Introduction to Heat Transfer

Introduction

 Thermodynamics
 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
  There are three modes of energy transfer
 Conduction
 Convection
 Radiation

Conduction Convection Radiation

T1 T1>T2 T2 Ts>T∞ T1
q1’’
Moving fluid T∞ T2
q’’ q2’’
q’’
Solid or Stagnant Fluid
Ts
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
  The basic equation for conductive heat transfer is defined by
Fourier’s law
 
q
  kT
A

Where q = 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)
 Or
 
q"  kT

Where q "= heat flux (W/m )2
 In one dimension equation becomes

dT
q x "  k
dx
 If the temperature distribution is linear
dT
becomes dx
T2  T1 T2  T1

x2  x1 L
T1

T2 qx "
x1 x2

L
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
1m x
24oC

0.5 m
0.5 cm
Solution

 Assumptions:
 Steady-State conditions
 One-dimensional conduction through the window
 Constant thermal conductivity
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
 Forced convection: Flow is caused by external means, fans, wind,
pumps etc.
 Free (natural) convection: Flow is induced by buoyancy forces, which
arise from density differences caused by temperature variation in the
fluid.

Cold

T ρ

T ρ

Hot
  Convection with latent heat exchange: Associated with a phase
change between liquid and vapor (boiling and condensation)

Moist Air

Water
q” Droplets

q”
Vapor Cold
Water Water
Bubbles

Hot plate
  The basic equation for convection heat transfer was
defined by Newton and is usually referred to as the
Newton rate equation:

q
 hT
A

Where q = 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)

 Or 
q "  hT

Where q "= heat flux (W/m )2
 When using above equation, tricky part is
the determination of h. It depends upon the
following factors:
 Typeof 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.
 Note1: The given convective equation is a
definition that simplifies the problem of
convective heat transfer, this is not a law.
 Note 2: Often it is necessary to distinguish
between the local and average convective
heat transfer coefficients (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
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
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
 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)

Eb   Ts4

Where Eb = emissive power (W/m2)

Ts = absolute temperature (K)
σ = Stefan-Boltzman constant 5.676x10-8 W/m2K4 or
0.1714x10-8 Btu/hr ft2 oR4
 For a real surface the emissive power is smaller and
can be calculated using

Eb    Ts4

Where ε = the emissivity of the surface 0    1

 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

0    1
 Radiation can also be reflected or transmitted

G E

G

G
 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


q
qrad "   EbTs  G   Ts  Tsur
A
4 4
 
 It is often convenient to linearize the
radiation rate equation and express it in a
manner similar to convection:


qrad  hr A Ts  Tsur 
Where  
hr   Ts  Tsur T  T s
2 2
sur 
 Note: hr depends strongly on temperature,
while the temperature dependence of the
convection heat transfer coefficient h is
generally weak.
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.
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
One-Dimensional Steady-State Conduction
 Conduction problems may involve multiple directions and time-
dependent conditions

 Inherently complex – Difficult to determine temperature

distributions

 One-dimensional steady-state models can represent accurately

numerous engineering systems
 The Plane Wall
Consider a simple case of one-
dimensional conduction in a plane wall, Cold fluid
separating two fluids of different T ,1 T , 2 , h2
temperature, without energy generation
 Temperature is a function of x Ts ,1
 Heat is transferred in the x-direction
Must consider Ts ,2
 Convection from hot fluid to wall T ,1 , h1 qx
T , 2
 Conduction through wall
 Convection from wall to cold fluid

 Begin by determining temperature Hot fluid x=0 x=L

distribution within the wall
x
 Temperature Distribution
 Heat diffusion equation (eq. 2.4) in the x-direction for steady-state
conditions, with no energy generation:

d  dT 
k 0  qx is constant
dx  dx 

• Boundary Conditions: T (0)  Ts ,1 , T ( L)  Ts ,2

• Temperature profile, assuming constant k:

x
T ( x)  (Ts,2  Ts,1 )  Ts,1
L
 Temperature varies linearly with x
Thermal Resistance
Based on the previous solution, the conduction hear transfer rate can be
calculated:

qx  kA
dT kA
 Ts,1  Ts,2   Ts,1  Ts,2 
dx L L / kA
Similarly for heat convection, Newton’s law of cooling applies:
(TS  T )
qx  hA(TS  T ) 
1 / hA
And for radiation heat transfer:
(Ts  Tsur )
qrad  hr A(Ts  Tsur ) 
1 / hr A
 Recall electric circuit theory - Ohm’s law for electrical resistance:
Potential Differenc e
Electric current 
Resistance
 Thermal Resistance
• We can use this electrical analogy to represent heat transfer problems
using the concept of a thermal circuit (equivalent to an electrical circuit).

Overall Driving Force Toverall

q 
Resistance R 
 The temperature difference is the “potential” or driving force for the heat
flow and the combinations of thermal conductivity, convection coefficient,
thickness and area of material act as a resistance to this flow:

L 1 1
Rt ,cond  , Rt ,conv  , Rt ,rad 
kA hA hr A
 Thermal Resistance
Cold fluid for Plane
Wall
T ,1 T , 2 , h2

Ts ,1
T,1  Ts,1 Ts,1  Ts,2 Ts,2  T,2
qx   
Ts ,2 1 / h1 A L / kA 1 / h2 A

In terms of overall
T ,1 , h1 qx
T , 2 temperature difference:

T,1  T,2
Hot fluid x=0 x=L qx 
Rtot
x
1 L 1
Rtot   
h1 A kA h2 A
 Composite Walls
 Express the following
geometry in terms of a
an equivalent thermal
circuit.
 Composite Walls
What is the heat transfer rate for this system?

q x  UA T
T

Alternatively 1
Rtot  Rt  
q UA

where U is the overall heat transfer coefficient and T the overall

temperature difference.
1 1
U 
Rtot A [(1 / h1 )  ( LA / k A )  ( LB / k B )  ( LC / kC )  (1 / h4 )]
 Composite Walls

(a) Surfaces normal to the x-

direction are isothermal

 For resistances in series:

Rtot=R1+R2+…+Rn
 For resistances in parallel:
(1/Rtot)=1/R1+1/R2+…+1/Rn
(b) Surfaces parallel to x-
direction are adiabatic