Chapter two 2.1 Study state conduction in one dimension: 2.1.1 The Plane Wall a. temperature destribution FIGURE 2.1 Heat transfer through a plane wall. (a) Temperature distribution. (b) Equivalent thermal circuit. Assumption: Temperature varies with x-direction only no heat generation. So, the general equation is reduces to: π2 π ππ₯ 2 =0 First integration: ππ ππ₯ = πΆ1 Second integration: π(π₯) = πΆ1 π₯ + πΆ2 2.1 this is the general solution πΆ1 and πΆ2 are constant of integration. Baundary condition: Substitute in general solution: 2.2 the heat transfer rate: 2.3 b. Thermal resistance The thermal resistance for conduction is: 2.4 The thermal resistance for convection is: 2.5 Circuit representations provide a useful tool for both conceptualizing and quantifying heat transfer problems. The equivalent thermal circuit for the plane wall with convection surface conditions is shown in Figure 2.1b. The heat transfer rate may be determined from separate consideration of each element in the network. Since qx is constant throughout the network, it follows that 2.1.2 The Composite Wall: a. Material in series: Equivalent thermal circuits may also be used for more complex systems, such as composite walls. Such walls may involve any number of series and parallel thermalresistances due to layers of different materials. Consider the series composite wall of Figure 2.2. The one-dimensional heat transfer rate for this system may be expressed as FIGURE 2.2 Equivalent thermal circuit for a series composite wall. *Overall heat transfer coefficient With composite systems it is often convenient to work with an overall heat transfer coefficient, U, which is defined by an expression analogous to Newton’s law of cooling. Accordingly, Or ππ = πΌπ¨(π»∞π − π»∞π ) That means: b. Material in parallel: ππ₯ = π1 + π2 π1 = π1 −π2 βπ₯ π1 π΄1 π1 π 1 2 βπ₯ π2 π2 = π1 −π2 βπ₯ π2 π΄2 ππ₯ = (π1 − π2 ) οΏ½ 1 π π 1 1 =π +π 1 1 βπ₯ π1 π΄1 +οΏ½ + οΏ½ 1 βπ₯ π2 π΄2 οΏ½= βπ π π 2 Where: π π : Equavelant resistance. Example 2.1: Find the heat transfer per unit area through the composite wall in Figure below. Assume one-dimensional heat flow. Solusion: βπ π = ∑π βπ₯ π = ππ΄ 0.025 π π΄ = (150)(0.1) = 1.667 ∗ 10−3 0.075 π π΅ = (30)(0.05) = 0.05 0.05 π πΆ = (50)(0.1) = 0.01 0.075 π π· = (70)(0.05) = 0.02143 π = π π΄ + π πΆ + 1 1 1 + π π΅ π π· = 2.667 ∗ 10−2 π= βπ π 370−66 = 2.667∗10−2 = 11400 W H.W CH2:1,2,3,4 1.1.2 Radial system a. Cylindrical Consider a long cylinder of inside radius r i , outside radius ro, and length L, such as the one shown in Figure 2-3. We expose this cylinder to a temperature differential T i − T o and ask what the heat flow will be. For a cylinder with length very large compared to diameter, it may be assumed that the heat flows only in a radial direction, so that the only space coordinate needed to specify the system is r. The general conduction equation in cylindrical coordinate: 2.6 Assumptions: οΆ steady state. οΆ One dimension with radias only. οΆ No heat generation. Equation 2.6 will reduced to: 1 π ππ οΏ½ππ ππ οΏ½ = 0 π ππ 2.7 We may determine the temperature distribution in the cylinder by solving Equation 2.7 and applying appropriate boundary conditions. Assuming the value of k to be constant, Equation 2.7 may be integrated twice to obtain the general solution π ππ ππ = πΆ1 FIGURE 2-3: Hollow cylinder with convective surface conditions. To obtain the constants of integration C 1 and C 2 , we introduce the following boundary conditions: Applying these conditions to the general solution, we then obtain Solving for C 1 and C 2 πΆ1 = ππ ,1 −ππ ,2 π ln( 1οΏ½π2 ) πΆ2 = ππ ,2 − ππ ,1 −ππ ,2 π πποΏ½ 1οΏ½π2 οΏ½ ∗ ππ(π2 ) and substituting into the general solution, we then obtain 2.8 The rate at which energy is conducted across any cylindrical surface in the solid may be expressed as ππ ππ ππ = −ππ΄ ππ = −π(2πππΏ) ππ 2.9 2.10 From this result it is evident that, for radial conduction in a cylindrical wall, the thermal resistance is of the form K/W, ΛC/W 2.11 *Composite Cylinder Wall FIGURE 2-4: Temperature distribution for a composite cylindrical wall. Consider now the composite system of Figure 2.4. Recalling how we treated the composite plane wall and neglecting the interfacial contact resistances, the heat transfer rate may be expressed as 2.12 *Overall Heat Transfer Coefficient The foregoing result may also be expressed in terms of an overall heat transfer coefficient. That is, 2.13 If U is defined in terms of the inside area, π΄1 = ππ ππ π³, Equations 2.12 and 2.13 may be equated to yield 2.14 This definition is arbitrary, and the overall coefficient may also be defined in terms of A 4 or any of the intermediate areas. Note that 2.15 Example 2.2: Water flows at 50β¦C inside a 2.5-cm-inside-diameter tube such that h 1 = 3500 W/m2 · ΛC. The tube has a wall thickness of 0.8 mm with a thermal conductivity of 16 W/m · ΛC. The outside of the tube loses heat by free convection with h 2 = 7.6 W/m2 · ΛC. Calculate the overall heat-transfer coefficient and heat loss per unit length to surrounding air at 20β¦C. Solution There are three resistances in series for this problem, with L = 1.0 m, d 1 = 0.025 m, and d 2 = 0.025 + (2)(0.0008) = 0.0266 m, the resistances may be calculated as: π 1 = 1 2ππ1 πΏβ1 π π‘,ππππ = π 2 = 1 = π ππ( 2 ) π1 2πππΏ 2ππ2 πΏβ2 = 1 (3500)π(0.025)(1) = 0.0266 ) 0.025 ππ( 2π(16)(1) 1 = 0.00364 ΛC/W. = 0.00062 ΛC/W. (7.6)π(0.0266)(1) = 1.575 ΛC/W. ππ = π1 π΄1 οΏ½π∞,1 − π∞,2 οΏ½ = π2 π΄2 οΏ½π∞,1 − π∞,2 οΏ½ π1 = π1 = 1 π΄1 π π‘ππ‘ππ = 1 (ππ1 πΏ)π π‘ππ‘ππ 1 π(0.025)(1)(0.00364+0.00062+1.575) = 8.064 π2 = π2 = 1 π΄2 π π‘ππ‘ππ = 1 (ππ2 πΏ)π π‘ππ‘ππ 1 π(0.0266)(1)(0.00364+0.00062+1.575) = 7.577 ππ = 8.064(π)(0.025)(1)(50 − 20) = 19 W ππ = 7.577(π)(0.0266)(1)(50 − 20) = 19 W Example 2.3: The heat flow is given by *Critical Thickness of Insulation Let us consider a layer of insulation which might be installed around a circular pipe, as shown in Figure 2-5. The inner temperature of the insulation is fixed at T i , and the outer surface is exposed to a convection environment at T∞. From the thermal network the heat transfer is Figure 2-5 Critical insulation thickness. 2.16 Now let us manipulate this expression to determine the outer radius of insulation ro, which will maximize the heat transfer. The maximization condition is which gives the result 2.17 Equation (2.17) expresses the critical-radius-of-insulation concept. If the outer radius is less than the value given by this equation, then the heat transfer will be increased by adding more insulation. For outer radii greater than the critical value an increase in insulation thickness will cause a decrease in heat transfer. Example 2.4: Calculate the critical radius of insulation for asbestos [k = 0.17 W/m · β¦C] surrounding a pipe and exposed to room air at 20β¦C with h = 3.0 W/m2 · β¦C. Calculate the heat loss from a 200β¦C, 5.0-cm-diameter pipe when covered with the critical radius of insulation and without insulation. Solution: From Equation (2-17) we calculate r o as The inside radius of the insulation is 5.0/2 = 2.5 cm, so the heat transfer is calculated from Equation (2.16) as Without insulation the convection from the outer surface of the pipe is So, the addition of 3.17 cm (5.67 − 2.5) of insulation actually increases the heat transfer by 25 percent. As an alternative, fiberglass having a thermal conductivity of 0.04 W/m · β¦C might be employed as the insulation material. Then, the critical radius would be Now, the value of the critical radius is less than the outside radius of the pipe (2.5 cm), so addition of any fiberglass insulation would cause a decrease in the heat transfer. b. Spherial Figure 2.6 show a hollow sphere of radius R1 at Ts,1 and R2 at Ts,2. For sphere the heat equation is gives by equation 1.9c Assumption: οΆ One dimension heat flow r-direction. οΆ Steady state. οΆ No heat generation. For the control volume in the figure energy conservation required that The approperiate form of Fouries low is The general heat equation reduced to 1 π π 2 ππ π ππ οΏ½ππ 2 ππ οΏ½ = 0 ππ οΏ½ππ 2 ππ οΏ½ = 0 ππ First integration: π2 ππ ππ ππ ππ = = πΆ1 πΆ1 Figure 2.6: conduction in spherical shell π2 Second integration: πΆ1 π»(π) = − π + πΆ2 ..........a B.C: at π = π1 , π = ππ ,1 at π = π2 , π = ππ ,2 apply B.C in equation a πΆ ..........b πΆ .............c ππ ,1 = − π 1 + πΆ2 π ππ ,2 = − π 1 + πΆ2 π Solve equation b and c we get: πΆ1 = (ππ ,1 −ππ ,2 ) 1 1 − π2 π1 πΆ2 = ππ ,1 − 1 (ππ ,1 −ππ ,2 ) π1 1 1 − π1 π2 Sub. πΆ1 & πΆ2 in equation a, for temperature distribution: π(π)−ππ ,1 = ππ , π΄ = 4ππ 2 ππ ,2 −ππ ,1 ππ = πΆ1 π2 π2 π ππ = −4πππ 2 ππ = 4ππ π π‘,ππππ = π−π οΏ½π −π1 οΏ½ 2 1 (π»π,π −π»π,π ) π π ππ ππ ππ ( − ) (π»π,π −π»π,π ) π π ππ ππ ( − ) 1 1 − π1 π2 4ππ 2.2 Conduction with thermal energy generation: 2.2.1 Plane wall with heat generation: To investigate the effect of uniform internal heat generation in one dimension temperature distribution within the plane wall. Figure 2.7 shows the plane wall with internal uniform heat generation, πΜ . Figure 2.7 Conduction in a plane wall with uniform heat generation with Symmetrical boundary conditions. The general conduction equation is: For study, one dimension: π2 π ππ₯ 2 πΜ + =0 π First integration gives: ππ ππ₯ =− ππ₯Μ + πΆ1 π Second integration gives: πΜ π(π₯) = − 2π π₯ 2 + πΆ1 π₯ + πΆ2 ................a B.C: at π₯ = πΏ, π = ππ at π₯ = −πΏ, π = ππ : Apply B.C to equation a: ππ = − ππ = − So: πΜ 2π πΜ 2π πΏ2 + πΆ1 πΏ + πΆ2 .........b πΏ2 − πΆ1 πΏ + πΆ2 πΆ2 = ππ + πΜ 2π ...........c πΏ2 and πΆ1 = 0 Sub. πΆ1 &πΆ2 in equation a gives: πΜ π(π₯) − ππ = 2π Thus at x=0, ππ πΏ2 οΏ½1 − π₯2 πΏ2 οΏ½ ..........d This is a parabolic temperature distribution that is symmetry about x=0 ππ₯ =0 And the maximum temperature axist at x=0 ππππ₯ − ππ = πΜ 2π πΏ2 From equation d πΜ 2π πΏ2 = π(π₯)−ππ π₯2 οΏ½1− 2 οΏ½ πΏ .........e Sub. In equation e we get: π(π₯)−ππ ππππ₯ −ππ =1− π₯2 πΏ2 in case the temperature distribution is asymmetrical as shown in Figure 2.8 the temperature distribution is as below: Figure 2.8: Conduction in a plane wall with uniform heat generation with Asymmetrical boundary conditions. 2.2.2 Cylinder with heat generation: Heat generation may occur in a variety of radial geometries. Consider the long, solid cylinder of Figure 2.9, which could represent a current-carrying wire or a fuel element in a nuclear reactor. For steady-state conditions the rate at which heat is generated within the cylinder must equal the rate at which heat is convected from the surface of the cylinder to a moving fluid. This condition allows the surface temperature to be maintained at a fixed value of T s . FIGURE 2.9 Conduction in a solid cylinder with uniform heat generation. To determine the temperature distribution in the cylinder, we begin with the appropriate form of the heat equation. For constant thermal conductivity k, the heat conduction Equation reduces to: 1 π π ππ ππ πΜ οΏ½π ππ οΏ½ + π = 0 Separating variables and assuming uniform generation, this expression may be integrated to obtain π ππ ππ =− πΜ 2π π 2 + πΆ1 Repeating the procedure, the general solution for the temperature distribution becomes π(π) = − πΜ 4π π 2 + πΆ1 πππ + πΆ2 .........a To obtain the constants of integration C 1 and C 2 , we apply the boundary conditions ππ ππ = 0 ππ‘ π = 0 , π = ππ ππ‘ π = ππ Therefore: πΆ1 = 0 πΆ2 = ππ + πΜ π2 4π π The temperature distribution is therefore π(π) − ππ = π2 πΜ ππ2 οΏ½1 − π 2οΏ½ 4π π = ππππ₯ ππ‘ π = 0 π ...........b Sub in equation ππππ₯ − ππ = πΜ ππ2 ..............c 4π Devided equation (b) to equation (c) we obtained π(π)−ππ π 2 ππππ₯ −ππ =1−οΏ½ οΏ½ π ππ πΜ From equation (b) ππ =− 2π π π ππ π(π) = −π2πππ ππ πΜ π(π) = −π2πππΏ(− 2π π) π(π) = ππΏπΜ π 2 (heat conduction in the radius direction) .........d π(ππ ) = ππΏπΜ ππ2 To relate the surface temperature, ππ , to the temperature of the cold fluid, π , either a surface energy balance or an overall energy balance may be used. πππππ = πππππ£ ππΏπΜ ππ2 = β(2πππ πΏ)(ππ − π∞ ) ππ = π∞ + πΜ ππ 2β H.W a similar analysis applies to a hollow cylinder as shown in the figure below: Baundary condition: at π = π2 π(π) = ππ ,2 βΆ π = π2 Example: ππ ππ =0 A current of 200 A is passed through a stainless-steel wire [k = 19 W/m · β¦C] 3 mm in diameter. The resistivity of the steel may be taken as 70 μβ¦ · cm, and the length of the wire is 1 m. The wire is submerged in a liquid at 110 β¦C and experiences a convection heat-transfer coefficient of 4kW/m2 · β¦C. Calculate the center temperature of the wire. Solution: All the power generated in the wire must be dissipated by convection to the liquid: π = πΌ 2 π = π = βπ΄(ππ − π∞ ) ..........a The resistance of the wire is calculated from πΏ π = ππ΄ = π (70π₯10−6 )(100) π(0.15)2 = 0.099β¦ where ρ is the resistivity of the wire. The surface area of the wire is πdL, so from Equation (a) (200)2 (0.099) = 4000π(3π₯10−3 )(1)(ππ − 110) ππ =215ΛC. The heat generated per unit volume πΜ is calculated from π = πΜ π = πΜ πππ2 πΏ So that, πΜ = 3960 π(1.5π₯10−3 )2 (1) = 560.2 ππ/π3 Finally, the center temperature of the wire is calculated from Equation below: ππππ₯ − ππ = ππππ₯ = πΜ ππ2 4π (5.605π₯108 )(1.5π₯10−3 )2 (4)(19) + 215 = 231.6 β 2.2.3 sphere with heat generation The general heat conduction equation For one dimension (r direction only) study state the general equation has been reduced to: Baundary condition: at π = 0 π ππ ππ πΜ οΏ½π 2 ππ οΏ½ = − π π 2 1 π 2 ππ πΜ οΏ½π οΏ½+ =0 2 π ππ ππ π ππ ππ = 0 & ππ‘ π = ππ π = ππ First integration gives: ππ πΜ π 2 ππ = − 3π π 3 + πΆ1 ππ ππ πΜ πΆ = − 3π π + π 12 Second integration gives: π(π) = − πΜ π 2 6π − πΆ1 π + πΆ2 H.W find the temperature distribution and heat transfer. 3.3 Contact Resistance Although neglected until now, it is important to recognize that, in composite systems, the temperature drop across the interface between materials may be appreciable. This temperature change is attributed to what is known as the thermal contact resistance, R t,c . The effect is shown in Figure 2.10, and for a unit area of the interface, the resistance is defined as FIGURE 2.10 Temperature drop due to thermal contact resistance.
