TP1:Example Problem 26

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Freezing of underground pipes

You may think that underground water lines never freeze, that being it is never a real concern for Floridians. In other parts of the world that experience cold winters, freezing pipes can be a real issue. When water freezes it expands which can cause major damage to the pipes in which carries the water and other structural components around it. In most cases, a proper installation, and knowledge of the local climate, will prevent pipes from freezing that are buried underground. Essentially, before construction one would want to research the area, focusing on the frost level and how deep the ground will freeze during the coldest points of the year.

Learning Outcome

To describe unsteady-state momentum, heat, and mass transfer into a semi-infinite medium.

Example Question 1

Problem: The ground at a particular location in Northern Canada is covered by snow at -15°C for four months. at this location, the average soil properties are K = 0.35 W/(m∙K) and α = 0.15×10^(-6) m^2/s. Assuming the initial uniform temperature is 18°C for the ground, determine the minimum burial depth to prevent the water pipes from freezing.

(Hint: assume the ground is a semi-infinite medium and that 4 months= 120 days)

Question 1 Solution

Given Information
Symbol	Value	Units	Description
$T_{0}$	-15	°C	Temperature of the ground with snow on it
$T_{i}$	18	°C	Initial uniform temperature of the ground
$T_{x}$	1	°C	The minimum temperature of water before it freezes
$k$	0.35	$[\frac{W}{m * K}]$	Thermal conductivity coefficient
$a$	$0.15 * 1 0^{-} 6$	$[\frac{m^{2}}{s}]$	alpha
$t$	$1.04 * 1 0^{7}$	sec.	4 months= 120 days= 2880 hours= $1.04 * 1 0^{7}$ seconds

Using the following equations to solve for the depth, x, in which the pipe should be buried:

$\frac{T_{x} - T_{0}}{T_{i} - T_{0}} = e r f (z)$

(1)

$z = (\frac{x}{2 \sqrt{a * t}})$

(2)

plug in the given values and solve for erf(Z):

$\frac{1 - (- 15)}{18 - (- 15)} = e r f (z)$

erf(z)= 0.4848

using the erf equation solve for z:

erf(z)= 0.4848 → z= 0.5070

using the second equation plug in all relevant values and solve for x:

$0.5070 = (\frac{x}{2 \sqrt{(0.15 * 1 0^{-} 6) * (1.04 * 1 0^{7})}})$

$x = 0.5070 * 2 \sqrt{(0.15 * 1 0^{-} 6) * (1.04 * 1 0^{7})}$

$x = 1.265$ meters

TP1:Example Problem 26

Freezing of underground pipes

Learning Outcome

Example Question 1

Question 1 Solution

Example Question 2

Question 2 Solution