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Transient Conduction: The Lumped Capacitance Method

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Presentation on theme: "Transient Conduction: The Lumped Capacitance Method"— 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 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 ( ), surface radiation conditions ( ), a surface temperature or heat flux, and/or internal energy generation. Solution Techniques The Lumped Capacitance Method Exact Solutions The Finite-Difference Method

4 Lumped Capacitance Method
The Lumped Capacitance Method Based on the assumption of a spatially uniform temperature distribution throughout the transient process. Hence, 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

5 Lumped Capacitance Method (cont.)
First Law: Assuming energy outflow due to convection and radiation and inflow 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 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) Note: (5.6) The thermal time constant is defined as (5.7) Thermal Resistance, Rt Lumped Thermal Capacitance, Ct The change in thermal energy storage due to the transient process is (5.8)

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 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: See Fig. 5.4. Criterion for Applicability of Lumped Capacitance Method:

10 The Lumped Capacitance Method
T(t) Liquid T < Ti t≥0 T = T(t) Ti t<0, T=Ti Ėst Ėout = qconv

11 The Lumped Capacitance Method
Energy Balance: Ėin + Ėg - Ėout = Ėst -Ėout = -qconv= -hAs(T-T ) = Ėst = Vc(dT/dt) Define: θ ≡ T-T

12 The Lumped Capacitance Method
Thermal Time Constant: τt

13 The Lumped Capacitance Method
Where: Rt is convection resistance, Ct 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 X Ts,1 Ts,2 T, h Bi<<1 Bi≈1 Bi>>1 qcond qconv L T < Ts,2 < Ts,1 Under steady-state conditions:

15 Validity of the Lumped Capacitance Method
Biot (Bi) number is the ratio of conduction resistance to convection resistance Characteristic Length Lc ≡ V/As Lc = ½ L for plane wall in Fig. 5.4 Lc = r0/2 for long cylinder Lc =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 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/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 5.1 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.

20 Example 5.1 Schematic: Leads Thermocouple T = 200 ºC Junction
h=400 W/m2K k=20 W/mK c=400 J/kgK =8500 kg/m3 Gas stream

21 Example 5.1 Assumptions: Analysis:
Uniform Temperature inside the junction Radiation 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 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 m2K/W. the opposite surface is well insulated 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, providing a convection coefficient h=25 w/m2K 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 Ts,i=1200 K? What is the temp. Ts,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(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:

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

29 Problem: Furnace Start-up (cont.)

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