Transient Conduction:Spatial Effects and the Role of

Analytical Solutions

Chapter 5

Sections 5.4 through 5.8

Plane Wall Solution to the Heat Equation for a Plane Wall withSymmetrical Convection Conditions

• If the lumped capacitance approximation can not be made, consideration mustbe given to spatial, as well as temporal, variations in temperature during thetransient process.

• For a plane wall with symmetrical convectionconditions and constant properties, the heatequation and initial/boundary conditions are:

2

2

1T T

x tα∂ ∂

=∂ ∂ (5.26)

( ),0 iT x T= (5.27)

0

0x

T

x =

∂=

∂(5.28)

( ),x L

Tk h T L t T

x ∞=

∂⎡ ⎤− = −⎣ ⎦∂

(5.29)

• Existence of seven independent variables:

( ), , , , , ,iT T x t T T k hα∞= (5.30)

How may the functional dependence be simplified?

Plane Wall (cont.)

• Non-dimensionalization of Heat Equation and Initial/Boundary Conditions:

Dimensionless temperature difference: *

i i

T T

T T

θθθ

∞

∞

−≡ =

−*

xx

L≡Dimensionless coordinate:

Dimensionless time: *2

tt Fo

L

α≡ ≡

Fourierthe Nu mberFo →

The Biot Number:solid

hLBi

k≡

( )* *, ,f x Fo Biθ =

• Exact Solution:

( ) ( )* 2 *

1exp cosn n n

nC Fo xθ ζ ζ

∞

== −∑ (5.39a)

( )4sin

tan2 sin 2

nn n n

n n

C Biζ ζ ζ

ζ ζ= =

+(5.39b,c)

See Appendix B.3 for first four roots (eigenvalues ) of Eq. (5.39c)1 4,...,ζ ζ

Plane Wall (cont.)

• The One-Term Approximation :( )0.2Fo >

Variation of midplane temperature (x*= 0) with time : ( )Fo

( )( ) ( )* 2

1 1expoo

i

T TC Fo

T Tθ ζ∞

∞

−≡ ≈ −

−(5.41)

1 1Table 5.1 and as a function of C Biζ→

Variation of temperature with location (x*) and time : ( )Fo

( )* * *1coso xθ θ ζ= (5.40b)

Change in thermal energy storage with time:

stE QΔ = − (5.43a)

1 *

1

sin1o oQ Q

ζ θζ

⎛ ⎞= −⎜ ⎟

⎝ ⎠(5.46)

( )o iQ cV T Tρ ∞= − (5.44)

Can the foregoing results be used for a plane wall that is well insulated on oneside and convectively heated or cooled on the other?

Can the foregoing results be used if an isothermal condition is instantaneously imposed on both surfaces of a plane wall or on one surface ofa wall whose other surface is well insulated?

( )s iT T≠

Heisler Charts

Graphical Representation of the One-Term ApproximationThe Heisler Charts, Section 5 S.1

• Midplane Temperature:

Heisler Charts (cont.)

• Temperature Distribution:

• Change in Thermal Energy Storage:

Radial Systems

Radial Systems• Long Rods or Spheres Heated or Cooled by Convection.

2

/

/o

o

Bi hr k

Fo t rα=

=

• One-Term Approximations:Long Rod: Eqs. (5.49) and (5.51)

Sphere: Eqs. (5.50) and (5.52)

1 1, Table 5.1C ζ →

• Graphical Representations:Long Rod: Figs. 5 S.4 – 5 S.6

Sphere: Figs. 5 S.7 – 5 S.9

Semi-Infinite Solid

The Semi-Infinite Solid• A solid that is initially of uniform temperature Ti and is assumed to extend

to infinity from a surface at which thermal conditions are altered.

• Special Cases:Case 1: Change in Surface Temperature (Ts)

( ) ( )0, ,0s iT t T T x T= ≠ =

( ), xerf

2 ts

i s

T x t T

T T α− ⎛ ⎞

= ⎜ ⎟− ⎝ ⎠ (5.57)

( )s is

k T Tq

tπα−

′′ =(5.58)

Semi-Infinite Solid (cont.)

( ) ( )1

2 22 /, exp

4

erfc2

oi

o

q t xT x t T

k t

q x x

k t

α πα

α

′′ ⎛ ⎞− = −⎜ ⎟⎝ ⎠

′′ ⎛ ⎞− ⎜ ⎟⎝ ⎠ (5.59)

Case 2: Uniform Heat Flux ( )s oq q′′ ′′=

( )0

0,x

Tk h T T t

x ∞=

∂⎡ ⎤− = −⎣ ⎦∂

( )

2

2

,erfc

2

exp erfc2

i

i

T x t T x

T T t

hx h t x h t

k k kt

α

α αα

∞

− ⎛ ⎞= ⎜ ⎟− ⎝ ⎠⎡ ⎤⎛ ⎞⎡ ⎤⎛− + +⎢ ⎥⎜ ⎟⎜⎢ ⎥ ⎜⎝⎣ ⎦ ⎢ ⎥⎠⎝⎣ ⎦ (5.60)

Case 3: Convection Heat Transfer ( ),h T∞

Multidimensional Effects

Multidimensional Effects• Solutions for multidimensional transient conduction can often be expressed

as a product of related one-dimensional solutions for a plane wall, P(x,t),an infinite cylinder, C(r,t), and/or a semi-infinite solid, S(x,t). See Equations (5 S.1) to (5 S.3) and Fig. 5 S.11.

• Consider superposition of solutions for two-dimensional conduction in ashort cylinder:

( ) ( ) ( )

( ) ( )

, ,, ,

,

i

Plane Infinitei iWall Cylinder

T r x t TP x t C r t

T T

T x t T T r,t T

T T T T

∞

∞

∞ ∞

∞ ∞

−= ×

−

− −= ×

− −

Objects with Constant Ts or qs

• Transient response of a variety of objects to a step change in surface temperature or heat flux can be unified by defining the dimensionless conduction heat rate:

Objects with Constant Surface Temperatures or Surface Heat Fluxes

where Lc is a characteristic length that depends on the geometry of the object.

• Consider the variation of q* with time, or Fo, for

– Interior heat transfer: Heat transfer inside objects such as plane walls, cylinders, or spheres,

– Exterior heat transfer: Heat transfer in an infinite medium surrounding an embedded object.

( )*

′′=

−s c

s i

q Lq

k T T(5.64)

• When q* is plotted versus Fo in Figure 5.10, we see that:

– All objects behave the same as a semi-infinite solid for short times.

– q* approaches a steady state for exterior objects.

– q* does not reach a steady state for interior objects, but decreases continually with time (Fo).

Objects with Constant Ts or qs, Cont.

Constant Ts Constant qs

• Why do all objects behave the same as a semi-infinite solid for short times?

Objects with Constant Ts or qs, Cont.

• Easy-to-use approximate solutions for q*(Fo) are presented in Table 5.2 for all the cases presented in Figure 5.10.

• As an example of the use of Table 5.2, consider:

– Infinite cylinder initially at Ti has constant heat flux imposed at its surface.

– Find its surface temperature as a function of time.

• Look in Table 5.2b for constant surface heat flux, Interior Cases, Infinite cylinder.

– Length scale is Lc = ro, the cylinder radius.

– Exact solution for q*(Fo) is a complicated infinite series.

– Approximate solution is given by:

Approximate Solutions for Objects with Constant Ts or qs

• It is then a simple matter to find Ts from the definition,

11 1

* for 0.2 or * 2 for 0.22 8 4

−⎡ ⎤= − < = + ≥⎢ ⎥⎣ ⎦

q Fo q Fo FoFo

π π

( )*

′′=

−s c

s i

q Lq

k T T

Problem: Thermal Energy Storage

Problem 5.63: Charging a thermal energy storage system consisting ofa packed bed of Pyrex spheres.

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

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

SCHEMATIC:

Problem: Thermal Energy Storage

ASSUMPTIONS: (1) One-dimensional radial conduction in sphere, (2) Negligible heat transfer to or from a sphere by radiation or conduction due to contact with adjoining spheres, (3) Constant properties.

ANALYSIS: With Bi ≡ h(ro/3)/k = 75 W/m2⋅K (0.0125m)/1.4 W/m⋅K = 0.67, the lumped capacitance method is inappropriate and the approximate (one-term) solution for one-dimensional transient conduction in a sphere is used to obtain the desired results.

To obtain the required time, the specified charging requirement ( )/ 0.9oQ Q = must first be used to obtain the dimensionless center temperature,

*.oθ

From Eq. (5.52),

( ) ( )

31

oo1 1 1

Q1

Q3 sin cos

ζθζ ζ ζ

∗ ⎛ ⎞= −⎜ ⎟−⎡ ⎤ ⎝ ⎠⎣ ⎦

With Bi ≡ hro/k = 2.01, 1 2.03ζ ≈ and C1 ≈ 1.48 from Table 5.1. Hence,

( )

( )

3

o0.1 2.03 0.837

0.1555.3863 0.896 2.03 0.443

θ ∗ = = =− −⎡ ⎤⎣ ⎦

Problem: Thermal Energy Storage

From Eq. (5.50c), the corresponding time is

2o o

211

rt ln

C

θαζ

∗⎛ ⎞= − ⎜ ⎟⎜ ⎟

⎝ ⎠

( )3 7 2k / c 1.4 W / m K / 2225 kg / m 835 J / kg K 7.54 10 m / s,α ρ −= = ⋅ × ⋅ = ×

( ) ( )

( )

2

27 2

0.0375m ln 0.155/1.48t 1,020s

7.54 10 m /s 2.03−= − =

×

From the definition of *,oθ the center temperature is ( )o g,i i g,iT T 0.155 T T 300 C 42.7 C 257.3 C= + − = ° − ° = °

The surface temperature at the time of interest may be obtained from Eq. (5.50b) with r 1,∗ =

( ) ( )o 1s g,i i g,i

1

sin 0.155 0.896T T T T 300 C 275 C 280.9 C

2.03

θ ζζ

∗ ×⎛ ⎞= + − = ° − ° = °⎜ ⎟⎝ ⎠

Is use of the one-term approximation appropriate?

Problem: Thermal Response Firewall

Problem: 5.79: Use of radiation heat transfer from high intensity lampsfor a prescribed duration (t=30 min) to assess

ability of firewall to meet safety standards corresponding tomaximum allowable temperatures at the heated (front) andunheated (back) surfaces.

( )4 210 W/msq′′ =

KNOWN: Thickness, initial temperature and thermophysical properties of concrete firewall. Incident radiant flux and duration of radiant heating. Maximum allowable surface temperatures at the end of heating.

FIND: If maximum allowable temperatures are exceeded.

SCHEMATIC:

Problem: Thermal Response of Firewall

ASSUMPTIONS: (1) One-dimensional conduction in wall, (2) Validity of semi-infinite medium approximation, (3) Negligible convection and radiative exchange with the surroundings at the irradiated surface, (4) Negligible heat transfer from the back surface, (5) Constant properties.

ANALYSIS: The thermal response of the wall is described by Eq. (5.59)

( ) ( )1/ 2 2o o

i2 q t / q xx x

T x, t T exp erfck 4 t k 2 t

α πα α

⎛ ⎞′′ ′′− ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

where, 7 2pk / c 6.92 10 m / sα ρ −= = × and for

( )1 / 2ot 30 min 1800s, 2q t / / k 284.5 K.α π′′= = = Hence, at x = 0,

( )T 0,30 min 25 C 284.5 C 309.5 C 325 C= ° + ° = ° < °

At

( ) ( )1 / 22ox 0.25m, x / 4 t 12.54, q x / k 1, 786K, and x / 2 t 3.54.α α′′= − = − = =

Hence,

( ) ( ) ( )6T 0.25m, 30 min 25 C 284.5 C 3.58 10 1786 C ~ 0 25 C−= ° + ° × − ° × ≈ °

Problem: Thermal Response of Firewall

Both requirements are met.

Is the assumption of a semi-infinite solid for a plane wall of finite thickness appropriate under the foregoing conditions?

COMMENTS: The foregoing analysis may or may not be conservative, since heat transfer at the irradiated surface due to convection and net radiation exchange with the environment has been neglected. If the emissivity of the surface and the temperature of the surroundings are assumed to be ε = 1 and Tsur = 298K, radiation exchange at Ts = 309.5°C would be

( )4 4 2rad s surq T T 6, 080 W / m K,εσ′′ = − = ⋅

which is significant (~ 60% of the prescribed radiation). However, under actual conditions, the wall would likely be exposed to combustion gases and adjoining walls at elevated temperatures.

Problem: Microwave Heating

Problem: 5.87: Microwave heating of a spherical piece of frozen ground beef using microwave-absorbing packaging material.

KNOWN: Mass and initial temperature of frozen ground beef. Rate of microwave power absorbed in packaging material.

FIND: Time for beef adjacent to packaging to reach 0°C.

SCHEMATIC: Beef, 1kg

Ti = -20°C

Packaging material, &q

Beef, 1kg

Ti = -20°C

Packaging material, &q

ASSUMPTIONS: (1) Beef has properties of ice, (2) Radiation and convection to environment are neglected, (3) Constant properties, (4) Packaging material has negligible heat capacity.

Problem: Microwave Heating

PROPERTIES: Table A.3, Ice (≈ 273 K): ρ = 920 kg/m3, c = 2040 J/kg·K, k = 1.88 W/m·K.

ANALYSIS: Neglecting radiation and convection losses, all the power absorbed in the packaging material conducts into the beef. The surface heat flux is

s 2s

q 0.5Pq = =

A 4πR′′

&

The radius of the sphere can be found from knowledge of the mass and density: 3o

1/31/3

o 3

4m = ρV = ρ π r

3

3 m 3 1 kgr = = = 0.0638 m

4π ρ 4π 920 kg/m

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

Thus

2s 2

0.5(1000 W)q = = 9780 W/m

4π×(0.0638 m)′′

The beef can be seen as the interior of a sphere with a constant heat flux at its surface, thus the relationship in Table 5.2b, Interior Cases, sphere, can be used. We begin by calculating q* for Ts = 0°C.

2

s o

s i

q r 9780 W/m × 0.0638 mq* = = = 16.6

k(T - T ) 1.88 W/m K(0°C - ( - 20°C))

′′⋅

Problem: Microwave Heating

We proceed to solve for Fo. Assuming that Fo < 0.2, we have

* 1 π πq -

2 Fo 4≅

-2π

Fo = π 2(q* + ) = 0.00264

⎡ ⎤⎢ ⎥⎣ ⎦

Since this is less than 0.2, our assumption was correct. Finally we can solve for the time:

2 2o o

2 3

t = Fo r /α = Fo r ρc/k

= (0.0026 × (0.0638 m) × 920 kg/m × 2040 J/kg K)/(1.88 W/m K) ⋅ ⋅

= 10.6 s COMMENTS: At the minimum surface temperature of -20°C, with T∞ = 30°C and h = 15 W/m2·K from Problem 5.33, the convection heat flux is 750 W/m2, which is less than 8% of the microwave heat flux. The radiation heat flux would likely be less, depending on the temperature of the oven walls.