boundary Conditions. The two boundary conditions are: (1) Specified flux at r = Ri . Frictional heat generated at the interface is proportional to the tangential force and tangential velocity. Assuming that frictional energy is conducted through the disk, Fourier’s law gives k (2) Finite temperature at r d ( Ri ) P Ri dr ( ) finite (j) (k) (4) Solution. Equation (g) is a second order differential equation with variable coefficient. It may be a Bessel equation. Comparing (g) with eq. (2.26) gives A0 B0 C 1 D i, p D / i n0 Since n is zero and D is imaginary the solution to (g) is given by eq. (2.29) ( r ) T ( r ) T C1 I 0 ( r ) C2 K 0 ( r ) (l) where C1 and C2 are constants of integration. Since I 0 () (Table 2.1) boundary condition (k) gives (m) C1 0 Thus solution (l) becomes T ( r ) T C2 K 0 ( r ) (n) Using eq. (2.46) to determine the derivative of K 0 ( r ) , boundary condition (j) gives C2 P Ri k K1 ( Ri ) C2 (o) Substituting (o) into (n) and using (i) gives T (r ) T P Ri K 0 ( r ) 2hk / K1 ( Ri ) 2-62 (p) PROBLEM 2.22 (continued) To determine interface temperature we set r Ri in (p) T ( Ro ) T P Ri K 0 ( Ri ) 2hk / K1 ( Ri ) (q) Expressing (p) in dimensionless form gives K ( r) T (r ) T 0 P Ri K1 ( Ri ) 2hk / (r) (5) Checking. Dimensional checks: (i) The argument of the Bessel function, ( r ) is dimensionless. (ii) Each term in (p) has units of temperature. Boundary conditions check: Equation (p) satisfies boundary conditions (j) and (k). Differential equation check: Direct substitution of (p) into (g) confirms that the governing equation is satisfied. Limiting check: If no frictional energy is generated, the disk temperature distribution should be uniform equal to T . Setting any of the quantities , P or equal to zero in (p) gives T (r) T . (6) Comments. (i) Since K o ( r ) decreases monotonically with r (see Fig. 2.15), equation (p) shows that the maximum disk temperature occurs at the interface. (ii) The solution is characterized by a single dimensionless parameter Ri . 2-63 PROBLEM 2.23 A circular disk of thickness and outer radius Ro is mounted on a shaft of radius Ri . The shaft and disk rotate inside a stationary sleeve with angular velocity . Because of friction between the disk and sleeve, heat is generated at a flux q o . The disk exchanges heat with the surroundings by convection. The ambient temperature is T . Due to radial variation in the tangential velocity, the heat transfer coefficient insulated varies with radius according to sleeve h, T h, T h h (r / R ) 2 i i Assume that no heat is conducted to the sleeve and shaft, use a fin model to determine the steady state temperature distribution. r h, T Ri h, T Ro (1) Observations. (i) This is a variable area annular fin with uniform thickness. (ii) Frictional heat generated at the sleeve is conducted through the disk. (iii) Heat exchange at the surface is by convection. (iv) Temperature distribution can be assumed one-dimensional. (v) The heat transfer coefficient varies over the disk. (vi) Eq. (2.5) should be used to formulate the heat equation for the disk. (2) Origin and Coordinates. The origin is selected at the center of shaft and the coordinate r is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4) Bi 1, (5) constant properties, (6) uniform ambient temperature T , (7) all frictional energy is conducted inwardly through the disk, (8) the shaft is insulated and (9) negligible radiation. (ii) Governing Equations. Temperature distribution in the disk is obtained from the solution to the fin heat equation. Based on the above assumptions eq. (2.5b) gives dA d 2T 1 dAc dT h (T T ) s 0 2 Ac ( r ) dr dr kAc ( r ) dr dr (a) The cross-sectional area Ac (r ) for an annular fin of constant thickness is Ac ( r ) 2 r (b) Eq. (2.6a) is used to formulate dAs / dr dAs C (r ) 1 (dys / dr) 2 dr 1/ 2 (c) where C(r) is the circumference and y s is the disk half thickness given by C ( r ) 4 r (d) ys / 2 (e) The heat transfer coefficient varies with radius according to 2-64 PROBLEM 2.23 (continued) h hi ( r / Ri ) 2 (f) d 2 d r 4 2 r 4 0 2 dr dr (g) Substituting (d), (e) and (f) into (a) r2 where T T (h) and 2 hi (i) 2k Ri2 (iii) Boundary Conditions. The two boundary conditions are (1) Insulated surface at r = Ri (2) Specified flux at r = Ro k d ( Ri ) 0 dr (j) d ( Ro ) qo dr (k) (4) Solution. Equation (g) is a second order differential equation with variable coefficient. It may be a Bessel equation. Comparing (g) with eq. (2.26) gives A0 B0 C2 D i, p D / i n0 Since n is zero and D is imaginary the solution to (g) is given by eq. (2.29) ( r ) T ( r ) T C1 I 0 ( r 2 ) C2 K 0 ( r 2 ) (l) where C1 and C2 are constants of integration. Application of boundary conditions (j) and (k) and using eqs. (2.45) and (2.46) to determine the derivatives of the Bessel functions in (l) give the constants C1 and C2 I 1 ( Ri2 ) 2 C1 I ( R ) K1 ( Ro2 ) 1 o 2 K1 ( Ri ) 2hi kRo2 / Ri2 qo C2 I 1 ( Ri2 ) 2 I ( R ) K1 ( Ro2 ) 1 o 2 K 1 ( Ri ) 2hi kRo2 / Ri2 qo 1 1 I 1 ( Ri2 ) K 1 ( Ri2 ) Substituting (m) and (n) into (l) gives the temperature distribution in dimensionless form 2-65 (m) (n) PROBLEM 2.23 (continued) T (r ) T qo I ( Ri2 ) I1 ( Ro2 ) 1 K1 ( Ro2 ) 2 K1 ( Ri ) 1 I1 ( Ri2 ) 2 I ( r ) K o ( r 2 ) (o) o 2 K1 ( Ri ) 2hi kRo2 / Ri2 (5) Checking. Dimensional checks: (i) The argument of the Bessel function, ( r ), is dimensionless. (ii) The quantity qo / 2hi kRo2 / Ri2 in (o) has units of temperature. Thus each term in (o) is dimensionless. Boundary conditions check: Equation (o) satisfies boundary conditions (j) and (k). Differential equation check: Direct substitution of (o) into (g) confirms that the governing equation is satisfied. Limiting check: If no frictional energy is generated, the disk temperature distribution should be uniform equal to T . Setting q o equal to zero in (o) gives T (r) T . (6) Comments. (i) Since K o ( r 2 ) decreases with r and I o ( r 2 ) increases with r (see Fig. 2.15), equation (o) shows that the maximum disk temperature occurs at the interface r Ro . (ii) The solution is characterized by two dimensionless parameter: Ri2 and Ro / Ri . 2-66 PROBLEM 2.24 A specially designed heat exchanger consists of a tube with fins mounted on its inside surface. The fins are circular disks of inner radius Ri and variable thickness given by y s (r / Ro ) 3 / 2 where Ro is the tube radius. The disks exchange heat with the ambient fluid by convection. The heat transfer coefficient is h and the fluid temperature is T . The tube surface is maintained at To . Modeling each disk as a fin, determine the steady state heat transfer rate. Neglect heat loss from the disk surface at Ri and assume that ( / Ro ) 1. h, T ys r 0 h, T Ri Ro To (1) Observations. (i) This is a variable area annular fin with variable thickness. (ii) The base is maintained at a specified temperature and the tip is insulated. (iii) Heat exchange at the surface is by convection. (iv) Temperature distribution can be assumed one-dimensional. (v) Eq. (2.5) should be used to formulate the heat equation for the fin. (2) Origin and Coordinates. The origin is selected at the center of tube and the coordinate r is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4) Bi 1, (5) constant properties, (6) uniform heat transfer coefficient h and ambient temperature T , and (7) no radiation. (ii) Governing Equations. Temperature distribution in the fin is obtained from the solution to the fin heat equation. Based on the above assumptions eq. (2.5b) gives dA d 2T 1 dAc dT h (T T ) s 0 2 Ac ( r ) dr dr kAc ( r ) dr dr (a) The cross-sectional area Ac (r ) for an annular fin of is Ac ( r ) 4 r y s (b) where y s is the local fin thickness given by Substituting (c) into (b) y s ( r / Ro ) 3 / 2 (c) Ac ( r ) 4 r 5 / 2 / Ro3 / 2 (d) Eq. (2.6a) is used to formulate dAs / dr dAs C (r ) 1 (dys / dr) 2 dr 2-67 1/ 2 (e) PROBLEM 2.24 (continued) Using (c) into (e) and noting that C ( r ) 4 r dAs 4r 1 (9 2 r / 4 Ro3 ) dr 1/ 2 (f) However, since / Ri 1, it follows that 9 2 r / 4 Ro3 1 . Thus the above simplifies to dAs 4 r dr (g) Substituting (d) and (g) into (a) and rearranging r2 d 2 d (5 / 2)r 2 r 1/ 2 0 2 dr dr where (h) T T (i) hRo3 / 2 k (j) and 2 (iii) Boundary Conditions. The two boundary conditions are (1) Insulated surface at r = Ri d ( Ri ) 0 dr (k) (2) Specified temperature at r = Ro ( Ro ) To T (l) (4) Solution. Equation (h) is a second order differential equation with variable coefficient. It may be a Bessel equation. Comparing (h) with eq. (2.26) gives A 3/ 4 B0 C 1/ 4 D 4 i, p D / i 4 n3 Since n is and integer and D is imaginary the solution to (h) is given by eq. (2.29) ( r ) T ( r ) T r 3 / 4 C1 I 3 (4 r1/ 4 ) C2 K 3 (4 r1/ 4 ) (m) where C1 and C2 are constants of integration. Application of boundary conditions (k) and (l) and using eqs. (2.45) and (2.46) to determine the derivatives of the Bessel functions in (k) give the constants C1 and C2 : (T T ) Ro3 / 4 ( 2 / 3) Ri1 / 4 I 2 (4 Ri1 / 4 ) I 3 ( 4 Ri1 / 4 ) I 3 (4 Ro1 / 4 ) C1 o K 3 (4 Ro1 / 4 ) ( 2 / 3) Ri1 / 4 K 2 (4 Ri1 / 4 ) K 3 ( 4 Ri1 / 4 ) K 3 ( 4 Ro1 / 4 ) and 2-68 1 (n) PROBLEM 2.24 (continued) (To T ) Ro3 / 4 ( 2 / 3) Ri1 / 4 I 2 ( 4 Ri1 / 4 ) I 3 ( 4 Ri1 / 4 ) I 3 ( 4 Ro1 / 4 ) C2 K 3 ( 4 Ro1 / 4 ) ( 2 / 3) Ri1 / 4 K 2 ( 4 Ri1 / 4 ) K 3 ( 4 Ri1 / 4 ) K 3 ( 4 Ro1 / 4 ) 1 ( 2 / 3) Ri1 / 4 I 2 ( 4 Ri1 / 4 ) I 3 ( 4 Ri1 / 4 ) (n) ( 2 / 3) Ri1 / 4 K 2 ( 4 Ri1 / 4 ) K 3 ( 4 Ri1 / 4 ) Fin heat transfer rate q is determined by applying Fourier’s law at r Ro q 4 kRo dT ( Ro ) dr (o) Using (m) into (o) 3 q 4 kRo Ro7 / 4 C1 I 3 (4 R1o / 4 ) C 2 K 3 (4 R1o / 4 ) 4 1 Ro3 / 2 4 C1 I 2 (4 R1o / 4 ) 3 R o1 / 4 C1 I 3 (4 R1o / 4 ) 4 3 4 C 2 K 2 (4 R1o / 4 ) R o1 / 4 C 2 K 3 (4 R1o / 4 ) 1 (p) (5) Checking. Dimensional checks: (i) The argument of the Bessel function, ( r 1 / 4 ), is dimensionless. (ii) C1 and C2 have units of ( o C m 3/4 ) . Thus each term in (m) has units of temperature. (6) Comments. (i) Neglecting the term (9 2 r / 4 Ro3 ) reduces the fin equation to a Bessel differential equation and facilitates obtaining a solution. (ii) The solution is characterized by two dimensionless parameter: Ri2 and Ro / Ri . 2-69 PROBLEM 2.25 A disk of radius Ro and thickness Ro Ri is mounted on a steam pipe of outer Rb radius Ri . The disk’s surface is h, T h, T insulated from r Ri to r Rb . The To 0 r remaining surface exchanges heat h, T h, T with the surroundings by convection. The heat transfer coefficient is h and the ambient temperature is T . The temperature of the pipe is To and the surface at r Ro is insulated. Formulate the governing equations and boundary conditions and obtain a solution for the temperature distribution in terms of constants of integration. (1) Observations. (i) This is an annular disk with uniform thickness. (ii) One section of the disk’s surface is insulated while the other exchanges heat by convection. Thus, two governing equations are needed for the temperature distribution. (iii) Assuming that Bi 1 the disk can be modeled as a fin with inner surface at a specified temperature and the outer surface insulated. (iv) Temperature distribution can be assumed one-dimensional. (v) Use eq. (2.5) to formulate the heat equation for each section of the disk. (2) Origin and Coordinates. The origin is selected at the center of shaft and the coordinate r is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4) Bi 1, (5) constant properties, (6) uniform heat transfer coefficient h and ambient temperature T and (7) negligible radiation. (ii) Governing Equations. Since one section is insulated while the other exchanges heat by convection, two fin equations are needed. Let T1 ( r ) = temperature distribution in the first section, Ri r Rb T2 ( r ) = temperature distribution in the second half, Rb r Ro The fin equation is obtained from eq. (2.5b) dA d 2T 1 dAc dT h (T T ) s 0 2 Ac ( r ) dr dr kAc ( r ) dr dr (a) The cross-sectional area Ac (r ) for an annular fin of constant thickness is Ac ( r ) 2 r (b) Eq. (2.6a) is used to formulate dAs / dr dAs C ( r ) 1 ( dys / dr) 2 dr 1/ 2 (c) where y s and C(r) is the circumference given by C ( r ) 4 r 2-70 (d) PROBLEM 2.25 (continued) In the first section the fin is insulated along its surface and thus the convection term vanishes. Substituting (b) into (a) and setting h = 0 gives d 2T1 1 dT1 0, dr 2 r dr Ri r Rb (e) Similarly, substituting (b), (c) and (d) into (a) gives the fin equation for the second section d 2 d r r 2 r 2 0 , 2 dr dr 2 Rb r R0 (f) where 2 and 2h k (r) T2 (r) T (g) (h) (iii) Boundary Conditions. Four boundary conditions are needed. (1) Specified temperature at r Ri T1 (0) To (i) T1 ( Rb ) T2 ( Rb ) ( Rb ) T (j) (2) Equality of temperature at r Rb (3) Equality of flux at r Rb dT1 ( Rb ) d ( Rb ) dr dr (k) (4) Insulated surface at r Ro d ( Ro ) 0 dr (4) Solution. Separating variables and integrating (e) twice T1 (r ) C1 ln r C2 (l) (m) where C1 and C2 are constants of integration. Equation (f) is a second order differential equation with variable coefficient. It may be a Bessel equation. Comparing (f) with eq. (2.26) gives A0 B0 C 1 D i, p D / i n0 Since n is zero and D is imaginary the solution to (f) is given by eq. (2.29) ( r ) T2 ( r ) T C3 I 0 ( r ) C4 K 0 ( r ) 2-71 (n) PROBLEM 2.25 (continued) where C3 and C4 are constants of integration. (5) Checking. Dimensional check: The argument of the Bessel function is dimensionless. Differential equation check: Direct substitution of (i) into (e) and (j) into (f) confirms that the governing equations are satisfied. (6) Comments. When conditions along a fin are not uniform, it is necessary to use more than one fin equation. 2-72 PROBLEM 2.26 Consider the moving plastic sheet of Example 2.4. In addition to cooling by convection at the top surface, the sheet is cooled at the bottom at a constant flux qo. Determine the temperature distribution in the sheet. h, T x U To furnace qo dx W t (1) Observations. (i) This is a constant area moving fin problem. (ii) Temperature distribution can be assumed one-dimensional. (iii) Along the upper surface heat is exchanged with the surroundings by convection. At the lower surface heat is removed at uniform flux. (iv) The temperature is specified at the outlet of the furnace. (v) The sheet is semi-infinite. (2) Origin and Coordinates. A rectangular coordinate system is used with the origin at the exit of furnace. The coordinate x is directed as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant pressure, (4) constant properties, (5) constant conveyor speed, (6) uniform heat transfer coefficient h and ambient temperature T , (7) uniform flux qo , (8) no radiation, (9) insulated sides and (10) Bi 1. (ii) Governing Equations. Formulation of the heat equation for this moving fin must take into consideration energy transport due to material motion, surface convection and heat loss at uniform flux from the lower surface. Consider a constant area fin moving with constant velocity U. For steady state, conservation of energy for the element dx gives dq dhˆ q x m hˆ q x x dx m hˆ m dx dqc dqo (a) dx dx is mass flow rate and q x and q c are heat transfer rates where ĥ is enthalpy per unit mass, m by conduction and convection, respectively and q o is heat loss from the bottom surface. Mass flow rate is given by m UAc (b) where Ac is the cross-sectional area and is density. For a constant pressure process the change in enthalpy is dhˆ c p dT (c) where c p is specific heat. Using Fourier’s and Newton’s laws, we obtain q x kAc and dT dx dqc h (T T )dAs (d) (e) where Ac is the heat conduction area, dAs is the surface area of the element through which heat is convected and T is the ambient temperature. The conduction area is given by Ac W t 2-73 (f) PROBLEM 2.26 (continued) where W is the sheet width and t is its thickness. Since the sides are assumed to be insulated the convection area dAs is dAs Wdx (g) Heat removed from the lower surface of the element, dqo , is given by dqo qoWdx (h) Substituting (b)-(h) into (a) q d 2T c p U dT h (T T ) o 0 2 k dx k t kt dx This is the fin equation for the moving sheet. (i) (iii) Boundary Conditions. Since equation (i) is second order with a single independent variable x, two boundary conditions are needed. (1) The temperature is specified at x = 0 T (0) To (j) (2) Far away from the furnace the temperature approaches T . Thus T ( ) = finite (k) (4) Solution. Equation (i) is a linear, second order differential equation with constant coefficients. Its solution is presented in Appendix A. Equation (i) is rewritten as d 2T dT 2b m 2T c 2 dx dx (l) where b, c and m are defined as b c pU 2k , c qo h T , kt kt m2 h kt (m) Noting that b 2 > m 2 , the solution is given by equation (A-6c) T C1 exp( bx b 2 m 2 x ) C2 exp( bx b 2 m 2 x ) c m2 (n) where C1 and C2 are constants of integration. Since m 2 0, equation (k) requires that C1 0 (o) Equation (j) gives C2 C2 To c m2 (p) Substituting (m), (o) and (p) into (n) gives the temperature distribution in the sheet c pU t (q o ) (q o ) T ( x) T 1 exp To T h(To T ) h(To T ) 2k ( c pU t 2k 2 ) ht x k t (5) Checking. Dimensional check: (i) The exponent in (q) must be dimensionless 2-74 (q) PROBLEM 2.26 (continued) ( kg/m 3 ) c p ( J/kg o C)U ( m/s) x ( m) k ( W/m o C) = dimensionless and h( W/m2 o C) o k ( W/m C)t ( m) 1/ 2 x ( m) = dimensionless (ii) The term qo / h(To T ) must be dimensionless qo ( W/m 2 ) h( W/m 2 o C)(To T )( o C) = dimensionless Boundary conditions check: Substitution of (q) into (j) and (k) shows that the two boundary conditions are satisfied. Differential equation check: Direct substitution of (q) into (i) confirms that the governing equation is satisfied. Limiting checks: (i) If h = 0 and qo 0 no energy can be exchanged with the sheet after leaving the furnace and therefore its temperature must remain constant. Setting h = 0 and qo 0 in (q) gives T ( x ) To . (ii) If the velocity U is infinite, the time it takes the fin to reach a distance x vanishes and therefore energy added or removed also vanishes. Thus for U the temperature remains constant. Setting U in (q) gives T ( x ) To . (iii) For the special case of qo 0 the solution should agree with the result of Example 2.4. Setting qo 0 in (q) gives the same result as eq. (2.22) for the case of insulated sides. (iv) Far away from the exit of the furnace ( x ) the temperature becomes uniform and thus energy removed from the bottom must be equal to energy added by convection at the top. Setting x in (q) gives T () T ( qo / h ) 0 or h T T () qo This is recognized as Newton’s law of cooling applied to the upper surface. (6) Comments. (i) The temperature decays exponentially with distance x. (ii) Sheet motion has the effect of slowing down the decay. In the limit, U , no decay takes place and the temperature remains constant. (iii) The solution is characterized by three dimensionless parameter: c pU t qo ht , , k h(To T ) k Note that the first parameter is the Biot number. The second is the ratio of the heat removed at the lower surface to the heat convected at the upper surface at x = 0. The third parameter is the Peclet number (Reynolds times Prandtl) represents the effect of motion (axial convection). 2-75 PROBLEM 2.27 A cable of radius ro , conductivity k, density and specific heat c p moves through a furnace with velocity U and leaves at temperature To . Electric energy is dissipated in the cable resulting in a volumetric energy generation rate of q . Outside the furnace the cable is cooled by convection. The heat h, T To transfer coefficient is h and the ambient x U temperature is T . Model the cable as a 0 q dx fin and assume that it is infinitely long. Determine the steady state temperature furnace distribution in the cable. (1) Observations. (i) This is a constant area moving fin problem. Temperature distribution can be assumed one-dimensional. (ii) Heat is exchanged with the surroundings by convection. (iii) The temperature is specified at the outlet of the furnace. (iv) The fin is semi-infinite. (v) The cable generates volumetric energy. (2) Origin and Coordinates. A rectangular coordinate system is used with the origin at the exit of furnace, as shown. (3) Formulation. dq c (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant pressure, (4) constant properties, (5) constant speed, (6) constant h, (7) no radiation, (8) uniform energy generation, and (9) Bi 1. (ii) Governing Equations. Consider an element dx of the cable. Energy balance for the element gives ˆ h m qx ˆ ˆdhdx) (h m dx dq qx x dx dx qAcdx dx dq dhˆ q x m hˆ q Ac dx q x x dx m hˆ m dx dqc dx dx (a) is mass flow rate, Ac is conduction area, and q x , q c where ĥ is enthalpy per unit mass, m and q are heat transfer rates by conduction, convection and energy generation, respectively. is given by Mass flow rate m m UAc (b) where is density. For constant pressure process the change in enthalpy is dhˆ c p dT (c) where c p is specific heat. Using Fourier’s and Newton’s laws, we obtain q x kAc dT dx (d) and dqc h (T T )dAs 2-76 (e) PROBLEM 2.27 (continued) where dAs is the surface area of the element and T is the ambient temperature. Surface area is expressed in terms of fin circumference C as dAs C dx (f) Substituting (b)-(f) into (a) d 2T c p U dT hC q ( T T ) 0 k dx k Ac k dx 2 (g) The cross-sectional area and circumference are given by Ac ro2 (h) C 2 ro (i) T T (j) d 2 d 2b m 2 c 2 dx dx (k) and Define Substituting (h)-(j) into(g) where the constants b and m 2 are defined as b c pU 2k , m2 2h q , c kro k (l) (iii) Boundary Conditions. Since equation (c) is second order with a single independent variable x, two boundary conditions are needed. The temperature is specified at x = 0. Thus (0) To T (m) Far away from the furnace the temperature is T . Thus, the second boundary condition is () = finite (n) (4) Solution. Equation (c) is a linear, second order differential equation with constant coefficients. Its solution is presented in Appendix A. Noting that b 2 m 2 , the solution is given by equation (A-6c) C1 exp(bx b 2 m 2 x) C2 exp(bx b 2 m 2 x) c m2 (o) where C1 and C2 are constants of integration. Since m 2 0, boundary condition (f) requires that C1 0 Equation (m) gives C2 2-77 (p) PROBLEM 2.27 (continued) c m2 Substituting (p) and (q) into (o) gives the temperature distribution in the cable C 2 To T (q) c c 2 2 (r) ) exp( bx b m x ) m2 m2 Substituting the definitions of c, b and m, given in equation (l), solution (r) is rewritten in dimensionless form (To T c pUro q ro T ( x) T 1 exp To T 2k 2h(To T ) ( c pUro 2k 2 ) 2hro k x q ro ro 2h(To T ) (s) (5) Checking. Dimensional check: (1) The exponent in (r) must be dimensionless: (kg/m 3 ) c p (J/kg o C)U (m/s)ro (m) k ( W/m o C) = dimensionless and h( W/m 2 o C)ro (m) k ( W/m o C) = dimensionless (2) The coefficient in (r) must be dimensionless: q ( W/m3 )ro (m) 2h( W/m 2 o C)(To T )( o C) dimensionless Boundary conditions check: Substitution (s) into (m) and (n) shows that the two boundary conditions are satisfied. Differential equation check: Direct substitution of (s) into (g) confirms that the governing equation is satisfied. Limiting checks: (i) If h = 0, no energy can be removed from the cable after leaving the furnace and therefore its temperature must continue to increase due to energy generation until it becomes infinite. Setting h = 0 in (s) gives T ( x ) . (ii) At x the cable reaches a uniform temperature T () representing a balance between energy generation and surface convection. Note that motion plays not role. Setting x in (s) gives q ro (t) T () T 2h Equation energy generated to surface energy convected for an cable element x at x gives 2ro xh[T () T ] ro2 xq (u) Solving (u) for T () gives the result shown in (t). (6) Comments. (i) The temperature decays exponentially with distance x. (ii) Cable motion has the effect of slowing down the decay. In the limit, U , no decay takes place and the 2-78 PROBLEM 2.27 (continued) temperature remains constant. (iii) The solution is characterized by three dimensionless parameter: c pU ro q ro hro , , k k h(To T ) Note that the first parameter is the Biot number. The second is the ratio of the heat generated at the lower surface to the heat convected at the upper surface at x = 0. The third parameter represents the effect of motion. 2-79 PROBLEM 2.28 A cable of radius ro , conductivity k, density and specific heat c p moves with velocity U through a furnace and leaves at temperature To . Outside the furnace the cable is cooled by radiation. The surroundings temperature is Tsur Tsur and surface emissivity is . In certain cases To axial conduction can be neglected. Introduce this x U 0 simplification and assume that the cable is infinitely long. Use a fin model and determine furnace the steady state temperature distribution in the cable. (1) Observations. (i) This is a constant area moving fin with surface radiation problem. Temperature distribution can be assumed one-dimensional. (ii) Heat is exchanged with the surroundings by radiation only. (iii) The temperature is specified at the outlet of the furnace. (iv) The fin is semi-infinite. (v) Axial conduction can be neglected. (2) Origin and Coordinates. A rectangular coordinate system is used with the origin at the exit of furnace, as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant pressure, (4) constant properties (5) constant speed, (6) no surface convection, (7) negligible axial conduction, and (8) Bi 1. (ii) Governing Equations. The heat equation for a moving fin with surface radiation is given by eq. (2.19) C 4 d 2T c p U dT hC 4 (a) (T T ) (T Tsur )0 2 k dx k Ac k Ac dx where Ac ro2 C 2 ro (b) (c) Neglecting axial conduction and surface convection and using (b) and (c), equation (c) simplifies to c p U dT k dx 2 4 (T 4 Tsur )0 k ro (d) (iii) Boundary Conditions. Since equation (d) is first order equation with a single independent variable x, one boundary condition is needed. The temperature is specified at x = 0. Thus T (0) To (e) (4) Solution. Separating variables, (d) becomes dT T 4 4 Tsur 2 dx c pU ro 2-80 (f) PROBLEM 2.28 (continued) Direct integration of (f) gives the solution T (x) T T 2 1 2 T x ln sur 3 tan 1 C 3 c p U ro Tsur 4Tsur Tsur T Tsur (g) where boundary condition (e) gives the constant of integration C C 1 3 4Tsur ln Tsur To T 2 3 tan 1 o Tsur To Tsur Tsur (h) Substituting (h) into (g) and rearranging the resulting equation in dimensionless form gives T T 1 1 o c pU 1 T Tsur Tsur x 1 To ln ln 8 tan tan 3 To ro 8 Tsur Tsur Tsur 1 T 1 Tsur Tsur (i) (5) Checking. Dimensional check: (1) Each term in (i) is dimensionless: Boundary conditions check: Solution (i) satisfies boundary condition (e). Limiting checks: at x cable temperature must be equal to Tsur . Setting T Tsur in (i) gives x . (6) Comments. The solution is characterized by two dimensionless parameter: c pU To and 3 Tsur Tsur 2-81 PROBLEM 2.29 A thin wire of diameter d, specific heat c p , conductivity k and density moves with velocity U through a furnace of length L. Heat is added to the wire in the furnace at a uniform surface flux qo . Far away from the inlet of the furnace the wire is at temperature Ti . Assume that no heat is exchanged with the wire before it enters and after it leaves the furnace. Determine the temperature of the wire leaving the furnace. L Ti T1 ( x ) 0 x T2 ( x ) T3 ( x ) U d qo (1) Observations. (i) This is a constant area moving fin. (ii) Temperature distribution can be assumed one dimensional. (iii) The fin has three sections. Surface and boundary conditions are different for each section. Thus three equations are needed. (iv) The surface of two sections is insulated while heat is added at uniform flux at the third (middle section). (2) Origin and Coordinates. The origin is selected at the start of the middle section and the coordinate x points in the direction of motion. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area, (4) Bi 1, (5) constant properties, (6) uniform surface flux, (7) no energy generation and (8) no radiation. (ii) Governing Equations. Three fin equations are needed. Let T1 ( x ) = temperature distribution in the first section, x 0 T2 ( x ) = temperature distribution in the second (middle) section, 0 x L T3 ( x ) = temperature distribution in the third section, L x We begin by formulating the fin equation for the second section. Consider a constant area fin moving with constant velocity U. For steady state, conservation of energy for an element dx gives dq dhˆ q x m hˆ q x x dx m hˆ m dx dqo (a) dx dx is mass flow rate, q x is conduction heat transfer rate where ĥ is enthalpy per unit mass, m conduction and q o is heat added at the surface. Mass flow rate is given by m UAc (b) where Ac is the cross-sectional area and is density. For a constant pressure process the change in enthalpy is dhˆ c p dT2 (c) where c p is specific heat. Using Fourier’s law we obtain 2-82 PROBLEM 2.29 (continued) q x kAc dT2 dx (d) where Ac is the cross-sectional area given by Ac d 2 / 4 where d is the wire diameter. Heat added at the surface of the element, dqo , is given by dqo ( d )qo dx Substituting (b)-(f) into (a) d 2T2 dx 2 c pU dT2 k dx (f) 4qo 0, kd (e) 0 xL (g) Equation (g) is the fin equation for the middle section. To obtain the fin equations for the first and third sections, set qo 0 in equation (g) d 2T1 dx 2 c pU dT1 k and d 2T3 dx 2 dx 0, c pU dT3 k dx 0, x 0 (h) xL (i) (iii) Boundary Conditions. Six boundary conditions are needed. (1) Specified temperature at x = T1 ( ) Ti (j) T1 (0) T2 (0) (k) (2) Equality of temperature at x = 0 (3) Equality of flux at x = 0 dT1 (0) dT2 (0) dx dx (4) Equality of temperature at x = L T2 ( L) T3 ( L) (5) Equality of flux at x = L dT2 ( L) dT3 ( L) dx dx (6) Finite temperature at x = T3 ( ) finite (l) (m) (n) (o) (4) Solution. The solutions to (g), (h) and (i) are T1 A1 exp( c pUx / k ) B1 2-83 (p) PROBLEM 2.29 (continued) T2 A2 exp( c pUx / k ) 4qo x B2 c pU d T3 A3 exp( c pUx / k ) B3 (q) (r) where A1 , A2 , A3 , B1 , B2 and B3 are constants of integration. Application of boundary conditions (j) to (o) give six equations for the six constants B1 Ti A1 Ti A2 B2 c pU k A2 exp( c pUL / k ) A2 c pU k A1 c pU k A2 4qo c pU d 4qo L B2 A3 exp( c pUL / k ) B3 c pU d exp( c pUL / k ) c pU 4qo A3 exp( c pUL /) c pU d k A3 0 The above six equations are solved for the six constants. Substitution into equations (p), (q) and (r) gives 4q (s) T1 ( x) (k / c pU ) 2 o 1 exp( c pUL / k ) exp( c pUx / k ) Ti kd T2 ( x) (k / c pU ) 2 4qo 4qo 4q exp ( c pU / k )( x L) x (k / c pU ) 2 o Ti (t) kd c pU d kd T3 ( x) Ti 4qo L c pU d (u) Expressing (s)-(u) in dimensionless form c pUL( x / L) c pUL T1 ( x) Ti exp 1 exp( ) 4kq o k k (v) ( c pU ) 2 d c pUL x c pUL T2 ( x) Ti ( x / L) 1 1 exp 4kqo k L k ( c pU ) 2 d T3 ( x) Ti 1 4q o L c p Ud 2-84 (w) (x) PROBLEM 2.29 (continued) (5) Checking. Dimensional checks: (i) The exponent of the exponential, c pUx / k , is dimensionless. (ii) Each term in equations (s), (t) and (u) has units of temperature and each term in (v), (w) and (x) is dimensionless. Boundary conditions check: Substitution of solutions (s), (t) and (u) into equations (j)-(o) shows that the six boundary conditions are satisfied. Differential equations check: Direct substitution of (s), (t) and (u) into (g), (h) and (i), respectively, shows that the governing equations are satisfied. Global energy conservation: energy added to the wire at the middle section must be axially convected by the wire. Thus conservation of energy between x and x gives c p T3 () Ti qo Ld m Substituting (b), (e) and (u) into the above gives U (d 2 / 4)c p Ti or 4qo L Ti qoLd c pUd qoLd qoLd Thus energy is conserved. Limiting check: (i) If wire velocity vanishes, the temperature of the wire will continue to rise. Thus for U 0 the wire temperature should be infinite. Setting U 0 in (s), (t) and (u) gives T1 ( x ) T2 ( x ) T3 ( x ) (ii) If the wire is not heated at the middle section, its temperature should be uniform equal to Ti . Setting qo 0 in (s), (t) and (u) gives T1 ( x ) T2 ( x ) T3 ( x ) Ti (iii) If the wire moves with infinite velocity, no heating will take place at the middle section and consequently its temperature should be uniform equal to Ti . Setting U in (s), (t) and (u) gives T1 ( x ) T2 ( x ) T3 ( x ) (6) Comments. (i) Wire temperature increases with distance along the first and second sections. However, in the third section the temperature is uniform. (ii) a simpler approach to solving this problem is to recognize that the temperature in the third section must be uniform, use global conservation of energy to determine it and solve the temperature distribution in the first and second sections. (iii) The solution is characterized by a single parameter c pUL / k. 2-85 PROBLEM 2.30 A wire of diameter d is formed into a circular loop of radius R. The loop rotates with constant angular velocity . One half of the loop passes through a furnace where it is heated at a uniform surface flux q o . In the remaining half the wire is cooled by convection. The heat transfer coefficient is h and the ambient temperature is T . The specific heat of the wire is c p , conductivity k and density . Use fin approximation to determine the steady state temperature distribution in the wire. d h, T x R 0 furnace qo (1) Observations. (i) This is a constant area moving fin problem. (ii) Temperature distribution can be assumed one dimensional. (iii) The fin has two sections. Since surface conditions are different for each section, two fin equations are needed. (iv) In the furnace section heat is added to the wire at uniform flux. Outside the furnace heat is removed by convection. (2) Origin and Coordinates. The origin is selected at the outlet end of the furnace. The coordinate x points in the direction of motion. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area, (4) Bi 1, (5) constant properties, (6) uniform surface flux, (7) uniform heat transfer coefficient and ambient temperature, (8) negligible curvature effect ( d / R 1), (9) no energy generation and (10) no radiation. (ii) Governing Equations. Two fin equations are needed. Let T1 ( x) = wire temperature distribution outside furnace, 0 x Rc T2 ( x) = wire temperature distribution inside furnace, Rc x 2 Rc where Rc is the distance from the center of loop to the wire center, given by Rc R d / 2 (a) We begin by formulating the fin equation outside the furnace. Consider a constant area fin moving with constant velocity. For steady state, conservation of energy for an element dx gives dq dhˆ q x m hˆ q x x dx m hˆ m dx dqc (b) dx dx where ĥ is enthalpy per unit mass, m is mass flow rate, q x is conduction heat transfer rate and q c is heat removed by convection. Mass flow rate is m UAc (c) where Ac is the cross-sectional area and is density. For a constant pressure process the change in enthalpy is 2-86 PROBLEM 2.30 (continued) dhˆ c p dT1 (d) where c p is specific heat. Using Fourier’s law we obtain dT1 dx (e) Ac d 2 / 4 (f) q x kAc The cross-sectional area Ac is given by where d is the wire diameter. Heat removed from the element by convection, dqc , is dqc ( d )h(T1 T )dx (g) Substituting (c)-(g) into (b) d 2T1 dx 2 c pU dT1 k dx 4h (T1 T ) 0 kd 0 x Rc (h) where U is the tangential velocity given by U Rc Equation (h) is the fin equation for the wire outside the furnace. To obtain the fin equation inside the furnace consider an element of the wire dx and apply conservation of energy dq dhˆ q x m hˆ dqo q x x dx m hˆ m dx dx dx (i) where dqo is the heat added at the surface given by dqo ( d )q o dx (j) Substituting (c)-(f) and (j) into (i) d 2T2 dx 2 c pU dT2 k dx 4qo 0 kd Rc x 2 Rc (k) This is the fin equation for the wire inside the furnace. (iii) Boundary Conditions. Four boundary conditions are needed. (1) Equality of temperature at x = 0 T1 (0) T2 (0) (l) dT1 (0) dT2 (0) dx dx (m) (2) Equality of flux at x = 0 (3) Equality of temperature at x Rc 2-87 PROBLEM 2.30 (continued) T1 ( Rc ) T2 ( Rc ) (n) dT1 ( Rc ) dT2 ( Rc ) dx dx (o) (4) Equality of flux at x Rc (4) Solution. The solutions to (h) and (k) are T1 ( x) A1 exp bx b 2 m 2 B1 exp bx b 2 m 2 and c m2 T2 A2 B2 exp( c pUx / k ) Qo x (p) (q) where A1 , A2 , B1 and B2 are constants of integration and b, c, m 2 and Qo are constant defined as c pU 4qo 4h 4h T , Qo , c b , m2 (r) kd kd 2k c pU d Application of the four boundary conditions give b A1 B1 (c / m 2 ) A2 B2 m ) R B exp(b (s) b 2 m 2 A1 b b 2 m 2 B1 ( c pUU / k ) B2 Qo A1 exp (b b 2 2 c 1 b 2 m 2 ) Rc (c / m 2 ) A2 B2 exp( Rc c pU / k ) Rc Qo A1 (b b 2 m 2 ) exp (b b 2 m 2 ) Rc (t) (u) B1 (b b 2 m 2 ) exp (b b 2 m 2 ) Rc B2 (U / ) exp( Rc c pU / k ) Qo (v) Equations (s)-(v) are solved for the four constants. (5) Checking. Dimensional check: (i) Each term in (h) and (k) has units of ( o C/m 2 ). The exponents of the exponentials in (p) and (q) are dimensionless. (ii) Each term in (p) and (q) has units of temperature. (6) Comments. The rotating wire can be thought of as a heat exchanger removing heat from the furnace and adding it to the ambient fluid outside the furnace. The net heat transfer rate can be determined by making an energy balance for the wire. For steady state energy added to the surface inside the furnace is equal to energy removed by convection outside the furnace. Considering the energy added in the furnace, we obtain q d (Rc )qo This shows that the rate of heat transfer is independent of angular velocity! 2-88 PROBLEM 2.31 A wire of radius ro moves with a velocity U through a h ,T U furnace of length L. It enters the furnace at Ti where it is heated by convection. The furnace temperature x 0 T furnace i is T and the heat transfer coefficient is h. Assume that no heat is removed from the wire after it leaves the furnace. Using fin approximation determine the wire temperature leaving the furnace. (1) Observations. (i) This is a constant area moving fin. (ii) Temperature distribution can be assumed one dimensional. (iii) The wire has two sections. Surface and boundary conditions are different for each section. Thus two equations are needed. (iv) In the furnace section convection takes place at the surface. The section outside the furnace is insulated. (v) To determine the wire temperature at the furnace outlet one must determine the temperature distribution in the wire. (2) Origin and Coordinates. The origin is selected at the exit of the furnace section (start of the insulated section) as shown. (3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area, (4) Bi 1, (5) constant properties, (6) uniform heat transfer coefficient and furnace temperature, (7) no energy generation and (8) no radiation. (ii) Governing Equations. Two fin equations are needed. Let T1 ( x ) = temperature distribution in the first section, L x 0 T2 ( x ) = temperature distribution in the second (middle) section, 0 x The heat equation for both sections is obtained from eq. (2.19) C 4 d 2T c p U dT hC 4 (T T ) (T Tsur )0 2 k dx k A k A dx c c (2.19) Neglecting radiation in eq. (2.19) gives the heat equation in the furnace section d 2T1 dx 2 c p U dT1 k dx hC (T1 T ) 0 k Ac (a) Similarly, neglecting radiation and convection in eq. (2.19) gives the heat equation in the insulated section d 2T2 c p U dT2 0 (b) k dx dx 2 (iii) Boundary Conditions. Four boundary conditions are needed. (1) Specified temperature at x = L T1 ( L) Ti 2-89 (c) PROBLEM 2.31 (continued) (2) Equality of temperature at x = 0 T1 (0) T2 (0) (d) dT1 (0) dT2 (0) dx dx (e) T2 () finite (f) (3) Equality of flux at x = 0 (6) Finite temperature at x = (4) Solution. The solutions to (a) is given by eq. (A-6c) of Appendix A. Rewriting (a) as d 2T1 dx 2 2b dT1 m 2T1 c dx where b (g) c pU (h) 2k 2h m2 kro 2h c T kro (i) (j) The solution is T1 A1 exp( bx b 2 m 2 x) B1 exp( bx b 2 m 2 x) c m2 (k) The solution to (b) is T2 A2 exp(2bx) B2 (l) where A1, B1, A2 and B2 are constants of integration. The four boundary conditions give Ti T A1 exp(bL b 2 m 2 L) b b m 2 b b m 2 b b2 m2 b b m 2 B1 exp(bL b m L) 2 2 2 2 2 b b2 m2 2-90 exp(bL b 2 m 2 L) (Ti T ) b b2 m2 A2 0 (m) (n) exp(bL b m L) 2 2 (o) PROBLEM 2.31 (continued) (Ti T ) B2 T exp( bL b 2 m 2 L) b b2 m2 b b2 m2 b b2 m2 1 b b2 m2 exp( bL b 2 m 2 L) (p) The wire temperature outlet temperature is obtained by evaluating solution (k) or (l) at x 0 T1 (0) T2 (0) B2 (q) (5) Checking. Dimensional check: (i) The argument of the exponentials in equations (k)-(n) is dimensionless. (ii) Each term in solutions (k) and (l) has units of temperature. Boundary conditions check: Solutions (k) and (l) satisfy boundary conditions (c)-(f). Differential equations check: Direct substitution shows that (k) satisfies (a) and (l) satisfies (b). Limiting check: If furnace temperature is the same as inlet temperature the entire wire will be at the inlet temperature. Setting T Ti into (m), (n) and (p) gives A1 B1 0 and B2 T Equations (i) and (j) give c T m2 Substituting these results into (k) and (l) gives T1 ( x) T2 ( x) T (6) Comments. (i) Because the wire has two sections having different surface conditions it was necessary to use two heat equations. (ii) The wire remains at uniform temperature once it exits the furnace. (iii) Rewriting solutions (k) and (l) in dimensionless form shows that the problem is characterized by two dimensionless parameters: bL c pUL 2k 2hL2 and m L kro 2 2 2-91 PROBLEM 2.32 . The water is Hot water at To flows downward from a faucet of radius ro at a rate m cooled by convection as it flows. The ambient temperature is T and the heat transfer coefficient is h. Suggest a model for analyzing the temperature distribution in the water. Formulate the governing equations and boundary conditions. (1) Observations. (i) If the Biot number, h ro / k , is small compared to unity a fin model can be used to analyze the temperature distribution in the water. (ii) Since water velocity increases with distance x, conservation of mass requires that the cross-sectional area decreases with distance. Thus this is a variable area fin which is moving with variable velocity. (iii) Temperature distribution can be assumed one dimensional. (2) Origin and Coordinates. The origin is selected at the outlet of the faucet and the coordinate x points in the direction of motion and gravity. (3) Formulation. (i) Assumptions. (1) One-dimensional temperature distribution, (2) steady state, (3) Bi 1, (4) constant conductivity, (5) constant density, (6) uniform heat transfer coefficient and ambient temperature, m (7) uniform velocity at any given section, (8) circular cross T o ro section, (9) no energy generation, (10) negligible air resistance and (11) no radiation. 0 (ii) Governing Equations. Consider an element dx of the moving fluid. Conservation of energy for steady state gives dq dhˆ q x m hˆ q x x dx m hˆ m dx dqc (a) dx dx x g h ,T where ĥ is enthalpy per unit mass, m is mass flow rate, q x is conduction heat transfer rate and q c is heat exchange by convection. For a constant pressure process the change in enthalpy is dhˆ c p dT where c p is specific heat. Using Fourier’s law we obtain dT q x kAc (c) dx where Ac (x) is the cross-sectional area. Heat exchange by convection is given by dqc (T T )dAs qx (b) (d) m c pT Ac dx dAs dq qx x dx dx where dAs is surface area of the element. Substituting (b)-(d) into (a) 2-92 dx dqc m c p T (dT / dx)dx PROBLEM 2.32 (continued) d dT m c p dT h dAs A ( x ) (T T ) 0 c dx dx k dx k dx (e) It remains to determine Ac (x) and dAs / dx . The conduction area is given by Ac ( x) r 2 (f) where r r (x) is the local radius. To determine r (x ) we apply conservation of mass U o ro2 U ( x) r 2 m (g) where U o is the discharge velocity and U(x) is the local velocity. Solving (g) for U o gives U o m / ro2 (h) where is density. Neglecting air resistance Bernoulli’s equation gives the local velocity U(x) U ( x) U o2 2 gx (m / ro2 ) 2 2 gx (i) where g is the gravitational acceleration. Substituting (i) into (g) and solving for r(x) r ( x) ro 1 ( ro2 / m ) 2 2 gx Using (j) into (f) 1 / 4 (j) Ac ( x) ro2 1 ( ro2 / m ) 2 2 gx 1 / 2 (k) Next we determine dAs / dx using eq. (2.6a) dAs C ( x) 1 (dr / dx) 2 dx (l) C (r ) 2 r (m) where and 5 / 4 dr 1 g ro ( ro2 / m ) 2 1 ( ro2 / m ) 2 2 gx dx 2 Substituting (j), (m) and (n) into (l) (n) 1 / 4 5 / 2 dAs ) 2 2 gx ) 4 1 ( ro2 / m ) 2 2 gx 2ro 1 ( ro2 / m 1 ( gro / 2) 2 ( ro2 / m dx 1/ 2 (o) Substituting (k) and (o) into (e) gives the heat equation for the flowing water. (5) Checking. Dimensional check: (i) Each term in (e) has units of temperature. (ii) Each term in (i) has units of velocity. (iii) The term ( ro2 / m ) 2 2 gx is dimensionless. (iv) The slope in (n) is dimensionless. (v) Equation (o) has units of length. 2-93 PROBLEM 2.32 (continued) Limiting check: For the special case of zero gravity the velocity and cross sectional area remain constant. The fin equation for this case is given by eq. (2.19). Setting g equal to zero in (i), (k) and (o) and substituting the result into (e) and using (g) gives d 2T dx 2 c pU o dT k dx 2h (T T ) 0 kro This is identical to eq. (2.19) with the radiation term set equal to zero. (6) Comments. The problem can be simplified by neglecting the dr / dx term in (l) and using an average value for the circumference C. 2-94 PROBLEM 3.1 Three sides of a rectangular plate L H are maintained at uniform temperature To while the fourth side exchanges heat by convection. The heat transfer coefficient is non-uniform, being h1 over half the surface and h2 over the remaining half. The ambient temperature is T . Determine the twodimensional steady state temperature distribution. h2 , T h1 , T y L/2 H 2 HBC To To L 0 To x (1) Observations. (i) This is a steady state twodimensional conduction problem. (ii) Four boundary conditions are needed. (iii) Three conditions can be made homogeneous by defining a new temperature variable ( x, y ) T ( x, y ) To . (iv) The heat transfer coefficient along boundary (x,H) is not uniform. This requires special consideration. (2) Origin and Coordinates. The origin and coordinate axes are selected as shown. (3) Formulation. (i) Assumptions. (1) Two-dimensional, (2) steady, (3) no energy generation and (4) constant thermal conductivity. (ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives the heat equation where 2 2 0 x2 y2 (a) ( x, y ) T ( x, y ) To (b) (iii) Independent Variable with Two Homogeneous Boundary Conditions. The two boundaries at x = 0 and x = L have homogeneous conditions. Thus, the x-variable has two homogeneous boundary conditions. (iv) Boundary Conditions. The boundary conditions are listed in the following order: starting with the two boundaries in the variable having two homogeneous conditions and ending with the non-homogeneous condition. Thus, the first three conditions in this order are homogeneous and the fourth is non-homogeneous. Therefore, we write (1) (0, y ) 0 , homogeneous (2) ( L, y ) 0 , homogeneous (3) ( x,0) 0 , homogeneous The fourth boundary condition is first written in terms of the variable T and then transformed in terms of the variable . Using Fourier’s law and Newton’s law k T ( x, H ) h1 T ( x, H ) T , y h2 T ( x, H ) T , 3-1 0 x L/2 L/2 x L PROBLEM 3.1 (continued) Expressed in terms of , the above gives the fourth boundary condition (4) k h1 ( x, H ) (T To ), 0 x L / 2 ( x, H ) f ( x) = y h2 ( x, H ) (T To ), L / 2 x L non-homogeneous This condition represents non-uniform convection. (4) Solution. (i) Assumed Product Solution. Assume a solution in the form ( x, y ) = X(x) Y(y) (c) Substituting into (a), separating variables and setting the resulting equation equal to constant 1 d2X 1 d 2Y 2n 2 2 X dx Y dy Assuming that n is multi-valued, the above gives d2Xn 2n X n 0 2 dx (d) d 2Yn 2nYn 0 2 dy (e) and (ii) Selecting the Sign of the 2n Terms. Since the x-variable has two homogeneous boundary conditions, the 2n X n term in (d) takes the positive sign. Therefore, (d) and (e) become d2Xn (f) 2n X n 0 2 dx and d 2Yn (g) 2nYn 0 2 dy For the important case of n 0, equations (f) and (g) become d2X0 0 d x2 (h) d 2Y0 0 d y2 (i) and (iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (f)-(i) are X n ( x ) An sin n x Bn cos n x 3-2 (j) PROBLEM 3.1 (continued) Yn ( y ) Cn sinh n y Dn coshn y X 0 ( x ) A0 x B0 Y0 ( y ) C0 y D0 (k) (l) (m) Corresponding to each value of n there is a temperature solution n ( x, y ). Thus and n ( x, y ) X n ( x )Yn ( y ) (n) 0 ( x, y ) X 0 ( x )Y0 ( y ) (o) The complete solution becomes ( x, y ) X 0 ( x )Y0 ( y ) X n 1 n ( x )Yn ( y ) (p) (iv) Application of Boundary Conditions. Boundary condition (1) applied to solutions (j) and (l) gives Bn B0 0 (q) Boundary condition (2) applied to (j) gives the characteristic equation for n sin n L 0 (r) Thus n n n L (n = 1,2,3,….) (s) Similarly, boundary condition (2) applied to (l) gives A0 0 With A0 B0 0, the solution corresponding to n 0 vanishes. Application of boundary condition (3) to (k) gives Dn 0 Solution (p) becomes ( x, y ) a (sin x )(sinh y ) n 1 n n n (t) where an An Cn . The only remaining unknown is an. Application of condition (4) gives f(x) = k (a n 1 n n cos hn H ) sin n x (u) where f ( x ) is defined in boundary condition (4). (v) Orthogonality. To determine a n in equation (u) we apply orthogonality. Note that the characteristic functions n ( x ) sin n x in (u) are solutions to equation (f). Comparing (f) with eq. (3.5a) shows that it is a Sturm-Liouville equation with 3-3 PROBLEM 3.1 (continued) a1 ( x) a2 ( x) 0 and a3 ( x ) 1 Thus eq. (3.6) gives p ( x ) w( x ) 1 and q( x ) 0 Since the boundary conditions at x = 0 and x = L are homogeneous, it follows that the characteristic functions n ( x ) sin n x are orthogonal with respect to the weighting function w( x ) 1. Multiplying both sides of (u) by w( x ) sin m x dx and integrating from x = 0 to x = L L f ( x ) w( x ) sin m x dx = k 0 0 (a n coshn H ) sin n x w( x )(sinm x ) dx n1 L Using the definition of f ( x ) in condition (4) and noting that w( x ) 1, the above gives h1 L/2 L 0 ( x, H ) (T To )sin m x dx h2 L / 2 ( x, H ) (T To )sin m x dx k L ( a n n coshn H ) sinn x sin m x dx n 1 0 (v) Evaluating (t) at y = H and substituting into (v) h1 L / 2 ( a n sinh n H ) sinn x sin m x dx h2 n 1 0 h1 (T To ) 0 L/ 2 sin m x dx h2 (T To ) L L/2 ( a n sinh n H ) sinn x sin m x dx L/2 n 1 L sin m x dx k L (a n n 0 n 1 cosh n H ) sinn x sin m x dx Interchanging the integration and summation signs and applying orthogonality, eq. (3.7), h1a n sinh n H h1 (T To ) L/ 2 0 L/2 0 sin2 n x dx h2 a n sinh n H sin n x dx h2 (T To ) L L / 2 L L / 2 sin n x dx sin n x dx kan n coshn H Evaluating the integrals and solving for a n we obtain an 2 L 0 sin 2(T To ) (h1 L / k ) 3 ( 1) n1 (h2 L / k ) 1 ( 1) n1 2( 1) n n L(h1 h2 )( L / k ) sinh n H 2(n L) coshn H 2 n x dx (w) (5) Checking. Dimensional check: (i) The arguments of sin, sinh and cosh are dimensionless. (ii) The coefficient a n in (t) must have units of temperature. Equation (w) shows that a n has units of temperature. Limiting check: (i) If h1 h2 0 , no heat can be exchanged at the (x,H) boundary and consequently the entire plate should be at a uniform temperature To . Setting h1 h2 0 in equation (w) gives a n 0. When this is substituted into (t) we obtain 3-4 PROBLEM 3.1 (continued) ( x, y ) T ( x, y ) To 0 which gives T ( x, y ) To . (ii) If T To , no heat can be exchanged at the (x,H) boundary and thus the entire plate should be at a uniform temperature To . Setting T To in equation (w) gives a n 0. When this is substituted into (t) we obtain T ( x, y ) To . Boundary conditions check: Direct substitution into solution (t) shows that boundary conditions (1), (2) and (3) are satisfied. Special case: For the special case of h1 h2 h equation (w) gives an 2( Lh / h)(T To ) 1 ( 1) n (n L)(hL / k ) sinh n L (n L) coshn L Formal determination of a n for this case gives the same result. (6) Comments. (i) A non-uniform non-homogeneous boundary condition along a boundary does not introduce added complications. (ii) The solution is expressed in term of hL/k which appears in a n . This dimensionless parameter is the Biot number. Equation (s) gives the dimensionless roots n L . However, the argument k H is determined by multiplying and dividing by L according to k H = (k L)( H / L) . This introduces an additional dimensionless parameters H / L . 3-5 PROBLEM 3.2 Heat is generated in a rectangular plate 3L H at a y T2 T1 T3 uniform volumetric rate of q . The surface at (0,y) is L L L maintained at To . Along surface (3L,y) the plate h exchanges heat by convection. The heat transfer q H T coefficient is h and the ambient temperature is T . To Surface (x,H) is divided into three equal segments 3L 0 x which are maintained at uniform temperatures T1 , T2 , and T3 , respectively. Surface (x,0) is insulated. Determine the two-dimensional steady state temperature distribution. (1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) Four boundary conditions are needed. (iii) Only one boundary condition is homogeneous. (iv) The energy generation term makes the differential equation non-homogeneous. (2) Origin and Coordinates. The origin and coordinate axes are selected as shown. (3) Formulation. (i) Assumptions. (1) Two-dimensional, (2) steady, (3) uniform energy generation and (4) constant thermal conductivity. (ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives the heat equation for this problem 2T 2T q 0 (a) x2 y2 k (iii) Independent Variable with Two Homogeneous Boundary Conditions. Neither variable has two homogeneous conditions. Only the x-variable can have two homogeneous conditions. (iv) Boundary Conditions. (1) T (0, y ) To , non-homogeneous (2) k (3) T (3L, y ) hT (3L, y ) T , non-homogeneous x T ( x,0) 0 , homogeneous y T1 , 0 x L (4) T ( x, H ) f ( x ) T2 , L x 2 L , non-homogenous T3 , 2 L x 3L This condition represents a boundary with non-uniform temperature given by f ( x ). (4) Solution. Equation (a) is non-homogeneous due to the heat generation term. Thus we can not proceed directly with the application of the method of separation of variables. Instead, we assume a solution of the form 3-6 PROBLEM 3.2 (continued) T ( x, y ) ( x, y ) ( x ) (b) Note that ( x, y ) depends on two variables while ( x ) depends on a single variable. Substituting (b) into (a) 2 2 d 2 q 0 (c) x2 y2 d x2 k The next step is to split (c) into two equations, one for ( x, y ) and the other for (x ). We select 2 2 0 (d) x2 y2 Thus, the second part must be d 2 q 0 (e) d x2 k Boundary conditions on ( x, y ) and ( x ) are obtained by substituting (b) into the four boundary conditions on T . In this process the function ( x, y ) is assigned homogeneous conditions whenever possible. Using (b) and boundary condition (1) (0, y ) (0) To Let Thus (0, y ) 0 (d-1) (0) To (e-1) Boundary condition (2) and (b) give k (3L, y ) d (3L) k h (3L, y ) h (3L) T x dx Let k (3L, y ) h (3L, y ) x (d-2) Thus d (3L) h (3L) T dx Boundary condition (3) and (b) give ( x,0) 0 y Finally, boundary condition (4) gives k ( x, H ) ( x ) f ( x ) Or ( x, H ) f ( x ) ( x ) = F(x) 3-7 (e-2) (d-3) PROBLEM 3.2 (continued) Using the definition of f ( x ) in boundary condition (4) the above gives T1 ( x ) , ( x, H ) f ( x ) ( x ) F ( x ) T2 ( x ) , T ( x ) , 3 0 x L L x 2 L , non-homogeneous (d-4) 2 L x 3L Thus non-homogeneous equation (a) is replaced by two equations: (d), which is a homogeneous partial differential equation and (e) which is a non-homogeneous ordinary differential equation. Equation (d) has three homogeneous conditions, (d-1), (d-2) and (d-3) and one non-homogeneous condition, (d-4). The solution to (e) is ( x) q 2 x ExF 2k (f) where E and F are constants of integration. Application of boundary conditions (e-1) and (e2) gives the two constants E (3q L / k )1 (3hL / 2k ) (h / k )(To T ) , 1 (3hL / k ) F To (g) (i) Assumed Product Solution. To solve partial differential equation (d) we assume a product solution. Let ( x, y ) X ( x )Y ( y ) (h) Substituting (h) into (d), separating variables and setting the resulting two equations equal to a constant, 2n , we obtain d2Xn 2n X n 0 2 dx (i) d 2Yn 2nYn 0 2 dy (j) and (ii) Selecting the Sign of the 2n Terms. Since the x-variable has two homogeneous boundary conditions, the plus sign is selected in (i). Thus (i) and (j) become d2Xn 2n X n 0 2 dx (k) d 2Yn 2nYn 0 2 dy (l) and For the special case of n 0 the above equations take the form d 2X0 0 d x2 and 3-8 (m) PROBLEM 3.2 (continued) d 2Y0 0 d y2 (n) (iii) Solutions to the Ordinary Differential Equations. The solutions to equations (k)(n) are X n ( x ) An sin n x Bn cosn x (o) Yn ( y ) Cn sinh n y Dn coshn y (p) X 0 ( x ) A0 x B0 (q) Y0 ( y ) C0 y D0 (r) and Corresponding to each value of n there is a solution n ( x, y ). Thus and n ( x, y ) X n ( x )Yn ( y ) (s) 0 ( x, y ) X 0 ( x )Y0 ( y ) (t) The complete solution to T ( x, y ) becomes T ( x, y) ( x) X 0 ( x)Y0 ( y) X n ( x)Yn ( y ) (u) n 1 (iv) Application of Boundary Conditions. Applying boundary condition (d-1) to equation (s) and (t) gives Bn B0 0 (v) Condition (d-2) applied to (s) gives the characteristic equation for n 3n L (3hL / k ) tan 3n L (w) Condition (d-2) applied to (t) A0 0 (x) Cn 0 (y) Boundary conditions (d-3) and (s) give With A0 B0 0, the solution X 0Y0 vanishes. Substituting into (s) and using (v) and (y), the solution to ( x, y ) becomes ( x, y ) a n (coshn y ) sin n x (z) n 1 where a n An Dn . Thus the only remaining unknown is a n . Application of condition (d-4) gives F ( x) a n (coshn H ) sin n x n 1 3-9 (z-1) PROBLEM 3.2 (continued) where F(x) is defined in (d-4). To determine a n from this result we apply orthogonality. (v) Orthogonality. The characteristic functions, sin n x, in (z-1) are solutions to equation (k). Comparing (k) with eq. (3.5a) shows that it is a Sturm-Liouville equation with a1 ( x) a2 ( x) 0 and a3 ( x ) 1 Thus eq. (3.6) gives p( x ) w( x ) 1 and q( x ) 0 Since boundary conditions (d-1) and (d-2) at x = 0 and x = 3L are homogeneous, it follows that the characteristic functions sin n x are orthogonal with respect to w = 1. Multiplying both sides of (aa) by sin m x dx , integrating from x = 0 to x = 3L 3L F ( x)w( x) sin m x dx = 0 0 ( a cosh L ) sin x n n n w( x )(sin m x ) dx n1 3L Interchanging the integration and summation signs, noting that w( x ) 1 and applying orthogonality, the above gives 3L F ( x ) sin n x dx = a n coshn H 0 3L sin2 n x dx 0 Evaluating the integral on the right-hand-side and solving the above for a n an 3L 0 F ( x ) sin n xdx (z-2) (1 / 2n ) coshn H 3n L (1 / 2) sin 6n L The integral in (z-2) is evaluated using the definition of F(x) in (d-4) and the solution to ( x ) given in (f). Thus 3L 0 F ( x ) sin n xdx T T (q / 2k ) x Ex F )sin xdx T (q / 2k ) x Ex F )sin xdx L 0 2L L 3L 2L 1 ( q / 2k ) x 2 Ex F ) sin n xdx 2 2 n 2 3 n where E and F are given in equation (g). Evaluating each of the three integrals in the above T L 1 0 (q / 2k ) x 2 Ex F sin n xdx ( F T1 )(1 / n )(cos n L 1) (q / 2k3n ) 2 n L sin n L ( 2n L2 2)(cos n L) 2 ( E / 2n )sin n L n L cos n L (z-3) 3-10 PROBLEM 3.2 (continued) T 2L 2 (q / 2k ) x 2 Ex F sin n xdx ( F T2 )(1 / n )(cos 2 n L cos n L) L (q / 2k3n ) 4 n L sin 2 n L (42n L2 2)(cos 2 n L) 2 n L sin n L (2n L2 2)(cos n L) sin 2 n L 2 n L cos 2 n L sin n L n L cos n L ( E / 2n ) (z 4) and T (q / 2k )x 3L 3 2L 2 Ex F sin n xdx ( F T3 )(1 / n )(cos 3 n L cos 2 n L) (q / 2k3n ) 6 n L sin 3 n L (92n L2 2)(cos 3 n L) 4 n L sin 2 n L (42n L2 2)(cos 2 n L) ( E / 2n )sin 3 n L 3 n L cos 3 n L sin 2 n L 2 n L cos 2 nL (z 5) (5) Checking. Dimensional checks: (i) The coefficient a n in (z-2) should have units of temperature. Since F(x) has units of temperature and n is measured in (1/m), (z-2) has units of temperature. (ii) Each term in (z-3), (z-4) and (z-5) has units of ( o C m ) . Limiting check: For the special case of q 0 and T1 T2 T3 T To , physical consideration requires that the plate be at a uniform temperature To . For this case E = 0 and ( x ) To . Substitution into (z-2) gives a n 0. Equation (u) gives T ( x, y ) To . Boundary conditions check: Boundary conditions (1) and (3) are readily shown to be satisfied. (6) Comments. (i) The same approach can be used to solve the more general problem where the heat generation rate is variable, q q (x ) . This will affect the solution of ( x ) and the integral in (z-2). (ii) The solution is expressed in terms of the Biot number, hL/k, which appears in equation (w). 3-11 PROBLEM 3.3 A rectangular plate L H is maintained at uniform temperature To along three sides. Half the fourth side is insulated while the other half is heated at uniform flux qo. Determine the steady state heat transfer rate through surface (0,y). qo y L/2 2 HBC To H To L 0 To x (1) Observations. (i) To determine the heat transfer through surface (0,y) it is necessary to first determine the temperature distribution in the plate. (ii) This is a steady state two-dimensional conduction problem. (iii) Four boundary conditions are needed. (iv) Three conditions can be made homogeneous by defining a new temperature variable ( x, y ) T ( x, y ) To . (v) The heat flux along boundary (x,H) is nonuniform. This requires special consideration. (2) Origin and Coordinates. The origin and coordinate axes are selected as shown. (3) Formulation. (i) Assumptions. (1) Two-dimensional, (2) steady, (3) no energy generation and (4) constant thermal conductivity. (ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives the heat equation for this case as 2 2 0 (a) x2 y2 where ( x, y ) T ( x, y ) To (b) Once the temperature distribution is determined, Fourier’s law give the heat transfer rate through surface (0,y). Thus H T (0, y ) q(0) k dy (c) x 0 where q (0) is the heat transfer rate through surface (0,y) (iii) Independent Variable with Two Homogeneous Boundary Conditions. The two boundaries at x = 0 and x = L have homogeneous conditions. Thus, the x-variable has two homogeneous boundary conditions. (iv) Boundary Conditions. The required four boundary conditions are (1) (0, y ) 0 , (2) ( L, y ) 0 , (3) ( x,0) 0 , homogeneous homogeneous homogeneous The fourth boundary condition is first written in terms of the variable T and then transformed in terms of the variable . Using Fourier’s law 3-12 PROBLEM 3.3 (continued) k T ( x, H ) 0 , y qo , 0 x L/2 L/2 x L Expressed in terms of , the above gives the fourth boundary condition (4) k 0, ( x, H ) f ( x) = y qo , 0 x L/2 L/2 x L non-homogeneous This condition represents a heat flux which varies along the boundary according to f ( x ). (4) Solution. (i) Assumed Product Solution. Assume a solution in the form ( x, y ) = X(x) Y(y) (d) Substituting into (a), separating variables and setting the resulting equation equal to constant 1 d2X 1 d 2Y 2n 2 2 X dx Y dy Assuming that n is multi-valued, the above gives d2Xn 2n X n 0 2 dx (e) d 2Yn 2nYn 0 d y2 (f) and (ii) Selecting the Sign of the 2n Terms. Since the x-variable has two homogeneous boundary conditions the 2n X n term in (e) takes the positive sign. Therefore, (e) and (f) become d2Xn (g) 2n X n 0 2 dx and d 2Yn (h) 2nYn 0 2 dy For the important case of n 0, equations (g) and (h) become d2X0 0 d x2 (i) d 2Y0 0 d y2 (j) and 3-13 PROBLEM 3.3 (continued) (iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (g)-(j) are X n ( x ) An sin n x Bn cos n x (k) Yn ( y ) Cn sinh n y Dn coshn y (l) (m) (n) X 0 ( x ) A0 x B0 Y0 ( y ) C0 y D0 Corresponding to each value of n there is a temperature solution n ( x, y ). Thus and n ( x, y ) X n ( x )Yn ( y ) (o) 0 ( x, y ) X 0 ( x )Y0 ( y ) (p) The complete solution becomes ( x, y ) X 0 ( x )Y0 ( y ) X n 1 n ( x )Yn ( y ) (q) (iv) Application of Boundary Conditions. Boundary condition (1) applied to solutions (k) and (m) gives B n Bo 0 (r) Boundary condition (2) applied to (k) gives the characteristic equation for n sin n L 0 (s) Or n n = 1,2,3…. L Similarly, boundary condition (2) applied to (m) gives n (t) A0 0 With A0 B0 0, the solution corresponding to n 0 vanishes. Application of boundary condition (3) to (l) gives Dn 0 Solution (q) becomes ( x, y ) a n (sin n x )(sinh n y ) (u) n 1 where an An Cn . The only remaining unknown is the set of constants an. Application of condition (4) gives f(x) = k (a n 1 n n cos hn H ) sin n x where f ( x ) is defined in boundary condition (4). 3-14 (v) PROBLEM 3.3 (continued) (v) Orthogonality. To determine a n in equation (v) we apply orthogonality. Note that the characteristic functions n ( x ) sin n x in (v) are solutions to equation (g). Comparing (g) with eq. (3.5a) shows that it is a Sturm-Liouville equation with a1 ( x) a2 ( x) 0 and a3 ( x ) 1 Thus eq. (3.6) gives p ( x ) w( x ) 1 and q( x ) 0 Since the boundary conditions at x = 0 and x = L are homogeneous, it follows that the characteristic functions n ( x ) sin n x are orthogonal with respect to the weighting function w( x ) 1. Multiplying both sides of (v) by w( x ) sin m x dx and integrating from x = 0 to x = L L L f ( x ) w( x ) sin m x dx = k (a n n coshn H ) sin n x w( x ) sin m x dx 0 0 n 1 Interchanging the summation with integration and invoking orthogonality gives L f ( x ) w( x ) sin n x dx = k an n coshn H 0 L 0 w( x ) sin2 n x Solving for a n and recalling that w(x) = 1 the above gives L an L 0 f ( x) sin n x dx 2 L f ( x) sin n x dx kn coshn H sin2 n x dx kn L coshn H 0 0 (w) Using the definition of f ( x ) in condition (4) the integral in (w) is evaluated L 0 f ( x ) sin n x dx L/2 0 (0) sin n x dx L L/2 qo sin n x dx ( q / n )(cosn L cosn L / 2) Substituting into (w) and using (t) to eliminate n L and (cosn L cosn L / 2) , we obtain an qoL / k ( 1) n 1 2( 1) n 1 (n ) (coshn H ) 2 (x) Substituting (x) into (u) and rearranging the result gives the dimensionless temperature distribution in the plate T ( x, y ) To qoL / k n 1 ( 1) n 1 2( 1) n 1 (sin n x ) sinh n y ( n ) 2 (coshn H ) To determine the heat transfer rate through surface (0,y) we substitute (y) into (c) 3-15 (y) PROBLEM 3.3 (continued) q(0) k (q o L / k ) H 0 n 1 q o L H 0 n 1 (1) n 1 2(1) n 1 (n ) 2 (cosh n H ) (1) n 1 2(1) n 1 (n ) 2 (cosh n H ) n (cos n 0) sinh n ydy H sinh n yd ( n y ) 0 Evaluating the integral q(0) q o L H 0 n 1 (1) n 1 2(1) n 1 (n ) 2 (cosh n H ) cosh n H 1 (z) (5) Checking. Dimensional check: (i) The arguments of the sin, sinh and cosh are dimensionless. (ii) Each term in solution (y) is dimensionless.(iii) Equation (z) has the correct units of heat transfer rate per unit length. Limiting check: If qo 0 the entire plate should be at a uniform temperature To . Setting qo 0 in (y) gives T ( x, y ) To . Boundary conditions check: Conditions (1), (2) and (3) can be shown to be satisfied by direct substitution. (6) Comments. (i) The same approach can be used to solve the more general problem where the heat flux at the boundary is variable, qo qo (x ). This will affect the integral in (w). (ii) Equation (t) gives the dimensionless roots n L . However, the argument k H is determined by multiplying and dividing by L according to k H = (k L)( H / L) . Thus the solution is characterized by a single dimensionless parameters H / L . 3-16 PROBLEM 3.4 A semi-infinite plate of width 2L is maintained at uniform temperature To along its two semi-infinite sides. Half the surface (0,y) is insulated while the remaining half is heated at a flux qo. Determine the two-dimensional steady state temperature distribution. y To qo L 2 HBC L 0 To x (1) Observations. (i) This is a steady state twodimensional conduction problem. (ii) Four boundary conditions are needed. (iii) Far away from the heated end the temperature should be uniform equal to To . (iv) Three conditions can be made homogeneous by defining a new temperature variable ( x, y ) T ( x, y ) To . (v) The heat flux along boundary (0,y) is non-uniform. This requires special consideration. (2) Origin and Coordinates. The origin and coordinate axes are selected as shown. (3) Formulation. (i) Assumptions. (1) Two-dimensional, (2) steady, (3) no energy generation and (4) constant thermal conductivity. (ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives the heat equation for this as where 2 2 0 x2 y2 (a) ( x, y ) T ( x, y ) To (b) (iii) Independent Variable with Two Homogeneous Boundary Conditions. The two boundaries at y = 0 and y = 2L have homogeneous conditions. Thus, the y-variable has two homogeneous boundary conditions. (iv) Boundary Conditions. The required four boundary conditions are (1) ( x,0) 0 , homogeneous (2) ( x,2 L) 0 , homogeneous (3) ( , y ) 0 = finite, (4) k homogeneous 0, (0, y ) f ( y) = x qo , 0 x L L x 2L non-homogeneous This condition represents a heat flux which varies along the boundary according to f ( y ). (4) Solution. (i) Assumed Product Solution. Assume a solution in the form ( x, y ) = X(x) Y(y) 3-17 (c) PROBLEM 3.4 (continued) Substituting into (a), separating variables and setting the resulting equation equal to constant 1 d2X 1 d 2Y 2n 2 2 X dx Y dy Assuming that n is multi-valued, the above gives d2Xn 2n X n 0 2 dx (d) d 2Yn 2nYn 0 2 dy (e) and (ii) Selecting the Sign of the 2n Terms. Since the y-variable has two homogeneous boundary conditions the 2nYn term in (e) takes the positive sign. Therefore, (d) and (e) become d2Xn (f) 2n X n 0 2 dx and d 2Yn (g) 2nYn 0 2 dy For the important case of n 0, equations (f) and (g) become d2X0 0 d x2 (h) d 2Y0 0 d y2 (i) and (iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (f)-(i) are X n ( x ) An exp( n x ) Bn exp( n x ) (j) Yn ( y ) Cn sin n y Dn cosn y X 0 ( x ) A0 x B0 (k) (l) Y0 ( y ) C0 y D0 (m) Corresponding to each value of n there is a temperature solution n ( x, y ). Thus and n ( x, y ) X n ( x )Yn ( y ) (n) 0 ( x, y ) X 0 ( x )Y0 ( y ) (o) 3-18 PROBLEM 3.4 (continued) The complete solution becomes ( x, y ) X 0 ( x )Y0 ( y ) X n 1 n ( x )Yn ( y ) (p) (iv) Application of Boundary Conditions. Boundary condition (1) applied to solutions (k) and (m) gives D n Do 0 (q) Boundary condition (2) applied to (j) gives the characteristic equation for n sin 2n L 0 (r) Thus n n n 2L (n = 1,2,3….) (s) Similarly, boundary condition (2) applied to (m) gives C0 0 With C0 D0 0, the solution corresponding to n 0 vanishes. Application of boundary condition (3) to (j) gives Bn 0 Solution (p) becomes ( x, y ) a n exp( n x )sin n y (t) n 1 where an An Cn . The only remaining unknown is the set of constants an. Application of condition (4) gives f(y) = k an n sin n y (u) n 1 where f ( y ) is defined in boundary condition (4). (v) Orthogonality. To determine a n in equation (u) we apply orthogonality. Note that the characteristic functions n ( y ) sin n y in (u) are solutions to equation (g). Comparing (g) with eq. (3.5a) shows that it is a Sturm-Liouville equation with a1 ( y ) a2 ( y ) 0 and a 3 ( y ) 1 Thus eq. (3.6) gives p ( y ) w( y ) 1 and q( y ) 0 Since the boundary conditions at y = 0 and y = 2L are homogeneous, it follows that the characteristic functions n ( y ) sin n y are orthogonal with respect to the weighting function w( y ) 1. Multiplying both sides of (u) by w( y ) sin m y dy and integrating from y = 0 to y = L 3-19 PROBLEM 3.4 (continued) 2L f ( y ) w( y ) sin m y dy = k 0 a n n sin n y w( y ) sin m y dy n 1 2L 0 Interchanging the summation with integration and invoking orthogonality gives 2L f ( y ) w( y ) sin n y dy = k an n 0 2L 0 w( y ) sin2 n y Solving for a n and recalling that w(y) = 1 the above gives an 2L 0 kn f ( y ) sin n y dy 2L 0 sin n y dy 2 1 kn L 2L f ( y ) sin n y dy 0 Evaluating the integral in the denominator, the above gives an 1 kn L 2L f ( y ) sin n y dy (v) 0 Using the definition of f ( y ) in condition (4) the integral in (v) is evaluated 2L 0 f ( y ) sin n y dy L 0 (0) sin n y dy 2L L qo sin n y dy ( q / n )(cos2n L cosn L) Substituting into (v) and using (s) to eliminate n L and (cosn L cosn L / 2) , we obtain an 2qo L / k (n ) 2 (1) n 1 2(1) n 1 (w) Substituting (w) into (t) and rearranging the result gives T ( x, y ) To ( 1) n 1 2( 1) n 1 n x 2 e sin n y qoL / k (n ) 2 n 1 (5) Checking. Dimensional check: According to (t) the coefficient a n should have units of temperature. The term qoL / k in (w) has units of temperature. Limiting check: If qo 0 , the entire plate should be at a uniform temperature To . Setting qo 0 in (w) gives a n 0. When this is substituted into (t) gives 0 , or T ( x, y ) To . Boundary conditions check: Conditions (1), (2) and (3) can be shown to be satisfied by direct substitution. (6) Comments. The same approach can be used to solve the more general problem where the heat flux at the boundary is variable, qo qo ( y ). This will affect the integral in (v). 3-20 PROBLEM 3.5 Consider two-dimensional steady state conduction in a rectangular plate L H . The sides (0,y) and (x,H) exchange heat by convection. The heat transfer coefficient is h and ambient temperature is T . Side (x,0) is insulated. Half the surface at x = L is cooled at a flux q1 while the other half is heated at a flux q2. Determine the heat transfer along (0,y). y h T 0 h, T H /2 q2 H /2 q1 2 HBC L x (1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) Heat transfer rate along surface (0,y) is determined using temperature distribution and Fourier’s law. (iii) Four boundary conditions are needed. (iv) Two conditions can be made homogeneous by defining a new temperature variable ( x, y ) T ( x, y ) T . (v) The heat flux along boundary (L,y) is not uniform. This requires special consideration. (2) Origin and Coordinates. The origin and coordinate axes are selected as shown. (3) Formulation. (i) Assumptions. (1) Two-dimensional, (2) steady, (3) no energy generation and (4) constant thermal conductivity. (ii) Governing Equations. To determine the rate of heat transfer we must first determine the temperature distribution in the plate. Introducing the above assumptions into eq. (1.8) gives the heat equation for this case 2 2 0 x2 y2 where (a) ( x, y ) T ( x, y ) T (b) (iii) Independent Variable with Two Homogeneous Boundary Conditions. The two boundaries at y = 0 and y = H have homogeneous conditions. Thus, the y-variable has two homogeneous boundary conditions. (iv) Boundary Conditions. The boundary conditions are (1) ( x,0) 0, y homogeneous (2) k ( x, H ) h ( x , H ) , y homogeneous (3) k (0, y ) h (0, y ) , x homogeneous (4) k q , ( L , y ) f(y) = 1 x q2 , 0 y H /2 H /2 y H non-homogeneous This condition represents a heat flux, which varies along the boundary according to f ( y ). 3-21 PROBLEM 3.5 (continued) (4) Solution. (i) Assumed Product Solution. Assume a solution in the form ( x, y ) = X(x) Y(y) (c) Substituting into (a), separating variables and setting the resulting equation equal to constant 1 d2X 1 d 2Y 2n 2 2 X dx Y dy Assuming that n is multi-valued, the above gives d2Xn 2n X n 0 2 dx (d) d 2Yn 2nYn 0 2 dy (e) and (ii) Selecting the Sign of the 2n Terms. Since the y-variable has two homogeneous boundary conditions the 2nYn term in (e) takes the positive sign. Therefore, (d) and (e) become d2Xn (f) 2n X n 0 2 dx and d 2Yn (g) 2nYn 0 2 dy For the important case of n 0, equations (f) and (g) become d2X0 0 d x2 (h) d 2Y0 0 d y2 (i) and (iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (f)-(i) are X n ( x ) An sinh n x Bn coshn x (j) Yn ( y ) Cn sin n y Dn cosn y (k) (l) (m) X 0 ( x ) A0 x B0 Y0 ( y ) C0 y D0 Corresponding to each value of n there is a temperature solution n ( x, y ). Thus n ( x, y ) X n ( x )Yn ( y ) 3-22 (n) PROBLEM 3.5 (continued) and 0 ( x, y ) X 0 ( x )Y0 ( y ) (o) The complete solution becomes ( x, y ) X 0 ( x )Y0 ( y ) X n 1 (p) n ( x )Yn ( y ) (iv) Application of Boundary Conditions. Boundary condition (1) applied to solutions (k) and (m) gives C n C0 0 (q) Boundary condition (2) applied to (k) gives the characteristic equation for n n H tan n H (hH / k ) (r) Similarly, boundary condition (2) applied to (m) gives D0 0 With C0 D0 0, the solution corresponding to n 0 vanishes. Application of boundary condition (3) to (j) gives Bn ( n H )(k / hH ) An Solution (p) becomes ( x, y ) a sinh x ( H )(k / hH ) cosh xcos n 1 n n n n ny (s) where a n An Dn . The only remaining unknown is the set of constants an. Application of condition (4) gives f(y) = k a cosh L ( H )(k / hH ) sinh Lcos n 1 n n n n n ny (t) where f ( y ) is defined in boundary condition (4). (v) Orthogonality. To determine a n in equation (t) we apply orthogonality. Note that the characteristic functions n ( y ) cosn y in (t) are solutions to equation (g). Comparing (g) with eq. (3.5a) shows that it is a Sturm-Liouville equation with a1 ( y ) a2 ( y ) 0 and a 3 ( y ) 1 Thus eq. (3.6) gives p ( y ) w( y ) 1 and q( y ) 0 Since the boundary conditions at y = 0 and y = H are homogeneous, it follows that the characteristic functions n ( y ) cosn y are orthogonal with respect to the weighting function w(y) = 1. Multiplying both sides of (t) by w( y ) cosm ydy and integrating from y 0 to y H 3-23 PROBLEM 3.5 (continued) H f ( y ) w( y ) cos m ydy k 0 0 ann cosh n L (n H )(k / hH ) sinh n Lcos n y w( y ) cos m y dy n 1 H Interchanging the summation with integration, recalling that w( y ) 1 and invoking orthogonality, eq. (3.7), gives f ( y ) cos n ydy kan n cosh n L n H (k / hH ) sinh n L H H cos2 n y dy 0 0 Solving for a n the above gives an H 0 f ( y ) cos n y dy k n cosh n L ( n H )(k / hH ) sinh n L H 0 cos2 n y dy Evaluating the integral in the denominator, the above becomes 4 an H 0 f ( y ) cos n y dy k cosh n L ( n H )(k / hH ) sinh n L(2 n H sin 2 n H ) (u) Using the definition of f ( y ) in condition (4) the integral in (u) is evaluated H 0 H /2 0 f ( y ) sin n y dy q1 cosn y dy q2 H H / 2 cosn y dy ( q1 / n ) sin n H / 2 ( q2 / n )(sinn H sin n H / 2) Substituting into (u) an 4(q1H / k ) sin n H / 2 (q 2 H / k )(sin n H sin n H / 2) ( n H )cosh n L ( n H )(k / hH ) sinh n L(2 n H sin 2 n H ) (v) Substituting (v) into (s) and rearranging the result gives T ( x, y ) T 4 q1H / k n 1 sin(n H / 2) (q2 / q1)(sin n H sin n H / 2)sinh n x (n H )(k / hH ) coshn x cos y n (n H )cosh n L (n H )(k / hH ) sinh n L (2n H sin 2n H ) (w) Having determined the temperature distribution in the plate, the heat transfer rate through surface (0,y) can be determined using Fourier’s law as follows q (0) k H o T (0, y ) dy x where q (0) is heat transfer rate through (0,y) per unit plate depth. Substituting (s) into the above 3-24 PROBLEM 3.5 (continued) H q (0) k a n n cos n ydy k 0 a n sin n H n 1 (5) Checking. Dimensional check: (i) The arguments of sin, cos, sinh and cosh are dimensionless. (ii) The Biot number, hH/k, is dimensionless. (iii) According to (s), the coefficient a n should have units of temperature. The term q / n k in (v) gives a n the correct units. Limiting check: If q1 q2 0 , the entire plate should be at a uniform temperature T . Setting q1 q2 0 in (w) gives T ( x, y) T . Boundary conditions check: Conditions (1), (2) and (3) can be shown to be satisfied by direct substitution. (6) Comments. (i) If the direction of q1 and/or q 2 is reversed, solution (w) is still applicable. However, the signs of q1 and q 2 must be changed accordingly. The same approach can be used to solve the more general problem where the heat flux at the boundary is variable, qo qo ( y ). (ii) Equation (r) gives the dimensionless roots k H in terms of the Biot number hH/k. However, the argument k L is determined by multiplying and dividing by L. This introduces an additional parameter H / L. Thus examining solution (w) we note that the temperature distribution depends on three parameters: hH/k, H/L and q2 / q1 . 3-25 PROBLEM 3.6 Radiation strikes one side of a rectangular element L H and is partially absorbed. The absorbed energy results in a non-uniform volumetric energy generation given by y T1 q Ao e y H qo L q Ao e y 0 T2 x radiation where Ao and are constant. The surface which is exposed to radiation, (x,0), is maintained at T2 while the opposite surface (x,H) is at T1. The element is insulated along surface (0,y) and cooled along (L,y) at a uniform flux qo . Determine the steady state temperature of the insulated surface. (1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) Four boundary conditions are needed. (iii) Only one boundary condition is homogeneous. (iv) The energy generation term makes the differential equation non-homogeneous. (v) Energy generation varies in the y-direction. (2) Origin and Coordinates. The origin and coordinate axes are selected as shown. (3) Formulation. ` (i) Assumptions. (1) Two-dimensional, (2) steady and (3) constant thermal conductivity. (ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives 2T 2T q 0 x2 y2 k (a) q Ao e y (b) 2T 2T Ao y e 0 x2 y2 k (c) where Substituting (b) into (a) (iii) Independent Variable with Two Homogeneous Boundary Conditions. Neither variable has two homogeneous conditions. Only the y-variable can have two homogeneous conditions. (iv) Boundary Conditions. (1) T ( x,0) T2 , non-homogeneous (2) T ( x, H ) T1 , non-homogeneous T (0, y ) 0 , homogeneous x T ( L, y ) qo , non-homogeneous (4) k x (3) 3-26 PROBLEM 3.6 (continued) (4) Solution. Equation (c) is non-homogeneous due to the heat generation term. Thus we can not proceed directly with the application of the method of separation of variables. Noting that the heat generation term is a function of y, we assume a solution of the form T ( x, y ) ( x, y ) ( y ) (d) Note that ( x, y ) depends on two variables while ( y ) depends on a single variable. Substituting (d) into (c) 2 2 d 2 Ao y e 0 (e) x2 y2 d y2 k The next step is to split (c) into two equations, one for ( x, y ) and the other for ( y ). We select 2 2 0 (f) x2 y2 Thus, the second part must be d 2 Ao y e 0 (g) d y2 k Boundary conditions on ( x, y ) and ( y ) are obtained by substituting (d) into the four boundary conditions. In this process the function ( x, y ) is assigned homogeneous conditions whenever possible. Using (d) and boundary condition (1) ( x,0) (0) T2 Let (0, y ) 0 (f-1) (0) T2 (g-1) Thus Boundary condition (2) and (b) give ( x, H ) ( H ) T1 Let ( x, H ) 0 (f-2) ( H ) T1 (g-2) (0, y ) 0 x (f-3) ( L, y ) qo x (f-4) Thus Boundary condition (3) and (b) give Boundary condition (4) and (b) give k Thus, non-homogeneous equation (c) is replaced by two equations: (f), which is a homogeneous partial differential equation and (g) which is a non-homogeneous ordinary 3-27 PROBLEM 3.6 (continued) differential equation. Equation (f) has three homogeneous conditions, (f-1), (f-2) and (f-3) and one non-homogeneous condition, (f-4). The solution to (g) is ( y) Ao k 2 e y E y F (h) where E and F are constants of integration. Application of boundary conditions (g-1) and (g-2) gives the two constants A A E1 (1 / H ) (T1 T2 ) 2o (e H 1) and F T2 2o k k Substituting (i) into (h) ( x) Ao y A A e (T1 T2 ) 2o (e H 1)( y / H ) + T2 2o 2 k k k (i) (j) (i) Assumed Product Solution. To solve partial differential equation (f) we assume a product solution. Let ( x, y ) X ( x )Y ( y ) (k) Substituting (k) into (f), separating variables and setting the resulting two equations equal to a constant, 2n , we obtain d2Xn 2n X n 0 2 dx (l) d 2Yn 2nYn 0 2 dy (m) and (ii) Selecting the Sign of the 2n Terms. Since the y-variable has two homogeneous boundary conditions, the plus sign is selected in (m). Thus (l) and (m) become d2Xn 2n X n 0 2 dx (n) d 2Yn 2nYn 0 2 dy (o) and For the special case of n 0 the above equations take the form d 2X0 0 d x2 (p) d 2Y0 0 d y2 (q) and 3-28 PROBLEM 3.6 (continued) (iii) Solutions to the Ordinary Differential Equations. The solutions to equations (n)(q) are X n ( x ) An sinh n x Bn coshn x (r) Yn ( y ) Cn sin n y Dn cosn y (s) X 0 ( x ) A0 x B0 (t) Y0 ( y ) C0 y D0 (u) and Corresponding to each value of n there is a solution n ( x, y ). Thus and n ( x, y ) X n ( x )Yn ( y ) (v) 0 ( x, y ) X 0 ( x )Y0 ( y ) (w) The complete solution to T ( x, y ) becomes T ( x, y ) ( x) X 0 ( x)Y0 ( y ) X n ( x)Yn ( y ) (x) n 1 (iv) Application of Boundary Conditions. Applying boundary condition (f-1) to equation (s) and (u) gives Dn D0 0 Condition (f-2) applied to (s) gives the characteristic equation for n sin n H 0 Thus n n / H , (n 1, 2, 3....) (y) Condition (f-2) applied to (u) C0 0 Boundary condition (f-3) and (r) give An 0 With C0 D0 0, the solution X 0Y0 vanishes. Substituting into (x) and using (r) and (s), the solution to ( x, y ) becomes ( x, y ) a n (coshn x ) sin n y (z) n 1 where a n Bn Cn . Thus the only remaining unknown is a n . Application of boundary condition (f-4) gives qo k a n ( n sinh n L) sin n y n 1 To determine a n from this result we apply orthogonality. 3-29 (z-1) PROBLEM 3.6 (continued) (v) Orthogonality. The characteristic functions, sin n y, in (z-1) are solutions to equation (o). Comparing (o) with eq. (3.5a) shows that it is a Sturm-Liouville equation with a1 ( y ) a2 ( y ) 0 and a 3 ( y ) 1 Thus eq. (3.6) gives p ( y ) w( y ) 1 and q( y ) 0 Since boundary conditions (f-1) and (f-2) at y = 0 and y = H are homogeneous, it follows that the characteristic functions sin n y are orthogonal with respect to w(y) = 1. Multiplying both sides of (z-1) by w(y) sin m y dy and integrating from y = 0 to y = H H qw( y) sin o m y dy =k 0 0 (a n n sinhn L) sin n y w( y )(sinm y ) dy n1 H Interchanging integration the integration and summation signs, noting that w( y ) 1 and applying orthogonality, the above gives H q o sin n ydy = a n n sinh n H 0 H sin 2 n y dy 0 Evaluating the integrals, solving the above for a n and using (y) gives 2( qoH / k ) ( 1) n 1 an (n ) 2 sinh n L (z-2) The solution to the temperature distribution is obtained by substituting (j), (z) and (z-2) into (d) A A A T ( x, y ) 2o e y (T1 T2 ) 2o (e H 1)( y / H ) + T2 2o + k k k 2( qoH / k ) (1) 1 (n ) n 1 n 2 sinh n L (coshn x ) sin n y (z-3) Expressed in dimensionless form the becomes (T1 T2 ) T ( x, y ) T2 H y ( e 1 ) 1 e ( y / H ) + 2 ( Ao / 2 k ) Ao / k n q H ( 1) 1 2 o 2 (coshn x ) sin n y Ao / n 1 (n ) 2 sinh n L (z-4) (5) Checking. Dimensional checks: (i) The arguments of sin, sinh and cosh are dimensionless. (ii). Each term in (z-4) is dimensionless. Limiting check: For the special case of qo 0 temperature distribution should be onedimensional. Setting qo 0 in (z-3) or (z-4) gives the one-dimensional solution to the problem. If we further assume that there is no energy generation, (z-3) gives 3-30 PROBLEM 3.6 (continued) T ( y ) T2 (T1 T2 )( y / H ) This is recognized as the linear one-dimensional solution in a plate with two boundaries at specified temperatures. Boundary conditions check: Conditions (1), (2) and (3) can be shown to be satisfied by direct substitution. (6) Comments. (i) Selecting in the assumed solution (d) to be a function of y rather than x is motivated by the fact that the energy generation term is a function of y. (ii) Equation (y) gives the dimensionless roots k H . However, the argument k L is determined by multiplying and dividing by L. This introduces H / L as a parameter, Thus examining solution (z-4) we note that the temperature distribution depends on four parameters: T1 T2 , Ao / 2 k qoH H , H and 2 L Ao / 3-31 PROBLEM 3.7 An electric strip heater of width 2b dissipates energy at a flux q o . The heater is mounted symmetrically on one side of a plate of height H and width 2L. Heat is exchanged by convection along surfaces (0,y) and (2L,y). The ambient temperature is T and the heat transfer coefficient is h. Surfaces (x,0) and (x,H) are insulated. Determine the temperature of the plateheater interface for two-dimensional steady state conduction. y b b h T 0 h H T L L x (1) Observations. (i) To determine the plate-heater interface temperature it is necessary to first determine the temperature distribution in the plate. (ii) This is a steady state twodimensional conduction problem. (iii) Four boundary conditions are needed. (iv) Temperature distribution is symmetrical about the vertical centerline. This makes it possible to analyze half the plate. (v) Only the x-variable can have two homogeneous conditions. (vi) Defining a new temperature variable ( x, y ) T ( x, y ) T makes the convection condition at (L,y) homogeneous. (vii) The heat flux along (x,H) is non-uniform. This requires special consideration. y (2) Origin and Coordinates. Because half the plate is to be considered, the origin is selected at the centerline and the coordinate axes are directed as shown. b (3) Formulation. 2HBC (i) Assumptions. (1) Two-dimensional, (2) steady, (3) no energy generation and (4) constant thermal conductivity. (ii) Governing Equations. Introducing assumptions into eq. (1.8) gives the heat equation where qo the L 0 H Th x above 2 2 0 x2 y2 (a) ( x, y ) T ( x, y ) T (b) (iii) Independent Variable with Two Homogeneous Boundary Conditions. The two boundaries at x = 0 and x = L have homogeneous conditions. Thus, the x-variable has two homogeneous boundary conditions. (iv) Boundary Conditions. The required four boundary conditions are (0, y ) 0, homogeneous x ( L, y ) h ( L, y ), homogeneous (2) x ( x ,0) 0, homogeneous (3) y (1) 3-32 PROBLEM 3.7 (continued) (4) k q , ( x, H ) f(x) = o y 0, 0 xb b x L This condition represents a heat flux which varies along the boundary according to f ( x ). (4) Solution. (i) Assumed Product Solution. Assume a solution in the form ( x, y ) = X(x) Y(y) (c) Substituting into (a), separating variables and setting the resulting equation equal to constant 1 d2X 1 d 2Y 2n 2 2 X dx Y dy Assuming that n is multi-valued, the above gives d2Xn 2n X n 0 2 dx (d) d 2Yn 2nYn 0 2 dy (e) and (ii) Selecting the Sign of the 2n Terms. Since the x-variable has two homogeneous boundary conditions the 2n X n term in (d) takes the positive sign. Therefore, (d) and (e) become d2Xn (f) 2n X n 0 2 dx and d 2Yn (g) 2nYn 0 2 dy For the important case of n 0, equations (f) and (g) become d2X0 0 d x2 (h) d 2Y0 0 d y2 (i) and (iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (f)-(i) are X n ( x ) An sin n x Bn cos n x (j) Yn ( y ) Cn sinh n y Dn coshn y (k) 3-33 PROBLEM 3.7 (continued) X 0 ( x ) A0 x B0 (l) (m) Y0 ( y ) C0 y D0 Corresponding to each value of n there is a temperature solution n ( x, y ). Thus and n ( x, y ) X n ( x )Yn ( y ) (n) 0 ( x, y ) X 0 ( x )Y0 ( y ) (o) The complete solution becomes ( x, y ) X 0 ( x )Y0 ( y ) X n 1 n ( x )Yn ( y ) (p) (iv) Application of Boundary Conditions. Boundary condition (1) applied to solutions (j) and (l) gives An A0 0 (q) Boundary condition (2) applied to (j) gives the characteristic equation for n n L tan n L hL / k Bi (r) Similarly, boundary condition (2) applied to (l) gives B0 0 With A0 B0 0, the solution corresponding to n 0 vanishes. Application of boundary condition (3) to (k) gives Cn 0 (s) Solution (p) becomes ( x, y ) a n (cosn x )(coshn y ) (t) n 1 where a n Bn Dn . The only remaining unknown is the set of constants an. Application of condition (4) gives f(x) = k (a n 1 n n sinh n H ) cosn x (u) where f ( x ) is defined in boundary condition (4). (v) Orthogonality. To determine a n in equation (u) we apply orthogonality. Note that the characteristic functions n ( x ) cosn x in (u) are solutions to equation (f). Comparing (f) with eq. (3.5a) shows that it is a Sturm-Liouville equation with a1 ( x) a2 ( x) 0 and a3 ( x ) 1 Thus eq. (3.6) gives p ( x ) w( x ) 1 and q( x ) 0 3-34 PROBLEM 3.7 (continued) Since the boundary conditions at x = 0 and x = L are homogeneous, it follows that the characteristic functions n ( x ) cosn x are orthogonal with respect to the weighting function w( x ) 1. Multiplying both sides of (u) by w( x ) sin m x dx and integrating from x = 0 to x = L L f ( x ) w( x ) cosm x dx = k 0 (a n n sinh n H ) cosn x w( x ) cosm x dx n 1 L 0 Interchanging the summation with integration and invoking orthogonality gives L f ( x ) w( x ) cosn x dx = k a n n sinh n H 0 L 0 w( x ) cos2 n x Solving for a n , recalling that w(x) = 1 and evaluating the integral on the right-hand-side, the above gives L an 4 0 f ( x) cosn x dx (v) k sinh n H ( 2n L sin 2n L) Using the definition of f ( x ) in condition (4) the integral in (v) is evaluated L 0 f ( x ) cosn x dx b 0 ( qo ) cosn x dx L b (0)n x dx ( qo / n ) sin n b Substituting into (v) an 4 qo sin n b kn sinh n H ( 2n L sin 2n L) (w) The solution to the temperature distribution is obtained by substituting (b) and (v0 into (t) sin n b T ( x, y ) T 4 (cosn x ) coshn y ( qoL / k ) L(sinh n H )(2n L sin 2n L) n 1 n (x) To determine the plate-heater interface temperature set y = H in (x) (cothn H )(sin n b) T ( x, H ) T 4 cosn x ( qoL / k ) L ( 2 L sin 2 L ) n n n n 1 0 xb (y) (5) Checking. Dimensional check: (i) The arguments of sin, cos, sinh and cosh are dimensionless. (ii) The term qoL / k in (x) has units of temperature. Thus both sides of (x) are dimensionless. Limiting check: If qo 0 , the entire plate should be at a uniform temperature T . Multiplying (x) through by ( qoL / k ) and setting qo 0 gives T ( x, y ) To . Boundary conditions check: Conditions (1), (2) and (3) can be shown to be satisfied by direct substitution. 3-35 PROBLEM 3.7 (continued) (6) Comments. (i) The same approach can be used to solve the more general problem where the heat flux at the boundary is variable, qo qo (x ). This will only affect the integral in (v). (ii) Taking advantage of symmetry simplifies the problem. (iii) Equation (r) gives the dimensionless roots n L . However, the argument k H is determined by multiplying and dividing by L according to k H = (k L)( H / L) . This introduces an additional dimensionless parameters H / L . Similarly, the argument k b introduces the parameter b / L. Thus, three parameters characterize the problem: Bi, b / L, and H / L. 3-36 PROBLEM 3.8 The cross section of a long prism is a right angle isosceles triangle of side L. One side is heated with uniform flux q o while the second side is maintained at zero temperature. The hypotenuse is insulated. Determine the temperature Tc at the mid-point of the hypotenuse. (1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) The geometry does not lend itself to analytic solution by the method of separation of variables. However, due to symmetry the temperature distribution in the right-angled isosceles triangle is the same as that in a square with symmetrical boundary conditions about the diagonal as shown. (iii) To determine the temperature at the mid-point of the hypotenuse (center of square) it is necessary to determine the temperature distribution in the square. (iv) There are two non-homogeneous boundary conditions. Since no direction has two homogeneous conditions a modified procedure for applying the separation of variables method is needed. qo L Tc T 0 y qo qo x Tc L L T 0 0 T 0 (2) Origin and Coordinates. The origin and coordinate axes for the square geometry are selected as shown. (3) Formulation. (i) Assumptions. (1) Two-dimensional, (2) steady, (3) no energy generation and (4) constant thermal conductivity. (ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives 2T 2T 0 x2 y2 (a) (iii) Independent Variable with Two Homogeneous Boundary Conditions. No variable has two homogeneous conditions. Thus the method of separation of variables can not be applied directly to obtain a solution. A modified procedure will be used in which a two-dimensional problem with two homogeneous conditions in the x-variable will represent one part of the solution. A second part which is one-dimensional will satisfy the nonhomogeneous conditions in the x-variable. (iv) Boundary Conditions. Consider the square formed by the triangle and its mirror image. The four boundary conditions are (1) T (0, y ) 0 , homogeneous T ( L, y ) qo , non-homogeneous x (3) T ( x,0) 0 , homogeneous (2) k (4) k T ( x, L) qo , non-homogeneous y 3-37 PROBLEM 3.8 (continued) (4) Solution. Since there are two non-homogenous boundary conditions we assume a solution of the form T ( x, y ) ( x, y ) ( x ) (b) Substituting (b) into (a) 2 2 d 2 0 x2 y2 x2 (c) 2 2 0 x2 y2 (d) Let Thus d 2 =0 d x2 (e) Boundary conditions on (d) and (e) are obtained by substituting (b) into the four conditions. Substituting (b) in condition (1) (0, y ) (0) 0 Let (0, y ) 0 (d-1) Thus ( 0) 0 (e-1) Condition (2) gives ( L, y ) d ( L) qo x dx Let ( L, y ) 0 (d-2) x Thus d ( L ) qo (e-2) dx Condition (3) gives ( x,0) ( x ) 0 Or ( x,0) ( x ) (d-3) Boundary condition (4) gives ( x, L) k qo (d-4) y Thus partial differential equation (c) has two homogeneous boundary conditions in the xvariable, (d-1) and (d-2). The solution to (e) with boundary conditions (e-1) and (e-2) is q ( x) o x (f) k We proceed now to determine the solution to (d) using separation of variables. 3-38 PROBLEM 3.8 (continued) (i) Assumed Product Solution. Assume a solution in the form ( x, y ) = X(x) Y(y) (g) Substituting into (d), separating variables and setting the resulting equation equal to constant 1 d2X 1 d 2Y 2n 2 2 X dx Y dy (h) Assuming that n is multi-valued, the above gives d2Xn 2n X n 0 2 dx (i) d 2Yn 2nYn 0 2 dy (j) and (ii) Selecting the Sign of the 2n Terms. Since the x-variable has two homogeneous boundary conditions the 2n X n term in (i) takes the positive sign. Therefore, (i) and (j) become d2Xn 2n X n 0 (k) 2 dx and d 2Yn 2nYn 0 (l) d y2 For the important case of n 0, equations (k) and (l) become d2X0 0 d x2 (m) d 2Y0 0 d y2 (n) and (iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (k)-(n) are X n ( x ) An sin n x Bn cosn x (o) Yn ( y ) Cn sin n y Dn cosn y (p) (q) (r) X 0 ( x ) A0 x B0 Y0 ( y ) C0 y D0 Corresponding to each value of n there is a temperature solution ( x, y ). Thus ( x, y ) X n ( x )Yn ( y ) 3-39 (s) PROBLEM 3.8 (continued) and 0 ( x, y ) X 0 ( x )Y0 ( y ) The complete solution becomes (t) ( x, y ) X 0 ( x )Y0 ( y ) X n ( x )Yn ( y ) (u) n 1 (iv) Application of Boundary Conditions. Boundary condition (d-1) applied to solutions (o) and (q) gives (v) Bn Bo 0 Boundary condition (d-2) applied to (o) gives the characteristic equation for n cosn L 0 Or ( 2n 1) 2 n L (w) (n = 1,2,3….) (x) Similarly, boundary condition (d-2) applied to (q) gives A0 0 With A0 B0 0, the solution corresponding to n 0 vanishes. Thus (u) becomes ( x, y ) ( an sinh n y bn coshn y ) sinn x (y) n 1 where an AnCn and bn An Dn . Condition (d-3) applied to (y) and using (f) gives qox k b sin x n 1 n (z-1) n Similarly, condition (d-4) and (y) give qo k (a cosh L b sinh L) sin x n n n n n n (z-2) n1 (v) Orthogonality. To determine a n and bn in equations (z-1) and (z-2) we apply orthogonality to each equation. Note that the characteristic functions n ( x ) sin n x in (z) are solutions to equation (k). Comparing (k) with eq. (3.5a) shows that it is a Sturm-Liouville equation with a1 ( x) a2 ( x) 0 and a3 ( x ) 1 Thus eq. (3.6) gives p ( x ) w( x ) 1 and q( x ) 0 Since the boundary conditions at x = 0 and x = L are homogeneous, it follows that the characteristic functions n ( x ) sin n x are orthogonal with respect to the weighting function w( x ) 1. Multiplying both sides of (z-1) by w( x ) sin m x dx and integrating from x = 0 to x = L gives 3-40 PROBLEM 3.8 (continued) L qo x w( x ) sin m x dx = k 0 bn sin n x w( x ) sin m x dx n 1 L 0 Interchanging the summation with integration and noting that w(x) = 1, orthogonality gives L qo x sin n x dx = k bn 0 L 0 sin 2 n x dx Evaluation the integrals, using equation (x) and solving for bn gives 8( qoL / k ) ( 1) n (z-3) 2 ( 2n 1) 2 Similarly, multiplying both sides of (z-2) by w( x ) sin m x dx and integrating from x = 0 to x = L gives bn L qo w( x ) sin m x dx = k 0 L 0 n (an coshn L bn sinh n L) sin n x w( x ) sin m x dx n 1 Interchanging the summation with integration and noting that w(x) = 1, orthogonality gives L qo sin n x dx = kn ( an coshn L bn sinh n L) 0 L 0 sin2 n x dx Evaluation the integrals using equation (x) and (z-3) and solving for a n gives an 8( qoL / k ) 1 ( 1) n sinh n L 2 ( 2n 1) 2 coshn L (z-4) The solution to the temperature distribution is obtained by substituting (f), (y), (z-3) and (z4) into (b) and rearranging the result T ( x, y ) x 8 ( qoL / k ) L 2 1 2 n 1 ( 2n 1) 1 ( 1) n sinh n L sinh n y ( 1) n coshn y sin n x (z-5) coshn L Evaluating (z-5) at the center, x = L/2 and y = L/2, gives Tc Tc 1 8 2 ( qoL / k ) 2 1 ( 1) n sinh n L 1 sinh n L / 2 ( 1) n coshn L / 2 sin n L / 2 2 coshn L n 1 ( 2n 1) (z-6) (5) Checking. Dimensional check: (i) The arguments of sin, cos, sinh and cosh are dimensionless. (ii) Each term in (z-5) and (z-6) is dimensionless. Limiting check: If qo 0 , the entire plate should be at zero temperature. Setting qo 0 in (z-5) gives T ( x, y ) 0. 3-41 PROBLEM 3.8 (continued) Boundary conditions check: Conditions (1), (2) and (3) can be shown to be satisfied by direct substitution. (5) Comments. (i) The same approach can be used to solve the more general problem where the heat flux at the side of the right-angled isosceles triangle is non-uniform as long as the corresponding square problem is assigned the same non-uniform flux on two of its adjacent sides such that there is symmetry with respect to the diagonal. (ii) The method used to solve this problem is limited to a right-angled isosceles triangle which forms a square with its mirror image. It does not work for a right-angled triangle which forms a rectangle with its mirror image because of asymmetry. (iii) An alternate approach to solving the square problem is to use the method of superposition by decomposing the problem into two simpler problems each having a single non-homogeneous condition as illustrated be below. Each problem is solved using separation of variables directly. y qo qo x L Tc T 0 T ( x, y ) T 0 0 T 0 L Tc1 = L x y y qo + L T1 ( x, y ) T 0 3-42 qo 0 x T 0 L Tc 2 L T2 ( x, y ) T 0 0 3-43 PROBLEM 3.9 The cross section of a long prism is a right angle isosceles triangle of side L. One side exchanges heat by convection with the surroundings while the second side is maintained at zero temperature. The heat transfer coefficient is h and ambient temperature is T . The hypotenuse is insulated. Determine the temperature Tc at the mid-point of the hypotenuse. h T L (1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) The geometry does not lend itself to analytic solution by the method of separation of variables. However, due to symmetry the temperature distribution in the right-angled isosceles triangle is the same as that in a square with h boundary conditions symmetrical about the diagonal as shown. T (iii) To determine the temperature at the mid-point of the x hypotenuse (center of square) it is necessary to determine the temperature distribution in the square. (vi) There are two nonhomogeneous boundary conditions. Since no direction has two homogeneous conditions the method of superposition should be used. Tc L T 0 y h, T Tc T 0 L L T 0 0 (2) Origin and Coordinates. The origin and coordinate axes for the square geometry are selected as shown. (3) Formulation. (i) Assumptions. (1) two-dimensional, (2) steady, (3) no energy generation, (4) constant conductivity and (5) uniform heat transfer coefficient h and ambient temperature T . (ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives 2T 2T 0 x2 y2 (a) (iii) Independent Variable with Two Homogeneous Boundary Conditions. No variable has two homogeneous conditions. Thus the method of separation of variables can not be applied directly to obtain a solution. Since there are two non-homogenous boundary conditions, the problem will be decomposed into two simpler problems each having a single non-homogeneous condition. (iv) Boundary Conditions. Consider the square formed by the triangle and its mirror image. The four boundary conditions are (1) T (0, y ) 0 , homogeneous (2) k T ( L, y ) hT ( L, y ) T , non-homogeneous x (3) T ( x,0) 0 , homogeneous (4) k T ( x, L) hT ( x, L) T , non-homogeneous y 3-43 PROBLEM 3.9 (continued) (4) Solution. Since there are two non-homogenous boundary conditions, the problem will be decomposed into two simpler problems each having a single non-homogeneous condition. Let (b) T ( x, y ) T1 ( x, y ) T2 ( x, y ) where T1 ( x, y ) and T2 ( x, y ) are the solutions to two problems each governed by (a) with boundary conditions as shown. y h, T L Tc h T T 0 h 0 = T ( x, y ) x h, T L Tc1 T 0 L 0 y y x h,0 T 0 T1 ( x , y ) T 0 h T + L T 0 L 0 x Tc 2 L T2 ( x , y ) T 0 0 Substituting (b) into (a) 2T1 2T1 2T2 2T2 0 x2 y2 x2 y2 (c) 2T1 2T1 0 x2 y2 (d) 2T2 2T2 0 x2 y2 (e) Let Thus The temperature at the center is given by Tc Tc1 Tc 2 (f) We recognize that the center temperature is the same for each problem. Therefore Tc 2 Tc1 (g) Thus it is only necessary to solve the temperature distribution in one problem. Working with the first problem, T1 ( x, y ), the boundary conditions are (1) T1 (0, y ) 0 , homogeneous (2) k T1 ( L, y ) hT1 ( L, y ) , homogeneous x (3) T1 ( x,0) 0 , homogeneous T1 ( x, L) hT1 ( x, L) T , non-homogeneous y Thus the x-variable has two homogeneous boundary conditions. (4) k 3-44 PROBLEM 3.9 (continued) (i) Assumed Product Solution. Assume a solution in the form T1 ( x, y ) = X(x) Y(y) (h) Substituting into (d), separating variables and setting the resulting equation equal to constant 1 d2X 1 d 2Y 2n X d x2 Y d y2 Assuming that n is multi-valued, the above gives d2Xn 2n X n 0 d x2 (i) d 2Yn 2nYn 0 2 dy (j) and (ii) Selecting the Sign of the 2n Terms. Since the x-variable has two homogeneous boundary conditions the 2n X n term in (i) takes the positive sign. Therefore, (d) and (e) become d2Xn (k) 2n X n 0 2 dx and d 2Yn (l) 2nYn 0 2 dy For the important case of n 0, equations (k) and (l) become d2X0 0 d x2 (m) d 2Y0 0 d y2 (n) and (iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (k)-(n) are X n ( x ) An sin n x Bn cos n x (o) Yn ( y ) Cn sinh n y Dn coshn y (p) (q) (r) X 0 ( x ) A0 x B0 Y0 ( y ) C0 y D0 Corresponding to each value of n there is a temperature solution T1n ( x, y ). Thus 3-45 PROBLEM 3.9 (continued) T1n ( x, y ) X n ( x )Yn ( y ) (s) T10 ( x, y ) X 0 ( x )Y0 ( y ) (t) and The complete solution becomes T1 ( x, y ) X 0 ( x )Y0 ( y ) X n 1 n ( x )Yn ( y ) (u) (iv) Application of Boundary Conditions. Boundary condition (1) applied to solutions (o) and (q) gives (v) Bn B0 0 Boundary condition (2) applied to (o) gives the characteristic equation for n n L (hL / k ) tan n L (w) Similarly, boundary condition (2) applied to (q) gives A0 0 With A0 B0 0, the solution corresponding to n 0 vanishes. Application of boundary condition (3) to (p) gives Dn 0 Solution (u) becomes T1 ( x, y ) a n 1 n (sinh n y )(sin n x ) (x) where an An Cn . The only remaining unknown is the set of constants an. Application of condition (4) gives k (a n 1 n n cos hn L) sin n x h Rearranging the above (a n 1 n sinh n L) sin n x hT hT a h sinh L k cosh Lsin x n n n n n (y) n1 (v) Orthogonality. To determine a n in equation (y) we apply orthogonality. Note that the characteristic functions n ( x ) sin n x in (y) are solutions to equation (k). Comparing (k) with eq. (3.5a) shows that it is a Sturm-Liouville equation with a1 ( x) a2 ( x) 0 and a3 ( x ) 1 Thus eq. (3.6) gives p ( x ) w( x ) 1 and q( x ) 0 3-46 PROBLEM 3.9 (continued) Since the boundary conditions at x = 0 and x = L are homogeneous, it follows that the characteristic functions n ( x ) sin n x are orthogonal with respect to the weighting function w( x ) 1. Multiplying both sides of (y) by w( x ) sin m x dx and integrating from x = 0 to x = L L 0 hT w( x ) sin m xdx L a (h sinh L k n 1 0 n n n coshn L) sin n x w( x ) sin m xdx Interchanging the summation with integration and invoking orthogonality give hT L 0 w( x ) sin n xdx a n ( h sinh n L kn coshn L) L 0 w( x ) sin2 n xdx Solving for a n , recalling that w(x) = 1 and evaluating the integrals in the above gives a n 1 cosn L a n 2T (z) sinh n L (k / hL)(n L) coshn L)n L (cosn L) sin n L Substituting into (x) (1 cos n L)(sin n x ) sinh n y T1 ( x, y ) 2 (z-1) sinh n L (k / hL)(n L) coshn L)n L (cosn L) sin n L T n 1 Evaluating (s-1) at the center, x = L/2 and y = L/2 and substituting into (g) gives Tc Tc (1 cosn L)(sin n L / 2) sinh n L / 2 4 sinh n L (k / hL)(n L) coshn L)n L (cosn L) sin n L T n 1 (z-2) (5) Checking. Dimensional check: (i) The arguments of sin, cos, sinh and cosh are dimensionless. (ii) Each term in solution (z-1) is dimensionless. Limiting check: If T 0 , the entire plate should be at zero temperature. Setting T 0 in (z-1) gives T1 ( x, y) 0. Boundary conditions check: Conditions (1) and (3) can be readily shown to be satisfied by direct substitution in equation (z-1). (6) Comments. (i) The solution is expressed in terms of the Biot number hL/k. This is characteristic of all problems involving surface convection, (ii) Equation (z-2) gives the temperature at the center of a square with convection along two adjacent sides while the other two sides are at zero temperature. If three sides exchange heat by convection, equation (z-2) still applies. However, the constant 4 is replaced by 6. This results from decomposing the problem into three simpler problems whose solutions are given by (z-1). (iii) The method used to solve this problem is limited to a right-angled triangle which forms a square with its mirror image. It does not work with a triangle which forms a rectangle with its mirror image because of asymmetry. 3-47 PROBLEM 3.10 2 HBC y By neglecting lateral temperature variation in the h, T analysis of fins, two-dimensional conduction is H modeled as a one-dimensional problem. To examine 0 x this approximation, consider a semi-infinite plate of H thickness 2H. The base is maintained at uniform To h, T temperature To . The plate exchanges heat by convection at its semi-infinite surfaces. The heat transfer coefficient is h and the ambient temperature is T . Determine the heat transfer rate at the base. (1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) At x the plate reaches ambient temperature. (iii) All four boundary conditions are nonhomogenous. However, by defining a temperature variable T T , three of the four conditions become homogeneous. (iv) Temperature distribution is symmetrical about y 0 and thus the boundary (x,0) is insulated. Therefore, consideration is given to half the plate. (2) Origin and Coordinates. The origin and coordinate axes are selected as shown. (3) Formulation. (i) Assumptions. (1) Two-dimensional, (2) steady, (3) uniform heat transfer coefficient and ambient temperature, (4) constant conductivity and (5) no energy generation. (ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives 2 2 0 x2 y2 where is defined as (a) ( x, y ) T ( x, y ) T (iii) Independent Variable with Two Homogeneous Boundary Conditions. In terms of , the y-variable has two homogeneous conditions. (iv) Boundary Conditions. We consider the upper half of the plate. The boundary conditions for this half are ( x,0) 0, homogeneous y ( x, H ) h ( x, H ), homogeneous (2) k y (3) ( , y ) = 0, homogeneous (1) (4) (0, y ) To T , non-homogeneous (4) Solution. (i) Assumed Product Solution. Let (ii) 3-48 PROBLEM 3.10 (continued) ( x, y ) X ( x )Y ( y ) Substituting into (a), separating variables and setting the resulting equation equal to constant 1 d2X 1 d 2Y 2n 2 2 X dx Y dy Assuming that n is multi-valued, the above gives d2Xn 2n X n 0 2 dx (b) d 2Yn 2nYn 0 d y2 (c) and (ii) Selecting the Sign of the 2n Terms. Since the y-variable has two homogeneous conditions, the 2n Yn term in (c) takes the positive sign. Thus d2Xn 2n X n 0 2 dx (d) d 2Yn 2nYn 0 2 dy (e) and For the special case of n 0, equations (d) and (e) become d2X0 d x2 0 (f) 0 (g) and d 2 Y0 d y2 (iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (d)-(g) are X n ( x ) An exp( n x ) Bn exp( n x ) (h) Yn ( y ) Cn sin n y Dn cosn y (i) X 0 ( x ) A0 x B0 (j) Y0 ( y ) C0 y D0 (k) Corresponding to each value of n there is a temperature solution n ( x, y ). Thus n ( x, y ) X n ( x )Yn ( y ) and 3-49 (l) PROBLEM 3.10 (continued) 0 ( x, y ) X 0 ( x )Y0 ( y ) (m) The complete solution becomes X n 1 ( x, y ) X 0 ( x )Y0 ( y ) n ( x )Yn ( y ) (n) (iv) Application of Boundary Conditions. Boundary condition (1) applied to (i) and (k) gives C n C0 0 (o) Applying boundary condition (2) to (i) gives the characteristic equation for n n H tan n L hH / k Bi (p) where Bi hH / k is the Biot number. Boundary conditions and (2) applied to (k) gives D0 0 Thus, the solution corresponding to n 0 vanishes. Boundary condition (3) gives Bn 0 The temperature solution (n) becomes ( x, y ) a n exp( n x ) cosn y (q) n 1 where a n An Dn . Finally, we apply boundary condition (4) to (q) To T a n cosn y (r) n 1 (v) Orthogonality. To determine a n in equation (r) we apply orthogonality. Note that the characteristic functions n ( y ) cosn y in (r) are solutions to equation (e). Comparing (e) with equation (3.5a) shows that it is a Sturm-Liouville equation with a1 ( y ) a2 ( y ) 0 and a 3 ( y ) 1 Thus eq. (3.6) gives p ( y ) w( y ) 1 and q( y ) 0 Since the boundary conditions at y = 0 and y = H are homogeneous, it follows that the characteristic functions n cosn y are orthogonal with respect to w(y) = 1. Multiplying both sides of (r) by w(y) cosm y dy and integrating from y = 0 to y = H, we obtain (To T ) H 0 w( y ) cosm ydy H a n cosn y w( y ) cosm ydy n1 0 3-50 PROBLEM 3.10 (continued) Interchanging the summation with integration, setting w(y) = 1 and applying orthogonality, the above gives (To T ) H cos n ydy a n 0 H cos2 n ydy 0 Evaluating the integrals and solving for a n 2(To T ) sin n H n H (sin n H ) cos n H (s) sin n H T ( x, y ) T 2 e n x cosn y To T n H (sin n H ) cosn H n 1 (t) an Substituting (s) into (r) (5) Checking. Dimensional check: (ii) The arguments of sin, cos and the exponential are dimensionless. (ii) Each term in solution (t) is dimensionless. Limiting check: If To T , no heat transfer takes place within the plate and consequently the entire plate should be at T . Setting To T in (t) gives T ( x, y ) T . Boundary conditions check: Conditions (1) and (3) can be readily shown to be satisfied by direct substitution into solution (q). (6) Comments. (i) By introducing the definition T T it was possible to transform three of the four boundary conditions from non-homogeneous to homogeneous. (ii) The solution is in terms of the Biot number hH/k which appears in the characteristic equation (p). This parameter appears in problems with surface convection. (iii) In the fin model, temperature variation in the y-direction is neglected and a one-dimensional solution is obtained. This approximation is valid for small Biot number compared to unity. That is for Bi hH / k 1 (u) Thus the two-dimensional solution (t) should reduce to the one-dimensional fin solution for small Biot number. Numerical solution to the characteristic equation (p) shows that for Bi 1 the first root is 1 H 1 . Thus for Bi 1 1 H 1 , sin 1L 1L, cos1H cos1 y 1 (v) When this is substituted into (p) we obtain 1 H hH / k (w) Substituting (v) and (w) into (t) and taking the first term of the series gives T1 ( x) T (To T ) e hk / H x (x) This result is identical to the one-dimensional temperature solution for a semi-infinite fin with surface convection and a specified temperature at the base. 3-51 PROBLEM 3.11 A very long plate of thickness 2H leaves a furnace at temperature T f . The plate is cooled as it moves with velocity U through a liquid tank. Because of the high heat transfer coefficient, the surface of the plate is maintained at temperature To . Determine the steady state two-dimensional temperature distribution in the plate. y 2 HBC To H U 0 x H Tf To (1) Observations. (i) This is a two-dimensional steady state conduction problem. (ii) Temperature distribution is symmetrical about the x-axis. (iii) At x the plate reaches a uniform temperature To . (iv) All four boundary conditions are non-homogenous. However, by defining a new variable, T To , three of the four boundary conditions become homogeneous. (v) Temperature distribution is symmetrical about the x-axis. Thus no heat crosses the boundary (x,0). (2) Origin and Coordinates. The origin and coordinate axes are selected as shown. (3) Formulation. (i) Assumptions. (1) Two-dimensional, (2) steady, (3) constant properties, (4) uniform velocity and (5) no energy generation. (ii) Governing Equations. Introducing the definition ( x, y ) T ( x, y ) To , eq. (1.7) gives 2 2 2 0 (a) 2 2 x x y where is defined as c pU (b) 2k (iii) Independent Variable with Two Homogeneous Boundary Conditions. direction has two homogeneous conditions in the variable. The y- (iv) Boundary Conditions. Because of symmetry we consider half the plate. The boundary conditions for the upper half are (1) ( x,0) 0 , homogeneous y (2) ( x, H ) 0 , homogeneous (3) ( , y ) = 0, homogeneous (4) (0, y) T f To , non-homogeneous (4) Solution. (i) Assumed Product Solution. 3-52 PROBLEM 3.11 (continued) Assume a product solution X ( x)Y ( y ). Substituting into (a), separating variables and setting the resulting equation equal to constant gives d2Xn d Xn 2 2n X n 0 2 dx dx (c) d 2Yn 2nYn 0 2 dy (d) and (ii) Selecting the Sign of the 2n Terms. Since the y-variable has two homogeneous conditions, the correct signs in (c) and (d) are d2Xn d Xn 2 2n X n 0 2 d x dx (e) d 2Yn 2nYn 0 2 dy (f) and For n 0, equations (e) and (f) become d 2X0 dX 0 2 0 2 dx dx (g) and d 2 Y0 d y2 0 (h) (iii) Solutions to the Ordinary Differential Equations. The solutions to equations (e)(h) are X n ( x ) [ An exp( x 2 2n x ) Bn exp( x 2 2n x )] (i) Yn ( y ) Cn sin n y Dn cosn y (j) X 0 ( x ) A0 exp( 2 x ) B0 (k) and Y0 ( y ) C0 y D0 The complete solution becomes (l) ( x, y ) X 0 ( x )Y0 ( y ) X n ( x )Yn ( y ) n 1 (iv) Application of Boundary Conditions. Condition (1) applied to (j) and (l) gives C n C0 0 Boundary condition (2) applied to (j) gives the characteristic equation 3-53 (m) PROBLEM 3.11 (continued) cosn H 0 This gives 2n 1 2 n H (n = 1, 2, 3….) (n) Boundary condition (2) applied to (l) gives D0 0 Condition (3) gives An 0 With C0 D0 0, it follows that X 0Y0 0. Thus equation (m) becomes ( x, y ) a n exp( 2 2n ) x cos n y (o) n 1 where a n Bn Dn . Finally, we apply boundary condition (4) to determine a n T f To a n cos n y (p) n 1 (v) Orthogonality. Note that the characteristic functions n cos n y are solutions to equation (f). Comparing (f) with equation (3.5a) shows that it is a Sturm-Liouville equation with a1 ( y ) a2 ( y ) 0 and a 3 ( y ) 1 Thus eq. (3.6) gives p ( y ) w( y ) 1 and q( y ) 0 Since the boundary conditions at y = 0 and y = H are homogeneous, it follows that the characteristic functions n cos n y are orthogonal with respect to w(y) = 1. Multiplying both sides of (p) by w( y ) cos m ydy and integrating from y = 0 to y = H gives (T f To ) H w( y ) cos m ydy 0 H a n cos n y w( y ) cos m ydy n1 0 Interchanging summation with integration, noting that w(y) = 1 and invoking orthogonality, the above becomes (T f To ) H cos n ydy an 0 H cos2 n ydy 0 Evaluating the integrals and solving for a n a n 2(T f T ) Substituting into (o) 3-54 sin n H n H (q) PROBLEM 3.11 (continued) T ( x, y ) To sin n H 2 exp( 2 2n ) x cos n y T f To n H n 1 (r) (5) Checking. Dimensional check: (i) The arguments of sin and cos are dimensionless. (ii) Each term in solution (r) is dimensionless. (iii) Solution (r) requires that units of be the same as that of n . According to (d), n is measured in (1/m). From the definition of in (b) we have ( kg / m 3 )c p ( J/kg o C)U ( m/s) k (W/m-o C) (1 / m) Limiting check: If T f To , no heat transfer takes place in the plate and consequently the temperature remains uniform. Setting T f To in (r) gives T ( x, y ) To . Boundary conditions check: conditions (1), (2) and (3) are readily shown to be satisfied by direct substitution in (o). (6) Comments. (i) For a stationary plate ( U 0 ), the solution to this special case is obtained by setting 0 in (r) T ( x, y ) To sin n H 2 exp ( n x) cos n y T f To n H n 1 (ii) The solution depends on two parameters. The first parameter is the Biot number hH / k which appears in the characteristic equation for n . The second is H c pUH / k which appears in the exponent of the exponential in equation (r) when the exponent is multiplied and divided by H. 3-55 PROBLEM 3.12 Liquid metal flows through a cylindrical channel of width H and inner radius R. It enters at Ti and is cooled by convection along the outer surface to To . The ambient temperature is T and heat transfer coefficient is h.. The inner surface is insulated. Assume that the liquid metal flows with a uniform velocity U and neglect curvature effect (H/R << 1), determine the steady state temperature of the insulated surface. To Ti 0 x H y R U (1) Observations. (i) This is a two-dimensional steady state conduction problem. (ii) To determine the temperature of the insulated surface it is necessary to determine the temperature h, T distribution in the moving liquid metal. (iii) The velocity is uniform and inlet and outlet temperatures are specified. (iv) Three boundary conditions are non-homogenous. However, by defining a new variable, T T , the y-direction will have two homogeneous conditions. (2) Origin and Coordinates. The origin and coordinate axes are selected as shown. y (3) Formulation. h, T (i) Assumptions. (1) Steady T H U i x state, (2) Two-dimensional, (3) 0 constant properties, (4) uniform L 2 R velocity, (5) negligible curvature effect (H/R<<1) and (6) no energy generation. To (ii) Governing Equations. Introducing the definition ( x, y ) T ( x, y ) T , eq. (1.7) gives 2 2 2 0 (a) 2 2 x x y where is defined as c pU (b) 2k (iii) Independent Variable with Two Homogeneous Boundary Conditions. direction has two homogeneous conditions in the variable. (iv) Boundary Conditions. The boundary conditions for the upper half are ( x,0) 0 , homogeneous y ( x, H ) h ( x, H ) , homogeneous (2) k y (1) (3) (0, y ) Ti T , non-homogeneous (4) ( L, y ) To T , non-homogeneous where L is the length of channel given by 3-56 The y- PROBLEM 3.12 (continued) L 2R (c) (4) Solution. (i) Assumed Product Solution. Assume a product solution X ( x )Y ( y ). Substituting into (a), separating variables and setting the resulting equation equal to constant gives d2Xn d Xn 2 2n X n 0 2 dx dx (d) d 2Yn 2nYn 0 d y2 (e) and (ii) Selecting the Sign of the 2n Terms. Since the y-variable has two homogeneous conditions, the correct signs in (d) and (e) are d2Xn d Xn 2 2n X n 0 2 dx dx (f) d 2Yn 2nYn 0 2 dy (g) and For n 0, equations (f) and (g) become d 2X0 dX 0 2 0 2 dx dx (h) and d 2 Y0 d y2 0 (i) (iii) Solutions to the Ordinary Differential Equations. The solutions to equations (f)(i) are X n ( x) An exp( x 2 2n x) Bn exp( x 2 2n x) (j) Yn ( y ) Cn sin n y Dn cosn y (k) X 0 ( x ) A0 exp( 2 x ) B0 (l) Y0 ( y ) C0 y D0 (m) and The complete solution is ( x, y ) X 0 ( x )Y0 ( y ) X n ( x )Yn ( y ) n 1 (iv) Application of Boundary Conditions. Condition (1) applied to (k) and (m) gives 3-57 (n) PROBLEM 3.12 (continued) C n C0 0 Boundary condition (2) applied to (k) gives n H tan n H hH / k Bi (o) where Bi is the Biot number. This is the characteristic equation whose roots give n . Boundary condition (2) applied to (m) gives D0 0 With C0 D0 0, it follows that X 0Y0 0. Thus equation (n) becomes ( x, y ) ( an e n 1 2 2n ) x bn e ( 2 2n ) x cosn y (p) where an An Dn and bn Bn Dn . Application of condition (3) and (4) to (p) gives Ti T (a n bn ) cosn y (q) n 1 and To T a e ( 2 2n ) L n bn e ( 2 2n ) L cosn y n 1 (r) (v) Orthogonality. Note that the characteristic functions n cosn y in (q) and (r) are solutions to equation (g). Comparing (g) with equation (3.5a) shows that it is a SturmLiouville equation with a1 ( y ) a2 ( y ) 0 and a 3 ( y ) 1 Thus eq. (3.6) gives p( y ) w( y ) 1 and q( y ) 0 Since the boundary conditions at y = 0 and y = H are homogeneous, it follows that the characteristic functions n sin n y are orthogonal with respect to w(y) = 1. Multiplying both sides of (q) by w( y ) sin m ydy and integrating from y = 0 to y = H gives (Ti T ) H 0 w( y ) cosm ydy H (an bn ) cosn y w( y ) cosm ydy n 1 0 Interchanging summation with integration, noting that w(y) = 1 and invoking orthogonality, the above becomes (Ti T ) H 0 cosn ydy ( an bn ) Evaluating the integral and solving for (an bn ) 3-58 H 0 cos2 n ydy PROBLEM 3.12 (continued) 2(Ti T ) sin n H n H (sin n H ) cosn H ( an bn ) (s) Similarly, multiplying both sides of (r) by w( y ) sin m ydy and integrating from y = 0 to y = H gives (To T ) H 0 w( y ) cosm ydy H ( a n e n 1 0 2 2n ) L bn e ( 2 2n ) L 2 w( y ) cos m ydy (t) Interchanging summation with integration, noting that w(y) = 1 and invoking orthogonality, the above becomes (To T ) H 0 ( cos n ydy an e 2 2n ) L bn e ( 2 2n ) L H 0 cos2 n ydy Evaluating the integrals and rearranging an e ( 2 2n ) L bn e ( 2 2n ) L 2(To T ) sin n H n H (sin n H ) cos n H (u) Equation (s) and (u) are solved for the coefficients a n and bn (To T ) /(Ti T ) e ( n ) L an 2 sin n H ( 2 2n ) L ( 2 2n ) L (Ti T ) n H (sin n H ) cosn H e e 2 2 (v) and 2 2 (To T ) /(Ti T ) e ( n ) L bn 2 sin n H 1 ( 2 2n ) L ( 2 2n ) L Ti T n H (sin n H ) cosn H e e (w) Rewriting solution (r) in dimensionless form gives T ( x, y ) T Ti T an T T n 1 e ( 2 2n ) x i bn ( e Ti T 2 2n ) x cosn y (x) where a n /(Ti T ) and bn /(Ti T ) are given in (v) and (w). The temperature of the insulated surface is obtained by evaluating (x) at y = 0 T ( x,0) T Ti T T T n 1 an i e ( 2 2n ) x bn ( e Ti T 2 2n ) x (y) (5) Checking. Dimensional check: (i) The arguments of the sin, cos and exponentials are dimensionless. (ii) Solution (x) requires that units of be the same as that of n . According to (g), n is measured in (1/m). From the definition of in (b) we have ( kg / m 3 )c p ( J/kg o C)U ( m/s) k (W/m-o C) (1 / m) 3-59 PROBLEM 3.12 (continued) Limiting check: If Ti To T , no heat transfer takes place and consequently the temperature in the entire region remains uniform. Setting Ti To T in (v) and (w) gives an bn 0. When this is substituted into (p) gives T ( x, y ) T . Boundary conditions check: Solution (p) can be readily shown to satisfy condition (1). (6) Comments. (i) For a stationary material ( U 0 ), the solution to this special case is obtained by setting 0 in (p) ( x, y ) a e n n x bn e n x cosn y n 1 where a n and bn are obtained by setting 0 in (v) and (w). (ii) Examination of equations (o), (v), (w) and (x) shows that the solution depends on four parameters. The first parameter is the Biot number hH / k which appears in the characteristic equation for n . The second is (To T ) /(Ti T ) shown in (v) and (w). The third parameter is H c pUH / k and the fourth is L/H which appears in n L n H ( L / H ) . 3-60 PROBLEM 3.13 A solid cylinder of radius ro and length L is maintained at T1 at one end. The other end is maintained at T2 and the cylindrical surface at T3 . Determine the steady state two-dimensional temperature distribution in the cylinder. T 3 r 0 T r o z L T 2 1 (1) Observations. (i) This is a steady state twodimensional conduction problem. (ii) There are three non-homogeneous boundary conditions. Defining a new temperature variable, T T3 , makes the boundary condition at the cylindrical surface homogeneous. This gives the radial direction two homogeneous conditions. (iii) A cylindrical coordinate system should be used. (2) Origin and Coordinates. The origin and coordinate axes are selected as shown. (3) Formulation. (i) Assumptions. (1) Two-dimensional, (2) steady, (3) constant conductivity and (4) no energy generation. (ii) Governing Equations. Defining T T3 and applying the above assumptions to eq. (1.11) gives 2 1 2 0 (a) r2 r r z2 (iii) Independent Variable with Two Homogeneous Boundary Conditions. The rvariable has two homogeneous conditions. (iv) Boundary Conditions. The boundary conditions on ( r , z ) are (1) (0, z ) 0, or (0, z ) finite, homogeneous r (2) ( ro , z ) = 0, homogeneous (3) (r ,0) (T1 T3 ), non-homogeneous (4) (r , L) (T2 T3 ), non-homogeneous (4) Solution. (i) Assumed Product Solution. To solve equation (a) we assume a product solution. Let ( r, z ) R( r ) Z ( z ) (b) Substituting (b) into (a), separating variables and setting the resulting two equations equal to a constant, 2k , we obtain d 2Zk 2k Z k 0 2 dz and 3-61 (c) PROBLEM 3.13 (continued) r2 d 2 Rk dR r k 2k r 2 Rk 0 2 dr dr (d) (ii) Selecting the Sign of the 2k Terms. Since the r-variable has two homogeneous boundary conditions, the plus sign is selected in (d). Thus (c) and (d) become d 2Zk 2k Z k 0 2 dz (e) and r2 d 2 Rk dR r k 2k r 2 Rk 0 2 dr dr (f) For the special case of k 0 the above equations take the form d 2Z0 0 d z2 (g) and r d 2 R0 dr2 dR0 0 dr (h) (iii) Solutions to the Ordinary Differential Equations. The solution to (e) is Z k ( z ) Ak sinh k z Bk coshk z (i) Equation (f) is a Bessel differential equation. Comparing (f) with eq. (2.26) shows that A B n 0 , C = 1 and D k . Thus the solution to (f) is given by eq. (2.27) Rk (r ) C k J 0 ( k r ) Dk Y0 ( k r ) (j) The solutions to (g) and (h) are Z 0 ( z ) A0 z B0 (k) R0 ( r ) C0 ln r D0 (l) and The complete solution to ( r , z ) becomes ( r, z ) R0 ( r ) Z 0 ( z ) Rk ( r ) Z k ( z ) (m) k 1 (iv) Application of Boundary Conditions. Applying boundary condition (1) to equation (j) and (l) gives Dk C0 0 Condition (2) applied to (j) gives the characteristic equation for k J 0 (k ro ) 0 3-62 (n) PROBLEM 3.13 (continued) Condition (2) applied to (l) gives D0 0 With C0 D0 0 the solution R0 Z 0 vanishes and the temperature solution (m) becomes ( r, z ) ak sinh k z bk coshk z J 0 ( k r ) (o) k 1 where ak Ak Ck and bk Bk Ck . Application of boundary conditions (3) and (4) to (o) gives T1 T3 bk J 0 (k r ) k 1 (p) and T2 T3 a k sinh k L bk coshk L J 0 ( k r ) (q) k 1 (v) Orthogonality. Note that the characteristic functions J 0 (k r ) in (p) and (q) are solutions to equation (f). Comparing (f) with eq. (3.5a) shows that it is a Sturm-Liouville equation with a1 (r ) 1/ r, a2 (r) 0 and a3 ( r ) 1 Thus eq. (3.6) gives p( r ) w( r ) r and q( r ) 0 Since the boundary conditions at r = 0 and r ro are homogeneous, it follows that the characteristic functions J 0 (k r ) are orthogonal with respect to w(r) = r. Multiplying both sides of (p) by J 0 ( i r ) w( r )dr and integrating from r = 0 to r ro (T1 T3 ) ro 0 J 0 (i r ) w( r )dr ro bk J 0 (k r )J 0 (i r ) w( r )dr k 1 0 Interchanging summation with integration, noting that w(r) = r and invoking orthogonality, eq. (3.7), the above gives (T1 T3 ) ro 0 J 0 (k r )rdr bk ro 0 J 02 (k r )rdr Evaluating the integrals using Appendix B and Table 3.1 gives bk 2(T1 T3 ) k ro J 1 ( k ro ) (r) Similarly, multiplying both sides of (q) by J 0 (i r ) w( r )dr and integrating from r = 0 to r ro (T2 T3 ) ro 0 J 0 (i r ) w( r )dr ro ak sinh k L bk coshk LJ 0 (k r )J 0 (i r ) w( r )dr k 1 0 3-63 PROBLEM 3.13 (continued) Interchanging summation with integration, noting that w(r) = r and invoking orthogonality, eq. (3.7), the above gives (T2 T3 ) ro 0 J 0 (k r ) w( r )dr (ak sinh k L bk coshk L) ro 0 J 02 (k r )rdr Evaluating the integrals using Appendix B and Table 3.1 2(T1 T3 ) k ro J 1 ( k ro ) (s) (T2 T3 ) (T1 T3 ) coshk L (t) ( ak sinh k L bk cosh k L) Solving for a k and using (r) to eliminate bk ak 2 (sinh k L)k ro J 1 ( k ro ) Substituting (r) and (t) into (o) gives the temperature solution in the cylinder T (r , z ) T3 1 2 (T1 T3 ) r k 1 k o sinh k z (T2 T3 ) J ( r ) cosh k L cosh k z 0 k sinh k L (T1 T3 ) J 1 ( k ro ) (u) (5) Checking. Dimensional checks: (i) The arguments of sinh, cosh, J 0 and J 1 are dimensionless. (ii) Each term in (u) is dimensionless. Limiting check: If T1 T2 T3 , no heat transfer can take place and the entire cylinder should be at uniform temperature. Setting T1 T2 T3 in (u) gives T ( r, z ) T . Boundary conditions check: Setting r = 0 and r ro in solution (u) shows that boundary conditions (1) and (2) are satisfied. (6) Comments. (i) An alternate approach is to assume a solution of the form T ( r, z ) ( r, z ) ( z ) The function (z ) can be assigned two non-homogeneous boundary conditions in the zdirection leaving ( r, z ) with two homogeneous conditions in the z-direction and one in the r-direction. Another approach is to apply the method of superposition by decomposing the problem into two simpler problems each having one non-homogeneous boundary condition. (ii) The solution depends on two dimensionless parameter: (T2 T3 ) /(T1 T3 ) and L / ro . 3-64 PROBLEM 3.14 A solid cylinder of radius ro and length L is maintained at T1 at one end and is cooled with uniform flux q o at the other end. The cylindrical surface is maintained at a uniform temperature T2 . Determine the steady state two-dimensional temperature distribution in the cylinder. r 0 T1 T2 2 HBC ro z qo L (1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) There are three non-homogeneous boundary conditions. Defining a new temperature variable, T T2 makes the boundary condition at the cylindrical surface homogeneous. This gives the radial direction two homogeneous conditions. (iii) Use a cylindrical coordinate system. (2) Origin and Coordinates. The origin and coordinate axes are selected as shown. (3) Formulation. (i) Assumptions. (1) Two-dimensional, (2) steady, (3) constant conductivity and (4) no energy generation. (ii) Governing Equations. Defining T T2 and applying the above assumptions to eq. (1.11) gives 2 1 2 0 (a) r2 r r z2 (iii) Independent Variable with Two Homogeneous Boundary Conditions. The rvariable has two homogeneous conditions. (iv) Boundary Conditions. The boundary conditions on ( r , z ) are (1) (0, z ) 0, or (0, z ) finite, r (2) ( ro , z ) = 0, homogeneous homogeneous (3) (r,0) (T1 T2 ), non-homogeneous (4) k (r , L) q o z non-homogeneous (4) Solution. (i) Assumed Product Solution. To solve equation (a) we assume a product solution. Let ( r, z ) R( r ) Z ( z ) (b) Substituting (b) into (a), separating variables and setting the resulting two equations equal to a constant, 2k , we obtain d 2Zk 2k Z k 0 2 dz 3-65 (c) PROBLEM 3.14 (continued) and r2 d 2 Rk dR r k 2k r 2 Rk 0 2 dr dr (d) (ii) Selecting the Sign of the 2k Terms. Since the r-variable has two homogeneous boundary conditions, the plus sign is selected in (d). Thus (c) and (d) become d 2Zk 2k Z k 0 2 dz (e) and r2 d 2 Rk dR r k 2k r 2 Rk 0 2 dr dr (f) For the special case of k 0 the above equations take the form d 2Z0 0 d z2 (g) and r d 2 R0 dr2 dR0 0 dr (h) (iii) Solutions to the Ordinary Differential Equations. The solution to (e) is Z k ( z ) Ak sinh k z Bk coshk z (i) Equation (f) is a Bessel differential equation. Comparing (f) with eq. (2.26) shows that A B n 0 , C = 1 and D k . Thus the solution to (f) is given by eq. (2.27) Rk (r ) C k J 0 ( k r ) Dk Y0 ( k r ) (j) The solutions to (g) and (h) are Z 0 ( z ) A0 z B0 (k) R0 ( r ) C0 ln r D0 (l) and The complete solution to ( r , z ) becomes ( r, z ) R0 ( r ) Z 0 ( z ) Rk ( r ) Z k ( z ) (m) k 1 (iv) Application of Boundary Conditions. Applying boundary condition (1) to equation (j) and (l) gives Dk C0 0 Condition (2) applied to (j) gives the characteristic equation for k J 0 ( k ro ) 0 3-66 (n) PROBLEM 3.14 (continued) Condition (2) applied to (l) gives D0 0 With C0 D0 0 the solution R0 Z 0 vanishes and the temperature solution (m) becomes ( r, z ) ak sinh k z bk coshk z J 0 ( k r ) (o) k 1 where ak Ak Ck and bk Bk Ck . Application of boundary conditions (3) and (4) to (o) gives b J 0 ( k r ) (p) coshk L bk sinh k L J 0 ( k r ) (q) T1 T2 k k 1 and qo k a k k k 1 (v) Orthogonality. Note that the characteristic functions J 0 (k r ) in (p) and (q) are solutions to equation (f). Comparing (f) with eq. (3.5a) shows that it is a Sturm-Liouville equation with a1 (r ) 1/ r, a2 (r) 0 and a3 ( r ) 1 Thus eq. (3.6) gives p( r ) w( r ) r and q( r ) 0 Since the boundary conditions at r = 0 and r ro are homogeneous, it follows that the characteristic functions J 0 (k r ) are orthogonal with respect to w(r) = r. Multiplying both sides of (p) by J 0 ( i r ) w( r )dr and integrating from r = 0 to r ro (T1 T2 ) ro 0 J 0 (i r ) w(r )dr ro bk J 0 ( k r )J 0 (i r ) w(r )dr k 1 0 Interchanging summation with integration, noting that w(r) = r and invoking orthogonality, eq. (3.7), the above gives (T1 T2 ) ro 0 J 0 ( k r )rdr bk ro 0 J 02 ( k r )rdr Evaluating the integrals using Appendix B and Table 3.1 the above gives bk bk 2(T1 T2 ) k ro J 1 ( k ro ) (r) Similarly, multiplying both sides of (q) by J 0 (i r ) w( r )dr and integrating from r = 0 to r ro qo ro 0 J 0 (i r ) w( r )dr k ro a 0 k k 1 k coshk L bk sinh k LJ 0 (k r )J 0 (i r ) w( r )dr 3-67 PROBLEM 3.14 (continued) Interchanging summation with integration, noting that w(r) = r and invoking orthogonality, eq. (3.7), the above gives ro qo J 0 (k r )rdr kk (ak cosh k L bk sinh k L) ro J 02 (k r )rdr 0 0 Evaluating the integrals using Appendix B and Table 3.1 the above gives (ak cosh k L bk sinh k L) 2 ( qoro / k ) (k ro ) 2 J 1 (k ro ) (s) Solving for a k and using (r) to eliminate bk ak ( qro / k ) 2 (T1 T2 ) sinh k L (cosh k L)( k ro ) J 1 ( k ro ) k ro (t) substituting (s) and (t) into (o) and rearranging the result T (r , z ) T2 T2 T1 k 1 sinh k z 2 (q ro / k ) J ( r ) sinh k L cosh k z 0 k ( k ro ) cosh k L k ro (T2 T1 ) J 1 ( k ro ) (u) (5) Checking. Dimensional check: (i) The arguments of sinh, cosh, J 0 and J 1 are dimensionless. (ii) Each term in (u) is dimensionless. Limiting check: If T1 T2 , and qo 0 no heat transfer can take place and the entire cylinder should be at uniform temperature. Setting T1 T2 in (u) gives T ( r , z ) T2 . Boundary conditions check: Setting r = 0 and r ro in solution (u) shows that boundary conditions (1) and (2) are satisfied. (6) Comments. (i) One of parameters which appears in the solution is qoro / k (T2 T1 ) . In addition, roots of (n) give k ro . However, the argument k L k ro ( L / ro ) introduces a third parameter which is ( L / ro ) . (ii) An alternate approach is to assume a solution of the form T ( r, z ) ( r, z ) ( z ) The function (z ) can be assigned two non-homogeneous boundary conditions in the zdirection leaving ( r, z ) with two homogeneous conditions in the z-direction and one in the r-direction. Another approach is to apply the method of superposition by decomposing the problem into two simpler problems each having one non-homogeneous boundary condition. 3-68 PROBLEM 3.15 The volumetric heat generation rate in a solid cylinder of radius ro and length L varies along the radius according to r q qo r To 0 2 HBC z q ro qo where qo is constant. One end is insulated while L the other end is cooled with uniform flux qo. The cylindrical surface is maintained at uniform temperature To . Determine the steady state temperature of the insulated surface. (1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) There are two non-homogeneous boundary conditions. (iii) Energy generation makes the heat equation non-homogeneous. (iv) A cylindrical coordinate system should be used. (2) Origin and Coordinates. The origin and coordinate axes are selected as shown. (3) Formulation. (i) Assumptions. (1) Two-dimensional, (2) steady and (3) constant conductivity. (ii) Governing Equations. Applying the above assumptions to eq. (1.11) gives 2T 1 T 2T q 0 r2 r r z2 k Energy generation q is given by q qo r where qo is constant. Substituting into the above heat equation gives 2T 1 T 2T qo r0 r2 r r z2 k (a) (iii) Independent Variable with Two Homogeneous Boundary Conditions. Neither variable has two homogeneous conditions. However, the problem can be decomposed into two problems one of which has two homogeneous conditions in the r-direction. (iv) Boundary Conditions. The boundary conditions are (1) T (0, z ) 0, or (0, z ) finite, homogeneous r (2) T ( ro , z ) To , non-homogeneous (3) T ( r,0) 0 , homogeneous z (5) k T (r , L) qo non-homogeneous z 3-69 PROBLEM 3.15 (continued) (4) Solution. Equation (a) is non-homogeneous due to the heat generation term. Thus we can not proceed directly with the application of the separation of variables method. Instead, we assume a solution of the form T ( r, z ) ( r, z ) ( r ) (b) Note that ( r, z ) depends on two variables while (r ) depends on a single variable r. the function (r ) is introduced to satisfy the non-homogeneous part of equation (a) which also depends on the variable r. Substituting (b) into (a) 2 1 2 d 2 1 d qo r0 r2 r r z2 d r2 r d r k (c) The next step is to split (c) into two equations, one for ( r, z ) and one for (r ) . We let 2 1 2 0 r2 r r z2 (d) d 2 1 d qo r0 d r2 r d r k (e) Thus Boundary conditions on (d) and (e) are obtained by substituting (b) into the four boundary conditions on T. In this process the function ( r, z ) is assigned homogeneous conditions whenever possible. Using (b) and condition (1) (0, z ) d (0) 0 r dr Let (0, z ) 0 r (d-1) d (0) 0 dr (e-1) Thus Boundary condition (2) and (b) give Let ( ro , z ) ( ro ) To ( ro , z ) 0 (d-2) ( ro ) To (e-2) ( r,0) 0 z (d-3) (r , L) qo z (d-4) Thus Boundary condition (3) and (b) give Boundary condition (4) and (b) give k 3-70 PROBLEM 3.15 (continued) Thus non-homogeneous equation (a) is replaced by two equations: (d) which is a homogeneous partial differential equation and (e) which is a non-homogeneous ordinary differential equation. Equation (d) has three homogeneous conditions, (d-1), (d-2) and (d-3) and one non-homogeneous condition (d-4). The solution to (e) is (r) qo 3 r E ln r F 9k (f) where E and F are constants of integration. Boundary conditions (e-1) and (e-2) give E and F. The above solution becomes (r) qo 3 3 ( ro r ) To 9k (g) (i) Assumed Product Solution. To solve equation (d) we assume a product solution. Let ( r, z ) R( r ) Z ( z ) (h) Substituting (h) into (d), separating variables and setting the resulting two equations equal to a constant, 2k , we obtain d 2Zk 2k Z k 0 2 dz (i) and r2 d 2 Rk dR r k 2k r 2 Rk 0 2 dr dr (j) (ii) Selecting the Sign of the 2k Terms. Since the r-variable has two homogeneous boundary conditions, the plus sign is selected in (j). Thus (i) and (j) become d 2Zk 2k Z k 0 d z2 (k) and r2 d 2 Rk dR r k 2k r 2 Rk 0 2 dr dr (l) For the special case of k 0 the above equations take the form d 2Z0 0 d z2 (m) and r d 2 R0 dr 2 dR0 0 dr (iii) Solutions to the Ordinary Differential Equations. The solution to (k) is 3-71 (n) Z k ( z ) Ak sinh k z Bk coshk z (o) PROBLEM 3.15 (continued) Equation (l) is a Bessel differential equation. Comparing (l) with eq. (2.26) shows that A B n 0 , C = 1 and D k . Thus the solution to (l) is given by eq. (2.27) Rk (r ) Ck J 0 (k r ) Dk Y0 (k r ) (p) The solutions to (m) and (n) are Z 0 ( z ) A0 z B0 (q) R0 ( r ) C0 ln r D0 (r) and The complete solution to (r , z ) becomes (r , z ) R0 (r ) Z 0 ( z ) R (r ) Z ( z ) k k (s) k 1 (iv) Application of Boundary Conditions. Applying boundary condition (d-1) to equation (p) and (r) gives Dk C0 0 Condition (2) applied to (p) gives the characteristic equation for k J 0 ( k ro ) 0 (t) Condition (2) applied to (r) gives D0 0 With C0 D0 0 the solution R0 Z 0 vanishes. Boundary condition (d-3) gives Ak 0 Substituting into (s) gives ( r, z ) (b cosh z) J k k 0 ( k r ) (u) k 1 where bk Bk Ck . Application of boundary conditions (4) to (u) gives qo k b (sinh L) J k k 0 ( k r ) (v) k 1 (v) Orthogonality. Note that the characteristic functions J 0 (k r ) in (v) are solutions to equation (l). Comparing (l) with eq. (3.5a) shows that it is a Sturm-Liouville equation with a1 (r ) 1/ r, a2 (r) 0 and a3 ( r ) 1 Thus eq. (3.6) gives p( r ) w( r ) r and q( r ) 0 3-72 Since the boundary conditions at r = 0 and r ro are homogeneous, it follows that the characteristic functions J 0 (k r ) are orthogonal with respect to w(r) = r. Multiplying both sides of (v) by J 0 ( i r ) w( r )dr and integrating from r = 0 to r ro PROBLEM 3.15 (continued) qo ro 0 J 0 (i r ) w( r )dr k ro bk (sinh k L) J 0 (k r )J 0 (i r ) w( r )dr k 1 0 Interchanging summation with integration, noting that w(r) = r and invoking orthogonality, eq. (3.7), the above gives qo ro 0 J 0 ( k r ) rdr k bk (sinh k L) ro 0 J 02 ( k r ) rdr Evaluating the integrals using Appendix B and Table 3.1 the above gives bk bk 2( qoro / k ) ( k ro ) (sinh k L) J 1 (k ro ) (w) 2 Substituting (w) into (u) gives the solution ( r, z ) ( 1r ) (r , z ) 2(qo ro / k ) k 1 k o 2 coshk z J 0 (k r ) sinh k L J1 (k ro ) (x) The complete solution to the temperature distribution in the cylinder is obtained by substituting (g) and (x) into (b) and rearranging the result 2qo T (r , z ) To 1 1 (r / ro ) 3 3 9 qo ro qoro2 k ( r ) k 1 1 k o 2 cosh k z J 0 ( k r ) sinh k L J 1 ( k ro ) (y) (5) Checking. Dimensional checks: (i) The arguments of sinh, cosh, J 0 and J 1 are dimensionless. (ii) Each term in (y) is dimensionless. Limiting check: If T1 T2 and qo qo 0 no heat transfer can take place and the entire cylinder should be at uniform a temperature To . Setting qo qo 0 in (y) gives T ( r , z ) To . Boundary conditions check: Setting r = 0 and r ro in solution (y) shows that boundary conditions (1) and (2) are satisfied. (6) Comments. (i) The key to solving this problem is to recognize that the nonhomogeneous term in differential equation (a) is a function of r. This suggests that the function in the assumed solution (b) should also be a function of r. (ii) Two parameters q L characterize the solution: o 2 and . ro q oro 3-73 PROBLEM 3.16 Heat is added at a uniform flux q o over a circular area of radius b at one end of a solid cylinder of radius ro and length L. The remaining surface of the heated end and the opposite surface are insulated. The cylindrical surface is maintained at uniform temperature To . Determine the steady state two-dimensional temperature distribution. r To 2 HBC ro 0 z qo L (1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) There are two non-homogeneous boundary conditions. Defining a new temperature variable, T To , makes the boundary condition at the cylindrical surface homogeneous. This gives the radial direction two homogeneous conditions. (iii) Part of boundary (r,L) is heated while the rest is insulated. Thus the heat flux along this boundary is non-uniform. (iv) A cylindrical coordinate system should be used. (2) Origin and Coordinates. The origin and coordinate axes are selected as shown. (3) Formulation. (i) Assumptions. (1) Two-dimensional, (2) steady, (3) constant conductivity and (4) no energy generation. (ii) Governing Equations. Defining T To and applying the above assumptions to eq. (1.11) gives 2 1 2 0 (a) r2 r r z2 (iii) Independent Variable with Two Homogeneous Boundary Conditions. The rvariable has two homogeneous conditions. (iv) Boundary Conditions. The boundary conditions on ( r , z ) are (1) (0, z ) 0, or (0, z ) finite, homogeneous r (2) ( ro , z ) = 0, homogeneous (3) ( r,0) 0, homogeneous z (4) k qo , ( r, L) f (r) z 0 0rb b r ro non-homogeneous This represents a specified heat flux which varies with r according to f(r) which is defined in condition (4). (4) Solution. 3-74 (i) Assumed Product Solution. To solve equation (a) we assume a product solution. Let ( r, z ) R( r ) Z ( z ) PROBLEM 3.16 (continued) (b) Substituting (b) into (a), separating variables and setting the resulting two equations equal to a constant, 2k , we obtain d 2Zk 2k Z k 0 d z2 (c) and d 2 Rk dR r r k 2k r 2 Rk 0 2 dr dr 2 (d) (ii) Selecting the Sign of the 2k Terms. Since the r-variable has two homogeneous boundary conditions, the plus sign is selected in (d). Thus (c) and (d) become d 2Zk 2k Z k 0 d z2 (e) and r2 d 2 Rk dR r k 2k r 2 Rk 0 2 dr dr (f) For the special case of k 0 the above equations take the form d 2Z0 0 d z2 (g) and r d 2 R0 dr 2 dR0 0 dr (h) (iii) Solutions to the Ordinary Differential Equations. The solution to (e) is Z k ( z ) Ak sinh k z Bk coshk z (i) Equation (f) is a Bessel differential equation. Comparing (f) with eq. (2.26) shows that A B n 0 , C = 1 and D k . Thus the solution to (f) is given by eq. (2.27) Rk (r ) C k J 0 ( k r ) Dk Y0 ( k r ) (j) The solutions to (g) and (h) are Z 0 ( z ) A0 z B0 (k) R0 ( r ) C0 ln r D0 (l) and The complete solution becomes 3-75 ( r , z ) R0 ( r ) Z 0 ( z ) R (r ) Z k k ( z) (m) k 1 PROBLEM 3.16 (continued) (iv) Application of Boundary Conditions. Applying boundary condition (1) to equation (j) and (l) gives Dk C0 0 Condition (2) applied to (j) gives the characteristic equation for k J 0 ( k ro ) 0 (n) Condition (2) applied to (l) gives D0 0 With C0 D0 0 the solution R0 Z 0 vanishes. Boundary condition (3) gives Ak 0 Substituting into (m) gives ( r, z ) (b cosh z) J k k 0 ( k r ) (o) k 1 where bk Bk Ck . Application of boundary conditions (4) to (u) gives f (r) k b (sinh L) J k k k 0 ( k r ) (p) k 1 where f(r) is defined in boundary condition (4). (v) Orthogonality. To determine bk in (p) we apply orthogonality. Note that the characteristic functions J 0 (k r ) in (p) are solutions to equation (f). Comparing (f) with eq. (3.5a) shows that it is a Sturm-Liouville equation with a1 (r ) 1/ r, a2 (r) 0 and a3 ( r ) 1 Thus eq. (3.6) gives p( r ) w( r ) r and q( r ) 0 Since the boundary conditions at r = 0 and r ro are homogeneous, it follows that the characteristic functions J 0 (k r ) are orthogonal with respect to w(r) = r. Multiplying both sides of (p) by J 0 ( i r ) w( r )dr and integrating from r = 0 to r ro ro 0 f ( r ) J 0 (i r ) w( r )dr k ro k bk (sinh k L) J 0 (k r )J 0 (i r ) w( r )dr k 1 0 Interchanging summation with integration, noting that w(r) = r and invoking orthogonality, eq. (3.7), the above gives 3-76 ro 0 f ( r ) J 0 ( k r ) rdr k k bk (sinh k L) ro 0 J 02 ( k r ) rdr Using the definition of f(r), the above becomes PROBLEM 3.16 (continued) qo b 0 J 0 ( k r ) rdr ro b (0) J 0 ( k r ) rdr k k bk (sinh k L) ro 0 J 02 ( k r ) rdr Noting that the second integral vanishes, the above gives qo b 0 J 0 ( k r ) rdr k k bk (sinh k L) ro 0 J 02 ( k r ) rdr Evaluating the integrals using Appendix B and Table 3.1 the above gives bk bk 2( qoro / k ) J 1 ( k b) 2 ( k ro ) (sinh k L) J 12 (k ro ) (q) Substituting (q) into (o) and rearranging gives the solution to the temperature distribution in the cylinder T (r , z ) To J 1 ( k b ) 2 (cosh k z ) J 0 ( k r ) 2 qo ro k 1 ( k ro ) (sinh k L) J 1 ( k ro ) k (r) (5) Checking. Dimensional checks: (i) The arguments of sinh, cosh, J 0 and J 1 are dimensionless. (ii) Each term in (r) is dimensionless. Limiting check: If qo 0 , no heat transfer can take place and the entire cylinder should be at uniform temperature To . Setting qo 0 in (r) gives T ( r , z ) To . Boundary conditions check: Setting r = 0 and r ro in solution (r) shows that boundary conditions (1) and (2) are satisfied. (6) Comments. Equation (r) shows that the temperature solution depends on two dimensionless parameters L / ro and b / ro . 3-77 3-78