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1. Introduction:
Heat Transfer: is that science which seeks to predict the energy transfer that may take place between material bodes as a result of a temperature difference.
Heat: the form of energy that can be transferred from one system to another as a result of temperature difference.
The transfer of energy as heat is always from the higher-temperature medium to the lower-temperature, and heat transfer stops when the two mediums reach the same temperature. Thermodynamics deals with equilibrium states and changes from one equilibrium state to another. Heat transfer, on the other hand, deals with systems that lack thermal equilibrium, and thus it is a nonequilibrium phenomenon. Heat transfer equipment such as heat exchangers, boilers, condensers, heaters, and refrigerators are designed primarily on the basis of heat transfer analysis.
Fig. (1) Heat flows in the direction of decreasing temperature.
2. Modes of Heat Transfer:
There are three fundamental modes of heat transfer; Conduction, Convection, and Radiation.
Fig.(2) heat transfer modes
2.1 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. The rate of heat conduction through a medium depends on the geometry of the medium, its thickness, and the material of the medium, as well as the temperature difference across the medium.
Fourier’s Law of Heat Conduction: - The law states that the rate of heat flow by conduction in any medium in any direction is proportional to the area normal to the direction of heat flow and also proportional to the temperature gradient in that direction.
Fig.(3) Heat conduction through a plane wall of thickness △ x and area A.
Qcond. = kA
T 1 −T 2
∆x = −kA^
∆T
∆x (Watt)
Where: k: is the thermal conductivity of the material (W/mK) A: is normal cross section area (𝑚^2 ) ΔT: is temperature difference between surfaces (K) Δx: is the wall thickness (m) The negative sign ensures that heat transfer in the positive x direction is a positive quantity. Thermal Conductivity: The thermal conductivity of a material is a measure of the ability of the material to conduct heat.
The convection heat transfer coefficient: is not a property of the fluid. It is an experimentally determined parameter whose value depends on all the variables influencing convection such as the surface geometry, the nature of fluid motion, the properties of the fluid, and fluid velocity.
2.3 Radiation: is energy transport due to emission of electromagnetic waves or photons from a surface or volume. The radiation does not require a heat transfer medium, and can occur in a vacuum. The heat transfer by radiation is proportional to the fourth power of the absolute material temperature.
Fig.(5) Radiation heat transfer between a surface and the surfaces surrounding it
Stefan–Boltzmann law:
𝑄𝑟𝑎𝑑. = 𝜀𝜎𝐴𝑠 (𝑇^4 𝑆 − 𝑇^4 𝑠𝑢𝑟𝑟)^ (Watt)
Where: 𝜎 = 5.67 × 10−8^ 𝑊/𝑚^2. 𝐾^4 is the Stefan–Boltzmann constant 𝜀 is the emissivity of the material, its value depend on the material 𝑇𝑆 is the surface temperature 𝑇𝑠𝑢𝑟𝑟 is the surrounding surface temperature
Example (1)/ One face of a copper plate 3 cm thick is maintained at 400◦C, and the other face is maintained at 100◦C. How much heat is transferred through the plate? the thermal conductivity for copper is 370W/m· ◦C. Solution/ q A = −k
∆T
∆x
= 3.7 MW/m^2
Example (2)/ Air at 20◦C blows over a hot plate 50 by 75 cm maintained at 250◦C. The convection heat-transfer coefficient is 25 W/𝑚^2 · ◦C. Calculate the heat transfer. Solution/ q = h A (TS − T∞) = (25)(0.05)(0.75)(250 − 20) = 2.156 𝑘𝑊
Example (3)/ Consider a person standing in a breezy room at 20°C. Determine the total rate of heat transfer from this person if the exposed surface area and the average outer surface temperature of the person are 1.6 𝑚^2 and 29°C, respectively, and the convection heat transfer coefficient is 6 W/𝑚^2. The emissivity of a person is 𝜀 = 0.95. Solution/ The total rate of heat transfer from a person by both convection and radiation to the surrounding air and surfaces at specified temperatures is to be determined.
The Thermal Resistance:
Heat conduction through a plane wall can be rearranged as:
R𝑤𝑎𝑙𝑙: is the thermal resistance of the wall against heat conduction or simply the conduction resistance of the wall. Note that the thermal resistance of a medium depends on the geometry and the thermal properties of the medium.
If more than one material is present, as in the multilayer wall shown in Fig. (7)
Fig. (7) One-dimensional heat transfer through a composite wall
The analysis would proceed as follows: The temperature gradients in the three materials are shown, and the heat flow may be written
Note that the heat flow must be the same through all sections.
Consider the composite wall shown in Fig(8), which consists of two parallel layers. The thermal resistance network, which consists of two parallel resistances, can be represented as shown in the figure.
Noting that the total heat transfer is the sum of the heat transfers through each layer, we have
Series and parallel one-dimensional heat transfer through a composite wall:
Fig. ( 8 ) Thermal resistance network for two parallel layers.
fiberglass combination is 44 kW. What is the area of the slab? 𝑘𝐶𝑜𝑝𝑝𝑒𝑟 =
386 𝑤/ 𝑚℃ 𝑘𝑓𝑖𝑏𝑒𝑟𝑔𝑙𝑎𝑠𝑠 = 0.038 𝑤/ 𝑚℃
Solution:
Example(7): A section of a composite wall with the dimensions shown below has uniform temperatures of 200°C and 50°C over the left and right surfaces, respectively. If the thermal conductivities of the wall materials are: kA = 70 W/m K , kB = 60 W/m K, kC = 40 W/m K, and kD = 20 W/m K , determine the rate of heat transfer through this section of the wall and the temperatures at the interfaces.
Solution:
The thermal circuit for the composite wall is
(a) Each of these thermal resistances has a form given by
Evaluating the thermal resistance for each component of the wall
The total thermal resistance of the wall section,
(b) The average temperature at the interface between material A and materials B and C ( TABC ) can be determined by examining the conduction through material A alone
The average temperature at the interface between material D and materials B and C ( TBCD ) can be determined by examining the conduction through material D alone
572 W (fora0.12m 1 msection)
0. 349 C/W
( 300 100 ) C
0. 04 0. 025 0. 034 0. 25 0. 349 C/W
0. 034 C/W
0. 025 C/W
1 2
1 , 1 , 2 7
, 2 , 2 5 6
, 1 , 1 2 3 4
total
total mid mid
mid mid
mid mid
R
T T
Q
R R R R R
R
R R R
R
R R R R
4. Steady – State Heat Conduction in Cylinders:
Consider a long cylinder of inside radius ri , outside radius ro , and length L , such as the one shown in Figure(10) We expose this cylinder to a temperature differential Ti − To and ask what the heat flow will be. For a cylinder with length very large compared to diameter, it may be assumed that the heat flows only in a radial direction. Again, Fourier’s law is used. The area for heat flow in the cylindrical system is
so that Fourier’s law is written
with the boundary conditions
The solution is:
Fig.( 10 ) One-dimensional heat flow through a hollow cylinder
and the thermal resistance in this case is
The thermal-resistance concept may be used for multiple-layer cylindrical walls just as it used for plane walls. For the three-layer system shown in Figure (11) the solution is
Fig.(11) One-dimensional heat flow through multiple cylindrical sections
Example(9): A stainless steel k = 14.4 W/(m.K) pipe carries water condensate from a condenser to a pump. The inside-wall temperature is 40 C, and the outside-wall temperature is 38°C. Determine the heat transfer through the pipe wall per unit length of pipe , D2 = 32.39cm, D1 = 29.53 cm Solution:
Example(12): A hot steam pipe having an inside surface temperature of 250◦C has an inside diameter of 8 cm and a wall thickness of 5.5 mm. It is covered with a 9-cm layer of insulation having k =0.5 W/m· ◦C, followed by a 4-cm layer of insulation having k =0.25W/m· ◦C. The outside temperature of the insulation is 20◦C. Calculate the heat lost per meter of length. Assume k =47 W/m· ◦C for the pipe. Solution:
Homework
- Hot air at 80°C is blown over a 2m * 4m flat surface at 30°C. If the average convection heat transfer coefficient is 55 W/m2 · °C, determine the rate of heat transfer from the air to the plate, in kW.
- Consider a 3m high, 5m wide, and 0.3m thick wall whose thermal conductivity is k = 0.9 W/m · °C .On a certain day, the temperatures of the inner and the outer surfaces of the wall are measured to be 16°C and 2°C, respectively. Determine the rate of heat loss through the wall on that day.
- Calculate the rate of heat loss from a furnace wall per unit area. The wall is constructed from an inner layer of 0.5cm thick steel ( k = 40 W/m K) and an outer layer of 10 cm zirconium brick ( k = 2.5 W/m K) as shown in Figure. The inner surface temperature is 900 K and the outside surface temperature is 460 K. What is the temperature at the interface?
- Determine the rate of heat transfer through this section of the wall. And draw the thermal circuit.
𝑘𝐴 = 60 𝑊 ⁄𝑚^ 𝐾, 𝑘𝐵 = 70 𝑊 ⁄𝑚^ 𝐾, 𝑘𝐶 =
20 𝑊 ⁄𝑚 𝐾, 𝑘𝐷 = 40 𝑊 ⁄𝑚 𝐾, 𝑇 1 = 250 ℃, 𝑇 2 =
100 ℃, ∆𝑋𝐴 = 0. 1 𝑚, ∆𝑋𝐵 = ∆𝑋𝐶 = ∆𝑋𝐷 = 0. 095 𝑚,
Wide= 1 m, ℎ𝐴 = 0. 9 𝑚, ℎ𝐵 = ℎ𝐶 = ℎ𝐷 = 0. 3 𝑚.
5. Convection and Conduction in Series for Plane wall:
In the preceding section we treated conduction through composite walls when the surface temperatures on both sides are specified. The more common problem is heat being transferred between two fluids of specified temperatures separated by a wall. In such a situation the surface temperatures are not known, but they can be calculated if the convection heat transfer coefficients on both sides of the wall are known. Newton’s law for convection heat transfer rate Qconv. = h A (TS − T∞) Can be rearranged as
Fig.(12) Schematic for convection resistance at a surface. the thermal resistance for convection heat transfer is
Figure (13) shows a situation in which heat is transferred between two fluids separated by a wall. According to the thermal network shown below.
Fig.(13) The thermal resistance network for heat transfer through a plane wall subjected to convection on both sides
Example(13): A 0.1-m-thick brick wall ( k = 0.7 W/m K) is exposed to a cold wind at 270 K through a convection heat transfer coefficient of 40 W/m2 K. On the other side is calm air at 330 K, with a natural-convection heat transfer coefficient of 10 W/m2 K. Calculate the rate of heat transfer per unit area (i.e., the heat flux). Solution: The three resistances are
the rate of heat transfer per unit area is
Example(14): Consider a 0.8-m-high and 1.5-m-wide double-pane window consisting of two 4-mm-thick layers of glass (k = 0.78 W/m · °C) separated by a 10- mm-wide stagnant air space (k = 0.026 W/m · °C). Determine the steady rate of heat transfer through this double-pane window and the temperature of its inner surface for a day during which the room is maintained at 20°C while the temperature
