Materials Selection Case Study 2 Advance Properties and ... Study 2.pdf · Deflection-limited...
Transcript of Materials Selection Case Study 2 Advance Properties and ... Study 2.pdf · Deflection-limited...
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Monday, November 13, 2017
Materials Selection – Case Study 2Advance Properties and Concepts
Professors: Anne Mertens and Davide Ruffoni
Assistant:Tommaso Maurizi Enrici
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Materials Selection 2Monday, November 13, 2017
Outline
• Case Study 13: Load-limited design, Energy-limited design and Deflection-limited design
• Case Study 14: Safe Pressure VesselsMultiple constraints• Case Study 14’: Light pressure Vessels with MULTIPLE CONSTRAINT
APPROACHTheory: Method of the weight factorTheory: Enhanced Digital Logic (EDL)• Case Study 15: Precision devices• Case Study 16: Long Span Transmission line• Case Study 17: Light Cable• Case Study 18: Kiln Walls• Case Study 19: Insulation for Short-Term isothermal containers• Case Study 20: Process for a CanSecret slides : Kubota case
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Materials Selection 3Monday, November 13, 2017
The Fracture-limited design
Fracture-limited design is importantto control crack propagation
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Materials Selection 4Monday, November 13, 2017
Ashby Diagrams
<15 MPa.m1/2
Materials to avoid among mechanical engineers
But they use polymers all the timeWhy???
Ceramics (1-6 MPa.m1/2) view with deep suspicion
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Materials Selection 5Monday, November 13, 2017
The Fracture-limited designThe resistance of a material to the propagation of a crack is measuredby its plane-strain fracture toughness value (K1c)
𝜎𝑓 =𝐾1𝑐 ∙ 𝐶
𝜋 ∙ 𝑎𝑐
When a brittle material is deformed,
it deflects elastically until it fractures.
The stress at which this happens is:
Usually C=1
ac = the length of the largest crack
𝐾1 = stress intensity factor
𝐾1𝑐 = 𝐾1 crack propagates
NDT = non-destructive testing
𝜎𝑓 𝑏𝑦 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛,
To consider all the stresses
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Materials Selection 6Monday, November 13, 2017
Case Study 13: Load-limited design
Objective • Minimize the volume (mass, cost)
Constraints • Design load specified
Free Variables • Choice of material
𝜎𝑓 =𝐾1𝑐 ∙ 𝐶
𝜋 ∙ 𝑎𝑐
𝐶𝑎𝑟𝑟𝑦 𝑎 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑 𝑙𝑜𝑎𝑑𝑜𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑓𝑟𝑎𝑐𝑡𝑢𝑟𝑖𝑛𝑔?
𝑬𝒂𝒔𝒚F, 𝑝 𝐾1𝑐
𝑴𝒆𝒕𝒂𝒍𝒔,𝒑𝒐𝒍𝒚𝒎𝒆𝒓 −𝒎𝒂𝒕𝒓𝒊𝒙 𝒄𝒐𝒎𝒑𝒐𝒔𝒊𝒕𝒆𝒔
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Materials Selection 7Monday, November 13, 2017
Case Study 13: Energy-limited design
𝑊𝑒𝑙 =𝜎𝑦
2
2𝐸
Objective • Minimize the volume (mass, cost)
Constraints • Design energy specified
Free Variables • Choice of material
Springs FlywheelsContainment systemsfor turbines
𝜎𝑓 =𝐾1𝑐 ∙ 𝐶
𝜋 ∙ 𝑎𝑐
𝑊𝑚𝑎𝑥 =𝐶2
2𝜋𝑎𝑐∙𝐾1𝑐
2
𝐸
Elastic energy
𝑊𝑚𝑎𝑥 𝐾1𝑐2
𝐸𝑀2 =
𝐾1𝑐2
𝐸≈ 𝐺𝑐 Toughness
for a given flaw size
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Materials Selection 8Monday, November 13, 2017
Ductile-Brittle Temperature
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Materials Selection 9Monday, November 13, 2017
Case Study 13: Energy-limited design
𝑊𝑚𝑎𝑥 𝐾1𝑐2
𝐸
𝑴𝒆𝒕𝒂𝒍𝒔, 𝒄𝒐𝒎𝒑𝒐𝒔𝒊𝒕𝒆𝒔𝒂𝒏𝒅 𝒔𝒐𝒎𝒆 𝒑𝒐𝒍𝒚𝒎𝒆𝒓𝒔
Jc > 1 kJ/m2
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Materials Selection 10Monday, November 13, 2017
Case Study 13: Deflection-limited design
Objective • Minimize the volume (mass, cost)
Constraints • Design deflection specified
Free Variables • Choice of material
Snap on bottle tops Snap-together fasteners𝑊𝑒𝑙 =
𝜎𝑦2
2𝐸
𝜎𝑓 =𝐾1𝑐 ∙ 𝐶
𝜋 ∙ 𝑎𝑐
𝜀 =𝜎
𝐸 𝜀𝑓 =𝐶
𝜋𝑎𝑐∙𝐾1𝑐𝐸
𝜀𝑓𝐾1𝑐
𝐸
𝑀2 =𝐾1𝑐𝐸
We want a large failure strain
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Materials Selection 11Monday, November 13, 2017
Case Study 13: Deflection-limited design
𝜀𝑓𝐾1𝑐
𝐸
𝑷𝒐𝒍𝒚𝒎𝒆𝒓𝒔, 𝒆𝒍𝒂𝒔𝒕𝒐𝒎𝒆𝒓𝒔𝒂𝒏𝒅 𝒕𝒐𝒖𝒈𝒉𝒆𝒔𝒕 𝒎𝒆𝒕𝒂𝒍𝒔
𝑲𝟏𝒄
𝑬> 10-3 m1/2
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Materials Selection 12Monday, November 13, 2017
The Fracture-limited design
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Materials Selection 13Monday, November 13, 2017
Case Study 14: Safe Pressure Vessels
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Materials Selection 14Monday, November 13, 2017
Case Study 14: Safe Pressure Vessels
Two different approaches
Leak-Before-Break (LBB)Yield-Before-Break (YBB)
cc*
σ
σffσf = σy
No failure, but distortion
Small vessels are designed
to allow general yield
Safe design
Smallest crack has a thickness greater
than the thickness of the vessel wall
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Materials Selection 15Monday, November 13, 2017
Case Study 14: Safe Pressure VesselsYBB
Two different approaches
Yield-Before-Break (YBB)
cc*
σ
σffσf = σy
No failure, but distortion
Small vessels are designed
to allow general yield
a
F , pOn use
Distortion
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Materials Selection 16Monday, November 13, 2017
Case Study 14: Safe Pressure Vessels YBB
Objective • Maximize safety (YBB)
Constraints • R radius specified
Free Variables • Choice of material
𝜋 ∙ 2 ∙ 𝑐∗ ≤ 𝐶2 ∙𝐾1𝑐𝜎𝑦
2
But notFail-safe
𝑀1 =𝐾1𝑐𝜎𝑦
𝜎 =𝑝 ∙ 𝑅
2𝑡
𝑝 ≤2𝑡 ∙ 𝐾1𝑐
𝑅 ∙ 𝜋 ∙ 𝑎∗𝑐𝑝 𝐾1𝑐
𝑐∗𝐾1𝑐𝜎𝑦
𝑎𝑐 Tolerable crack size
𝜎 ≤𝐶 ∙ 𝐾1𝑐
𝜋 ∙ 𝑎∗𝑐
If the inspection is faulty for some reasons the crack length is greater
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Materials Selection 17Monday, November 13, 2017
Case Study 14: Safe Pressure Vessels YBB
𝑡 =𝑝 ∙ 𝑅
2𝜎Since 𝑡 𝜎𝑦
𝐹𝑜𝑟 𝑇ℎ𝑖𝑛𝑛𝑒𝑟 𝑡
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Materials Selection 18Monday, November 13, 2017
Case Study 14: Safe Pressure Vessels YBB
𝑆𝑚𝑎𝑙𝑙 𝑏𝑜𝑖𝑙𝑒𝑟𝑠(𝐻𝑎𝑟𝑑 𝑑𝑟𝑎𝑤𝑛 𝑐𝑜𝑝𝑝𝑒𝑟)
𝐿𝑜𝑤 𝑎𝑙𝑙𝑜𝑦𝑠 𝑠𝑡𝑒𝑒𝑙 = 𝐶𝑙𝑎𝑠𝑠𝑖𝑐𝑎𝑙 𝑐ℎ𝑜𝑖𝑐𝑒
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Materials Selection 19Monday, November 13, 2017
Case Study 14: Safe Pressure VesselsLBB
Two different approaches
Two different approaches
cc*
σ
σffσf = σy
b
F , pOn use
Leak-Before-Break(LBB)
Safe design
Smallest crack has a thickness greater
than the thickness of the vessel wall
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Materials Selection 20Monday, November 13, 2017
Case Study 14: Safe Pressure Vessels LBB
Objective • Maximize safety (LBB)
Constraints • R radius specified
Free Variables • Choice of material
𝑀2 =𝐾1𝑐
2
𝜎𝑦𝜎𝑦 =
𝑝 ∙ 𝑅
2𝑡𝑝 ≤
4𝐶2
𝜋𝑅∙𝐾1𝑐
2
𝜎𝑦
𝑝𝐾1𝑐
2
𝜎𝑦
𝑎𝑐 Tolerable crack size
𝜎 ≤𝐶 ∙ 𝐾1𝑐
𝜋 ∙ 𝑡/2
𝒃 =𝒕
𝟐
𝑡 =𝑝 ∙ 𝑅
2𝜎𝑦
𝑡 =𝑝 ∙ 𝑅
2𝜎𝑦
𝑡 𝜎𝑦
𝑇ℎ𝑖𝑛𝑛𝑒𝑟 𝑡
Hp:
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Materials Selection 21Monday, November 13, 2017
Case Study 14: Safe Pressure Vessels LBB
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑡𝑎𝑛𝑘𝑠𝑜𝑓 𝑟𝑜𝑐𝑘𝑒𝑡𝑠
𝑁𝑢𝑐𝑙𝑒𝑎𝑟𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 vessels
(316 Stainless Steel)
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Materials Selection 22Monday, November 13, 2017
Minimize the mass
Multiple constraints
Minimize the volume
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Materials Selection 23Monday, November 13, 2017
Case Study 14’: Materials for Light pressure VesselsWith MULTIPLE CONSTRAINTAPPROACH
𝜎 =𝑝 ∙ 𝑅
2𝑡≤
𝐾1𝑐
𝜋 ∙ 𝑐𝑚1 = 2𝜋𝑅3 ∙ 𝑝 ∙ 𝜋 ∙ 𝑐 ∙
𝜌
𝐾1𝑐
𝑚 = 4𝜋𝑅2 ∙ 𝑡 ∙ 𝜌
𝜎 =𝑝 ∙ 𝑅
2𝑡≤ 𝜎𝑦
𝑚2 = 2𝜋𝑅3 ∙ 𝑝 ∙𝜌
𝜎𝑦
𝑀1 =𝜌
𝐾1𝑐
𝑀2 =𝜌
𝜎𝑦
Objective Constraints Performance equation Index
Yield constraint
Fracture constraint
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Materials Selection 24Monday, November 13, 2017
Case Study 14’: Materials for Light pressure VesselsWith MULTIPLE CONSTRAINTAPPROACH
𝜌
𝜎𝑦= 𝜋 ∙ 𝑐 ∙
𝜌
𝐾1𝑐𝑀2 = 𝜋 ∙ 𝑐 ∙ 𝑀1
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Materials Selection 25Monday, November 13, 2017
Case Study 14’: Materials for Light pressure VesselsWith MULTIPLE CONSTRAINTAPPROACH
C = 10 mm
C = 10 µm
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Materials Selection 26Monday, November 13, 2017
Theory: Method of the weight factor
𝑍𝑝 = 𝛼𝐻 ∙H
max(H)+ 𝛼𝑀 ∙
M
max(M)𝐸𝑥𝑎𝑚𝑝𝑙𝑒 𝑜𝑓 𝑖𝑡𝑠 𝑎𝑝𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛
𝑜𝑛 𝑎 𝑝𝑟𝑜𝑐𝑒𝑠𝑠
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒
𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒
𝒀𝒐𝒖𝒏𝒈′𝒔 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 𝑫𝒆𝒏𝒔𝒊𝒕𝒚
The most important point is to understand if you must
maximize or minimize a property
Example:
𝐸
1/𝐸
1/𝜌
𝜌
With the normalization and the performance function you
can consider all the properties that you want and with
various order of magnitude.
But the importance coefficients? EDL
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Materials Selection 27Monday, November 13, 2017
Theory: Enhanced Digital Logic
𝑍𝑝 = 𝛼𝐻 ∙H
max(H)+ 𝛼𝑀 ∙
M
max(M)
A really more important than B 3-1
A more important than B 2-1
A as imp. as B 1-1
Addition
σ 2 2 3 7 0,389 0,39
W 1 1 2 4 0,222 0,22
C 1 1 2 4 0,222 0,22
F 1 1 1 3 0,167 0,17
18 1
𝑰𝒎𝒑𝒐𝒓𝒕𝒂𝒏𝒄𝒆 𝑺𝒆𝒒𝒖𝒆𝒏𝒄𝒆σ > W = C > F
σ Yield Strength
W Weldability
C Corrosion Resistance
F Fatigue Resistance
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Materials Selection 28Monday, November 13, 2017
Theory: Enhanced Digital Logic
And if I want to have a property value similar to another material?
Normalisation MANCINI-MAURIZI
𝐼𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑖𝑠 𝑏𝑖𝑔𝑔𝑒𝑟 𝑡ℎ𝑎𝑛 𝑡ℎ𝑒 𝑡𝑎𝑟𝑔𝑒𝑡 𝑣𝑎𝑙𝑢𝑒 𝑁𝑛 = 1 −𝐶𝑇𝐸𝑛 − 𝐶𝑇𝐸𝑡𝑎𝑟𝑔.
𝐶𝑇𝐸𝑚𝑎𝑥 − 𝐶𝑇𝐸𝑡𝑎𝑟𝑔.
Example : Coefficient of thermal expansion α (CTE)
𝐼𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑡ℎ𝑎𝑛 𝑡ℎ𝑒 𝑡𝑎𝑟𝑔𝑒𝑡 𝑣𝑎𝑙𝑢𝑒 𝑁𝑛 = 1 −𝐶𝑇𝐸𝑡𝑎𝑟𝑔. − 𝐶𝑇𝐸𝑛
𝐶𝑇𝐸𝑡𝑎𝑟𝑔. − 𝐶𝑇𝐸𝑚𝑖𝑛
Again, a value close to 1 = the two CTEs are similar (the contrary to 0)
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Materials Selection 29Monday, November 13, 2017
Using materials at high temperatures
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Materials Selection 30Monday, November 13, 2017
Case Study 15: Materials for Precision devices
Es. Sub micrometer displacement gauge
Objective • Minimize distortion to maximize positional accuracy
Constraints • Tolerate heat flux • Tolerate vibration
Free Variables • Choice of material
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Materials Selection 31Monday, November 13, 2017
Case Study 15: Materials for Precision devices
The distortion is proportional to the gradient of strain𝑑𝜀
𝑑𝑥= 𝛼 ∙
𝑑𝑇
𝑑𝑥==
𝛼
λ∙ 𝑞
𝜀 = 𝛼(𝑇 − 𝑇0)
𝑞 = −λ𝑑𝑇
𝑑𝑥
Fourier’s Law in steady state 𝑞 Heat fluxλ 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦
𝑑𝑇
𝑑𝑥𝑟𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡
𝑑𝜀
𝑑𝑥
λ
𝛼
𝑉𝐸1/2
𝜌Moreover, we want to avoid flexural vibrations at lowest
frequencies; proportional to (finally must be stiff)
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Materials Selection 32Monday, November 13, 2017
Case Study 15: Materials for Precision devices
𝑑𝜀
𝑑𝑥
λ
𝛼
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Materials Selection 33Monday, November 13, 2017
Case Study 15: Materials for Precision devices
Carbides? Excellent performances, but difficult to shapeCopper High density gives a low M2Al alloys the cheapest and most easily shaped choiceSilicon the best choice
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Materials Selection 34Monday, November 13, 2017
Case Study 15: Materials for Precision devices
With some researches you can find better materials!Ex. Invar Add Record IF IT HAS A SENSE
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Materials Selection 35Monday, November 13, 2017
Conductors, insulators and dielectrics
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Materials Selection 36Monday, November 13, 2017
Case Study 16: Materials for Long Span Transmission line
Objective • Maximize current flux
Constraints • Easy to manufacture• Must be strong?• Dimensions fixed
Free Variables • Choice of material
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Materials Selection 37Monday, November 13, 2017
Case Study 16: Materials for Long Span Transmission line
[http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thercond.html]
𝐿 ∝ 𝐾 =λ
𝜌𝑒
Wiedemann-Franz Law𝐿 Lorenz number
𝐾 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦λ 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦𝜌𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑖𝑡𝑦
𝐾λ
𝜌𝑒
Index from General Knowledge
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Materials Selection 38Monday, November 13, 2017
Case Study 16: Materials for Long Span Transmission line
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Materials Selection 39Monday, November 13, 2017
Case Study 16: Parenthesis about real life
On a standard shape
How the shape can change the properties Decrease the I
𝑛𝜋𝑟2 = 𝑏2 = 𝐴
𝐼0 =𝐴2
12=
𝑛𝜋𝑟2 2
12
𝐼 = 𝐼0 ∙ 𝑛 =𝑛𝜋𝑟4
4
On the single wire
𝑰
𝑰𝟎=
𝒏𝝅𝒓𝟒
𝟒𝒏𝝅𝒓𝟐 𝟐
𝟏𝟐
=𝟑
𝒏𝝅∝𝟏
𝒏
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Materials Selection 40Monday, November 13, 2017
Case Study 17: Materials for Light Cable
Objective • Minimize the mass
Constraints • Good conductor • Easy to manufacture• Dimensions fixed
Free Variables • Choice of material
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Materials Selection 41Monday, November 13, 2017
Case Study 17: Materials for Light Cable
http://toelatingsexamen.org/en/app/course/71/pouillets-law-electrodynamics?chapterID=84§ionID=3
Pouillet’s Law
𝑚 = 𝜌 ∙ 𝐿 ∙ 𝐴
𝑟 = 𝜌𝑒𝐿
𝐴
Index from « Equations »
𝐴 = 𝜌𝑒𝐿
𝑟
𝜌𝑒 Electrical resistivityr Electrical resistance of a uniform specimen of the material
𝑚 = (𝜌 ∙ 𝜌𝑒) ∙𝐿2
𝑟
𝑚1
𝜌 ∙ 𝜌𝑒
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Materials Selection 42Monday, November 13, 2017
Excellent M1 and it is the lighter material
Case Study 17: Materials for Light Cable
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Materials Selection 43Monday, November 13, 2017
Case Study 18: Materials for Kiln Walls
Objective • Minimize energy consumed in firing cycle
Constraints • Maximum and minimum operating temperature
• Possible limit on W for space reasons
Free Variables • Choice of material
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Materials Selection 44Monday, November 13, 2017
Case Study 18: Materials for Kiln Walls
𝑄1 = −λ𝑇𝑖 − 𝑇0𝑤
∙ 𝑡
q.
𝑇𝑖
𝑇0
𝑇0+𝑖
𝑡 = 0
𝑡 > 0
q. = Heat fluxλ = Thermal conductivity𝐶𝑝 = Specific heat capacity
t = time Heat loss by conduction
Heat (loss) absorbed by walls
𝑄2 =∆𝑇 ∙ 𝑤 ∙ 𝜌 ∙ 𝐶𝑝
2
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑑𝑢𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 𝑢𝑠𝑒 𝑄 = 𝑄1+𝑄2
w
Energy-efficient kiln walls? Reduce 𝝀 and w
𝑇ℎ𝑢𝑠, 𝑤𝑒 𝑤𝑎𝑛𝑡 𝑎 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑄 𝑤 −→ a w that correspond to𝑑𝑄
𝑑𝑤= 0
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Materials Selection 45Monday, November 13, 2017
Case Study 18: Materials for Kiln Walls
q.
𝑇𝑖
𝑇0
𝑇0+𝑖
𝑡 = 0
𝑡 > 0
w
q. = Heat fluxλ = Thermal conductivity𝐶𝑝 = Specific heat capacity
t = time
a = Thermal diffusivity
𝑄 = 𝑄1+𝑄2= −λ𝑇𝑖 − 𝑇0𝑤
∙ 𝑡 +∆𝑇 ∙ 𝑤 ∙ 𝜌 ∙ 𝐶𝑝
2
𝑑𝑄
𝑑𝑤= 0 = −λ
𝑇𝑖 − 𝑇0𝑤2 ∙ 𝑡 +
∆𝑇 ∙ 𝑤 ∙ 𝜌 ∙ 𝐶𝑝2
For an optimum thickness
𝑤∗ =2λ𝑡
𝜌 ∙ 𝐶𝑝= 2𝑎𝑡
So, if we consider
𝑤∗ = 2𝑎𝑡
𝑄 = 𝑄1+𝑄2 𝑄 =∆𝑇 ∙ 2𝑡 ∙ λ
𝑎
𝑄𝑎
λ
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Materials Selection 46Monday, November 13, 2017
Case Study 18: Materials for Kiln Walls
𝑄𝑎
λ
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Materials Selection 47Monday, November 13, 2017
Case Study 18: Materials for Kiln Walls
𝑄𝑎
λ
+ Limit on max service temperature200 < Tmax <1000°C
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Materials Selection 48Monday, November 13, 2017
Case Study 19: Materials for Insulation forShort-Term isothermal containers
Objective • Maximize the time t before internal temperature changes when external temperature suddenly drops
Constraints • Wall thickness (W)
Free Variables • Choice of the material
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Materials Selection 49Monday, November 13, 2017
Case Study 19: Materials for Insulation forShort-Term isothermal containers
Fick’s Law for the temperatureat Steady state
𝜕𝑇
𝜕𝑡= a
𝜕2𝑇
𝜕𝑥2
𝑞. = −λ(𝑇𝑖 − 𝑇0)
𝑤
q.
WE ARE NOT MINIMIZING THE HEAT FLUX!!!WE WANT TO MAXIMIZE THE TIME tBEFORE 𝑻𝟎 CHANGE CONSIDERABLY
w
𝑇𝑖
𝑇0
𝑇0+𝑖
𝑡 = 0
𝑡 > 0
x
q. = Heat fluxt = time x = distance of propagation of thermal equilibrium
a = thermal diffusivity
Fick’s Law for the temperatureat not-steady state
Too complex, sowe use 𝑥 ≅ 2𝑎𝑡 ≤ 𝑤
𝑡 =𝑤2
2𝑎
𝑡 𝑎𝑎 =λ
𝜌∙𝐶𝑝
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Materials Selection 50Monday, November 13, 2017
Case Study 19: Materials for Insulation forShort-Term isothermal containers
𝑡 𝑎
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Materials Selection 51Monday, November 13, 2017
Case Study 20: Process for a Can
Objective • Find an High production + Cheap process
Constraints • Shape : Dished Sheet• Physical aspect :
1. Mass range max 1 kg2. Tolerance 0,2-0,5 mm
• Primary Process
Free Variables • Choice of process
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Materials Selection 52Monday, November 13, 2017
Case Study 20: Process for a Can
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Materials Selection 53Monday, November 13, 2017
Case Study 20: Process for a Can
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Materials Selection 54Monday, November 13, 2017
Case Study 20: Process for a Can
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Materials Selection 55Monday, November 13, 2017
Results : Deep drawing (deep drawing is a particular stamping)
Case Study 20: Process for a Can
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Materials Selection 56Monday, November 13, 2017
Deep drawing (deep drawing is a particular stamping)Case Study 20: Process for a Can
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Materials Selection 57Monday, November 13, 2017
Secret slides : Kubota case
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Materials Selection 58Monday, November 13, 2017
Secret slides : Kubota case
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Materials Selection 59Monday, November 13, 2017
Secret slides : Kubota case