Chemical Engineering October 2015 - 61

Engineering Practice
Pipe Insulation: Finding the Optimal Thickness
Avoid tedious numerical calculation by using this quick, shortcut method
Alejandro Anaya Durand, Allan
Osornio Maldonado, Alexander
Álvarez Holly, Itzayana Montesinos
Aguilar, Adalberto Fuentes
Martínez and María Daniela Ávalos
Galván
National Autonomous University of
Mexico, Faculty of Chemistry
I
nsulated pipes are important in
many applications in the chemical
process industries (CPI). When it
comes to the insulation itself, the
two main questions are: which material
should be used; and how much
of it (thickness)? Three variables are
considered in order to find the most
economical thickness of an insulated
pipe: the inner radius of pipe, the
cost of energy and the cost of the
insulation material itself.
Determining the most economical
thickness of insulation for a pipeline
is a problem of finding the right
balance between using the least
amount of insulating material with the
least amount of energy loss. When
increasing the thickness of the insulating
material, the energy loss decreases,
but the cost of the insulating
material rises. When decreasing the
thickness of the insulating material,
the material cost decreases, but the
energy lost rises - that translates
into monetary losses.
The equation that solves this problem
is complicated. It is a function of
many variables, such as inside and
outside temperatures, cost of energy,
conductivities, nominal sizes of
Ca
d0
FT
FN
FM
FD
h2
hc
hr
K
L
M
Insulation
Figure 1. The problem is to determine the optimal
insulation thickness, R2 - R1
pipelines, the cost of insulating materials
and more.
In order to find a system's minimal
cost and therefore the optimum insulation
thickness, numerical methods
are required. This process involves
finding an equation describing the
cost as a function of the variables,
then setting the derivative of that
equation equal to zero. Although it
is possible to solve such a problem
numerically, using software such as
Solver for Excel, it is tedious and
time consuming.
This article first outlines the numerical
approach, and then presents a
simple shortcut method that can be
used as a quick, practical tool to determine
the most economical thickness
of different insulating materials.
An example calculation using both
methods is then given, which shows
the same results are achieved.
Fundamentals
To demonstrate the complexity of the
problem, we first develop the equaNomeNclature
Least
squares fit of mineral wool costs (Table 2)
Outer diameter of insulation = 2r2, in.
Adjustment factor for temperature
Adjustment factor for pipe size
Adjustment factor for temperature
Adjustment factor for pipe size
hc + hr
Surface heat-transfer coefficient, radiation
Surface heat-transfer coefficient, convection
Thermal conductivity, Btu/h.ft.°F
Length of pipe insulation, ft
Cost of heat, $/million Btu
mp Cost of heat loss, $/ft.yr
np
qR
Cost of insulation per year, $/ft.yr
Heat transfer rate in radial direction, Btu/h
R1, R2 Insulation inner and outer radius, ft
r1, r2
T1, T2 Temperature at R1 and R2, °F
Ta
Ambient temperature, °F
U2
U
Y
yp
ξ
Heat flowrate from R2 to outside air, Btu/h
Heat flowrate (per foot) from R2 to outside air, Btu/h.ft
Number of hours operation, h/yr
Total cost (insulation plus energy), $/ft.yr
Emissivity of insulation surface
ChemiCal engineering www.Chemengonline.Com oCtober 2015
61
Insulation inner and outer radius, in.
Where:
qi = the heat-transfer rate in the i direction,
Btu/h
A = the area normal to the direction i
(direction of the heat flow), ft2
dt/dxi = the temperature gradient in
the i direction, °F/ft
K = the thermal conductivity, Btu/h.
ft.°F
For an insulated pipe of length L,
(Figure 1) the heat transfer is in the
radial direction, R, and the area is
2πRL
(2)
Equation (2) is integrated from R1
(inside radius of pipe insulation) to
R2 (outside radius of pipe insulation),
and from T1 to T2, where T1
is the temperature in the inside radius
of insulation. For practical purposes,
T1 is considered the same
L
R1
R2
Pipe
tions that need to be solved numerically.
After this, a shortcut method is
presented, with examples to compare
the results of using the numerical
versus the shortcut method.
Heat transfer. The general equation
for the rate of heat transmission
through a pipe having a homogeneous
insulation material can be derived
by starting with Fourier's law,
represented by Equation (1):
(1)
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