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Calculating Thermal Expansion in Pipes

July 29, 2011

In the Building Systems exam there several math problem you may encounter, one of them is calculating the thermal expansion of piping. All materials expand when exposed to heat and contract when exposed to cold. We typically think of building materials expanding and contracting due to the changing weather, but piping inside of the building can be exposed to thermal changes as well. A building could have miles of piping carrying hot water to toilet rooms and kitchens, this expansion needs to be accounted for or the pipes may fail.

 

Equation: 

The equation for calculating thermal expansion in piping is:

ΔL = aLT

ΔL is the change in length of the pipe.

L is the length of piping being used.

T is the temperature difference.

a is the coefficient of expansion for the piping. Different materials will have a different coefficient. Most likely any questions using this equation will use copper as the piping as it is the most commonly used in construction. Click here for a table showing the thermal expansion coefficients for various materials.

The coefficient of thermal expansion is sometimes expressed as a #(10-6) or 10 to the negative 6 power. If you get this form of the coefficient you will need to convert it to a number you can easily input into your equation. To do this just move the decimal point 6 places in the negative direction. Ex. 17(10-6) would be .000017

 

Variations on this equation:
Note that in a word problem they might not give you the information as neatly as you might like. You may have to solve T first. In this case the equation would look like this.

ΔL = aL(T1-T2)

T1 being the beginning temperature and T2 being the final temperature. Note that this temperature difference should always be a positive number.

The multiple choice answerer’s you have to choose from may be in inches when the length of piping is given in feet. In this case you can either convert your answer to inches after you have solved the problem or you can convert the length of piping from feet to inches in the beginning of the equation. In that case the equation would look like this.

ΔL = a(L12)T

 

Example problem:
What is the thermal expansion for a 5 ft copper pipe that is exposed to a 30 degree temperature change?

L = 5 ft
T = 30
a = 0.000017

ΔL=aLT

ΔL = (0.000017)(5)(30) Solving right to left

ΔL = (0.000017)(150)

ΔL = .0025 ft

Don’t forget that the length is expressed in feet because the original length of piping is in feet. You may need to convert to inches.

 

One last thing, I wouldn’t bother memorizing the different coefficient’s for piping materials. If you get a question about this on your Building Systems exam, they will give you the coefficients in a reference table, and just in case if you forget it, they should give you the equation also.

 

Good luck on your next exam!

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