HEAT TRANSFER

Category: Professional Core Course

(Theory and Practice)
Course code: 21ME54
Faculty: Dr P R Venkatesh, Associate Professor, Mech Dept, RVCE, Bengaluru
Room No 2, Ground Floor, Department of Mechanical Engineering, RVCE
Mob: 9620117755 Email: venkateshpr@rvce.edu.in

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Unit III
Forced Convection:
Dimensional analysis, Physical significance of Reynolds, Prandtl, Nusselt and Stanton
numbers. External forced convection: Dimensional analysis, flow over flat plates, and flow
across cylinders, Spheres; Internal forced convection: Laminar and turbulent flow in tubes
with entry length concepts. Problems
Natural Convection:
Physical mechanism of convection, classification of fluid flow, concepts of velocity
boundary layer; General expressions for drag coefficient and drag force; thermal boundary
layer, general expression for local heat transfer coefficient, Average heat transfer coefficient
Physical mechanism of natural convection, dimensional analysis, natural convection over
surfaces - Vertical plates, cylinders, horizontal and inclined plates. Numerical problems
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
Forced Convection

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

• In convection heat transfer, there is a flow of fluid associated with heat
transfer and the energy transfer is mainly due to bulk motion of the fluid.
When the flow of fluid is caused by an external agency such as a fan or
pump or due to atmospheric disturbances, the resulting heat transfer is
known as Forced convection heat transfer.
• When the flow of fluid is due to density differences caused by temperature
differences, the heat transfer is said to be by Natural (or free) convection.
• For example, if air is blown on a hot plate by a blower, heat transfer occurs
by forced convection, whereas, a hot plate simply hung in air will lose heat
by natural convection. Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Dimensional analysis applied to Forced convection
Dimensional analysis is a mathematical technique which makes use of the study of the dimesnions for several
engineering problems.

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Dimensional analysis applied to forced convection heat transfer
Let the heat transfer coefficient in a fully developed forced convection in a tube is a function
of the following variables; h = f (  , D,V ,  , c p , k )
Then, we can write f1 ( h,  , D,V ,  , c p , k ) = 0 (i ) Here, the total number of variables n = 7
Fundamental dimensions in the problem are M , L, T &  ; hence, m = 4
Number of dimensionless  -terms ( n − m) = (7 − 4) = 3
The equation (i) may be written as f1 ( 1 ,  2 ,  3 ) = 0 Choosing h,  , D,V as repeating variables;
 1 = h a . b .D c .V d .
1 1 1 1
 2 = h a . b .D c .V d .c p
2 2 2 2
 3 = h a . b .D c .V d .k
3 3 3 3

π1 : M 0 L0T 0 = ( MT −3 −1 ) a1 .( ML−3 )b1 .Lc1 .( LT −1 ) d1 .( ML−1T −1 )

Equating the exponents of M, L, T and θ respectively we get;
For M : 0 = a1 + b1 + 1 For L : 0 = −3b1 + c1 + d1 − 1 For T : 0 = −3a1 − d1 − 1 For θ : 0 = − a1
 μ 
Solving the above equations, a1 = 0, b1 = −1, c1 = −1, d1 = −1  π1 =  
 ρDV 
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
π2 : M 0 L0T 0 = ( MT −3 −1 ) a2 .( ML−3 )b2 .Lc1 .( LT −1 ) d2 .( L2T −2 −1 )
Equating the exponents of M, L, T and θ respectively we get;
For M : 0 = a2 + b2 For L : 0 = −3b2 + c2 + d 2 + 2 For T : 0 = −3a2 − d 2 − 2 For θ : 0 = − a2 − 1
 c p ρV  k
Solving the above equations, a2 = −1, b2 = 1, c2 = 0, d 2 = 1  π2 =   As h =
 h  D
 c p ρVD 
 π2 =  
 k 
π3 : M 0 L0T 0 = ( MT −3 −1 ) a3 .( ML−3 )b3 .Lc3 .( LT −1 ) d3 .( MLT −3 −1 )
Equating the exponents of M, L, T and θ respectively we get;
For M : 0 = a3 − 3b3 + 1 For L : 0 = −3b3 + c3 + d 3 + 1 For T : 0 = −3a3 − d 3 − 3 For θ : 0 = − a3 − 1
 k 
Solving the above equations, a3 = −1, b3 = 0, c3 = −1, d 3 = 0  π3 =  
 hD 
According to Buckingham  - theorem, π3 =  ( π1 , π2 )
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
According to Buckingham  - theorem, π3 =  ( π1 , π2 )
    c p VD 
m' n'
 k 
  =C    where m ' and n ' ( m '  n ' ) are constants.
 hD    DV   k 
m ' −n ' m ' −n '
    c p VD  c p VD 
n' n' n'
 k         
  = C   DV    DV   k  = C   DV    DV  
 hD           k 
m ' −n '
 c p    DV    c p 
n' m n
 k      hD 
  = C   DV   k  Rearranging the terms;   =C   k 
 hD       k      
Nu = C(Re)m (Pr)n where C, m and n are constants evaluated experimentally.
 hD   ρDV   μc p 
Also, Nu = Nusselt number =   , Re = Reynold's number =   , Pr = Prandtl number =  
 k   μ   k 
Note : If the repeating variables are chosen as V,  ,  and c p ; the analysis will result in the following
 ρDV   μc p   h 
dimensionless groups Re =   , Pr =  k  , St =  Vc  where St = Stanton number =  (Re, Pr)
 μ     p 

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Dimensional numbers and their physical significance
Refer Table of dimensionless numbers in P112, Data hand book (7th edition)
1)Reynold's number (Re) : It is defined as the ratio of the inertia force to the viscous force.
uL  ud   ud  
Symbol → Re =   or   or  
        
where u = velocity, d = diameter, L = length,  = kinematic viscosity,  = dynamic viscosity
Significance : Reynold's number signifies the relative predominance of inertia force compared
to viscous force occuring in the flow systems. It is taken as an important criterion of kinemtic
and dynamic similarities in forced convection heat transfer.
2)Prandtl number (Pr) : It is defined as the ratio of the kinematic viscosity to thermal diffusivity.
  k         cp     cp 
Symbol → Pr =   But  =  Pr =     =  (  =  )
  cp    k   k 
where k = therml conductivity, c p = specific heat at constant pressure, ic viscosity
Significance : Prandtl number provides a measure of the relative effectiveness of the momentum and
energy transport by diffusion.  Movement of molecules from a region of higher concentration to lower concentration 
Prandtl number relates the velocity field and temperature field and its value strongly influences the relative
growth of the velocity and the thermalDrboundary layers.
P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Dimensional numbers and their physical significance...
3)Nusselt number (Nu) : It is defined as the ratio of temperature gradients by conduction and
hL  hd 
convection at the surface. Symbol → Nu =   or   where;
 k   k 
h = convective heat transfer coefficient and k = thermal conductivity
Significance : Nusselt number is a convenient measure of the convective heat transfer coefficient.
Nusselt number indicates the ratio of convective heat transfer to conductive heat transfer within a fluid,
providing insight into the efficiency of heat exchange between a solid surface and the surroundings.
  h × L  . But the difference is Biot number uses thermal conductivity 
Note : Biot number is also equal to  
  k  
 
 of the body ( not fluid ) , whereas Nusselt number uses thermal conductivity of the fluid. 
2) Stanton number (St) : It is a dimensionless number that represents the ratio of heat transfer coefficient
 h   Nu 
to heat capacity of a fluid.It is expressed with the following equation : Symbol → St =  =
 c  u   Re× Pr 
 p 
Significance : Stanton number indicates the amount of heat delivered by the fluid when there is heat transfer
between solid surface and fluid. The greater Stanton number is, the more effectively heat is transferred.
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Flow over a flat plate:
Hydrodynamic boundary layer
(Laminar flow)

Above fig represents the formation of boundary layers.Hydrodynamic boundary layer is formed due to
the viscosity effect of the fluid. Less viscous fluid will have very small hydrodynamic boundary layer
and highly viscous fluid will form very thick boundary layer. The thickness of the hydrodynamic
5x
boundary layer is given by δhx = = 5xRe x -0.5 where x represents the distance from the leading
Re x
edge and Re x represents Reynolds number at that point.Hydrodynamic boundary layer is an imaginary
curve which is the locus of the all points where the velocity of the fluid affected by the viscosity effect
is about 99% ofthe free stream velocity U  .
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Flow over a flat plate:
Thermal boundary layer
(Laminar flow)

Ts  T
Thermal boundary layer is formed due to thermal resistance of the fluid.Highly conducting fluid will
create very small thermal boundary layer. Less conducting fluid will form very large boundary layer.
Thermal boundary layer is the imaginary curve formed by collecting all points where temperature
difference between the free stream fluid T and the surface Ts is about 1%.
The thickness of the thermal boundary layer is given by δTx = δhx Pr -0.333

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Reynolds numbers for laminar flow, as shown in the table below.
Geometry Critical Reynolds number for laminar flow
Flow in a pipe 2300
Flow along a flat plate 500,000
Flow across a sphere 24
Flow across a symmetric wedge 200-2,000, depending on orientation angle
Flow across a cylinder 30

Note:
1. The flow of a fluid in a pipe or duct is considered as internal flow, while the flow over a
flat plate, cylinder, sphere are considered as external flow.
2. The equations for boundary layer thickness, Nusselt number, friction factor, etc. are
listed in pages 113 to 120 for external flow (Laminar and turbulent) and in pages 124 to
129 for internal flow (laminar & turbulent)
3. The above page numbers are as per Heat transfer data hand book 7th edition
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 NUMERICAL PROBLEMS ON EXTERNAL FORCED CONVECTION
EXTERNAL FLOW: For empirical formulae, refer Page 113 to 120
(Data book 7th Edition)

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Problem 1
Air at atmospheric pressure and 200 C flows over a plate with a velocity of
5 m/sec. The plate is 15 mm wide and is maintained at a temperature of
1200 C . Calculate the thickness of hydrodynamic and thermal boundary
layers and the local heat transfer coefficient at a distance of 0.5 m from
the leading edge. Assume that flow is on one side of the plate.  = 0.815
kg/m3,  = 24.5 x 10-6 m2, Pr = 0.7, k = 0.0364 W/m-K.
 Tw +T 
Data : Here, the properties of air are given. If not given, refer p 34 at T f =  
 2 
Free stream temperature T = 20 0 C, Plate surface temperature Tw = 120 0 C, u = 5 m / sec
x = 0.5 m,  = 0.815 kg / m 3 ,  = 24.5× 10 -6 N − s / m 2 , Pr = 0.7, k = 0.0364 W / m - K
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
Solution : To ascertain the type of flow (laminar or turbulent) Reynold' number is used.
ux   24.5×10 -6 
Rex = Here, u = 5m / sec, L = x = 0.5 m,  = =   = 3.01×10 -5
m 2
/ sec.
   0.815 
 5  0.5 
 Rex =  -5 
= 0.8316 × 1 0 5
< 5 × 10 5
 The flow is laminar (page 113, data book)
 3.01×10 
 For flow over a flat plate, the critical Reynold's number is 5 × 10 5 
(i) Thickness of the hydrodynamic boundary layer :  hx = 5 xRex −0.5 → P 113, I row
δhx = 5  0.5  (0.8316×10 )
5 −0.5
= 8.67 × 10 -3 m = 8.67 mm
(ii) Thickness of the thermal boundary layer :  Tx =  hx Pr −0.333 → P 113, I row
δTx = 8.67× 10 -3  (0.7 )
−0.333
= 9.763 × 10 -3 m = 9.763 mm
(iii) Local heat transfer coefficient (h) : From II row, page 113, for constant wall temperature;
= 0.332  (0.8316×10 )
5 0.5
(0.7 ) = 85.02
0.5 0.333 0.333
Nusselt number Nu = 0.332Rex Pr
 hx   h  0.5 
Also, from page 112, Nu =    85.02 =   or h = 6.2 W / m 2
-K
 k   0.0364 
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Problem 2
Air at atmospheric pressure and 400 C flows over a plate with a velocity of 5
m/sec. The plate is 2 m long and is maintained at a uniform temperature of
1200 C . Calculate the average heat transfer coefficient over the 2 m length of
the plate. Also find the rate of heat transfer between the plate and the air per
1 m width of the plate.

Data : Free stream temperature T = 40 0 C, Plate surface temperature Tw = 120 0 C, u = 5 m / sec

L = 2m, B = 1 m
 T +T   120 + 40 
For the properties of air, refer p 34, for film temperature T f =  w   =   = 80 0
C
 2   2 

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

Solution : From Page 34, at t = T f = 80 0 C , the properties of dry air ( at atmospheric pressure) are;
 = 1.0 kg / m 3 ,  = 21.08×10 -6 N − s / m 2 ,  = 21.09×10 -6 m 2 / sec, Pr = 0.692, k = 0.03047 W / m - K
To ascertain the type of flow (laminar or turbulent) Reynold' number is used.
uL 5 2 
Re = =  = 4.74 × 10 5
< 5 × 10 5
 The flow is laminar (page 113, data book)
  21.09×10 -6 

(i) Average heat transfer coefficient over the 2 m length of the plate :
To determine the average value of heat transfer coefficient, the average value of Nusselt number is used.
From P 114, II row, NuL = 2 NuL = 2  0.332Re Pr 0.5 0.333
 = 2  0.332  ( 4.74  10 )  0.6920.333  = 404.4
 5 0.5
 
 hL   h2 
Also, from page 112, Nu =    404.4 =   or h = 6.16 W / m 2
-K
 k   0.03047 

(ii) Rate of heat transfer between the plate and the air per 1 m width of the plate :
By Newton's law of cooling, q = hAs (Tw − T ) = h  ( L  B )  (Tw − T ) = 6.16  (2  1)  (120 − 40) = 986 W

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Problem 3
In a certain glass making process, a square plate of glass 1 m2 area and 3 mm
thick heated uniformly to 900C is cooled by air at 200C flowing over both sides
parallel to the plate at 2 m/sec. Calculate the rate of cooling of the plate.
Neglect the temperature gradient in the glass plate and consider only forced
convection. The properties of glass are  = 2500 kg/m3 and c=0.67 kJ/kg-K.

Data : Free stream temperature T = 20 0 C, Plate surface temperature Tw = 90 0 C, u = 3 m / sec

A = 1 m 2  L = 1m, B = 1 m, Thickness of glass plate H = 3 mm = 0.003 m
 glass = 2500 kg / m 3  Mass of glass plate m =   Volume of plate = 2500  1  1  0.003 = 7.5 kg
Specific heat of glass, c = 0.67 kJ / kg
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Solution :
 T +T   90 + 20 
For the properties of air, refer p 34, for film temperature T f =  w   =   = 55 0
C
 2   2 

From Page 34, at t = T f = 55 0 C , properties of dry air ( averaging the values for 50 0 C & 60 0 C ) are;
 = 1.076 kg / m 3 ,  = 18.46×10 -6 m 2 / sec, Pr = 0.697, k = 0.0286 W / m - K
Rate of cooling of the glass plate : It is obtained by equating the heat absorbed by the air stream
to the heat lost by the glass plate.
The heat absorbed by the air stream is; q = hAs (Tw − T ) (i )
The heat lost by the glass plate is; q = mc( T ) (ii ) where ΔT is the change in temperature of glass
and m = Mass of glass plate =  glass  Volume of glass plate = 2500  1  1  0.003 = 7.5 kg
and c is the specifc heat of glass = 0.67 kJ / kg = 670 J / kg
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
To ascertain the type of flow (laminar or turbulent) Reynold' number is used.
uL 2 1 
Re = =  = 1.083 × 10 5
< 5 × 10 5
 The flow is laminar (page 113, data book)
  18.46×10 -6 

Average heat transfer coefficient over the 1 m length of the plate :

To determine the average value of heat transfer coefficient, the average value of Nusselt number is used.
From P 114, II row, NuL = 2 NuL = 2  0.332Re Pr 0.5 0.333
 = 2  0.332  (1.083  10 )  0.697 0.333  = 193.8
 5 0.5
 
 hL   h 1 
Also, from page 112, Nu =    193.8 =   or h = 5.543 W / m 2
-K
 k   0.0286 
Rate of heat transfer between the plate and the air (from both sides) of the plate :
By Newton's law of cooling, q = h  2  As  (Tw − T ) = 5.543  (2  1)  (90 − 20) = 776 W
 As it is given that air flows over both sides of the plate, surface area = 2A  s

Equating the above heat transfer to heat lost by the glass plate ; q = mc( T )
where m = mass of the glass = 7.5 kg and c = specific heat of glass = 0.67kJ / kg - K = 670 J / kg - K
 776 = 7.5  670  ( T ) or the rate of cooling of the glass plate (ΔT) = 0.154 0 C / sec
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Problem 4
Air at –10°C flows over a flat surface at 10°C with a free stream velocity of 80
m/s. The length of the plate is 3.1 m. Determine the location at which the
flow turns turbulent. Also determine the local and average value of
convection coefficient assuming that the flow is turbulent all through.

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Assignment Problem
Atmospheric air at a pressure of 1.01325 bar and temperature 30°C flows over a flat
surface at 10°C with a free stream velocity of 5 m/s. The entire plate surface is maintained
at a temperature of 700C. Assuming the transition occurs at a Reynold’s number of 5x105,
find the distance from the leading edge at which the flow changes from laminar to
turbulent . At that location, determine;
(i) Thickness of the hydrodynamic boundary layer
(ii) Thickness of the thermal boundary layer
(iii) Local and average convective heat transfer coefficients
(iv) Heat transfer from both sides for unit width of the plate
(v) The skin friction coefficient.
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Solution to assignment problem
Data : Free stream temperature T = 30 0 C, Plate surface temperature Tw = 70 0 C, u = 5 m / sec
Width of the plate B = 1 m, Critical value of Re = 5  10 5 , Pressure = 1 .01325 bar = 1 atm
 Tw +T   70 + 30 
For film temperature T f =  =
   = 50 0
C, From p 34, the properties of air at 1 atm
 2   2 
 = 1.093 kg / m 3 , = 17.95×10 -6 m 2 / sec, Pr = 0.698, k = 0.02826 W / m - K
ux 5 x
Solution : At the transition point, = 5  105  = 5  105
or x = 1.795 m
 17.95×10 -6

Upto x = 1.795 m from the leading edge, the flow is laminar and beyond that it is turbulent.

(i) Thickness of the hydrodynamic boundary layer :  hx = 5 xRex −0.5 → Page113,I row,II column
 hx = 5  1.795  ( 5  10 )
5 −0.5
= 0.0127 m
(i) Thickness of the thermal boundary layer :  Tx =  hx Pr −0.333 → Page113,I row,II column
 Tx = 0.0127  ( 0.698)
−0.333
= 0.0143 m
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
(iii) Local and average convective heat transfer coefficients :
As the wall temperature is constant, Nusselt number Nu x = 0.332 Re0.5 Pr 0.333
→ Page113,II row,II column (data hand book)
hx  x
Nux = 0.332  ( 5  10 )
5 0.5
(0.698)0.333 = 208.27. Also, Nu x 
k
hx  1.795
208.27  or hx = 3.28 W / m 2 - K → Local heat transfer coefficient
0.02826
From page 114, IIrow, II column, the avearge Nusselt number Nu = 2Nu  h = 2hx
 h = 2  hx = ( 2  3.28) = 6.56 W / m 2 - K → Average heat transfer coefficient
( iv ) Heat transfer from both sides for unit width of the plate :
q = hA(Tw - T ) = h  ( x  B)  (Tw - T ) = 6.656  (1.795  1)  (70 − 30) = 477.9 W
( v ) The skin friction coefficient : C fx = 0.664 Re x −0.5
= 0.664  ( 5  10 )
5 −0.5
= 9.39 × 10 -4
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Problem 5
Calculate the rate of heat loss from a human body which may be considered
as a vertical cylinder 30 cm in diameter, and 175 cm high while standing in a
30 km/hr wind at 15°C. The surface temperature of the human body is 35°C.

Data :
Free stream temperature T = 15 0 C, Cylinder surface temperature Tw = 35 0 C, d = 30 cm = 0.3 m
 30  1000 
Length of the cylinder L = 175 cm = 1.75 m, u = 30 kmph =   = 8.333 m / sec
 3600 

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

  35 + 15  0 0 0
Solution : For t = T f =   = 25 C , properties of dry air ( averaging the values for 20 C & 30 C ) are;
 2 
 = 1.185 kg / m 3 ,  = 15.53×10 -6 m 2 / sec, Pr = 0.702, k = 0.02634 W / m - K
ud 8.333  0.3 
Re = =  = 1.61× 10 5
→ It is in the range 40,000 - 400,000
  15.53×10 -6 

Heat transfer coefficient of the cylinder (human body) : For cylinders, page 116, I row;
The Nusselt number based on diameter of the cylinder is NuD = C ( ReD ) Pr 0.333
m

From the table, in II column p 116, for ReD = 40,000 - 400,000; C = 0.0266, m = 0.805
 NuD = 0.0266  (1.61  105 ) (0.702 )
0.805
= 367.46
0.333

 hd   h  0.3 
Also, from page 112, Nu =    367.46 =  2
 or h = 32.26 W / m - K
 k   0.02634 
Rate of heat transfer between the human body and the air :
By Newton's law of cooling, q = h   dL  (Tw − T ) = 32.26  (  0.3  1.75)  (35 − 15) = 1064 W

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Problem 6
Calculate the value of convection coefficient for flow of the following fluids at
10°C across a pipe 20 mm dia at 30°C, the flow velocity being 5 m/s;
(a) Air (b) Water (c) engine oil (d) liquid ammonia.

Data :
Free stream temperature T = 10 0 C, Cylinder surface temperature Tw = 30 0 C, d = 20 mm = 0.02 m
u = 5 m / sec

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

  30 + 10  0
Solution : Film temeprature t = T f =   = 20 C
 2 
(a) Air : From page 34, (Data book, 7th edition), properties of dry air at 20 0 C are;
 = 1.205 kg / m 3 ,  = 15.06×10 -6 m 2 / sec, Pr = 0.703, k = 0.02593 W / m - K
ud 5  0.02 
ReD = =  = 6640 → Range (4000 to 40,000), From the table, C = 0.193, m = 0.618
  15.06×10 -6 
(Page 116, II column flow over cylinders) Also, Nusselt number is Nu = C ( ReD ) .( Pr )
m 0.333

Nu = 0.193  ( 6640 )  ( 0.703) = 39.51

0.618 0.333

 hd   h  0.02 
Also, from page 112, Nu =    39.51 =   or h = 51.22 W / m 2
-K
 k   0.02593 
(b) Water : From page 22, (Data book, 7th edition), properties of water at 20 0 C are;
 = 1000 kg / m 3 ,  = 1.006×10 -6 m 2 / sec, Pr = 7.02, k = 0.5978 W / m - K
ud 5  0.02 
ReD = =  = 99403.6 → Range (40,000 to 400,000), From the table, C = 0.0266, m = 0.805
  1.006×10 -6 

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

(Page 116, II column flow over cylinders) Also, Nusselt number is Nu = C ( ReD ) .( Pr )
m 0.333

Nu = 0.0266  ( 99403.6 )  ( 7.02 ) = 536.56

0.805 0.333

 hd   h  0.02 
Also, from page 112, Nu =    536.56 =   or h = 16038 W / m 2
-K
 k   0.5978 
(c) Engine oil : From page 25, (Data book, 7th edition), properties of engine oilat 20 0 C are;
 = 888 kg / m 3 ,  = 901×10 -6 m 2 / sec, Pr = 10400, k = 0.1454 W / m - K
 5  0.02
ud 
ReD = =  = 111 → Range (40 to 400), From the table, C = 0.683, m = 0.466
  901×10 -6 
(Page 116, II column flow over cylinders) Also, Nusselt number is Nu = C ( ReD ) .( Pr )
m 0.333

Nu = 0.683  (111)  (10400 ) = 133.42

0.466 0.333

 hd   h  0.02 
Also, from page 112, Nu =    133.42 =   or h = 969.95 W / m 2
-K
 k   0.1454 

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

(d) Liquid Ammonia : From page 22, (Data book, 7th edition), properties of Ammonia at 20 0 C are;
 = 612 kg / m 3 ,  = 0.358×10 -6 m 2 / sec, Pr = 2.02, k = 0.5210 W / m - K
ud  5  0.02 
ReD = = -6 
= 2.7933 × 10 5
→ Range (40,000 to 400,000)
  0.358×10 
From the table, C = 0.0266, m = 0.805 (Page 116, II column flow over cylinders)
Also, Nusselt number is Nu = C ( ReD ) .( Pr )
m 0.333

Nu = 0.0266  ( 2.7933×10 )
5 0.805
 ( 2.02 ) = 814.12
0.333

 hd   h  0.02 
Also, from page 112, Nu =    814.12 =   or h = 21, 208 W / m 2
-K
 k   0.5210 

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Problem 7
Air stream at 270C is moving at 0.3 m/s across a 100 W electric bulb at 1270C. If
the bulb is approximated by a 60 mm diameter sphere, estimate the heat
transfer rate and the percentage of power lost due to convection.
Data :
Free stream temperature T = 27 0 C, Bulb surface temperature Tw = 127 0 C, d = 60 mm = 0.06 m
u = 0.3m / sec, Power rating of the bulb P = 100 W

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

  127 + 27 
Solution : For t = T f =   = 77 0
 80 0
C , properties of dry air are;
 2 
 = 1.0 kg / m 3 ,  = 21.09×10 -6 m 2 / sec, Pr = 0.692, k = 0.03047 W / m - K
ud 0.3  0.06 
Re = =  = 853.5 < 70000 (Refer page 120, I row flow over spheres)
  21.09×10 -6 
Nusselt number is Nu = 0.37Re0.6 = 0.37  ( 853.5 ) = 21.23
0.6

 hd   h  0.06 
Also, from page 112, Nu =    21.23 =   or h = 10.78 W / m 2
-K
 k   0.03047 
(i) Rate of heat transfer b / n the bulb and the air :
Surface area of a sphere As = 4 r 2 = π(2r)2 =  d 2 ; By Newton's law of cooling, q = h  As  (Tw − T )
 q = 10.78  (  0.062 )  (127 − 27) = 12.2 W
q   12.2 W 
(ii) Percentage of power lost due to convection =   100  =   100  = 12.2%
P   100 W 
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Problem 8
Liquid sodium at 200°C flows over a sphere of 5 cm dia at a velocity of 1.2 m/s.
The surface is at 400°C. Determine the rate of heat loss from the sphere.

Data :
Free stream temperature T = 200 0 C, Surface temperature Tw = 400 0 C, d = 5 cm = 0.05 m
u = 1.2 m / sec

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

Solution : Liquid Sodium is a liquid metal.
 400 + 200  0
For t = T f =   = 300 C , From page 33, (Data book, 7th edition)
 2 
properties of Sodium are;  = 878 kg / m 3 ,  = 0.394×10 -6 m 2 / sec, Pr = 0.0063, k = 70.94 W / m - K
ud 1.2  0.05 
Re = =  = 1.52 × 10 5
< 1.525 × 10 5
(Refer page 120, last row for liquid metals)
  0.394×10 -6 
Nusselt number is Nu =  2 +0.386(Re× Pr)0.5  =  2 +0.386(1.52×10 5 ×0.0063)0.5  = 13.96
 hd   h  0.05 
Also, from page 112, Nu =    13.96 =   or h = 19806.45 W / m 2
-K
 k   70.94 

Rate of heat transfer between the sphere and the liquid sodium :
Surface area of a sphere = 4 r 2 = π(2r)2 =  d 2
By Newton's law of cooling, q = h   d 2  (Tw − T ) = 19806.45  (  0.052 )  (400 − 200) = 31112 W

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Assignment Problems on forced convection (External flow):
1) Air at a temperature of 200C flows through past a 1 m long with a velocity 45
m/sec. If the surface of the plate is maintained at 3000C. Determine (i) Heat
transferred from the entire plate length considering both laminar & turbulent
portion. (ii) % error if boundary layer is assumed to be turbulent through out.
Assume unit width of the plate and critical Reynold’s number as 5 x 105.
Ans: L1=0.33 m, L2=0.67 m, q1=4166.42W, q2=20172.5W, q = 23259.4 W,
% error=4.43%
2) Air at 1 atm pressure and temperature of 250C flowing at a velocity of 50 m/sec
crosses an industrial heater made of long solid rod of diameter 20 mm. The
surface temperature of the heater is 4750C. Determine the allowable electrical
power density (W/m3) within the heater per meter length.
Ans: Nu=87.7, h=187.16 W/m2-K, q =5292 W Power density =q/Vol= 16.84 MW/m3
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 NUMERICAL PROBLEMS ON INTERNAL FORCED CONVECTION
INTERNAL FLOW: For empirical formulae, refer Page 124 to 129
(Data book 7th Edition)

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

• The internal flow configuration is the most convenient and popularly used geometry
for heating or cooling of fluids in various thermal and chemical processes. There are
basic differences in the development of boundary layer between the external flow
geometry and internal flow geometry.
• In the case of internal flow, the fluid is confined by a surface, and the boundary layer
after some distance cannot develop further. This region is called entrance region. The
region beyond this point is known as fully developed region.
• Another important difference is that the flow does not change over at a location from
laminar to turbulent conditions, but is laminar or turbulent from the start, depending
upon the value of Reynolds number (based on diameter) being greater or less than
about 2300.
• A third difference is that in many cases there is no well defined free stream velocity
as in the case of flow over a flat plate. So, the mean velocity using the mass flow is to
be adopted in place of free stream velocity. . Bengaluru-59
Dr P R Venkatesh, Mech Dept, RVCE,
 The alternate expression for Reynold’s number used in internal flow is;
 uD   D2  4m
Re D = If the mass flow rate is m =  Au =     u  or u =
  4   D 2

4m
 D
 D 2
 4m 
 Reynold's number based on diameter ReD = =  → Page124, III column
  πDμ 
• A fourth difference is that as the fluid flows through the pipe, its mean temperature
increases and there is no free stream temperature as such.
• Let T be the mean temperature difference between the wall and the air stream.
However, temperature of air stream goes on changing from entry to exit in the heat
exchanger. So, we use a ‘mean temperature difference’ called LMTD (log mean
temperature difference).
• Expression for LMTD is derived in the chapter on heat exchangers. For the present, let
us take for LMTD

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Problem 9
Water at a bulk mean temperature of 40°C flows through a tube of 0.05 m
inner diameter with a velocity of 0.025 m/s. The tube is 2 m long and its surface
is at a constant temperature of 60°C. Determine the heat transfer and the mean
temperature rise during the flow. Assume hydrodynamic boundary layer already
developed.
Data : Bulk temperature Tm = 30 0 C, Surface temperature Tw = 60 0 C, d = 0.05 m
u = 0.025 m / sec, L = 2 m
Note : (Refer page 124, Data book 7th edition, for Internal flow)
 Tmi + Tmo 
Bulk mean temperature is Tm =   where; Tmi = Mean temperature at the inlet
 2 
and Tmo = Mean temperature at the outlet
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
Solution : For Tm = 40 0 C , From page 22, (Data book, 7th edition)
properties of water are;
 = 995 kg / m 3 ,  = 0.657×10 -6 m 2 / sec, Pr = 4.34, k = 0.628 W / m - K, c = 4178 kJ / kg - K

ud  0.025  0.05 
Reynold's number Re = = -6 
= 1903 < 2300  flow is laminar (Refer page 124, 1.1)
  0.657 ×10 
Using row 1.2, page 124; Nusselt number for fully developed hydrodynamic boundary layer;
D  0.05 
0.0668   (Re× Pr) 0.0668   (1903× 4.34)
Nusselt number is Nu = 3.66 + L = 3.66 +  2 
0.67 0.67
 D    0.05  
1+ 0.04   (Re× Pr) 1+0.04   (1903× 4.34)
 L    2  
 Nu = 9.354
 hd   h  0.05 
Also, from page 112, Nu =    9.354 =   or h = 117.5 W / m 2
-K
 k   0.628 
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
(i) Rate of heat transfer between the tube surface and water :
By Newton's law of cooling, q = h  As  (Tw − Tm )
Surface area of a tube = As =  DL =   0.05  2 = 0.3142 m 2
q = 117.5  0.3142  (60 − 40) = 19806.45  (  0.052 )  (400 − 200) = 738.3 W

(ii) The mean temperature rise of water during the flow :

The heat absorbed by water is also given by q = m  c  ΔT where;
 2
m = mass flow rate of water =  Acu = 995    0.05   0.025 = 0.0488 kg / sec
4 
c = specific heat of water = 4178 J / kg - K
738.3 = 0.0488  4178  ΔT or ΔT = 3.62 0 C

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Problem 10
Engine oil enters at 35°C into a tube of 10 mm ID at the rate of 0.05 kg/s and is
to be heated to 45°C. The tube wall is maintained at 100°C by condensing
steam. Determine the length of the tube required.
Data :
Wall temperature of the pipe Tw = 1000 C , Mean inlet temperature Tmi = 350 C
Mean outlet temperature Tmo = 450 C
 T +T   35 + 45 
Bulk mean temperature is Tm =  mi mo  =   = 40 0
C , c = 1964 kJ / kg - K
 2   2 
Mass flow rate of oil m = 0.05 kg / sec, Inner dia of the pipe D = 10 mm = 0.01 m
From page 25 ,7 th edition data hand book, properties of engine oil at 40 0 C are;
 = 876 kg / m 3 , ν = 241  10 −6 m 2 / sec, Pr = 2870, k = 0.1442 W / m - K
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 4m
Solution : Reynold's number ReD = → Page124, III column DDHB
 D
Also,  =    = ( 241  10−6  876 ) = 0.211 Pa - sec
4  0.05
ReD = = 30.17  2300  Laminar flow
  0.01  0.211
Assuming long tube, L >>> D, Nu = 3.66 (Page 124, Sl No 1.2.2)
 hd   h  0.01 
Also, from page 112, Nu =    3.66 =   or h = 52.78 W / m 2
-K
 k   0.1442 
Heat transferred to water Q = m  c  T = 0.05×1964×(45 - 35) = 982 W

Tmax − Tmin ( Tw − Tmi ) − (Tw − Tmo )

LMTD = =
 Tmax   Tw − Tmi 
log e   log e  

 min 
T −
 w mo 
T T
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Tmax − Tmin ( Tw − Tmi ) − (Tw − Tmo ) (100 − 35) − (100 − 45) ( 65) − (55)
LMTD = = = = = 59.86 0 C
 Tmax   Tw − Tmi   100 − 35   65 
log e   log e   log e   log e  
 Tmin   Tw − Tmo   100 − 45   55 

Heat gained by water = Heat lost by the tube by convection

982 = h   DL  LMTD = 52.78    0.01  L  59.86
 Length of the tube required L = 9.89 m

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Problem 11
Air at 20°C flows inside a pipe 18 mm ID having a uniform heat flux of 150
W/m2 on the surface, the average flow velocity at entry being 1.0 m/s. The air
pressure is 2 bar. Determine the value of convection coefficient. If the pipe is
2.5 m long, determine the air exit temperature and the wall temperature at the
exit. Assume fully developed hydrodynamic boundary layer.
Data :
Mean inlet temperature Tmi = 200 C , Q = 150 W / m 2 , pressure p = 2 bar
Inner dia of the pipe D = 18 mm = 0.018 m
Length of the pipe L = 2.5 m, Average velocity of flow u = 1m / sec
h = ?, Wall temperature of the pipe Tw = ?, Mean outlet temperature Tmo = ?
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
Solution :
(i) Exit temperature of air :
For air, gas constant R = 0.287 kJ / kg - K, Specific heat c p = 1.005 kJ / kg
p 200
From the gas equation, = RT  = 0.287  ( 20 + 273) or ρ = 2.38 kg / m 3
 
Mass flow rate of air = density  c / s area of pipe  velocity
  
m =   Ac  V =  2.38   0.018  1.0 = 6.05 × 10 -4 kg / sec
2

 4 
Equating the heat lost from the pipe to the heat gained by the air;
Heat flux / m 2  surface area of the pipe = mass flow rate of air  spheat  Temp rise
q   DL = mcT i.e. 150    0.018  2.5 = 6.05×10 -4  1005  T  ΔT = 34.86
But ΔT = (Tmo - Tmi )  34.86 = (Tmo - 20) or Exit temperature of air Tmo  55 0 C
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
  Tmi +Tmo   20 + 55 
Bulk mean temperature is Tm =  =
   = 37.50
C  40 0
C
 2   2 
From page 34 ,7 th edition data hand book, properties of air at 40 0 C are;
(Take only the values of  , c p , Pr and k which are independent of pressure)
 = 19.12  10−6 Pa − sec, c p =1005, Pr = 0.699, k = 0.02756
  19.12  10−6 
 = =   = 8.034  10 −6
m 2
/ sec
  2.38 
4m 4×6.05×10 -4
Reynold's number ReD = = = 2238  2300  Laminar
 D Dr PπR Venkatesh,
 0.018  19.12  10 −6
Mech Dept, RVCE, Bengaluru-59
As the flow is fully developed, (constant heat flux) from page 125, Row 1.2.7, Nu = 4.36 for pr > 0.6
 hd   h  0.018 
Nu =    4.36 =  2
 or h = 6.68 W / m - K
 k   0.02756 
Heat lost per unit surface area of the pipe Q = h  T  150 = 6.68× T
 T = 22.45  Wall temperature at the exit Two = Tmo + T = (55 + 22.45) = 77.45 0 C

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Problem 12
Consider the flow of water at a rate of 0.051 kg/sec through a square duct 2 cm
x 2 cm whose walls are maintained at a uniform temperature of 1000C.
Assuming that the flow is thermodynamically and thermally developed,
determine the length of the duct required to heat water from 300C to 700C.

Note : For flow through non cirular sections, the hydraulic diameter Dh is to be calculated.
4A
Dh = where A = Flow area & P is the wetted perimeter. → Row2.5, Page 127, DHB 7th ed
P
(The value of Dh must be used only in calculation of dimensionless numbers such as Nusselt, Reynold's
numbers, but not for calculating the areas)

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

Data :
Mean inlet temperature Tmi = 300 C , Mean exit temperature Tmo = 700 C
Wall temperature Tw = 1000 C , Size of the duct B = H = 2 cm = 0.02 m
Mass flow rate of air m = 0.015 kg / sec
 Tmi +Tmo   30 +70 
Solution : Bulk mean temperature is Tm =  =
   = 50 0
C
 2   2 
From page 22, DHB, properties of water at 500 C (averaging between 400C & 600C) are;
ρ = 990 kg / m 3 ,  = 0.5675  10−6 m 2 / sec, Pr = 3.68, c = 4180.5 kJ / kg - K
k = 0.6397 W / m - K
Mass flow rate of water = density  c / s area of duct  velocity
m =   Ac  V  0.015 =  990  0.022  V  or V = 0.0379 m / sec
4 A 4(bH ) 4  0.02  0.02
Hydraulic diameter Dh = = = = 0.02 m
P 2(b + H ) 2(0.02 + 0.02)
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 uDh  0.0379  0.02 
Reynold's number Re = = -6 
= 1336 < 2300  flow is laminar
  0.5675×10 
(Refer page 124, 1.1)
Using Table, page 128; Nusselt number for fully developed internal flow (laminar);
For square geometry, Nusselt number Nu2 = 2.976 (for constant wall temperature)
hDh h  0.02
Also, Nu =  2.976 = or h = 95.18 W / m 2 - K
k 0.6397
Heat transfer rate q = mc p (Tw − T ) = 0.015  4180.5  (70 − 30) = 2508.3 W
Also, q = hAs (Tw − Tm ) where As = Perimeter  Length = 2( B + H )  L
As = 2(0.02 + 0.02) L = 0.08L
q = hAs (Tw − Tm )  2508.3 = 95.18  0.08L  (100 − 50) or L = 6.59 m
Required length of the duct L = 6.59 m
[Note : If the flow is turbulent, use the equation Nu = 0.023Re D 0.8 Pr n → 2.3.1, Page 126]
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Assignment Problems on forced convection (Internal flow):
1) Air at a temperature of 200C flows through a rectangular duct with a velocity 10
m/sec. The duct is 30 cm x 20 cm in size and air leaves at 340C. Find the heat
transfer coefficient, heat gained by the air and the wall temperature of the duct
which is 10 m long.
Ans: Nu=280.79, h=32 W/m2-K, q = 9835 W, Tw=57.70C
2) Water at a velocity of 1.5 m/sec enters a 2 cm dia heat exchanger tube at 400C.
The heat exchanger tube wall is maintained at 1000C. If the water is heated to a
temperature of 800C in the heat exchanger tube , find the length of the exchanger
tube required.
Ans: Nu=246.53, h=8028.25 DrW/m 2-K, q =77665 W L= 3.85 m
P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
Natural or Free Convection

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
Dimensional analysis; Free convection

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Grashoff number (Gr) and its significance :
 gL3T
Grashoff number indicates the relative strength of the buoyant force. Gr =
2
  2  gL3T  V 2 2
L
Putting  = , Gr = Multiplying & dividing by V L ; Gr =  gL T 
2 2 3

  ( VL )
2 2

Inertia force
But V 2 L2 = Inertia force and VL = Viscous force  Gr = Buoyant force 
Viscous force 2
Grashoff number in free convection is analogous to Reynold's number in forced convection.
At low Grashoff number, free convection is supressed, and it begins at some critical value of
Gr and becomes more and more effective as Gr increases.
 1 
Note :  is the coefficient of volume expansion, given by  = 
 T + 273 
0
where T f is in C
 f 
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
The mechanism of heat transfer in natural convection involve the motion of the fluid particles past
a solid boundary which is a result of the density differences resulting from the energy exchange.
As result, the heat transfer coefficient will vary with the geometry of the system.
The following geometries are important in free convection studies.
1)Vertical plates & cylinders
2) Horizontal plates
3) Horizontal cylinders
4) Spheres
5) Rectangular enclosures
The relationships for the above mentioned geometries are listed from Page 135 to 142, Heat transfer
data hand book, 7 th edition.

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Numerical problems on
Natural or Free Convection

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Problem 1
Calculate the convection heat loss from a radiator 0.5 m wide and 1 m high
maintained at a temperature of 840C in a room at 200C. Treat the radiator as a
vertical plate.

Data : Wall temperature Tw = 840 C , T = 200 C , B = 0.5 m, L = 1.0 m, Q = ?

 Tw + T   84 + 20 
Solution : Mean film temperature T f =  =
   = 52 0
C
 2   2 
From page 34, DHB 7 th edition, for T f = 520 C  500 C the propeties of air are;
 = 17.95  10−6 m 2 / sec, Pr = 0.698, k = 0.02826 W / m - K
 1   1 
 = Coefficient of volume expansion =  = =  −3
 T + 273   52 + 273 
3.077 10 /K
 f 
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
  gL3T 3.077  10−3  9.81  13  (84 − 20)
Grashoff number Gr = = = 5.996  10 9

2 (17.95  10 )−6 2

The product of Gr & Pr is known as Rayleigh number, Ra [Page 112, DHB]

Ra = Gr  Pr = ( 5.996  109  0.698 ) = 4.185  109 > 10 9 → Turbulent flow
[From data book (7th edition) page No 135, for a vertical surface; I column]
For turbulent flow, [page 136, Sl No 2.1] Nusselt number Nu = 0.1 ( Gr  Pr )
0.333

Nu = 0.1 ( 4.185  10 )
9 0.333 hL
= 160 Also, from page 112, Nu =
k
h 1
 160 = or h = 4.52 W / m 2 - K
0.02826
The heat lost from the radiator q = hAs (Tw − T ) = 4.52  (1  0.5)  (84 − 20) = 144.66 W
[Assuming that the convection occurs from one face of the plate only]
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Problem 2
A square plate (0.5mx0.5m) with one surface insulated and the other surface
maintained at a temperature of 385 K is placed in ambient air at a temperature
of 315 K. Calculate the average heat transfer for free convection for the
following orientations of the hot surface:
(i) The plate is horizontal and the hot surface faces up
(ii) The plate is horizontal and the hot surface faces down
Note : For horizontal plates, the Grashoff number is to be calculated based on
Area LB
the characteristic length given by Lc = =
Perimeter 2( L + B )
[Row 2.1.1, Page 136, DHB]
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
Data : Wall temperature Tw = 885K , T = 315K , B = 0.5 m, L = 0.5 m, h = ?
 T + T   385 + 315 
Solution : Mean film temperature T f =  w   =   = 350 K
 2   2 
From page 34, DHB 7 th edition, for T f = 350 K = (350 − 273) = 770 C the properties of air are;
[Approximaing the values for 80 0 C]
 = 21.09  10−6 m 2 / sec, Pr = 0.692, k = 0.03047 W / m - K
1  1 
 = Coefficient of volume expansion = =  = 2.86  10 −3
/K
T f  350 
Area LB 0.5  0.5
Characteristic length Lc = = = = 0.125 m
Perimeter 2( L + B ) 2(0.5 + 0.5)
 gLc 3T 2.86  10−3  9.81  0.1253  (385 − 315)
Grashoff number Gr = = = 8.624  10 6

2 ( 21.09  10 )
−6 2

The product of Gr & Pr is known as Rayleigh number, Ra [Page 112, DHB]

Ra = Gr  Pr = (8.624  106  0.675Dr
) =P R5.97  10 6
(Lies between
Venkatesh, Mech Dept, RVCE, Bengaluru-59
2  10 4
and 8  10 6
) → P 136,2.1.1
(i) When the hot surface is facing upwards :
From data book (7th edition) page No 136, 2.1.1, Nusselt number Nu = 0.27 ( Gr  Pr )
0.25

[Hot surface up,with constant wall temperature] Nu = 0.54 ( 5.97  10 )

6 0.25
= 26.7
hL h  0.125
Also, from page 112, Nu =  26.7 = or h = 6.51 W / m 2 - K
k 0.03047

(ii) When the hot surface is facing downwards :

From data book (7th edition) page No 137, 2.1.2, Nusselt number Nu = 0.27 ( Gr  Pr )
0.25

[Hot surface down,with constant wall temperature] Nu = 0.27 ( 5.97  10 )

6 0.25
= 13.35
hL h  0.125
Also, from page 112, Nu =  13.35 = or h = 3.254 W / m 2 - K
k 0.03047

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Problem 3
A tube of 0.036 m outer diameter and 0.4 m length is maintained at 1000C. It is
exposed to air at a temperature of 200C. Determine the rate of heat transfer
from the surface when; (i) tube is vertical (ii) tube is horizontal

Note :
(i) For vertical cylinders, the Grashoff number is to be calculated based on
 gL3T
the length of the cylinder (similar to the vertical plate). Gr =
2
(ii) For horizontal cylinders, the Grashoff number is to be calculated based on
 gD 3T
the diameter. GrD =
2
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
Data : Wall temperature Tw = 1000 C , T = 200 C , D = 0.036 m, L = 0.4 m, q = ?
 Tw + T   100 + 20 
Solution : Mean film temperature T f =  =
   = 60 0
C
 2   2 
From page 34, DHB 7 th edition, for T f = 600 C , the properties of air are;
 = 18.97  10−6 m 2 / sec, Pr = 0.696, k = 0.02896 W / m - K
1  1 
 = Coefficient of volume expansion = =  = 3  10 −3
/K
T f  60 + 273 
(i) When the tube is vertical :
 gL3T 3  10−3  9.81  0.43  (100 − 20)
Grashoff number Gr = = = 418.72  10 6

2 (18.97  10 )
−6 2

The product of Gr & Pr is known as Rayleigh number, Ra [Page 112, DHB]

Ra = Gr  Pr = ( 418.72  106  0.696 ) = 0.291  109 104  (Gr Pr)  109
II row, 1.2, Page 136 Nu =Dr0.59(G rPr)
P R Venkatesh,
0.25
= 0.59(
Mech Dept, RVCE, 0.291  109 )0.25 = 77.06
Bengaluru-59
Nu = 0.59(GrPr)0.25 = 0.59( 0.291  109 )0.25 = 77.06
hL h  0.4
Also, from page 112, Nu =  77.06 = or h = 5.58 W / m 2 - K
k 0.02896
The heat lost q = hAs (Tw − T ) = 5.58  (  0.036  0.4)  (100 − 20) = 20.2 W
(ii) When the tube is horizontal : ( Gr is based on diameter)
 gD 3T 3  10−3  9.81  0.0363  (100 − 20)
Grashoff number GrD = = = 0.305  10 6

2 (18.97  10 )
−6 2

The product of Gr & Pr is known as Rayleigh number, Ra [Page 112, DHB]

Ra = Gr  Pr = ( 0.305  106  0.696 ) = 0.212  106 104  (GrD Pr)  107
I row, 3.1, Page 138 , C = 0.48, m = 0.25. Nu = C(GrD Pr)m = 0.48  ( 0.212  106 )0.25
hD h  0.036
Nu = 10.3 Also, from page 112, Nu =  10.3 = or h = 8.286 W / m 2 - K
k 0.02896
The heat lost q = hAs (Tw − T ) = 8.286  (  0.036  0.4)  (100 − 20) = 30 W
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Problem 4
Considering the body of a man as a vertical cylinder of 300 mm diameter and
170 cm height, calculate the heat generated by the body in one day. Take the
body temperature as 360C and the atmospheric temperature as 140C.
Data : Wall temperature Tw = 360 C , T = 140 C , D = 0.3 m, L = 1.7 m, q = ?
 T + T   36 + 14 
Solution : Mean film temperature T f =  w   =   = 25 0
C
 2   2 
From page 34, DHB 7 th edition, for T f = 250 C , the properties of air are;
[Averaging the values for 20 0 C and 30 0 C]
 = 15.53  10−6 m 2 / sec, Pr = 0.702, k = 0.02634 W / m - K
1  1 
 = Coefficient of volume expansion = =  = 3.356  10 −3
/K
TRVCE,
Dr P R Venkatesh, Mech Dept, f  25 + 273 
Bengaluru-59
The man's body is considered as a vertical cylinder :
 gL3T 3.356  10−3  9.81  1.73  (36 − 14)
Grashoff number Gr = = = 1.475  1010

2 (15.53  10 )−6 2

The product of Gr & Pr is known as Rayleigh number, Ra [Page 112, DHB]

Ra = Gr  Pr = (1.475  1010  0.702 ) = 10.36  109  109 → Turbulent flow
V row, 2.1, Page 136 Nu = 0.1(GrPr)0.333 = 0.1  ( 10.36  109 )0.333 = 216.33

hL h  1.7
Also, from page 112, Nu =  216.33 = or h = 3.352 W / m 2 - K
k 0.02634
The heat lost q = hAs (Tw − T ) = 3.352  (  0.3  1.7)  (36 − 14) = 118.15 W
Heat generated in one day = (118.15  24  3600) = 10.21× 10 6 Joules

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59

 Problem 5
Calculate the heat transfer from a 60 W incandescent bulb at 1250C to ambient
air at 250C. Assume the bulb as sphere of 50 mm diameter. Also find the
percentage of power lost by free convection.
Data : Wall temperature Tw = 1250 C , T = 250 C , D = 0.05 m, P = 60 W , q = ?
 Tw + T   125 + 25 
Solution : Mean film temperature T f =  =
   = 75 0
C
 2   2 
From page 34, DHB 7 th edition, for T f = 750 C , the properties of air are;
[Averaging the values for 70 0 C and 80 0 C]
 = 20.555  10−6 m 2 / sec, Pr = 0.693, k = 0.0301 W / m - K
1  1 
 = Coefficient of volume expansion = =  = 2.8734  10 −3
/K
T f  75 + 273 
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
  gD 3T 2.8734  10−3  9.81  0.053  (125 − 25)
Grashoff number GrD = = = 0.834  10 6

2 ( 20.555  10 )
−6 2

For free convection from a sphere :

The product of Gr & Pr is known as Rayleigh number, Ra [Page 112, DHB]
Ra = Gr  Pr = ( 0.834  106  0.693) = 5.78  105 ; 3  105  (Gr× Pr) < 8  108
IV row, 4.1, Page 136 Nu = 2 +0.5(GrD Pr)0.25 = 2 +0.5( 5.78  105 )0.25 = 15.79
hD h  0.05
Also, from page 112, Nu =  15.79 = or h = 9.51 W / m 2 - K
k 0.0301
The heat lost q = hAs (Tw − T ) = 9.51  (  0.052 )  (125 − 25) = 7.467 W
q   7.467 
% of power lost by convection =   100  =   100  = 12.44 %
P   60 
Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59
 Assignment Problems on free convection:
1) A circular plate 15 cm in diameter is maintained at 1500C in atmospheric air at 200C.
Calculate the free convection heat loss from both the surface or the plate, when the
upper surface is heated and the lower surface is cooled and the plate is kept in
horizontal position.
Ans: Nu=12.12, h=9.92 W/m2-K, q = 45.6 W
A d2 d
Characteristic length Lc = = =
P 4  d 4

2) Water is heated in a tank by dipping a vertical square plate of side 300 mm. The surface
temperature of the plate on both sides is maintained at 1400C. Assuming the initial
temperature of water surrounding the plate as 200C, find the heat loss from the plate
per hour.
Ans: Nu=692.37, h=1543.3 W/m2-K, q (one side)=16667.5 W=1.44x109 Joules
Note: The properties of water are to be taken at the mean film temperature from page 22. Also, the
value of  for water has to be taken from page 30 at the film temperature.

Dr P R Venkatesh, Mech Dept, RVCE, Bengaluru-59