More Case Studies in Materials Selection � Material for a pressure vessel � Short term thermal insulation � Energy efficient kilns More info: “ Materials Selection in Mechanical Design ”, Chapters 5 and 6 ME 474-674 Winter 2008 Slides 6 -1
Safe pressure vessels � Cylindrical pressure vessels are containers for a fluid under pressure � A safe design will be based on one of two factors Detectable plastic deformation (small pressure vessels) � “Leak before break” (larger pressure vessels) � � The maximum principal stress is the hoop stress pR σ = t p 2a R t ME 474-674 Winter 2008 Slides 6 -2
Safe pressure vessels Function Pressure vessel – contain pressure p safely Objective Maximize safety • Yield before break or • Leak before break Constraints •Radius R is specified Free variables: Material ME 474-674 Winter 2008 Slides 6 -3
Safe pressure vessels � Pressure vessels are usually examined for any flaws that may be present � Ultrasonic or X-ray techniques have a detection limit of “ 2a * c ” � There are no flaws larger than 2a * c � Have to assume flaws up to size 2a * c are present � The stress required to catastrophically propagate a crack in the presence of a flaw of size 2a * c is CK σ = 1 C π * a c where K iC is the fracture toughness of the material and C ( ≈ 1) is a constant that depends upon the shape and location of the crack ME 474-674 Winter 2008 Slides 6 -4
Safe pressure vessels � Therefore, for safety t t K = σ ≤ 1 C p π R R * a c � The corresponding material index to be maximized is 1 = M K 1 C ME 474-674 Winter 2008 Slides 6 -5
Safe pressure vessels � However, if one wanted to ensure that the material yielded before fracture, then it should be possible to reach the failure stress or yield stress even when the flaw size is greater than the detection limit of the NDE technique 2 ⎡ ⎤ K π ≤ ⎢ ⎥ 2 1 C a C σ c ⎢ ⎥ ⎣ ⎦ f � In order to maximize the flaw size for with “yield before break” occurs, the material index to be maximized is K 2 = 1 C M σ f ME 474-674 Winter 2008 Slides 6 -6
Safe pressure vessels � It may not be possible to subject a large pressure vessels to complete X- ray or ultrasonic examination to locate pre-existing flaws � Therefore, if the vessel is designed such that critical flaw size ( 2a c ) is at least equal to the thickness of the wall the even when the stress reaches the yield stress, then the vessel will “leak before break” 2 ⎛ ⎞ 2 ⎛ ⎞ t C K ⎜ ⎟ = ⎜ ⎟ 1 C ⎜ ⎟ π σ ⎝ ⎠ 2 ⎝ ⎠ f pR σ = t or 2 ⎛ ⎞ 2 2 C K = ⎜ ⎟ 1 C p π σ ⎝ ⎠ R f � Under this situation, the material index to be maximized is 2 K 3 = 1 C M σ f ME 474-674 Winter 2008 Slides 6 -7
Safe pressure vessels � If one wanted to make a thin walled pressure vessel, the thinnest wall is obtained by having a high value of the yield strength. � Therefore, there is a fourth index that needs to be maximized. Namely M 4 = σ f � The following slides show the successive application of each of the indices to select a material ME 474-674 Winter 2008 Slides 6 -8
Safe pressure vessels � Summary of Material Performance parameters Parameter Equation Objective M 1 Maximize K 1 C K M 2 Maximize 1 C σ (minimize σ f ?) f 2 M 3 Maximize K 1 C σ (minimize σ f ?) f σ f M 4 Maximize ME 474-674 Winter 2008 Slides 6 -9
Safe pressure vessels 100 M 1 Fracture toughness (MPa.m^ 1/2) K 1C > 10 MPa.m 0.5 10 30 of 95 Materials All metals 1 Ferrous and non- Ferrous 0.1 0.01 0.01 0.1 1 10 100 1000 Yield strength (elastic limit) (MPa) ME 474-674 Winter 2008 Slides 6 -10
Safe pressure vessels Stainless steel Nickel Copper 100 Non age-hardening wrought Al-alloys M 2 = 0.4m 0.5 Commercially pure lead Fracture toughness (MPa.m^ 1/2) 10 Leather 15 of 95 Materials Metal foam Including 1 � Lead Flexible Polymer Foam (MD) � Polymer Foam � Metal Foam 0.1 � Leather 0.01 0.01 0.1 1 10 100 1000 Yield strength (elastic limit) (MPa) ME 474-674 Winter 2008 Slides 6 -11
Safe pressure vessels Nickel Copper 100 Lead alloys Fracture toughness (MPa.m^ 1/2) Commercially pure lead M 3 = 4 MPa.m 10 Leather Metal foam 22 of 95 Materials 1 Lead is still an Flexible Polymer Foam (MD) option 0.1 0.01 0.01 0.1 1 10 100 1000 Yield strength (elastic limit) (MPa) ME 474-674 Winter 2008 Slides 6 -12
Safe pressure vessels Medium carbon steel Nickel Copper 100 CFRP, epoxy matrix (isotropic) M 4 = 100 MPa Fracture toughness (MPa.m^ 1/2) 36 of 95 Materials 10 Low alloy steel Aluminum nitride Silicon nitride Lead and foams 1 are gone but we Tungsten carbides have picked up a bunch of ceramic Silicon materials 0.1 0.01 0.01 0.1 1 10 100 1000 Yield strength (elastic limit) (MPa) ME 474-674 Winter 2008 Slides 6 -13
Safe pressure vessels Stainless steel Zinc die-casting alloys Nickel 100 Copper All stages Bronze Non age-hardening wrought Al-alloys Fracture toughness (MPa.m^ 1/2) 10 8 materials Cast Al-alloys 1 Zinc die-casting alloys 0.1 0.01 0.01 0.1 1 10 100 1000 Yield strength (elastic limit) (MPa) ME 474-674 Winter 2008 Slides 6 -14
Safe pressure vessels � Select Materials - All Stages Brass � Cast Al-alloys � Commercially pure zinc � Copper � Nickel � Non age-hardening wrought Al-alloys � Stainless steel � Zinc die-casting alloys � ME 474-674 Winter 2008 Slides 6 -15
Short term thermal insulation � An application for short term thermal insulation is the rescue beacons for military aircraft pilots � These electronic devices do not function if the temperature drops below a critical value � Therefore, to give the rescue operation the greatest chance of being effective, the temperature of the electronics in the radio beacon must not fall below a critical value for the longest period of time even when exposed to cold temperatures The temperature of most of the earth’s oceans is around 4ºC � � The electronics have to be wrapped in an insulating blanket ME 474-674 Winter 2008 Slides 6 -16
Short term thermal insulation Function Short term thermal insulation Objective Maximize time before which internal temperature drops below critical value Constraints Wall thickness must not exceed w Free variables: Material Insulating material of wall thickness w Electronic circuits packaged in this space ME 474-674 Winter 2008 Slides 6 -17
Short term thermal insulation � Model 1 � Minimize heat flux out of the containment area � First law of heat conduction ( ) i − dT T T = − λ ≈ λ o q dx w � Where q is heat flux, λ is thermal conductivity � Therefore, minimize λ to minimize heat flow � Best materials are polymer foams ME 474-674 Winter 2008 Slides 6 -18
Short term thermal insulation Ceramics 100 Foams and Thermal conductivity (W/m.K) Hybrids 10 Metals Polymers 1 Rigid Polymer Foam (HD) 0.1 Flexible Polymer Foam (MD) Rigid Polymer Foam (LD) Rigid Polymer Foam (MD) ME 474-674 Winter 2008 Slides 6 -19
Short term thermal insulation � But is this the answer we are looking for? � The answer is no! � The problem requires that the time that it takes for the electronic package to cool down be maximized. � This is not a steady state problem. � Therefore use 2 nd law of heat conduction � If the temperature at the surface is decreased suddenly, as in dropping the pilot and his radio beacon into a cold ocean, the distance x from the surface at which a certain temperature is reached changes with time t as ∝ x 2 at � Where a is the thermal diffusivity λ = a ρ C p ME 474-674 Winter 2008 Slides 6 -20
Short term thermal insulation � ρ is the density and C p is the specific heat of the material. � We can replace x in the above equation by the wall thickness to get 2 w ≈ t 2 a � Therefore, we seek the material with the smallest a to maximize the time t , if the thickness of the insulation w is fixed � The best materials are therefore elastomers ME 474-674 Winter 2008 Slides 6 -21
Short term thermal insulation 100 Thermal conductivity (W/m.K) 10 Isoprene (IR) Polychloroprene (Neoprene, CR) 1 0.1 Butyl Rubber Isoprene (IR) 1e-7 1e-6 1e-5 1e-4 Thermal Diffusivity ME 474-674 Winter 2008 Slides 6 -22
Energy efficient kiln � Kilns used for firing pottery are heated up from room temperature to the firing temperature during each cycle Unbaked pottery is placed in the furnace � The heating mechanism, electric or gas, is turned on and the kiln is � heated up to the firing temperature After the requisite time at temperature, the kiln is allowed to cool down � Once cooled, the pottery is removed and the cycle is repeated � � There are two major factors that consume energy The energy to heat up the kiln � The energy lost through conduction through the walls � � The first can be minimized by reducing the thermal mass of the system, i.e. minimize the wall thickness � The second can be minimized by reducing the heat loss through the wall by increasing its thickness ME 474-674 Winter 2008 Slides 6 -23
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