‫بسم هللا الرحمن الرحيم‬
Thermal Insulation
1.1
Introduction
Thermal insulation of certain system is
to minimize the heat transferred from or to this
system. Thermal insulations are materials or
combinations of materials that are used to
provide resistance of heat flow. Most of
insulations are heterogeneous materials made
of low thermal conductivity materials, and they
involve air pockets.
Temperature difference is the driving
force for heat flow. To decrease heat flow
between to mediums (media) at different
temperatures, barriers are put on the path of
heat flow. Thermal insulations serve as such
barriers.
Ti
Ti
Qin
Qin
Q
Q
To
To
A
k
l
k
T=Ti-To
Ti R=l/kA To
Ti
Q = T / R
kin
l
lin
R=l/kA Rin=lin/kinA
To
Qin=T / (R+Rin)
Q > Qin
Figure (1) Insulation reduces lost
heat
1.2 Reasons of Insulations
The main purpose of using insulations is the saving of energy. Beside this
important target, there are many reasons to
Boundary layer
take thermal insulation in account, they are
T
listed as follows:
a) Energy conservation: conserving
energy by reducing the rate of heat
Q
flow is the primary reason for
insulating
surfaces.
Insulation
materials of working temperatureT
T
range from -268oC to 1000oC are
available.
b) Personal Protection and Comfort:
Insulation layer
A surface that is too hot poses a
Metalic layer
danger to people who are working
Figure (2) Insulation decreases
in that area of accidentally touching
surface-temperature
hot
surface
and
burning
themselves. To prevent this danger,
the temperature of hot surfaces should be reduced to below 60oC by
insulating them. Also the excessive heat coming off the hot surfaces
creates an unpleasant environment in which to work.
c) Maintaining Process Temperature: Some processes in chemical
industry and temperature-sensitive, and hence it become necessary to
insulate the process tanks and flow sections heavily to maintain the same
temperature throughout.
i
o
oo
1
d) Reducing Temperature Variation and fluctuations: Insulation
minimizes temperature non-uniformity in an enclosure and slows down
fluctuations.
e) Condensation and Corrosion Prevention: Insulation of cold surfaces
prevents water vapor in atmospheric air to be condensed. Hence the
metal tanks or pipes become safe from erosion to be occurring and algae
to be growth.
f) Fire Protection: Damage during a fire may be minimized by keeping
valuable combustibles in a safety box that is well insulated.
g) Freezing Protection: Adequate insulation will slow down the heat loss
from the water and prevent freezing during limited exposure to subfreezing
temperatures, hence the bursting of water pipes can be prevented.
h) Reducing Noise and vibration: An added benefit of thermal insulation is
its ability to dampen noise and vibration.
1.2
Insulating Materials
The most important property of insulating materials is the low thermal
conductivity of such materials, of course beside some mechanical and physical
properties .Most of insulation materials are primarily made of fiber glass, mineral
wool, polyethylene, foam, or calcium silicate.
The various forms of fiber glass insulation are widely used in process
industries and heating and air-conditioning applications because of their low cost,
light weight, resiliency, and versatility. But they are not suitable for some applications
because of their low resistance to moisture and fire and their limited maximum
service temperature. Fiberglass insulations come in various forms such as unfaced
fiberglass insulation, vinyl-faced fiberglass insulation, foil-faced fiberglass insulation,
and fiberglass insulation sheet. Mineral wool is resilient, lightweight, fibrous, woollike, thermally efficient, fire resistant up to 1100 oC, and forms a sound barrier.
Mineral wool insulation comes in form of blankets, rolls, or blocks. Calcium silicate
is a solid material that is suitable for use at temperatures, but it is more expensive
Refractories are the materials which resist change of shape, weight, or
physical properties at high temperatures. Fire-clay, silica, kaolin, alumina are
examples of such materials. These are mostly
Thin metal sheets
used in the form of bricks. A fire brick is built
primarily to withstand high temperatures and
T1
T2
secondarily to have insulating value. Unfortunately,
since high refractory quality is incompatible with
low conductivity, a wall to withstand extremely high
furnace-side temperatures is built of more than one
material. As in furnaces walls, these walls are
multilayer walls, where the high refractory layer is
the inside layer.
Superinsulators
are
very
effective
Vacuum
insulations, which are commonly used in space
Insulating
material
applications and cryogenics. Cryogenics is the
branch of heat transfer dealing with temperatures Figure (3) Superinsulator
2
below 100K, such as those encountered in liquefaction, storage, and transportation
of gases (helium, hydrogen, nitrogen, and oxygen). These superinsulations are
obtained by using closely packed layers of highly reflective thin metal sheets such as
aluminum foil (usually 25 sheets per cm) separated by fibers made of insulating
material such as glass fiber. Further, the space between the layers is evacuated to
form a vacuum under 10-6 atm. pressure to minimize conduction and convection
heat transfer through the air space between the layers. This insulating material has
apparent thermal conductivity below 2×10-5 W/m.K (one thousand times less than
the conductivity of air or common insulating materials).
4. The R-value of Insulation
The effectiveness of insulation materials is given by some manufacturers in
terms of their R-value [m2.K/W], which is the thermal resistance of material per unit
surface area and sometimes is known as the overall unit thermal resistance, which
can be determined according to the following relation:
𝑅 − 𝑣𝑎𝑙𝑢𝑒 = 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑐𝑒 × 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎
Accordingly, for flat insulation the R-value is given by the following relation as:
𝑅 − 𝑣𝑎𝑙𝑢𝑒 = 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 × 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 =
𝑙
𝑙
×𝐴=
𝑘𝐴
𝑘
Where l is the thickness [m], k is the thermal conductivity of the material [W/m.K] and
A is surface area [m2]. For pipe insulation, R-value is determined using thermal
resistance relation as:
𝑅 − 𝑣𝑎𝑙𝑢𝑒 = 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 × 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 =
𝑟
𝑙𝑛 (𝑟2 )
1
2𝜋𝑘𝐿
× 2𝜋𝑟2 𝐿 =
𝑟2
𝑟2
× 𝑙𝑛 ( )
𝑘
𝑟1
Where r2 is outer radius, r1 is inner radius, L is the length of the pipe and k is thermal
conductivity of the material. Once the R-value is available, the rate of heat transfer
through the insulation can be determined from the following equation:
𝑄̇ =
∆𝑇
× 𝐴𝑟𝑒𝑎
𝑅 − 𝑉𝑎𝑙𝑢𝑒
It is noted, that the unit of R-value [m2.K/W] is the reciprocal of the unit of the
coefficient of heat transfer h [W/m2.K].
5. Optimum Thickness of Insulation
It is clear that, the thicker the insulation is, the lower the rate of heat transfer
becomes. On the other hand as the thickness of insulation increases, the cost of it
becomes higher. In accordance, there is an optimum thickness of insulation that is
corresponding to a minimum combined cost of insulation and lost heat. Knowing the
3
expected life time of the insulation, one can
determine the annual cost of insulation. The
annual cost of lost heat can be determined
knowing the annual cost of consumed fuel to
produce this quantity heat. Plotting annual cost
of insulation and cost of lost heat versus
insulation thickness, the minimum total annual
cost can be obtained as shown in figure. To
choose one of several suitable available
insulations, the curves of total annual cost of
each versus its own thickness are plotted and
that of the lowest annual total cost is the
selected thickness and type of insulation.
Cost per year
Total cost
Lost heat cost
Insulation Cost
Optimum
thickness
Insulation Thickness
Figure (4) Determination of
the optimum insulation
thickness
6. Thermal Resistance for Different Modes of Heat Transfer
6.1 Thermal Resistance in Conduction Heat transfer
a) Plane wall
𝑙
𝐴𝑘
𝑅=
Where l is the wall thickness [m], A is the surface area [m2] and k is the
thermal conductivity [W/m.K].
b) Cylindrical wall
𝑅=
𝑟
𝑙𝑛 ( 𝑟2 )
1
2𝜋𝐿𝑘
Where r2 is the outer radius [m], r1 is the inner radius of the cylinder [m], L
is the length of cylinder [m] and k is the thermal conductivity of cylinder
material [W/m.K].
c) Spherical wall
𝑅=
𝑟2 − 𝑟1
4 𝜋 𝑟1 𝑟2 𝑘
Where r2 is the outer radius [m], r1 is the inner radius of the sphere [m] and
k is thermal conductivity of sphere material [W/m.K].
4
6.2 Thermal Resistance in Convection Heat transfer
𝑄̇ = ℎ 𝐴 (𝑇2 − 𝑇1 ) =
And thus;
𝑅𝑐𝑜𝑛 =
∆𝑇 𝑇2 − 𝑇1
=
𝑅
𝑅
1
𝐴ℎ
Where As is the surface area adjacent to fluid film [m2], h is coefficient of heat
transfer due to convection [w/m2K].
6.3 Thermal Resistance in Radiation Heat transfer
Heat transfer from a surface of area As [m2] and temperature T1 [K] to
surrounding of temperature T2 [K] is given by the following Stefan-Boltzmann
equation as:
𝑄̇ = 𝜀 𝜎 𝐴𝑠 (𝑇24 − 𝑇14 )
Where  is the emissivity of the surface, σ is a constant of value of 5.67×10-8
W/m2.K4.The foregoing equation can be written, also, as:
𝑄̇ =
And thus;
𝑅=
∆𝑇 𝑇2 − 𝑇1
=
= 𝜀 𝜎 𝐴𝑠 (𝑇24 − 𝑇14 )
𝑅
𝑅
𝑇2 − 𝑇1
1
=
4
4
𝜀 𝜎 𝐴𝑠 (𝑇2 − 𝑇1 ) 𝜀 𝜎 𝐴𝑠 (𝑇2 + 𝑇1 )(𝑇22 + 𝑇12 )
Accordingly;
𝑅𝑟𝑎𝑑 =
1
𝜀 𝜎 𝐴𝑠 (𝑇2 + 𝑇1 )(𝑇22 + 𝑇12 )
And Coefficient of heat transfer by radiation is given by:
ℎ𝑟𝑎𝑑 = 𝜀𝜎(𝑇2 + 𝑇1 )(𝑇22 + 𝑇12 )
6.4 Combined Thermal Resistance of Heat Transfer by both Convection and
Radiation
Combined radiation convection thermal resistance can be determined by the
following relation as:
5
1
𝑅𝑐𝑜𝑚
=
1
𝑅𝑟𝑎𝑑
+
1
Ts
Rcon
Ts > T
𝑅𝑐𝑜𝑛
Air
T
And
Rrad
Surface area As
𝑅𝑐𝑜𝑚 =
𝑅𝑟𝑎𝑑 × 𝑅𝑐𝑜𝑛
𝑅𝑟𝑎𝑑 + 𝑅𝑐𝑜𝑛
Figure (5) Combined radiation
convection heat transfer
The coefficient of heat transfer of combined radiation convection is given by the
following equation, taking in account that the area for both is the same:
ℎ𝑐𝑜𝑚 = ℎ𝑟𝑎𝑑 + ℎ𝑐𝑜𝑛
Example 1
A 3-m internal diameter spherical tank made of 2-cm thick stainless steel (k= 15
W/m.K) is used to store ice water at T ∞,1 = 0oC. The tank is located in a room whose
temperature is T∞,2 = 22oC. The walls of the room are also at 22oC. The outer
surface of the tank is black and heat transfer
between the outer surface of the tank and
22oC
surroundings by natural convection and
radiation. The convection heat transfer
coefficients at the inner and outer surfaces of
m
the tank are h1 = 80 W/m2.K and h2 = 10
2C
Iced water
W/m2.K; respectively. Knowing that Latent
m
heat of ice melting hif = 333.7 kJ/kg,
1.5
determine: (a) the rate of heat transfer to the
0oC
iced water in the tank, (b) the amount of ice at
h1
h2
0oC that melts during a 24-h period.
Given:
Sphere, r1 = 1.5 m, r2 = 1.52 m, r3 = 1.57 m,
T∞1 = 0oC, T∞2 = 22oC, h1 =80 W/m2.K,
h2 =10 W/m2.K, kst = W/m.K
Required: (a) 𝑄̇ [W],
Rrad
T
(b) m/day [kg/day],
Solution:
(a)Starting with calculation of thermal
resistance of inside convection Ri as:
6
1
Ri
T1
R1 T
2
T
Ro
2
𝑅𝑖 =
1
1
1
𝐾
=
=
= 0.000442
2
2
𝐴1 ℎ1 (4𝜋𝑟1 ) × ℎ1 (4𝜋 × 1.5 ) × 80
𝑊
And thermal resistance of conduction through the metallic wall of the tank R 1 is:
𝑅1 =
𝑟2 − 𝑟1
1.52 − 1.5
𝐾
=
= 0.000047
4𝜋𝑘𝑟2 𝑟1 4𝜋 × 15 × 1.5 × 1.52
𝑊
Thermal resistance associated with convection heat transfer at outer surface Ro is:
𝑅𝑜 =
1
1
1
𝐾
=
=
= 0.00345
2
2
𝐴2 ℎ2 (4𝜋𝑟2 ) × ℎ2 (4𝜋 × 1.52 ) × 10
𝑊
Assuming the temperature of outer surface of the sphere is equal to 5 oC, one can,
first, calculate the radiation heat transfer coefficient according to the relation:
2 )
ℎ𝑟𝑎𝑑 = 𝜀𝜎(𝑇22 + 𝑇∞2
× (𝑇2 + 𝑇∞2 )
And accordingly, the thermal resistance of radiation is given by:
𝑅𝑟𝑎𝑑 =
1
1
1
𝐾
=
=
= 0.00646
2
2
𝐴2 ℎ𝑟𝑎𝑑 (4𝜋𝑟2 ) × ℎ𝑟𝑎𝑑 (4𝜋 × 1.52 ) × 5.34
𝑊
The two parallel resistances Ro and Rrad have an equivalent value, according to the
following relation:
1
𝑅𝑒𝑞𝑢𝑖𝑣
=
1
𝑅𝑟𝑎𝑑
+
1
1
1
𝑊
=
+
= 444.7
𝑅𝑜 0.00646 0.00345
𝐾
Therefore,
𝑅𝑒𝑞𝑢𝑖𝑣 =
1
𝐾
= 0.00225
444.7
𝑊
Since all resistances are, know, in series, then total resistance Rtotal can be
calculated as:
𝐾
𝑅𝑡𝑜𝑡𝑎𝑙 = 𝑅𝑖 + 𝑅1 + 𝑅𝑒𝑞𝑢𝑖𝑣 = 0.000442 + 0.000047 + 0.00225 = 0.00274
𝑊
Accordingly the rate of heat transferred to ice is equal to:
𝑄̇ =
𝑇∞2 − 𝑇∞1 22 − 0
𝑘𝐽
=
= 8029 𝑊 = 8.029
𝑅𝑡𝑜𝑡𝑎𝑙
0.00274
𝑠
To check the validity of our assumption of T2 = 5oC, one uses the flowing relation:
7
𝑄̇ =
𝑇∞2 − 𝑇2
𝑅𝑒𝑞𝑢𝑖𝑣
And
𝑇2 = 𝑇∞2 − 𝑄̇ 𝑅𝑒𝑞𝑢𝑖𝑣 = 22 − (8029 × 0.00225) = 4℃
Which is sufficiently close to the assumed 5oC, otherwise further assumptions must
be carried out, till we obtain a proper value.
(b) The amount of melt ice can be calculated according to the following:
𝑄 = 𝑄̇ ∆𝑡 = (8.029 × 24 × 3600) = 673,700 𝑘𝐽
And
𝑄
𝑚𝑖𝑐𝑒 = ℎ =
𝑖𝑓
673,700
333.7
= 2079 𝑘𝑔
N. B.
Another approach to determine the value of temperature T 2 is to apply the numerical
technique known as “Simple Iteration method. To explain this technique, it is applied
to the foregoing example. Since the rate of heat transferred through total resistance
(Rtotal) is the same of that transfers through the equivalent resistance ( Requiv.), then
one can write the following relation:
𝑇∞2 − 𝑇∞1 𝑇∞2 − 𝑇2
𝑄̇ =
=
𝑅𝑡𝑜𝑡𝑎𝑙
𝑅𝑒𝑞𝑢𝑖𝑣
And thus;
𝑅𝑒𝑞𝑢𝑖𝑣
𝑇2 = 𝑇∞2 − (𝑇∞2 − 𝑇∞1 ) ×
= 𝑓(𝑇2 )
𝑅𝑡𝑜𝑡𝑎𝑙
One can note that all terms of right hand side of this equation are constant except
Requiv., which is a function of T2, hence the previous equation can be written in the
following mathematical form as:
𝑥 = 𝑓(𝑥)
And the recursive relation of the solution is:
𝑥𝑗+1 = 𝑓(𝑥𝑗 )
According to this technique and assuming the first guess of T2 = 273 K, the solution
are shown in the following table;
i
T2 [K]
Requiv. [K/W]
f(T2) [K]
1
273
5.966×10-5
292.6
2
292.6
2.190×10-3
277.02
8
3
277.02
2.253×10-3
276.92
Example 2
Referring to the previous example, now the ice-tank is covered with 5-cm thick glass
wool insulation with k= 0.05 W/m.K, determine: (a) the temperature of the outer
surface and (b) the percentage energy saving.
Required:
(a) T3 oC,
(b) ∆Q %
T
2
Solution:
𝑅𝑖 =
𝑅1 =
𝑅2 =
𝑅𝑜 =
1
1
1
=
=
𝐴1 ℎ1 (4𝜋𝑟12 ) × ℎ1 (4𝜋 × 1.52 ) × 80
𝐾
= 0.000442
𝑊
1.5
0oC
𝑟2 − 𝑟1
1.52 − 1.5
=
4𝜋𝑘𝑟2 𝑟1 4𝜋 × 15 × 1.52 × 1.5
𝐾
= 0.000047
𝑊
h2
Rrad
T
1
Ri
T1
R1
T2
R2 T3
1
1
1
=
=
𝐴3 ℎ2 (4𝜋𝑟32 ) × ℎ2 (4𝜋 × 1.572 ) × 10
𝐾
= 0.00323
𝑊
(a) Assume T3 =280 K, then
2 )
ℎ𝑟𝑎𝑑 = 𝜀𝜎(𝑇32 + 𝑇∞2
× (𝑇3 + 𝑇∞2 ) = 1 × (5.67 × 10−8 ) × (2802 + 2952 )
× (280 + 295) = 5.393 𝑊/𝑚2 . 𝐾
1
1
1
𝐾
=
=
=
0.00646
2
𝐴3 ℎ𝑟𝑎𝑑 (4𝜋𝑟3 ) × ℎ𝑟𝑎𝑑 (4𝜋 × 1.572 ) × 5.393
𝑊
𝑅𝑒𝑞𝑢𝑖𝑣. =
m
157 Cm
h1
𝑟3 − 𝑟2
1.52 − 1.5
=
4𝜋𝑘𝑟3 𝑟2 4𝜋 × 0.05 × 1.57 × 1.52
𝐾
= 0.0333
𝑊
𝑅𝑟𝑎𝑑 =
m
2C
Iced water
𝑅𝑜 × 𝑅𝑟𝑎𝑑 0.00323 + 0.00646
𝐾
=
= 2.153 × 10−3
𝑅𝑜 + 𝑅𝑟𝑎𝑑 0.00323 × 0.00646
𝑊
9
T
Ro
2
𝑅𝑡𝑜𝑡𝑎𝑙 = 𝑅𝑖 + 𝑅1 + 𝑅2 + 𝑅𝑒𝑞𝑢𝑖𝑣 = 0.000442 + 0.000047 + 0.0333 + 0.002153
𝐾
= 0.035942
𝑊
𝑄̇ =
𝑇∞2 − 𝑇∞1 𝑇∞2 − 𝑇3
=
𝑅𝑡𝑜𝑡𝑎𝑙
𝑅𝑒𝑞𝑢𝑖𝑣
𝑇3 = 𝑇∞2 − (𝑇∞2 − 𝑇∞1 ) ×
𝑅𝑒𝑞𝑢𝑖𝑣
= 𝑓(𝑇3 )
𝑅𝑡𝑜𝑡𝑎𝑙
𝑇3 = 295 − (295 − 273) ×
0.002153
= 293.7 𝐾 = 20.7°𝐶 (1𝑠𝑡 𝑖𝑚𝑝𝑟𝑜𝑣𝑒𝑑 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)
0.035942
Repeating the previous step, the obtained 2nd improved solution:
T3 = 293.75 K = 20.75oC
(b)This 2nd iteration is acceptable, and thus one can calculate the lost heat as the
following:
𝑄̇ =
𝑇∞,2 − 𝑇3 295 − 293.75
=
= 580.6 𝐽/𝑠
𝑅𝑒𝑞𝑢𝑖𝑣.
2.153 × 10−3
∆𝑄̇ = 𝑄̇𝑤𝑖𝑡ℎ 𝑜𝑢𝑡 𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛 − 𝑄̇𝑤𝑖𝑡ℎ 𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 8029 − 580.6 = 7448.4 𝐽/𝑠
Percentage of saved heat = 𝑄̇
∆𝑄̇
𝑤𝑖𝑡ℎ 𝑜𝑢𝑡 𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛
10
=
7448.4
8029
× 100 = 92.8%