Transcript Thermal Expansion - MFSacedon's website

Thermal Expansion
Created by: Marlon Flores Sacedon
Physics section, DMPS
June 2010
MFS
Thermal Expansion
Most materials expand when their temperature increase.
• The decks of bridges need special joints and supports to allow for
expansion.
• A completely filled and tightly capped bottle of water cracks when
it is heated
• You can loosen a metal jar lid by running hot
water over it. .
These are examples of Thermal Expansion
Two Kinds of Thermal Expansion
• Linear Expansion
• Volume Expansion
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Linear
Expansion
Volume
Expansion
T1
Lo
T2
L
T2>T1
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Linear
Expansion
Volume
Expansion
Vo @ T1
T1
Lo
L
T2
T2>T1
L
L  L o T
V@ T2
T2>T1
V  Vo T
L   L o T
V   Vo T
 
 
L  Lo
L

L o T L o T2  T1 
Where:  = coef. of linear expansion (1/K)
L = change in length (m)
L & Lo = final & initial length (m)
T2&T1= final & initial temperature (oC)
V  Vo
V

Vo T Vo T2  T1 
Where:  = coef. of volume expansion (1/K)
V = change in volume (m3)
V & Vo = final & initial volume (m3)
T2&T1= final & initial temperature (oC)
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Coefficients of Linear Expansion
Material
 [K-1 or (Co)-1]
Coefficients of Volume Expansion
 [K-1 or (Co)-1]
Material
Aluminum
2.4x10-5
Aluminum
7.2x10-5
Brass
2.0x10-5
Brass
6.0x10-5
Copper
1.7x10-5
Copper
5.110-5
Glass
0.4-0.9x10-5
Glass
1.2-2.7x10-5
Invar
0.09x10-5
Invar
0.27x10-5
Quartz
0.04x10-5
Quartz
0.12x10-5
Steel
1.2x10-5
Steel
3.6x10-5
Ethanol
75x10-5
Carbon Disulfide
115x10-5
Glycerine
49x10-5
Mercury
18x10-5
Relationship
between coefficient
of volume
expansion &
coefficient of linear
expansion
  3
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Example 1: A surveyor uses a steel measuring tape that is exactly
50.000 m long at a temperature of 20 oC. What is its length on a hot
summer day when the temperature is 35 oC? Answer: 50.009 m
Lo
Solution
50 m
Temperature of tape @ 20oC
L=?
Temperature of tape @ 35oC
From table, the coefficient of linear expansion
From the formula:
Transforming
  1.2x105 (Co ) 1
L  Lo
L


L o T L o T2  T1 
L  Lo  Lo T2  T1   Lo 1  T2  T1  = 50.0009 m Ans
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Example 2: The surveyor uses the measuring tape (in Example 1) to
measure a distance when the temperature is 35 oC; the value that she
reads off the tape is 35.794m. What is the actual distance?
Answer: 35.800 m
Solution
Lo = 50 m
Temperature of tape @ 20oC
L = 50.009 m
Temperature of tape @ 35oC
Let: x = actual distance @ T=35 oC
x
50.009m

35.794
50m
x  35.8m
Ans
Example 3: A glass flask with volume 200 cm3 is filled to the brim with
mercury at 20 oC. How much mercury overflows when the
temperature of the system is raised to 100 oC? The coefficient of
linear expansion of the glass is 0.40x10-5 K-1. Answer: 2.7 cm3
Solution
Mercury overflows
( Vover)
Mercury,
expanded
volume( VHg)
Glass flask, expanded
volume ( Vglass)
Mercury column
Vover  VHg  Vglass
Vover  Hg Vo T2  T1   glassVo T2  T1 
Glass flask

Vover  Vo T2  T1  Hg   glass
 

1
Glass flask filled w/
mercury
@ T1=20oC
@ T2=100oC
 glass  0.40 x10
From table:  Hg  18x105 Co
1 
1

glass  3 0.4x105 Co   1.2x105 Co


 

 
5

C 
o 1
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Vover  200100 2018x105 1.2x105
Seat Work1: The Humber Bridge in England has the world’s longest
single span, 1410m in length. Calculate the change in length of steel
deck of the span when the temperature increases from -5.0oC to
18.0oC.
Answer: 0.39 m
Seat Work2: A metal rod is 40.125 cm long at 20.0oC and 40.148 cm
long at 45.0oC. Calculate the average coefficient of linear expansion of
the rod for this temperature range.
Answer: 2.3x10-5 (Co)-1
Seat Work3: A glass flask whose volume is 1000.00 cm3 at 0.0oC is
completely filled with mercury at this temperature. When flask and
mercury are warmed to 55.0oC, 8.95 cm3 of mercury overflow. If the
coefficient of volume expansion of mercury is 18.0x10-5 K-1, compute
the coefficient of volume expansion of the glass.Answer: 1.7x10-5 (Co)-1
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Assignment
1) A Pendulum shaft of a clock is made of brass. What is the fractional
change in length of the shaft when it is cooled from 19.50oC to 5.00oC?
Answer: -2.9x10-4
2) An underground tank with a capacity of 1700L (1.70m3) is filled with
ethanol that has an initial temperature of 19.0oC. After the ethanol has
cooled off to the temperature of the tank and ground, which is 10.0oC,
how much air space will there be above the ethanol in the tank?
(Assume that the volume of the tank doesn’t change.)
Answer: 2.3x10-5 (Co)-1
3) A metal rod that is 30.0 cm long expands by 0.0650 cm when its
temperature is raised from 0oC to 100oC. A rod of a different metal and
of the same length expands by 0.0350 cm for the same rise in
temperature. A third rod, also 30.0 cm long is made up of pieces of
each of the above metals placed end-to-end and expands 0.0580 cm
between 0oC and 100oC. Find the length of each portion of the
Answer: 23.0cm, 7.0cm
composite bar.
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Thermal Stress
A
Y
F
L
A
Lo
 L 
F


so 

L
o

 tension AY
 L 


  T
L 
 o  thermal
F
Tensile Stress 
Force( F)
Area( A )
 L 
 L 
F








T

0
L 
L 
AY
 o  thermal  o  tension
F
L
Tensile Strain 
Lo
Change in length (L)
InitialLength (L o )
F
Stress
A
Young' s Modulus(Y) 

Strain L
Lo
F
  Y T
A
(thermal stress)
Where:
• F = Tensile force, (N)
• A = cross-section area, (m2)
• Y = Young’s Modulus, (Pa or N/m2)
•  = coef. of linear expansion, (K-1)
• T = change in tempersture, (K)
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Approximate Young’s Modulus
Substance
Young’s
Modulus, Y
(Pa)
Aluminum
7.0x1010
Brass
9.0x1010
Copper
11x1010
Crown glass
6.0x1010
Iron
21x1010
Lead
1.6x1010
Nickel
21x1010
Steel
20x1010
Example 1: An aluminum cylinder 10 cm long, with a cross-section
area of 20 cm3, is to be used as a spacer between two steel walls. At
17.2 oC it just slips in between the walls. When it warms to 22.3oC,
calculate the stress in the cylinder and the total force it exerts on each
wall, assuming that the walls are perfectly rigid and a constant distance
apart.
6
4
Answer: -8.6x10 and -1.7x10 N
F=0
Lo=10cm
@ T1=17.2
oC
Area A=20cm3
Stress 
F
  Y T
A
F
  0.70 x1011 Pa (2.4 x10 5 K 1 )( 22.3  17.2) K
A
F
  8.6 x10 6 Pa (or  1200 lb 2 )
in
A
F>0
Lo=10cm
@ T1=22.3 oC
Stress is - 8.6x106 Pa
Negative sign indicates compression
F
  8.6 x106 Pa
2
20in
F = -1.7x104 N
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(compression)
Assignment: a) A wire that is 1.50 m long at 20oC is found to increase
in length by 1.9 cm when warmed to 420oC. Compute its average
coefficient of linear expansion for this temperature range. b) The wire is
stretched just taut (zero tension) at 420oC. Find the stress in the wire if
it is cooled to 20oC without being allowed to contact. Young’s modulus
for the wire is 20.x1011 Pa.
Answer: a) 3.2x10-5 (Co)-1; b)2.5x109 Pa
1.50 m
F
  Y T
A
@ 20oC
0.019 m
@ 420oC
L
0.019


 3.17x105 K 1
Lo  T 1.50(420 20)
F
  20x1011 (3.17x105 )(20  420)
A
 2.5 x109 Pa ( stress) Ans
1.519 m
@ 420oC
1.519 m
@ 20oC
F>0
F>0
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Assignment
1) A brass rod is 185 cm long and 1.60 cm in diameter. What force
must be applied to each end of the rod to prevent it from contracting
Answer: 4.0x104 N
when it is cooled from 120oC to 10oC?
2) Steel train rails are laid in 12.0 m-long segments placed end-to-end.
The rails are laid on a winter day when their temperature is -2.0oC. a)
How much space must be left between adjacent rails if they are to just
touch on a summer day when their temperature is 33.0oC? b) If the rails
are originally laid in contact, what is the stress in them on a summer
day when their temperature is 33.0oC?
Answer: a) 5.0x10-3 m; b) 8.4x107 Pa
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