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HW #3 /Tutorial # 3WRF Chapter 17; WWWR Chapter 18
ID Chapter 5
• Tutorial # 3
• WWWR #18.12, (additional data: h = 6W/m2-K); WRF#17.1; WRF#17.10 ; WRF#17.14.
• To be discussed on Jan 31, 2017.
• By either volunteer or class list.
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Unsteady-State Conduction
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Transient Conduction Analysis
pC
qT
t
T
2
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Spherical metallic specimen, initially at uniform temperature, T0
Energy balance requires
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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 conductionSmall value of Bi•Internal resistance is negligibly small•More capacity to transfer heat by conduction than by convection
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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.
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Example 1
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Heating a Body Under Conditions of Negligible Surface Resistance
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BC (1) -> C1=0BC (2) -> = n/LFo = t/(L/2)2
IC -> Fourier expansion of Yo(x) …..> Equation (18-12) Engineering Mathematics: PDE
BC(1)
BC(2)
IC
V/A = (WHL)/(2WH)=L/2
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Detailed Derivation for Equations 18-12, 18-13
Courtesy by all CN5 students, presented by Lim Zhi Hua, 2003-2004
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Detailed Derivation for Equations 18-12, 18-13
Courtesy by all CN5 students, presented by Lim Zhi Hua, 2003-2004
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Example 2
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Heat Transfer to a Semi-Infinite Wall
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Temperature-Time Charts for Simple Geometric Shapes
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Example 3
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or Figure F.4
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Example 4
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Transient (Unsteady – State) Conduction Summary
i) Calculate Biot Modulus (Bi)
kA
VhBi
if Bi ≤ 0.1 → Lumped Parameter Analysis
TT
TT
tA
Vch op ln
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
TT
TTo or 21x
t
And read off 21x
t or
TT
TTo
To find t or T if 0.1 ≤ Bi ≤ 100 → Use appendix F of W3R ( refer to examples 18.3 and 18.4)
Using Y =
oTT
TT X =
21x
t
n = 1x
x m =
1hx
k
Courtesy contribution by ChBE Year Representative, 2006.
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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
0x
T (At centerline) and
TTk
h
x
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
xerf
TT
TT
oS
S
2 or
t
xerf
TT
TT
oS
o
21
For Heat transfer into a semi – infinite medium with finite surface resistance
t
x
k
therf
k
th
k
hx
t
xerf
TT
TT
o
21exp
2 2
2
(18 – 21)