THE MAHARAJA SAYAJIRAO UNIVERSITY OF BARODA

FACULTY OF TECHNOLOGY AND ENGINEERING

P.G. MECHANICAL ENGINEERING DEPARTMENT

Submitted by:-
Name :- PATEL PRATIKKUMAR BHARATBHAI
ROLL NO :- 191
EXAM NO:- 267012
SUBJECT:- ADVANCED HEAT AND MASS TRANSFER
Branch :- M.E. Thermal Science

Question :- 1
Prove that Nu = 4.36 for fully developed laminar flow through a tube
with constant heat flux.
In order to estimate the distribution of temperature let us consider the flow of
heat through an elementary ring of thickness dr as shown in Fig. 1. Considering
the radial conduction (neglecting axial conduction) and axial enthalpy transport
in the annular element, we have:
Heat conducted into the annular element,

Qr  k (2 r.dx) t
r
Heat Conducted out of the annular element,

 k 2 (r  dr).dx   t  t dr 
 
dQ
r dr r  r 
Net Heat convected out of the annular element,

dQconv.   2 r dr u c p t dx
r
Considering energy balance on the annular element, we obtain
(Heat conducted in)net = (Heat convected out)net

dQr  dQ  (dQconv.)net
r dr

 k (2 r.dx) t    k 2  r  dr dx   t  t dr    2 r dr u c p t dx

  
r  r  r  x

 2t 

 k (2 r.dx)  k 2  r  dr dx
t  t   dr    2 r dr u c p t dx
r  r
r 2  x
 

 2t 
t  t  
 k (2 r.dx)  k  2 r dx   k  2 r dx  dx   k (2 dr dx) t
r  r   r 2  r
 
 2 
 k  2 dr dx  t dr    2 r dr u c p t dx
 r 2  x


Neglecting second order terms, we get


k  2t  dx dr   r u c t dx dr
 t  r
 r p x
 r 2 
1   r t   u  c p t
r r  r  k x
1   r t   u t
r r  r   x
  2 
 r
Inserting the value of u  umax 1  2  we get,
 R 

 2 
1   r t   1 t u 
1 r
r r  r   x max  R2 
 

  r t   umax t  r  r3 
r  r   x  R3 
 

Let us consider the case of uniform heat flux along the wall, where we can take t
x
as a constant, Integrating above equation, we have

t umax t  r 2 r 4 
r   C
r  x  2 4R2  1
 
 3  C
t  umax t r
 r   1
r  x  2
4R2  r


Integrating again, we have

 2 4 
t  max t  r  r   C ln r  C .
u
.......... ......1
 x  4 16R2  1 2
 

(where C1 and C2 are the constants of integration)

At r  0, t  0
r
At r  R, t  ts
Applying the above boundary conditions, we get

C1 = 0, 1 t 3R 2
C  ts  u
2  x max 16

Substituting the values of C1 and C2 in eqn. (1),we have

umax t  r 2 r 4   1 t 3R 2
t   t  u  .......... .......( 2)
 x  4 16R2   s  r max 16 
   

umax t  3R2 r 2 r 4 
ts  t   
 x  16 4 16R2 
 

For determining the heat transfer coefficient for fully developed pipe flow,
it is imperative to define a characteristic temperature of the fluid. It is the bulk
temperature (tb) or the mixing up temperature of the fluid which is an average
taken so as to yield the total energy carried by the fluid and is defined as the ratio
of flux of enthalpy at a cross-section to the product of the mass flow rate and the
specific heat of the fluid. Thus,

R
  (2 r dr) u c p t
t 0
b R
  (2 r dr) u c p
0

For an incompressible fluid having constant density and specific heat

R
 u r t dr
t 0 . .......... .......... ...(3)
b R
 u r dr
0
The average/mean velocity (u) also known as the bulk mean velocity is
calculated from the following definition:

R
u  1  2 r dr u
 R2 0

R
u  2  u r dr
R2 0

Substituting this value of u in eqn. (3), we get

R
t  2  u t r dr .......... .......... 4
b u R2
0
 2
Substituting the value of u  umax 1 r  and eqn.(2) from eqn. (4), we get
 R2 


2 R  r2   u t 
 3R 2 r 2 r 4 
t  
2u 1  ts  max   
b u R2   
. 
 x  16 4 16R2   rdr
0  R2    

4 R   r 3  umax t 3R2 r 7 3 5 r 5 r 7  
 t r    r   
  s   
 x  16 16 16 R2 16 R4   dr 
R2 0   R2    


R R
 2
r
4  4 umax t  3R2 r 2 7 r 4 5 r 6 8 
 4 t
 
r 
     r 
R2 s
  2 R 2 
0  R2 x  32 16 4 16 6 R2 816 R4 
 0

  2 2  4 umax t  3R 4 7 R 4 5 R 4 R 4 
4 R R
 t s           
2 
R   2 4  R 2 x  32 16 4 16 6 128
  
  2  4 umax t 11 4
4  R
  t s  4    R
2
R  R 2 x 96


t  t s  11 max R 2 t
u
or, b 96  x

The heat transfer coefficient is calculated from the relation

k A  t 
 
Q  r  r  R
h 
A  t s  t  A (t s  tb )
 b

From eqn. (2),we have

 t 
 max t   R  R 
u  
  
 r 
  r R  x  2 4 

 t  umax R t
  
 r 
  r R 4 x

k  max t
u R
h 4 x  24 k  48 k
11 umax R 2 t 11 R 11 D
96  x

Where D is the diameter of the tube.

The Nusselt number is given by

Nu  h D  48 k  D  48  4.364
k 11 D k 11
This shows that the Nusselt number for the fully developed laminar tube
flow is constant and is independent of the Reynolds number and Prandtl number.
Question :- 2
Explain flow across a bank of tubes and also explain its types?
Some typical applications of flow across the tubes are:
1. Air-conditioning application;
2. Water tube boiler (where water flows through the tubes and hot gases flow
across these tubes);
3. Waste heat recovery systems;
4. Different types of heat exchangers.

Tube bank arrangements:

Fig. shows two types of tube bank arrangements frequently used:
(a) “In-line” and
(b) Staggered

Fig.2.1 Types of tube bank arrangements

The geometry of tube bank is characterised by:

i. The tube diameter, D,

ii. The longitudinal pitch SL ,
iii. The transverse pitch ST , and
iv. The diagonal pitch SD (between the centres of the tubes in the diagonal
row), sometimes used for the staggered arrangement:
 The maximum / largest velocity (Umax) of the fluid flowing through the
bank of tubes in the above arrangements in given as:

S
U max inline  S T U
D
.......... .......... (1)
T

And, S
Umax staggered  2(S T
 D)
U .......... .........( 2)
T
Where, U is the velocity of fluid as it approaches the tube bank,

Thukauskas (1972) recommended the following correlation for Nusselts number:

 Pr 
0.25
Nu  h D  C Re  Pr 
m 0.36  
 Pr 
.......... .......( 3)
k  w
Where, Nu  The average Nusselts number,
h  The average heat transfer coefficien t ,
 U max D
Re  ,

Pr The bulk prandtl number , and
Prw The wall prandtl number ,

 In case of a staggered arrangement, while calculating U max , calculate

with both the Eqns.(1) and (2) and adopt the large value so obtained.
 For “gases”, Prandtl number ratio may be ‘ ignored ’ , since it does not
influence the result much.
 All properties, except Prw, are to be evaluated at the temperature of free
stream.
When the number of tube rows in the bank, N > 20, and 0.7< Pr < 500, and
1000 < Remax < 2 × 106, the Eqn. (3) provides very good prediction. However,
this equation can be used even when N < 20, by applying a ‘correction factor’;
the error involved in prediction is about 25% if N=4.

Forms of the Eqn. (3) for different flow regimes:

1. Laminar Flow (10 Re < 100):

(i) In – line tubes:
 Pr 
0.25
Nu  0.8Re Pr 
0.4 0.36  
 Pr 
.......... .........( 4)
 w

(ii) Staggered tube:

 Pr 
0.25
Nu  0.9 Re Pr 
0.4 0.36  
 Pr 
.......... .......... (5)
 w

The Eqns. (4) and (5) have been validated also with range: 50 < Re <1000.

2. Transition regime (1000 < Re < 2 × 105):

(i) In-line tubes, ST/SL > 0.7:

 Pr 
0.25
Nu  0.27 Re0.63 Pr   Pr 
0.36
 w

(ii) Staggered tubes, ST/SL ≥ 2:

 Pr 
0.25
Nu  0.40 Re Pr 
0.6 0.36  
 Pr 
 w
 3. Turbulent regime (Re < 2 × 105)
(i) In-line tubes:
 Pr 
0.25
Nu  0.021Re0.84 Pr   Pr 
0.36
 w

(ii) Staggered tubes:

 Pr 
0.25
Nu  0.8Re Pr 
0.4 0.36  
 Pr 
…for Pr > 1
 w

Nu  0.019 Re0.84 …for Pr = 0.7

Achenbach recommended the following relation for staggered arrangement, with

ST/D = 2 and ST/D = 1.4:

Nu  0.0131Re0.883Pr 0.36

This eqn. is also valid in the range 4.5 × 105 < Re < 7 ×104
When N (number of tube rows) < 20, the correction factor is applied to the
calculated Nusselts number as follows.

Nu aN  C  Nu

Where, Nu aN = Nusselts number for the actual tube bank with N < 20,
and
C = Correction factor (taken from table 2.1)
 Table . 2.1 Correction for C in Eqn. (5) for N < 20.

“Pressure drop” for flow of gases over a tube bank:

Pressure drop (∆p) in Pascal, for flow of gases over a bank of tubes is given as
follows:
 
0.14
2 f `Gmax N  w 
2
p   
  
 b

Where f’ = Friction factor

Gmax = Mass velocity of minimum flow area (= ρ Umax),
N = Number of transverse rows,
Ρ = Density, evaluated at free stream conditions, and
µb = Average free stream viscosity.

F’ (friction factor) is calculated by using the following relation:

 
 S  


0.08 L  

 D 
' 
f  0.044   

Re0.15 … (for in-line tubes)
  S D 
0.431.13 D 
  T  S 
  D 
L 
   

 
 
 
 
f'  0.25 
0.118
1.08
 Re0.16 … (for staggered tubes)
 
 S D 
 T 
   
 D  
 