21-D Transient Conduction 1-D conduction in X direction, Not internal heat generation.
3Transient ConductionTransient ConductionA heat transfer process for which the temperature varies with time, as wellas location within a solid.It is initiated whenever a system experiences a change in operating conditions.It can be induced by changes in:surface convection conditions ( ),surface radiation conditions ( ),a surface temperature or heat flux, and/orinternal energy generation.Solution TechniquesThe Lumped Capacitance MethodExact SolutionsThe Finite-Difference Method
4Lumped Capacitance Method The Lumped Capacitance MethodBased on the assumption of a spatially uniform temperature distributionthroughout the transient process.Hence,Why is the assumption never fully realized in practice?General Lumped CapacitanceAnalysis:Consider a general case,which includes convection,radiation and/or an appliedheat flux at specifiedsurfacesas well as internal energygeneration
5Lumped Capacitance Method (cont.) First Law:Assuming energy outflow due to convection and radiation andinflow due to an applied heat flux(5.15)May h and hr be assumed to be constant throughout the transient process?How must such an equation be solved?
6Special Case (Negligible Radiation) Special Cases (Exact Solutions, )Negligible RadiationThe non-homogeneous differential equation is transformed into ahomogeneous equation of the form:Integrating from t = 0 to any t and rearranging,(5.25)To what does the foregoing equation reduce as steady state is approached?How else may the steady-state solution be obtained?
7Special Case (Convection) Negligible Radiation and Source Terms(5.2)Note:(5.6)The thermal time constant is defined as(5.7)ThermalResistance, RtLumped ThermalCapacitance, CtThe change in thermal energy storage due to the transient process is(5.8)
8Special Case (Radiation) Negligible Convection and Source TermsAssuming radiation exchange with large surroundings,(5.18)This result necessitates implicit evaluation of T(t).
9The Biot Number and Validity of The Lumped Capacitance Method The Biot Number: The first of many dimensionless parameters to beconsidered.Definition:Physical Interpretation:See Fig. 5.4.Criterion for Applicability of Lumped Capacitance Method:
12The Lumped Capacitance Method Thermal Time Constant: τt
13The Lumped Capacitance Method Where: Rt is convection resistance,Ct is lumped thermal capacitanceTotal energy transfer Q:
14Validity of the Lumped Capacitance Method This method is simple and convenient, but it requiresBi < 0.1TXTs,1Ts,2T, hBi<<1Bi≈1Bi>>1qcondqconvLT < Ts,2 < Ts,1Under steady-state conditions:
15Validity of the Lumped Capacitance Method Biot (Bi) number is the ratio of conduction resistance to convection resistanceCharacteristic Length Lc ≡ V/AsLc = ½ L for plane wall in Fig. 5.4Lc = r0/2 for long cylinderLc =r0/3 for sphere
16Validity of the Lumped Capacitance Method Bi <<1 (Bi < 0.1)Conduction resistance within solid is much less than the resistanceto convection across the fluid boundary layer. The T distribution insidethe solid is negligible. Lumped capacitance method is valid when Bi<0.1
17Validity of the Lumped Capacitance Method Where Fo is Fourier number
18Example 5.1Example 5.1 A thermocouple junction, which may be approximated as a sphere, is to be used to measure the temperature of a gas stream. The convection coefficient is h=400 W/m2·K, k is 20 W/m·k, c=400 J/kg·K, and =8500 kg/m3. Determine the junction diameter needed for the thermocouple to have a time constant of 1 s. If the junction is at 25 ºC and placed in a gas stream that is at 200 ºC, how long will it take for the junction to reach 199 ºC ?
19Example 5.1Known:Thermal properties of the thermocouple junction, time constant.Find:1. Diameter needed to achieve 1s time constant;2. Time required to reach 199 ºC in a gas stream at 200 ºC.
21Example 5.1 Assumptions: Analysis: Uniform Temperature inside the junctionRadiation is negligible;No conduction through the leads;Constant properties.Analysis:D is unknown, Bi number can’t be determined. But we can assume Bi<0.1, so that Lumped Capacitance method can be applied.
22Example 5.1 For sphere, V = πD3/6, As = πD2, Lc =V/As= D/6 Time constant:
23Example 5.1 Bi = (hLc/k) =(hD/6k) = 2.35 x 10-4 <<1 Original assumption (Bi<0.1) is valid, so lumped capacitance method is a good approximation.From Eqn. 5.6
24Example 2 (Problem 5.12)A plane wall of a furnace is fabricated from plain carbon steel(k=60 W/mK, ρ=7850kg/m3, c=430J/kgK) and is of thicknessL=10 mm. To protect it from the corrosive effects of the furnacecombustion gases, one furnace surface of the wall is coated witha thin ceramic film that, for a unit surface area, has a thermalresistance of R”t,f=0.01 m2K/W. the opposite surface is wellinsulated from the surroundings.At furnace start-up the wall is at initial temperature of Ti=300 K,and the combustion gases at T∞=1300K enter the furnace, providinga convection coefficient h=25 w/m2K at the ceramic film. Assumingthe film to have negligible thermal capacitance, how long will ittake for the inner surface of the steel to achieve a temperature ofTs,i=1200 K? What is the temp. Ts,o of the exposed surface of theceramic film at this time?
25Problem: Thermal Energy Storage Problem 5.12: Charging a thermal energy storage system consistingof a packed bed of aluminum spheres.Schematic:
26Problem: Thermal Energy Storage (cont.) ANALYSIS: To determine whether a lumped capacitance analysis can be used, first computeBi = h(ro/3) = 75 W/m2∙K ( m)/150 W/m∙K = << 1.From Eq. 5.8, achievement of 90% of the maximum possible thermal energy storage corresponds to<<
27Problem: Furnace Start-up Problem 5.22: Heating of coated furnace wall during start-up.Schematic: