Mechanical Engineering Principles

THERMAL EXPANSION 223

In practice, allowances are made for such expan-
sions.U-shaped expansion joints are connected into
pipelines carrying hot fluids to allow some ‘give’ to
take up the expansion.

Problem 2. An electrical overhead

transmission line has a length of 80.0 m

between its supports at 15°C. Its length

increases by 92 mm at 65°C. Determine the

coefficient of linear expansion of the material

of the line.

LengthL 1 = 80 .0m,L 2 = 80 .0m+92 mm =
80 .092 m, temperaturet 1 = 15 °C and temperature
t 2 = 65 °

LengthL 2 =L 1 [1+α(t 2 −t 1 )]

i.e. 80. 092 = 80 .0[1+α( 65 − 15 )]

80. 092 = 80. 0 +( 80. 0 )(α)( 50 )

i.e. 80. 092 − 80. 0 =( 80. 0 )(α)( 50 )

Hence, the coefficient of linear expansion,

α=

0. 092

( 80. 0 )( 50 )

= 0. 000023

i.e.α= 23 × 10 −^6 K−^1 (which is aluminium- see
above)

Problem 3. A measuring tape made of

copper measures 5.0 m at a temperature of

288 K. Calculate the percentage error in

measurement when the temperature has

increased to 313 K. Take the coefficient of

linear expansion of copper as 17× 10 −^6 K−^1.

LengthL 1 = 5 .0 m, temperaturet 1 =288 K,t 2 =
313 K andα= 17 × 10 −^6 K−^1
Length at 313 K is given by:

LengthL 2 =L 1 [1+α(t 2 −t 1 )]

= 5 .0[1+( 17 × 10 −^6 )( 313 − 288 )

= 5 .0[1+( 17 × 10 −^6 )( 25 )]

= 5 .0[1+ 0 .000425]

= 5 .0[1.000425]= 5 .002125 m

i.e. the length of the tape has increased by
0.002125 m

Percentage error in measurement at 313 K

=

increase in length

original length

×100%

=

0. 002125

5. 0

× 100 = 0 .0425%

Problem 4. The copper tubes in a boiler are

4.20 m long at a temperature of 20°C.

Determine the length of the tubes (a) when

surrounded only by feed water at 10°C,

(b) when the boiler is operating and the

mean temperature of the tubes is 320°C.

Assume the coefficient of linear expansion of

copper to be 17× 10 −^6 K−^1.

(a) Initial length,L 1 = 4 .20 m, initial tempera-

ture,t 1 = 20 °C, final temperature,t 2 = 10 °C

andα= 17 × 10 −^6 K−^1

Final length at 10°C is given by:

L 2 =L 1 [1+α(t 2 −t 1 )]

= 4 .20[1+( 17 × 10 −^6 )( 10 − 20 )]

= 4 .20[1− 0 .00017]= 4 .1993 m

i.e. the tube contracts by 0.7 mm when the

temperature decreases from 20°Cto10°C.

(b) Length,L 1 = 4 .20 m,t 1 = 20 °C,t 2 = 320 °C

andα= 17 × 10 −^6 K−^1

Final length at 320°C is given by:

L 2 =L 1 [1+α(t 2 −t 1 )]

= 4 .20[1+( 17 × 10 −^6 )( 320 − 20 )]

= 4 .20[1+ 0 .0051]= 4 .2214 m

i.e.the tubes extend by 21.4 mm when the

temperature rises from 20°C to 320°C

Now try the following exercise

Exercise 102 Further problems on the co-

efficient of linear expansion

1. A length of lead piping is 50.0 m long at
   a temperature of 16°C. When hot water
   flows through it the temperature of the
   pipe rises to 80°C. Determine the length