CH3 Convection PDF
CH3 Convection PDF
CH3 Convection PDF
CONVECTION
•• MEC
MEC551
551Thermal
ThermalEngineering
Engineering
1
Convection
Outlines for Chapter 3
Free Convection
Newton’s
law of
cooling
hLc
Nu Lc - Characteristic Length
k - Thermal conductivity of fluid
k
qconv h T h L
k T Nu
qcond L k
13
• MEC 551 Thermal Engineering
Convection
Convection boundary layer : velocity boundary layer
• Velocity boundary development on a flat plate:
u 0.99 u
constant.
Ts+0.99·(T-Ts)
• The thickness of the thermal boundary layer (dt) at any location along
the surface is defined as the distance from the surface at which:
dT=T-Ts=0.99·(T-Ts)
• Reynolds Number
– Derived by Osbourne Reynolds (1842-1912) of
Britain
– The transition from laminar to turbulent flow
depends on the surface geometry, surface
roughness, free stream velocity, surface
temperature, and type of fluid (among other
things).
22
• MEC 551 Thermal Engineering
Convection
Convection Boundary layer : Reynolds Number
Inertia Forces V L V L
Re
Viscous Forces
V – upstream velocity
L – characteristic length
= / – kinematic viscosity of fluid
Inertia Forces V L V L
Re
Viscous Forces
Flat Plate:
u xcrit
Recrit 5 105
where: xcrit= Distance between the leading edge of the plate to the
transition point from laminar to turbulent flow takes place.
dx
u
u dy u dx
x
v 27
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
u v
u dy v dx u dx dy v dy dx
x y
u v
u dy v dx u dy dx dy v dx dx dy
x y
u v
0 ~ 2-D Continuity Equation
x y
• MEC 551 Thermal Engineering
28
Convection
Forced Convection in Laminar Flow : External Flow
• Conservation of Momentum
From newton’s second law, dy
Σm·a = Net Force
y
P
Mass X Pressure, P P dx
Acceleration
dy x
dx
u u P 2u 2u
u v 2 2 gx
x y x x
Net Body force
y
ax Net effect of viscous per unit
pressure volume
force and shear forces
• In the y-direction:
v v P 2v 2v
u v 2 2 g y
x
x y y Body force
y
ay Net Net effect of viscous per unit
pressure and shear forces volume
force
30
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
• Conservation of Energy
T T T T
2 2 u v
2 2
u v
2
C p u v k 2 2 2
x y x y x y y x
T T 2T 2T
C p u v k 2 2
x y x y
32
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate
T, u Boundary layer
y dy
dx
x
u(x,0)= 0
v(x,0)= 0
T(x,0)= Ts
u v
Continuity: 0
x y
u u 2u
Momentum: u v 2
x y y
T T 2T
Energy: u v 2
x y y
34
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate
• Boundary conditions:
– At x= 0: u(0,y)= u, T(0,y)= T
– At y= 0: u(x,0)= 0, v(x,0)= 0, T(x,0)= Ts
u
y
x
• Dependent variable:
f
x u y
u
u
36
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate
• Therefore:
x df u df
u u u
y y u d x d
x df u
v u f
x x u dx 2 u x
1 u df
f
2 x d
37
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate
• So:
u u d 2 f
2
x 2 x d
u u 2
d f
u 2
y x d
2u u2 d 3 f
3
y 2
x
38
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – momentum equation
• Substituting these into the momentum equation and simplifying gives:
d3 f d2 f
2 3f 0
d d 2
39
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – momentum equation
• Using the definitions for f and η, the boundary equations
in terms of the similarity variables can be found.
f 0 0
• However, the transformed equation
df with its similarity variable cannot be
0 solved analytically.
d 0
• Therefore, an alternative solution is
df
1 necessary.
d
40
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – momentum equation
• The non-dimensional velocity profile can be obtained by
plotting u/u vs. η. The results agree experimentally.
df u
• A value of: 5.0 corresponds to: d u 0.992
41
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – momentum equation
• So substituting these values into the definition for η, gives the
boundary layer thickness for a flat plate.
5.0; y
u
y
x
u
5
x
5.0 5.0 x u x
For laminar
where : Re
flat plate:
u Re
x 42
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – energy equation
Energy Equation
• Knowing the velocity profile, we can now solve the energy
equation.
• Introduce dimensionless temperature:
T x, y Ts
x, y
T Ts
43
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – energy equation
• Substituting θ into the energy equation gives:
2
u v 2
x y y
• Again using the similarity variable, η, so θ= θ(η)
u
y
x
• So the energy equation becomes:
2
df d d 1 u df d d d d 2
u f
2
d d dx 2 x d d dy d dy
• MEC 551 Thermal Engineering
44
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – energy equation
• Since:
u
y
x
3
d 1 u 2 1 u
y y
dx 2 x 2 u x
d u
dy x
45
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – energy equation
• and:
f
x u y
u
u
df
d u y
46
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – energy equation
• Substituting these in gives:
2
df d d 1 u df d d d d
2
u f
2
d d dx 2 x d d dy d dy
d y u 1 u u d u
u
y
u y d 2u x 2 x x u y u d
y x
d 2 u
d 2 x
47
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – energy equation
1 d u u u 1 u u
2 d u x x x u x u y x
d 2 u
2
d x
d u 1 u x d 2
2
d u x x x xy u d 2
d x u u d 2
2
d u x u x u y d 2
Prandtl number d d 2
2 2
Pr d u y d
Pr
f
d 2 d
2 2 Pr f 0
d d
d
0.332 Pr 3
1
d 0
50
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – energy equation
• The temperature gradient at the surface is:
T
T Ts T Ts
y y 0
y y 0
0 y y 0
u u
• Since: y then:
x y x
hx
q s
k k 0.332 Pr T T
T
y y 0
1
3
s
u
x
Ts T Ts T Ts T
u
hx 0.332 Pr k
1
3
x
52
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – laminar flow
• The local Nusselt number is the dimensionless temperature gradient at
the surface. This is defined as:
hx x
Nu x
k
• Thus for Pr > 0.6, the local Nusselt number for laminar flow is:
Nu x 0.332 Pr Re
1 1
3 2
53
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – laminar flow
0.332 Pr 3 k
L 1 L
1 u
h hx dx
L0
h
L
0
x
dx
0.332 Pr 3 k u
1
L
2 x
L 0
0.664 Pr 3 k u L
1
L
0.664 k Pr Re
1 1
3 2
L
55
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – laminar flow
h L
Nu 0.664 Re L Pr 3
0.5 1
56
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – laminar flow
• Solving numerically for temperature profile
for different Prandtl numbers, and using the
definition of the thermal boundary layer, it
is determined that for laminar flow over a
flat plate:
13 13
t Pr Pr
1.026
57
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – laminar flow
• Example 3.1a Calculate the heat transfer and the
thermal boundary layer thickness ¼ of the way
along a flat plate that is 50 m long. Liquid (Tsat =
40 ºC) flows over it at 4 m/s. The plate is kept at a
surface temperature (Ts= 80 ºC).
Ts= 80ºC
40 ºC
4 m/s
y
x
50 m
• MEC 551 Thermal Engineering
58
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – laminar flow
40 ºC
y
80 ºC
x
50 m
• The first step is to calculate the mean film temperature of the fluid
flowing along the plate.
• This is just the average of the surface temperature and the fluid bulk
temperature.
Ts T 80C 40C
T film 60C
2 2
59
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – laminar flow
40 ºC
y
80 ºC
x
50 m
4.67 kg
m s
k 0.654 W
m C
Pr 2.99
• MEC 551 Thermal Engineering
60
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – laminar flow
• First calculate the Reynolds number to determine whether the flow is
laminar or turbulent.
u x 983.3 4 50 m
kg m 1
Re m3
10,527.8
s 4
4.67 kg
ms
5 x
5 504m
0.609 m
Re 10,527.8
t
Pr 13
0.609 m 2.99 13
0.412 m
1.026 1.026
61
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – laminar flow
u L 983.3 4 50 m
kg m
Re m3
42,111.3
s
4.67 kg
ms
62
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – laminar flow
u u
h 0.332 Pr k 0.332 Pr k
1 1
3 3
x x
0.619 W
m 2 C
63
• MEC 551 Thermal Engineering
Convection
Forced Convection in Laminar Flow : External Flow
Convection over a Flat Plate – laminar flow
• Using this h, we can now find the convection heat transfer:
q h (Ts T )
0.619 W
m 2 C
80C 40C
24.8 mW2
64
• MEC 551 Thermal Engineering
Convection
Forced Convection : External Flow
Convection over a Flat Plate – turbulent and mixed flow
0.6 Pr 60
Nu 0.037 Re Pr 0.8 1
3
L
5 105 Re 107
• For a mixed combination of laminar and turbulent flow over the plate:
Nu 0.037 Re 871 Pr 0.8
L 1
3
0.6 Pr 60
5 105 Re L 107
67
• MEC 551 Thermal Engineering
Convection
Forced Convection : External Flow
Convection over a Flat Plate – turbulent and mixed flow
• Example 3.2 Oil flows over a 40-m long heated plate at free
stream conditions of 5 m/s and 25ºC. If the plate is held at 45ºC.
Ts= 45ºC
u= 5 m/s
T= 25ºC
40 m
a) Determine the velocity and thermal boundary layer
thicknesses at the middle of the plate.
k 0.2864 W
mC
1,255 mkg3
69
• MEC 551 Thermal Engineering
Convection
Forced Convection : External Flow
Convection over a Flat Plate – turbulent and mixed flow
a) At the middle of the plate:
40 m
x 20 m
2
u x 5 ms 20 m
Re mid 2.86 10 5
Re mid Re crit
point
70
• MEC 551 Thermal Engineering
Convection
Forced Convection : External Flow
Convection over a Flat Plate – turbulent and mixed flow
• The hydrodynamic (or velocity) boundary layer is:
5 x 5 20 m
x 20 0.187 m or 18.7 cm
Re 2.86 105
1.026
3,711 0.0118 m or 11.8 mm
0.187 m 13
1.026
71
• MEC 551 Thermal Engineering
Convection
Forced Convection : External Flow
Convection over a Flat Plate – turbulent and mixed flow
b) At the end of the plate:
Re end
u L
5 ms 40 m
5.714 105
3.5 10 4 m2
s
Since Re > Recrit the flow is turbulent at the end
The critical distance (transition point from laminar to
turbulent is:
Re crit
xcrit
u
5 10 3.5 10
5 4 m2
35 m
s
5 ms
0.037 5.7110
L
5 0.8
871 3,711 3
1
9,600.7
Nu k
h
L
9,600.7 0.2864 W
68.7
mC W
m 2 C
40 m
73
• MEC 551 Thermal Engineering
Convection
Forced Convection : External Flow
Convection over a Flat Plate – turbulent and mixed flow
74
• MEC 551 Thermal Engineering
Convection
Forced Convection : External Flow
Convection over a cylinder or sphere
0.62 Re Pr Re
1 1 5 5
8
hD 2 3
0.3 1
Nucyl
1 0Pr.4 3
1
k 282,000
2 4
1
0.4
4
hD
Nu sph 2 0.4 Re 0.06 Re Pr
1 2
2 3
k s
76
• MEC 551 Thermal Engineering
Convection
Forced Convection : External Flow
Convection over a cylinder or sphere
Additionally the following empirical correlations have been made by
Zukauskas and Jakob for the average Nusselt number for flow over circular
and non-circular cylinders (Table 7-1 in text):
77
• MEC 551 Thermal Engineering
Convection
Forced Convection : External Flow
Convection over a cylinder or sphere
10 cm
1m
Q h As Ts T
k
h Nu
D
80
• MEC 551 Thermal Engineering
Convection
Forced Convection : External Flow
Convection over a cylinder or sphere
The properties of air at the average film temperature of:
Ts T 110C 10C
T film 60C
2 2
k 0.02808 W
m C ; Pr 0.7202
1.896 10 5 m 2
s
81
• MEC 551 Thermal Engineering
Convection
Forced Convection : External Flow
Convection over a cylinder or sphere
The Reynold’s number is:
Re
V D
8 ms 0.10 m
4.219 104
1.896 10 5 m 2
s
Nu 0.153 Re Pr
0.638 3
122.5
82
• MEC 551 Thermal Engineering
Convection
Forced Convection : External Flow
Convection over a cylinder or sphere
Therefore:
k
h Nu
D
0.02808 mWC
122.5 34.4 W
m C
2
0.10 m
The surface area of the hexagon is:
D
As 6 L
2 sin 60
3 0.10 m 1 m
sin 60
0.346 m 2
83
• MEC 551 Thermal Engineering
Convection
Forced Convection : External Flow
Convection over a cylinder or sphere
Therefore,
Q h As Ts T
34.4 W
m C
2 0.346 m 110C 10C
2
1,191.7 W
V2
T Ts Surface temperature
Free stream temperature
85
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Non-dimensionalized convection equations
Introducing these variables these equations become:
u * v*
Continuity: * 0
x y
*
u *
u *
1 2 *
u dP *
Momentum: u * * v* * *2 *
x y Re L y dx
Energy: T *
T *
1 2
T
u *
v *
*2
x *
y *
Re L Pr y
86
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Non-dimensionalized convection equations
• For a plate, the boundary conditions are:
u * 0, y * 1
v* x* ,0 0
T * 0, y * 1
u x ,0 0
* *
T x ,0 0
* *
u x , 1
* *
T x , 1
* *
u, T
y*
Ts
x*
87
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Non-dimensionalized convection equations
• Where:
V L
Re L Pr
• For a given geometry, the solutions of problems
with the same Re and Nu are similar, thus Re and
Nu are called similarity parameters.
• Since: *
u f1 x , y , Re L
* *
u V u * V
s
y
L y *
L
f 2 x* , Re L
y 0 y * 0
96
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Forced convection-Reynolds Analogy
2
Re L
f 2 x* , Re L
f 3 x* , Re L
k Ty
y 0
h
Ts T
98
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Forced convection-Reynolds Analogy
• Since:
T Ts y y
T *
y
*
Ts T L x
for local
• Then:
T Ts T T * Ts Ts T T *
*
y y x
*
x y
99
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Forced convection-Reynolds Analogy
Therefore:
k T Ts T *
h *
x Ts T y y * 0
k T *
*
L y y * 0
100
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Forced convection-Reynolds Analogy
• Substituting this into the local Nusselt number
equation gives:
h x x k T * T *
Nu x *
k k x y y*0 y * y*0
h
101
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Forced convection-Reynolds Analogy
• The average friction and heat transfer coefficients are
determined by integrating the local CF,x and Nux over the
surface of the given body with respect to x* (from 0 to 0.1),
which removes the dependence on x* and thus gives:
102
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Forced convection-Reynolds Analogy
Nu C Re Pr
m
L
n
– Where m and n are constant exponents (normally between 0 and 1), and the
value of C depends on geometry.
P *
0 (true when u = u = V = constant)
x*
For Pr = 1, the
thermal and velocity
boundary layers
coincide
Momentum: u *
u *
1 2 *
u
u *
v *
*2
x *
y *
Re L y
Energy: T *
T *
1 2 *
T
u *
v *
*2
x *
y *
Re L y
• Note: These two equations are exactly in the same form for
u* and T*.
105
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Forced convection-Reynolds Analogy
u x , 1
* *
T x , 1
* *
u * T * Equation
• Then: * *
y * y * 0
y y * 0
106
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Forced convection-Reynolds Analogy
• As previously derived:
2 u * k T *
CF * and h *
Re y L y
u * CF , x Re T * h L
and Nu x
y * y *0
2 y * y *0
k
107
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Forced convection-Reynolds Analogy
h Nu
St
CP V Re Pr
110
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Forced convection-Chilton-colburn Analogy
Pr = 1
P*
0
x *
111
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Forced convection-Chilton-colburn Analogy
• Recall as previously derived:
12
CF , x 0.664 Re Nu x 0.332 Pr Re x2
1 1
x and 3
2 C p V 112
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Forced convection-Chilton-colburn Analogy
• The Chilton-Colburn Analogy is derived using:
– Laminar flow
P
– Over a flat plate ( 0)
x
– Also the analogy above can be used for local or average quantities.
113
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Example 3.4
Laminar flow profile over a vertical Air Flow
T= 15ºC
plate. A 2 x 3 m plate is suspended in a V = 7 m/s
room and subject to air flow parallel to
its surfaces along its 3 m side. The total
drag force acting on the plate is 0.86 N.
Determine the average heat transfer
coefficient (h) for the plate: 3m
Ts=25C
2m
114
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Example 3.4
C F V C p = from table
h 2
2 Pr 3 = given
Ffriction D 12 CF As V2
Ffriction D CF As V
1
2
2
• Therefore:
2 D 2 0.86 N
CF 0.00243
2
As V 1.204 mkg3 12 m 2 7 s
m 2
117
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Example 3.4
• Then from the modified Reynold’s analogy
(Chilton-Colburn) the average heat transfer
coefficient (h) can be calculated:
C F V C p
h 2
2 Pr 3
0.00243 1.204
kg
m3
7 1007
m
s
J
kg C
2
2 0.7309 3
12.7 m 2WC
118
• MEC 551 Thermal Engineering
Convection
Principle of dynamic similarity
Internal flow
Internal flow relates to flow through fixed conduits
such as pipes or ducts.
m Ac Vm
m
Vm
Ac
121
• MEC 551 Thermal Engineering
Convection
Internal Flow – circular tubes
• In a circular tube:
D2
4 Ac 4 4
Dh D
p D
Vm D
Re
L= 2 mm u(y) Oil
k= 0.145 W/(m·K)
μ= 0.8 kg/(m·s)
L= 2 mm u(y) Oil
k= 0.145 W/(m·K)
μ= 0.8 kg/(m·s)
126
• MEC 551 Thermal Engineering
Convection
Internal Flow – Internal flow equation
• x-momentum equation:
0 0 0 0 0
u v u u P
2 2
u v 2 2 g x
x y x y x
u 2
0
y 2
u(0)= 0
u(L)= V= 12 m/s
0 C1 0 C2 V C1 L 0
C2 0 V
C1
L
128
• MEC 551 Thermal Engineering
Convection
Internal Flow – Internal flow equation
Therefore the equation becomes:
y
u V
L
129
• MEC 551 Thermal Engineering
Convection
Internal Flow – Internal flow equation
So the energy equation for this system is:
0 0 0
T T 2T 2T
C p u v k 2 2
x y x y
0 0 02
u v u v
2 2
2
x y y x
2
T u
2
0k
y 2
y
130
• MEC 551 Thermal Engineering
Convection
Internal Flow – Internal flow equation
• Since:
y
u V
L
u V
y L
• Therefore the equation becomes:
T
2
V
2
k 2
y L
y
2
T V C3 y C4
2k L
T(0) = T0 y 0 : T0 C4
L
2
T(L) = T0 y L : T0
V C3 L T0
2k L
C3 V2
2kL
132
• MEC 551 Thermal Engineering
Convection
Internal Flow – Internal flow equation
y 2
2 y 2
T 2 V V T0
2k L 2kL
V y y 2 2
T0 2
2k L L
133
• MEC 551 Thermal Engineering
Convection
Internal Flow – Internal flow equation
V 2 2
L L 2
2
Tmax T0 2
2k L L
V 2
T0
8k
20C
0.8 12 N s
m2
m 2
s 1W
N m 119C
8 0.145 mWC 1 s
135
• MEC 551 Thermal Engineering
Convection
Internal Flow – Internal flow equation
dT V 2 y
q0 k k 1 2
dy y 0
2kL L
V 2
0.8 12
N s
m2
m 2
s 1W
N m
2L 2 0.002 m 1 s
W
28,800 2
m
136
• MEC 551 Thermal Engineering
Convection
Internal Flow – Internal flow equation
V 2
0.8 12
N s
m2
m 2
s 1W
N m
2L 2 0.002 m 1 s
W
28,800 2
m
137
• MEC 551 Thermal Engineering
Convection
Internal Flow – Internal flow equation
• Discussion of example
T=20ºC Upper plate moving
V= 12 m/s
L= 2 mm T=119ºC
L= 2 mm
kW
q 28.8
m2
Heat flux is equivalent to the mechanical energy rate of dissipation.
Therefore, mechanical energy is being converted into thermal
energy to overcome friction in oil. This accounts for the
temperature flux.
139
• MEC 551 Thermal Engineering
Convection
Free convection
Warm air
Hot air rises due to the buoyancy
effect.
Cold
can Cold air
141
• MEC 551 Thermal Engineering
Convection
Free convection-volume expansion coefficient
In natural convection studies, the condition of the fluid
sufficiently far from the hot or cold surface is indicated by the
subscript “” to indicate that the presence of the surface is not
felt.
1 1
T T T
T T
142
• MEC 551 Thermal Engineering
Convection
Free convection-volume expansion coefficient
1 1 RTP 1
P
T P RT T T
143
• MEC 551 Thermal Engineering
Convection
Free convection-Grashof number
• The velocity and temperature for natural
convection over a vertical plate are shown in
the figure.
u u 2u P
u v 2 g
x y y x
P
g
x
145
• MEC 551 Thermal Engineering
Convection
Free convection-Grashof number
• Since:
P Px P x
P P
g
x x
• Then the momentum equation becomes:
u v 2u
u v 2 g
x y y
u v 2u
u v 2 g T T
x y y
u v 2u
u v 2 g T T
x y y
146
• MEC 551 Thermal Engineering
Convection
Free convection-Grashof number
* u
*
v * u
*
g Ts T Lc T
3
*
1 u 2 *
u 2 *2
x y Re L Re L y
* * 2
Grashof Number
147
• MEC 551 Thermal Engineering
Convection
Free convection-Grashof number
g Ts T L3c
GrL
2
148
• MEC 551 Thermal Engineering
Convection
Free convection-Grashof number
g Ts T L 3
Ra L c
Pr
2
150
• MEC 551 Thermal Engineering
Convection
Free convection- example 3.6
A 6-m long section of 8-cm diameter horizontal hot
water pipe passes through a large room. The pipe
surface temperature is 70 ºC. Determine the heat
loss from the pipe by natural convection.
Ts= 70 ºC T= 20 ºC
D= 8 cm
L= 6 m
????
It is a natural convection in pipe (cylinder)
Table – T film
k
h Nu
D g Ts T D 3
Ra D Pr
Table – natural conv. and cylinder 2
• Assume:
– Steady operating conditions
– Air is an ideal gas
– The local atmospheric pressure is 1 atm
Ts T 70C 20C
T film 45C
2 2
• From Table A-15, the properties of air are:
k 0.02699 W
mC ; 1.749 10 5 m2
sec ; Pr 0.7241
153
• MEC 551 Thermal Engineering
Convection
Free convection- example 3.6
154
• MEC 551 Thermal Engineering
Convection
Free convection- example 3.6
g Ts T D 3
Ra D Pr
2
9.81
m
s2
343 K 293 K 0.08 m 0.7241
1
318 K
3
1.749 10 5 m 2 2
s
1.869 10 6
155
• MEC 551 Thermal Engineering
Convection
Free convection- example 3.6
0.387 Ra D6
1
Nu D 0.60 8
0.559 16
9 27
1
Pr
2
0.60
0.387 1.869 106 6
1
17.4
8
0.559 16
9 27
1
0.7241
157
• MEC 551 Thermal Engineering
Convection
Free convection- example 3.6
• Then:
k
h Nu
0.02699 mWC
17.4 5.869 m2WC
D 0.08 m
As D L
0.08 m 6 m 1.508 m 2
158
• MEC 551 Thermal Engineering
Convection
Free convection- example 3.6
Q h As Ts T
5.869 W
mC
1.508 m 70C 20C
2
442.5 W
159
• MEC 551 Thermal Engineering