Chapter 20 – THERMAL PROPERTIES AND PROCESSES

Thermal Expansion

When the temperature goes up, the lengh of an object increases,

ΔL/L = αΔT,

where the proportionality coefficient α is the coefficient of linear expansion.

It depends on pressure and temperature; a stricter definition is

α = lim [ΔL/LΔT

] ΔT→0 = �

�

��

��

The coefficient of volume expansion

β = lim [ΔV/V

ΔT] ΔT→0 =

�

�

��

��

For a given material

β = 3α.

Example 1. Suppose we have an object with a circular hole. Will the radius of

the hole decrease or increase with rising temperature?

To understand the answer, let us assume that we have a circular hole on a1

ruler with the radius 2 cm and centered on a 4 cm mark on the ruler. Then the

edges of the hole are at 2cm and 6 cm marks. Since the distance between the

marks on the ruler increase with rising temperature, the radius of the hole

increases as well.

Thus the size of the holes always increases with temperature.

Water is an important exception:

In a narrow interval of temperatures it actually

contracts when heated!

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Example 2. What is the change in length of a 500 mi steel pipeline

after it temperature raises by 1800F?

ΔT = � �

1800� = 100�.

Then

∆� = ��∆� = 11 ∙ 10 − 6�− 1 × 100� × 500� = 0.55�

Example 3. How much water will be spilled after you heat a 100L Pyrex can with

100L of water and heat the water from 200C to500C?

The volume change of water

∆! = "#!∆� = 0,21 ∙ 10 − 3�− 1 × 100� × 30� = 0.63L

For the glass,

∆! = "'()*+!∆� = 3�'()*+!∆� = 3 × 3.25 ∙ 10 − 6�− 1 × 100� × 30�

= 2.95 ∙ 10− 2�

The amount of spilled water is

0.63L - 2.95 ∙ 10− 2� = 0.60L

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Example 4. Young modulus of copper is Y = 110GN/m2 and its breaking stress is

230MN/m2. By how much can you cool a copper rod with fixed ends before it

breaks?

The stress F/A is related to stretching as

-=

.∆�

�= /�∆�. Solving for ∆�,

∆8 =

9

:

.;=

<=>?@

A<

BB>C@

A< ×�D∙�E

FG�

F�

= 122K

The van der Waals Equation and Liquid –Vapor Isotherms

Often, the ideal gas equation H! = IJ� should be replaced by a more accurate

van der Waals equation of state,

H +LIM

!M! − NI = IJ�,

Where b is the volume of one mole of gas molecules and a describes the

interaction between molecules.

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The value of the constant b can be used to estimate the size of a molecule.

For example, since the volume of 1mol of nitrogen molecules occupies the

volume 38.7cm3, the volume of one molecule

O =N

P-

=38.7R�S/�UV

6.02 × 10MS�UV*RWV*X/�UV= 6.43 × 10FMSR�S/�UV*RWV*

and the size of a molecule Z = O�/S = 0.4I�.5

Example 5. A 50L tank contains 1000 mol of helium at H = 500L[�. What

fraction of the pressure is the termLIM

!M and what fraction of the volume

is NI?

gh<

�<=

>.>=ijk<lmA

Ano< �EEEpqr M

(�E�)<= 13.8atm; 13.8/500 = 2.8%

NI = (0.0238�

pqr)1000mol =23.8 l; 23.8/50 = 47.6%

Temperature of the gas

� =(H +

LIM

!M )(! − IN)

IJ=

513.8L[� × 26.2�

1000�UV × 0.082L[� ∙ � ∙ �UVF� ∙ �F�

= 164�

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Phase Diagrams

P-T phase diagram of H2O

At the critical point density of the

condensed vapor becomes equal

to that of the liquid and the phase

difference disappears (the end

point). The curve O-B (liquid –

solid equilibrium) does not have

an end point

The direct change from a solid to a

vapor below the curve A-O is called

sublimation

Liquid water cannot exist below the temperature of the triple point

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The Transfer of Heat

Heat is the transfer of energy due to a temperature difference.

There are three mechanisms of this energy transfer: conduction, convection, and radiation

For all mechanisms of heat transfer the rate of cooling is approximately proportional to

the temperature difference between the body and its surrounding (the Newton law of

cooling)

During conduction, the energy is transferred due to interaction between molecules while

the molecules themselves are not transported. The thermal current I

u =�v

�w= −xy

��

�z,

where dT/dx is the temperature gradient, A is the cross-section of the conductor, and k is

the coefficient of thermal conductivity

Temperature gradually changes

along the systemThe equation for the thermal current

can be rewritten as

∆� = u|∆+|

xy= uJ

where R is the thermal resistance and

∆� is the temperature drop.8

If the thermal resistors are connected in series,

�� − �M = uJ�, �M − �S = uJM,

and

∆� = �� − �S = u J� + JM = uJ|}

as for electric currents through the resistors

connected in series

If the resistors are connected in parallel,

uwqwgr = u� + uM + ⋯ = ∆�1

J�

+1

JM

+ ⋯ =∆�

J|}

,

�

���=

�

�B+

�

�<+ ⋯,

as for electric currents through the resistors

connected in parallel

Example 6. Find thermal current in each bar in the

figure, the total current, and the equivalent thermal

resistance of the two-bar system. The bars of length

5 cm have a rectangular cross-section 2cm×3cm.

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The resistance of a lead bar

J�� =|∆+��|

x��y��

=0.05�

35.3�/(� ∙ �) × (0.02� ∙ 0.03�)

=2.36�

�For a silver bar,

J-� =|∆+-�|

x-�y-�

=0.05�

429�/(� ∙ �) × (0.02� ∙ 0.03�)

=0.194�

�The currents are

u�� = ∆�

J��

=100�

2.36� ∙ �F�= 42.4�,

u-� = ∆�

J-�

=100�

0.194� ∙ �F�= 515.5�

The equivalent thermal resistance and the total current

are1

J|}

=1

J��

+1

J-�

, J|} = 0.179� ∙ �F�,

uwqw = u�� + u-� = 558�10

In the building industry the thermal resistance of a square foot of cross-sectional area

of material is called its R factor, Rf. The R-factor is related to the thermal resistance R

of the system measured in 0F/(Btu/h) as Rf = AR where the cross-section A is measured

in square feet. The R-factor can be also expressed as

J� = Jy =|∆+|

xwhere |Δx| is the thickness of the sheet of material. The R-factors of various building

materials are in the Table. Pay attention to the units!

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Example 7. Find the equivalent R-factor of a roof consisting of a triple layer of

asphalt shingles, 4in of roof insulation, and 1.5in-thick pine board.

The R-factor of triple layer of shingles is Rf sh = 3∙(0.44h∙ft2∙0F/Btu)

The R-factor of 4in of roof insulation is Rf in = 4∙(2.8h∙ft2∙0F/Btu)

The R-factor of 1.5in of pine should be obtained from the conductivity table,

Rf p =|Δx|/kp = 1.5in/(0.78Btu∙in/h∙ft2∙0F) = 1.92 h∙ft2∙0F/Btu

Since all these layers are assembled in series, the equivalent R-factor of the roof

Rf eq = Rf sh + Rf in + Rf p = 14.4 h∙ft2∙0F/Btu

The thermal conductivity of air is very low. The effective way of using air for

thermal insulation is to trap air in small pockets to prevent convection.

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Convection

Convection is the transfer of heat by the transport material itself.

Convection takes place mostly in fluids and gases and in responsible for

ocean currents, air circulation, wind. Convection is used to heat (cool) the

building by circulating the hot (cold) air. Mathematics of convection is rather

complicated.

Radiation

All objects can emit and absorb electromagnetic radiation. The rate of

radiation of energy is given by the Stefan-Boltzmann law,

H� = *�y�� ,

Where is the radiated power, A is the surface area of the radiating body,

0 ≤ e ≤ 1 is the emissivity of the surface, and the Stefan’s constant

� = 5.6703 ∙ 10F� �

�M ∙ ��

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The rate at which an object absorbs radiation is given by

Hg = *�y��4

where Ts is the temperature of the source of radiation. If the object both

emits and absorbs the radiation, the net radiated power is

Hh|w = *�y(�� − ��4)

An object that absorbs all incoming radiation e = 1 is called a blackbody

which is also an ideal radiator.

Power radiated by a blackbody as a function of

the wavelength at different temperatures

The positions of the maxima

on the curves are given by the

Wien’s law:

λ�L+ = 2.898 �� ∙ �

�

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Radiation from the Sun

The radiation emitted by the surface of the Sun has the wavelength about

500 nm. According to the Wien’s law this means that the temperature of

the surface of the Sun is approximately 5800 K.

For a body at room temperature T = 300 K, λ�L+ = 9.66 μm.

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Review of Chapter 20

Coefficient of linear thermal expansion � = ∆�

� ∆�

Coefficient of volume thermal expansion " = ∆�

� ∆�= 3�

The van der Waals equation of state H +gh<

�<! − IN = IJ�

The triple point is the unique point at which liquid, solid, and gas phases

are at equilibrium with each other, For water �w���r| = 273.16�

Vapor pressure is the pressure at which liquid and gas phases are in

equilibrium with each other.

The critical point is the point in which the density of liquid and gas phases

become equal to each other. For water �� = 647.4�

Heat transfer occurs by conduction, convection, and radiation16

Thermal current u =�v

��= −xy

��

�z(k is the coefficient of thermal

conductivity, A is the cross-section)

Thermal resistance J =∆�

�=

∆z

�-

Thermal conductors in series J|} = J� + JM + JS + ⋯

Thermal conductors in parallel 1/J|} = 1/J� + 1/JM + 1/JS + ⋯

The R-factor is the thermal resistance for a square foot of a slab of material

in units of in∙ft2∙0F/(Btu/h), J� = Jy =∆z

�

The Stefan-Boltzmann law of radiation H� = *�y�� where the Stefan’s

constant � = 5.6703 × 10F� �

p<∙�iand the emissivity e is between 0 and 1

Net power radiated by a body at T to its environment at T0

Hh|w = *�y(�� − �E4)

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Wien’s law for the wavelength for the maximal radiation λpgz =M.�� pp∙�

�

A blackbody absorbs all radiation, is a perfect radiator, and has the emissivity

e = 1

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