# FME 521 Chapter 5

```CHAPTER 5 Natural convection heat transfer
Course outline
5.1 Introduction
5.2 Objectives
5.3 Free convection over flat vertical surfaces
5.3.1 Flow equations
5.3.2 Energy equations
5.4 Integral method for laminar flow
5.5 Turbulent flow
5.6 Combined free and forced convection.
5.7 Simplified correlations for air
5.8 Summary
5.1 Introduction:
Welcome to chapter 6. In the previous chapters in this unit, we have considered forced
convection heat transfer where the fluid motion is forced over a surface or through a
tube by an external device such as a fan or pump. In this chapter we consider heat
transfer by natural convection where fluid motion is due to differences in density of
the fluid as a result of a heating process. Practical examples where heat transfer by
natural convection occur include; cooling of transmission lines, electric transformers,
rectifiers, heating rooms by radiators, heat transfer from hot pipes and ovens.
We start by describing the physical mechanism of natural convection where the
characteristic parameters are identified. Then we discuss free convection on a heated
vertical plate by examining momentum and energy equations in laminar flow.
Relations in turbulent flow are briefly highlighted and finally combined free and
forced convection is briefly considered.
5.2 Objectives:
The objectives of this chapter are:
1) To describe the physical mechanism of Natural convection.
2) To examine and discuss the flow equations for a vertical plate in
laminar flow
3) To discuss briefly turbulent flow in free convection and the
combined free and forced convection
5.3 Physical mechanism in Free convection:
When a heated surface is in a fluid whose temperature is lower, the fluid adjacent to
the surface gets heated. This results in thermal expansion of the fluid and reduction in
density as compared to the fluid away from the surface. Because of the density
differences, a buoyancy force acts on the low density fluid causing it to flow up the
surface. As a result, hydrodynamic and thermal boundary layers are set up in the
neighbourhood of the surface. Since the flow velocity is developed here as a result of
1
temperature differential, the two boundary layers are of the same order irrespective of
the Prandtl number.
The fluid motion in natural convection however is characterized by low velocities
(typically less than 1 m/s) compared to forced convection. Hence, the convection heat
transfer coefficients in natural (free) convection are lower.
An important property in natural convection is the coefficient of thermal expansion of
the fluid, . The coefficient of thermal expansion is defined by,
1  v 
1   
=   =  
v  T  p   T  p
(5.1)
For ideal or perfect gases,
R
RT
1
 v 
and 
 = and therefore,  =
P
T
 T  p P
The temperature being absolute
v=
In natural convection, Grashof number is the dimensionless grouping that plays a
major role in natural convection just as Reynolds number plays in forced convection.
It incorporates the coefficient of thermal expansion β, in the expression.
In physical terms, the Grashof number represents the ratio of buoyancy force on the
fluid to viscous force. It is expressed as;
Gr =
g  2 (Tw − T ) L3
(5.2)
2
L – is the characteristic length
 - is the fluid density
 - is the dynamic viscosity of the fluid.
Tw – is the surface or wall temperature
T - is the temperature of the surrounding fluid.
For the flow of air over a flat plate, the critical Grashof number has been observed to
be approximately 4 × 108.
Free convection relationship is found to have the following dependence,
Nu = f (Gr , Pr)
In many cases, the relationship simplifies to the form
(5.3)
Nu = C (Gr  Pr) m .
(5.4)
A dimensionless group that is often used is the Rayleigh number,
Ra = Gr  Pr
(5.5)
The product is also a criterion of laminar or turbulent character of the flow as
determined by its values.
For laminar flow
104 < GrPr < 109.
For turbulent flow,
GrPr > 109.
2
The values of Gr and Pr are evaluated at the mean film temperature.
Tw + T
(5.6)
2
When average Nusselt number and therefore average convection coefficient is known,
the rate of heat transfer by natural convection from a solid surface at a uniform
temperature Tw to the surrounding fluid at T∞ is expressed by the Newton’s law of
cooling,
Tf =
Q = hAs (Tw − T ) ,
(5.7)
where h is the average heat transfer coefficient and A s is the heat transfer surface area.
5.4 Free convection over vertical flat surface.
Although we understand the mechanism of natural convection, the complexities of
fluid motion make it very difficult to obtain simple analytical relations for heat
transfer by solving the governing equations of motion and energy. The analytical
solutions that exist are obtained for simple geometries under simplifying assumptions
and lack generality and are normally for laminar flow. Turbulent flow and other
geometries are too complex and empirical and relations and experimental data are
used to estimate heat transfer rates. Therefore with exception of some simple cases,
heat transfer relations in natural convection are based on experimental studies. In this
section we briefly discuss flow over a vertical surface – a simple geometry.
Consider a flat vertical surface at a temperature higher than the surrounding fluid.
Because of this temperature difference, a natural convection boundary layer will be
formed on the surface. The figure below shows a convection boundary layer formed
on the surface.

Turbulent
GrPr > 109
Tw
T
u
FB
x
Laminar
104 < GrPr < 109
g
y
Figure 5.1 Natural convection on a vertical plate.
The velocity of the fluid at the surface is zero and increases to some maximum value
and then decreases to zero again at the edge of the boundary.
3
The initial boundary-layer development is laminar, but some distance from the
leading edge depending on the fluid properties and the temperature difference
between the plate and the surrounding, turbulent eddies are formed and transition to
turbulent boundary layer begins. Further up the surface the boundary layer becomes
fully turbulent.
5.4.1 Flow equations:
The continuity equation in the boundary layer is the same as that for forced
convection.
For the case of momentum equation, there is an additional body force due to gravity,
g. Hence, the momentum equation becomes,
 u
u 
p
 2u
+ v  = − − g +  2
y 
x
y
 x
 u
g – weight force on the element and the pressure gradient in the x- direction is
hydrostatic due to the height of the plate.
p
= −  g
x
 u
u 
 2u
+ v  = g (  −  ) +  2
Therefore,   u
(5.8)
y 
y
 x
Further, the density difference (-) may be expressed in terms of the volume
coefficient of expansion .
1   
1   −  


 = 
  T  p   T − T 
The momentum equation becomes,
 =−
 u
u 
 2u
+ v  =  2 +  g  (T − T )
(5.9)
y 
y
 x
The momentum equation is coupled to the energy equation in that the solution for
velocity profile demands knowledge of the temperature distribution.
 u
5.4.2 Energy equation:
The energy equation for natural convection is similar to that for forced convection,
especially at low velocities. Hence the energy equation still remains,
 dT
dT 
d 2T
cp  u
+v
=k 2
dy 
dy
 dx
1
For ideal gases,  =
T
(5.10)
4
The viscous dissipation has been neglected in the energy equation. The coupling of
the momentum and energy equations requires that both equations be solved
simultaneously.
Numerous solutions to laminar free convection boundary layer equations have been
obtained. Free convection from an isothermal vertical surface has received special
attention. In this case the boundary layer equations must be solved with boundary
conditions,
y = 0:
u=v=0
T = Tw
y→
u→0
T → T
A similarity solution to the foregoing problem was obtained by Ostrach, using
transformation of variables. A numerical solution gives local Nusselt number as a
function of x as,
1/ 4
 Gr 
(5.11)
Nu x =  x  f ( Pr )
 4 
The function of Prandtl number f(Pr) is determined from the results of the numerical
solution, as shown in the table 5.1.
Table 5.1 The function of Prandtl number
Pr
0.01
0.72
1
f(Pr)
0.081
0.505
0.567
2
0.716
10
1.169
100
2.191
1000
3.966
The average coefficient for surface of length L is then,
L
1
k  g  (Tw − T ) 
h =  hx dx = 

L0
L
4 2

1/ 4
L
f ( Pr ) 
0
dx
x1/ 4
(5.12)
Integrating,
1/ 4
hL 4  GrL 
Nu L =
= 

k
3 4 
For x = L,
NuL =
f ( Pr )
(5.13)
4
Nu L
3
(5.14)
The above equations are applicable whether the surface temperature is lower or higher
than the surrounding fluid. Further, the results may be used with reasonable accuracy
for constant heat flux as well as constant surface temperature.
5.4.3 Integral equations for laminar flow:
Free convection laminar flow on a vertical flat plate has also been solved by use of
integral equations. The momentum equation for flat vertical plate is derived as,
5



d 
2

u
dy
=
−

+
 g  (T − T )dy


w

dx  0
0


 du 
= −    +   g  (T − T ) dy
 dy  y =0 0
The energy equation is,
(5.15)

 dT 
d t
(T − T )udy =   

dx 0
 dy  y =0
(5.16)
Viscous work is neglected. For simplicity in analysis it will be assumed that the
thermal and velocity layers are virtually the same ( = t).
The boundary conditions are,
At y = 0. u = 0, T = Tw
At y = , u = 0, T = T,
And at y = 0,
dT du
=
=0
dy dy
d 2u
 T −T 
= −g  w  
2
dy
  
The dimensionless temperature profile is assumed to be in the form,
T − T  y 
= 1 − 
Tw − T   
2
(5.17)
The selection of the velocity profile is more difficult because an obvious characteristic
velocity for non-dimensionalizing does not exist. The velocity profile is however
assumed to be of the form,
u y y
= 1 − 
u*    
2
(5.18)
where u* is an arbitrary function with dimensions of velocity.
Substitution of the temperature and velocity profiles in the integral equations, (6.15),
(6.16) and with algebraic manipulation gives the boundary layer thickness as,

1/ 4
= 3.93Pr −1/ 2 ( 0.952 + Pr ) Grx−1/ 4
x
The heat transfer coefficient may then be evaluated from,
6
(5.19)
 dT 
qw = −kA 
 = hA (Tw − T )
 dy  y =0
h=
hx
2k
x
or x = 2  
k


(5.20)
Therefore the Nusselt number is given as,
−1/ 4
Nu x = 0.508 Pr1/ 2 ( 0.952 + Pr ) Grx1/ 4
(5.21)
The average Nusselt number for a plate of length L is then readily found as,
L
h=
1
4
hx dx = hL

L0
3
NuL =
hL
−1/ 4
= 0.678GrL1/ 4 Pr1/ 2 ( 0.952 + Pr )
k
NuL = 0.678Ra
1/ 4
L
 0.952 
1 +

Pr 

−1/ 4
(5.22)
For air, Pr = 0.7, the relations simplify to,
Nu x = 0.378Grx1/ 4 and NuL = 0.504GrL1/ 4
(5.23)
Properties are evaluated at mean film temperature, Eq (5.6).
5.5 Turbulence considerations:
Transition in a free convection boundary layer depends on the relative magnitude of
the buoyancy and viscous forces in the fluid. The transition is often correlated in
terms of Rayleigh number, which is the product of Grashof and Prandtl numbers. For
vertical plates, the critical Rayleigh number has been found to be,
Rax ,c = Grx ,c Pr =
g  (Tw − T  ) x 3

 109
(5.24)
For turbulent boundary layer flows, that is when Rayleigh number RaL  109, reliance
is placed on experimental results. Some of the correlations based on experimental data
include the following.
Nu = 0.68 +
0.67 Ra1/L 4
1 + ( 0.492 / Pr )9 /16 


-1
9
for 10 < RaL < 10 .
(5.25)
4/9
7
Nu1/ 2 = 0.825 +
0.387 Ra1/L 6
1 + ( 0.492 / Pr )6 /16 


(5.26)
8/ 27
For 10-1 < RaL < 1012 For the entire range.
For constant wall heat flux, an integral solution obtained similar to the laminar flow is
given to be,
Nu x = 0.295
9
For Grx > 10 ,
Pr 7 /15 Gr 2 / 5
(1 + 0.492 Pr )
2/3 2/5
(5.27)
For uniform wall heat flux, the recommended correlation for vertical plates is,
Nu x = 0.60 ( Grx* Pr )
1/ 5
(
Nu x = 0.17 Grx• Pr
)
1/ 4
for 105  Grx* Pr  1011 , laminar flow.
for 2 103  Grx* Pr  1016 , turbulent flow.
(5.28)
(5.29)
Gr = Grx Nu x .
*
x
When hx =
qw
g  x 4 qw
, then Grx* =
Tw − T
k 2
(5.30)
Properties evaluated at the mean film temperature.
Example 5.1:
A vertical metallic plate exposed to the sun, receives 600 W/m2 of sun’s energy. The
plate is 4m high and 2m wide. It is insulated from the back side, and is blankened on
the front side so that all incident radiation is absorbed by the plate, which in turn
losses heat to the surrounding air at 40°C by natural convection. Determine the
equilibrium temperature of the plate.
Soln.
This is a problem with constant heat flux qw.
To estimate film temperature and air properties, it is necessary to make an
approximation of the plate temperature.
First approximation:
An approximation value of h for free convection in air can be taken as 10 W/m 2 K, so
that we have approximately,
T =
q w 600
T
60

= 60 0 C. Hence, T f = T +
= 40 +
= 70 0 C .
h
10
2
2
Properties of air at 700C, are k = 0.0295 W/mK, ν = 2.005 × 10-5 m2/s, Pr = 0.7
1
=
= 2.92  10 −3 / K
Tf
8
Modified Grashof number, equation 6.30,
gq w x 4 9.81(2.92  10 −3 )(600)(4) 4
Grx• =
=
= 3.71 × 1014. Turbulent flow.
2
−5 2
k
(0.0295)(2.005  10 )
(
From equation 6.29, Nu x = 0.17 3.71  1014  0.7
h=
)
1/ 4
= 682.44
k
0.0295
Nu x =
 682.44 = 5.03 W/m2K.
x
4
Second Approximation:
q
600
119
T = w =
= 119 K, T f = 40 +
 100 0 C
h 5.03
2
Properties of air at 1000C,
k = 0.0318 W/mK
ν = 2.311 ×10-5 m2/s
Pr = 0.694
β = 1/373 = 2.681 × 10-3
Grx =
9.81(2.681  10 −3 )(600)(4) 4
= 2.379  1014
(0.0318)(2.311  10 −5 ) 2
0.0318
 (0.17)(2.379  1014  0.694)1 / 4 = 4.84 W/m2K
4
The change is small and the value of h may be accepted.
h=
q w 600
=
= 124 K.
h
4.84
Average plate temperature, Tw = 40 + 124 = 164°C.
Temperature difference, T = Tw − T =
Other Relations:
Vertical plate;
Lc = L,
Ra, 104 – 109
Ra, 10 – 10
9
13
Nu = 0.59 Ra 1L/ 4 .
Nu = 0.1Ra 1L/ 3
(5.31)
(5.32)
Inclined plate;
Lc = L,
Use the vertical plate equations for the upper surface of a cold
plate and the lower surface of a hot plate.
Replace g by gcosθ for Ra < 109. θ is the inclined angle from the normal.
Horizontal plates;
A
Lc = s , where surface area As and perimeter P.
P
Upper surface of a hot plate, Ra, 104 – 107 Nu = 0.54 Ra 1L/ 4
1/ 3
Ra, 107 – 1011 Nu = 0.15Ra L
9
(5.33)
(5.34)
Lower surface of a hot plate or upper surface of a cold plate;
Ra, 105 – 1011 Nu = 0.27 Ra 1L/ 4
(5.35)
Vertical cylinder; A vertical cylinder can be treated as a vertical plate when,
D
35L
.
GrL1 / 4
(5.36)
Horizontal cylinder;
Lc = D, and



Nu = 0.6 +



Ra d  1012 .
2


1/ 6

0.387 Ra d
.
8 / 27 
  0.559  9 / 16 

 
1 + 


  Pr  
Ra d  1011 ,
Sphere, Lc = D, and
Nu = 2 +
0.589 Ra 1D/ 4
  0.469  9 / 16 
 
1 + 
  Pr  
4/9
(5.37)
Pr  0.7
.
(5.38)
Example 5.2:
A 6m long section of an 8cm diameter horizontal hot-water pipe passes through a
large room whose temperature is 20°C. If the outer surface temperature of the pipe is
70°C, determine the rate of heat loss from the pipe by natural convection.
Soln.
Tw + T (70 + 20)
=
= 45 0 C .
2
2
Properties of air, k = 0.02699 W/m °C, Pr = 0.7241, ν = 1.750 × 10-5 m2/s.
1
1
=
=
.
T 318
Lc = d = 0.08m,
1
9.81 
 (70 − 20)  0.08 3 0.7241
3
g (Tw − T )d Pr
318
Ra d =
=
2
(1.750  10 −5 ) 2
= 1.867 × 106.
Mean film temperature, T f =
10
Nusselt number from equation 5.37,



Nu = 0.6 +



2


1/ 6

0.387 Ra d
= 17.39
8 / 27 
  0.559  9 / 16 

 
1 + 


  Pr  
k
0.02699
Nu =
 17.39 = 5.867 W/m 2  C .
d
0.08
As = dL =   0.08  6m = 1.508 m 2 .
h=
Heat Transfer, Q = hAs (Tw − T ) = 5.867  1.508  (70 − 20) = 442 W.
Question:
If the thermal emissivity of the pipe surface is assumed ε = 1, can you
show that the heat loss by radiation is 553 W. And it is more than that of
convection.
Activity:
Consider a 0.6m × 0.6m thin square plate maintained at a temperature of
90°C, while the other side is insulated. Determine the rate of heat transfer
from the plate by natural convection if the plate is a) vertical b)
horizontal with hot surface facing up and c) horizontal with hot surface
facing down.
[115 W, 128 W, 64.2 W].
Take Note:
The natural convection heat transfer is the lowest in the case of the hot
surface facing down. The hot air is trapped under the plate in this case
and cannot get away from the plate easily. As a result, the cooler air in
the vicinity of the plate will have difficulty reaching the plate, which
results in a reduced rate of heat transfer.
11
Activity:
Consider a thin 16cm long and 20cm wide horizontal plate suspended in
air at 20°C. The plate is equipped with electrical resistance heating
elements with a rating of 20W. Now the heater is turned on and the plate
temperature rises. Determine the temperature of the plate when steady
operating conditions are reached. The plate has an emissivity of 0.9 and
the surrounding surfaces are at 17°C.
Soln: (Complete this solution)
At steady operating conditions, the plate surface temperature Tw. Mean
T + T
film temperature, T f = w
.
2
Natural convection heat transfer from upper face – Qcon1.= h1 A1 (Tw − T )
Natural convection heat transfer from lower face - Qcon2. =
h2 A2 (Tw − T )
(
)
Total heat transfer, QT = Qcon1 + Qcon2 + Qrad = 20W
Take Note
To obtain air properties at Tf, plate temperature Tw must be known. But
this is what the question requires. We therefore take properties at an
assumed temperature say at T∞ and obtain an approximation temperature
of the plate Tw. We then use this to obtain improved Tf, obtain properties
and get a new Tw and so on until the differences in Tw are negligible.
This is iteration.
5.6 Combined free and forced convection:
Buoyancy has the effect of distorting the mean velocity and temperature profiles in
forced convection, which in turn affects the heat transfer coefficient. The relative
importance of free and forced convection is expressed as the ratio of body to inertia
forces as below.
Gr
 1 - Free convection predominates
Re 2
Gr
1 - forced and free convection important
Re 2
Gr
 1 - Forced convection predominates
Re 2
For vertical isothermal flat plate with both free and forced convection in the same
Grx
direction, neglecting free convection results in an error of about 5% or less for
Re 2
given in table 5.2.
12
Table 5.2 Effect of free convection.
Pr
100
10
0.24
0.13
Grx
2
Re x
For flow inside horizontal tubes and for
0.72
0.08
Gr
Re
0.03 – 0.003
0.056 – 0.05
1 , the recommended correlation is
4/3
Nud = 1.75 Gz + 0.012 ( GzGr1/ 3 ) 


1/ 3
 b 


 w 
0.14
(5.39)
For laminar flow. And,
D
Gz = Re Pr , - D = Pipe diamter, L = pipe length
L
Re, Gr, and Nu are all based on D – the pipe diameter, b and w refer to bulk
and wall temperatures respectively.
The correlation for turbulent flow is recommended to be
D
Nud = 4.69 Re0.27 Pr 0.21 Gr 0.07  
L
0.36
(5.40)
5.7 Simplified correlations for air:
Some simplified equations for free convection from various surfaces to air at
atmospheric pressure that have been found very useful are as follows.
Vertical plane or cylinder to air.
h = 0.95(T )
1/ 2
1/ 4
- Laminar (5.41)
- Turbulent.
Horizontal cylinder to air
 T 
h = 1.24

 D 
 T 
h = 1.42

 L 
(5.42)
 T 
h = 1.32

 D 
1/ 4
- Laminar
(5.43)
1/ 3
- Turbulent.
(5.44)
Horizontal plate to air:
Heated surface facing upward or cooled surface facing downward.
 T 
h = 1.32

 L 
1/ 4
- Laminar
(5.45)
13
h = 1.43(T )
1/ 2
- Turbulent
(5.46)
Heated facing downward or cooled surface facing upward
 T 
h=

 L 
1/ 5
(5.47)
Example 5.3:
a) The suction line of a refrigeration system has refrigerant 134a flowing inside a
copper tube, 1.25cm OD and 2m long. The temperature of the tube wall is 22°C. The surrounding air is at a temperature of 40°C. Calculate the heat gain
of the line by natural convection.
b) If air is blown across the line at a velocity of 5 m/s what will be the heat gain?
Soln:
a) Still air, (Natural convection)
40 − 22
= 90 C .
2
1
1
=
=
= 3.546  10 −3 / K
273 + 9 282
k = 0.02481 W/mK
Pr = 0.714
ν = 13.39 × 10-6 m2/s.
Tf =
Ra = GrPr =
g (T )D 3
2
Pr .
(9.81)(3.546  10 −3 )(0.0125) 3 (62)(0.714)
 10 9
(13.39  10 −6 ) 2
The boundary layer is therefore laminar. Applying equation 6.43,
=
 T 
h = 1.32

 D 
1/ 4
 62 
= 1.32

 0.0125 
1/ 4
= 11.1
Q = hAT = 11.1(  0.0125  2)(62) = 54.1W
Heat loss to still air,
b) Blowing air (Forced convection)
Reynolds number, instead of Grashof number, the criterion is,
Du m
(0.0125)(5)
= 4668 Laminar flow.

13.39  10 −6
For flow over tubes where Re = 4000 – 40,000,
Re d =
=
14
Nu = 0.693 Re 0.618 Pr 1 / 3
= 0.693(4668) 0.618 (0.714)1 / 3 = 31
31  0.02481
h=
= 61.5
0.0125
Heat loss to blowing air, Q = 61.5(  0.0125  2)(62) = 299W
5.8 Summary:
In this chapter we have considered natural convection heat transfer where
the fluid motion is caused by changes in density due to a heating process.
It is the resulting buoyancy force that is responsible for fluid movement.
The heat transfer coefficient (or Nusselt number) is a function of Grashof
number (Gr) and the Prandtl number (Pr). The Grashof number is a
dimensionless grouping that plays a major role in natural convection and
incorporates the coefficient of thermal expansion – an important property
in natural convection.
Free convection relationship is found to have the form, Nu = C (Gr  Pr) m
. Due to complexity of fluid motion, it is difficult to obtain simple
analytical relations. Hence most heat transfer relations are largely based
on experimental studies. However, for the simple geometry such as a
vertical plate and with simplifying assumptions, analytical heat transfer
relations have been presented; but in laminar flow. Turbulent flow
relations are based on experimental measurements. The velocities in free
convection are lower when compared to forced convection. Therefore the
heat transfer coefficient and thus heat transfer are lower in natural
convection.
Activity:
1. A vertical hot Oven door, 0.5m high is at 200C and is exposed to
atmosphere at 20C. Estimate the average heat transfer coefficient
at the surface of the door.
2. A 350mm long glass plate is hung vertically in the air at 24C
while its temperature is maintained at 80C. Calculate the
boundary layer thickness at the trailing edge of the plate. If a
similar plate is placed in a wind tunnel and air is blown over it at
a velocity of 5 m/s, find the boundary layer thickness at its
trailing edge. Also, determine the average heat transfer coefficient
for natural and forced convection for the above data.
3. Gas at 350C is conveyed through 250mm diameter pipe laid in
an atmosphere of quiescent air at 20C. The convection heat
transfer coefficient from a hot cylindrical surface freely exposed
1/ 4
 T 
to still air is h = 1.55 

 D 
(i)
Calculate the heat loss per m length of the bare pipe
15
(ii)
Estimate the percentage reduction in heat loss if the
pipe is covered with 75mm thick layer of material
whose thermal conductivity is 0.072 W/m C. Any
temperature drop in the metal may be neglected.
4. What heat load is generated in a restaurant by 1.0m  0.8m grill
which is maintained at 134C? The room temperature is 20C.
(hint: find the expression for Nusselt number for a horizontal
heated plate facing up)
5. An uninsulated 10.75 in OD steam line passes through a 9ft high
room in which the air is at 120F. What is the heat loss per foot of
pipe if its surface is at 760F when, (a) Horizontal, (b) vertical.
6. A glass-door firescreen, used to reduce exfiltration of room air
through a chimney, has a height of 0.71m and a width of 1.02m
and reaches a temperature of 232C. If the room temperature is
23C, estimate the convection heat rate from the fireplace to the
room.
7. A horizontal, high-pressure pipe of 0.1m outside diameter passes
through a large room whose walls and air temperature are 23C.
The pipe has an outside surface temperature of 165C. Estimate
the heat loss from the pipe per unit length.
8. Consider a large vertical plate with uniform surface temperature
of 130C suspended in quiescent air at 25C and atmospheric
pressure.
(a) Estimate the boundary layer thickness at a
location 0.25m measured from the lower edge.
(b) What is the maximum velocity in the boundary
layer at this location and what position in the
boundary layer does the maximum occur?
(c) At what location on the plate measured from the
lower edge will the boundary layer become
turbulent?
9. Air flow through a long, 0.2m square air conditioning duct
maintains the outer duct surface temperature at 10C. If the duct
is uninsulated and exposed to air at 35C in the crawl space
beneath a home, what is the heat gain to the duct per unit length
of the duct?
10. A flat plate 10cm wide and 18cm high at 250C is placed
vertically in still air at 1 atm and 20C. What is the total loss from
the two sides of the plate?
Because of changes in solar radiation, the total heat loss from the
plate may reach 100 W. At this loss and assuming uniform heat
flux, determine the variation of surface temperature and the
average plate temperature.
Natural convection over a flat plate Gr =
g  x 3 (Tw − T )
2
Average Nusselt number for constant wall temperature – laminar
flow
16
NuL
Pr1/ 2
=
GrL1/ 4 ( 2.435 + 4.884 Pr1/ 2 + 4.953Pr )1/ 4
Local Nusselt number for constant heat flux – laminar flow
Nu x
Pr 2 / 3
=
( Grx* ) ( 3.91 + 9.32 Pr1/ 2 + 9.95 Pr )1/ 5
Grx* = Grx Nu x
17
, where
```

– Cards

– Cards