Mechanical Engineering Principles

THERMAL EXPANSION 225

20.6 Coefficient of cubic expansion

The amount by which unit volume of a mate-
rial increases for a one degree rise of temperature
is called thecoefficient of cubic (or volumetric)
expansionand is represented byγ(Greek gamma).
If a material having an initial volume V 1 at
temperaturet 1 and having a coefficient of cubic
expansionγ, has its temperature raised tot 2 ,then
the new volumeV 2 of the material is given by:

New volume=initial volume

+increase in volume

i.e. V 2 =V 1 +V 1 γ(t 2 −t 1 )

i.e. V 2 =V 1 [1+γ(t 2 −t 1 )] ( 20. 3 )

It is shown in Problem 6 below that the coeffi-
cient of cubic expansion is three times the coeffi-
cient of linear expansion, i.e.γ = 3 α,toavery
close approximation. A liquid has no definite shape
and only its cubic or volumetric expansion need be
considered. Thus with expansions in liquids, equa-
tion (3) is used.

Problem 6. Show that for a rectangular

block of material having dimensionsL,b

andh, the coefficient of cubic expansion

γ≈ 3 α,whereαis the coefficient of linear

expansion.

Initial volume,V 1 =Lbh. For a temperature rise of
1 K, sideLexpands to(L+Lα),sidebexpands to
(b+bα)and sidehexpands to(h+hα)
Hence the new volume of the blockV 2 is given by:

V 2 =(L+Lα)(b+bα)(h+hα)

=L( 1 +α)b( 1 +α)h( 1 +α)

=Lbh( 1 +α)^3 =Lbh( 1 + 3 α+ 3 α^2 +α^3 )

≈Lbh( 1 + 3 α)

since terms inα^2 andα^3 are very small
HenceV 2 ≈V 1 ( 1 + 3 α)
For a temperature rise of(t 2 −t 1 )K,

V 2 ≈V 1 [1+ 3 α(t 2 −t 1 )]

Thus from equation (20.3),γ≈ 3 α

Some typical valuesfor the coefficient of cubic

expansion measured at 20°C (i.e. 293 K) include:

Ethyl alcohol 1. 1 × 10 −^3 K−^1 Mercury 1. 82 × 10 −^4 K−^1

Paraffin oil 9 × 10 −^2 K−^1 Water 2. 1 × 10 −^4 K−^1

The coefficient of cubic expansionγis only constant

over a limited range of temperature.

Problem 7. A brass sphere has a diameter

of 50 mm at a temperature of 289 K. If the

temperature of the sphere is raised to 789 K,

determine the increase in (a) the diameter

(b) the surface area (c) the volume of the

sphere. Assume the coefficient of linear

expansion for brass is 18× 10 −^6 K−^1.

(a) Initial diameter,L 1 =50 mm, initial tempera-

ture,t 1 =289 K, final temperature,t 2 =789 K

andα= 18 × 10 −^6 K−^1.

New diameter at 789 K is given by:

L 2 =L 1 [1+α(t 2 −t 1 )]

from equation (20.1)

i.e. L 2 =50[1+( 18 × 10 −^6 )( 789 − 289 )]

=50[1+ 0 .009]= 50 .45 mm

Hence the increase in the diameter is

0.45 mm.

(b) Initial surface area of sphere,

A 1 = 4 πr^2 = 4 π

(

50

2

) 2

= 2500 πmm^2

New surface area at 789 K is given by:

A 2 =A 1 [1+β(t 2 −t 1 )]

from equation (20.2)

i.e. A 2 =A 1 [1+ 2 α(t 2 −t 1 )]

since β= 2 α,

to a very close approximation

Thus A 2 = 2500 π[1+ 2 ( 18 × 10 −^6 )( 500 )]

= 2500 π[1+ 0 .018]

= 2500 π+ 2500 π( 0. 018 )