2. Focal Point in Conduction Heat Transfer
What drives heat transfer is temperature difference
and not temperature;
Conduction heat transfer problems involve the
determination of temperature distribution in a
region;
How is the rate of heat transfer related to
temperature distribution?
What governs temperature distribution in a region?
Fourier’s law of conduction provides the answer to
the first question while the principle of conservation
of energy gives the answer to the second.
3. Conservation of Energy
What determines temperature distribution in a
region is that the temperature at each point is
adjusted such that the principle of conservation of
energy is satisfied everywhere;
By so doing, volumetric energy generation;
Examples of volumetric energy generation include,
among others, heat conduction in nuclear elements,
metabolic heat production in tissue, electrical
energy loss in transmission lines etc
4. Boundary Conditions
Since boundary conditions involve thermal interaction
with the surroundings, it is necessary to first describe
two common modes of surface heat transfer:
convection and radiation.
Surface Convection: Newton's Law of Cooling.
In this mode of heat transfer, energy is exchanged
between a surface and a fluid moving over it. The flux
in convection is directly proportional to the difference
in temperature between the surface and the streaming
Fluid. i.e
5. This result is known as Newton's law of cooling. The constant of
proportionality h is called heat transfer coefficient. This coefficient
depends on geometry, fluid properties, motion, and in some cases
temperature difference (Ts, T). Thus, unlike thermal conductivity, h is not
a property of material.
Surface Radiation: Stefan-Boltzmann Law
radiation energy is transmitted by electromagnetic waves, which travel
best in a vacuum.
The maximum possible radiation is described by the Stefan-Boltzmann
law, which gives surface radiation flux for an ideal body called
blackbody as
6. Radiation flux qr
To determine the radiation flux qr emitted from a real
surface, a radiation property called emissivity, ε, is
defined as
Thus
Radiation energy exchange between two surfaces
depends on the geometry, shape, area, orientation and
emissivity of the two surfaces.
7. Classification of conduction heat
transfer problems
The driving force for any form of heat transfer is
the temperature difference. This can be classified
as:
• steady versus transient heat transfer,
• multidimensional heat transfer,
• heat generation.
18. During steady one-dimensional heat conduction in a
spherical (or cylindrical) container, the total rate of
heat transfer remains constant, but the heat flux
decreases with increasing radius.
Under steady conditions,
The entire heat generated within the medium is
conducted through the outer surface of the cylinder,
The heat generated within this inner cylinder must
be equal to the heat conducted through the outer
surface of this inner cylinder
22. Heat transfer in Composite materials
Consider sketch below, in which heat will tend to flow
between materials B and C, and this heat flow will be
normal to the primary direction of heat transfer.
In general, when thermal resistances occur in
parallel, heat will flow in more than one direction;
In this case, the one-dimensional calculation of q
represents an approximation, that is generally quite
acceptable for process engineering purposes.
23.
24. The resistance to heat transfer is termed the thermal resistance,
and is denoted by Rth. Thus, the general rate equation may
be written as:
In this equation, all quantities take on positive values only, so that q
and T represent the absolute values of the heat-transfer rate and
temperature difference, respectively.
The thermal resistance concept permits some relatively complex heat-
transfer problems to be solved in a very simple manner. The reason is
that thermal resistances can be combined in the same way as electrical
resistances. Thus, for resistances in series, the total resistance is the sum
of the individual resistances:
Likewise, for resistances in parallel:
25. The effect of the additional resistance is to decrease the rate of
heat transfer.
Since the contact resistance is difficult to determine, it is often
neglected or a rough approximation is used. For example, a value
equivalent to an additional 5 mm of material thickness is
sometimes used for the contact resistance between two pieces
of the same material.
A slightly modified form of the thermal resistance, the R-value, is
commonly used for insulations and other building materials. The R-value
is defined as:
26. where B is the thickness of the material and k is its
thermal conductivity. The R-value is the thermal
resistance, in English units, of a slab of material
having a cross-sectional area of 1 ft2. Since the R-
value is always given for a specified thickness, the
thermal conductivity of a material can be obtained
from its R-value.
Also, since R-values are essentially thermal
resistances, they are additive for materials arranged in
series.
27. The summary of procedures
Write down the conduction equation in the appropriate
coordinate system;
Impose any restrictions dictated by the physical situation to
eliminate terms that are zero or negligible;
Integrate the resulting differential equation to obtain the
temperature profile;
Use the boundary conditions to evaluate the constants of
integration;
Use the appropriate form of Fourier’s law to obtain the heat
flux;
Multiply the heat flux by the cross-sectional area to obtain
the rate of heat transfer.
28. Questions
Q1. What drives heat transfer?
Q2. How is the rate of heat transfer related to temperature distribution?
Q3. What governs temperature distribution in a region?
Q4. In order to size the compressor of a new refrigerator, it is desired to
determine the rate of heat transfer from the kitchen air into the
refrigerated space through the walls, doors and the top and bottom
section of the refrigerator. In your analysis, would you treat this as a
transient or steady-state heat transfer problem? Also would you consider
the heat transfer to be one-dimensional or multidimensional? Explain.
29. Questions Cont.
Q5. Determine the heat transfer through a copper tube
5m long with inner diameter 80mm and outer
diameter 100mm. The inside temperature is 2000C
and the outside temperature is 700C. (The thermal
conductivity of copper is 385W/m. K)
Q6. A spherical steel reaction vessel has an outer
radius of 1.5m and is covered in lagging 200mm
thick. The thermal conductivity of the lagging is
0.1W/m. K. the temperature at the surface of the
steel is 3400C and the surface temperature of the
lagging is 450C. Taking thermal conductivity of
steel as 60W/m. K, calculate the heat loss.
30. Q7. A wall of an area of 25m2 is made up of four
layers. On the inside is plaster 15mm thick, then
there is brick 100mm thick, then insulation 60mm
thick and finally brick 100mm thick. The thermal
conductivities are : of plaster is 0.1W/m. K, brick is
0.6W/m. K and insulation is 0.08W/m. K. The inner
surface temperature of the wall is 180C and the outer
surface temperature is -20C. Calculate:
(a)The heat loss
(b) Temperature at the interface between the plaster
and the brick.
31. Q8. A steel pipe 120mm inside diameter has a wall
10mm thick. It is covered with insulation 20mm
thick. The thermal conductivity of steel is 60W/m. K
and for the insulation is 0.09W/m. K. The pipe
carries steam at 1500C and the outer surface
temperature of the insulation is 00C. Determine:
(i) The heat loss per meter length
(ii) The temperature at the pipe’s outer surface
