2 1-D Transient Conduction 1-D conduction in X direction, Not internal heat generation.
3 Transient 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
4 Lumped 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
5 Lumped 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?
6 Special 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?
7 Special 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)
8 Special Case (Radiation) Negligible Convection and Source TermsAssuming radiation exchange with large surroundings,(5.18)This result necessitates implicit evaluation of T(t).
9 The 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:
12 The Lumped Capacitance Method Thermal Time Constant: τt
13 The Lumped Capacitance Method Where: Rt is convection resistance,Ct is lumped thermal capacitanceTotal energy transfer Q:
14 Validity 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:
15 Validity 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
16 Validity 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
17 Validity of the Lumped Capacitance Method Where Fo is Fourier number
18 Example 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 ?
19 Example 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.
20 Example 5.1 Schematic: Leads Thermocouple T = 200 ºC Junction h=400 W/m2Kk=20 W/mKc=400 J/kgK=8500 kg/m3Gas stream
21 Example 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.
22 Example 5.1 For sphere, V = πD3/6, As = πD2, Lc =V/As= D/6 Time constant:
23 Example 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
24 Example 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?
25 Problem: Thermal Energy Storage Problem 5.12: Charging a thermal energy storage system consistingof a packed bed of aluminum spheres.Schematic:
26 Problem: 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<<
27 Problem: Furnace Start-up Problem 5.22: Heating of coated furnace wall during start-up.Schematic: