Thermal expansion of a copper pipe

  • Thread starter slaw155
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  • #1
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Homework Statement


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 63.1 m of copper pipe whose inside radius is 7.69 x 10^-3 m. When the water and pipe are heated from 20.3 to 60.2 °C, what must be the minimum volume of the reservoir tank to hold the overflow of water?


Homework Equations


change in volume = initial volume x change in temp x volumetric expansion coefficient
coefficient as provided by teacher = 51 x 10^-6 /degC


The Attempt at a Solution


change in vol = 63.1m^3 x ∏ x (7.69x10^-3)^2 x (60.2-20.3) x 51 x 10^-6 = 2.3854 x 10^-5 m^3
Now would I add this change in volume to the volume of the pipe initially or is this in itself the correct answer?
 
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Answers and Replies

  • #2
SteamKing
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You should always include units in your calculations. How else can you tell what these numbers mean?
 
  • #3
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The units are in the question.
 
  • #4
CWatters
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The expansion vessel only has to accommodate the extra volume.

Domestic models are typically specified in Litres rather than cubic meter. Google them.
 
  • #5
I think you should calculate change in water volume too, use volumetric coefficient as 0.000214/deg C. initial volume of water is same as copper pipe.
From the values we can understand change in volume of water will be much greater compared to copper , also liquid expands more than solid.
If you subtract the volume change of water - volume change of copper, you will get the overflowing water volume i.e the answer.
 
  • #6
haruspex
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Now would I add this change in volume
You have calculated a change in interior volume of pipe, correct? Is it an increase or a decrease? How does that relate to overflow?
 
  • #9
If you as the initial volume to the change in volume, it won't change the results. Because initial volume will get added to both water and copper, so they will cancel each other

V initial +chage in V of water =V initial + change in V of copper

Thus, the initial volume will get cancelled on each side
 

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