Transient Conduction:Spatial Effects and the

Role ofAnalytical Solutions

Chapter 5Sections 5.4 to 5.7

Lecture

10

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

• If the lumped capacitance approximation cannot be made, consideration must be given to spatial, as well as temporal, variations in temperature during the transient process.

• For a plane wall with symmetrical convection conditions and constant properties, the heat equation and initial/boundary conditions are:

2

2

1T T

x t

(5.29)

,0 iT x T (5.30)

0

0x

T

x

(5.31)

,x L

Tk h T L t T

x

(5.32)

• Existence of eight independent variables:

, , , , , , ,iT T x t T T L k h (5.33)

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

*

xx

LDimensionless space coordinate:

Dimensionless time: *2

tt Fo

L

Fourier the Nu mberFo

The Biot Number:solid

hLBi

k

* *, ,f x Fo Bi • Exact Solution:

* 2 *

1exp cosn n n

nC Fo x

(5.42a)

4sin

tan2 sin 2

nn n n

n n

C Bi

(5.42b,c)

See Appendix B.3 for first four roots (eigenvalues ) of Eq. (5.42c).1 4,...,

Plane Wall (cont.)

• The One-Term Approximation : Valid for 0.2Fo Variation of midplane temperature (x*= 0) with time : Fo

* 2

1 1expoo

i

T TC Fo

T T

(5.44)

1 1Table 5.1 and as a function of C Bi

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

* * *1coso x (5.43b)

Change in thermal energy storage with time:

stE Q (5.46a)

1 *

1

sin1o oQ Q

(5.49)

o iQ cV T T (5.47)

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 r

• One-Term Approximations:Long Rod: Eqs. (5.52) and (5.54) Sphere: Eqs. (5.53) and (5.55)

1 1, Table 5.1C

• 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

,erf

2s

i s

T x t T x

T T t (5.60)

s is

k T Tq

t

(5.61)

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

(5.62)

Case 2: Constant 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

(5.63)

Case 3: Surface Convection ,h T

Example 1 (Problem 5.34)

A 100-mm-thick steel plate (=7830 kg/m3, c=550 J/kgK,

k=48 W/mK) that is initially at uniform temperature of

Ti=200C is to be heated to a minimum temperature of 550C.

Heating is affected in a gas-fired furnace, where the products

of combustion at T = 800 C maintain a convection coefficient

of h=250 W/m2K on both surfaces of the plate, how long should the

plate be left in the furnace ? What is the surface temperature?

Example 1

Known: Dimension and properties of steel plate, convection condition.

Find: Time needed to reach Tmin of 550C, and T surface

Schematic:

Example 1

Assumptions: 1-D conduction in the plate, radiation negligible Constant properties.

Analysis:

1. Lumped Capacitance Method (Need Bi<0.1)

1.026.0)48/05.0*250( k

hLBi c

sh

cLct 3.861)250/05.0*550*7830(

sTT

TTt

it 0.754)]800200/()800550ln[(*3.861ln

Example 1

Analysis:

2. Approximate Solution (Need Fo>0.2)

The minimum T is at the center of the plate

26.0)48/05.0*250( k

hLBi c

)exp(800200

800550 211

0*0 FoC

TT

TT

i

From Table 5.1, Bi=0.26, ζ1=0.488 rad, C1=1.0396

Solve Fo=3.839 >0.2

Example 1

Example 1

Analysis:

Fo=3.839 =(αt/L2)

α = k/(c) = 48/(7830*550)=1.1146x10-5 m2/s

t =Fo*L2/α

t= Fo*L2/α = 3.839*(0.05)2/1.11x10-5 = 861 s

Example 1

Analysis:

)cos( 1*

0*

TT

TT

i

LL

4167.0800200

8005500*0

TT

TT

i

368.0)488.0cos(*4167.0800200

800*

radTL

L

Solve TL =579.2 C

Example 1

Analysis:

3. Heisler Chart (Figure D.1, page 948)

The minimum T is at the center of the plate

26.0)48/05.0*250( k

hLBi c

4167.0800200

8005500*0

TT

TT

i

Get Fo value from Figure D.1, page 948

Fo ≈ 3.8, t = 852 s

85.31 Bi

Example 1

Example 1

Fo

Bi-1

*0

Example 1

Analysis:

4. The surface T can be determined by using Heisler Chart

26.0Bi

800550

80087.0

00

1

LL T

TT

TT

Get θ1/ θ0 value from Figure D.2, page 949

Solve TL =582.5 C

85.31 Bi

Example 1

Methods for Solving Transient Problems

1. Check Bi <0.1? Or Assume Bi<0.1, analytical solution, then

check Bi<0.1?

2. Bi>0.1, Fo>0.2?, Approximate solution (P257), Table 5.1 (P274)

or Bi>0.1, Fo>0.2, Approximate solution , Heisler Chart (Fig. D1-

D9, 948-952) (Pay attention to the B.C. and I.C. in Figure 5.6)

3. Bi>0.1, Fo<0.2, Exact solution (Eqn. 5.39)

SCHEMATIC:

Pyrex sphere D = 75 m m ,

T = 25 Ci o

Gas

T Cg,i o= 300

h = 75 W /m -K2

= 2225 kg/m 3

k = 1.4 W /m -Kc = 835 J/kg-K

Problem: Thermal Energy Storage

Problem 5.80: 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.

Problem: Thermal Energy Storage (cont.)

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/m2K (0.0125m)/1.4 W/mK = 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,

*.o

From Eq. (5.55),

31

1 1 11

3 sin cosoo

Q

Q

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

3

o0.1 2.03 0.837

0.1553 0.896 2.03 0.443 5.386

Problem: Thermal Energy Storage (cont.)

From Eq. (5.53c), the corresponding time is

2

211

lno ort

C

3 7 2/ 1.4 W/m K / 2225kg/m 835J/kg K 7.54 10 m /s,k c

2

27 2

0.0375 ln 0.155 /1.481,020 s

7.54 10 m /s 2.03

mt

From the definition of *,o the center temperature is , ,0.155 300°C 42.7°C 257.3°Co g i i g iT T T T

The surface temperature at the time of interest may be obtained from Eq. (5.53b)

with 1,r

1, ,

1

sin 0.155 0.896300°C 275°C 280.9°C

2.03o

s g i i g iT T T T

Is use of the one-term approximation appropriate?

<

<

<

Problem: Thermal Response Firewall

Problem: 5.93: Use of radiation heat transfer from high intensity lamps for a prescribed duration (t=30 min) to assessability 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.

q s = 10 W /m 24

L = 0.25 m x

Concrete, T = 25ioC

= 2300 kg/m 3

c = 880 J/kg-Kk = 1.4 W /m -Ks = 1.0

T Cm ax o= 325 T Cm ax

o= 25

SCHEMATIC:

Problem: Thermal Response of Firewall (cont.)

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. (5.62)

1/2 22 /, exp erfc

4 2o o

iq t q xx x

T x t Tk t k t

where, 7 2/ 6.92 10 m /spk c and for

1/230 min 1800 s, 2 / / 284.5 K.ot q t k Hence, at x = 0,

0,30 min 25°C 284.5°C 309.5°C 325°CT At

1/220.25 m, / 4 12.54; / 1,786 K, and / 2 3.54.ox x t q x k x t

Hence, 60.25 m, 30 min 25 284.5°C 3.58 10 1786°C ~ 0 25°CT C

Problem: Thermal Response of Firewall (cont.)

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 = 1 and Tsur = 298K, radiation exchange at Ts = 309.5C would be

4 4 2rad sur 6080 W/m K,sq T T

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.

<

Beef, 1kg

Ti = -20°C

Packaging material, q

Beef, 1kg

Ti = -20°C

Packaging material, q

Problem: Microwave Heating

Problem: 5.101: 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:

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 (cont.)

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

2

0.5 = =

4πs

s

q Pq

A R

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

1/31/3

3

4 = = π 3

3 3 1 kg = = = 0.0638 m4π 4π 920 kg/m

o

o

m rV r

mr

Thus

2

20.5(1000W)

9780 W/m4 0.0638 m

sq

29780 W/m 0.0638 m

* 16.61.88 W/m K 0 C - -20 C

s o

s i

q rq

k T T

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.

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

1 π π* -

2 4q

Fo

-2π

= π 2( * + ) = 0.00264

Fo q

Problem: Microwave Heating (cont.)

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

2 2

2 3

= / =

= (0.0026 × (0.0638 m) × 920 kg/m × 2040 J/kg K)/(1.88 W/m K) o ot Fo r Fo r c / k

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.

= 10.6 s <