MECHANISM OF HEAT TRANSFER
There are three modes of heat transfer: Coduction , convection and radiation heat transfer.
Mechanism of heat conduction:
Conduction is the transfer of energy from the more energetic particles of a substance to the
adjacent less energetic ones as a result of interactions between the particles. Conduction can take
place in solids, liquids, or gases. In gases and liquids, conduction is due to the collisions and
diffusion of the molecules during their random motion. In solids, it is due to the combination of
vibrations of the molecules in a lattice and the energy transport by free electrons. A cold canned
drink in a warm room, for example, eventually warms up to the room temperature as a result of
heat transfer from the room to the drink through the aluminum can by conduction.
The mechanism of heat conduction in a liquid is complicated by the fact that the molecules are
more closely spaced, and they exert a stronger intermolecular force field. The thermal
conductivities of liquids usually lie between those of solids and gases. The thermal conductivity
of a substance is normally highest in the solid phase and lowest in the gas phase.
In solids, heat conduction is due to two effects: the lattice vibrational waves induced by the
vibrational motions of the molecules positioned at relatively fixed positions in a periodic manner
called a lattice, and the energy trans-ported via the free flow of electrons in the solid (Fig. 1–27).
The expression for heat conduction for one dimensional is
𝑄𝑥 = −𝐾𝐴
𝜕𝑇
𝜕𝑥
Here K is the thermal conductivity of the material. Unit is W/m.0C.
The thermal conductivity of a substance is normally highest in the solid phase and lowest in the
gas phase. Unlike gases, the thermal conductivities of most liquids decrease with increasing
temperature, with water being a notable exception. Like gases, the conductivity of liquids
decreases with increasing molar mass. Liquid metals such as mercury and sodium have high
thermal conductivities and are very suitable for use in applications where a high heat transfer rate
to a liquid is desired, as in nuclear power plants.
Unlike metals, which are good electrical and heat conductors, crystalline solids such as diamond
and semiconductors such as silicon are good heat conductors but poor electrical conductors. As a
result, such materials find wide spread use in the electronics industry. Despite their higher price,
diamond heat sinks are used in the cooling of sensitive electronic components because of the
excellent thermal conductivity of diamond. Silicon oils and gaskets are commonly used in the
packaging of electronic components because they provide both good thermal contact and good
electrical insulation.
Define Thermal Conductivity and thermal diffusivity and their physical
significances
Thermal Conductivity: Thermal conductivity of a material can be defined as the
rate of heat transfer through a unit thickness of the material per unit area per unit
temperature difference. The thermal conductivity of a material is a measure of the
ability of the material to conduct heat. A high value for thermal conductivity
indicates that the material is a good heat conductor, and a low value indicates that
the material is a poor heat conductor or insulator.
𝑊𝑒 𝑘𝑛𝑜𝑤, 𝑄 = 𝐾𝐴
𝑇1 − 𝑇2
𝑄𝐿
𝑊
≫𝐾=
𝑢𝑛𝑖𝑡 0
𝐿
𝐴(𝑇1 − 𝑇2 )
𝑚 𝐶
Thermal conductivities of some common materials at room temperature are shown in Table 1-1.
The relatively high thermal conductivities of pure metals are primarily due to the electronic
component. The lattice component of thermal conductivity strongly depends on the way the
molecules are arranged. For example, diamond, which is a highly ordered crystalline solid, has
the highest known thermal conductivity at room temperature.
Pure metals have high thermal conductivities, but The thermal conductivity of an alloy of two
metals is usually much lower than that of either metal, as shown in Table 1–2. The thermal
conductivities of materials vary with temperature (Table1–3).The variation of thermal
conductivity over certain temperature ranges is negligible for some materials, but significant for
others, as shown in Fig. 1–28.
h
Fig 1-28: The variation of the thermal conductivity of various solids, liquids,and gases with
temperature (from White, Ref. 10).
The temperature dependence of thermal conductivity causes considerable complexity in
conduction analysis. Therefore, it is common practice to evaluate the thermal conductivity k at
the average temperature and treat it as a constant in calculations.
Thermal Diffusivity:
It’s a material property that appears in the transient heat conduction analysis is the thermal
diffusivity, which represents how fast heat diffuses through a material and is defined as
𝛼=
𝐻𝑒𝑎𝑡 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑒𝑑
𝐻𝑒𝑎𝑡 𝑠𝑡𝑜𝑟𝑒𝑑
=
𝑘
𝜌𝐶𝑝
( 𝑚2 /s)
Note that the thermal conductivity k represents how well a material conducts heat, and the heat
capacity Cp represents how much energy a material stores per unit volume. Therefore, the
thermal diffusivity of a material can be viewed as the ratio of the heat conducted through the
material to the heat stored per unit volume. A material that has a high thermal conductivity or a
low heat capacity will obviously have a large thermal diffusivity. The larger the thermal
diffusivity, the faster the propagation of heat into the medium. A small value of thermal
diffusivity means that heat is mostly absorbed by the material and a small amount of heat will be
conducted further.
The thermal diffusivities of some common materials at 20°C are given in Table 1-4.
Fourier Law of Heat Conduction
Heat conduction through a medium in a specified direction (say, in the x-direction) is
proportional to the temperature difference across the medium and the area normal to the direction
of heat transfer, but is inversely proportional to the distance in that direction. This was expressed
in the differential form by Fourier’s law of heat conduction for one-dimensional heat conduction
as
where k is the thermal conductivity of the material, which is a measure of the ability of a
material to conduct heat, and dT/dx is the temperature gradient, which is the slope of the
temperature curve.
Why is negative sign?
Heat is conducted in the direction of decreasing temperature, and thus the temperature gradient is
negative when heat is conducted in the positive x-direction. The negative sign in Eq. 2–1 ensures
that heat transfer in the positive x-direction is a positive quantity.
One Dimensional Heat Conduction Equation
Through Plane wall (one dimensional) heat conduction
Consider heat conduction through a large plane wall such as the wall of a house, the glass of a
single pane window, the metal plate at the bottom of a pressing iron, a cast iron steam pipe, a
cylindrical nuclear fuel element, an electrical resistance wire, the wall of a spherical container, or
a spherical metal ball that is being quenched or tempered. Heat conduction in these and many
other geometries can be approximated as being one-dimensional since heat conduction through
these geometries will be dominant in one direction and negligible in other directions.
Consider a thin element of thickness x in a large plane wall, as shown in Figure 2–13. Assume
the density of the wall is, the specific heat is Cp, and the area of the wall normal to the direction
of heat transfer is A. An energy balance on this thin element during a small time interval t can
be expressed as
But the change in the energy content of the element and the rate of heat generation within the
element can be expressed as
Substituting into Equation 2–6, we get
Dividing by A∆x the above equation will be
Taking the limit as
∆x→ 0 and
∆t→ 0 yields
since, from the definition of the derivative and Fourier’s law of heat conduction,
Noting that the area A is constant for a plane wall, the one-dimensional transient heat conduction
equation in a plane wall becomes
However, the thermal conductivity in most practical applications can be assumed to remain
constant at some average value. The equation above in that case reduces to
Where α the property =k/ρC the thermal diffusivity, C is the thermal diffusivity of the material
and represents how fast heat propagates through a material. It reduces to the following forms
under specified conditions:
For Long Cylinder (one dimensional)
Figure 2.1
Now consider a thin cylindrical shell element of thickness δr in a long cylinder, as
shown in Figure 2–1. Assume the density of the cylinder is ρ, the specific heat is C,
and the length is L. The area of the cylinder normal to the direction of heat transfer
at any location is A = 2rπL where r is the value of the radius at that location. Note
that the heat transfer area A depends on r in this case, and thus it varies with
location.
Like previous way we find the following equation for a cylinder
Using A = 2πrL, we have the following relation
Where α the property =k/ρC the thermal diffusivity, C is the thermal diffusivity of the material
and represents how fast heat propagates through a material. It reduces to the following forms
under specified conditions:
For Sphere
Here A = 4πr2, S0 the heat conduction equation will be
Where α the property =k/ρC the thermal diffusivity, C is the thermal diffusivity of the material
and represents how fast heat propagates through a material. It reduces to the following forms
under specified conditions:
Combined One-Dimensional Heat Conduction Equation
where n = 0 for a plane wall, n = 1 for a cylinder, and n = 2 for a sphere. In the
case of a plane wall, it is customary to replace the variable r by x.
Three Dimensional heat conduction (General) Equation
Most of the cases one dimensional heat conduction is occurred. In some cases three
dimensional problems are found.
In such cases heat conduction is said to be multidimensional, and in this section we will develop
the governing differential equation in such systems in rectangular, cylindrical, and spherical
coordinate systems.
Rectangular Coordinates
Considering thermal conductivity k is constant and using thermal diffusivity we get
Cylindrical Co-ordinates
Using the following relation into the rectangular coordinates systems we find the general
expression for cylindrical system as follows:
Spherical Coordinates:
Again after lengthy manipulations, we obtain
Problems:
1. Steady state heat conduction condition in a wall of thermal conductivity 50 W/m.0C is
observed to be T(0C) = a + bx2, where a = 200 0C and b = -2000 0C/m2, x is in meters.
Wall thickness is 50 mm.
a) What is the heat generation rate g in the walls?
b) Determine the heat flux at the wall faces.
Solution:
a) We know T = a + bx2, Given K = 50 W/m. 0C.
= 200 -2000 x2.
We know for one dimensional steady state heat conduction equation for constant k
𝑑2 𝑇
𝑑𝑥 2
𝑔
+ 𝐾 = 0 (1)
𝑇 = 200 − 2000 𝑥 2
𝑑2 𝑇
= −2000 × 2𝑥 = −4000𝑥
𝑑𝑥 2
𝑑2 𝑇
= −4000
𝑑𝑥 2
Putting this value into Equation (1) we get
𝑔
−4000 + 50 = 0 ≫ 𝑔 = 200000 𝑊/𝑚3 (Ans).
b)
𝑑𝑇
𝑞́ = −𝐾 𝑑𝑥
= -50 × (-4000 x)
= 200000 x
At the wall, x = 50 mm,
𝑞́ = 200000 × 50 × 10−3
= 10000 W/m2(Ans)
2. Given, 𝑇 = 900 − 300𝑥 − 50 𝑥 2 Heat generation rate= 1000 W/m3
A = 10 m 2, density = 1600 kg/m3, K= 40 W/m2. 0C. Cp = 4.0 KJ/kg. k
Determine the heat transfer rate entering the wall at x = 0 and x= 1.0m leaving the wall.
𝒅𝑻
Solution: 𝑸𝒙 = −𝑲𝑨 𝒅𝒙 = -KA ( -300-100x)
At x = 0 m, 𝑸𝒙 = -40 x10 (-300-100 x 0) = 120,000 W
At x = 1.0 m 𝑸𝒙 = −𝟒𝟎 × 𝟏𝟎 × (−𝟑𝟎𝟎 − 𝟏𝟎𝟎 × 𝟏) = 𝟏𝟔𝟎, 𝟎𝟎𝟎 𝑾
6
Fig. 5: Steady, one‐dimensional heat conduction in a cylindrical layer.
After integration:
The convection resistance remains the same in both cylindrical and spherical coordinates, Rconv
= 1/hA. However, note that the surface area A = 2πrL (cylindrical) and A = 4πr2
(spherical) are functions of radius.
Example 2: Multilayer cylindrical thermal resistance network
Steam at T∞,1 = 320 °C flows in a cast iron pipe [k = 80 W/ m.°C] whose inner and outer
diameter are D1 = 5 cm and D2 = 5.5 cm, respectively. The pipe is covered with a 3‐cm
thick
glass wool insulation [k = 0.05 W/ m.°C]. Heat is lost to the surroundings at T∞,2 = 5°C by natur
al convection and radiation, with a combined heat transfer coefficient of h2 = 18 W/m2.
°C.
Taking the heat transfer coefficient inside the pipe to be h1 = 60 W/m2
K, determine the rate of heat loss from the steam per unit length of the pipe. Also determine the t
emperature drop across the pipe shell and the insulation.
Assumptions:
Steady‐state and one‐dimensional heat transfer.
Solution:
Taking L = 1 m, the areas of the surfaces exposed to convection are:
A1 = 2πr1L = 0.157 m2
A = 2πr L = 0.361 m2
Critical Radius of Insulation
To insulate a plane wall, the thicker the insulator, the lower the heat transfer rate (since the area
is constant). However, for cylindrical pipes or spherical shells, adding insulation results in
increasing
the
surface area which
in
turns
results
in
increasing
the convection heat transfer. As a result of these two competing trends the heat transfer may incr
ease or decrease.
Heat Generation in Solids
Conversion of some form of energy into heat energy in a medium is called
heat generation. Heat generation leads to a temperature rise throughout the medium. Some exam
ples of heat generation are resistance heating in wires, exothermic chemical reactions in solids,
and nuclear reaction. Heat generation is usually expressed per unit volume (W/m3).
In most applications, we are interested in maximum temperature Tmax and surface
temperature Ts of solids which are involved with heat generation. The maximum temperature T
max in a solid that involves uniform heat generation will occur at alocation furthest away from th
e outer surface when the outer surface is maintained at a constant temperature, Ts.
Consider a solid medium of surface area A, volume V, and constant thermal conductivity k, wher
heat is generated at a constant rate of g• per unit volume. Heat is transferred from the solid to the
surroundings medium at T∞. Under steady conditions, the energy balance for the solid can be
expressed as:
Using the above relationship, the surface temperature can be calculated for a plane wall of thickn
ess 2L, a long cylinder of radius r0, and a sphere of radius r0, as follows:
Thermal Contact Resistane
Temperature distribution and heat flow lines along two solid plates pressed against each other for
the case of perfect and imperfect contact.
•
When two such surfaces are pressed against each other, the peaks form good material
contact but the valleys form voids filled with air.
•
These numerous air gaps of varying sizes act as insulation because of the low thermal
conductivity of air.
•
Thus, an interface offers some resistance to heat transfer, and this resistance per unit
interface area is called the thermal contact resistance, Rc.
Heat flux can be expressed through contact surface, q = ℎ𝑐 (𝑑𝑇𝑐 ) =
𝑑𝑇𝑐
1
ℎ𝑐
𝑑𝑇
= 𝑅 𝑐, where hc is the
𝑐
surface conductance, dTc is the temperature drop at the contact point, Rc is the contact
resistance.
The value of thermal contact resistance depends on:
•
surface roughness,
•
material properties,
•
ttemperature and pressure at the interface
•
type of fluid trapped at the interface.
Physical Significance of Thermal Contact resistance
Thermal contact resistance is significant and can even dominate the heat transfer for good heat
conductors such as metals, but can be disregarded for poor heat conductors such as insulations. If
thermal conductance increases then thermal contact resistance will increase.
Figure: Effect of metalic coating on thermal conductance
The thermal contact resistance can be minimized by applying
•
a thermal grease such as silicon oil
•
a better conducting gas such as helium or hydrogen
•
a soft metallic foil such as tin, silver, copper, nickel, or aluminum.
•
Increasing contact pressure.
Some Problems
1. A composite wall is made up of an external thickness of brick work 110 mm thick and
inside layer of fibre-glass of 75 mm thick. The fibre glass is faced internally by an
insulating board of 25 mm thick. The coefficient of thermal conductivity for three
materials are as follows:
Brick work= 1.15 W/m.k, fibre glass = 0.04 W/m.k, insulating board = 0.06 W/m.k.
The surface heat transfer coefficient for inside and out wall are 2.5 W/m2 .k and 3.1
W/m2.k.
If the size of the wall is 10 m high and 4 m wide, determine the total thermal insulation
and overall heat transfer coefficient. For outside temperature 40 0C and inside
temperature 27 0C, find the heat lost per sec.
Solution:
1
𝐿1
𝐿2
𝐿3
1
We know, 𝑅𝑡𝑜𝑡𝑎𝑙 = ℎ1 + 𝐾1 + 𝐾2 + 𝐾3 + ℎ2
1
0.110
0.075
0.025
1
= 2.5 + 1.15 + 0.04 + 0.06 + 3.1 = 3.115 𝑚2 . 𝐾/𝑊
Overall heat transfer coefficient , U = 1/R total= 1/3.115 = 0.32 W/m2.K.
Total heat Transfer rate, Q = UA (∆𝑇) =0.32 x 10x 4 x (40-27) = 217.5 W.
Heat Transfer from Finned Surfaces
From the Newton’s law of cooling, Q• conv = h A (Ts ‐ T∞), the rate of convective heat
transfer from a surface at a temperature Ts can be increased by two methods:
1) Increasing the convective heat transfer coefficient, h
2) Increasing the surface area A.
Increasing the convective heat transfer coefficient may not be practical and/or adequate.
An
increase in surface area by attaching extended surfaces called fins to the surface
is more convenient.
Finned surfaces are commonly used in practice to enhance heat transfer. In the analysis of
the fins, we consider steady operation with no heat generation in the fin. We also assume
that the convection heat transfer coefficient h to be constant and uniform over the entire
surface of the fin.
Problems: A very long copper rectangular plate (k = 380 W/m0C) of size 2 cm wide and 2 mm
thickness having base temperature of 300 0C is attached with a hot surface. The air temperature
is 25 0C where h = 10 W/m2 0C. Find the heat transfer rate from the fin. Also find the
effectiveness of that fin.
Solution: Given: h = 10 W/m2 0C , K = 380 W/m0C, p = 2(2x10-2+2x10-3) m =0.044 m
Area, A = zt= 2x10-2x 2x10-3 m2 = 0.00004 m2, θb = 300-25 = 275 K
Heat transfer by fin, 𝑄 = √ℎ𝑝𝑘𝐴 θb = √10 × 0.044 × 380 × 0.00004 × 275 𝑊
= 22.48 W
𝑘𝑝
380×0.044
Fin effectiveness, 𝜀 = √ℎ𝐴 = √10×.00004 = 204.45
Radiation Problems:
A black sphere of diameter 3 cm has temperature of 500 0C. What is the emissive power of the body?
Q = AσT4 =4/3 π r3 x 5.67 x 10-8 x 7734 =0. 0504 W.
Nodal problem There are four nodal points in a square. The temperature of the top side is 400 0C,
bottom is 500 0C, left side is 300 0C and right side is 200 0C. Each node has same size. Find the nodal
points temperature by using any method.