Mechanical Engineering Principles

224 MECHANICAL ENGINEERING PRINCIPLES

of the hot pipe if the coefficient of linear

expansion of lead is 29× 10 −^6 K−^1.

[50.0928 m]

2. A rod of metal is measured at 285 K
   and is 3.521 m long. At 373 K the rod
   is 3.523 m long. Determine the value of
   the coefficient of linear expansion for the
   metal. [6. 45 × 10 −^6 K−^1 ]


3. A copper overhead transmission line has
   a length of 40.0 m between its supports at
   20 °C. Determine the increase in length at
   50 °C if the coefficient of linear expansion
   of copper is 17× 10 −^6 K−^1. [20.4 mm]


4. A brass measuring tape measures 2.10 m
   at a temperature of 15°C. Determine
   (a) the increase in length when the tem-
   perature has increased to 40°C

(b) the percentage error in measurement

at 40°C. Assume the coefficient of

linear expansion of brass to be

18 × 10 −^6 K−^1.

[(a) 0.945 mm (b) 0.045%]

5. A pendulum of a ‘grandfather’ clock is
   2.0 m long and made of steel. Determine
   the change in length of the pendulum if
   the temperature rises by 15 K. Assume the
   coefficient of linear expansion of steel to
   be 15× 10 −^6 K−^1. [0.45 mm]


6. A temperature control system is operated
   by the expansion of a zinc rod which is
   200 mm long at 15°C. If the system is
   set so that the source of heat supply is
   cut off when the rod has expanded by
   0.20 mm, determine the temperature to
   which the system is limited. Assume the
   coefficient of linear expansion of zinc to
   be 31× 10 −^6 K−^1 .[47. 26 °C]


7. A length of steel railway line is 30.0 m
   long when the temperature is 288 K. Det-
   ermine the increase in length of the line
   when the temperature is raised to 303 K.
   Assume the coefficient of linear expansion
   of steel to be 15× 10 −^6 K−^1.
   [6.75 mm]


8. A brass shaft is 15.02 mm in diameter
   and has to be inserted in a hole of diam-
   eter 15.0 mm. Determine by how much
   the shaft must be cooled to make this

possible, without using force. Take the

coefficient of linear expansion of brass as

18 × 10 −^6 K−^1 .[74K]

20.5 Coefficient of superficial

expansion

The amount by which unit area of a material

increases when the temperature is raised by one

degree is called thecoefficient of superficial (i.e.

area) expansionand is represented byβ (Greek

beta).

If a material having an initial surface areaA 1 at

temperaturet 1 and having a coefficient of superficial

expansionβ, has its temperature increased tot 2 ,then

the new surface areaA 2 of the material is given by:

New surface area

=original surface area+increase in area

i.e. A 2 =A 1 +A 1 β(t 2 −t 1 )

i.e. A 2 =A 1 [1+β(t 2 −t 1 )] ( 20. 2 )

It is shown in Problem 5 below that the coefficient

of superficial expansion is twice the coefficient of

linear expansion, i.e.β = 2 α, to a very close

approximation.

Problem 5. Show that for a rectangular area

of material having dimensionsLbybthe

coefficient of superficial expansionβ≈ 2 α,

whereαis the coefficient of linear expansion.

Initial area,A 1 =Lb. For a temperature rise of

1 K, sideLwill expand to(L+Lα)and sideb

will expand to(b+bα). Hence the new area of the

rectangle,A 2 , is given by:

A 2 =(L+Lα)(b+bα)

=L( 1 +α)b( 1 +α)=Lb( 1 +α)^2

=Lb( 1 + 2 α+α^2 )≈Lb( 1 + 2 α)

since α^2 is very small (see typical values in

Section 20.4)

HenceA 2 ≈A 1 ( 1 + 2 α)

For a temperature rise of(t 2 −t 1 )K

A 2 ≈A 1 [1+ 2 α(t 2 −t 1 )]

Thus from equation (20.2),β≈ 2 α