Lecture 9 Conduction with Internal Heat Generation

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Oct 30, 2023

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9/19/23 1 MECE 4364 Heat Transfer Prof. Dong Liu Department of Mechanical Engineering University of Houston 1 Lecture 7 1 Last Lecture ¨ Heat conduction equation in a cylindrical wall ¨ The solutions ¨ Thermal resistance ¨ Thermal network 2 ࠵? ࠵?࠵? ࠵? ࠵?࠵? ࠵?࠵? = 0 ࠵? ! = −࠵?࠵? ࠵?࠵? ࠵?࠵? = 2࠵?࠵?࠵? ࠵? " − ࠵? # ࠵?࠵? ࠵? # /࠵? " ࠵? ࠵? = $ ! %$ " &’ ! ! /! " ࠵?࠵? ! ! " + ࠵? # r 1 r 2 T 1 T 2 ࠵? )*& = ࠵? " − ࠵? # ࠵? ! = ࠵?࠵? ࠵? # /࠵? " 2࠵?࠵?࠵? ࠵? ! = ࠵? +," − ࠵? +,# ࠵? -.-/& ࠵? -.-/& = ࠵? ).’0," + ࠵? )*&," + ࠵? )*&,# + ࠵? )*&,1 + ࠵? ).’0,1 2
9/19/23 2 Example 1 ¨ Steam at ࠵? +," = 320℃ flows in a cast iron pipe ( ࠵? = 80 ࠵?/࠵?࠵?) whose inner and outer diameters are ࠵? " = 5 ࠵?࠵? and ࠵? # = 5.5 ࠵?࠵?. The pipe is covered with 3-cm-thick glass wool insulation with ࠵? = 0.05 2 34 . Heat is lost to the surroundings at ࠵? +,# = 5℃ by natural convection and radiation, with a combined heat transfer coefficient of # = 18 2 3 " 4 . Taking the heat transfer coefficient inside the pipe to be " = 60 2 3 " 4 . Determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drops across the pipe shell and the insulation. 3 3 Critical Radius of Insulation ¨ Adding more insulation to a plane wall or to the attic always decreases heat transfer. The thicker the insulation, the lower the heat transfer rate ¤ This is expected, since the heat transfer area A is constant, and adding insulation always increases the thermal resistance of the wall without increasing the convection resistance ¨ However, this may not be the case for a cylindrical pipe 4 ¨ Consider a cylindrical pipe of outer radius r 1 whose outer surface temperature T 1 is maintained constant ¨ The pipe is now insulated with a material whose thermal conductivity is k and outer radius is r 2 ( the larger r 2 , the thicker the insulation ) ¨ Heat transfer from the insulated pipe to the surrounding air is ¨ What if we want to minimize the heat loss? ࠵? ! = ࠵? " − ࠵? + ࠵? 5’6 + ࠵? ).’0 = ࠵? " − ࠵? + ࠵?࠵? ࠵? # /࠵? " 2࠵?࠵?࠵? + 1 2࠵?࠵? # ࠵? 4
9/19/23 3 Critical Radius of Insulation ¨ The variation of q r with the outer radius of the insulation r 2 shows a maximum at which ¨ The critical radius of insulation is ¨ Should we worry about r cr when insulating hot- water pipes? ¤ r cr is the largest when k is large and h is small ¤ For ࠵? !"# = 0.05 $ !% and !&’ = 5 $ ! " % 5 ࠵? ! = ࠵? " − ࠵? + ࠵? 5’6 + ࠵? ).’0 = ࠵? " − ࠵? + ࠵?࠵? ࠵? # /࠵? " 2࠵?࠵?࠵? + 1 2࠵?࠵? # ࠵? ࠵?࠵? ! ࠵?࠵? # = 0 ࠵? )!,)*&5’78! = ࠵? ࠵? ࠵? ࠵? !"# ࠵? (")* 5 Example 2 6 6
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9/19/23 4 Example 2 7 ࠵? ࠵? 7 Example 2 8 8
9/19/23 5 Example 3 9 9 Example 3 10 10
9/19/23 6 Example 3 11 11 Conduction with Heat Generation (Section 3.4) 12 12
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9/19/23 7 Heat Generation in A Solid ¨ Many practical heat transfer applications involve the conversion of some form of energy into thermal energy in the medium ¤ Such mediums are said to involve internal heat generation, which manifests itself as a rise in temperature throughout the medium ¤ Some examples of heat generation are resistance heating in wires, exothermic chemical reactions in a solid, and nuclear reactions in nuclear fuel rods where electrical, chemical, and nuclear energies are converted to heat 13 ¨ Heat generation is usually expressed per unit volume of the medium, whose unit is W/m 3 ¨ Example ¤ Heat generation in an electrical wire of outer radius r 0 and length L ̇࠵? = ̇ ࠵? +*’,*-*./)&. ࠵? = ࠵? 0 ࠵? * ࠵?࠵? 1 0 ࠵? 13 1D Steady-State Heat Generation A large plane wall of thickness 2L experiences a uniform heat generation, determine the temperature variation within the wall ¨ Begin with the general heat conduction equation ¨ Assumptions ¤ 1D ¤ Steady state ¤ Constant properties ¨ Integrating twice gives the general solution 14 ࠵? 0 ࠵? ࠵?࠵? 0 + ̇࠵? ࠵? = 0 ࠵? 0 ࠵? ࠵?࠵? 0 + ࠵? 0 ࠵? ࠵?࠵? 0 + ࠵? 0 ࠵? ࠵?࠵? 0 + ̇࠵? ࠵? = 1 ࠵? ࠵?࠵? ࠵?࠵? x = -L x = L T T 1 T 2 ! q ࠵? ࠵? = − ̇࠵? 2࠵? ࠵? 0 + ࠵? 2 ࠵? + ࠵? 0 Making a concrete slab 14
9/19/23 8 ¨ Consider the boundary conditions ࠵? = −࠵? : ࠵? −࠵? = ࠵? " = − ̇: #; ࠵? # − ࠵? " ࠵? + ࠵? # ࠵? = ࠵? : ࠵? ࠵? = ࠵? # = − ̇: #; ࠵? # + ࠵? " ࠵? + ࠵? # ¨ Solving for C 1 and C 2 ¨ The temperature field is 1D Steady-State Heat Generation 15 ࠵? ࠵? = ̇࠵? 2࠵? ࠵? 0 − ࠵? 0 ࠵? 2 − ࠵? 0 2࠵? ࠵? + ࠵? 2 + ࠵? 0 2 Parabolic profile x = -L x = L T T 1 T 2 ! q ࠵? " = ࠵? # − ࠵? " 2࠵? ࠵? # = ̇࠵? 2࠵? ࠵? # + ࠵? " + ࠵? # 2 (what if ̇࠵? = 0 ?) 15 ¨ Special case ¤ ࠵? 2 = ࠵? 0 = ࠵? 3 n Maximum temperature n At x = 0 n There is no heat transfer through this plane n Heat transfer at other locations? n It may be represented by an adiabatic surface n Share the same solution 1D Steady-State Heat Generation 16 ࠵? ࠵? = ̇࠵? 2࠵? ࠵? 0 − ࠵? 0 + ࠵? 0 − ࠵? 2 2࠵? ࠵? + ࠵? 0 + ࠵? 2 2 ࠵? ࠵? = ̇࠵? 2࠵? ࠵? 0 − ࠵? 0 + ࠵? 3 ࠵? ࠵? = 0 = ̇ ࠵?࠵? 0 2࠵? + ࠵? 3 ࠵?࠵? ࠵?࠵? #41 = 0 16
9/19/23 9 ¨ Heat flux ¨ Special case ¤ ࠵? 2 = ࠵? 0 = ࠵? 3 1D Steady-State Heat Generation 17 ࠵? 55 ࠵? = −࠵? ࠵?࠵? ࠵?࠵? = ̇࠵?࠵? − ࠵? ࠵? 0 − ࠵? 2 2࠵? Linear heat flux profile ࠵? ࠵? = ̇࠵? 2࠵? ࠵? 0 − ࠵? 0 + ࠵? 0 − ࠵? 2 2࠵? ࠵? + ࠵? 0 + ࠵? 2 2 ࠵? 55 ࠵? = −࠵? ࠵?࠵? ࠵?࠵? = ̇࠵?࠵? 17 ¨ A solid wall of thickness L is subject to the following boundary conditions ¨ What is the surface temperature at x = L? ¨ What is the surface temperature at x = 0? Example 1 18 ࠵? ࠵? = ̇࠵? 2࠵? ࠵? 0 − ࠵? 0 + ࠵? 6 + ̇ ࠵?࠵? ࠵? ࠵? = ࠵? = ࠵? 6 + ̇ ࠵?࠵? = ࠵? 3 ࠵? ࠵? = − ̇࠵? 2࠵? ࠵? 0 + ࠵? 2 ࠵? + ࠵? 0 Standard approach ࠵? ࠵?࠵? ࠵?࠵? ࠵?࠵? = − ̇࠵? ࠵? ࠵? ࠵? = 0 = ̇࠵?࠵? 0 2࠵? + ࠵? 6 + ̇ ࠵?࠵? D ࠵?࠵? ࠵?࠵? <=> = 0 D −࠵? ࠵?࠵? ࠵?࠵? <=? = ℎ ࠵? ࠵? = ࠵? − ࠵? + 18
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9/19/23 10 Example 1 19 A different approach to find T s : ࠵? 3 = ࠵? 6 + ̇ ࠵?࠵? −࠵? .7’8 55 ࠵? + ̇࠵? ࠵?࠵? = 0 For the solid wall, energy balance yields ̇ ࠵? 79/ + ̇ ࠵? + = 0 ℎ࠵? ࠵? 3 − ࠵? 6 = ̇࠵? ࠵?࠵? 19 Example 2 ¨ A composite plane wall consists of materials A and B. ¨ A has uniform heat generation ̇࠵? = 1.5×10 : ࠵?/࠵? ; . The inner surface of A is insulated ¨ The outer surface of B ( ࠵? < = 150 ࠵?/࠵?࠵? ) is cooled by a water stream ࠵? 6 = 30℃, ℎ = 1000 ࠵?/࠵? 0 ࠵? . ¤ Sketch the steady-state temperature distribution ¤ Determine T 0 and T 2 20 20
9/19/23 11 Example 2 ¨ Sketch the steady-state temperature distribution 21 ¨ A: conduction with heat generation ¤ BCs: ¤ Parabolic temperature profile ¨ B: conduction without heat generation ¤ BCs: ¤ Linear temperature profile 21 Example 2 22 22
9/19/23 12 Example 2 23 ! " = 0 = %& ! ̇ 2) + ! " 23 Heat Generation in A Cylinder ¨ A long homogeneous resistance wire of radius ࠵? > = 0.2 ࠵?࠵?࠵?ℎ in and thermal conductivity ࠵? = 7.8 @-A B D- ℉ is being used to boil water at atmospheric pressure by the passage of electric current. Heat is generated in the wire uniformly as a result of resistance heating at a rate of ̇࠵? = 2400 ࠵?࠵?࠵?/ℎ K ࠵?࠵? 1 . If the outer surface temperature of the wire is measured to be ࠵? 6 = 226℉ , obtain a relation for the temperature distribution, and determine the temperature at the centerline of the wire when steady operating conditions are reached. 24 This heat transfer problem is similar to what we have solved, except that ࠵? = ࠵? ࠵? Boundary conditions 1 ࠵? ࠵? ࠵?࠵? ࠵? ࠵?࠵? ࠵?࠵? + ̇࠵? ࠵? = 0 ࠵? ࠵? > = ࠵? 6 = 226℉ D ࠵?࠵? ࠵?࠵? !=> = 0 24
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9/19/23 13 Heat Generation in A Cylinder ¨ Integrating once gives ¨ Applying the surface temperature BC 25 1 ࠵? ࠵? ࠵?࠵? ࠵? ࠵?࠵? ࠵?࠵? + ̇࠵? ࠵? = 0 ࠵? ࠵? > = ࠵? 6 = 226℉ D ࠵?࠵? ࠵?࠵? !=> = 0 ࠵? ࠵?࠵? ࠵?࠵? = − ̇࠵? 2࠵? ࠵? # + ࠵? " ࠵? 2 = 0 ࠵?࠵? ࠵?࠵? = − ̇࠵? 2࠵? ࠵? ࠵? = − ̇࠵? 4࠵? ࠵? # + ࠵? # ࠵? 6 = − ̇࠵? 4࠵? ࠵? > # + ࠵? # ࠵? # = ࠵? 6 + − ̇࠵? 4࠵? ࠵? > # ࠵? ࠵? = ࠵? 6 + ̇࠵? 4࠵? ࠵? > # − ࠵? # 25 Heat Generation in A Two-Layer Cylinder ¨ This time the wire of radius ࠵? " is wrapped by a layer of insulation material. Determine ¤ The temperature at the center of the wire ࠵? 1 ¤ The temperature at the interface of the wire and the insulation layer ࠵? & ¨ Solution ¤ The temperature in the wire can be formulated as with ¤ The problem was already solved 26 1 ࠵? ࠵? ࠵?࠵? ࠵? ࠵?࠵? ࠵?࠵? + ̇࠵? ࠵? = 0 ࠵? ࠵? " = ࠵? 5 D ࠵?࠵? ࠵?࠵? !=> = 0 ࠵? ࠵? = ࠵? 5 + ̇࠵? 4࠵? ࠵? " # − ࠵? # ࠵? & is unknown 26
9/19/23 14 Heat Generation in A Two-Layer Cylinder ¤ Note that the insulation layer does not involve any heat generation ¤ The boundary conditions are ¤ The solution was obtained from last lecture ¨ How to find ࠵? & ? ¤ At the interface, heat flux in the wire and the insulation must be the same 27 ࠵? ࠵? " = ࠵? 5 ࠵? ࠵?࠵? ࠵? ࠵?࠵? ࠵?࠵? = 0 ࠵? ࠵? # = ࠵? 6 ࠵? ࠵? = $ # %$ $ &’ ! ! /! " ࠵?࠵? ! ! " + ࠵? 6 ࠵? & is still unknown −࠵? F5!8 D ࠵?࠵? ࠵?࠵? ! ! ,F5!8 = −࠵? 5’6 D ࠵?࠵? ࠵?࠵? ! ! ,5’6 ࠵? 5 = ̇:! ! " #; %#&’ ࠵?࠵? ! " ! ! + ࠵? 6 27 Heat Generation in A Solid Sphere 1D, steady-state conduction in a sphere of constant k , uniform generation , and convection cooling : General Solution: T r ( ) = ! q 6 k r 2 C 1 r + C 2 Specific Solution: What are the two BCs? T r o ( ) = T s dT dr r = 0 = 0 1 r 2 d dr kr 2 dT dr + ! q = 0 centerline temperature/thermal condition for radial systems is the same as an insulated wall T r ( ) = ! qr o 2 6 k 1 r 2 r o 2 + T s 28 28

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