EXPERIMENT 4: THERMAL COEFFICIENT OF EXPANSION
What is Temperature???
y
Consider N atoms confined in a
box.
x
z
Force on wall = ma = m dv/dt =
d(mv)/dt
F
L
dp
dt
Each time a particle hits a wall, the component
of the velocity that is normal to that wall reverses directions. Consider the left wall
normal to the x direction of velocity. The change in momentum when the particle his this
wall is (-mvx)-mvx = -2mvx. The time between collisions, or the time it takes for the
momentum of the particle to change |2mvx| is the time is takes to travel between opposite
walls of the box. With the velocity vx, this time is 2L/vx. Substituting in 2mvx for change
in momentum in time 2L/vx for force, and L2 for the area of the wall that the force is
acting over, pressure (P) can be calculated by summing over all of the particles in the
box:
N
 2mv x
F
1 1
Nmv x2
P 2  2

L
L 2L / vx
L3
v2 = vx2 + vy2 + vz2 and vx = vy = vz, it can be seen that vx2 = 1/3 v2 Replacing vx with v
and volume (V) for L3 gives:
Nmv 2
P
3V
then, using the ideal gas law: PV = NKT
NkT Nmv2
P

V
3V
1 2 3
mv  kT
2
2
So… T is a measure of the kinetic energy of atoms!
Each atom does not have the exact same velocity, so t accurately
describe T a velocity distribution can be found.
12%
Maxwellian velocity
distribution :
10%
p(V)
8%
6%
4%
2%
0%
0
10
20
V
30
 m 
P(v )  4 v 

2
k
T
 b 
2
40
3/ 2
e
 mv 2 / 2 kbT
P(v) = probability of finding an atom with velocity V if temperature
is T
Interesting side note: The distribution of the x, y, and z velocity
components follow a random, or Normal distribution, however the
magnitude of the total velocity of each particle, v, (v2 = vx2 + vy2 + vz2)
follows the Maxwell distribution – which is not random. One step
farther, if you measure the temperature of a room consisting of billions
of randomly moving atoms, you get a very precise number. So…. How
does all this randomness add up to create an overall ordered system? I
don’t know – you tell me!
Why do most solids expand when they warm up?
Atoms in a solid are connected to each other with bonds
As T increases, atomic velocities increase
Vibrating atoms push their neighbors aside
This increases the volume of the solid!
Eventually the vibrations become so violent
the bonds break, and the solid melts.
THERMAL STRESS IN MATERIAL SPECIMEN AND STRAIN GAGES
coefficient of thermal expansion =  =  L/( T*Lo)
where:
 T = the change in temperature
 L = the change in dimension
Lo = the initial dimension
Thermal expansion coef for engineering materials:
http://www.vishay.com/brands/measurements_group/guide/tn/tn504/504t1.htm
(we are using Titanium rather than glass – this is the heaver of the two metals)
Strain gage resistance change is due to
 the change in resistivity of the material
 the thermal expansion of the strain gage, and
 the thermal expansion of the test specimen.
 R / R  i = [ G + (  i -  G) FG]  T
Where:
i = either material to be tested or the reference material
 G = temperature coefficient for resistivity of the strain gage (This should drop
out of the final equ)
 = thermal expansion coefficient for the test specimen (i) or the Gage, Go
FG = gage factor = 2.09
Note that there is a differential expansion between the grid and the tested
material. Plus the resistivity of the grid material varies with temperature.
Circuit:
10 ohm resistor
measure voltage across
this to find
current and power to
circuit… Don’t put this
resistor in until Q8.
R
Thermiste
r
Thermister
Vout
Helpful Hints:
Vout will be VERY small so…
Balance your bridge
Choose the smallest input V range
Ground LabVIEW
This is a pretty complex circuit so….
Be sure to organize your wires well!
Use your gray twisted pair to connect circuit to LabView
Color code your wires (black = ground, red = +V supply)
Known (Ti)
strain gage
Experiment 4 Preparation Results
To get you started….
1. Determine how to connect the two material specimens in the bridge circuit so the
thermal effects on the strain gages are compensated for or cancelled. Attach the
circuit schematic (20 pts).
-Use a ½ wheatstone bridge for this one. Hint – you want Vout to be a function of
(Runknown - Rknown) and R, or, if both materials were the same,( Runknown = Rknown),
Vout would be 0 for this arrangement…
2. a. For the bridge circuit from question 1, develop an equation for Vout in terms of
Vsupply, Runknown, Rknown, and R (6 pts).
At the very end of this one, you can make an assumption – assume that R <<< R, so
(R + R) ~ R.
Experiment 4 Report Questions: Help – and some additions!
1. What is the thermal coefficient of expansion for the unknown
specimen?_____________ Show your work completely and clearly below. (20 pts)
Addition – using the data listed below, what is the unknown metal made of?
8. Place 10ohm resistor between power supply and circuit.
Measure V drop across 10 ohm resistor (should be around 140mV)
I = V10ohm/R(10ohm) = current into your circuit
power = Vs * I = power into circuit
11. a. Multiply your bench number by 20. What is the available DAS hardware gain
closest to this number (5 pts)?
Ie if you are at bench 3, 3*20 = 60, closest gain = 50
11. d. At this gain, and with a quarter bridge, what is the resolution in strain (5 pts)?
Ie. Using a ¼ bridge, what strain would be represented by Vout if Vout =
Vresolution?
Thermistor with Yellow wire / Aluminum tag:
http://www.thermometrics.com/assets/images/sa.pdf
http://www.thermometrics.com/assets/images/d10_3.pdf
Thermistor with black wires / brass tag:
http://www.thermometrics.com/assets/images/ms.pdf
http://www.thermometrics.com/assets/images/d95.pdf
Help with T calculations (using Mathematica)
http://www.mines.edu/fs_home/jmoss/exp4.nb
(save into temp, then open…)
°C to °F
°C * 1.8 + 32 = °F
http://www.soc.duke.edu/~garyt/temp.html
Table 504.1
Thermal Expansion Coefficients of Engineering Materials
Material
Description
Expansion
Coefficients **
Alumina ,
fired
3.0 ppm/° F
(5.4 ppm/° C)
03
Aluminum Alloy ,
2024-T4*, 7075-T6
12.9 ppm/° F
(23.2 ppm/° C)
13*
Beryllium ,
pure
6.4 ppm/° F
(11.5 ppm/° C)
06
Beryllium Copper ,
Cu 75, Be 25
9.3 ppm/° F
(16.7 ppm/° C)
09
Brass, Cartridge ,
Cu 70, Zn 30
11.1 ppm/° F
(20.0 ppm/° C)
13
Bronze, Phosphor ,
Cu 90, Sn 10
10.2 ppm/° F
(18.4 ppm/° C)
09
Cast Iron ,
Gray
6.0 ppm/° F
(10.8 ppm/° C)
06
Copper ,
pure
9.3 ppm/° F
(16.7 ppm/° C)
09
Glass ,
Soda, Lime, Silica
5.1 ppm/° F
(9.2 ppm/° C)
05
Inconel ,
Ni-Cr-Fe alloy
7.0 ppm/° F
(12.6 ppm/° C)
06
Inconel X
Ni-Cr-Fe alloy
6.7 ppm/° F
(12.1 ppm/° C)
06
Invar
Fe-Ni alloy
0.8 ppm/° F
(1.4 ppm/° C)
00
Magnesium Alloy *,
AZ-31B
14.5 ppm/° F
(26.1 ppm/° C)
15*
Molybdenum *,
pure
2.2 ppm/° F
(4.0 ppm/° C)
03*
Monel ,
Ni-Cu alloy
7.5 ppm/° F
(13.5 ppm/° C)
06
Recommended
S-T-C
Nickel-A ,
Cu-Zn-Ni alloy
6.6 ppm/° F
(11.9 ppm/° C)
06
Quartz ,
fused
0.3 ppm/° F
(0.5 ppm/° C)
00
Steel, Alloy ,
4340
6.3 ppm/° F
(11.3 ppm/° C)
06
Steel, Carbon ,
1008, 1018*
6.7 ppm/° F
(12.1 ppm/° C)
06*
Steel, Stainless ,
Age Hardenable (17-4PH)
6.0 ppm/° F
(10.8 ppm/° C)
06
Steel, Stainless ,
Age Hardenable (17-7PH)
5.7 ppm/° F
(10.3 ppm/° C)
06
Steel, Stainless ,
5.0 ppm/° F
Age Hardenable (PH15-7Mo) (9.0 ppm/° C)
05
Steel, Stainless ,
Austenitic (304*)
9.6 ppm/° F
(17.3 ppm/° C)
09*
Steel, Stainless ,
Austenitic (310)
8.0 ppm/° F
(14.4 ppm/° C)
09
Steel, Stainless ,
Austenitic (316)
8.9 ppm/° F
(16.0 ppm/° C)
09
Steel, Stainless ,
Ferritic (410)
5.5 ppm/° F
(9.9 ppm/° C)
05
Tin ,
pure
13.0 ppm/° F
(23.4 ppm/° C)
13
Titanium ,
pure
4.8 ppm/° F
(8.6 ppm/° C)
05
Titanium Alloy ,
6Al-4V*
4.9 ppm/° F
(8.8 ppm/° C)
05*
Titanium Silicate *,
polycrystalline
0.0 ppm/° F
(0.0 ppm/° C)
00*
Tungsten ,
pure
2.4 ppm/° F
(4.3 ppm/° C)
03
Zirconium ,
pure
3.1 ppm/° F
(5.6 ppm/° C)
03
*Indicates type of material used in determining thermal output curves supplied
with Micro-Measurements strain gages.
**Nominal values at or near room temperature for temperature coefficient of
expansion values.
***ppm = parts per million = 10-6