Chapter 2 Introduction to Heat Transfer
2.1 Basic Concepts
2.1.1 Conduction, convection, and radiation
Heat transfer is the transfer of heat due to a temperature difference with different
mechanisms: conduction, convection, and radiation. Conduction refers to heat
transfer that occurs across a stationary solid or fluid in which a temperature gradient
exists. Convection refers to the heat transfer that occurs across a moving fluid in
which a temperature gradient exists. Radiation refers to the heat transfer between
two surfaces at different temperatures separated by a medium transparent to the
electromagnetic waves emitted by the surfaces.
1
2.1.2 Fourier’s law of conduction
2.1.2.1 One-dimensional
Consider the conduction of heat through a slab of thickness L, as shown in
Fig. 2.1-2. The lower and upper surfaces are kept at a constant temperature T1 and
T2, respectively. A steady-state temperature profile T(y) is established in the slab.
Consider two surface in slab separated with a infinitesimal distance dy, as shown
in Fig. 2.1-2. due to temperature gradient generated in the slab, heat flow from the
surface y to the surface y+dy. A heat flux is defined as the amount of heat transferred
per unit area per unit time, and can be expressed as
dT
q y  k
dy
[2.1-1]
where k is the thermal conductivity of the medium. This equation is Fourier’s law
of conduction for one-dimensional heat conduction in the y-direction. The mks units of
the heat flux and the thermal conductivity are W/m2 and Wm-1K-1, respectively.
2
The thermal conductivities of some common
materials are given in Fig. 2.1-3 and Table 2.1-1.
2.1.2.2 Three-dimensional
For heat transfer in a three-dimensional medium, the Fourier’s law can be
expressed for each of the three coordinate directions
q x  k
dT
[2.1-2]
dx
dT
q y  k
[2.1-3]
dy
dT
q z  k
[2.1-4]
dz
And can be expressed in a three-dimensional form of Fourier’s law of conduction.
q  kT
2.1.2.3 The Thermal Diffusivity
The thermal diffusivity, α , is defined as
k

 Cv
Where  and Cv are the density and specific heat of the material, respectively.
3
2.1.3 Thermal boundary layer
Consider a fluid of uniform temperature T∞ approaching a flat plate of constant
temperature Ts in the direction parallel to the plate. At the solid/liquid interface the
fluid temperature is Ts since the local fluid particles achieve thermal equilibrium at
the interface. The fluid temperature T in the region near the plate is affected by the
plate, varying from Ts at the surface to T∞ in the main stream. This region is called
the thermal boundary layer.
4
Definition of thermal thickness: The thickness of thermal boundary layer δT
is taken as the distance from the plate surface at which the dimensionless
temperature (T-TS)/(T∞-TS) reaches 0.99. In practice it is usually specified
that T=T∞ and T / y  0 at y=δT.
The effect of conduction is significant only in the boundary layer. Beyond it
the temperature is uniform and the effect of conduction is no longer significant.
A fluid of uniform temperature T entering a circular tube of inner diameter D and
uniform wall temperature TS, as illustrated in Fig. 2.1-5. A thermal boundary layer
begins to develop at the entrance , gradually expanding until the layers from
oppositive sides approach the centerline. This occurs at
 v D   C  
z
 0.05    v   0.05Re D Pr
D
   k 
where
v D
Re D  

Inertial force
Viscous force
(Reynolds number)


Cv 

Pr  

k


Cv
(Prandtl number)
[2.1-7]
Viscous diffusivity
[2.1-8]
thermal diffusivity
5
Define average temperature
Tav
C TvdA


 C vdA
A
A
v
[2.1-10]
v
The thermally fully developed temperature profile in a tube is one with a
dimensionless temperature (Ts-T)/(Ts-Tav) independent of the axial position,
that is
  TS  T 

0
z  TS  Tav 
[2.1-12]
2.1.4 Heat transfer coefficient
Consider the thermal boundary layer. At the solid/liquid interface heat transfer
occurs only by conduction since there is no fluid motion. Therefore, the heat flux
across the solid/liquid interface is
T
q y y 0  k
[2.1-13]
y 0
y
This equation cannot be used to calculate the heat flux when the temperature
gradient is an unknown. A convenient way to avoid this program is to introduce
a heat transfer coefficient, defined as follows:
q y y 0
k  T y  y 0
[2.1-14]
h

Ts  T 
Ts  T 
6
The absolute values are used to keep h always positive. From Eq. [2.1-14]
qy
y 0
 h (TS  T )
[2.1-15]
The equation is called Newton’s law of cooling. For fluid flow through a tube of
an inner radius R and wall temperature TS, a similar equation can be used:
h
qr r  R
TS  Tav 

k  T r 
TS  Tav 
r R
[2.1-16]
Where Tav is the average fluid temperature over the cross-sectional area pR2.
Consider the thermally fully developed region shown in Fig. 2.1-5. In the case of a
constant heat flux, the heat transfer coefficient h is constant in the thermally fully
developed region. From Eq. [2.1-16] we see that (TS-Tav) is also constant.
From this and Eq.[2.1-12], we have
Ts T Tav


z
z
z
(constant
qr
r R
)
[2.1-17]
Since TS and Tav are independent of r, T / zis also independent of r, Let us
consider the case of a constant wall temperature TS. Eq. [2.1-12] can be expanded
and solved for T / z to give
7
T  T Tav
T
( S
)
z
TS  Tav z
(constant TS)
[2.1-18]
Since T is dependent on r, T / z is also dependent on r.
2.2 Overall energy-balance equation
2.2.1 Derivation
Consider a control volume Ω bounded by control surface A through which a
moving fluid is flowing. As defined in previous chapter, the control surface is
composed by Ain , Aout, and Awall. Consider an infinitesimal area dA in vector form
is ndA, the inward and outward heat transfer rate through area dA is -q． ndA
and q． ndA, respectively.
The energy conservation law (first law of thermodynamics) written for an open
system under unsteady-state condition is
 rateof energy
 rateof energyin 
 rateof energyout 

(1)  
(2)  
(3)
 accumulation 
 by mass inflow 
 by mass outflow 
 rateof otherheat 
 rateof work 
 rateof heat 
[2.2-3]






 transfer ot system (4)   done by system (5)   generation (6)
 fromsurroundings 
 on surroundings 
 in system 






8
Term 1: Rate of energy accumulation
The thermal, kinetic, and potential energy per unit mass of the fluid are
CvT, v2/2, and ψ, respectively, where Cv, T, v are the specific heat, temperature,
and velocity of the fluid. The total energy per unit mass of the fluid
1 2
et  CvT  v  
2
[2.2-4]
The total energy in the differential volume element dΩis dEt=ρetdΩ. dm =ρdΩ
This can be integrated over Ω to obtain the total energy in the control volume Et

et d (integral)

dEt 
  et d
And the rate of energy change in Ω is
dt
t 
Et (overall) =
Terms 2 & 3: Rate of energy in by mass inflow
The inward energy flow rate is energy per unit mass, et, times inward mass flow
rate through dA and can be expressed as
  et v  ndA
A
Since v=0 at the wall, above term can be expressed as
 (  et v  ndA)in  (  et v  ndA) out  (0) wall
A
A
9
Term 4: Rate of heat transfer
The heat transfer by conduction dA is
Q    q  ndA
A
Term 5: The rate of work done by the fluid in the C.V. on the surroundings, includes:
(1) The rate of shaft work done
The rate of shaft work done by the fluid in the C.V. on the surroundings,
that is, through a turbine or compressor, is Ws
(2) The rate of pressure work done
To leave the C.V. through dA, the fluid has to work against the pressure of
the surrounding fluid. Since the pressure force is pndA, the rate of pressure
work required is dWp= pv‧ndA. Therefore, the rate of pressure work the fluid
has to do to go through the C.V. is
W p   pv  ndA
A
(3) The rate of viscous work done
To overcome the viscous force t‧ndA, the rate of viscous work required is
dWv= (t‧n)‧vdA. The rate of viscous work the fluid has to do is
Wv   (t  v)  ndA
A
10
Term 6: Heat generation rate
The heat generation rate per unit volume, such as that due to Joule heating,
phase transformation, or chemical reaction. The rate of heat generation in the
differential volume element dΩ is dS=s dΩ. The heat generation in the control
volume is S,
S   sd

Substituting the integral form of terms(1) through (6) into Eq.[2.2.3]

et d


t
=
  et v  ndA   q  ndA
A
A
[2.2-5]
  pv  ndA   (t  v)  ndA W s
A
A
In most problem, including those in materials processing, the kinetic and
potential energies are neglegible as compared to the thermal energy. Furthermore,
the pressure, viscous and shaft work are usually negligible or even absent. As such,
Eq.[2.2-5] reduces to

CvTd    CvTv  ndA   q  ndA   sd

A
A

t 
[2.2-6]
11
According to [2.2-3], Eq. [2.2-6] can be written as
ET
 (  CvTvdA)in  (  CvTvdA) out  Q  S
A
A
t
[2.2-7]
Where ET is the thermal energy in the control volume, substituting Eq. [2.1-11]
(definition of Tav) into this equation and assuming constant Cv, we obtain
ET
 ( CvTav vav A)in  ( CvTav vav A) out  Q  S
t
 (m CvTav ) in  (m CvTav ) out  Q  S
[2.2-8]
2.2.2 Bernoulli’s Equation
Consider the steady-state isothermal flow of an inviscid incompressible fluid
without heat generation, heat conduction, shaft work, and viscous work. Substituting
Eq. [2.2-4] into [2.2-5] and assuming uniform properties over the cross-sectional area
A, we have
12
0     (et 
A
p

) v  ndA




1
p
1
p
  (Cv T  v 2    )vA    (Cv T  v 2    )vA 
2

2


1

2
[2.2-9]
Since T1=T2 and (vA)1=(vA)2, Eq.[2.2-9] reduces to
1 2
p1 1 2
p2
v1  1   v 2  2 
2
 2

If the z direction is taken vertically upward,  =gz, where g is the gravitational
acceleration. As such, Eq/.[2.2-10], on multiplying by , becomes
1
1
v12  gz1  p1  v 2 2  gz 2  p2
2
2
or simply
1 2
v  gz  p  constant
2
Which is the Bernoulli equation.
13
Example 2.2.2 Conduction through cylindrical composite wall
14
Example 2.2.3 Heat transfer in fluid flow through a pipe
Based on the C.V.
selected in the figure
mCvTav  mCv (Tav  dTav )  dQ  0  dQ  mCv dTav  dQ
Based on the definition of overall
Heat transfer coefficient
dQ  2p RdzU (T  Tav )
(A)
B.C2: T  TL at z=L  ln
mCv dTav  2p RdzU (T  Tav )
d (T  Tav )
2p RU

dz 
(T  Tav )
mCv
2p RU
ln(T  Tav )  
z C
mCv
B.C.1: Tav  Tz at z=0  C=ln(T  T0 )
ln
(T  Tav )
2p RU

z
(T  T0 )
mCv
T  TL  (T  T0 ) e
-
(T  TL )
2p RU

L
(T  T0 )
mCv
2p RU
L
mCv
(T  T0 )  (TL  T0 )  (T  T0 ) e
TL  T0  (T  T0 ) (1-e
-
2p RU
L
mCv
-
2p RU
L
mCv
)
15
Example 2.2.4 Counterflow heat exchanger
Given:
Hot stream Th1, Th2, mh
cold stream inlet(Tc2), outlet(Tc1),
and mass flow rate mc
Overall heat transfer coefficient U, turbulent
Find Qe (steady state heat exchange rate Qe
in terms of U, Th and Tc).
 For hot stream
0  mhCvTh1  mhCvTh2  (Qe )
From 
view 
 For cold stream 0  mcCvTc2  mcCvTc1  Qe

of
Th2  Th1
1
overall  Therefore

[2.2-47]

m
C
Q
h v
e

and
Tc2  Tc1
1

[2.2-48]
mcCv
Qe
16
From view of C.V.
Because we want to express Qe in terms of U,
therefore, considering the C.V. in the inner pipe.
0  mhCvTh  mhCv (Th  dTh )  [(2p RdzU )(Th  Tc )]
energy in
energy out
dTh
2p RUdz

Th  Tc
mhCv
heat loss from the wall
[2.2-50]
Similarly, for the C.V. in the outer pipe
0  mhCv (Tc  dTc )  mcCvTc  (2p RdzU )(Th  Tc )
energy in
and
energy out
dTc
2p RUdz

Th  Tc
mcCv
gain heat the wall
[2.2-52]
Subtracting Eq.[2.2-52] from Eq. [2.2-50], we have
d (Th  Tc )
1
1
 2p RU (
)dz [2.2-53]
Th  Tc
mhCv mcCv
17
Integrating from z=0 to z=, and substituting [2.2-47] and [2.2-48] to [2.2-53]

Th 2 Tc 2
Th1 Tc1
or
d ln(Th  Tc ) 
L
2p RU
(Th 2  Th1  Tc 2  Tc1 )  dz
0
Qe
(2.2-54)


 (T  T )  (T  T ) 
h1
c1
 (2.2  55)
Qe  (2p RL)U  h 2 c 2
(T  T )


ln h 2 c 2


(Th1  Tc1 )
18
Example 2.2.5 Heat transfer in laminar flow over a flat plate
Given: Steady state, constant physical properties,
no heat generation
Find: dT and heat transfer coefficient
Approach:
1. Construct C.V.
2. Find the mass flow rate into the C.V. from surface 4
d dT
m4  [   vz dy ]dz
dz 0
3. Consider the energy balance

dT
0
dT
CvT vz dy z   CvT vz dy z z  qy
0
y 0
z  m4CvT  0
4. Substituting m4 into Eq. [2.2-57], and
according to Fourier’s law of conduction
qy
y 0
T
 k
y
y 0
19
we have

dT
0
CvT  vz dy z  
dT
0
T
CvT  vz dy z z k
y
dividing by Cv
d dT
T
 [  Tvz dy]dz 
dz 0
y
T

y
y 0
d dT
z  ( CvT vz dy )z  0
dz 0
y 0
d dT
z  (  Tvz dy )dz  0
dz 0
y 0
d dT
  vz (T  T )dy
dz 0
5. Assume
and
T  Ts 3 y
1 y 3
= ( ) - ( ) [2.2-61]
T  Ts 2 d T
2 dT
vz 3 y
1 y
= ( ) - ( )3 [2.2-64]
v 2 d
2 d
6. Assume
Please see the derivation in other pages
1

Cv 
dT
3
dT  d , we have
 Pr , [2.2-65] where Pr=
d
k
20
7. By the definition of heat transfer coefficient
k
h
T
y
y 0
Ts  T
we have h =
[2.2-67], substituting Eq[2.2-61] into [2.2-67]
3 k
[2.2  68]
2 dT
From Eq.[1.4-62] we have
z
d
=
4.64
[2.2  69]
Rez
From Eqs. [2.2-65], [2.2-68], and [2.2-69]
We have
1
1
1
hz
3 z 3 d z 3 13 1
=Nu z 

 Pr
Re z 2  0.323Pr 3 Re z 2
k
2 dT 2 dT d 2
4.64
[2.2-65]
[2.2-69]
21