Mechanical Engineering Principles

226 MECHANICAL ENGINEERING PRINCIPLES

Hence increase in surface area

= 2500 π( 0. 018 )= 141 .4mm^2

(c) Initial volume of sphere,

V 1 =

4

3

πr^3 =

4

3

π

(

50

2

) 3

mm^3

New volume at 789 K is given by:

V 2 =V 1 [1+γ(t 2 −t 1 )]

from equation (20.3)

i.e. V 2 =V 1 [1+ 3 α(t 2 −t 1 )]

since γ= 3 α,to a very close

approximation

Thus V 2 =

4

3

π( 25 )^3

×[1+ 3 ( 18 × 10 −^6 )( 500 )]

=

4

3

π( 25 )^3 [1+ 0 .027]

=

4

3

π( 25 )^3 +

4

3

π( 25 )^3 ( 0. 027 )

Hence the increase in volume

=

4

3

π( 25 )^3 ( 0. 027 )=1767 mm^3

Problem 8. Mercury contained in a

thermometer has a volume of 476 mm^3 at

15 °C. Determine the temperature at which

the volume of mercury is 478 mm^3 ,

assuming the coefficient of cubic expansion

for mercury to be 1. 8 × 10 −^4 K−^1.

Initial volume,V 1 =476 mm^3 , final volume,
V 2 =478 mm^3 , initial temperature,t 1 = 15 °Cand
γ= 1. 8 × 10 −^4 K−^1
Final volume,

V 2 =V 1 [1+γ(t 2 −t 1 )],

from equation (20.3)

i.e.V 2 =V 1 +V 1 γ(t 2 −t 1 ),

from which

(t 2 −t 1 )=

V 2 −V 1

V 1 γ

=

478 − 476

( 476 )( 1. 8 × 10 −^4 )

= 23. 34 °C

Hencet 2 = 23. 34 + 15 = 38. 34 °C

Hence the temperature at which the volume of

mercury is 478 mm^3 is 38. 34 °C

Problem 9. A rectangular glass block has a

length of 100 mm, width 50 mm and depth

20 mm at 293 K. When heated to 353 K its

length increases by 0.054 mm. What is the

coefficient of linear expansion of the glass?

Find also (a) the increase in surface area

(b) the change in volume resulting from the

change of length.

Final length,L 2 =L 1 [1+α(t 2 −t 1 )], from equa-

tion (20.1), hence increase in length is given by:

L 2 −L 1 =L 1 α(t 2 −t 1 )

Hence 0. 054 =( 100 )(α)( 353 − 293 )

from which, the coefficient of linear expansion is

given by:

α=

0. 054

( 100 )( 60 )

= 9 × 10 −^6 K−^1

(a) Initial surface area of glass,

A 1 =( 2 × 100 × 50 )+( 2 × 50 × 20 )

+( 2 × 100 × 20 )

= 10000 + 2000 + 4000

=16 000 mm^2

Final surface area of glass,

A 2 =A 1 [1+β(t 2 −t 1 )]

=A 1 [1+ 2 α(t 2 −t 1 )],

sinceβ= 2 αto a very close approximation

Hence,increase in surface area

=A 1 ( 2 α)(t 2 −t 1 )

=(16 000)( 2 × 9 × 10 −^6 )( 60 )

= 17 .28 mm^2