See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/329443308
Fundamentals of Heat and Mass Transfer, 8th Edition (2017) - Book review
Technical Report · December 2018
CITATIONS
READS
0
46,113
1 author:
Gur Mittelman
Afeka Tel-Aviv Academic College of Engineering
53 PUBLICATIONS 1,250 CITATIONS
SEE PROFILE
All content following this page was uploaded by Gur Mittelman on 17 April 2020.
The user has requested enhancement of the downloaded file.
Fundamentals of Heat and Mass Transfer, 8th Edition
(2017)* by Bergman, Lavine, Incropera and DeWitt – Review
Gur Mittelman, gur.mittelman@gmail.com
* The review is referred to the 7th edition (2011), but valid to the 8th edition as well.
Synopsis
The fundamental principles of heat transfer were developed for centuries now. Thus, if
we ever get a chance to challenge the very basic laws of this discipline, well, it could be
quite exciting. In the excellent textbook by Bergman et al., this material is delivered with
great detail and patience, while uncompromising the degree of clarity. However, heat
transfer is still a very tricky field, and the deep observations provided in this book give an
opportunity to think it over again. The current review comes across some of the
fundamental concepts, not just in the current textbook, but in the field as general (see for
example notes 2, 3 and 5). The following annotations are definitely not recommended for
the faint-hearted readers.
Review
1. Extended surfaces. Section 3.6.
The fin surface area notation Af is used in a dual meaning. In equation (3.78) and Table
3.5 it is considered as the entire fin surface area, including the fin tip. However, in
equations (3.91-3.92), for the adiabatic tip fin it is only considered as P∙L i.e. the
surface area excluding the tip (As). A possible resolution could be defining Af as “the
surface area of the fin exposed to convection”.
Likewise, the integration in equation (3.78) shall take place over the elemental area
dAf . A possible formulation of could be:
qf 
 h(T  T )dA

Af
f

 h(T  T )dA

s
 h L A c (TL  T )
As
In Table 3.5 the rectangular fin surface area is given as Af  2wL  wt but in equation
(3.104) the overall fin array surface area is given as At  NAf  Ab . Thus, if Af
represents the entire single fin surface area (see above), and A b is the entire array
base area, then the area w  t is considered twice, once in NAf and once in A b .
A possible resolution of could be the following formulation:
rectangular
A t  N  P  L  A b  N  (2w  2t)  L  A b
All rights reserved © 2018 Gur Mittelman
1
2. Viscous dissipation. Section 6.4.
It is highly instructive to notice how viscous dissipation is considered in the energy
equation, while referring to both integral and differential forms.
The energy equation in a differential form is given in the appendix by equation (E.4):
c p (u
T
T
 2T  2T
 v )  k( 2  2 )  
x
y
x
y
The energy equation in an integral form is given in chapter 1 as equation (1.12d):
1
1
m(u t  p  V 2  gz)in  m(u t  p  V 2  gz)out  Q  W  0
2
2
where the work term includes both shaft and viscous (shear) work e.g. W  Ws  Wf .
Now, consider a fully developed, incompressible flow in an adiabatic, horizontal pipe.
In the differential form, the fluid heating due to viscous dissipation is expressed
directly by the  term e.g. problem 8.10:
cp u
T
du
 ( ) 2
x
dr
For the integral form, if we select the control volume to be the pipe surface, we have
no viscous work transfer, Wf  0 because the velocity at the pipe (solid) surface is zero.
The integral form then becomes:
m(u t  p 
1 2
1
V  gz )in  m(u t  p  V 2  gz )out  Q  Ws  Wf  0
2
2
or
u t,out  u t,in  c v (Tout  Tin ) 
pin  pout

Thus, the fluid is heated due to the pressure drop, which is related directly to the wall
4Lw
shear stress (friction), p 
from the momentum balance. Hence, fluid heating
D
is related to friction despite the fact that viscous work transfer is obscured.
This is rather tricky.
3. Boundary layer similarity. Section 6.5 p.398.
For flat plate parallel to the incoming flow, we get the following hydrodynamic and
energy equations:
u *
u *
1  2u *
u*
 v*

x *
y * Re L y *2
All rights reserved © 2018 Gur Mittelman
2
u*
T *
T *
1  2T *
 v*

x *
y * Re L  Pr y *2
which are argued to be similar for the velocity and temperature fields, leading to
identical nondimensional solutions. However, the first convection term on the left side
T *
T *
does not seem similar, as we have u *
and not T *
. In the hydrodynamic
x *
x *
equation, this convection term is non-linear.
4. Local Nu number. Section 6.5.2 p. 401.
The heat transfer coefficient, h (or hx) in equation (6.5) is local. Accordingly, the
definition for the local Nu should be:
Nu x 
hx  x

kf
k f T *

x
L y * y*0
kf
 x*
T *
y * y*0
This is different than what is given in equation (6.48).
5. Normalized boundary layer equations. Section 6.5.2 p. 401.
The result of the governing equations normalization analysis is [equation (6.49)]:
Nu x  f (x*, ReL , Pr)
which implies that the local Nu number actually depends on 3 groups.
However, performing a dimensional analysis for the flat plate in steady, laminar, 2D
parallel flow, starting from (see for instance, the Lienhard & Leinhard textbook):
h x  f (x, u  , , , k f , c p )
we obtain 7-4=3 i.e. the local Nu number depends only on 2  groups:
 Nu x  f (Rex , Pr)
which is in agreement with analytical solutions e.g. Blasius/Pohlhausen.
Thus, the result obtained by normalizing the governing equations is probably false,
providing one extra  group.
A possible explanation for the discrepancy could be that the parameter L, which is
introduced in the normalization (e.g. x*=x/L), does not really affect the local solution
hx.
k
The local heat transfer coefficient scales as h x  f where t (x) is the thermal
t (x)
boundary layer thickness which develops from the leading edge (or elsewhere) further
downstream. The thermal boundary layer problem is similar to initial value problems,
where the solution is affected only from the past, but not by the future. Thus, any local
All rights reserved © 2018 Gur Mittelman
3
solution can’t be dependent on information available downstream such as the plate
length, L.
6. The Reynolds Analogy. Section 6.7.3 p. 416.
If my notes 3-5 above are correct, than equation (6.66) is probably not valid.
7. Enthalpy flow and specific heat. Section 8.2.1 p. 525.
The simplified steady-flow thermal energy equation (1.12e) is given as follows:
q  mcp (Tout  Tin )
where the right side of the equation represents the net rate of enthalpy outflow (or
advection).
Paragraph 1 in page 525 and equation (8.24) suggests that the enthalpy per unit mass
can be represented as
i  cpT .
However, the specific heat is a quantity representing a net change (differential) in
enthalpy, rather than absolute i.e.
cp  (
i
)p
T
and for incompressible liquid or ideal gas,
di  c p dT
Thus, even for a constant specific heat we get,
T
T
To
To
i   c p dT c p  dT c p (T  To )
where To is some reference temperature.
Only if we opt to evaluate the net enthalpy change between two cross sections, say 1
and 2, then, the reference temperature value could be ignored as
i 2  i1  cp (T2  To )  cp (T1  To )  c p (T2  T1 ) .
Note that evaluating the net change in enthalpy is also useful for the derivation of the
energy equation at the differential form [equation (E.4)] as indicated by the advection
terms framed in blue in Figure 1 below.
All rights reserved © 2018 Gur Mittelman
4
Figure 1. Energy flows in differential form (A. Bejan, Convection Heat Transfer, J.P.
Holman, Heat Transfer). The advection is highlighted blue.
Finally, the separation of inflow and outflow enthalpy streams while using the specific
heat is also apparent in the solution for problem 7.28, despite the fact that the
advection formulation is first presented in chapter 8.
8. Axial conduction in internal flow. Section 8.3.1 p.529.
Equations (8.34-8.35, and 8.47) are correct as long as axial conduction can be
neglected.
Not that this valid only when the Péclet number is large,
PeD  ReD Pr 
umD

1.
9. Velocity scale in free convection on a vertical surface. Section 9.3 p. 599.
Note that owing to the fact that the motion is driven by buoyancy, another way to
derive the velocity scale, u0, is by directly comparing between the first inertia term
and the buoyancy term in equation (9.7):
u
u
 2u
u v
 g(T  T )   2
x
y
y
u
u
x
u 02
L
g(T  T )
g(Ts  T )
All rights reserved © 2018 Gur Mittelman
5
Solution Manual
1. Solution manual problem 4.1
After equation (7), the boundary condition at y=0 should be,
(x, 0)  C5ex  C6ex  (...) .
The last equation should be,
x  L (L, y)  C5 (e L  C6e L )C8 sin y  0 .
2. Solution manual problem 5.95
The manual suggests that the temperature at the interface (x=L) remains constant,
even after thermal effects penetrate the bodies from two sides of the interface.
However, this assumption is not well established due to the following analysis.
Consider a general case of two surfaces, each at a different initial temperature but of
the same material, that are suddenly pressed together (see below). Using the method
of separation of variables, the temperature distribution is obtained as following:
a 2( T1  T2 )  1
na 
T ( x ,t )  T2  ( T1  T2 ) 
sin
e

L

L
n 1 n
n 2  2 t
L2
cos
nx
L
T
T1
T2
a
L
x
Which suggests that the interface temperature, T(x=a) is in general time dependent.
3. Solution manual problem 5.108
At Fo=1, the temperature should be 150oC, instead of 250oC.
All rights reserved © 2018 Gur Mittelman
6
4. Solution manual problem 12.54
The reflectivity calculation is presented with the notation of a blackbody radiation
function (or fraction), F, while in this case the incident radiation is not coming from a
black (or gray) body.
Maybe the following formulation could be clearer:
6
 q


1
q
"
6
"



 (1  

1
q
"
3
)q d
"


0
 (1  

200
6
0.5
200
) q d   (1   ) q" d
"

1
3
q
"
All rights reserved © 2018 Gur Mittelman
7
View publication stats

1
(1 200  2  0.5  200  3)  0.7
1000