Wednesday, June 8, 2011

Solutions to Thermal Expansion Problems (Pt. 16)

We had the following thermal expansion Problems:

1)A vertical steel antenna tower is 200 m high. Calculate the change in height of the tower that takes place when the temperature changes from -20 C on a winter day to 30 C on a summer day.

2) A metal rod is 200.00 cm long at 0C and 200.18 cm long at 60C. What is its coefficient of linear expansion?

3) A metal vessel has a volume of 500.00 cm^3 at 0 C. If the coefficient of linear expansion is 0.000011 /C, what is the volume at 50 C?

4) A round hole of 6.000 cm in diameter at 0 C is cut in a sheet of brass(alpha = 0.000018 /C). Find the new diameter of the hole at 50 C.

Solutions:

(1)Given: L1 = 200m

T2 = 30 C and T1 = (-20 C) then (T2 – T1) = (30C – (-20C)) = 50 C

alpha (steel) = 0.000011/C

then: change in length is

x = (0.000011 /C) [200m/ 50C]= 0.000044 m = (L2 – L1)

(2)Given: L1 = 200.00 cm and L2 = 200.18 cm

So: x = (L2 – L1) = 0.18 cm

(T2 – T1 ) = (60 C – 0C) = 60C

Then: alpha = x/ {L1 (T2 - T1)} = 0.18 cm/ [200.00 cm (60C)]

alpha = 0.000015 /C

(3)V = 500.00 cm^3 and (T2 – T1) = 50C

alpha= is 0.000011 /C

Beta = 3(alpha) = 3( 0.000011/C) = 0.000033 /C

delta V= (500.00 cm^3)(0.000033/C) (50C) = 0.825 cm^3

then the new volume = V + delta V = 500.00 cm^3 + 0.82 cm^3 = 500.82cm^3

(4)Given: alpha = 0.000018 /C

(T2 – T1)= 50 C

D = 6.000 cm so r = 3.000 cm

And Area A = πr^2 = 28.274 cm^2

But delta A = 2 (alpha) A ((T2 – T1)

= 2 (0.000018 /C) 28.274 cm^2 (50C) = 0.051 cm^2

New Area A’ = A + delta A = 28.274 cm^2 + 0.051 cm^2

A’ = 28.325 cm^2

Then the new radius r’ = (A’/π)^ ½ = (28.325 cm^2/ 3.141)^ ½

= 3.002 cm

So the new diameter D’ = 2 r’ = 2 (3.002 cm) = 6.004 cm