17.1.2 Assignment No. 2
TASK 1
a) Describe, briefly, what happens to plastic guttering and uPVC windows when the air temperature rises during the day and drops during the night.
Answer:
Plaster guttering and uPVC windows will expand during the day as the air temperature increases and contract during the night as the temperature drops. The expansion and contraction will not cause any problem in plastic guttering as it is not restrained, but in the case of uPVC windows an all around gap of 5 mm is left between the window frame and the wall to accommodate the expansion/contraction.
b) Describe, briefly, what happens to water when:
i) its temperature is lowered to – 3 °C
ii) its temperature is raised to 100 °C.
Answer:
Water becomes ice when its temperature is lowered to -3 ºC (change from liquid to solid) and it expands. At 100 ºC water starts to evaporate as steam is formed, i.e. change from liquid to gas state takes place.
c) Explain the effect of temperature change, in terms of latent heat and sensible heat, on the physical changes in the state of water.
Answer:
Refer to section 8.4.2 (chapter 8) for the answer to this question.
d) (i) A 5.0 m length of steel tube is heated from 10 °C to 40 °C. Calculate the thermal movement if the coefficient of linear expansion of steel is 0.000 012.
Answer:
Original length (l) = 5.0 m = 5000 mm
Change in temperature (Dt) = 40 - 10 = 30 °C
a = 0.000 012
a =
Change in length or thermal movement, Dl = a.l.Dt
= 0.000 012 × 5000 × 30
= 1.8 mm (expansion)
d) (ii) A 4.0 m length of steel tube is cooled from 45 °C to 20 °C. Calculate the thermal movement if the coefficient of linear contraction of steel is 0.000 012.
Answer:
Original length (l) = 4.0 m = 4000 mm
Change in temperature (Dt) = 45 - 20 = 25 °C
a = 0.000 012
Change in length or thermal movement, Dl = a.l.Dt
= 0.000 012 × 4000 × 25
= 1.2 mm (contraction)