Transcript Thermal Props 1 PPT

Thermal Behavior
Thermal Properties
• Heat capacity
•
Specific Heat
• Thermal Expansion
•
Thermal Conductivity
•
Thermal Shock
Heat Capacity
•
As a material absorbs heat, its temperature rises
• The HEAT CAPACITY is the amount of heat
required to raise its temperature by 1°C
• C = Q/∆T
•
C is the Heat Capacity (J/mol-K),
•
Q is the amount of heat (J/mol), and
• ∆T is the change in temperature (K or C)
Specific Heat
•
Often we use Specific Heat instead of Heat
Capacity because it is per unit mass instead of
mol
• c = q/(m ∆T)
•
where c is the specific heat (J/kg-K)
• and m is the mass of material being heated (kg)
How to measure?
•
There are two ways to measure specific heat:
• Volume is maintained constant (and pressure
thus builds up), cv
•
Pressure is maintained constant (and volume
increases), cp
cv is not a constant
But...
•
The Debye temperature is below room
temperature for many solids, so we can still use a
value that is useful and approximately constant
Material
cp (J/kg-K)
Aluminum
900
Copper
385
Gold
129
Iron
444
Lead
159
Nickel
444
Silver
237
Titanium
523
Tungsten
133
Al2O3
160
MgO
457
SiC
344
Carbon (diamond)
519
Carbon (graphite)
711
Nylon 6,6
1260-2090
Phenolic
1460-1670
Polyethylene (high density)
1920-2300
Polypropylene
1880
Polytetraflouroethylene
1050
Example
•
A passive house will include a trombe wall to
absorb and store heat. It will be built from 2 kg
bricks.
•
How many bricks are needed to absorb 50 MJ of
heat by increasing 10° C?
• (The specific heat of the brick is 850 J/kg-K)
Example (cont.)
• q = cp m ∆T => m = q/(c ∆T)
•
Recall that our target ∆T is 10C = 10K.
• m = 50 MJ/(850 J/kgK * 10K) = 5,880 kg
•
We have 2 kg bricks, so we need
•
2,940 bricks.
Example cont.
• Suppose we wanted to accomplish the same goal
using water?
•
The specific heat of water is 1 cal/g-K, with a
density of 1 Mg/m3. (1 liter = .001 m3)
Example cont
•
Well, we need to fix our units first. We have been
solving things with Joules, but our number is in
calories.
•
There are 0.2389 cals in a Joule. So:
•1 (cal/g-K)/ 0.2389 (cal/J) = 4.19 J/g-K
Example cont.
•
q = cp m ∆T => m = q/(c ∆T)
•
Recall that our target ∆T is 10C = 10K.
•
m = 50 MJ/ (4.19 J/g-K * 10 K) = 1.19 Mg
•
There are 1,000 grams in a liter of water, so
• we need
•
1,190 liters of water.
Example - compare
• 2,940 bricks at 2 kg each, means 5,980 kg of
bricks (almost 6 tons)
•
1,190 liters of water means 1,190 kg of water (a
little over 1 ton)
Thermal Expansion
QuickTime™ and a
decompressor
are needed to see this picture.
Thermal Expansion
• An increase in temperature leads to increased
thermal vibration of the atoms
•
This leads to greater seperation distance of the
atoms.
Thermal Expansion
Thermal Expansion
• The percent change in length is given by:
• ε = α ∆T
• ε = strain
•
α = coefficient of thermal expansion
•
∆T = change in temperature
CTE vs Temperature
Material
CTE @ 27C (10-6/C)
CTE @ 527C (10-6/C)
Aluminum
23.2
33.8
Copper
16.8
20.0
Gold
14.1
16.5
Nickel
12.7
16.8
Silver
19.2
23.4
Tungsten
4.5
4.8
Mullite
5.3
5.3
Porcelain
6.0
6.0
Fireclay refractory
5.5
5.5
Al2O3
8.8
8.8
MgO
7.6
7.6
SiC
4.7
4.7
Silica glass
0.5
0.5
Soda-lime-silica glass
9.0
9.0
Nylon 6,6
30
xx
Phenolic
30-45
xx
Polyethylene (high density)
149-301
xx
Polypropylene
68-104
xx
Polytetraflouroethylene
(PTFE)
99
xx
General correlations
Weakly bonded
solids
Strongly bonded
solids
Low melting point
High melting point
Low elastic modulus High elastic modulus
High CTE
Low CTE
Example
•
We have a tungsten pin that is just a LITTLE too
big to fit into the opening in a nickel bar.
•
The pin is 5.000 mm in diameter and the hole is
4.999 mm at room temperature (25 C)
•
We know that nickel has a higher CTE than
tungsten (12.7 x 10-6 vs 4.5 x 10-6), so we figure
we can heat them both up and the hole will
expand more than the pin and it should fit!
•
How much should we heat them up?
Example continued
•
Well, we want to heat it up enough so that the
diameter of the tungsten equals the diameter of
the nickel. If the tungsten diameter increases by
∆dt and the nickel by ∆dn, we want
•
dt + ∆dt = dn + ∆dn
•
We know dt = 5.000 and dn = 4.999 already, so we
want
•
5.000 + ∆dt = 4.999 + ∆dn
Example
•
We know that the change in diameter comes from
strain:
•
∆dt = ε dt
•
and we know that the strain comes from the
temperature change:
• ε = α ∆T
Example: Plugging together
•
put all the equations together:
•
dt + ∆dt = dn + ∆dn
•
dt + εt dt = dn + εn dn
• dt + αt ∆T dt = dn + αn ∆T dn
•
dt - dn =αn ∆T dn - αt ∆T dt = (αn dn - αt dt)∆T
•
∆T = (dt - dn)/(αn dn - αt dt)
•
∆T = (5 - 4.999)/(4.5 x 10-6 x 4.999 - 12.7 x 10-6 x
5.000) ∆T = 49.4 C = 121 F