transferencia de calor

1. FUNDAMENTALS
OF HEAT TRANSFER
Guide | Fundamentals of Heat Transfer

2. 2
FUNDAMENTALS OF HEAT TRANSFER
Contents Page
Steady State Heat Transfer
Steady State Conditions................................................................................................................................................................................ 2
Steady State Heat Flow................................................................................................................................................................................. 2
Definitions...................................................................................................................................................................................................3, 4, 5
Heat Flow Problems.......................................................................................................................................................................................... 6
Surface Film Factor........................................................................................................................................................................................... 7
Thermal Transmittance.................................................................................................................................................................................... 8
Surface Temperature........................................................................................................................................................................................ 9
Heat Transmission
Hot Flat Surfaces..........................................................................................................................................................................................10
Cold Flat Surfaces........................................................................................................................................................................................12
Round Surfaces............................................................................................................................................................................................13
Hot Round Surfaces.....................................................................................................................................................................................14
Cold Round Surfaces...................................................................................................................................................................................15
Vapor Retarders
Properties of Moist Air.................................................................................................................................................................................16
Condensation Control..................................................................................................................................................................................17
Appendix
Thermal Conductivity vs Mean Temperature, Fiberglas®
pipe insulation.........................................................................................18
Insulation Thickness Constants, Fiberglas®
pipe insulation ..............................................................................................................18
Insulation Thickness Constants................................................................................................................................................................19
Steady state heat transmission through
insulated surfaces can be calculated
in a relatively simple manner. Readers
who have had previous experience
with heat transfer calculations can use
this book as a refresher or as a guide
to help others understand the subject.
In any case it will be helpful to start at
the beginning and work through the
discussions and examples step by step.
There are two recognizable states of
heat flow: transient and steady state.
In transient heat flow, the temperature
varies with time. In steady state heat
flow all factors, including temperature,
have reached equilibrium; i.e., they
remain constant. As far as this book
STEADY STATE HEAT TRANSFER
is concerned, steady state conditions
are assumed for all discussions and
examples.
In addition these discussions will be
limited to steady state “series” heat
flow and not “parallel” heat flow. In
“series,” heat flows through continuous
layers and the total resistance is the
sum of the resistances of the layers.
Steady State Conditions
The discussion that follows and
examples shown are based on
the assumption of steady state
conditions. Under steady state
conditions, heat in equals heat out.
There is no accumulation of heat in
the system.
Steady State Heat Flow
The temperature and heat flow
remain constant with time. The
same heat flows through any plane
in the system. Heat flowing through
“A” is the same as heat flowing
through “B” and “C”.

3. 3
DEFINITIONS
Some of the terms defined below refer to both English and metric units of measurement.
However, only English units will be used in the discussions and examples in this book.
Heat Heat is a form of energy and is commonly stated in such units as
Btu, calories, or watt-hours.
Cold Cold is a relative term which denotes a body at a lower temperature
than another. It is a measure of the absence of heat.
Temperature Temperature is a measure of the level of the heat energy in a
material.
English units are: °F (Fahrenheit) °R (Rankine)
Metric units are: °C (Celsius) °K (Kelvin)
Fahrenheit (F) Temperature scale of the English units. The ice point of water is
defined as 32°F and the steam point is 212°F. There are 180 even
divisions between points. Absolute zero is-459. 7°F.
Celsius (C) Temperature scale where the ice point of water is 0°C, and the
steam point is 100°C. Absolute zero is-273.2°C.
Rankine (R) An absolute temperature scale where the ice point of water is
491.7°R, and the steam point is 671.7°R. The degree divisions of
Rankine are the same as those of Fahrenheit. Absolute zero is 0°R.
Kelvin (K) Temperature scale where the ice point of water is 273.2°K and the
steam point is 373.2°K. The degree divisions of Kelvin are the same
as those of Celsius. Absolute zero is 0°K.
Mean Temperature (tm) The arithmetic mean between the hot and cold surface
temperatures.
Mean temperature is defined as:
where t1 = hot surface temperature
t2 = cold surface temperature
Temperature Difference (∆t) The difference in temperature between the hot and cold surface
temperature
Temperature difference is defined as ∆t = t1-t2
where: t1 = hot surface temperature
t2 = cold surface temperature
Temperature difference (∆t) in degrees Celsius (C°) is the same as
temperature difference in Kelvin (°K).
The temperature difference (∆t) in Fahrenheit (°F) or Rankine (° R) is
9/5 the temperature difference in Celsius or Kelvin.
tm =
2
(t1+t2)

4. 4
DEFINITIONS
Heat Transfer Heat transfer is the movement of heat energy. Heat energy always
moves from a higher temperature to a lower temperature by
conduction, convection or radiation.
Conduction Conduction is the transfer of heat by direct molecular contact.
This contact occurs within a material or by two materials that are
touching.
Convection Convection is the transfer of heat by the movement of a mass
(liquid or gas). When a liquid or gas moves from one place to
another it takes heat energy with it.
Radiation Radiation is the transfer of energy by electro-magnetic waves. An
example is the transfer of heat from the sun to the earth.
Thermal Conductivity (k) Thermal conductivity (k) is the rate of heat flow through one inch of
a homogeneous material. Homogeneous is defined as a material
having thermal conductivity that does not change within the range
of thickness normally used.
Conductance (C) Conductance (C) is the rate of heat flow through a specific material,
either homogeneous or heterogeneous, at the construction or
thickness stated.
For homogeneous materials:
Thermal Resistance (R) Thermal resistance is the ability of a material to retard heat flow.
As the resistance increases, the heat flow is reduced. It is the
reciprocal of the conductance, C.
Total Resistance (Rt) To determine the resistance of any insulation system where the
resistances are in series, the individual resistances are added
together to obtain the total resistance, Rt
Total Resistance = Sum of Resistances Rt = R1 + R2 + R3 + R4
Thermal Transmittance (U) Thermal transmittance is the rate of heat flow through a system.
Note that thermal transmittance applies to heat flow through a
system; while conductance (C), having the same dimensional units,
applies to individual materials.
English Units =
hr•ft2
F
Btu
Thermal Conductivity
Thickness
or C=
k
L
English Units =
Btu
hr•ft2
F
English Units =
Btu
hr•ft2
F
English Units =
Btu
hr•ft2
F
1
conductance
=
1
C
,
where C =
k
L
,
so R =
L
k
U =
1
Rt

5. 5
DEFINITIONS
Heat Flow (Q) Heat Flow (Q) is the rate at which heat moves from an area of
higher temperature to an area of lower temperature.
Heat flow is generally used to quantify the rate of total heat gain or
loss of a system, i.e. building envelopes or storage tanks. However,
calculations of the type used in this book are made on components
of a system and are expressed in heat flow per square foot (heat
flux, q).
Heat Flux (q) Heat Flux (q) is the heat flow through a one square foot area.
The heat flux can be determined from: q =
where Rt is the total thermal resistance of the system.
Thermal conductivity (k), conductance (C) and thermal resistance
(R) are specific properties materials. The ASHRAE Handbook of
Fundamentals contains tables of these values for most materials
used in building construction. Many manufacturers of building
materials, particularly insulation products, provide very specific
information on thermal properties of their products. The thermal
values of the materials used in the examples in this book were
taken from ASHRAE, while the thermal insulation values represent
those of Owens Corning products.
English Units =
Btu
hr•ft2
English Units =
Btu
hr
heat flow
area
=
Q
A
Heat Flux =
∆t
R
or
∆t
Rt
1
Ct
Since R = the q = = Cx∆t
Δt
1
Ct

6. 6
HEAT FLOW PROBLEMS
Now that we have defined the terms involved in heat transfer calculations,
let’s work a few problems to see how these terms are used.
Example 1
Conditions: A metal faced panel is insulated with 3½" of a material
having a thermal conductivity of 0.318 Btu•in/hr•ft2
•F.
Hot and cold side temperatures as shown in the diagram
at the right. (Remember: heat flow is always the result of a
temperature difference.)
Problem 1 (a): What is the resistance value R of this material?
Solution:
Problem 1 (b): If the metal skins are 125°F and 25°F, what is the heat flux
q through the insulation?
Solution:
Problem 1 (c): What is the heat flux q through: The hot side metal skin?
The cold side metal skin? The total panel?
Solution: If your answers are other than 9.1 , go
back and review definitions on pages 3, 4 and 5.
Problem 1 (d): What is the conductance C of the insulation?
Solution:
Example 2
Conditions: A wall (see diagram, right) is composed of:
Wood siding, with thermal conductance of 1.25
Foam sheathing, with thermal resistance of 6.0
Insulation 3.5 in. thick, with thermal conductivity
of 0.318
Wall board inside finish with thermal conductance of
2.0
Problem 2: What is the total resistance value Rt of the wall?
Solution: Wood siding R1 R = = = 0.8
Foam sheathing R2 R = 6.0 = = 6.0
Insulation R3 R = = = 11.0
Wall board R4 R = = = 0.5
Total resistance Rt = R1 + R2 + R3 + R4 = 18.3
L
k
hr•ft2
•F
Btu
R = = = 11.0
3.5
0.318
Δt
R
Btu
hr•ft2
q = = = 9.1
125-25
11.0
Btu
hr•ft2
k
L
C = = = 0.09
0.318
3.5
Btu
hr•ft2
•F
1
R
Or: C = = = 0.09
1
11.0
Btu
hr•ft2
•F
Btu
hr•ft2
hr•ft2
•F
Btu
Btu•in
hr•ft2
Btu
hr•ft2
1
C
1
1.25
L
k
3.5
0.318
1
C
1
1.20

7. 7
SURFACE FILM FACTOR
Up to this point we have been concerned only with thermal properties of
materials. However, surfaces in contact with air have insulation values.
These values are expressed as surface air film resistances.
Sometimes a surface file insulation value is expressed as a conductance
with symbols of either h or f. In order to calculate total thermal
resistance or heat flux, this conductance must be converted to a thermal
resistance.
The term “surface emittance” or “emissivity” is sometimes encountered
in industrial heat transfer applications. This term is a measure of the
ability of a surface to radiate heat into the surrounding air, and is a
characteristic of the type of surface material covering the insulation.
Typical values for surface emittance are: 0.2 for bright metal jacket; 0.9
for a foil scrim kraft jacket. Surface emittance is one of several terms
used to determine the surface film coefficient or surface film factor.
Further discussion of surface film coefficient is rather complex and is
beyond the scope of this book. For convenience, surface film factors will
be expressed only as thermal resistances in this book. The table (right)
shows typical surface resistances for a variety of conditions.
Condition Resistance (R)
Still air (0 mph)
Heat flow up 0.61
Heat flow down 0.92
Heat flow horizontal 0.68
Moving air
7.5 mph (summer) 0.25
15.0 mph (winter) 0.17
Round pipe 0.65
Note: Surface resistances decrease as air veloc-
ities increase.
All values are taken from ASHRAE Handbook of
Fundamentals.
Some of these examples consider only the
insulation surface-to-ambient film factor. It is
assumed that the inside surface area is at the
same temperature as the contents (air, gas or
liquid), such as in a duct, pipe or tank. Generally,
the inside air film factor is used only for cases
involving occupied spaces (60F to 90F).
Example 3
To see the effect of film factors we will complete the same problem several ways.
Conditions: Vertical surface, horizontal heat flow, no film factor
Operating temperature t1 = 200°F
Ambient temperature t2 = 80°F
Insulation thickness L = 2"
Insulation thermal conductivity k = 0.27
Problem 3 (a): What is the insulation resistance R1? What is the heat flux q?
Solutions: 1. Insulation resistance
2. Heat flux q =
Problem 3 (b):
Conditions: Vertical surface, horizontal heat flow, film factor at 0 mph
Same as problem 3(a) except for the addition of the surface file
RS at 0 mph (RS = 0.68).
Solution: 1. Heat flux q =
Problem 3 (c):
Conditions: Vertical surface, horizontal heat flow, film factor at 15 mph
Same as problem 3(a) except for the addition of the surface file
RS at 15 mph (RS = 0.17).
Solution: 1. Heat flux q =
As these examples show, the proper choice of the surface film factor can have a significant effect on the calculation of heat flow.
L
k
R1 = = = 7.40
2
0.27
Btu•in
hr•ft2
•F
Δt
R
Btu
hr•ft2
= = = 16.2
t1-t2
R1
200-80
7.40
Δt
Rt
Btu
hr•ft2
= = = =14.9
t1-t2
R1+Rs
200-80
7.40+0.68
120
8.08
Δt
Rt
Btu
hr•ft2
= = = =15.9
t1-t2
R1+Rs
200-80
7.40+0.17
120
7.57

8. 8
THERMAL TRANSMITTANCE (U)
Example 4
Problem 4: Determine thermal transmittance U of the residential
wall construction shown at right.
Conditions: Typical R-values used in residential wall construction.
Building materials and conditions Resistance (R values)
Inside air film, 0 mph 0.68
½" wall board 0.45
3½" insulation 11.00
Wood fiber sheathing 1.32
Aluminum siding 0.00
Outside air film, 15 mph 0.17
Total resistance (Rt) 13.62
Solution: Thermal transmittance
Example 5
Problem 5: Determine thermal transmittance U of the typical roof
deck installation shown at right.
Conditions: Typical R-values used in residential wall construction.
Building materials and conditions Resistance (R values)
Outside air film, 15 mph 0.17
Built-up roof and aggregate 0.33
2¼" roof insulation (C=0.11) 9.09
Metal deck 0.00
Inside air film, heat flow up 0.61
Total resistance (Rt) 10.20
Question: What is the thermal transmittance of the roof?
Solution: Thermal transmittance
Rounded:
It is customary to describe the heat loss or gain of a system by its thermal transmittance, U, which includes material and
surface film resistances.
Thermal transmittance is the reciprocal of the total resistance of a system:
Referring to Example 1 on page 6, if we use an inside surface air film resistance of 0.68 (still air) and an outside air film
resistance of 0.25 (7.5 mph air), the total resistance (Rt) becomes:
And the thermal transmittance (U) is:
Let’s take a look at how film factors are used in calculating U values in various building constructions.
English Units =
Btu
hr•ft2
U =
1
Rt
Rt = 0.68 + 11.0 + 0.25 = 11.93
hr•ft2
•F
Btu
1
Rt
U = = = 0.08
1
11.93
hr•ft2
•F
Btu
1
Rt
U = = = 0.07
1
13.62
Btu
hr•ft2
•F
1
Rt
U = = = 0.098
1
10.20
Btu
hr•ft2
•F
U = 0.10
Btu
hr•ft2
•F

9. 9
SURFACE TEMPERATURE
Very often, when working heat flow problems, the surface temperature must be determined. To do this, the first step may be
to estimate a surface temperature. The mean temperature (hot surface to cold surface) can then be determined and the k
value found.
The Values of thermal conductivity, k, are a function of mean temperature, tm. As the mean temperature varies the thermal
conductivity changes. The graphs on page 18 show thermal conductivity, k, versus mean temperature tm for typical
insulation products.
Thermal resistance values (often referred to simply as R-values) are also a function of mean temperature (usually 75° F). For
most residential and commercial heat loss calculations, however, the use of the stated R-value at 75°F mean is sufficient.
After the heat flux is determined, the surface temperature is calculated. If it is possible to read a new k value based on the
new tm value, rework the problem.
Surface temperature of an insulation can be determined once the heat flux through the system is established.
Example 6 (a) illustrates the procedure for calculating surface temperature when the R-value is given; this is representative
of typical residential or commercial construction.
Example 6 (b) illustrates the same procedure for industrial type construction in which the k-value must be determined based
on mean temperature.
Example 6
Problem 6 (a): Vertical surface, horizontal heat flow.
Conditions: Inside surface temperature, ti = 125°F
Ambient temperature, ta = 20°F
Insulation thickness, L = 4"
Insulation R-value Rl = 13.0
Surface resistance, Rs (still air) = 0.68
Question: What is the cold surface temperature ts?
Solution: Heat flux q =
Since q =
∆ts = temperature drop across surface film.
Rs = 0.68
Using the heat flux from above ∆ts is determined:
∆ts = 7.7 x 0.68 – 5.2°F drop through air film.
This is the temperature drop across the
air film = ts- ta, so:
∆ts = 5.2°F = ts – 20°F,
And ts = 25.2°F.
Δt
Rt
Btu
hr•ft2
= = = =7.7
ti-ta
Rl+Rs
(125-20)
(13.0+0.68)
105
13.68
, ∆t = q x R; or ∆ts = q x Rs, where:
Δt
R

10. 10
HEAT TRANSMISSION, HOT FLAT SURFACES
Since the thermal resistance of the insulation, R, can also be written where L is thickness and k is thermal conductivity, the
heat flux formula can be written
q =
where: q = heat flux
ti = inside temperature
ta = ambient temperature
ts = surface temperature
L = insulation thickness
k = thermal conductivity
Rs = air film resistance
Problem 6 (b): Vertical surface, horizontal heat flow
Conditions: Operating temperature, ti = 400°F
Ambient temperature, ta = 80°F
Insulation thickness, L = 2"
Insulation: fibrous glass
What is the heat flux, q?
What is the surface temperature, ts?
Solution: Estimate surface temperature ts = 85°F.
Assume temperature at interface of equipment surface
is the same operating temperature ti.
Mean temperature =
Thermal conductivity, k at 242°F mean is 0.35 (refer to page 18)
Heat flux, q =
q =
= 12.8°F temperature drop through air film.
ts = ta + ∆ts = 80°F + 12.8°F = 92.3°F
This is close to our original estimate of surface temperature and will not appreciably change our answer,
so there is no need for recomputation.
L
k
ti-ta
+Rs
= = 242°F.
ti-ta
2
(400-85)
2
L
k
Btu
hr•ft2
= = = =18.8
(400-80)
2
0.35
L
k
ti-ta
+Rs
320
5.71 + 0.68
x 0.68
120
6.4
Δts
Rs
or ∆ts = q x Rs or ∆ts = 18.8 x 0.68

11. 11
TEMPERATURE GRADIENT
Example 7
Problem 7: Find the temperature drop across each component of the
wall system shown at right.
Conditions:
Building materials and conditions Resistance (R values)
Inside air film, 0 mph 0.68
½" wall board 0.45
3½" insulation 11.00
1" foam insulating sheathing 7.20
Aluminum siding 0.00
Outside air film, 15 mph 0.17
Total resistance (Rt) 19.50
What is the temperature drop across each component?
What is the total temperature gradient across the wall?
Solution: Find the heat flux q through the entire wall.
The heat flux is the same through each component of the
wall system, so:
∆t across inside air film =
q x R = 3.85 x 0.68 = 2.6°F
∆t across wall board =
q x R = 3.85 x 0.45 = 1.7°F
∆t across insulation =
q x R = 3.85 x 11.0 = 42.3°F
∆t across foam sheathing =
q x R = 3.85 x 7.20 = 27.7°F
∆t across aluminum siding =
q x R = 3.85 x 0.00 = 0.0°F
∆t across outside air film =
q x R = 3.85 x 0.17 = 0.7°F
Total temperature drop across wall = 75.0°F
When these temperature drops are shown on a graph such as shown at the right,
the result is known as the temperature gradient across the wall.
Δt
Rt
q = = = 3.85
Btu
hr•ft2
(75-0)
19.50
q = or ∆t = q x R for each component:
Δt
R

12. 12
COLD FLAT SURFACES
Example 8
Problem 8: What is the surface temperature ts?
Conditions: Operating temperature, ti = 40°F
Ambient temperature, ta = 80°F
Insulation thickness, L = 2"
Solution: Estimate surface temperature ts = 75°F.
Assume temperature at interface of equipment
surface and insulation is the same as inside
temperature ti.
Mean temperature =
Conductivity, k at 57°F mean temperature is 0.23
(Graph page 18, Figure 2)
Heat flux, q =
(Minus sign indicates heat gain)
=-2.9°F temperature drop through air film. Then:
ts = 80°F + (-2.9°F) – 77.1°F
This is close to our original estimate of surface temperature, so there is no need for recomputation.
= = = =-4.3
L
k
ti-ta
+Rs
-40
8.7 + 0.68
(40-80)
2
0.23
x 0.68
-40
9.38
= = 57°F.
t1-ts
2
40 + 75
2
Since q = , then ∆ts = q x Rs =-4.3 x 0.68
Δts
Rs
Btu
hr•ft2

13. 13
ROUND SURFACES
The equation we have been using is for flat surface calculations. When
we consider a curved surface, we have to take into account the fact that
the inner surface (A, diagram at right) and outer surface (B) have different
areas. We do this by using an equivalent insulation thickness.
This equivalent insulation thickness is the logarithmic mean L1.
This value, then, replaces the thickness value L in the formula for heat flux.
Heat flux is found from:
where: q = of outer surface insulation
r1 = inner radius of insulation in inches
r2 = outer radius of insulation in inches
t1 = temperature, inner surface of insulation
t2 = temperature, ambient
k = thermal conductivity at mean temperature
ts = surface temperature
Rs = air film resistance
In – natural logarithm (Loge)
To eliminate the time-consuming mathematics of solving for, r2 In
these have been calculated and put into tables on pages 18 and 19.
All you need to find the correct number is the Pipe Size and Insulation.
Once the relationship L1 = R2 In (equivalent thickness) is accepted as a
mathematical relation for the length of the path along which heat must flow
(designated L in the flat surface equations), the formula is not as complex
to use as it first appears. Tables in the appendix give values for L1 for all
usual pipe sizes and insulation thicknesses.
The value A in Tables 1 and 2 is he surface area in square feet per lineal
foot length of insulation of the specified size. To determine the heat loss
in , multiply the heat loss by the factor A shown in the table.
L1 = r2 In
r2
r1
( )
temperature difference
equivalent thickness
conductivity
+ air film
resistance
q = = (t1- t2)
r2 In
r2
r1
( ) + Rs
k
Btu
hr•ft2
r2
r1
( )
r2
r1
( )
Btu
hr•lin.ft
Btu
hr•ft2

14. 14
HOT ROUND SURFACES
Example 9
Problem 9: What is the heat loss per foot of length at conditions
shown below and in the diagram at right?
Conditions: Line (operating) temperature ti = 200°F
Ambient temperature ta = 80°F
Pipe size, NPS = 8"
Insulation = 8x2"
Find heat gain per foot length.
Solution: Estimate surface temperature ts = 85°F. Assume temperature
at interface of pipe surface and insulation is the same as line
temperature ti.
Mean temperature =
Conductivity, k at 142°F mean is 0.27 (graph, page 18)
L1 = r2In = 2.49 (Table 1, page 18)
Air film resistance RS = 0.65 (Page 7)
temperature drop through air film.
Surface temperature = ts x ∆ts = 80 x 8.2 = 88.2°F
This is close to our estimate, and will not appreciable change our heat loss answer.
The heat loss per foot of length (H) = q x A (A = 3.34, from Table 1).
H = 12.1 x 3.34 = 40.4
Btu
hr•lin.ft
q = or ∆ts = q x Rs =12.1 x 0.68 = 8.2°F
Δts
Rs
= = = 142°F.
t1 + ts
2
(200 + 85)
2
285
2
r2
r1
( )
Btu
hr•ft2
q =
(t1- t2)
r2 In
r2
r1
( ) + Rs
k
120
9.2 x 0.68
(200-80)
2.49
0.27
x 0.68
= = =
120
9.88
= 12.1

15. 15
COLD ROUND SURFACES
Example 10
Problem 10: What is the heat gain per foot of length at conditions
shown below and in the diagram at right?
Conditions: Line (operating) temperature ti = 40°F
Ambient temperature ta = 80°F
Pipe size, NPS = 8"
Insulation = 8x2"
Find heat loss per foot length.
Solution: Estimate surface temperature ts = 75°F.
Assume temperature at interface of pipe surface and
insulation is the same as line temperature ti.
Mean temperature =
Conductivity, k at 57°F mean is 0.23 (graph, page 18)
L1 = r2In = 2.49 (Table 1, page 18)
Air film resistance RS = 0.65 (Page 7)
(Minus sign indicates heat gain rather than loss.)
temperature drop through air film.
Surface temperature = ts + ∆ts = 80°F + (-2.3°F) = 77.7°F
The heat gain per foot of length (H) = q x A (A = 3.34, from Table 1).
H =-3.49 x 3.34 = 11.7 Btu
lin.•ft•hr
Btu
hr•ft2
q =
(ti- ta)
r2 In
r2
r1
( ) + Rs
k
-40
10.8 x 0.65
(40-80)
2.49
0.23
x 0.65
= = =
-40
11.45
= -3.49
= = = 57°F.
ti + ta
2
(40 + 75)
2
115
2
r2
r1
( )
q = or ∆ts = q x Rs =3.49 x 0.65 = 23.5°F
Δts
Rs

16. 16
VAPOR RETARDERS
Insulation requires a vapor retarder on the hot side when the surface temperature is below the ambient air temperature. This
will prevent moisture from entering the insulation and condensing. Without the vapor retarder the insulation can become
wet. This can lower thermal efficiency, cause dripping water, and add weight.
Condensation:
The presence of a vapor retarder will not prevent condensation of water on the outside of the barrier. Condensation on the
surface is controlled by insulation thickness. The thickness should be selected such that the surface temperature will be
above the dew point of the surround air.
The nomograph below can be used to determine the dew point temperature with a known air temperature and design
relative humidity. To use the nomograph, place a straight-edge across the vertical lines A, B and C. using the two known
values, the third value can be determined.
Properties of moist air

17. 17
CONDENSATION CONTROL, FLAT VERTICAL SURFACES
Example 11:
Problem 11 (a): Find the insulation thickness required to prevent condensation.
Conditions: A cooler room inside a building. Metal panel construction.
(See diagram, right.)
Ambient temperature t1 = 90°F
Inside temperature t2 = 20°F
Relative humidity 80%
Solution: Assume metal facings have no resistance value. Ignore temperature drops through air films for calculating
mean temperature. (This is valid in this case because wind velocity is the same—0 mph—on both sides of
the panel, making the air film temperature drops the same. Thus the air film temperature drops cancel each
other out of the mean temperature calculation.)
Mean temperature =
Conductivity, k at 55°F is 0.23 (see graph, Page 18). Try 1" of insulation.
temperature drop through air film.
ts = t1- ∆ts = 90°F – 8.4°F = 81.6°F surface temperature.
Check nomograph, page 16: 90°F ambient temperature, 80% relative humidity. Dew point temperature is
read 84°F. There will be condensation because the surface temperature is below the dew point temperature
at these conditions. Rework the problem using 2" of insulation as follows:
∆ts = q x Rs1 = 7.0 x 0.68 = 4.67°F temperature drop through air film.
ts = t1- ∆ts = 90 – 4.67 = 85.3°F surface temperature. No condensation will occur because the surface
temperature is above the dew point temperature of 84°F as indicated by the nomograph.
Problem 11 (b): To find the thickness to prevent condensation, proceed as follows: Using the nomograph on Page 16, find
the dew point for 90°F ambient temperature, 80% relative humidity. Dew point is 84°F.
Surface temperature must be just above dew point, so let ts = 85°F
Find the heat flux through the air film.
Enter known values into formulas to find correct thickness:
L = 1.88". Specify 2" of insulation.
= = 55°F.
t1 + t2
2
(90 + 20)
2
q = or ∆ts = q x Rs1 =12.3 x 0.68 = 8.4°F
Δts
Rs1
Btu
hr•ft2
q =
(t1- t2)
Rs2
L
k
70
0.68 + 4.35 + 0.68
(90-20)
= = = 70
5.71
= 12.3
Rs1 + + 0.68
1
0.23
0.68+
Btu
hr•ft2
q =
70
10.06
(90-20)
= = 7.0
+ 0.68
2
0.23
0.68+
Btu
hr•ft2
q =
(t1- t2)
R
(90-85)
0.68
= = 7.35
q =
(t1- t2)
Rs2
L
k
Rs1 +
(90-20)
7.35=
+ 0.68
L
0.23
0.68+
(at 55°F mean temperature, k = 0.23)

18. 18
APPENDIX
Figure 1.
Thermal conductivity vs. Mean temperature
Fiberglas®
Pipe Insulation (ASTM C 335)
Notes to Figures 1 and 2:
These figures represent conductivity values for typical
Owens Corning products that would be used in the
examples in this book. For specific applications, always
use the conductivity values for the product being used.
Symbols for Table 1:
NPS = nominal pipe size
L1 = equivalent thickness in inches
L1 = r2 In
where:
r1 = inner radius of insulation, inches
r2 = outer radius of insulation, inches
In = log to the base e (natural log)
A = Square feet of pipe insulation surface
per lineal foot of pipe
r2
r1
( )

19. 19
FUNDAMENTALS OF HEAT TRANSFER
Table 1.
Insulation Thickness Constants, Fiberglas pipe insulation-Equivalent Thicknesses in Inches
½ 1 1½ 2 2½ 3 3½ 4 4½ 5 5½ 5
NPS L1 A L1 A L1 A L1 A L1 A L1 A L1 A L1 A L1 A L1 A L1 A L1 A
½ 0.76 0.49 1.77 0.75 3.12 1.05 4.46 1.31 - - - - - - - - - - - - - - - -
¾ 0.75 0.56 1.45 0.75 2.68 1.05 3.90 1.31 - - - - - - - - - - - - - - - -
1 0.71 0.62 1.72 0.92 2.78 1.18 4.02 1.46 - - - - - - - - - - - - - - - -
1¼ 0.63 0.70 1.31 0.92 2.76 1.31 3.36 1.46 - - - - - - - - - - - - - - - -
1½ 0.60 0.75 1.49 1.05 2.42 1.31 4.13 1.73 - - - - - - - - - - - - - - - -
2 0.67 0.92 1.43 1.18 2.36 1.46 3.39 1.73 4.43 1.99 - - - - - - - - - - - - - -
2½ 0.66 1.05 1.38 1.31 2.75 1.73 3.71 1.99 4.73 2.26 - - - - - - - - - - - - - -
3 0.57 1.18 1.29 1.46 2.11 1.73 2.96 1.99 3.88 2.26 4.86 2.52 - - - - - - - - - - - -
3½ 0.92 1.46 1.67 1.73 2.46 1.99 3.31 2.26 4.22 2.52 5.31 2.81 - - - - - - - - - - - -
4 0.59 1.46 1.28 1.73 2.01 1.99 2.80 2.26 3.65 2.52 4.68 2.81 - - - - - - - - - - - -
4½ 0.94 1.74 1.61 1.99 2.35 2.26 3.15 2.52 4.11 2.81 5.02 3.08 5.97 3.34 7.21 3.67 - - - - - - - -
5 0.58 1.74 1.20 1.99 1.89 2.26 2.64 2.52 3.54 2.81 4.40 3.08 5.29 3.34 6.46 3.67 - - - - - - - -
6 0.54 2.00 1.13 2.26 1.79 2.52 2.60 2.81 3.36 3.08 4.17 3.34 5.23 3.67 6.12 3.93 7.05 4.19 8.00 4.45 8.99 4.71 10.00 4.97
7 - - 1.11 2.52 1.84 2.81 2.54 3.08 3.27 3.34 4.25 3.67 5.07 3.93 5.92 4.19 6.81 4.45 7.72 4.71 8.67 4.97 9.64 5.24
8 - - 1.18 2.81 1.81 3.08 2.49 3.34 3.39 3.67 4.15 3.93 4.94 4.19 5.76 4.45 6.62 4.71 7.50 4.97 8.40 5.24 9.35 5.50
9 - - 1.17 3.08 1.79 3.34 2.62 3.67 3.32 3.93 4.06 4.19 4.83 4.45 5.63 4.71 6.46 4.97 7.31 5.24 8.19 5.50 9.09 5.76
10 - - 1.09 3.34 1.85 3.67 2.50 3.93 3.18 4.19 3.90 4.45 4.64 4.71 5.41 4.97 621 5.24 7.03 5.50 7.88 5.76 8.75 6.02
12 - - 1.22 3.93 1.82 4.19 2.45 4.45 3.10 4.71 3.79 4.97 4.50 5.24 5.24 5.50 6.00 5.76 6.78 6.02 7.59 6.28 8.42 6.54
14 - - 1.07 4.19 1.65 4.45 2.26 4.71 2.90 4.97 3.57 5.24 4.26 5.50 4.97 5.76 5.71 6.02 6.47 6.28 7.25 6.54 8.05 6.81
16 - - 1.06 4.71 1.63 4.97 2.23 5.24 2.86 5.50 3.50 5.76 4.17 6.02 4.87 6.28 5.58 6.54 6,31 6.81 7.06 7.07 7.83 7.33
18 - - 1.05 5.24 1.62 5.50 2.21 5.76 2.82 6.02 3.45 6.28 4.11 6.54 4.78 6.81 5.47 7.07 6.19 7.33 6.92 7.59 7.66 7.85
20 - - 1.05 5.76 1.61 6.02 2.19 6.28 2.79 6.54 3.41 6.81 4.05 7.07 4.71 7.33 5.39 7.59 6.08 7.85 6.79 8.12 7.52 8.38
24 - - 1.04 6.81 1.59 7.07 2.16 7.33 2.74 7.59 3.35 7.85 3.97 8.12 4.60 8.38 52.5 8.64 5.92 8.90 6.60 9.16 7.30 9.42