Mechanical Engineering Exam > Mechanical Engineering Videos > Topper Handwritten Notes & Videos for GATE ME > Newton's Law of Cooling HMT Tutorial- 24, Heat and Mass Transfer, Mechanical Engineering, GATE
Newton's Law of Cooling HMT Tutorial- 24, Heat and Mass Transfer, Mechanical Engineering, GATE Video Lecture | Topper Handwritten Notes & Videos for GATE ME
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FAQs on Newton's Law of Cooling HMT Tutorial- 24, Heat and Mass Transfer, Mechanical Engineering, GATE Video Lecture - Topper Handwritten Notes & Videos for GATE ME
| 1. What is Newton's Law of Cooling? |  |
Ans. Newton's Law of Cooling states that the rate of heat loss or transfer from a hot object to its surroundings is directly proportional to the temperature difference between the object and its surroundings. This can be mathematically expressed as q = hA(T - Ts), where q is the heat transfer rate, h is the heat transfer coefficient, A is the surface area, T is the temperature of the object, and Ts is the temperature of the surroundings.
| 2. How is Newton's Law of Cooling used in heat and mass transfer? | |
Ans. Newton's Law of Cooling is a fundamental concept used in heat and mass transfer to analyze and predict the rate of heat transfer between a hot object and its surroundings. It is commonly applied in various engineering fields, including mechanical engineering, to design and optimize heat exchangers, cooling systems, and thermal processes. By understanding the principles of Newton's Law of Cooling, engineers can determine the heat transfer coefficient, calculate the rate of heat transfer, and make informed decisions in thermal system design.
| 3. What are some practical applications of Newton's Law of Cooling in mechanical engineering? | |
Ans. Newton's Law of Cooling finds several practical applications in mechanical engineering. Some examples include:
- Designing and optimizing cooling systems for electronic devices and engines to prevent overheating.
- Calculating the heat transfer rate in heat exchangers, such as radiators and condensers, used in HVAC systems, power plants, and refrigeration systems.
- Analyzing the cooling performance of thermal insulation materials to enhance energy efficiency in buildings.
- Predicting the cooling rates of heated components to determine the required cooling time in manufacturing processes, such as heat treatment and casting.
| 4. How does Newton's Law of Cooling relate to the concept of convective heat transfer? | |
Ans. Newton's Law of Cooling is closely related to the concept of convective heat transfer. Convective heat transfer refers to the transfer of heat between a solid surface and a fluid (gas or liquid) in motion. In Newton's Law of Cooling, the heat transfer coefficient (h) represents the effectiveness of convective heat transfer. It takes into account the flow characteristics of the fluid, such as velocity, viscosity, and thermal conductivity, and determines how efficiently heat is transferred from the surface to the fluid. Therefore, Newton's Law of Cooling provides a quantitative approach to analyze convective heat transfer phenomena.
| 5. How can Newton's Law of Cooling be applied to analyze cooling processes in everyday life? | |
Ans. Newton's Law of Cooling can be applied to analyze various cooling processes in everyday life. Some examples include:
- Determining the time it takes for a hot beverage to cool down to a drinkable temperature.
- Calculating the cooling rate of food items in refrigerators or freezers.
- Evaluating the cooling performance of air conditioning systems in maintaining comfortable indoor temperatures.
- Predicting the rate at which a heated room cools down after the heating source is turned off.
By applying Newton's Law of Cooling, one can understand and optimize these cooling processes for improved comfort, energy efficiency, and safety.
Text Transcript from Video
well my dear the surgeon Kumar sane and
in this video thought well we are going
to learn about Newton's law of cooling I
hope you are aware about it and today
you know if this is one of the law which
is really important if you are studying
convection just like conduction for
years law claim oast important rule in
case of convection this Newton's law of
cooling really play most important Road
so you need to understand its basic
approach in this thought when I am going
to share with you what is exactly
Newton's law of cooling as well as they
will learn a numerical problem which has
been asked in gate previous year and
that question was I think for two and
two pepto marks that's not bad I think
so right
so let's go for it but basically what it
says it's say at Q is equal to as a t as
minus TF this H is heat transfer
coefficient heat transfer coefficient
and you need unit play most important
all the heat heat transfer coffees
annual days worth per meter square
centigrade or sometimes we have seen
what permitted square Kelvin so don't
get confused if you see defends in
Kelvin M centigrade because other units
will be in the respective way right area
a means area TS surface temperature
surface temperature TF is fluid
temperature free or maybe gas or maybe
liquid and Q is heat transfer rate which
is always in case of heat on state and
mass transfer it is always in Joule per
second or what don't get confused my
dear keep all these things clear that's
why you will do much battery will not
confused so this is all about Newton's
law of cooling I hope you are now
interested to learn that numerical
problem which has been asked indeed
previous a paper for four bucks
so let's go boys yeah here is the
question in front of you right now Holly
so this is this is seems to be really
lengthy question I have wall real deals
leave this question you can pause this
video you can read it the portion even
you can get all these coefficients from
Internet you can easily download but I
am NOT interested in that I am NOT
interested to the pass your time I'm
interested to share with you the
question how to solve it
we're having two positions each
quotients of two mugs and both of them
asking KP visa and good IP if I'm not
wrong
I think this question has been asked in
Kate 2003 I think so
right so here we got two questions these
are two positions first questions what
it said the value of temperature
gradient in the glass and the water
glass interface even Kelvin per meter
first question is that Kelvin per meter
that means we have to find out
temperature in gradient right yeah so
here what we have given right now we
have to see that we have given all
things we require I have already seen
the question so what is the formula for
finding out temperature gradient we know
that you know it's a topic of conduction
I mean we have to use for this law for
solving this problem
okay so we know that Q is equal to minus
K DT by DX suppose that this is for
class and again we get me right Q is
equal to K DT by DX this is for suppose
that water because we are having two
components here water as well as glass
right
so theta cancels because here a is equal
to one meter square we are cancelling
that now caking msre is cancel KK and
having different values right so you
have given conductivity of water is this
and conductivity of glass is this so
this is K for water this is a for glass
don't get confused here right now would
we have to find out water glass
interface that means we have to find out
DT by D X class because we have already
even so it is why this given was so we
can consider it
as why right so we are given what depend
what we have even their specific event M
pressure in the water at water class
interface
DT by D Y Z so DT by dy its water glass
interface so water W and it is even
wanted to tend to differ okay now what
we have to do we have to find out DT by
dy DZ so DT by dy into G's is equal to
ke w kill G okay kW per kg small G and
so kW by KZ into DT by dy its 4w okay
now k WK w is 0.6 and then k z DS k is
1.2 and so 1 into 1 10 to the power 4 so
1 6 punches gets cancel 10 to the 1st 4
by 2 so we can say 0.5 into 10 to the
power 4 unit will be ki per meter so
let's check the answer
yes it's C it's C so what important
things we have use to solve this
question let me share we do that thing
because probably I have gone a little
bit fast so let me share with you what
exactly we have be used to swap this
quotient if I will say nothing we have
used because you know all of you
probably all of the students those
stated in a mass transfer they are aware
of these two formulas I mean at least
one formula that gives four years law
because doesn't matter where from where
you have this phase region mass transfer
boo taught you but definitely he shared
with you the formula of Fourier slow
that is Q is equal to minus K V DP by DX
so
what I have used here to solve the
problem have you ever seen this
I hope so we have seen this so I haven't
use anything else other than this just
approach I have approached to solve the
problem because I have used few problems
related to this formula that's why I got
approached so most important thing is
that you have to build approach you have
to build concern that's why you can
solve the problem only knowing the
formula cannot help you at all you have
to practice more you will practice
better you will learn how to solve the
problem okay
that's why most our students lack here
they got very good marks in the
university examination but they unable
to get good marks in the complete
examination their numerical problems ask
most of the time so you need to put your
hole concentration in it if you were to
pay them for you get like examination I
hope you are getting me right so let's
go for the another question which is
under this question only I am interested
I am sure you are interested to learn
that question right so let's go for it
so here we got that question that is the
heat transfer coefficient adds in what
per meter square Kelvin but I have
already shared with you just let me
share with you what I have share with
you just beginning of the video what is
that yeah I have share with you da you
need up as in what per meter square
centigrade or what parameters for
careful because here all units even in
this quotient in Kelvin in Kelvin in
Kelvin that's why they have asked the
here in camping so they don't want to
get you confused to solve the question
they could ask him to centigrade okay
now the heat transfer coefficient as you
know what per meter square sending it
and what per meter square Kelvin so we
have to find out H the formula I have
one it is filled with unit once law of
cooling right so what is that q is equal
to h EI e s minus t
okay we know this formula I hope you
know this formula but but we have to
find out you first for this right q is
key w KW e DT by D Y into W key is what
for water for water K is 0.6 0.6 area I
have already shared with you that we are
considering 1 meters we got one set
meter square because they haven't
profile the area of the water and all
those things so 0.6 and 1 into 1 into 10
to the power 4 so it becomes 6,000 what
per meter now we already know Q is equal
to this month so we have to find out H
so H is equal to 6 thousand area is 1 TS
minus TF it is 1 so what we caught
adamantly 6,000 by TS minus TF ok we got
it
so remember this that's why you can
understand it properly
yeah now now what we have given in the
question that is temperature I mean
surface temperature and fluid
temperature fluid is what in Goudas you
would easier word let's see this water
water is 48 degree and glass base
surface is exposed to expose to the
water add in water a a broken glass is
40 degree and you know what it is 40
degree so we got here what we got here
t s is equal to t s is equal to 48
degree and DF is equal to 40 degree so
we'll have to put both the values here
6040 8 minus 14
so ultimately what we got here 6,000
divided by 8 so
if you do this eight eight seven four
five zero so seven five zero what per
meter square Kelvin Titov obviously them
to be like this because all units we
have used in what meter as or as Kelvin
so this is all about Newton's law of
cooling and related numerical problem
which for gram problem has been asked in
gate original paper I hope you learn
something amazing from this video right
so hopefully you enjoy the video thanks
for watching see you soon in my next
video bye bye
in this video thought well we are going
to learn about Newton's law of cooling I
hope you are aware about it and today
you know if this is one of the law which
is really important if you are studying
convection just like conduction for
years law claim oast important rule in
case of convection this Newton's law of
cooling really play most important Road
so you need to understand its basic
approach in this thought when I am going
to share with you what is exactly
Newton's law of cooling as well as they
will learn a numerical problem which has
been asked in gate previous year and
that question was I think for two and
two pepto marks that's not bad I think
so right
so let's go for it but basically what it
says it's say at Q is equal to as a t as
minus TF this H is heat transfer
coefficient heat transfer coefficient
and you need unit play most important
all the heat heat transfer coffees
annual days worth per meter square
centigrade or sometimes we have seen
what permitted square Kelvin so don't
get confused if you see defends in
Kelvin M centigrade because other units
will be in the respective way right area
a means area TS surface temperature
surface temperature TF is fluid
temperature free or maybe gas or maybe
liquid and Q is heat transfer rate which
is always in case of heat on state and
mass transfer it is always in Joule per
second or what don't get confused my
dear keep all these things clear that's
why you will do much battery will not
confused so this is all about Newton's
law of cooling I hope you are now
interested to learn that numerical
problem which has been asked indeed
previous a paper for four bucks
so let's go boys yeah here is the
question in front of you right now Holly
so this is this is seems to be really
lengthy question I have wall real deals
leave this question you can pause this
video you can read it the portion even
you can get all these coefficients from
Internet you can easily download but I
am NOT interested in that I am NOT
interested to the pass your time I'm
interested to share with you the
question how to solve it
we're having two positions each
quotients of two mugs and both of them
asking KP visa and good IP if I'm not
wrong
I think this question has been asked in
Kate 2003 I think so
right so here we got two questions these
are two positions first questions what
it said the value of temperature
gradient in the glass and the water
glass interface even Kelvin per meter
first question is that Kelvin per meter
that means we have to find out
temperature in gradient right yeah so
here what we have given right now we
have to see that we have given all
things we require I have already seen
the question so what is the formula for
finding out temperature gradient we know
that you know it's a topic of conduction
I mean we have to use for this law for
solving this problem
okay so we know that Q is equal to minus
K DT by DX suppose that this is for
class and again we get me right Q is
equal to K DT by DX this is for suppose
that water because we are having two
components here water as well as glass
right
so theta cancels because here a is equal
to one meter square we are cancelling
that now caking msre is cancel KK and
having different values right so you
have given conductivity of water is this
and conductivity of glass is this so
this is K for water this is a for glass
don't get confused here right now would
we have to find out water glass
interface that means we have to find out
DT by D X class because we have already
even so it is why this given was so we
can consider it
as why right so we are given what depend
what we have even their specific event M
pressure in the water at water class
interface
DT by D Y Z so DT by dy its water glass
interface so water W and it is even
wanted to tend to differ okay now what
we have to do we have to find out DT by
dy DZ so DT by dy into G's is equal to
ke w kill G okay kW per kg small G and
so kW by KZ into DT by dy its 4w okay
now k WK w is 0.6 and then k z DS k is
1.2 and so 1 into 1 10 to the power 4 so
1 6 punches gets cancel 10 to the 1st 4
by 2 so we can say 0.5 into 10 to the
power 4 unit will be ki per meter so
let's check the answer
yes it's C it's C so what important
things we have use to solve this
question let me share we do that thing
because probably I have gone a little
bit fast so let me share with you what
exactly we have be used to swap this
quotient if I will say nothing we have
used because you know all of you
probably all of the students those
stated in a mass transfer they are aware
of these two formulas I mean at least
one formula that gives four years law
because doesn't matter where from where
you have this phase region mass transfer
boo taught you but definitely he shared
with you the formula of Fourier slow
that is Q is equal to minus K V DP by DX
so
what I have used here to solve the
problem have you ever seen this
I hope so we have seen this so I haven't
use anything else other than this just
approach I have approached to solve the
problem because I have used few problems
related to this formula that's why I got
approached so most important thing is
that you have to build approach you have
to build concern that's why you can
solve the problem only knowing the
formula cannot help you at all you have
to practice more you will practice
better you will learn how to solve the
problem okay
that's why most our students lack here
they got very good marks in the
university examination but they unable
to get good marks in the complete
examination their numerical problems ask
most of the time so you need to put your
hole concentration in it if you were to
pay them for you get like examination I
hope you are getting me right so let's
go for the another question which is
under this question only I am interested
I am sure you are interested to learn
that question right so let's go for it
so here we got that question that is the
heat transfer coefficient adds in what
per meter square Kelvin but I have
already shared with you just let me
share with you what I have share with
you just beginning of the video what is
that yeah I have share with you da you
need up as in what per meter square
centigrade or what parameters for
careful because here all units even in
this quotient in Kelvin in Kelvin in
Kelvin that's why they have asked the
here in camping so they don't want to
get you confused to solve the question
they could ask him to centigrade okay
now the heat transfer coefficient as you
know what per meter square sending it
and what per meter square Kelvin so we
have to find out H the formula I have
one it is filled with unit once law of
cooling right so what is that q is equal
to h EI e s minus t
okay we know this formula I hope you
know this formula but but we have to
find out you first for this right q is
key w KW e DT by D Y into W key is what
for water for water K is 0.6 0.6 area I
have already shared with you that we are
considering 1 meters we got one set
meter square because they haven't
profile the area of the water and all
those things so 0.6 and 1 into 1 into 10
to the power 4 so it becomes 6,000 what
per meter now we already know Q is equal
to this month so we have to find out H
so H is equal to 6 thousand area is 1 TS
minus TF it is 1 so what we caught
adamantly 6,000 by TS minus TF ok we got
it
so remember this that's why you can
understand it properly
yeah now now what we have given in the
question that is temperature I mean
surface temperature and fluid
temperature fluid is what in Goudas you
would easier word let's see this water
water is 48 degree and glass base
surface is exposed to expose to the
water add in water a a broken glass is
40 degree and you know what it is 40
degree so we got here what we got here
t s is equal to t s is equal to 48
degree and DF is equal to 40 degree so
we'll have to put both the values here
6040 8 minus 14
so ultimately what we got here 6,000
divided by 8 so
if you do this eight eight seven four
five zero so seven five zero what per
meter square Kelvin Titov obviously them
to be like this because all units we
have used in what meter as or as Kelvin
so this is all about Newton's law of
cooling and related numerical problem
which for gram problem has been asked in
gate original paper I hope you learn
something amazing from this video right
so hopefully you enjoy the video thanks
for watching see you soon in my next
video bye bye
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