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.)
<