1. Introduction to Heat Transfer
Chapter 1
Introduction to Heat Transfer

2. 1.0 Introduction Thermodynamics Heat Transfer
study the effects of adding or removing a quantity of heat (or energy) to or from a system.
Heat Transfer
study the rate at which the heat (or energy) is transferred.
When two systems are in contact and are at different temperatures, they will exchange thermal energy
Energy travels from the system of high temperature to the low one
The rate of exchange is proportional to the temperature difference

3. There are three modes of energy transfer
Conduction
Convection
Radiation
Conduction
T1>T2
Convection
Ts>T∞
Radiation T1 T2 T1
q1’’
Moving fluid T∞ T2
q’’
q2’’
q’’
Solid or Stagnant Fluid Ts

4. 1.1 Conduction
When there exists a temperature gradient within a body, heat energy will flow from the region of high temperature to the region of low temperature
This mode of heat transfer occurs at the molecular level via two processes:
The energy from one molecule is transferred to an adjacent molecule or
The energy is transferred by free electrons (Mostly encountered in pure metallic solids).
No bulk motion

5. The basic equation for conductive heat transfer is defined by Fourier’s law
Where
= heat transfer rate (W or Btu/hr)
A = area normal to direction of heat flux (ft2, m2)
k = thermal conductivity, property of material,
(Wm-1K-1, Btu hr-1 ft-1 oF-1)
Where
= heat flux (W/m2)

6. In one dimension equation 1.1 becomes
If the temperature distribution is linear becomes T1 T2 x1 x2 L

7. Example 1.1
Calculate the rate of heat transfer through a pane of window glass (k=0.78W/m K) 1 m high, 0.5 cm thick, and 0.5 m wide, if the outer surface temperature is 24oC and the inner surface temperature is 24.5oC. y
24.5oC
1 m x
24oC
0.5 m
0.5 cm

8. Solution Assumptions: Steady-State conditions
One-dimensional conduction through the window
Constant thermal conductivity

9. Solution

10. 1.2 Convection
Heat energy transfers between a solid and a fluid when there is a temperature difference between the fluid and the solid
This mode of heat transfer occurs both at the molecular level and macroscopic level:
The energy from one molecule is transferred to an adjacent molecule
The energy is transferred by the bulk or macroscopic motion of the fluid

11. Forced convection: Flow is caused by external means, fans, wind, pumps etc.
                                                      

12. Free (natural) convection:
Free (natural) convection: Flow is induced by buoyancy forces, which arise from density differences caused by temperature variation in the fluid. ρ T
Hot
Cold

13. Convection with latent heat exchange: Associated with a phase change between liquid and vapor (boiling and condensation)
Moist Air
Water
Droplets q” q”
Cold
Water
Vapor
Bubbles
Water
Hot plate

14. The basic equation for convection heat transfer was defined by Newton and is usually referred to as the Newton rate equation: Or
[Equation 1.2]
Where
= heat transfer rate (W or Btu/hr)
A = area normal to direction of heat flux (ft2, m2)
h =convection heat transfer coefficient,
(Wm-2K-1, Btu hr-1 ft-2 oF-1)
Where
= heat flux (W/m2)

15. When using equation 1.2, the key to solving convective heat transfer problems is the determination of h. We will devote several lectures on this problem later on. Basically h depends upon the following factors:
Type of convection: Free (natural) or
Forced
Geometry
Type of flow: Laminar: heat transfer is through conduction between streamlines.
Turbulent: heat transfer due to conduction and macroscopic movement of fluid in the direction of the heat transfer. Therefore the convective heat transfer coefficient is usually higher than that of laminar flow.

16. Note1: Equation 1.2 is a definition that simplifies the problem of convective heat transfer, this is not a law.
Note 2: We often have to distinguish between the local and average convective heat transfer coefficient(hx, h)
Note 3: The convective heat transfer coefficient is not an inherent property of the material. But it will depend on the density, viscosity , velocity and for free convection on the thermal coefficient of expansion of the fluid

17. Example 1.2
Calculate the rate of heat transfer by natural convection between a shed roof of area 20m x 20m and ambient air, if the roof surface temperature is 27oC the air temperature is -3oC and the average convection heat transfer coefficient is 10W/m2K.
20 m
Troof = 20 oC
20 m

18. Solution
Assumptions:
Steady-State conditions

19. 1.3 Radiation
Energy emitted by matter that is at a finite temperature.
This mode of heat transfer is attributed to changes in the atom configuration.
Does not require the presence of a medium
Most efficiently done in a vacuum

20. The basic equation for radiation heat transfer comes from Stefan-Boltzman law, which represents the upper limit to the emissive power (emissive power of a blackbody)
Where Eb = emissive power (W/m2)
Ts = absolute temperature (K)
σ = Stefan-Boltzman constant x10-8 W/m2K4 or
0.1714x10-8 Btu/hr ft2 oR4

21. For a real surface the emissive power is smaller and can be calculated using
Radiation may also be incident to the surface. G the irradiation, designates the rate of all radiation incident on a unit area of surface.
A portion or all of the incident radiation may be absorbed based on the surface radiative property termed absorptivity α
[Equation 1.3]
Where ε = the emissivity of the surface

22. Radiation can also be reflected or transmittedG aG
 G
 G E J

23. Special Case
Radiation exchange between small surface s at temperature Ts and large enclosing surface sur at temperature Tsur.
s is a gray surface (α = ε)
Here the net rate of radiation heat transfer from the surface can be
Expressed as

24. It is often convenient to linearize the radiation rate equation and express it in a manner similar to convection:
Note: hr depends strongly on temperature, while the temperature dependence of the convection heat transfer coefficient h is generally weak.
Where

25. Example 1.3
A long, cylindrical electrically heated rod, 2 cm in diameter, is installed in a vacuum furnace as shown below. The surface of the heating rod is maintained at 1000 K, while the interior walls of the furnace are black and are at 800 K. Calculate the net rate at which heat is lost from the rod per unit length and the radiation heat transfer coefficient.

26. Solution Assumptions: Steady-State conditions
Radiation exchange between the electrically heated rod and the furnace is between a small surface and in much larger enclosure
The surface emissivity and absorptivity are equal

27. Solution

28. 1.4 Conservation of Energy for a Control Volume (C.V.)
Application of the first law of thermodynamic.
Need to define a control volume bounded by a control surface through which energy and matter pass
Need to define an appropriate time basis

29. Accumulation of Energy = Energy In - Energy Out + Energy Generated
Energy in or out due to radiation
Energy in or out due to conduction
Generation of Energy
Accumulation of Energy
Energy in or out due to convection
Accumulation of Energy = Energy In - Energy Out + Energy Generated

30. General Form of the Energy Conservation Equation
Where
Rate of change of the energy stored in the C.V.
Rate at which energy enters the C.V.
Rate at which energy is generated in the C.V.
Rate at which energy leaves the C.V.

31. Example 1.4
A fluid of density ρ and specific heat Cp flows in a circular pipe. Derive an expression for the temperature of the fluid as a function of position, given that the pipe inner wall temperature Tw is constant and uniform.
Data: temperature of the fluid at the inlet T0 (T0 < Tw)
velocity of fluid v0 is constant
velocity profile is flat (plug shape)
convection heat transfer coefficient h is constant Tw T0 v0

32. Solution Assumptions: Steady-State conditions No radiation effect
The fluid is well-mixed (highly turbulent), so the temperature is uniform in the radial direction
Thermal conduction of heat along the axis is small relative to convection

33. Solution define a control volume z z+Δz Δz Convection
Rate of mass in (Tz)
Rate of mass out (Tz+Δz) z
z+Δz Δz

34. Solution
Accumulation of Energy = Energy In - Energy Out + Energy Generated
Energy In - Energy Out = 0
Energy In =
Energy Out =
Divide by