HW #3 /Tutorial # 3
WRF Chapter 17; WWWR Chapter 18
ID Chapter 5
• Tutorial # 3
• WWWR #18.12
(additional data: h =
6W/m 2 -K); WRF#17.1;
WWWR#18.4;
WRF#17.10 ; WRF#17.14.
• To be discussed during the week 2-6 Feb., 2015.
• By either volunteer or class list.
• Homework # 3 practice)
• WRF #17.9;
WRF#17.16.
• ID # 5.6, 5.9.
(Self

Unsteady-State Conduction

Transient Conduction Analysis

T
 t
   2
T

 q

C p

Spherical metallic specimen, initially at uniform temperature, T
0
Energy balance requires

Large value of Bi
•Indicates that the conductive resistance controls
•There is more capacity for heat to leave the surface by convection than to reach it by conduction
Small value of Bi
•Internal resistance is negligibly small
•More capacity to transfer heat by conduction than by convection

Example 1
(WWWR Page 266)
• A long copper wire, 0.635cm in diameter, is exposed to an air stream at a temperature of
310K. After 30 s, the average temperature of the wire increased from 280K to 297K.
Using this information, estimate the average surface conductance, h.

Example 1

Heating a Body Under Conditions of Negligible Surface Resistance

IC
BC(1)
BC(2) x
BC (1) -> C
F o
1
=0
BC (2) -> l
= n p
/L
=
 t/(L/2) 2
V/A = (WHL)/(2WH)=L/2
IC -> Fourier expansion of Y o
(x) …..>
Equation (18-12) Engineering
Mathematics: PDE

Detailed Derivation for Equations 18-12, 18-13
Courtesy by all CN5 students, presented by Lim
Zhi Hua, 2003-2004

Detailed Derivation for Equations 18-12, 18-13
Courtesy by all CN5 students, presented by Lim Zhi Hua, 2003-2004

Example 2

Heating a Body with Finite
Surface and Internal Resistance

Heat Transfer to a Semi-Infinite
Wall

Temperature-Time Charts for
Simple Geometric Shapes

Example 3

or Figure F.4

Example 4

WWWR 18-12; 18-13
WWWR 18-16
(a) T=Ts @ x =0 WWWR 18-20
(b) -k dT/dx = h (T-T∞) @ x =0
WWWR 18-21

Transient (Unsteady – State) Conduction Summary i) Calculate Biot Modulus (Bi)
Bi
 if Bi ≤ 0.1
h
 
A k
→
Courtesy contribution by ChBE Year
Representative, 2006.
Lumped Parameter Analysis h

 c p
V tA ln


T o
T


T

T


 if Bi ≥ 100
→
There is temperature variation within the object. If the geometry of the solid objects falls into the 6 shapes given in Fig. 18.3
→ Use figure 18.3 to calculate the temperature at the specific time.
Calculate


T o
T


T

T


 or
 t x
1
2
And read off
 t x
1
2 or


T o
T


T

T



To find t or T if 0.1 ≤ Bi ≤ 100
→
Use appendix F of W
3
R ( refer to examples 18.3 and 18.4)
Using Y =


T
T




T
T o

 X =
 t x
1
2 x n = x
1 k m = hx
1

ii) Slab Heating
Heating of Body under negligible surface resistance.
Check Bi no. and let m = 0.
Heating a body with finite surface and internal resistance

T
 x

0 (At centerline) and


T
 x
 k h 
T

T


(At surface) iii) Heat transfer into a semi – infinite wall
Different from (ii) because there is no defined length scale
Use Appendix L
For Heat transfer into a semi – infinite medium with negligible surface resistance
T
T
S
S


T
T o
 erf


2 x
 t

 or
T
T
S


T o
T o

1
 erf


2 x
 t


For Heat transfer into a semi – infinite medium with finite surface resistance
T
T




T
T o
 erf


2 x
 t


 exp

 hx k
 h
2
 k
2 t






1
 erf

 h k
 t

2 x
 t





(18 – 21)