UNIT II CONVECTION

Free and Forced Convection – Hydrodynamic and Thermal Boundary Layer. Free and Forced
Convection during external flow over Plates and Cylinders and Internal flow through tubes.

Part-A.
Free and Forced Convection
1. Define Convection?
When a fluid flows inside a duct or over a solid body and the temperatures of the fluid and solid
surface are different, heat transfer between the fluid and the solid surface will take place. This type of
heat transfer is called convection.
2. Differentiate between forced convection & free convection.
Free Convection
Forced Convection
 If the fluid motion is setup by  It the fluid motion is artificially
buoyancy effects resulting from created by means of external agency
the density variation caused by such as blower, fan etc, the mode of
the temperature difference in the heat transfer is termed as forced
fluid, the mode of heat transfer is said convection.
to be free or natural convection.
 It occurs naturally.  It is created artificially.
 The rate of heat transfer is slow when  The rate of heat transfer is high.
 compared to forced convection.  Eg : Cooling of a fluid by means of
 Eg : Heating of water in a vessel. overhead fan.
3. State
Newton’s law of convection.
Heat transfer from the moving fluid to soled surface is given by the equation

Q=hA ( T s −T ∞ )
Where,
h - Local heat transfer coefficient in W/m2K
A- Surface area in m2
Ts- Surface (or) Wall temperature in K
T ∞ - Temperature of fluid in K

4. What is meant by free of natural convection?

If the fluid motion is produced due to change in density resulting from temperature gradients,
the mode of heat transfer is said to be free or natural convection.

5. What is forced convection?

If the fluid motion is artificially created by means of an external force like a blower or fan, that
type of heat transfer is known as forced convection.

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UNIT II HEAT AND MASS TRANSFER
 6. What are the dimensionless parameters used in forced and free convection heat transfer
analysis?[May/June 2012]
Forced convection:
 Reynolds number (Re)
 Nusselt number (Nu)
 Prandtl number (pr)
Free convection:
 Grashoff number,
 Nusselt number (Nu)
 Prandtl number (pr)
7. In which mode of heat transfer coefficient is usually higher, natural or forced convection? Why?
[April/May-2011]
Convection heat transfer coefficient is usually higher in forced convection then in natural
convection, because it mainly depends upon the factors such as
 fluid density,
 velocity and
 viscosity
8. What do you understand by free and forced convection? [April/May-2010]
If the fluid motion is artificially created by means of an external force like a blower or fan, that
type of heat transfer is known as forced convection.
If the fluid motion is produced due to change in density resulting from temperature gradients,
the mode of heat transfer is said to be free or natural convection.

9. What is the physical meaning of Fourier number?[May/June 2009]

It is defined as the ratio of characteristic body dimension to temperature wave penetration depth
in time.It signifies the degree of penetration of heating or cooling effect of the solid.

Characteristicbodydimension
Fouriernumber=
Temperaturewavepenetration

10. What is overall heat transfer coefficient?

The heat transfer coefficient or film coefficient, in thermodynamics and in mechanics is the
proportionality coefficient between the heat flux and the thermodynamic driving force for the flow of
heat.
q
h=
ΔT
where:
q- amount of heat required (Heat Flux), W/m2
dQ
q=
i.e., thermal power per unit area, dA
h-heat transfer coefficient, W/(m2K)
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UNIT II HEAT AND MASS TRANSFER
 Hydrodynamic and Thermal Boundary Layer.
11. What is hydrodynamic boundary layer ?
In hydrodynamic boundary layer, velocity of the fluid is less than 99% of free stream velocity.

Laminar flow: D
( )
x fd , h
lam
≈0. 05 Re D

Laminar flow of an incompressible,

∂u
=0
Constant property fluid ∂ x

Turbulent flow:
10≤ (( ) )
x fd , h
D turb
≤60

12. Define the term ‘boundary layer’(thermal)[May/June 2012][April/May 2010]

In thermal boundary layer, temperature of the fluid is less than 99% of free stream temperature.

Laminar flow: D
( )
x fd , t
lam
≈0 .05 ReD Pr

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UNIT II HEAT AND MASS TRANSFER
 (( ) )
xf
d,t
=10
D turb

Turbulent flow:

13. Write about the term Velocity Boundary Layer.

Consider the flow of a fluid over a flat plate, the velocity and the temperature of the fluid approaching
the plate is uniform at U∞ and T∞. The fluid can be considered as adjacent layers on top of each others.

14. Define boundary layer thickness.

The thickness of the boundary layer has been defined as the distance from the surface at which the
local velocity of temperature reaches 99% of the velocity or temperature.

15. What is the significance of Dimensional Number? [May/June 2007]

Complete solution of all field problems is not possible of all cases. Under such instances, using
dimensional analysis number of independent variables is either reduced converted into another form.
This is a useful technique in all experimentally based areas of engineering.
16. What is dimensional analysis?
Dimensional analysis is a mathematical method which makes use of the dimensions for solving
several engineering problems. This method can be applied to all types of fluid resistances, heat flow
problems in fluid mechanics and thermodynamics.

17. State Buckingham π theorem. [Nov/Dec 2008]

Buckingham π theorem states as follows: “If there are n variables in a dimensionally
homogeneous equation and if these contain m fundamental dimensions, then the variables are arranged
into (n – m) dimensionless terms. These dimensionless terms are called π terms.

18. What are all the advantanges of dimensional analysis?

 Dimensional equations are used to validate the correctness of a physical equation.
 Dimensional equations are used to derive correct relationship between different physical quantities.
 Dimensional equations are used to convert one system of units to another.
 Dimensional equations are used to find the dimension of a physical constant.

4
UNIT II HEAT AND MASS TRANSFER
 19. What are all the limitations of dimensional analysis?
 The complete information is not provided by dimensional analysis.
 It only indicates that there is some relationship between the parameters.
 No information is given about the internal mechanism of physical phenomenon.
 Dimensional analysis does not give any clue regarding the selection of variables.
 Dimensional constant involved in the physical relation cannot be determined.
 It fails to give information about trigonometric function , logarithmic, exponential
Quantities involved in the physical relation.

20. Define Reynolds number (Re). [Nov/Dec 2006]

It is defined as the ratio of inertia force to viscous force.
Inertia force
Re =
Viscous force
uL
Re L=
υ
21. Define Prandtl number (Pr). [Nov/Dec 2008] [May/June 2014]
It is the ratio of the momentum diffusivity to the thermal diffusivity.
Momentum diffusivity
Pr =
Thermal diffusivity

22. Sketch the boundary development of a flow.

23. Define displacement thickness.

The displacement thickness is the distance, measured perpendicular to the boundary, by which
the free stream is displaced on account of formation of boundary layer.

(
δ ¿=∫ 1−
u
Us )
dy

24. Define momentum thickness.

momentum thickness is defined as the distance through which the total loss of momentum per
second is equal to if it were passing a stationary plate.

5
UNIT II HEAT AND MASS TRANSFER
 δ
θ=∫
0
u
Us
1−
u
Us (
dy
)
25. Define energy thickness.
The energy thickness can be ;defined as the distance, measured perpendicular to the boundary
of the solid body, by which the boundary should be displaced to compensate for the reduction in
kinetic energy of the flowing fluid on account of boundary layer formation.

( )
∞ u u
2
¿∗¿ = ∫ 1− dy
U U
0 s 2
s
δ ¿
26. Define Nusselt Number (Nu).
It is defined as the ratio of the heat flow by convection process under an unit temperature gradient to
the heat flow rate by conduction under an unit temperature gradient through a stationary (L) of metre.
Q conv
Nusselt Number (Nu) =
Q cond
27. Define Grashof number (Gr). [Nov/Dec 2008] [Nov/Dec 2006] [May/June 2014]
It is defined as the ratio of product of inertia force and buoyancy force to the square of viscous force.
Inertia force × Buoyancy force
Gr = 2
(Viscous force)

28. Define Stantan number (St).

It is the ratio of Nusselt number to the product of Reynolds number and Prandtl number.
Nu
St=
Re×Pr

29. What is meant by laminar flow and turbulent flow?

Laminar flow: Laminar flow is sometimes called stream line flow. In this type of flow, the fluid
moves in layers and each fluid particle follows a smooth continuous path. The fluid particles in each
layer remain in an orderly sequence without mixing with each other.

6
UNIT II HEAT AND MASS TRANSFER
 Turbulent flow: In addition to the laminar type of flow, a distinct irregular flow is frequently
observed in nature. This type of flow is called turbulent flow. The path of any individual particle is
zig-zag and irregular. Fig. shows the instantaneous velocity in laminar and turbulent flow.

30. Write about Sieder – Tate equation for laminar flow.

The Sieder-Tate result can be more accurate as it takes into account the change in viscosity ( μ and s
μ
due to temperature change between the bulk fluid average temperature and the heat transfer surface
temperature, respectively. The Sieder- and Tate equation for laminar flow:
Nu = 1.86
The above correlation is applicable when,
0.48 < Pr < 16700the viscosity ratio is within the range 0.0044 << 9.75> 108

31. Write Sieder and Tate equation for Turbulent flow?

μ
The Sieder-Tate result can be more accurate as it takes into account the change in viscosity ( μ and s
due to temperature change between the bulk fluid average temperature and the heat transfer surface
temperature, respectively
Sieder and Tate equation for Turbulent flow
Nu = 0.027 Re 0.8 Pr 0.33
The condition of applicability of this equation is
0.7 Pr 16700
Re = 10000
D /L= 0.1

T2 – Lower temperature level

32. Discuss about viscous sub-layer and buffer layer. [May/June 2014]

The turbulent flow near a flat wall can be divided up into four regimes. At the wall, the fluid
velocity is zero, and for a thin layer above this, the flow velocity is linear with distance from the wall.
This region is called the viscous sub layer, or laminar sub layer. Further away from the wall is a region
called the buffer layer.

7
UNIT II HEAT AND MASS TRANSFER
Free and ForcedConvection during external flow over Plates and Cylinders and
Internal flow through tubes
33. Define critical Reynolds number, what is its typical value for flow over a flat plate and flow
through a pipe? [May/June 2013]
The Reynolds number at which flow changes from laminar to turbulent is called critical reynold’s
number.
5
typical value for flow over For plate, 5×10 and
typical value for flow over for pipe 2300.

uL
Re L=
υ

34. What is the form of equation used to calculate heat transfer for flow through cylindrical pipes?
The equation used for calculation of heat transfer for flow through cylindrical pipes is

Nu = 0.023 (Re)0.8(Pr)n
35. An electrically heated plate dissipates heat by convection at a rate of 8000 W/m2 into the
ambient air at 25o C. If the surface of the hot plate is at 125 oC, calculate the heat transfer
coefficient for convection between the plate and air.
Given: Heat dissipation, Q = 8000 W / m2
Ambient temperature, T ∞ = 25oC+273 = 298 K
Surface temperature, Tw = 125oC+273 = 398 K
To find: Heat transfer coefficient, (h).
Solution: We know that,
Heat transfer, Q = hA (Tw – T ∞ )
= h (1)(398 -298)
8000 = h (100)
h = 80 W/m2K
Result: Heat transfer coefficient, h = 80 W/m2K

8
UNIT II HEAT AND MASS TRANSFER
 36. Write down the momentum equation for a steady, two dimensional flow of an incompressible,
constant property Newtonian fluid in the rectangular coordinate system and mention the
physical significance of each term.
Momentum equation,

[∂u ∂u
p u +v
∂x ∂y ]
∂P
=F x− +μ
∂x
∂2 u ∂2 u
+
∂ x2 ∂ y2 [ ]
Where,

pu
[ ∂u ∂u
+v =
]
∂ x ∂ y Inertia force.
F x = Body force
∂P
∂ x =Pressure force
∂2 u ∂2 u
+
∂ x 2 ∂ y 2 =Viscous forces
37. Draw the velocity and temperature profiles for free convection on a hot vertical plate.
[May/June 2009] [April/May 2008]

38. What is Colburn analogy?

The Reynolds analogy does not always give satisfactory results. Thus, Chilton and Colburn
experimentally modified the Reynolds’ analogy. The empirically modified Reynolds’ analogy is known
as Chilton-Colburn analogy and is given by eq.

39. What is Dittus- Boelter equation? When does it apply? [Nov/Dec 2015]
The Dittus-Boelter equation (for turbulent flow) is an explicit function for calculating the Nusselt
number. It is easy to solve but is less accurate when there is a large temperature difference across the
fluid. It is tailored to smooth tubes, so use for rough tubes (most commercial applications) is
cautioned. The Dittus-Boelter equation is:

9
UNIT II HEAT AND MASS TRANSFER
 Dittus – Boelter Equation for turbulent flow through a circular pipe:
Nu = 0.023 Re0.8 Prn
where n = 0.4 for heating ( Tw > T )
n = 0.3 for cooling ( Tw > T )
40. Define Grashof number and explain its significance in free convection heat transfer. [Nov/Dec
2015]
Grashof number is defined as the ratio of product of inertia force and buoyancy force to the square of
viscous force
Inertia force X Buoyancy force g x β x L3 x ΔT
Gr=
( Viscous force )2 or ν2

PART-B
1. In a straight tube of 50 mm diameter, water is flowing at 15m/s. The tube surface temperature is
maintained at 60oC and the flowing water is heated from the inlet temperature 15 oC to an outlet
temperature of 45oC. Calculate the heat transfer coefficient from the tube surface to water and length
of the tube.[may/june 2014] (16)
Given Data:
Length of the pipe, L=6m
Diameter of the pipe, d=50 mm=0.05m
Velosity, U=15 m/s
Surface temperature, Tw=60oC
Fluid temperature, @ inlet Tmi=15oC
@ outlet Tmo=45oC
To find:
a) the heat transfer coefficient from the tube surface to water and
b) length of the tube
Solution:
Film temperature,
T +T 15+ 45
T f = mi mo = =30o C
2 2
The properties of air at 44oC
ρ= 997 kg/m3
Pr= 5.5
k =0.61 W/m.K
υ= 0.857¿ 10-6 m2/s
Cp=4178J/kgK
UD
Re L=
Reynold’s Number, υ

10
UNIT II HEAT AND MASS TRANSFER
 0.857×10−6 ¿
=15×0.05¿ 5
¿
=4.927×10 ¿
5
Since, Re>5¿ 10 ,∴ The flow is turbulent
For turbulent flow (Internal flow):
0. 8 n
Nusselt number, Nu=0 . 023(Re ) (Pr)
This is heating process. So n=0.4 [ ∵ T mo >T mi ]
Nu=0.023(Re )0.8 (Pr)n
¿0.023(4 .927×105 )0 .8 (5.5)0 .4
¿1629.06
hD
Nu=
k
h×0 . 05
1629 . 06=
We know that, 0 . 61
h =19874.61 W/m2K
Heat transfer,
Q=mC p ΔT
=mC p (T mo −T mi )
¿1×4178(45−15 )
¿125340 W
We know that, Q=hA(T w −T m )
125340=19874 .61×π×0 . 05×L×(60−30)
L=1.338 m
Result:
The heat transfer coefficient from the tube surface to water, h =19874.61 W/m2K
Length of the tube,h =19874.61 W/m2K

2. (i).A horizontal heated plate measuring 1.5 m x 1.1 m and at 215 oC, facing upwards is placed in still
h=3 .05 ( T f ) 1/4 , T f
air at 25oC. Calculate the heat loss by natural conversion. Use the relation = Mean
flilm temperature.[may/june 2014] (8)
Given Data:
T ∞=25 o C
T s =215o C ,
A=1. 5×1 . 1m2
To find:
heat loss
Solution:
Given that
11
UNIT II HEAT AND MASS TRANSFER
 h=3 .05 ( T f ) 1/4 ,
T f = Mean flilm temperature
T w +T ∞ 215+25
Tf= = =120 o C
2 2
Propertics of water at 120oC[From HMT data book,page No.33]
ρ= 0.898 kg/m3
Pr= 0.686
k =0.03338 W/m.K
υ= 25.45¿ 10-6 m2/s
1 1
β= =
T f 120+273
¿ 2 .54×10−3 K−1
heat lossQ=hA ΔT
=( 3 . 05 ( T f )1/4 ) 1 .5×1. 1× (215−25 )
¿ ( 3 . 05 ( 120 ) ) 1. 5×1. 1×( 215−25 )
1/4

¿ 28 . 68 KW

(ii).Explain the velocity and thermal boundary layer for flow over a horizontal flat plate.
[may/june 2014][April/May-2011][May/june-2009] (8)
Velocity boundary layer
 The convective heat transfer coefficient the key is the determination of the temperature
gradient in the fluid at the solid-fluid interface. The velocity gradient at the surface is also
involved in the determinations.
 This is done using the boundary layer concept to solve for u = f(y), T = f’′(y). The simplest
situation is the flow over a flat plate.
 The fluid enters with a uniform velocity of u∞ as shown in Fig. When fluid articles touch the
surface of the plate the velocity of these particles is reduced to zero due to viscous forces.
 These particles in turn retard the velocity in the next layer, but as these two are fluid layers,
the velocity is not reduced to zero in the next layer.
 This retardation process continues along the layers until at some distance y the scale of
retardation becomes negligible and the velocity of the fluid is very nearly the same as free
stream velocityu∞ at this level. The retardation is due to shear stresses along planes parallel to
the flow.

The value of y where velocity u = 0.99 u∞is called hydrodynamic boundary layer thickness
denoted by δ. The velocity profile in the boundary layer depicts the variation of u with y, through
the boundary layer. This is shown in Fig.

12
UNIT II HEAT AND MASS TRANSFER
 THERMAL BOUNDARY LAYER
 Velocity boundary layer automatically forms when a real fluid flows over a surface, but
thermal boundary layer will develop only when the fluid temperature is different from the
surface temperature.
 Considering the flow over a flat plate with fluid temperature of T∞ and surface temperature
Ts the temperature of the fluid is T∞ all over the flow till the fluid reaches the leading edge
of the surface. The fluid particles coming in contact with the surface is slowed down to zero
velocity and the fluid layer reaches equilibrium with the surface and reaches temperature Ts.
 These particles in turn heat up the next layer and temperature gradient develops. At a
distance y, the temperature gradient becomes negligibly small.
The distance y at which the ratio [(Ts – T)/(Ts – T∞)] = 0.99 is defined as thermal boundary
layer thickness δt. The flow can now be considered to consist of two regions. A thin layer of
thickness δt in which the temperature gradient is large and the remaining flow where the temperature
gradient is negligible.

3. Water at 60oC and a velocity of 2cm/sec flows over a 5m long flate plate Which is maintained at a
temperature of 20oC. Determine the total drag foce and the rate of heat transfer per unit width of the
entire plate.[may/june2013] (16)
Given Data:
13
UNIT II HEAT AND MASS TRANSFER
 T ∞=60 o C
Velocity , u=2cm /s
¿0 . 02 m/ s
x=5 m
T s =20o C ,
L=5 m
To find:
i. Drag force, FD
ii. Rate of heat transfer per unit width of the entire plate
Solution:
T +T ∞ 60+20
Tf= w = =40o C
Film temperature, 2 2
Propertics of water at 40oC[From HMT data book,page No.33]
ρ= 995 kg/m3
Pr= 4.34
k =0.628 W/m.K
υ= 0.657¿ 10-6 m2/s
uL
Re L=
Reynold’s Number, υ

0.657×10−6 ¿ 5
=0.02×25¿ 6 =1.52 ×10 ¿ ¿
0.1×10
= ¿
0.657
5
Since, Re<5¿ 10 ,∴ The flow is laminar.
C fL=1 . 328×Re
Average friction coefficient, L−0 . 5

−0 .5
=1 .328×( 1 .522×105 )
1 .328
¿
390 .128
¿3 .4×10−3
To find Drag force, FD
FD=Area ¿ Average shear stress

14
UNIT II HEAT AND MASS TRANSFER
 ρu2
=1×5×C fL
2
995 ( 0 .02 )2
¿ 5× ×3 . 4×10−3
2
1 .99
¿ 3 . 4×10−3 ×
2
¿ 0 . 995×3 . 43×10−3
−3
=¿ 3 . 41×10 N
Local heat transfer coefficient,h x :

Local Nusselt number,

Nu x=0 . 332 ( Re )0. 5 ( Pr )0 . 333
0.5
=0.332 ( 1. 522×10 5 ) ( 4 . 34 )0.333
¿0 .332×390.128×1 .6303
Nu x=211. 168
h L
Nu x= x
k
11. 16×0 .628
¿ =h x
5
h x =26 .52 W /m2 K
h=2×h x
¿ 2×26 . 52
2
=¿ 53 . 04 W /m K
To find Rate of heat transfer:
Q=hA ( T s −T ∞ )
=( 53. 04×5×( 333−293 ) )
Q=10608 W
Result:
−3
Drag force, FD=3 . 41×10 N
Rate of heat transfer: Q=10608 W

4. A horizontal pipe of 6m length and 8 cm diameter passes through a large room in which the air and
walls are at 18oC. Find the rate of heat loss from the pipe by natural convection. [may/june2013]
(16)
Given Data:
Length of the pipe, L=6m
Diameter of the pipe, d=8cm=0.08m
Surface temperature, T s =70+273=343 K
Fluid temperature, T ∞=18+273=291 K

15
UNIT II HEAT AND MASS TRANSFER
 To find: Q/L
Solution:
T s +T ∞
Tf=
Film temperature, 2
70+18 o
= =44 C
2
¿ 317 K
The properties of air at 44oC
ρ= 1.11 kg/m3
Pr= 0.6985
k =0.02791 W/m.K
υ= 17.45¿ 10-6 m2/s

α=25 . 014×10−6 m2 /s
1 1
β= = =0. 00315 K −1
T f 317
gβΔTd 3
Gr=
υ2
3
9 . 81×0 . 00315× (343−291 ) ( 0 .08 )
¿
( 17 . 455×10−6 )2
8 . 229×10−4 ×1012
¿
304 . 677
8 . 277×10 8
¿ =2. 7×106
304 . 677
Gr . Pr=2. 7×10 6 ×0 .6985
¿ 1 .88×106
For horizontal cylinder,

{ [( ( ) ) ] }
0.167 2
Gr D .Pr
NuD = 0.6+0.387
0.5625 0 .296
0.559
1+
Pr

10−5 <Gr D Pr <1012

{ [( ( ) ) ] }
6 0 .167 2
1.88×10
= 0.6+0.387
0 .5625 0.296
0.559
1+
0.6985

16
UNIT II HEAT AND MASS TRANSFER
 { ] }
0 .167 2
= 0 . 6+0 .387 [
1 . 88×106
1 .2058

={ 0 .6 +0 .387 [ 1559130 . 86 ]
0. 167 2
}
={ 0. 6+ 0. 387 ( 10 . 819 ) }
2

={ 0 . 6+4 .186953 }
2

hd
NuD =22 .916 ⇒ NuD =
k
22. 916×0 . 02791
h= =7 . 99
0 .08
Q=hA ( T s −T ∞ )
¿ 7 . 99×π ×0 .08×6×( 343−291 )
¿ 626 . 529W

5. Caster oil at 30oC flows at a velocity of 1.5m/s part a flat plate, in a certain process. If the plate is 4
m long and is maintained at a uniform temperature of 90oC, calculate the following:
i. The hydrodynamic and thermal boundary layer thicknesses on one side of the plate.
ii. The total drag force per unit width on one side of the plate’
iii. The local heat transfer coefficient at the trailing edge it is heart transfer rate; properties ofoil at
−8 2
60oC are ρ= 956.8 kg/m3;k =0.213 W/m.K; υ= 0.6¿ 10-4 m2/s;α =7 . 2×10 m / s [may/june2012]
(16)
Given data:
Length of the pipe, L=4m
Velocity, U =1.5m/s
o
Surface temperature, T s =30 C
To find:
i. The hydrodynamic and thermal boundary layer thicknesses on one side of the plate.
ii. The total drag force per unit width on one side of the plate’
iii. The local heat transfer coefficient at the trailing edge and
iv. The heart transfer rate;
Solution:
uL
Re L=
Reynold’s Number, υ

0.65×10−4 ¿
=1.5×4¿ 4
¿
=9.23×10 ¿
5
Since, Re<5¿ 10 ,∴ The flow is laminar.
For plate plate, laminar flow:

17
UNIT II HEAT AND MASS TRANSFER
 [From HMT data book,page No.112]
To find Hydrodynamic boundary layer thickness,δ hx

¿δhx=5×x×(Re)−0.5 ¿ ( 4)−0.5 =0.065m¿¿

=5×04× 9.23×10 ¿
To find Thermal boundary layer thickness,δ Tx
δ Tx=δ hx ×( Pr )−0. 333
¿ 0 . 065×( 902. 77 )−0 . 333
δ Tx=6 . 74×10−3 m [ υ 0 . 65×10−4
Pr= =
α 7. 2×10−8
=902 .77
]
To find total drag force on one side of the plate:
Average skin friction coefficient,
C fL=1 . 328 ( Re )−0 .5
−0.5
C fL=1 . 328 ( 9 .23×10 )
4

C fL=4 .37×10−3
We know
τ
C fL=
ρU 2
2
−3 τ
4 .37×10 =
956 . 8 ( 1 .5 )2
2
τ =4 . 70 N /m2
2
Average shear stress τ =4 . 70 N /m
Drag force, FD=Area¿ Average shear stress
= ( L×W )×4 . 70
= ( 4×1 )×4 .70 W=1 m
Drag force, FD=18.8N
To find heat transfer rate:
Local heat transfer co-efficient,h x
Local Nusselt number
Nux=0.332(Re)0.5(Pr)0.333

18
UNIT II HEAT AND MASS TRANSFER
 0. 5
=0. 332 ( 9 .23×10 4 ) ×( 902 .77 )0. 333
Nu x=972 . 6
h x ×L
Nu x=
k
h ×4
972 . 6= x
0 . 213
[ ∵ x=L=4 m ]
2
h x =51. 7 W /m K
Average heat transfer coefficient
h=2×h x
¿ 2×51 . 7
¿ 103 .58 W /m2 K
Q=hA ( T s −T ∞ )
¿ 103 .58×4×1×( 90−30 )
¿ 24 . 859 W
Result:
i. Hydrodynamic boundary layer thickness,δ hx =0 .065 m
−3
ii. Thermal boundary layer thickness,δ Tx=6 . 74×10 m
iii. Drag force FD=18.8N
iv. Heat transfer Q=24 .859 KW

6. (i).A vertical plate of 0.7 m wide and 1.2 m height maintained at a temperature of 90 oC in a room at
30oC. Calculate the convective heat loss.[may/june2012][April/May-2010](8)
Given Data:
Wide, W=0.7 m
Height or Length, L=1.2 m
Wall temperature, Tw=90oC
o
Room temperature, T ∞=30 C
To find:
Convective heat loss(Q).
Solution: Velocity (U) is not given. So, this is natural convection type problem.
T =T ∞
Tf= w
Flim temperature, 2
90+30
=
2
T f =60 o C
Propertics of water at 60oC
[From HMT data book,page No.33]

19
UNIT II HEAT AND MASS TRANSFER
 ρ= 1.060 kg/m3
Pr= 0.696
k =0.02896 W/m.K
υ= 18.97¿ 10-6 m2/s
1 1
β= =
T f 60+273
¿ 3×10−3 K−1
Grashof Number,
[From HMT data book,page No.134]
3
g× β×L × ΔT
Gr=
υ2
3
9 . 81×3×10−3 ×( 1. 2 ) ×( 90−30 )
¿
( 18 . 97×10−6 )2
Gr=8 . 4×109
Gr . Pr=8. 4×10 9 ×0 .696
Gr . Pr=5 . 9×109
9
Since Gr Pr >10 , flow is turbulent.
For turbulent flow,
Nusselt Number,
[From HMT data book,page No.135]
0. 333
Nu=0 . 10 ( Grpr )
¿0 . 10(5. 9×109 )0 .333
Nu=179 . 3
hL
Nu=
k
h×1. 2
179 .3=
0 . 02896
h=4 . 32 W /m2 K
2
Convective heat transfer coefficient h=4 . 32 W /m K
Q=hA ( ΔT )
=h×W ×L×( T w −T ∞ )
¿4 .32×0 .7×1. 2× ( 90−30 )
Heat loss , Q=218. 16 W
Result:
Convective heat loss, Q=218.16W

20
UNIT II HEAT AND MASS TRANSFER
 (ii).Calculate the heat transfer from 60 W incandescent bulb at 115 oC to ambient air at 25oC. Assume
the bulb as a sphere of 50 mm diameter. Also find the % of power lost by free
convection.[may/june2012] (8)
Given Data:
Assume bulb as a sphere, D=50 mm
surface temperature, Tw=115oC
o
Ambient air temperature, T ∞=25 C
To find:
Heat transfer
% of power loss.
T =T ∞
Tf= w
Solution:Flim temperature, 2
115+25
=
2
T f =70 o C

Propertics of water at 70oC

[From HMT data book,page No.33]
ρ= 1.060 kg/m 3

Pr= 0.694
k =0.02966 W/m.K
υ= 20.02¿ 10-6 m2/s
1 1
β= =
T f 70
¿ 2 .915×10−3 K −1
Grashof Number,
[From HMT data book,page No.134]
3
g× β×D ×ΔT
Gr=
υ2
3
9 . 81×2 . 915×10−3× ( 0. 050 ) ×( 383−298 )
¿
( 20 . 02×10−6)2
Gr=7 . 58×105
Gr . Pr=7 . 58×105 ×0 .694
Gr . Pr=5 . 26×105
[From HMT data book,page No.137]
0. 25
Nusselt Number, Nu=2+0 . 5 ( Gr . Pr )
Nu=15.46

21
UNIT II HEAT AND MASS TRANSFER
 hD
Nu=
k
hD
15 . 46=
k
h=9 .15 W /m2 K
Heat transfer, Q=
h A(T s −T ∞ )
Q=9. 15×4 πr 2 ( 383−298 )
Q= 6.1W
% of power lost by free conversion
Q
×100
60
6.1
= ×100
60
==10 . 18 %
Result:
Heat transfer,Q= 6.1W
% of power lost by free conversion
=10 . 18 %

7. Atmospheric air at 30oC flows with a velocity of 4 m/s over a 1.5 m long flat plate whose temperature is
130oC. Determine the average heat transfer coefficient and the rate of heat transfer for plate width of 1m.
[April/May-2011] (16)
Given Data:
Wide, W=1 m
Height or Length, L=1.5 m
Plate temperature, Tw=130oC
o
Fuild temperature, T ∞=30 C
Velocity, U=4 m/s
To find:
i. Avearge heat transfer coefficient, h
ii. Heat transfer,Q
Solution:
T +T ∞
Tf= w
Film temperature, 2
130+ 30
= =80o C
2
Propertics of water at 80oC
[From HMT data book,page No.33]
ρ= 1kg/m3
Pr= 0.692
k =0.03047 W/m.K
22
UNIT II HEAT AND MASS TRANSFER
 υ= 21.09¿ 10-6 m2/s
uL
Re L=
Reynold’s Number, υ

21.09×10−6 ¿
=4×1.5¿ 5
¿
=2.84×10 ¿
5
Since, Re<5¿ 10 ,∴ The flow is laminar.
For flat plate, laminar flow,
[From HMT data book,page No.112]

Local Nusselt Number,

Nu x=0 . 332 ( Re )0. 5 ( Pr )0 . 333
0 .5
=0.332 ( 2.84×10 ) ( 0.692 )
5 0 .333

Nu x=156 .51
hx L
Nu=
Local Nusselt Number , k
h x×1 . 5
156 . 51=
0 . 03047
h x =3. 179 W /m2 K
2
Local heat transfer coefficient,h x =3. 179 W /m K
Avearge heat transfer coefficient, h=2¿ h x
=2¿ 3 .179
h = 6.358W/m2K
Heat transfer, Q= h A(T w −T ∞ )
=h×W ×L×( T W −T ∞ )
¿6 . 358×1×1. 5×( 130−30 )
Q=953 .7 W
Result:
Heat transfer coefficient,h = 6.358W/m2K
Q=953 .7 W
Heat transfer,

8. A steam pipe 80 mm in diameter is covered with 30 mm thick layer of insulation which has a surface
emissivity of 0.94. The insulation surface temperature is 85 oC and the pipe is placed in atmospheric
air at 15oC. If the heat is lost both by radiation and free convection, find
i. The heat loss from 5 m length of the pipe.
ii. The overall heat transfer coefficient.
iii. Heat transfer coefficient due to radiation.[April/May-2011] (16)
Given Data:
Diameter of the pipe =80 mm
23
UNIT II HEAT AND MASS TRANSFER
 Insulation thickness =30 mm
Actual diameter of pipe, D=0.08+2(0.03)=0.14m
Emissivity,ε =094
Tube surface temperature, Tw=85oC
o
Air temperature, T ∞=15 C
To find:
i. The heat loss from 5 m length of the pipe,Q
ii. The overall heat transfer coefficient,ht
iii. Heat transfer coefficient due to radiation,hr.
Solution:
T +T ∞
Tf= w
Film temperature, 2
85+15
= =50 o C
2
Propertics of water at 50oC
[From HMT data book,page No.33]
ρ= 1.093kg/m 3

Pr= 0.698
k =0.02826 W/m.K
υ= 17.95¿ 10-6 m2/s
Coefficient of thermal expansion,
1 1
β= =
T f 50+273
¿ 3 . 095×10−3 K−1
Grashof Number,
[From HMT data book,page No.134]
g× β×D3 ×ΔT
Gr=
υ2
3
9 . 81×3 . 095×10−3× ( 0. 14 ) × ( 85−15 )
¿
( 17 . 95×10−6 )2
Gr=18 .10×106
Gr . Pr=18 . 10×10 6 ×0 .698
Gr . Pr=1 . 263×107
[From HMT data book,page No.137]
m
Nu=C (Gr . Pr )
Nusselt Number,
Gr . Pr=1 . 263×107
Corresponding C=0.125 and m=0.333

24
UNIT II HEAT AND MASS TRANSFER
 Nu=C (Gr .Pr )m
0 .333
¿0.125 ( 1.263×10 7 )
Nu=28.952
hD
Nu=
k
hD
28 . 952=
k
h×0 .14
28 . 952=
0 . 02826
h=5 .84 W /m2 K

Convective heat transfer coefficient. hc=5.84 W/m2K

Heat lost by convection,

Qcov=
h A(T s −T ∞ )
Q=5 . 84×π DL ( 85−15 )
=5 . 84×π×0 . 14×5×( 85−15 )
Qcov=898.99W
Heat lost by radiation,
Qrad =εσ A T [ w
4 −T 4
∞ ]
Where,
ε = Emissivity
σ =Stefen Boltzmann constant

=5 . 67×10−8 W /m2 K 4
A=Area-m2
Tw=Surface temperature,K
T ∞= Fluid temperature,K
Tw=85+273=358K T ∞= 15+273=288K
[
Qrad =εσ A T 4 −T
w ∞
4 ]
¿ ε ×σ ×π DL× T 4 −T[ w ∞4 ]
¿ 0 . 94×5 .67×10 ×π×0. 14×5× [ 3584 −288 4 ]
−8

Qrad =1118 .90 W

Total heat loss, Qt=Qconv+Qrad
=898.99+1118.9
Qt=2017.89W

25
UNIT II HEAT AND MASS TRANSFER
 Total heat transfer,Qt =h t AΔT
=ht ×π DL×( T w−T ∞ )
2017 . 89=ht ×π×0 . 14×5×( 85−15 )
ht =13. 108 W /m2 K
Radiative heat transfer coefficient,
h r =ht −h c
¿ 13 .108−5 .84
h r =7 .268 W /m2 K
Result:
i. The heat loss from 5 m length of the pipe,
a. By convection,Qcov =898.99W
b. By radiation,Qrad =1118 .90 W
2
ii. The overall heat transfer coefficient,ht =13. 108 W /m K
2
iii. Heat transfer coefficient due to radiation,h r =7 .268 W / m K

9. Air at 30oC flows over a flat plate at a velocity of 2 m/s. The plate is long and 1.5 m wide. Calculate
the following:
i. Hydrodynamic and thermal boundary layer thickness at the trailing edge of the plate
ii. Total drag force,
iii. Total mass flow rate through the boundary layer between x=40 cm and x= 85 cm. [April/May-
2010] (16)
Given:
o
Fluid temperature, T ∞=30 C
Velocity, U=2m/s
Length, L=2m
Wide, W=1.5m
To find:
i. Hydrodynamic boundary layer thickness,δ hx
ii. Thermal boundary layer thickness,δ Tx
iii.Total drag force
iv.Total mass flow rate through the boundary layer between x=40 cm and x= 85 cm
Solution:
Propertics of water at 30oC
[From HMT data book,page No.33]
ρ= 1.165kg/m3
Pr= 0.701
k =0.02675 W/m.K
υ= 16¿ 10-6 m2/s
26
UNIT II HEAT AND MASS TRANSFER
 We know that,
UL
Reynolds number, Re= v
2×2
=16 ×10
-6 [ L=2 m ]
Re=2. 5×105
5
Since Re<5¿ 10 , flow is laminar.
For flat plate, laminar flow,
[Refer HMT data book, page No. 112]
Hydrodynamic boundary layer thickness,δ hx

¿δhx=5×x×(Re)−0.5 ¿ ( 5)−0.5 =0.02m¿¿

=5×2× 2.5×10 ¿ x=L=2m

Thermal boundary layer thickness,δ Tx

δ Tx=δhx ×( Pr )−0. 333
¿ 0 . 02×( 0 . 701 )−0. 333
δ Tx=0 . 0225 m
Average friction coefficient,
C fL=1 . 328 ( Re )−0 .5
−0. 5
C fL=1 . 328 ( 2. 5×10 )
5

C fL=2 . 65×10−3
We know
τ
C fL=
ρU 2
2
τ
2 .65×10−3 =
1 .165 ( 2 )2
2
τ =4 . 70 N /m2
−3 2
Average shear stress τ =6 .1×10 N /m
Drag force, FD=Area¿ Average shear stress
= ( L×W )×6. 1×10−3
−3
¿ ( 2×1 . 5 )×6 . 1×10 W=2 m
Drag force on two sides of the plate, FD=0.018 ¿ 2=0.036N
Drag force, FD=0.036N
27
UNIT II HEAT AND MASS TRANSFER
 Total mass flow rate between x=40 cm and x=85 cm.
Hydrodynamic boundary layer thickness
δ hx=0. 85=5×x×( Re )−0 . 5

¿ 5×0. 85×
υ ( )
U ×x −0. 5

¿ 5×0. 85×
(
16×10−6 )
2×0 . 85 −0 .5
[ ∵ x=85 cm=0 . 85 m ]
δ hx=0. 85= 0.013m

δ hx=0. 40=5×x×( Re )−0. 5

¿ 5×0. 40×
υ ( )
U×x −0 .5

¿ 5×0. 40×
(
16×10−6 )
2×0 . 40 −0 . 5
[ ∵ x=40 cm=0 . 40 m ]
δ hx=0. 85=8 . 9×10−3 m
5
Δm= ρU [ δ hx=85−δ hx=40 ]
8
5
= ×1.165×2 [ 0.013−8.9×10 ]
−3
8
Δm=5.97×10−3 kg/s
Result:
Hydrodynamic boundary layer thickness,δ hx == 0 . 02m
i. Thermal boundary layer thickness,δ Tx=0 . 0225 m
ii. Drag force, FD=0.036N
iii. Total mass flow rate through the boundary layer between x=40 cm and x= 85 cm
Δm=5 .97×10−3 kg /s

10. Air at atmospheric pressure and 200°C flows over a plate with velocity of 5 m/s. The plate is 15 mm
wide and is maintained at a temperature of 120°C. Calculate the thickness of hydrodynamic and
thermal boundary layers and the local heat transfer co-efficient at a distance of 0.5 m from the
leading edge. Assume that the flow is on one side of the plate. ρ = 0.815 kg/m 3 ; µ= 24.5x 10-6 Ns/m2
: Pr= 0.7 K=0.0364W/mK [April/May-2004] (16)
Given:
o
Fluid temperature, T ∞=200 C
Velocity, U=5m/s
Wide of the plate, W=15mm=0.015m
Plate surface temperature=Tw=120oC
28
UNIT II HEAT AND MASS TRANSFER
 Distance, x=0.5m
ρ = 0.815 kg/m3
µ= 24.5x 10-6 Ns/m2
Pr= 0.7 K=0.0364W/mK
To find:
i. Hydrodynamic boundary layer thickness,δ hx
ii. Thermal boundary layer thickness,δ Tx
iii. Local heat transfer co-efficient,h x
Solution:
UL
Reynolds number, Re= v
5×0 . 5
= ν [ x=L=0 . 5 m ]
5×0 . 5
U=
μ
ρ
5×0 .5
=
24 .5×10−6
0. 815
Re=8 .32×104
5
Since Re<5¿ 10 , flow is laminar.
For flat plate, laminar flow,
[Refer HMT data book, page No. 112]

Hydrodynamic boundary layer thickness,δ hx

¿δhx=5×x×(Re) ¿ ( 4)−0.5 =8.6 7×10−3m¿¿

−0.5
=5×0.5× 8.32×10 ¿
Thermal boundary layer thickness,δ Tx
δ Tx=δhx ×( Pr )−0. 333
¿ 8 . 667×10−3× ( 0 .7 )−0. 333
δ Tx=9. 76×10−3 m

Local heat transfer co-efficient,h x

Local Nusselt number
Nux=0.332(Re)0.5(Pr)0.333

29
UNIT II HEAT AND MASS TRANSFER
 0.5
=0. 332 ( 8 .32×104 ) × ( 0. 7 )0. 333
Nu x=85 . 03
h x ×L
Nu x=
k
h ×0 . 5
85 . 03= x
0. 0364
[ ∵ x=L=0. 5 m]

h x =6 .19 W /m2 K
Result:
−3
i. Hydrodynamic boundary layer thickness,δ hx =8 .667×10 m
−3
ii. Thermal boundary layer thickness,δ Tx=9. 76×10 m
2
iii. Local heat transfer co-efficient,h x =6 .19 W /m K

11. Cylindrical cans of 150mm length and 65 mm diameter are to be cooled from an initial temperature
of 20oC by placing them in a cooler containing air at a temperature of 1 oC and a pressure of 1 bar.
Determine the cooling rates when the cans are kept in:
(i).horizontal position,
(ii).Vertical position. [May/june2009][April/May-2008] (16)
Given Data:
Diameter =65 mm
Length =150 mm
surface temperature, Tw=20oC
o
Air temperature, T ∞=1 C
To find:

Cooling rates
(i).horizontal position,
(ii).Vertical position.
Solution:
T w +T ∞
Tf=
Film temperature, 2
20+1
= =10 . 5o C
2

Propertics of water at 10.5oC

[From HMT data book,page No.33]
30
UNIT II HEAT AND MASS TRANSFER
 ρ= 1.247kg/m3
Pr= 0.705
k =0.02512 W/m.K
υ= 14.16¿ 10-6 m2/s
Coefficient of thermal expansion,
1 1
β= =
T f 10. 5+273
¿ 3 .527×10−3 K −1
Grashof Number,
[From HMT data book,page No.134]
3
g× β×D ×ΔT
Gr=
υ2
3
9 . 81×3 . 527×10−3× ( 0. 150 ) ×( 20−1 )
¿
( 14 . 16×10−6 )2
Gr=11. 06×10 6
Gr . Pr=11. 06×106 ×0 . 705
Gr . Pr=7 . 8×106
Since Pr<109, flow is laminar.
For vertical position:
[From HMT data book,page No.137]
Nu=0 . 59 ( Gr . Pr )0. 25
Nusselt Number,
Gr . Pr=7 . 8×106
0 . 25
=0.59 ( 7.8×10 6 )
Nu=31.179
hD
Nu=
k
hD
31 .179=
k
h×0. 15
31 .179=
0 .02512
h=5 .22 W /m2 K

heat transfer coefficient. h=5.22 W/m2K

Heat transfer,

Q=hA (T s −T ∞ )
31
UNIT II HEAT AND MASS TRANSFER
 Q=5 . 22×π DL ( 20−1 )
=5 . 822×π ×0 .065×0 .15×( 20−1 )
Q =3.037W
For horizontal position:
Grashof Number,
[From HMT data book,page No.134]
3
g× β×D ×ΔT
Gr=
υ2
9 . 81×3 . 527×10−3× ( 0. 65 )3 ×( 20−1 )
¿
( 14 . 16×10−6 )2
Gr=9×106
Gr . Pr=9×106 ×0. 705
Gr . Pr=6. 347×105
[From HMT data book,page No.137]
m
Nu=C (Gr . Pr )
Nusselt Number,
Gr . Pr=6. 347×105
Corresponding C=0.48 and m=0.25
Nu=C (Gr .Pr )m
0 .25
¿0.48 ( 6.347×10 5 )
Nu=13.548
hD
Nu=
k
hD
13 .548=
k
h×0 .65
13 .548=
0 . 02512
h=5 .235 W /m2 K

Heat transfer,

Q=hA (T s −T ∞ )
Q=5 . 235×π DL ( 20−1 )
=5 . 235×π ×0 .065×0 . 15×( 20−1 )
Q =3.046W
o
12. Air at 20 C at atmospheric pressure flows over a flat plate at avelocity of 3.5 m/s. If the plate is 0.5
m wide and at 60oC, calculate the following at x=0.400 m.
a) Boundary layer thickness.
b) Local friction coefficient.
32
UNIT II HEAT AND MASS TRANSFER
 c) Average friction coefficient.
d) Shearing stress due to friction.
e) Thermal boundary layer thickness.
f) Local convective heat transfer coefficient.
g) Average convective heat transfer coefficient.
h) Rate of heat transfer by convection.
i) Total drag force on the plate.
j) Total mass flow rate through the boundary. [Nov/Dec-2008]
(16)
Given:
o
Fluid temperature, T ∞=20 C
Velocity, U=3.5m/s
Wide of the plate, W=0.5m
Plate surface temperature=Tw=60oC
Distance, x=0.4m
To find:
a) Boundary layer thickness,δ
b) Local friction coefficient,
C fx
C
c) Average friction coefficient, fL
d) Shearing stress due to friction,τ s
e) Thermal boundary layer thickness.δ Tx
f) Local convective heat transfer coefficient.h x
g) Average convective heat transfer coefficient,h
h) Rate of heat transfer by convection,Q
i) Total drag force on the plate.FD
j) Total mass flow rate through the boundary,m.
Solution:
T +T ∞
Tf= w
Film temperature, 2
60 +20
= =40o C
2

Propertics of water at 40oC

[From HMT data book,page No.33]
ρ= 1.128kg/m 3

Pr= 0.699
k =0.02756 W/m.K
υ= 16.96¿ 10-6 m2/s

33
UNIT II HEAT AND MASS TRANSFER
 UL
Reynolds number, Re= v
3. 5×0 . 4
= ν [ x=L=0 . 5 m ]
3 . 5×0. 4
=
16 . 96×10−6
Re=8 .25×104
5
Since Re<5¿ 10 , flow is laminar.
For flat plate, laminar flow,
[Refer HMT data book, page No. 112]

a) Boundary layer thickness or Hydrodynamic boundary layer thickness,δ hx

¿δhx=5×x×(Re) ¿ ( 4)−0.5 =6.96×10−3m¿¿

−0.5
=5×0.4× 8.25×10 ¿
b) Local friction coefficient,
C fx
C fx =0 .664 (Re )−0. 5
¿0 . 664(8 . 25×10 4 )−0 . 5
¿2 .31×10−3
C
c) Average friction coefficient, fL
C fL=1 . 328(Re )−0. 5
¿1 .328( 8. 25×104 )−0 . 5
¿4 .623×10−3
d) Shearing stress due to friction,τ s
τs
C fx =
ρU 2
2
ρU 2
τ s =C fx×
2
−3 1. 128×3. 52
¿ 2 .31×10 ×
2
¿ 0 . 0159 N /m2
e) Thermal boundary layer thickness.δ Tx
δ Tx=δhx ×( Pr )−0. 333
¿6 . 96×10−3 ×( 0 .699 )−0. 333
δ Tx=7 . 84×10−3 m

34
UNIT II HEAT AND MASS TRANSFER
 f) Local convective heat transfer coefficient.h x
Nux=0.332(Re)0.5(Pr)0.333
0. 5
=0. 332 ( 8 .25×10 4 ) ×( 0 .699 )0. 333
Nu x=84 . 7
h x ×L
Nu x=
k
h ×0. 4
84 . 7= x
0 .02756 [ ∵ x=L=0. 4 m]
h x =5. 83 W /m2 K
g) Average convective heat transfer coefficient,h
h=2×h x
¿ 11. 66 W /m2 K
h) Rate of heat transfer by convection,Q

Heat transfer,

Q=hA (T s −T ∞ )
Q=11. 66×WL ( 20−1 )
=11. 66×0 .5×0 . 4×( 60−20 )
Q =93.28W
i) Total drag force on the plate. FD
τ
C fL=
ρU 2
Average Friction coefficient, 2
τ
−3
4 .623×10 =
1 .128 ( 3 . 5 )2
2
τ =0 . 0319 N /m2
2
Average shear stress τ =0 . 0319 N / m
Drag force, FD=Area¿ Average shear stress
= ( L×W )×0. 0319
= ( 0. 5×0. 4 )×0. 0319
=6 . 38×10−3 N
−3
Drag force on two sides of the plate, FD=6 . 38×10 ¿ 2=0.0127N
Drag force, FD=0.0127 N
j) Total mass flow rate through the boundary,m.

35
UNIT II HEAT AND MASS TRANSFER
 5
m= ρU [ δ2 X −δ 1 x ]
8
δ 1 x=0 ,δ2 x =δ x =6. 96×10−3 m
5
m= ×1 .128×3 .5×[ 6 .96×10−3 ]
8
m=0 . 017 kg/s
Result:
−3
a) Boundary layer thickness,δ =6 . 96×10 m
b) Local friction coefficient,
C fx =0. 0159 N /m2
C
c) Average friction coefficient, fL =4 . 623×10
−3

d) Shearing stress due to friction,τ s =0. 0159 N /m

2

−3
e) Thermal boundary layer thickness.δ Tx δ Tx=7 . 84×10 m
2
f) Local convective heat transfer coefficient.h x =5. 83 W /m K
2
g) Average convective heat transfer coefficient,h=11. 66 W /m K
Rate of heat transfer by convection,Q =93.28W
h) Total drag force on the plate.FD=0.0127 N
i) Total mass flow rate through the boundary,m=0 . 017 kg /s

13. A vertical pipe of 12 cm outer diameter, 2.5 m long, at a surface temperature of 120 oC is in a room
where the air is at 20oC. Calculate the heat loss per metre length of the pipe.[Nov/Dec-2008]
(16)
Given Data:
Diameter =12 cm
Length =2.5 mm
surface temperature, Tw=120oC
o
Air temperature, T ∞=20 C
To find:
heat loss
Solution:
the heat loss per metre length of the pipe
T +T ∞
Tf= w
Film temperature, 2
120+ 20
= =170 o C
2

Propertics of water at 70oC

[From HMT data book,page No.33]
36
UNIT II HEAT AND MASS TRANSFER
 ρ= 1.029kg/m3
Pr= 0.694
k =0.02966 W/m.K
υ= 20.02¿ 10-6 m2/s
Coefficient of thermal expansion,
1 1
β= =
T f 70+273
¿ 2 .91×10−3 K −1
Grashof Number,
[From HMT data book,page No.134]
3
g× β×D ×ΔT
Gr=
υ2
3
9 . 81×2 . 91×10−3 ×( 2 .5 ) × (120−20 )
¿
( 20 . 02×10−6)2
Gr=1 .11×1011
Gr . Pr=1 . 11×10 11×0 . 694
Gr . Pr=7 . 72×10 10
Since Pr<109, flow is laminar.
[From HMT data book,page No.137]
0. 333
Nu=0 . 10 ( Gr . Pr )
Nusselt Number,
Gr . Pr=7 . 72×10 10
0 .333
=0.1 ( 7.72×1010 )
Nu=422.3
hD
Nu=
k
hD
4 22. 3=
k
h×2 .5
4 22. 3=
0. 02966
h=5 .01 W /m2 K

heat transfer coefficient. h=5.01 W/m2K

Heat transfer,

Q=hA (T s −T ∞ )

37
UNIT II HEAT AND MASS TRANSFER
 Q=5 . 22×π DL ( 20−1 )
=5 . 01×π ×0 .12×1×( 120−20 )
Q =188.8W
14. Atmospheric air at 275 K and a free stream velocity of 20 m/s flows over flat plate 1.5 m long that is
maintained at a uniform temperature of 325 K. Calculate the average heat transfer coefficient over a
region where the boundary layer is laminar, the average heat transfer coefficient over the entire
length of the plate and the total heat transfer rate from the plate to the air over the length 1.5 m width
5
1 m. Assume transition occurs at Rec=2¿ 10 [May/June-2006](16)
Given Data:
Width,W =1 mm
Length =1.5 mm
Velocity, U=20 m/s
surface temperature, Tw=325K=52oC
T ∞=275 K=2o C
Air temperature,
Re c=2×105
Critical Reynolds number,
To find:
a) average heat transfer coefficient,hl
(boundary layer is laminar)
b) average heat transfer coefficient,ht
(Entrie length of the plate)
c) Total heat transfer rate, Q.
Solution:
T +T ∞
Tf= w
Film temperature, 2
52+2
= =27 o C
2

Propertics of water at 27oC

[From HMT data book,page No.33]
ρ= 1.189kg/m 3

Pr= 0.702
k =0.02634 W/m.K
υ= 15.53¿ 10-6 m2/s
Case( i)
UL
v
Reynolds number, Re=
5
Transition Occurs at Rec=2¿ 10

38
UNIT II HEAT AND MASS TRANSFER
 20×L
2×10 5=
(15 . 53×10−6 )
L=0.155m
For flat plate, laminar flow,

Local Nusselt Number,

Nu x=0 . 332 ( Re )0. 5 ( Pr )0 . 333

0. 5
¿ 0 . 332 ( 2×10 ) ( 0 . 702 )
5 0. 333

¿ 131 .97
h x ×L
Nu x=
k
h x ×0 .155
131 .97=
0 . 0634
h x =22. 42 W /m2 K
Average heat transfer coefficient for laminar flow, hl
hl =2×h x
¿ 2×22 . 42
¿ 44 .84 W /m2 K
Case(ii)
UL
ReL= v
Reynolds number,
20×1 .5
=
(15 .53×10−6 )
¿ 1 .93×106
5
Since ReL>5¿ 10
For flat plate, laminar-Turbulent combined flow,

Average Nusselt Number,

Nu x=( Pr )0 . 333 ( 0 . 037 ( ReL ) 0. 8 −871 )
0. 333
¿ ( 0 . 702 ) (0 . 037 ( 1 . 93×106 )0 . 8−871 )
¿ 2737 . 18

39
UNIT II HEAT AND MASS TRANSFER
 h x ×L
Nu x=
k
h t ×0 .155
2737 . 18=
0 . 02634
ht =48 . 06 W /m2 K
We know that,
Total heat transfer rate, Q
Q=hA ( T s −T ∞ )
¿ 48 . 06×1. 5×1×( 52−2 )
¿ 3604 . 5 W
Result:
2
a) average heat transfer coefficient,hl=44 . 84 W /m K
(boundary layer is laminar)
2
b) average heat transfer coefficient,ht =48 . 06 W /m K
(Entrie length of the plate)
c) Total heat transfer rate, Q=3604 .5 W

15(ii) Water at 60°C enters a tube of 2.54 mm diameter at a mean flow velocity of 2 cm/s.
Calculate the exit water temperature if the tube is 3 m long and the wall temperature is
constant at 80°C. [April/May 2015] (12)

Given data
T∞ = 60°C
Tw = 80°C
D=2.54 mm = 0.00254m
U = 0.02 m/s
L=3m
To Find exit water temperature
As bulk mean temp. is not given take the values for 60°C from data book
pg no. 21

ρ=985 kg /m3
ν=0 .478 x 10−6 m2 /s
Pr = 3.020
K=0.6513W/mK
We know that

40
UNIT II HEAT AND MASS TRANSFER
 UD
Re=
ν

0 . 02∗0. 00254
Re=
0 . 478 x 10−6 = 106.276
Re<2400 So the flow is laminar

Nu=3 . 66 From Data book page No.123 6th edition

h∗D
Nu=
k
h∗0 .00254
3 . 66=
0. 6513

h=237.9 W/m2K

Q=h∗A∗ΔT
Q=h∗π DL∗ΔT

Q=338 . 57∗( π∗0. 00254∗3 )∗(80−60)

Q=113 .8 W

We know that
Q=m∗C p∗ΔT Q=1∗4 .183∗(T out −T in )

113. 8=1∗4 .183∗(T out −60 )

T out =87 °C
Result
T out =87 °C
15. (i) A long 10 cm diameter steam pipe whose external surface temperature is 110°C passes
through some area that is not protected against the winds. Determine the rate of heat loss from
the pipe per unit length when air is at 1 atm and 10°C and the wind is blowing across the pipe
at a velocity of 8 m/s.
[Nov/Dec 2015] (8)
Given data
T∞ = 10°C
Tw = 110°C
D=10 cm = 0.1m
U = 8 m/s
41
UNIT II HEAT AND MASS TRANSFER
 To Find heat loss from the pipe per unit length
T +T ∞
Tf= w
2
110 +10
Tf= =60° C
2

From Data book page No. 33 6th edition

ρ=1. 060 kg /m3

ν=18.97 x 10−6 m2 / s
Pr = 0.696
K=0.02896W/mK
We know that
UD
Re=
ν

8∗0 . 1
Re=
18 .97 x 10−6 = 42171.8
Flow over cylinder
Nu=C *Re m *Pr 0 .333 From Data book page No. 115 6th edition

C=0.0266
m= 0.805

Nu=0 .0266∗42171 . 80.805∗0. 6960 .333

Nu=124.6

h∗D
Nu=
k
h∗0 .1
124 . 6=
0 . 02896

h=36.09 W/m2K

Q=h∗π DL∗ΔT

Q=36 . 09∗π∗0 . 1∗1∗(60−20 )

Q/ L=1133. 6 W /m
42
UNIT II HEAT AND MASS TRANSFER
 Result
Q/ L=1133. 6 W /m

15.(ii) An air stream at 0°C is flowing along a heated plate at 90°C at a speed of 75 m/s. The
plate is 45 cm long and 60cm wide. Calculate the average values of friction coefficient for the
full length of the plate. Also calculate the rate of energy dissipation from the plate.
[Nov/Dec 2015] (8)

Given data
T∞ = 0°C
Tw = 90°C
L=45 cm = 0.45m
W=60 cm = 0.6m
U = 75 m/s
To Find
1. Average friction coefficient
2. Rate of heat energy dissipated from the plate
T +T ∞
Tf= w
2
90+ 0
Tf= =45 ° C
2

From Data book page No. 33 6th edition

ρ=1. 1105kg/m3
ν=17.45 x 10−6 m2 /s
Pr = 0.6835
K=0.02791W/mK
We know that
UL
Re=
ν

75∗0. 45
Re=
17 . 45 x 10−6 = 1934097.42

Flow is turbulent

From Data book page No. 114 6th edition

Average friction coefficient

C fL=0. 074 (Re)−0. 2 −1742(Re)−1 . 0

43
UNIT II HEAT AND MASS TRANSFER
 C fL=0. 074 (1934097)−0. 2 −1742(1934097)−1 . 0

C fL=3 . 1913 x 10−3

Average Nusselt Number

Nu=(Pr ). 333∗[ 0 .037 (Re)0 . 8 −871 ]

Nu=(0 .6835 ). 333∗[ 0 . 037(1934097 . 42 )0. 8 −871 ]

Nu=2718 . 86
h∗L
Nu=
k
h∗0 . 45
2718 . 86=
0 .0291

h= 175.8 W/m2K

Q=h∗A∗ΔT Q=h∗0 . 6∗0 . 45∗(90−0 )

Q=4272 . 4 W

Result
−3
1. C fL=3 . 1913 x 10

2. Q=4272 . 4 W

16.(ii) A 6 m long section of 8 cm diameter horizontal hot water pipe passes through a large
room whose temperature is 20°C. If the outer surface temperature and emissivity of the pipe
are 70°C and 0.8 respectively, determine the rate of heat loss from the pipe by
(1) Natural convection
(2) Radiation [Nov/Dec 2015] (10)

Given data
T∞ = 20°C
Tw = 70°C
L=6m
44
UNIT II HEAT AND MASS TRANSFER
 D=8 cm = 0.08m
ε =0 . 8

To Find heat loss

1. Natural Convection
2. Radiation
T +T ∞
Tf= w
2
70+20
Tf= =45° C
2

From Data book page No. 33 6th edition

ρ=1. 1105kg/m3
ν=17.45 x 10−6 m2 /s
Pr = 0.6835
K=0.02791W/mK

1 1
β= β= =3 .1446 x 10−3 K −1
T f inK 318 K

We know that
g∗β∗L3∗ΔT
Gr=
ν2
9 .81∗3 .1446 x 10−3∗6 3∗50
Gr=
(17 . 45 x 10−6 )2
Gr=1.09 x 1012
Gr Pr=1.09 x 1012∗0.6835
Gr Pr=7.478 x 10 11
From Data book page No. 137 6th edition

Nu=C (Gr Pr )m
C = 0.125 ; m= 0.333
Nu=0.125(7.478 x 1011)0 .333
Nu=1124 .29
h∗D
Nu=
k
h∗0 .08
1124 .29=
0. 02791
45
UNIT II HEAT AND MASS TRANSFER
 h= 392.2 W/m2K
Q=h∗π DL∗ΔT
Q=392 . 2∗π∗0 .08∗6∗(50 )
Q=29556. 24 W

Heat loss By Radiation

Q=ε∗A∗σ [T 4w −T 4∞ ]

Q=0.8∗π∗0.08∗6∗5.67 x 10−8 [243 4 −293 4 ]

Q=893. 7 W

Result
1. Q=29556. 24 W Heat loss due to natural convection

2. Q=893. 07 W Heat loss due to Radiation

46
UNIT II HEAT AND MASS TRANSFER