Material for a pressure vessel Short term thermal

ME 474-674 Winter 2008 Slides 6 -1

More info: “Materials Selection in Mechanical Design”, Chapters 5 and 6

More Case Studies in Materials Selection

Material for a pressure vessel

Short term thermal insulation

Energy efficient kilns


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

t

tpR

=σ

2a



MaterialFree variables:•Radius R is specifiedConstraints

Maximize safety• Yield before break or• Leak before break

ObjectivePressure vessel – contain pressure p safelyFunction



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 thepresence of a flaw of size 2a*

c is

where KiC is the fracture toughness of the material and C ( ≈ 1) is a constant that depends upon the shape and location of the crack

*1

c

C

aCKπ

σ =



Therefore, for safety

The corresponding material index to be maximized is

*1

c

C

aK

Rt

Rtp

πσ ≤=

CKM 11 =



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

In order to maximize the flaw size for with “yield before break” occurs, the material index to be maximized is

2

12

⎥⎥⎦

⎤

⎢⎢⎣

⎡≤

f

Cc

KCaσ

π

f

CKMσ

12 =



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 (2ac) 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”

Under this situation, the material index to be maximized is

f

C

f

C

KCR

p

ort

pR

KCt

σπ

σ

σπ

21

2

2

12

2

2

⎟⎠⎞

⎜⎝⎛=

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

f

CKMσ

21

3 =



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

M4 = σf

The following slides show the successive application of each of the indices to select a material



Summary of Material Performance parameters

MaximizeσfM4

Maximize(minimize σf ?)

M3

Maximize(minimize σf ?)

M2

MaximizeM1

ObjectiveEquationParameter

f

CKσ

21

f

CKσ

1

CK1



M1

K1C > 10 MPa.m0.5

30 of 95 MaterialsAll metalsFerrous and non-Ferrous

Yield strength (elastic limit) (MPa)0.01 0.1 1 10 100 1000

Frac

ture

tou

ghne

ss (

MPa

.m^

1/2)

0.01

0.1

1

10

100



M2 = 0.4m0.5

15 of 95 MaterialsIncluding

LeadPolymer FoamMetal FoamLeather


Frac

ture

tou

ghne

ss (

MPa

.m^

1/2)

0.01

0.1

1

10

100

Metal foam

Flexible Polymer Foam (MD)

Commercially pure lead

Leather

CopperStainless steel

Non age-hardening wrought Al-alloys

Nickel



M3 = 4 MPa.m

22 of 95 Materials

Lead is still an option


Frac

ture

tou

ghne

ss (

MPa

.m^

1/2)

0.01

0.1

1

10

100

Metal foam

Commercially pure lead

Lead alloys

CopperNickel

Leather




M4 = 100 MPa

36 of 95 Materials

Lead and foams are gone but we have picked up a bunch of ceramic materials


Frac

ture

tou

ghne

ss (

MPa

.m^

1/2)

0.01

0.1

1

10

100 CopperNickel

Tungsten carbides

Silicon

Silicon nitride

Aluminum nitride Low alloy steel

Medium carbon steel

CFRP, epoxy matrix (isotropic)



All stages

8 materials


Frac

ture

tou

ghne

ss (

MPa

.m^

1/2)

0.01

0.1

1

10

100 Copper


Zinc die-casting alloysNickel

Stainless steel

Zinc die-casting alloys

Cast Al-alloys

Bronze



Select Materials - All Stages

Brass

Cast Al-alloys

Commercially pure zinc

Copper

Nickel


Stainless steel

Zinc die-casting alloys



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



MaterialFree variables:

Wall thickness must not exceed wConstraints

Maximize time before which internal temperature drops below critical value

ObjectiveShort term thermal insulationFunction

Insulating material of wall thickness w

Electronic circuits packaged in this space



Model 1

Minimize heat flux out of the containment area

First law of heat conduction

Where q is heat flux, λ is thermal conductivity

Therefore, minimize λ to minimize heat flowBest materials are polymer foams

( )w

TTdxdTq oi −≈−= λλ



Ther

mal

con

duct

ivity

(W

/m.K

)

0.1

1

10

100

Rigid Polymer Foam (LD)

Rigid Polymer Foam (MD)


Rigid Polymer Foam (HD)

Ceramics

Polymers

Foams and Hybrids

Metals



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 2nd 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

Where a is the thermal diffusivityatx 2∝

pCa

ρλ

=



ρ is the density and Cp is the specific heat of the material.We can replace x in the above equation by the wall thickness to get

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

awt2

2

≈



Thermal Diffusivity1e-7 1e-6 1e-5 1e-4

Ther

mal

con

duct

ivity

(W

/m.K

)

0.1

1

10

100

Butyl RubberIsoprene (IR)

Isoprene (IR)

Polychloroprene (Neoprene, CR)


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



How can these apparently contradictory requirements be reconciled?

Is there a material index that can capture both requirements?

Wall thickness w

Insulation

T-con λ

Density ρ

Sp-heat CpTiTo



MaterialWall thickness

Free variables:

Hard: Max operating temp = 1000°CSoft: Wall thickness due to space limitation

ConstraintsMinimize energy consumed in each cycleObjective

Thermal insulation for kiln (cyclic heating and cooling)

Function



AnalysisThere are two sources of heat loss

Heat lost by conduction through walls

Heat required to increase temperature of insulating material

Total heat loss is

tw

TTtdxdTQ oi −=−= λλ1

( )22

oip

TTwCQ −= ρ

( )221

oip

oi TTwCtw

TTQQQ −+

−=+= ρλ



To minimize total heat loss, differentiate the above equation and set equal to zero and find w

Substituting back into the equation for Q gives

The material index to be maximized is

( ) 2/1

2/1

22 atC

twp

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

ρλ

( )( ) ( ) 2/12/12 ρλ poi CtTTQ −=

( )λ

ρλ2/1

)2/1( aCM p == −



Select Materials - Stage 1 – limit stageMin operating temperature - 1000°C14 materials

AluminaAluminum nitrideBoron carbideBrickCeramic foamGlass ceramicNickel-based superalloysNickel-chromium alloysSilica glassSilicon carbideSilicon nitrideTungsten alloysTungsten carbidesZirconia



Thermal diffusivity1e-7 1e-6 1e-5 1e-4

Ther

mal

con

duct

ivity

(W/m

.K)

0.1

1

10

100

Stage 2

High values of

λ

2/1aM =



Thermal diffusivity1e-7 1e-6 1e-5 1e-4

Ther

mal

con

duct

ivity

(W/m

.K)

0.1

1

10

100

Both Stages



Select Materials - All Stages5 materials

Brick

Ceramic foam

Glass ceramic

Silica glass

Zirconia

Switching to the larger database gives over 60 materials.



Thermal Diffusivity = Thermal conductivity / Specific heat / Density 1e-7 1e-6 1e-5 1e-4

Ther

mal

con

duct

ivity

(W

/m.K

)

0.1

1

10

100Graphite (perpendicular to plane)

Mullite (Al2O3-SiO2 alloys)

Carbon (Vitreous)

Graphite Foam (0.12)

Alumina Foam (99.8%)(0.4)

Carbon Foam (Reticulated, Vitreous)(0.05)

Glass Ceramic (N11)

Glass Ceramic - Slipcast

Plaster of Paris

Carbon Fiber Reinforced Carbon Matrix Composite (Vf:50%)

Additional criteria can be imposed such as cost, oxidation resistance, flammability, etc. to screen out certain materials like carbon

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Material for a pressure vessel Short term thermal