Basics of Radiation Video Lecture | Heat Transfer - Mechanical Engineering

welcome to a lecture online now we're
still talking about heat transfer and so
far we've talked about conduction we
talked about convection now we're going
to talk about radiation and let's start
let's talk about the basics of what heat
transfer through radiation is so what
you need is you need an object that has
certain size and they're there for a
certain surface area because the
radiation does depend upon how much
surface there is for it to be able to
radiate out so objects radiates heat
through the surface away from the
surface so the amount of dq/dt the
amount of heat being taken away from an
object does depend will be proportional
to the area the surface area of the
object the balls hitting the flag here
the next thing is it also depends on the
temperature so not only does this object
have a surface area it also will be kept
at some temperature right here and
therefore the DQ DT will depend on the
inter the temperature but not just T it
will depend on T to the fourth power
which is kind of interesting because if
you have two identical objects though
with the very same surface area and one
object is twice as hot as the other then
because of this this relationship here
the amount of heat radiated by the
hotter one bill will be sixteen times as
much because if it's 2t instead of one T
two T to the fourth power will give you
sixteen T and therefore sixteen times as
much heat will be generated if two
similar objects are at different
temperatures let's say if two to simple
objects are let me start it over again
if if you have two objects that are
identical in size but let's say that
this one is at temperature T and this
one is a temperature three T well three
t to the fourth power
three to the fourth power is 81
that means this object will radiate
radiated out 81 times as much heat as
this one in again to finish the example
let's say that one object is at ten
times the temperature is the other one
you can see that ten to the fourth power
is 10,000 then this object will radiate
out ten thousand times
much heat is this one simply by it being
ten times as hot so you can see that the
temperature is a very important
component to determine how much heat is
being radiated out from a hot object so
now how do we turn this into an equation
how do we write DQ DT is equal to well
obviously you will need an A and a t to
the fourth power but what else do you
need in here well you need two things
you need a constant which is the
constant Sigma and then you need the
emissivity of the material so we'll look
at these things in just a moment first
of all let's start with the constant the
constant Sigma is equal to five point
six seven times ten to the minus eight
that would be watts per meter squared
and Kelvin to the fourth power so
instead of using centigrade degrees
we're going to go to Kelvin because for
this equation to work the temperature
must be expressed in Kelvin and of
course the size of degrees is the same
but sin it's an important reminder that
we plug in the temperature in Kelvin so
in other words let's say that this
temperature is at ten degrees centigrade
and this one here is a twenty degree
centigrade would you say that the
temperature here is twice the
temperature here and that's not the case
because when we convert that to Kelvin
this would then be a 283 Kelvin and this
would be a 293 Kelvin so you can see
that the temperature is not twice as
much so be careful not to make that
mistake
the second constant that we need here is
what we call the emissivity and the
emissivity is the is dependent on the
property of the material emissivity I
hope I got that spelled correctly all
right and usually that number is
somewhere between zero and one so object
that readily radiates out energy they
have an emissivity close to one and most
materials are indeed close to one when
an object doesn't emit radiation readily
then it has an emissivity much closer to
zero for example aluminum fall has an
emissivity of less than 0.1 in other
words if you put aluminum fall around an
object it is less likely to lose heat
through through
radiation so if you want to keep food
hot and you wrap in aluminum foil
it will keep hot a lot longer than if
you didn't wrap it or aluminum foil but
most other materials are very readily
capable of releasing energy through
radiation and their emissivity is much
closer to one matter of fact I think
most materials have an emissivity of 0.9
or greater all right knowing all that
let's now take a look at an example
let's say that we have an an object here
where the cross-sectional or not the
cross-section air but the surface area
is equal to 2 meters squared and let's
say that the temperature of this object
is equal to let's say 20 degrees
centigrade which is equal to 293 Kelvin
all right so let's plug that into our
equation here let's say that emissivity
is 0.9 so in this case let's say that
the emissivity is equal to 0.9 what
would be the amount of heat radiated
from this object so that is equal to 0.9
times Sigma which is five point six
seven times ten to the minus eight and
that would be what's per square meter
times Kelvin to the fourth power all
right now we have the cross-sectional
area of two meters squared so you can
see that meter square cancels out and a
temperature of 293 it's right here so
293 Kelvin and that would have to be
raised to the fourth power now you can
see of course Kelvin to the fourth we'll
get rid of Kelvin to the fourth over
there which leaves us with what's and if
we then calculate this we have 0.9 times
five point six seven e to the eighth
minus x two and x to 93 square square
again equals n this is an equal to 752
watts so an object at this temperature
which is room temperature would be
radiated radiating out energy at a rate
of 7 or 252 watts that's seven or
fifty-two joules per second so it's a
while wouldn't that mean that the object
would cool and very quickly lose all of
it
heat and the answer is no because at the
same time it would also be receiving
radiation heat from its surroundings and
if the amount of heat that it receives
equals the amount of heat that it
radiates out then it stays at the same
temperature now if it radiates out more
heat and it receives and the temperature
will cool down and then if it receives
more energy and it radiates out then the
temperature would go up and so that
gives you a kind of a good overview of
how to look at heat transfer through
radiation