Chapter 9: Natural
Convection
Dr Ali Jawarneh
Department of Mechanical Engineering
Hashemite University
Objectives
When you finish studying this chapter, you should be able to:
• Understand the physical mechanism of natural convection,
• Derive the governing equations of natural convection, and obtain the
dimensionless Grashof number by nondimensionalizing them,
• Evaluate the Nusselt number for natural convection associated with
vertical, horizontal, and inclined plates as well as cylinders and
spheres,
• Examine natural convection from finned surfaces, and determine the
optimum fin spacing,
• Analyze natural convection inside enclosures such as double-pane
windows, and
• Consider combined natural and forced convection, and assess the
relative importance of each mode.
•
•
•
•
Buoyancy forces are responsible for the fluid motion in natural
convection.
Viscous forces appose the fluid motion.
Buoyancy forces are expressed in terms of fluid temperature
differences through the volume expansion coefficient
β: a property that represents the variation of the density of a fluid with
temp. at constant pressure
1 ⎛ ∂ν ⎞
1 ⎛ ∂ρ ⎞
β= ⎜
⎟ = ⎜
ν ⎝ ∂T ⎠ P ρ ⎝ ∂T ⎟⎠ P
Viscous
Force
(1 K )
(9-3)
Buoyancy
Force
volume expansion coefficient β
• The volume expansion coefficient can be
expressed approximately by replacing
differential quantities by differences as
1 Δρ
1 ρ∞ − ρ
β ≈−
=−
ρ ΔT
ρ T∞ − T
( at constant P )
or ρ∞ − ρ = ρβ (T − T∞ ) ( at constant P )
• β
Table A-9 to A-14
• For ideal gas
βideal gas
1
=
T
(1/K )
(9-4)
(9-5)
(9-6)
Equation of Motion and the Grashof
Number
• Consider a vertical hot flat plate
immersed in a quiescent fluid body.
• Assumptions:
–
–
–
–
–
steady,
laminar,
two-dimensional,
Newtonian fluid, and
constant properties, except the density
difference ρ-ρ∞ (Boussinesq
approximation).
g
• Consider a differential volume element.
• Newton’s second law of motion
(9-7)
δ m ⋅ ax = Fx
g
δ m = ρ ( dx ⋅ dy ⋅1)
• The acceleration in the x-direction is
obtained by taking the total
differential of u(x, y)
du ∂u dx ∂u dy
ax =
=
+
dt ∂x dt ∂y dt
∂u
∂u
ax =u
+v
∂x
∂y
(9-8)
in
m
Zoo
• The net surface force acting in the x-direction
Net pressure force
Gravitational force
⎛ ∂τ ⎞
⎛ ∂P ⎞
Fx = ⎜ dy ⎟ ( dx ⋅1) − ⎜
dx ⎟ ( dy ⋅1) − ρ g ( dx ⋅ dy ⋅1)
⎝ ∂x ⎠
⎝ ∂y ⎠
⎛ ∂ 2u ∂P
⎞
= ⎜μ 2 −
− ρ g ⎟ ( dx ⋅ dy ⋅1) (9-9)
∂x
⎝ ∂y
⎠
Net viscous force
• Substituting Eqs. 9–8 and 9–9 into Eq. 9–7 and
dividing by ρ·dx·dy·1 gives the conservation of
momentum in the x-direction
⎛ ∂u
∂u ⎞
∂ 2u ∂P
ρ ⎜u + v ⎟ = μ 2 −
− ρg
∂y ⎠
∂y
∂x
⎝ ∂x
(9-10)
• The x-momentum equation in the quiescent fluid
outside the boundary layer (setting u=0)
∂P∞
(9-11)
= − ρ∞ g
∂x
• Noting that
– v<<u in the boundary layer and thus ∂v/ ∂x≈ ∂v/∂y ≈0, and
– there are no body forces (including gravity) in the ydirection,
the force balance in the y-direction is
∂P
∂P ∂P∞
=0
=
= − ρ∞ g
∂x
∂x
∂y
Substituting into Eq. 9–10
⎛ ∂u
∂u ⎞
∂ 2u
ρ ⎜ u + v ⎟ = μ 2 + ( ρ∞ − ρ ) g
∂y ⎠
∂y
⎝ ∂x
(9-12)
• Substituting Eq. 9-5 it into Eq. 9-12 and dividing both
sides by ρ gives
∂u
∂u
∂ 2u
(9-13)
u
+v
= ν 2 + g β (T − T∞ )
∂x
∂y
∂y
• The momentum equation involves the temperature,
and thus the momentum and energy equations must
be solved simultaneously.
• The set of three partial differential equations (the
continuity, momentum, and the energy equations) that
govern natural convection flow over vertical
isothermal plates can be reduced to a set of two
ordinary nonlinear differential equations by the
introduction of a similarity variable.
The Grashof Number
• The governing equations of natural convection
and the boundary conditions can be
nondimensionalized
T − T∞
x
y
u
v
*
*
*
*
;y =
;u = ;v = ;T =
x =
Lc
Lc
V
V
Ts − T∞
*
• Substituting into the momentum equation and
simplifying give
3
*
*
*
2 *
⎡
⎤
−
g
β
T
T
L
(
)
∂
∂
∂
u
u
T
u (9-14)
1
s
∞
c
*
*
+v
=⎢
u
⎥ 2 +
*
*
2
*2
∂x
∂y
ν
⎣
⎦ Re L Re L ∂y
GrL
• The dimensionless parameter in the brackets represents
the natural convection effects, and is called the Grashof
number GrL
g β (Ts − T∞ ) L3c
(9-15)
GrL =
2
ν
Buoyancy force
GrL=
Viscous force
• The flow regime in natural convection is
governed by the Grashof number
GrL>109 flow is turbulent
Viscous
force
Buoyancy
force
Natural Convection over Surfaces
• Natural convection heat transfer on a surface depends on
–
–
–
–
geometry,
orientation,
variation of temperature on the surface, and
thermophysical properties of the fluid.
• The simple empirical correlations for the average
Nusselt number in natural convection are of the form
hLc
n
Nu =
= C ⋅ ( GrL ⋅ Pr ) = C ⋅ RaLn (9-16)
k
• Where RaL is the Rayleigh number
g β (Ts − T∞ ) L3c
(9-17)
Pr
RaL = GrL ⋅ Pr =
2
ν
• The values of the constants C and n depend on the
geometry of the surface and the flow regime (which
depend on the Rayleigh number).
• All fluid properties are to be evaluated at the film
temperature Tf=(Ts+T∞).
Ts=const
• The Nusselt number relations for the constant surface
temperature and constant surface heat flux cases are
nearly identical.
• The relations for uniform heat flux is valid when the
plate midpoint temperature TL/2 is used for Ts in the
evaluation of the film temperature.
• Thus for uniform heat flux:
qs L
hL
=
Nu =
(9-27)
k
k TL 2 − T∞
(
)
Empirical correlations for Nuavg
The first two relations are very simple. Despite, we suggest
using the third one
1) Constant Surface temp.:
TL/2 :Iteration till
2) Constant Heat Flux:
Nu =
qs L
hL
=
k
k TL 2 − T∞
(
)
(9-27)
(9-21) and (9-27) match
An outer surface of a vertical cylinder can be treated
as a vertical plate when the diameter of the cylinder is
sufficiently large so that the curvature effects are
negligible. This condition is satisfied if:
Net Force: F=g (ρB- ρ)
{the difference between
the buoyancy and gravity}
Driving Force: Fx=g (ρB- ρ) cosθ
Use vertical plate equations for
the upper surface of a cold plate
and the lower surface of a hot
plate
Where the B.L appears
Hot Plate {Lower Surface}: The force that drives the
motion is reduced, we expect the convection currents to be
weaker, and the rate of heat transfer to be lower relative
to the vertical plate case.
Hot Plate {Upper Surface}: force component Fy initiates
upward motion in addition to the parallel motion along
the plate, and thus the boundary layer breaks up and
forms plumes. As a result, the thickness of the boundary
layer and thus the resistance to heat transfer decreases,
and the rate of heat transfer increases relative to the
vertical orientation.
Cold Plate: the opposite occurs as expected: The boundary
layer on the upper surface remains intact with weaker
boundary layer flow and thus lower rate of heat transfer, and
the boundary layer on the lower surface breaks apart (the
colder fluid falls down) and thus enhances heat transfer.
Lc= a/4 for a horizontal square
surface of length a, and D/4 for a
horizontal circular surface of
diameter D
-The rate of heat transfer to or from a horizontal surface
depends on whether the surface is facing upward or
downward.
-For a hot surface in a cooler environment, the net force
acts upward, forcing the heated fluid to rise.
-If the hot surface is facing upward, the heated fluid rises
freely, inducing strong natural convection currents
and thus effective heat transfer.
- If the hot surface is facing downward, the plate blocks the
heated fluid that tends to rise (except near the edges),
impeding heat transfer.
- The opposite is true for a cold plate in a warmer
environment since the net force (weight minus
buoyancy force) in this case acts downward, and the
cooled fluid near the plate tends to descend.
1)The B.L over a hot horizontal cylinder
starts to develop at the bottom, increasing
in thickness along the circumference, and
forming a rising plume at the top. So, the
local Nu is highest at the bottom, and
lowest at the top of the cylinder when the
B.L flow remains laminar.
2)The opposite is true in the case of a cold
horizontal cylinder in a warmer medium,
and the B.L in this case starts to develop
at the top of the cylinder and ending with
a descending plume at the bottom.
Discussion: Similar to Horizontal Cylinder
EXAMPLE: A 10-m-long section of a 6-cm-diameter
horizontal hot water pipe passes through a large room
whose temperature is 22°C. If the temperature and the
emissivity of the outer surface of the pipe are 65°C and
0.8, respectively, determine the rate of heat loss from the
pipe by (a) natural convection and (b) radiation.
Pipe
Ts = 65°C
ε = 0.8
Air
T∞ = 22°C
D = 6 cm
L=10 m
Properties: The properties of air at 1 atm and the film temperature of
(Ts+T∞)/2 = (65+22)/2 = 43.5°C are (Table A-15)
k = 0.02688 W/m.°C
υ = 1.735 × 10−5 m 2/s
Pr = 0.7245
β=
1
1
=
= 0.00316 K -1
T f (43.5 + 273)K
Analysis (a):
Lc = D = 0.06 m.
Ra =
gβ (Ts − T∞ ) D 3
υ2
Pr =
(9.81 m/s 2 )(0.00316 K -1 )(65 − 22 K )(0.06 m ) 3
⎧
0.387 Ra 1 / 6
⎪
Nu = ⎨0.6 +
⎪⎩
1 + (0.559 / Pr )9 / 16
[
(1.735 × 10 −5 m 2 /s ) 2
2
(0.7245) = 692,805
⎫
⎧
0.387(692,805)1 / 6
⎪
⎪
= ⎨0.6 +
8 / 27 ⎬
⎪⎭
⎪⎩
1 + (0.559 / 0.7245)9 / 16
]
[
2
⎫
⎪
= 13.15
8 / 27 ⎬
⎪⎭
]
k
0.02688 W/m.°C
Nu =
(13.15) = 5.893 W/m 2 .°C
D
0.06 m
As = πDL = π (0.06 m )(10 m ) = 1.885 m 2
h=
Q = hAs (Ts − T∞ ) = (5.893 W/m 2 .°C)(1.885 m 2 )(65 − 22)°C = 477.6 W
(b)
[
]
Q rad = εAs σ (Ts 4 − Tsurr 4 ) = (0.8)(1.885 m 2 )(5.67 × 10 −8 W/m 2 .K 4 ) (65 + 273 K ) 4 − (22 + 273 K ) 4 = 468.4 W
Natural Convection from Finned
Surfaces
• Natural convection flow through a channel formed by
two parallel plates is commonly encountered in
practice.
• Long Surface
– fully developed channel flow.
• Short surface or large spacing
– natural convection from two
independent plates in a quiescent
medium.
Natural Convection Cooling of Finned Surfaces (Ts=
constant)
Finned surfaces of various shapes, called heat sinks, are
frequently used in the cooling of electronic devices. Energy
dissipated by these devices is transferred to the heat sinks by
conduction and from the heat sinks to the ambient air by natural or
forced convection.
vertical finned surfaces of
rectangular shape
S: spacing between adjacent fins
L: the fin length
t: fin thickness
W: width of the base of heat sink
• The recommended relation for the average Nusselt
number for vertical isothermal parallel plates is
−0.5
hS ⎡
576
2.873 ⎤
Nu =
=⎢
+
⎥ (9-31)
2
0.5
k ⎢ ( Ras S L ) ( Ras S L ) ⎥
⎣
⎦
• Closely packed fins
– greater surface area
– smaller heat transfer coefficient.
• Widely spaced fins
– higher heat transfer coefficient
– smaller surface area.
• Optimum fin spacing for a vertical heat sink
⎛S L⎞
= 2.714 ⎜
⎟
⎝ Ras ⎠
3
Sopt
0.25
L
= 2.714 0.25 (9-32)
RaL
Heat Transfer from the Fin Surface:
Note: if t is considerable
Afin=2nLH+ntL
Number of fins on the heat sink:
All fluid properties are to be evaluated at the average
temperature Tavg = (Ts + TB)/2.
Natural Convection Cooling of Vertical PCBs
Arrays of printed circuit boards
used in electronic systems can often be
modeled as parallel plates subjected to
uniform heat flux
The modified Rayleigh number for
uniform heat flux:
The Nusselt number at the upper edge of the plate where
maximum temperature occurs is determined from [Bar-Cohen
and Rohsenow (1984)]
Heat Transfer from the Fin Surface:
For Afin see the previous note at Eq. (9-34)
EXAMPLE: A 12.1-cm-wide and 18-cm-high vertical
hot surface in 25°C air is to be cooled by a heat sink
with equally spaced fins of rectangular profile. The fins
are 0.1 cm thick and 18 cm long in the vertical
direction, and they have a height of 0.00362 m from the
base. Determine the rate of heat transfer by natural
convection from the heat sink if the base temperature
is 65°C.
W = 12.1
cm
H
L = 18
cm
S
65°C
T∞= 25°C
Properties : The properties of air at 1 atm and 1 atm and the film
temperature of (Ts+T∞)/2 = (65+25)/2 = 45°C
k = 0.02699 W/m.°C
υ = 1.749 × 10 −5 m 2 /s
Pr = 0.7241
1
1
β=
=
= 0.003145 K -1
(45 + 273)K
Tf
Analysis: The characteristic length in this case is the height of the surface
Lc = L = 0.18 m
Ra =
gβ (Ts − T∞ ) L3
υ2
Pr =
(9.81 m/s 2 )(0.003145 K -1 )(65 − 25 K )(0.18 m ) 3
(1.749 × 10 −5 m 2 /s ) 2
(0.7241) = 1.703 × 10 7
The optimum fin spacing is
S = 2.714
L
0.18 m
=
2
.
714
= 0.007605 m = 7.605 mm
Ra1 / 4
(1.703 × 107 )1 / 4
The heat transfer coefficient for this optimum fin spacing case is
h = 1.307
k
0.02699 W/m.°C
= 1.307
= 4.638 W/m 2 .°C
S
0.007605 m
n=
w
w
0.121
≅ =
≅ 16 fins
s + t s 0.007605
As = 2nLH = 2 × 16 × (0.18 m)(0.00362 m) = 0.02085 m 2
Then the rate of natural convection heat transfer becomes
Q = hAs (Ts − T∞ ) = (4.638 W/m 2 .°C)(0.02085 m 2 )(65 − 25)°C = 3.87 W
Natural Convection Inside Enclosures
• In a vertical enclosure, the fluid adjacent to the hotter
surface rises and the fluid adjacent to the
cooler one falls, setting off a rotationary
motion within the enclosure that enhances
heat transfer through the enclosure.
• Heat transfer through a horizontal enclosure
– hotter plate is at the top ─ no convection
currents (Nu=1).
– hotter plate is at the bottom
• Ra<1708 no convection currents (Nu=1).
• 3x105>Ra>1708 Bénard Cells.
• Ra>3x105 turbulent flow.
> The Rayleigh number for an enclosure is determined from
where the characteristic length Lc is the distance between the hot and
cold surfaces, and T1 and T2 are the temperatures of the hot and cold
surfaces, respectively.
All fluid properties are to be evaluated at
Tavg =(T1 + T2)/2
> Effective Thermal Conductivity
> When the Nu is known, the rate of heat
transfer through the enclosure can be
determined from
- Radiation must be considered in natural convection problems that involve
a gas. This is especially the case for surfaces with high emissivities.
- Radiation heat transfer from a surface at temperature Ts surrounded
by surfaces at a temperature Tsurr (both in K) is determined from
- Radiation heat transfer between two large parallel plates at
temperatures T1 and T2 is expressed
EXAMPLE: Two concentric spheres of diameters 15 cm
and 25 cm are separated by air at 1 atm pressure. The
surface temperatures of the two spheres enclosing the air
are T1= 350 K and T2= 275 K, respectively. Determine the
rate of heat transfer from the inner sphere to the outer
sphere by natural convection.
D2 = 25 cm
T2 = 275 K
Lc =5 cm
D1 = 15 cm
T1 = 350 K
Properties : The properties of air at 1 atm and the average temp. of
(T1+T2)/2 = (350+275)/2 = 312.5 K = 39.5°C are (Table A-15)
k = 0.02658 W/m.°C
υ = 1.697 × 10 −5 m 2 /s
Pr = 0.7256
1
1
β=
=
= 0.003200 K -1
Tf
312.5 K
Analysis
D2 − D1 25 − 15
Lc =
=
= 5 cm.
2
2
Ra =
gβ (T1 − T2 ) L3c
υ2
Pr =
(9.81 m/s 2 )(0.003200 K -1 )(350 − 275 K )(0.05 m ) 3
(1.697 × 10 −5 m 2 /s ) 2
(0.7256) = 7.415 × 10 5
The effective thermal conductivity is
Geometric Factor
Fsph =
k eff
Lc
( Di D o ) ( Di − 7 / 5 + D o
4
Pr
⎛
⎞
= 0.74 k ⎜
⎟
⎝ 0.861 + Pr ⎠
1/ 4
−7 / 5 5
)
=
0.05 m
[(0.15 m)(0.25 m)]
[(0.15 m)
-7/5
+ (0.25 m)
]
-7/5 5
= 0.005900
( Fsph Ra ) 1 / 4
0.7256
⎛
⎞
= 0.74(0.02658 W/m.°C)⎜
⎟
⎝ 0.861 + 0.7256 ⎠
⎛D D
Q = k eff π ⎜⎜ i o
⎝ Lc
4
1/ 4
[(0.00590)(7.415 × 10 )]
5
1/ 4
= 1315 W/m.°C
⎞
⎡ (0.15 m )(0.25 m ) ⎤
⎟(Ti − To ) = (0.1315 W/m.°C)π ⎢
⎥ (350 − 275)K = 23.3 W
⎟
(
0
.
05
m
)
⎣
⎦
⎠
Combined Natural and Forced
Convection
• Heat transfer coefficients in forced convection are
typically much higher than in natural convection.
• The error involved in ignoring natural convection may
be considerable at low velocities.
• Nusselt Number:
– Forced convection (flat plate, laminar flow):
Nuforced convection ∝ Re
12
– Natural convection (vertical plate, laminar flow):
Nunatural convection ∝ Gr
14
• Therefore, the parameter Gr/Re2 represents the
importance of natural convection relative to forced
convection.
• Gr/Re2<0.1
– natural convection is negligible.
• Gr/Re2>10
– forced convection is negligible.
• 0.1<Gr/Re2<10
hot isothermal vertical plate
– forced and natural convection are not negligible.
• Natural convection may help or hurt forced convection
heat transfer
depending on the
relative directions
of buoyancy-induced
and the forced
convection motions.
Nusselt Number for Combined Natural
and Forced Convection
• A review of experimental data suggests a
Nusselt number correlation of the form
(
Nucombined = Nu
n
forced
± Nu
n
natural
)
1n
(9-66)
• Nuforced and Nunatural are determined from the
correlations for pure forced and pure natural
convection, respectively.
• n usually between 3 and 4
EXAMPLE: Consider a 5-m-long vertical plate at
85°C in air at 30°C. Determine the forced motion
velocity above which natural convection heat
transfer from this plate is negligible.
Plate,
Ts = 85°C
L=5m
Air
T∞ = 30°C
V∞
Properties The properties of air at 1 atm and 1 atm and the film
temperature of (Ts+T∞)/2 = (85+30)/2 = 57.5°C are (Table A-15)
υ = 1.871× 10−5 m 2/s
β=
1
1
=
= 0.003026 K -1
T f (57.5 + 273)K
Analysis The characteristic length is the height of the plate, Lc = L
= 5 m. The Grashof and Reynolds numbers are
Gr =
Re =
gβ (Ts − T∞ ) L3
υ2
V∞ L
υ
=
=
(9.81 m/s 2 )(0.003026 K -1 )(85 − 30 K )(5 m ) 3
(1.871× 10 −5 m 2 /s ) 2
V∞ (5 m )
1.871× 10 −5 m 2 /s
= 5.829 × 1011
= 2.67 × 10 5 V∞
and the forced motion velocity above which natural convection
heat transfer from this plate is negligible is
Gr
Re
2
= 0.1 ⎯
⎯→
5.829 × 1011
5
(2.67 × 10 V∞ )
2
= 0.1 ⎯
⎯→ V∞ = 9.04 m/s