hw #3 /tutorial # 3 wrf chapter 17; wwwr chapter 18 id chapter 5 tutorial # 3 wwwr #18.12...
DESCRIPTION
Transient Conduction AnalysisTRANSCRIPT
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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; 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.
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Unsteady-State Conduction
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Transient Conduction Analysis
pCqT
tT
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
x
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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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Heating a Body with Finite Surface and Internal Resistance
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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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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
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Transient (Unsteady – State) Conduction Summary
i) Calculate Biot Modulus (Bi)
kA
VhBi
if Bi ≤ 0.1 → Lumped Parameter Analysis
TTTT
tAVc
h 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
TTTTo or 2
1xt
And read off 21xt or
TTTTo
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 =
oTTTT
X = 21xt
n = 1xx m =
1hxk
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
0xT (At centerline) and
TTkh
xT (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
txerf
TTTT
oS
S
2 or
txerf
TTTT
oS
o
21
For Heat transfer into a semi – infinite medium with finite surface resistance
tx
ktherf
kth
khx
txerf
TTTT
o
21exp
2 2
2
(18 – 21)