Transient Conduction:The Lumped

Capacitance Method

Chapter FiveSections 5.1 thru 5.3

Lecture 9

1-D Transient Conduction

1-D conduction in X direction, Not internal heat

generation.

t

T

k

q

z

T

y

T

x

T

1

2

2

2

2

2

2

t

T

x

T

1

2

2

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 ( ), ,h T

• Solution Techniques

– The Lumped Capacitance Method– Exact Solutions– The Finite-Difference Method

– surface radiation conditions ( ),sur,rh T

– a surface temperature or heat flux, and/or

– internal energy generation.

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

, , ,, , ,s c s r s hA A A

Hence, . ,T r t T t

Lumped Capacitance Method (cont.)

First Law:

in outst

gdE dT

c E E Edt dt

• Assuming energy outflow due to convection and radiation and inflow due to an applied heat flux ,sq

, , , sur gs s h s c r s r

dTc q A hA T T h A T T E

dt

• May h and hr be assumed to be constant throughout the transient process?

• How must such an equation be solved?

(5.15)

Special Case (Negligible Radiation)

• Special Cases (Exact Solutions, ) 0 iT T

Negligible Radiation , / :T T b a

, ,/ /gs c s s ha hA c b q A E c

The non-homogeneous differential equation is transformed into a homogeneous equation of the form:

da

dt

Integrating from t = 0 to any t and rearranging,

/exp 1 exp

i i

T T b aat at

T T T T

(5.25)

To what does the foregoing equation reduce as steady state is approached?

How else may the steady-state solution be obtained?

Special Case (Convection)

Negligible Radiation and Source Terms , 0, 0 :gr sh h E q

,s c

dTc hA T T

dt (5.2)

, 0is c

tc d

hAdt

,exp s c

i i

hAT Tt

T T c

exp

t

t

The thermal time constant is defined as

,

1t

s c

chA

(5.7)

ThermalResistance, Rt

Lumped ThermalCapacitance, Ct

The change in thermal energy storage due to the transient process is

out0

t

stE Q E dt ,

0

t

s chA dt 1 expit

tc

(5.8)

Note: T T

(5.6)

Special Case (Radiation)

Negligible Convection and Source Terms , 0, 0 :gr sh h E q

Assuming radiation exchange with large surroundings,

4 4, surs r

dTc A T T

dt

,

4 4sur

0 i

s r T

T

tA

c

dTT T

dt

sur sur

3, sur sur

1n 1n4

i

s r sur i

T T T Tct

A T T T T T

(5.18)

This result necessitates implicit evaluation of T(t).

1 1

sur sur

2 tan tan iTT

T T

Biot Number The Biot Number and Validity ofThe Lumped Capacitance Method

• The Biot Number: The first of many dimensionless parameters to be considered.

Definition:chL

Bik

convection or radiation coefficienth

thermal conductivity of t soe d h lik

of the solid ( / or coordinate

associated with maximum spatial temperature differenc

characteristic length

e)c sL A

Physical Interpretation:

cond solid

conv solid/fluid

/

1 /c s

s

L kA R TBi

hA R T

Criterion for Applicability of Lumped Capacitance Method:

1Bi

See Fig. 5.4.

The Lumped Capacitance Method

T(t)

Liquid

T < Ti

t≥0T = T(t)

Tit<0, T=Ti

Ėst

Ėout = qconv

The Lumped Capacitance Method

Energy Balance:

Ėin + Ėg - Ėout = Ėst

-Ėout = -qconv= -hAs(T-T ) = Ėst = Vc(dT/dt)

Define: θ ≡ T-T

dt

d

hA

Vc

s

TTdtd

hA

Vcii

t

si

,

0

The Lumped Capacitance Method

thA

Vc

is

ln

])(exp[ tVc

hA

TT

TT s

ii

tts

t CRVchA

))(1

(

Thermal Time Constant: τt

The Lumped Capacitance Method

Where: Rt is convection resistance,

Ct is lumped thermal capacitance

Total energy transfer Q:

)]exp(1[)(00

ti

t

s

t tVcdthAqdtQ

]exp[tii

t

TT

TT

Validity of the Lumped Capacitance Method

This method is simple and convenient, but it requires

Bi < 0.1

T < Ts,2 < Ts,1

Under steady-state conditions:

)()( 2,2,1, TThATTL

kAsss

T

X

Ts,1 Ts,2

Ts,2

Ts,2

T, h

Bi<<1

Bi≈1

Bi>>1

qcond qconv

L

Validity of the Lumped Capacitance Method

)/1(

)/(

)(

)(

2,

2,1,

hA

kAL

TT

TT

s

ss

Bik

hL

R

R

conv

cond

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

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

k

hLBi c

h

cLct

Validity of the Lumped Capacitance Method

]][][[2c

c

ct L

t

c

k

k

hL

cL

htt

]][][[2c

c

L

t

c

k

k

hL

2cL

tFo

Where Fo is Fourier number

)*exp(]exp[ FoBit

TT

TT

tii

FoBiL

tBi

c

*]}][{[*2

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 ?

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

Example 5.1

Schematic:

T = 200 ºCh=400 W/m2K

D ?

Gas stream

ThermocoupleJunctionTi=25 ºC

k=20 W/mKc=400 J/kgK=8500 kg/m3

Leads

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 can’t be determined.

But we can assume Bi<0.1, so that Lumped

Capacitance method can be applied.

Example 5.1

For sphere, V = πD3/6, As = πD2, Lc

=V/As= D/6

Time constant:

h

cD

h

cL

hA

Vc c

st 6

mxc

hD t 41006.7

6

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

]exp[ti

t

TT

TT

sTT

TTt

it 16.5)}

20025

200199ln(*1{)ln(

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 furnacecombustion 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 ofTs,i=1200 K? What is the temp. Ts,o of the exposed surface of the ceramic film at this time?

Problem: Thermal Energy Storage

Problem 5.12: Charging a thermal energy storage system consistingof a packed bed of aluminum spheres.

KNOWN: Diameter, density, specific heat and thermal conductivity of aluminum spheres used in packed bed thermal energy storage system. Convection coefficient and inlet gas temperature.

FIND: Time required for sphere at inlet to acquire 90% of maximum possible thermal energy and the corresponding center temperature.

Alum inum sphere D = 75 m m ,

T = 25 Ci oGas

T Cg,i o= 300

h = 75 W /m -K2

= 2700 kg/m 3

k = 240 W /m -Kc = 950 J/kg-K

Schematic:

Problem: Thermal Energy Storage (cont.)

ASSUMPTIONS: (1) Negligible heat transfer to or from a sphere by radiation or conduction due to contact with other spheres, (2) Constant properties.

Hence, the lumped capacitance approximation may be made, and a uniform temperature may be assumed to exist in the sphere at any time.

st 0.90 1 exp / ti

Et

cV

ln 0.1 427 s 2.30 984 stt

From Eq. (5.6), the corresponding temperature at any location in the sphere is , ,984 s exp 6 /g i i g iT T T T ht Dc

2 3984 s 300°C 275°C exp 6 75 W/m K 984 s / 2700 kg/m 0.075 m 950J/kg KT

If the product of the density and specific heat of copper is (c)Cu 8900 kg/m3 400 J/kgK = 3.56

106 J/m

3K, is there any advantage to using copper spheres of equivalent diameter in lieu of aluminum spheres?

Does the time required for a sphere to reach a prescribed state of thermal energy storage change with increasing distance from the bed inlet? If so, how and why?

984 s 272.5°CT

3

2

2700 kg/m 0.075 m 950 J/kg K/ / 6 427 s.

6 75 W/m Kt sVc hA Dc h

ANALYSIS: To determine whether a lumped capacitance analysis can be used, first compute

Bi = h(ro/3) = 75 W/m2∙K (0.0125 m)/150 W/m∙K = 0.006 << 1.

From Eq. 5.8, achievement of 90% of the maximum possible thermal energy storage corresponds to

<

<

Problem: Furnace Start-up

Problem 5.22: Heating of coated furnace wall during start-up.

KNOWN: Thickness and properties of furnace wall. Thermal resistance of ceramic coating on surface of wall exposed to furnace gases. Initial wall temperature.

FIND: (a) Time required for surface of wall to reach a prescribed temperature, (b) Corresponding value of coating surface temperature.

Schematic:

Problem: Furnace Start-up (cont.)

ASSUMPTIONS: (1) Constant properties, (2) Negligible coating thermal capacitance, (3) Negligible radiation.

PROPERTIES: Carbon steel: = 7850 kg/m3, c = 430 J/kgK, k = 60 W/mK.

ANALYSIS: Heat transfer to the wall is determined by the total resistance to heat transfer from the gas to the surface of the steel, and not simply by the convection resistance.

11

1 2 2 2tot 2

1 110 m K/W 20 W/m K.

25 W/m KfU R R

h

220 W/m K 0.01 m0.0033 0.1

60 W/m K

ULBi

k

and the lumped capacitance method can be used.

(a) From Eqs. (5.6) and (5.7),

exp / exp / exp /t t ti

T Tt t R C Ut Lc

T T

3

27850 kg/m 0.01 m 430 J/kg K 1200 1300

ln ln300 130020 W/m Ki

Lc T Tt

U T T

3886 s 1.08 h.t

Hence, with

<

Problem: Furnace Start-up (cont.)

(b) Performing an energy balance at the outer surface (s,o),

/s,o s,o s,i fh T T T T R

2 -2 2

2

/ 25 W/m K 1300 K 1200 K/10 m K/W

1/ 25 100 W/m K

s,i fs,o

f

hT T RT

h R

1220 K.s,oT

How does the coating affect the thermal time constant?

<