Question

Many hot-water heating systems have a reservoir tank connected directly to the pipeline, so as to allow for expansion when the water becomes hot. The heating system of a house has 73.0 m of copper pipe whose inside radius is 7.87 x 10^{- 3}m. When the water and pipe are heated from 27.7 to 68.4 °C, what must be the minimum volume of the reservoir tank to hold the overflow of water?

Answer 1

Use the expression for volumetric expansion as -

∆V = Vo*3α*∆T

Given that -

r = 7.87 * 10 ^ -3 m
L = 73.0 m
Vo = A * L
Vo = pi * r ^ 2 * L m^3 = 3.141*(7.87*10^-3)^2 * 73.0 = 0.0142 m^3

Write the value for coefficients of linear expansion of water and copper from your text book. The values are -

αc = 17 * 10 ^ -6 /C
αw = 69 * 10 ^ -6 /C
∆T = 68.4 - 27.7 = 40.7 degrees C

Change in volume of water
∆Vw = Vo * 3 * αw * ∆T = 0.0142 * 3*69*10^-6 = 2.94 x 10^-6 m^3

Change in volume of copper:
∆Vc = Vo * 3 * αc * ∆T = 0.0142 * 3 * 17*10^-6 = 0.724 x 10^-6 m^3

Therefore, the water that spills out is:
∆Vw - ∆Vc = 2.94 x 10^-6 m^3 - 0.724 x 10^-6 m^3 = 2.216 x 10^-6 m^3

So, the minimum volume of reservoir tank to hold the overflow of water = 2.216 x 10^-6 m^3 (Answer)