Download Spherical metallic specimen, initially at uniform temperature, T 0

HW #3 /Tutorial # 3
WRF Chapter 17; WWWR Chapter 18
ID Chapter 5
• Tutorial # 3
• WWWR #18.12
(additional data: h =
6W/m2-K); WRF#17.1;
WWWR#18.4;
WRF#17.10 ; WRF#17.14.
• To be discussed during the
week 1-5 Feb., 2016.
• By either volunteer or
class list.
• Homework # 3 (Self
practice)
• WRF #17.9;
WRF#17.16.
• ID # 5.6, 5.9.
Unsteady-State Conduction
Transient Conduction Analysis

T
q
2
  T 
t
C p
Spherical metallic specimen, initially at uniform temperature, T0
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
V/A = (WHL)/(2WH)=L/2
BC (1) -> C1=0
BC (2) -> l = np/L
Fo = t/(L/2)2
IC -> Fourier expansion of Yo(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 

hV
A

Courtesy contribution by ChBE Year
Representative, 2006.
k
if Bi ≤ 0.1
→
Lumped Parameter Analysis
h
c pV
tA
 To  T
ln 
 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
And read off
To find
 To  T

 T T


t
2
1
x
t




or
t
x12
or
 To  T

 T T


or
T




if 0.1 ≤ Bi ≤ 100 →
Use appendix F of W3R ( refer to examples 18.3 and 18.4)
 T  T 
t
Using
Y= 
X=
T  T 

x12
o 
 
x
k
n=
m=
x1
hx1
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
 0 (At centerline)
x
iii)
and

T h
 T  T  (At surface)
x k
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
TS  T
 x
 erf 
TS  To
 2 t



or
T  To
 x
 1  erf 
TS  To
 2 t



For Heat transfer into a semi – infinite medium with finite surface resistance
T  T
 x
 erf 
T  To
 2 t
 h t
 hx h 2t  

x 

  exp  2  1  erf 

k  
2 t 

 k
 k
(18 – 21)