1 Safe pressure vessels Safe pressure vessels Therefore, for - - PowerPoint PPT Presentation

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ME 499-699 Fall 2006 Slides 6 -1

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

Case Studies in Materials Selection

Material for a pressure vessel Short term thermal insulation Energy efficient kilns

ME 499-699 Fall 2006 Slides 6 -2

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

p R t t pR = σ

2a

ME 499-699 Fall 2006 Slides 6 -3

Safe pressure vessels

Material Free variables:

• Radius R is specified

Constraints Maximize safety

• Yield before break or
• Leak before break

Objective Pressure vessel – contain pressure p safely Function

ME 499-699 Fall 2006 Slides 6 -4

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

The stress required to catastrophically propagate a crack in the

presence of a flaw of size 2a*

c is

where KiC is the fracture toughness of the material and C ( ≈ 1) is a constant

* 1 c C

a CK π σ =

2

ME 499-699 Fall 2006 Slides 6 -5

Safe pressure vessels

Therefore, for safety The corresponding material index to be maximized is

* 1 c C

a K R t R t p π σ ≤ =

C

K M

1 1 =

ME 499-699 Fall 2006 Slides 6 -6

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

In order to maximize the flaw size for with “yield before break” occurs,

the material index to be maximized is

2 1 2

⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ≤

f C c

K C a σ π

f C

K M σ

1 2 =

ME 499-699 Fall 2006 Slides 6 -7

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 (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

K C R p

• r

t pR K C t σ π σ σ π

2 1 2 2 1 2

2 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =

f C

K M σ

2 1 3 = ME 499-699 Fall 2006 Slides 6 -8

Safe pressure vessels

Material indices M2 and M3 indicate σf should be as small

as possible.

A material like lead would satisfy these indices However, if one wanted to make a thin walled pressure

vessel, the thinnest wall is obtained by having a high value

• f 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

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ME 499-699 Fall 2006 Slides 6 -9

Safe pressure vessels

Yield strength (elastic limit) (Pa)

10000 100000 1e6 1e7 1e8 1e9

Fracture toughness (Pa.m^1/2)

10000 100000 1e6 1e7 1e8

M1 K1C > 10 MPa.m0.5

ME 499-699 Fall 2006 Slides 6 -10

Safe pressure vessels

Yield strength (elastic limit) (Pa)

10000 100000 1e6 1e7 1e8 1e9

Fracture toughness (Pa.m^1/2)

10000 100000 1e6 1e7 1e8

M2

ME 499-699 Fall 2006 Slides 6 -11

Safe pressure vessels

Yield strength (elastic limit) (MPa)

0.01 0.1 1 10 100 1000

Fracture toughness (MPa.m^1/2)

0.01 0.1 1 10 100

M3

ME 499-699 Fall 2006 Slides 6 -12

Safe pressure vessels

M4

Yield strength (elastic limit) (Pa)

10000 100000 1e6 1e7 1e8 1e9

Fracture toughness (Pa.m^1/2)

10000 100000 1e6 1e7 1e8

Non age-hardening wrought Al-alloys Copper Cast Al-alloys Commercially pure zinc Stainless steel Zinc die-casting alloys Bronze

4

ME 499-699 Fall 2006 Slides 6 -13

Safe pressure vessels

Yield strength (elastic limit) (Pa)

10000 100000 1e6 1e7 1e8 1e9

Fracture toughness (Pa.m^1/2)

10000 100000 1e6 1e7 1e8

Non age-hardening wrought Al-alloys Copper Cast Al-alloys Commercially pure zinc Stainless steel Zinc die-casting alloys Bronze

All stages

ME 499-699 Fall 2006 Slides 6 -14

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 499-699 Fall 2006 Slides 6 -15

Short term thermal insulation

Material Free variables: Wall thickness must not exceed w Constraints Maximize time before which internal temperature drops below critical value Objective Short term thermal insulation Function

Insulating material

• f wall thickness w

Electronic circuits packaged in this space

ME 499-699 Fall 2006 Slides 6 -16

Short term thermal insulation

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 flow Best materials are polymer foams

( )

w T T dx dT q

• i −

≈ − = λ λ

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ME 499-699 Fall 2006 Slides 6 -17

Short term thermal insulation

Untitled

MaterialUniverse:\Ceramics and glasses MaterialUniverse:\Hybrids: composites, foams, natural materials MaterialUniverse:\Metals and alloys MaterialUniverse:\Polymers and elastomers

Thermal conductivity (W/m.K)

0.1 1 10 100

Rigid Polymer Foam (LD) Rigid Polymer Foam (MD) Cork Flexible Polymer Foam (MD)

ME 499-699 Fall 2006 Slides 6 -18

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 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 diffusivity

at x 2 ∝

p

C a ρ λ =

ME 499-699 Fall 2006 Slides 6 -19

Short term thermal insulation

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

a w t 2

2

≈

ME 499-699 Fall 2006 Slides 6 -20

Short term thermal insulation

T-diffusivity

1e-7 1e-6 1e-5 1e-4

Thermal conductivity (W/m.K)

0.1 1 10 100

Polymethyl methacrylate (Acrylic, PMMA) Isoprene (IR) Isoprene (IR) Isoprene (IR) Rigid Polymer Foam (HD)

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ME 499-699 Fall 2006 Slides 6 -21

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 499-699 Fall 2006 Slides 6 -22

Energy efficient kiln

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 Cp Ti To

ME 499-699 Fall 2006 Slides 6 -23

Energy efficient kiln

Material Wall thickness Free variables: Hard: Max operating temp = 1000°C Soft: Wall thickness due to space limitation Constraints Minimize energy consumed in each cycle Objective Thermal insulation for kiln (cyclic heating and cooling) Function

ME 499-699 Fall 2006 Slides 6 -24

Energy efficient kiln

Analysis

There are two sources of heat loss

• Heat lost by conduction through walls
• Heat required to increase temperature of insulating material
• Total heat loss is

t w T T t dx dT Q

• i −

= − = λ λ

1

( )

2

2

• i

p

T T w C Q − = ρ

( )

2

2 1

• i

p

• i

T T w C t w T T Q Q Q − + − = + = ρ λ

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ME 499-699 Fall 2006 Slides 6 -25

Energy efficient kiln

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

2 2 at C t w

p

= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ρ λ

( )( ) (

)

2 / 1 2 / 1

2 ρ λ

p

• i

C t T T Q − =

( )

λ ρ λ

2 / 1 ) 2 / 1 (

a C M

p

= =

−

ME 499-699 Fall 2006 Slides 6 -26

Energy efficient kiln

Select Materials - Stage 1 – limit stage

• Min operating temperature - 1000°C
• 14 materials

Alumina Aluminum nitride Boron carbide Brick Ceramic foam Glass ceramic Nickel-based superalloys Nickel-chromium alloys Silica glass Silicon carbide Silicon nitride Tungsten alloys Tungsten carbides Zirconia

ME 499-699 Fall 2006 Slides 6 -27

Energy efficient kiln

Thermal diffusivity

1e-7 1e-6 1e-5 1e-4

Thermal conductivity (W/m.K)

0.1 1 10 100

Stage 2

ME 499-699 Fall 2006 Slides 6 -28

Energy efficient kiln

Thermal diffusivity

1e-7 1e-6 1e-5 1e-4

Thermal conductivity (W/m.K)

0.1 1 10 100

Both Stages

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ME 499-699 Fall 2006 Slides 6 -29

Energy efficient kiln

Select Materials - All Stages 5 materials

• Brick
• Ceramic foam
• Glass ceramic
• Silica glass
• Zirconia

Switching to the larger database gives over 60 materials. Additional criteria

can be imposed such as cost, oxidation resistance etc. to identify a specific material