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 59.7 m of copper pipe whose inside radius is 9.20 x 10- 3 m. When the water and pipe are heated from 20.5 to 64.5 °C, what must be the minimum volume of the reservoir tank to hold the overflow of water

Answer 1

Answer:

$108.306\times 10^{-6}m^3$

Explanation:

According to volume thermal expansion the expansion in volume due to temperature is given by $\Delta V=V_0\beta \Delta T$ here $\beta$ is coefficient of volume expansion

The volume of copper pipe before expansion is $V_0=\pi r^2L=3.14\times (9.20\times 10^{-3})^2\times 59.7=0.0158m^3$

Now the increase of copper pipe due to increase in temperature = $\Delta V_c=V_0\beta _c\Delta T=0.0158\times 51\times 10^{-6}\times (64.5-20.5)=35.6\times 10^{-6}m^3$  

As $\beta$ for copper is $51\times 10^{-6}$

Now the increase of water due to increase in temperature = $\Delta V_w=V_0\beta _c\Delta T=0.0158\times 207\times 10^{-6}\times (64.5-20.5)=143.906\times 10^{-6}m^3$

As $\beta$ for water is $207\times 10^{-6}$

So the minimum volume of reservoir tank to hold the overflow of water = $\Delta V=\Delta V_w-\Delta V_c=143.906\times 10^{-6}-35.6\times 106{-6}=108.306\times 10^{-6}m^3$