Free Convection Theory Video Lecture | Heat Transfer - Mechanical Engineering

natural convection or free convection in
case of natural convection or free
convection the heat transfer the heat
transfer rate is due to molecules which
gain very low velocity due to the
temperature difference the density
changes are very less the difference
between the molecules near the solid
wall and the surrounding fluid in case
of free convection no external agencies
use like pumps fan or blower which are
very commonly used in the forced
convection so thus the movement is not
taking place due to the external force
rather the motion of the fluid takes
place due to the buoyancy action in the
natural convection or free convection
the heat transfer the flow is classified
into two categories on the basis of
grashof number as laminar and turbulent
flow whereas in the forced convection we
can identify laminar or turbulent flow
on the basis of Reynolds number the
grashof number is defined as the ratio
of buoyant force to the viscous force
the expression for grashof number is
beta multiplied by G multiplied by theta
characteristic length Q divided by
kinematic viscosity square G is the
acceleration due to gravity and is fixed
value nine point eight one meter per
second square theta is called as excess
temperature is the difference between
the wall temperature and the surrounding
fluid temperature always take positive
value LC is a characteristic line and is
measured parallel to the boundary layer
as the fluid grows as the boundary layer
grows along the surface beta is called
coefficient of volume expansion for
gases the beta is equals to 1 upon T M
where TM should be in Kelvin or to add
273 and the mean temperature is TW plus
infinity but it Y to remember it must be
in Kelvin for liquids beta or to obtain
from the tables properties of fluid you
have to consider at mean temperature
that is TW plus T infinity W plus T
infinity divided by 2
normally a properties are given on page
number 34 your page number may change
and water properties are on page number
22 please check your tables
now how to use the correlations so I
will give you a brief idea about this
one you have to check the position
geometry and then select the table
normally data is available in two cases
one eg are into P R to the power 10 to
the power 9 and greater than 10 to the
power less than 10 to the power 9 and
greater than 10 to the power 9 in the
product of grashof number and the
prandtl number is sometimes called as
relays number so RA equal to gr to PR R
a is less than 10 to the power of 9 the
natural convection is considered as
laminar and if RA is greater than 10 to
the power 9 it is considered as
turbulent for vertical cylinder or a
vertical plate LC equals to L if where
RA is less than 10 to the power 9 the
table number is 1.1 if RA is greater
than 10 to the power 9 use table number
one point to point one for natural
convection table for horizontal cylinder
LC is taken as diameter d we have a
variety of correlations for different
ranges of arrays that is the product of
GR PR for this one use table number
three point one on page number 136
eighth please check your table numbers
and page number for horizontal plate we
have two cases the top plate is hot and
bottom plate is cold or bottom plate is
hot and cold upper bilities the top
plate is cold in this case LC is given
by a s that is the surface area of the
plate divided by a perimeter you have to
select table number two point one point
one which is on page number 136 a
horizontal pipe 30 point 5 cm in
diameter is maintained at temperature of
250 degree Celsius in a room where
ambient air is at 20 researchers
calculate convection heat loss per meter
length of pipe properties are given to
you in this numerical velocity is not
given so it's a problem of natural
convection check the position we have a
horizontal position and we have pipe it
means that we have a free convection
case our geometry is pipe and the pipe
is horizontal so we take a case of
horizontal cylinder we have to just
check out our Raleigh's number which is
the product of what gr and PR find out
gr and P are we required characteristic
length L
see now in this one we will see how the
bond dealer will grow we have a diameter
of cylinder equals to five length equals
two is not given so we'll assume one
meter diameter is 30 0.5 cm now this is
the molecules of steel air which is at
20 degree Celsius and right now the
cylinder is at 250 degree Celsius so
molecules near the wall will get heated
up and gain the velocity because they
become the lighter as compared to the
surrounding fluid now this boundary
layer will grow as we move from the
bottom to top so as compared to the
steel air the cylinder is hot and the
air is cold so there is zero boundary
layer at the bottom surface and the
boundary layer will grow along the
diameter so this color will indicate the
boundary layer growth over the solid
cylinder so you have to measure
dimension parallel to this that will be
called as characteristic length inside
this one there is a moment of molecule
and outside the molecules are stationary
so this region is called as boundary
layer so characteristic length will be
equal to LC parallel to boundary layer
so we have to first calculate grashof
number for this we collect the property
so mean temperature is T wall plus T
infinity divided by 2 that is 250 plus
20 divided by 2 which is 270 divided by
2 is 135 so to calculate beta as 1 upon
1 1 upon p.m. that should be in Kelvin
so I am adding here plus 273 to convert
degree Celsius into Kelvin so 1 upon 135
plus 273 will be 4:08 Kelvin it is for
calivita is called excess temperature is
T wall minus T infinity so we are 250
minus 20 degree Celsius is 230 degree
celsius so grashof number equals to beta
multiplied by G multiplied by theta
multiplied by LCQ divided by nu square
put all this value I will add just later
on first of all we'll start with G 9.81
it I equals to 230 and C is a diameter
which is equal to 0.3 0-5 cube of this
quantity now we will write the bitter
term here that equals to 4 0 8 so when
beta is adjusted as 1 upon 4 0 8 into
kinetic viscosity square 26.2 5 10 to
the power minus 6 square normally
grashof number is very large and used
you prefer to write in 10 to the power
something 9 or 11 so grashof number is
2.2 7 into 10 to the power 8 who select
a table we are to find out the product
of GR into PR then we can use the table
on 3.1 page number 138 every time I will
say the page number but as for your
Edition the page number may vary the
correlation was available in the form of
C multiplied by G R P R to the power M
you have to first check the range of G R
P R then you can select the
corresponding number C and corresponding
number M so if we check out the product
of G R into PR you will get a C equals
to 0.125
and M equals to 0.33 so you have to
write your correlation as 0.125 into G R
into P R power M that is 0.33 so stood
all this value G R as 2.27 10 to the
power 8 Prandtl number is point 6 a to
the power point 3 3 nusselt number come
out to be sixty six point seven but
nusselt number is HLC by K where LC you
know and K you know so H equals to sixty
six point seven divided by LC is point
three zero five K is point zero two
three four it transfer is five point one
one coefficient this values are very
small as compared to force convection
because the fluid is moving with a very
low velocity heat transfer is given by s
into multiplied by H multiplied by T
wall minus T infinity so surface area is
perimeter melted by length multiplied by
H into TW minus to infinity perimeter is
PI D so this is PI into point three zero
five we had to calculate per meter
length so we assumed L equals to one H
value we have five point one one wall
temperature is 250 and the fluid
temperature is 20 degree Celsius so heat
transfer is one one to seven point eight
that's a vertical pipe of 10 cm diameter
at the outer surface of 100 is kept in a
room at 20b celsius the pipe is three
meter long and you have to calculate the
heat loss from one meter so velocity is
not given so this is a numerical of free
convection the position of the pipe is
vertical
characteristic length will be equals to
Elsie but remember one thing you want to
calculate the heat transfer from only
one meter whereas the length is given as
3 meters diameter is 10 cm that equals 2
point 1 meter the total length is 3
meter the surface of cylinder is
maintained at 100 degree Celsius the
surrounding fluid is at a temperature of
20 degree Celsius so this blue dot
indicates the stationary air and which
is cold as compared to the wall of
cylinder so in this case the boundary
layer will grow from the bottom to top
at bottom the Monda layer thickness will
be equals to 0 and as we move along the
top surface the boundary layer will go
on increasing so inside this region
there is a movement of fluid molecules
so this line indicates the boundary
layer inside this one we have continuous
movement of the fluid particles that is
the air molecules due to temperature
difference outside the air is stationary
in case of natural convection the
characteristic length is measured
parallel to the boundary layer thickness
so LC will be same as 3 meters we are
mean temperature equals 2
kewal plus T infinity divided by 2 that
equals 200 plus 20 divided by 2 equal to
60 degree Celsius since the surrounding
fluid is air we have to calculate the
coefficient of thermal expansion that is
beta is given by 1 upon TM plus 273 must
be in Kelvin so 273 plus 60 is 1 by 3 33
when it is per Kelvin theta is called as
excess temperature is given by the
difference of hot fluid - hot - cold
these 100 - 20 that equals to 80 degree
Celsius so everything is known to us so
we can calculate grashof number reshef
number is given by beta multiplied by G
multiplied by theta multiplied by LC q
mica nematic viscosity square so put
this value j is 9.81 theta is 80 LC is 3
cubed when you had to calculate for 1
meter you have to calculate
LC you ought to put LC equal to 3 meter
kinetic viscosity is given to us is 18.9
7 10 to the power minus 6 and square of
this term so garage number is 1 point 7
6 10 to the power 11 let define the
problem we have a natural convection
a vertical cylinder now you to select
the correlation we have to first find
out gr PR now we are given the value of
G are equal to 1 point 7 6 10 to the
power 11 parental number is given as
point six nine six so the product will
come as 1 point 2 3 into 10 to the power
11 now if you go to the table you will
find that this number is more than 10 to
the power 9 so the flow is turbulent you
have to probably go to the table number
2 so go to the table number 2 point 1 as
per my data book page number is 137 6
most commonly used correlation is
mentioned there for vertical is given by
nusselt number number length is given by
0.1 GRP r2 power point 3 so can put 4 g
RP r so we get point 1 into 1 point 2 3
10 to the power 11 and hole power is
0.33 so nusselt number based on total
length that is 3 meter is equals to or
ninety three point one now in the case
of vertical plate if you see check your
table at the bottom you are given the
correlation to convert from anywhere to
a new bar that is a 4 calculate average
heat transfer coefficient so we have a
new bar equals to 5 by 4 nu for laminar
flow this ratio Li 4 by 3 and for
turbulent this ratio is 4 by 5 by 4 so
you have to first calculate what is the
average value of nusselt number using
this relation as 5 by 4 into 4 93.1 so
average nusselt number come out to be 6
1 6 point 4 now this average number
represents the average heat transfer
coefficient multiplied by LC divided by
K so average heat transfer will be
equals to 6 1 6 point 4 but and the
value of L is 3 meters don't put one
meter still here to calculate for 3
meters only so average heat transfer
coefficient over a length of 3 meter is
five point nine five watts per meter
square Kelvin now for calculation of
heat transfer for one meter will
consider
area of one meter now that is we will
take only 1 meter in so of 3 minutes so
we have a equation of it transfer by
convection as X multiplied by area
multiplied by T wall minus T infinity
the value of H is you have to use as
average heat transfer coefficient
and surface area is perimeter multiplied
by Lang and the difference is TW minus T
infinity perimeter you take for only
length you take for one meter only so we
have h equals to five point nine three
circumferential perimeter is PI into D D
is point one length we are interested
only for one meter so multiplied by 1 so
the wall temperature is 100 and T
infinity equal to 20 so all this
equation to obtain the value of Q final
answer is one 49.54 watts divided by k
is point zero two eight nine six a
rectangular plate point 2 meter by a
point four meter is maintained at a
uniform temperature of 80 it is placed
in atmosphere air at 24 compare the heat
transfer from the plate for the two
cases when the vertical height is point
two meter and other time is point for me
it is not given in the numerical hence
this represents the free convection
numerical the wall temperature is given
to us is 80 degree celsius fluid
temperature is 24 degree celsius so
excess temperature is TW minus T
infinity minus TW plus T infinity
divided by two so a plus 24 is zero is
52 degree Celsius surrounding fluid is
air so we have to take the coefficient
of thermal expansion as 1 upon p.m. plus
273 so is 1 upon 52 plus 273 is 325 per
Kelvin properties are not given so we
will collect the property of air at mean
temperature of 52 degree Celsius so you
can approximately take 250 degree
Celsius the property collected our first
is kinematic viscosity at 50 degree
Celsius is point 0 2 8 6 per meter
kelvin prandtl number is 0.698 animatic
viscosity required for grashof number
seventeen point nine five 10 to the
power minus 6 meter square per second
both cases we require the heat transfer
so initially will calculate the area
available is w into L is
point-to-multipoint 4 is point zero
eight 2 square so this time you have to
solve for two different cases like the
first case is vertical height is 0.2
meter in this time the Bond dealer will
grow parallel to the what
side so LC will be same as point two so
let's calculate grashof number using
beta multiplied by G multiplied by theta
multiplied by LC q divided by nu square
so put all this value and we can
calculate grashof number beta or to put
as 1 by 325 g is 9.81 ETA equals to 6 so
your number is come out to be four point
one nine 10 to the power 7 to use table
we first calculate GRP our Prandtl
number is given as 0.698 so we have an
answer of four point one nine 10 to the
power 7 multiplied by 0.698 GRP are come
out to be two point nine seven 10 to the
power 7 so this number is less than 10
to the power 9 so expected flow is
laminar now this define of a problem
very clearly away natural convection we
have a plate and the plate is vertical
and gr PR is less than 10 to the power 9
this table also we are given the two
equation one is annual and one is a new
bar the anybody question is been at the
bottom so take care of this problem so
let's calculate first a new held is
based on the total length that is LC so
model is constant wall temperature so
we'll use a new X equal to 0.5 0 8 P R
to the power 0.5 0.9 Phi 2 in plus P R
to the power minus 0.25 and gr X to the
power point two five so stood for PR
equal to 0.6 9 8 and gr 2 four point one
nine 10 to the power 7 all for local
nusselt number we are putting for X
equal to L so we will get n ul equals to
annual come out to be 30 now we have
equation of n UL and nu bar at the
bottom or table for laminar flow this
equation will be nu bar equals to 4 by 3
of anywhere so 4 by 3 multiplied by 30
30 mins 10 is answer is 40 now any bar
is defined as H bar multiplied by L C
divided by K we want to calculate the
average heat transfer coefficient over
the vertical length of 0.2 meter so
average value will be 40 multiplied by K
point 0 to H 6 and LC is point 2 so
average heat transfer name is 5 point 6
now we can calculate Q equal to surface
area multiplied by H bar x TW minus T
infinity surface area already calculated
is point 0 it is body is 5 point 6 7 or
temperature is 80 and the fluid timber
is 24 when the smaller side is kept
vertical the heat transfer come out to
be 50 point 8 acts right Oh repeat this
exercise for that size that equal to 0.4
so vertical height this time is 0.4
meter so naturally the Monda layer will
also grow parallel to the height LC will
be equal to 0.4 and other dimension is
0.23 to repeat for gr calculation use
the same value except L C equals 2.4 we
get G R equals to 0.35 10 to the power
calculate G R P are having same 3.35 10
to the power 8 multiplied by 0.6 9 8 so
this time also g RP r will come less
than 10 to the power 9 so you have to
use the same correlations so it is
slightly in the problem now because the
correlation we have selected is very
linear repeat this procedure
put all this value recalculate and check
your annual will be equals to is fifty
point four seven it is known that a new
bar will be equals to four by three of
any well so a new bar will be equals to
67 so we can calculate H bar equals to a
new bar multiplied by K divided by L C
so 67 multiplied by a point zero two
eight divided by LC LC is point four so
we get value of H bar equals to four
point seven five so we have a same
equation of Q that was given to us so we
our surface area multiplied by H bar
multiplied by T wall minus T infinity
surface area is point zero eight ivali
is 80 and 5024 you can substitute all
this is forty two point seven five