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Transient Conduction: The Lumped Capacitance Method Chapter Five Sections 5.1 thru 5.3 Lecture 9.

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Presentation on theme: "Transient Conduction: The Lumped Capacitance Method Chapter Five Sections 5.1 thru 5.3 Lecture 9."— Presentation transcript:

1 Transient Conduction: The Lumped Capacitance Method Chapter Five Sections 5.1 thru 5.3 Lecture 9

2 1-D Transient Conduction 1-D conduction in X direction, Not internal heat generation.

3 Transient Conduction A heat transfer process for which the temperature varies with time, as well as 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 ( ), Solution Techniques – The Lumped Capacitance Method – Exact Solutions – The Finite-Difference Method – surface radiation conditions ( ), – a surface temperature or heat flux, and/or – internal energy generation.

4 Lumped Capacitance Method The Lumped Capacitance Method Based on the assumption of a spatially uniform temperature distribution throughout the transient process. Why is the assumption never fully realized in practice? General Lumped Capacitance Analysis: Consider a general case, which includes convection, radiation and/or an applied heat flux at specified surfaces as well as internal energy generation Hence,.

5 Lumped Capacitance Method (cont.) First Law: Assuming energy outflow due to convection and radiation and inflow due to an applied heat flux May h and h r be assumed to be constant throughout the transient process? How must such an equation be solved? (5.15)

6 Special Case (Negligible Radiation) Special Cases (Exact Solutions, ) Negligible Radiation The non-homogeneous differential equation is transformed into a homogeneous 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) The thermal time constant is defined as (5.7) Thermal Resistance, R t Lumped Thermal Capacitance, C t The change in thermal energy storage due to the transient process is (5.8) Note: (5.6)

8 Special Case (Radiation) Negligible Convection and Source Terms Assuming radiation exchange with large surroundings, (5.18) This result necessitates implicit evaluation of T(t).

9 Biot Number The Biot Number and Validity of The Lumped Capacitance Method The Biot Number: The first of many dimensionless parameters to be considered. Definition: Physical Interpretation: Criterion for Applicability of Lumped Capacitance Method: See Fig. 5.4.

10 The Lumped Capacitance Method T(t) Liquid T < T i t0 T = T(t) TiTi t<0, T=T i Ė st Ė out = q conv

11 The Lumped Capacitance Method Energy Balance: Ė in + Ė g - Ė out = Ė st -Ė out = -q conv = -hA s (T-T ) = Ė st = Vc(dT/dt) Define: θ T-T

12 The Lumped Capacitance Method Thermal Time Constant: τ t

13 The Lumped Capacitance Method Where: R t is convection resistance, C t is lumped thermal capacitance Total energy transfer Q:

14 Validity of the Lumped Capacitance Method This method is simple and convenient, but it requires Bi < 0.1 T < T s,2 < T s,1 Under steady-state conditions: T X T s,1 T s,2 T, h Bi<<1 Bi1 Bi>>1 q cond q conv L

15 Validity of the Lumped Capacitance Method Biot (Bi) number is the ratio of conduction resistance to convection resistance Characteristic Length L c V/A s L c = ½ L for plane wall in Fig. 5.4 L c = r 0 /2 for long cylinder L c =r 0 /3 for sphere

16 Validity of the Lumped Capacitance Method Bi <<1 (Bi < 0.1) Conduction resistance within solid is much less than the resistance to convection across the fluid boundary layer. The T distribution inside the 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 5.1 Example 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/m 2 ·K, k is 20 W/m·k, c=400 J/kg·K, and =8500 kg/m 3. 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 Known: 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. Example 5.1

20 Schematic: T = 200 ºC h=400 W/m 2 K D ? Gas stream Thermocouple Junction T i =25 ºC k=20 W/mK c=400 J/kgK =8500 kg/m 3 Leads

21 Example 5.1 Assumptions: Uniform Temperature inside the junction Radiation is negligible; No conduction through the leads; Constant properties. Analysis: D is unknown, Bi number cant be determined. But we can assume Bi<0.1, so that Lumped Capacitance method can be applied.

22 Example 5.1 For sphere, V = π D 3 /6, A s = π D 2, L c =V/A s = D/6 Time constant:

23 Example 5.1 Bi = (hL c /k) =(hD/6k) = 2.35 x <<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/m 3, c=430J/kgK) and is of thickness L=10 mm. To protect it from the corrosive effects of the furnace combustion gases, one furnace surface of the wall is coated with a thin ceramic film that, for a unit surface area, has a thermal resistance of R t,f =0.01 m 2 K/W. the opposite surface is well insulated from the surroundings. At furnace start-up the wall is at initial temperature of T i =300 K, and the combustion gases at T =1300K enter the furnace, providing a convection coefficient h=25 w/m 2 K at the ceramic film. Assuming the film to have negligible thermal capacitance, how long will it take for the inner surface of the steel to achieve a temperature of T s,i =1200 K? What is the temp. T s,o of the exposed surface of the ceramic film at this time?

25 Problem: Thermal Energy Storage Problem 5.12: Charging a thermal energy storage system consisting of 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 compute Bi = h(r o /3) = 75 W/m 2 K ( m)/150 W/mK = << 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:

28 Problem: Furnace Start-up (cont.) Hence, with <

29 Problem: Furnace Start-up (cont.) <


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