[Music] so in the last class we have derived the heat diffusion equation which is the fundamental equation for conduction heat transfer and then we have considered the flow of heat net flow of heat by conduction in the X direction in the Y and in the Z directions and then we invoked the conservation of energy which tells us that for a control volume the rate net rate of energy in the rate of energy in minus the rate of energy out plus any heat that can be generated that that is generated inside the system the algebraic sum of these three terms should result in a change in the energy stored of the system or in other words it's expressed as Y dot g e dot in that is a rate of energy in minus y dot out plus e dot g is equal to is equal to the rate of energy stored in the system and we time rate of energy stored in the system and we know that the energy of a system can be expressed as Rho the density CP it's the the heat capacity times the temperature minus a reference temperature so when we express that for our control volume and from this equation which is a difference equation we have obtained the governing equation for conduction in Cartesian coordinate systems and this is the equation which we have derived the last class where for Cartesian coordinate systems the temperature change now so in all these cases stick would be a function of X Y Z and time so the delta T by Del X square and for y for Z and this Q dot is the amount of heat generated per unit volume k is the thermal conductivity and this alpha as we have as we have seen it is units of meter square per second it's defined as K by Rho CP and it's known as the thermal diffusivity so this is the equation that one has to use one this is the starting point for any annal any conduction analysis for a system where the temperature could be a function of location where there could be some amount of heat generation and as a result of all these the temperature can also vary with time at fixed XY and Z so this is the fundamental equation which is also known as the heat diffusion equation similarly the similar type of equation can be derived for cylindrical systems as well as for spherical systems I did not write the spherical systems which fundamentally conceptually there is no difference between these two except for the coordinate system so here you see that T is essentially a function of R Phi and axial location Z and it could also be a function of time so the same way the the equation for the spherical coordinate systems can also be written it's there in your textbook so I am NOT writing it over here once again and we have all discussed that how this equation let's say this equation can be simplified when we have four different conditions for example let's say we have a steady state system in a steady state system the temperature does not vary with time okay so the right-hand side would be zero and let's also assume that we have a situation in which there is no heat generation in the system so this term would also be zero so therefore temperature is a function of XY and z only under certain can under some conditions it can also happen the temperature is a function only of X and not of Y or a Zed so if we think about steady state condition with no heat generation and temperature being a function of only one special coordinates then your this equation would simply equation can simply be written as d to t by DX square equals 0 I do not need to use the partial sign anymore since T is a function of time in which would provide which would give this T is a linear function of X so this we have we have discussed and we also understand that this governing equation needs to be can be solved with certain boundary conditions so in today's class we will see what those boundary conditions are which could be used to which get which it may be used to solve for to obtain the temperature profile starting at the governing equation so now we are in a position to know what is the governing equation what kind of simplifications I can make to that governing equation depending on whether it is a steady state or an unsteady state temperature is a function of X or temperature is a function of both x and y and so on so for all these cases if you look at the equation once again UT that you would see that you would require two conditions on X two conditions on Y two on Z and one initial condition on time to solve for the temperature profile so in the simplest possible situation of one-dimensional steady state conduction where the governing equation is d 2 t DX square equals 0 so you require two conditions on temperature when you have it it's an unsteady state situation however the temperature is still a function only of one special coordinates then again if you look at the equation you would see only the first term on the left hand side and the right hand side will remain in the equation so therefore your governing equation would be del 2 T del X square equal to 1 by alpha del temperature by Del x so this governing equation would then require two conditions on X since it's Delta T del X square on the left hand side and it would require one condition on time that is the initial condition when you look at the right hand side so for that specific case you need one inch initial condition and 2 reconditions so it's it's important to understand in important to identify what could those boundary conditions be and it can the boundary conditions can also be coupled with some other heat transfer process which is taking place in at the boundary for example you have you have a let's a cuboid which is at a hot at a temperature high temperature hundred degree centigrade and which is exposed to a room where the temperature is maintained at 25 so inside the solid inside the solid key void the heat transfer mode is conduction and if we neglect radiation since the radiation is going to be important only when the temperature is sufficiently high so at the boundary of this cuboid what you get is convection so any the the cuboid is going to lose heat to the ambient air by a process which is which all of us would know as convection so let's say as a small breeze is blowing over the cuboid making it cooler and in that condition at the interface between the solid and air and the moving air what you have is convection so inside the solid its conduction once you reach the interface between the solid and the air what you have is convection so in some of the some of the heat transfer cases conditions you will have this simultaneous present of convection and conduction at the interface so whenever we talk about conviction or whenever it we discuss about some other modes of heat transfer we need to identify what would be the governing equation for for that mode of heat transfer because we understand that in conduction the the flow of heat is governed by four years law of conduction so what governs the convective process the conviction is much more complicated then conduction because he in conviction you are also going to have flow of the fluid over the hot surface so you need to know what is the velocity profile what kind of velocity you have on the solid plate what is the combined momentum and heat transfer process that is taking place close to the interface so you not only have to have an idea of the fluid mechanics the momentum transfer part of it you would also have to have something which is similar to navier-stokes equation Appling applicable for energy transfer which we would cover later on when we study or when we have a more thorough study of conviction but it is suffice to say that the process of conviction is complicated because of the because of the presence of fluid motion and the associated momentum and heat transfer process with it so in conviction it is customary to define a convective heat transfer coefficient and it has been found experimentally that the heat loss from a heated surface exposed to a moving stream of fluid of different temperature the amount of heat loss is proportional to temperature difference this temperature difference is temperature difference between the solid and the fluid which is moving over it so the cue of conviction the cue the heat loss or gain depending on whether the solid is at a higher temperature or the liquid is at a higher temperature this loss or gain of heat in conviction is proportional to temperature difference so mark the difference with the conduction heat transfer in conduction heat transfer it's the thermal gradient which is simple which is which is deciding how much of heat is getting transferred between the B to between two points whereas in conviction the heat flow the heat rate is proportional to temperature differ so not the gradient and this law which is essentially obtained by looking at experimental results of conviction over a wide variety of substrates and fluids at different temperatures where the fluid is flowing with different velocities on the surface and so on this is known as the Newton's law of cooling so Newton's law of cooling is a fundamental relation of convective heat transfer which again cannot be derived which is the which is a result of the observer result of a large number of experimental observations generally in in a generalized form so the Newton's or convection we are going to use Newton's law of cooling and this Newton's law of cooling simply tells us that the heat loss or gain by convection is proportional to temperature difference TS minus T infinity where TS is the temperature of the substrate the solid and T infinity is the fluid which is in in which is with which the solid is in contact so if this is the solid and if the temperature on the solid side of the interface is TS and let's say the temperature of the fluid is T infinity at a distance far from it then this Q is equal to H times 8 where a is the area which is in contact with the fluid and TS minus T infinity this H is known as the convection heat transfer coefficient so this H is the convective heat transfer coefficient convection heat transfer coefficient and most of the studies of conviction hit conviction is to find what would be the expression of this convective heat transfer now let's think what this eight is going to be a function of it's it's our common observation that if you want to cool a surface cool as solid faster you have to blow air at a higher velocity so the velocity of this system velocity of the fluid which is flowing over the solid is definitely going to dictate what is going to be an important parameter in deciding what is the value of the heat transfer coefficient convective heat transfer coefficient it's not only the velocity it would also depend on the thermo physical properties of the fluid what's its density what is its thermal conductivity and since it's a case of flow and heat transfer whenever there is a mention of flow we cannot neglect we we cannot set aside the important parameter of fluid flow which is viscosity so and it's also going to depend on what kind of a thermal conductivity the fluid which is flowing will have so it's going to depend on the on the on the density of the fluid on the thermal conductivity of the fluid on the specific heat of the fluid and also on the viscosity of the fluid so Rho CP mu and K these are the thermo physical properties which would be relevant to decide what is the convective heat transfer coefficient there is an operating parameter which is the velocity with which these the fluid is made to flow over the solid surface so that velocity let's call it as U is also going to be in the functional form of H so if we think of the the expression of H this functional form of H should contain therefore it should contain the following parameters so H is going to be a function what is the velocity which is which is the fluid velocity it's going to be a function of Rho CP mu K among others and it would in in some cases there are going to be certain other parameters for example it's going to depend on what is the geometry of the surface is it is the flow taking place or a flat surface is it on a rounded surface is it a rough surface or is it a smooth surface so those extraneous factors that define the nature of the surface will also will also decide what is the value of the heat transfer coefficient so if you look at the the parameters which I have written over there that there is it's V Rho mu and and K there's also going to be this geometric parameter let's I am representing representing it by L where L is the characteristic length and I do not talk about anything else right now like for example the nature of the surface the roughness and and all so if you think of these V Rho mu and L so obviously when you think of these four terms V Rho mu and L you can you can you can you can clearly see that they are they can be combined in the form of Reynolds number where Reynolds number is simply Rho VL by mu so H is going to be a function of Reynolds number of the system and when you think of CP CP mu and K when you think of these three CP mu and K you know that CP mu and K they are called Prandtl number where the Prandtl number is defined as CP mu by K this spear is Prandtl number one of the famous scientists who has done extraordinary work not only on momentum transfer the concept of boundary layer but also on on heat transfer so most likely the any correlation in a relation that you would see for forced convection where you the fluid ease by an external agencies made to flow is going to be a function of Reynolds and Prandtl number so we will see this more when we discussed conviction but right now it would be sufficient to know that the Newton's law of cooling will decide how heat gets transferred between the solid and a moving fluid which would help us in deciding what are the boundary conditions for this case so let's see what are the boundary and initial conditions that one would one would get when you come across this so let's talk about this is the solid and on this side I have a fluid and in in the solid let's say this is my x equals 0 and this is x equals L so through the solid you only have conduction and at this point from the interface you have convection so the first condition possible condition is that the surface temperature is constant known it's known or constant which would tell you the t at x equals 0 at any time T is maintained at some value of TS so this if the surface temperature at this point is constant then the temperature over here would also be constant and let's also assume that some amount of heat is being added to this surface which we call as Q s double prime so this is the flux and the second condition could be that constant surface heat flux this constant surface heat flux let's call it as q is double prime as again double prime refers to the two the flux not the total amount now when this heat comes to the solid when this this could be a hitter this could be some sort of radiation which is falling on the surface and it's getting absorbed once it gets absorbed delivered at x equals zero the heat then travels through the solid by a conduction process so Q s double prime which is the heat flux at x equals 0 must be equal to minus K del T del X at x equals zero so this can this is the this is the second type of boundary condition which you may see where the amount of heat flux amount of delivered heat flux which is to be carried into the solid by a conduction process and by invoking Frias law I can write that the heat flow at x equals zero by conduction is minus KDT DX and this is x equals 0 is the control surface the control surface doesn't have any mass of its own so for a control surface the conservation equation takes the form that in is equal to out and therefore at the control surface the amount of heat which comes in is Q s double prime and the heat that gets transformed that gets transported by the conduction is simply going to be minus K DT DX at x equals 0 so how would it look like let's say this is my Q s double prime this is the temperature profile and if I draw a tangent to this at x equals 0 okay so this temperature could be a function of X and time so this tangent is essentially your del T del X at x equals 0 multiply that with minus K and what you get is keyways double Prime so this is by a conservation by invoking conservation at the control surface of the inn input heat flux and the heat which gets transported in the x-direction the third condition could be an adiabatic and areaway adiabatic wall or insulated wall so if it's an adiabatic or an insulated wall what you have then here is that this is a surface on this side it's perfectly insulated so if it is perfectly insulated then Q is Q s double prime must be equal to zero if it is insulated then no heat can cross this surface so if no heat can cross this surface I am again going to write the the equality of the heat that goes out and the Equality of a and and the heat that reaches the surface by conduction so this side is solid in the I on the on the other other side I have perfectly insulated this so since Q s double Prime due to insulation is equal to zero so this must be equal to K del T del X at x equals zero so this would give you that del T del X at x equals zero is zero so perfect insulation or an adiabatic surface would simply show you tell you that the gradient of temperature the special gradient of temperature at that plane is equal to zero which comes directly again by a heat balance on the cat on the on the surface on the control surface since no heat can cross the control surface this k dt/dx on the other side has to be equal to zero which would give rise to since ki cannot be zero DT DX is zero so the temperature gradient vanishes temperature gradient becomes equal to zero on a surface which is perfectly insulated or which is area and I will give you examples of a geometric surface later on when we discuss about how what is the what is the amount of heat general for a systems in which you have heat generation a system in which you have symmetric generation of heat everywhere then you would see that at certain point inside this earth inside the solid object the temperature reaches a maximum and when temperature reaches a maximum then obviously dtdx at that point would be equal to 0 so if dtdx is zero in that plane then no heat can cross this plane in either direction and that such such a surface is known as the adiabatic surface and the required boundary condition the prerequisite for having no heat crossing this adiabatic plane from either side is dtdx to be zero so dtdx to be zero can be a valid boundary condition if we know that at certain location the it's either an adiabatic surface or it's a surface which is perfectly insulated not allowing any heat to escape so that could be the third boundary condition and the fourth boundary condition we would see that it's it's a case of conviction surface condition conviction surface condition it would tell you that if you have a so if you have a solid on this surface side and let's say a fluid on this side in what you have is the fluid is moving with some velocity and let's say it's temperature is T infinity and the thermal and the convective heat transfer coefficient is H and over here the temperature varies as a function of X and time so if we take this as my control surface and let's say that T infinity is more than T so it's the solid is exposed to a hot environment so heat is going to move in this direction from the fluid to the interface by convection from the interface to the inside of the solid by a conduction process and at the interface the conviction must be equal to the conduction so what is conduction conduction is minus K del T del X at x equals zero so this is the conductive flow of heat conductive heat flux which must be equal to the convict convective heat flux which is T at x equals zero minus T in minus T infinity so this is T I'll put a minus sign over here or rather it's t at infinity minus minus T at a different page so what you have then I'll draw it once again this is the solid side this is the fluids fluid the fluid is moving with a velocity with some velocity having a temperature T infinity and with a convection heat transfer coefficient as H and let's say that the fluid temperature is more than that of the solid so this is how the temperature inside the solid would change with time so if we take this to be our control surface then the amount of heat which reaches by convection the same amount has to be transported by conduction since this is a control surface and therefore conservation demands that conviction and conduction are equal and if we write the conduction equals convection at x equals 0 this is let's say this is the location of x equals 0 then what you have is minus K del T del X at x equals 0 must be equal to this is the conduction conductive heat flux is equal to H times T infinity where T infinity is the temperature of the fluid at a point far from it minus T at 0 T so this is at the at x equals 0 at any given time T so this kind of conduction convection mixed boundary conditions are also prevalent in the study of heat transfer so once again if you look at the the possible boundary conditions one is the temperature is known that could be one condition the heat flux is known where the heat flux is equated to the conductive heat flow so that could be another boundary condition the third could be it's a third would be it could be an adiabatic or insulated wall condition in both cases Q dot s Q double prime s is 0 which would tell you that del T del X is equal to 0 and the fourth condition is this is the candidate condition boundary condition fourth which is a convection surface convection surface condition where the equality between conduction and convection at x equals 0 would simply give you that minus K Delta at X equal zero is this the convective flux convective heat flux now depending on what is the condition what is the specific problem that you are dealing with one or more of these conditions can be used so any the solution of any heat transfer conductive heat transfer problem is first to identify what is the what is the coordinate system that needs to be used is it a cylindrical system or is it a planar system or a spherical system then write the governing equation write the governing equation for that situation and based on the physics of the problem you cancel the terms which are not relevant if it is one dimensional problem with heat generation but at steady state then your equation would be if you look at the equation once again you would see that temperature is a function only of X it's not a function of Y it's not a function of Z and it's not a function of time if that is that NQ dot is not zero if that is the case then the equation would be D 2 T DX squared the first term on the lower left hand side the second and the third term would be 0 since T is not a function of Y and T is not a function of Z the fourth term is Q dot by K where Q dot is the generated heat per unit volume and the right hand side temperature is not a function of time so therefore the only term on the right hand side can also be equated to zero so you have governing equation in that case would be D 2 T DX square plus Q dot by K is equal to 0 so this way looking at the problem at hand you'd be able to simplify the governing equation to arrive at the final form of the governing equation then C out of the four boundary conditions that I have discussed so far which are relevant in for for this specific case is temperature at a given location known is the heat known or does the physics of the problem tell you that the flux at some X location such some location would be zero because either it is perfectly insulated or it's an adiabatic surface so you know which boundary condition to use or is it that at some location the solid is exposed to a fluid and the solid is losing or gaining heat by convection so using Newton's law and the Equality of conduction and convection at that specific location you should be in a position to write the mixed boundary condition which is the boundary condition for that I have discussed so that is how a conduction problem is to be attempted and if you can solve the governing equation with the boundary conditions what you get as a result is temperature as a function of location as well as as as well as a function of time and once I have temperature known at any point inside the control volume then I can do all other possible analysis to obtain the flux at a point and so on so what I do now is I will give you just to work a problem for you to practice and then the the in the in the tutorial part of it the problem can be analyzed in greater detail so very quickly the problem that I would like to give you is about about a solid wall which is generating some amount of heat so this solid wall there is some amount of heat which is generated so a dot G and it's not a steady state so there is some not stored in it some amount of heat comes into the system some amount of heat goes out of this system the X starts from here and X and this is from zero to L and it is known that the temperature inside this is as a function of X T as a function of X is equal to a plus BX plus C X square so this L is equal to one meter and this is the temperature profile at a certain instant of time so that's why you would see that the temperature function that temperature expression does not contain any time because it is measured and expressed in this form at a given given instant and the values of ABC which are constant these are a is 900 degree centigrade B is minus 300 degree centigrade per meter and C is minus 50 degree centigrade per meter square the value of Q dot which is the amount of heat generated inside the system is thousand watt per meter cube so this is heat generation per unit volume in the the area of the wall is 10 meter square the material of the wall the row is 1,600 kg per meter cube the thermal conductivity K is 40 watt per meter Kelvin and CP the heat capacity is 4 kilo Joule per kg Kelvin what you need to find out is what is the rate of heat entering the wall entering the wall which is at x equals 0 and leaving the wall which is at x equals L what is the rate of energy storage in the system at this instant so what is the value of e dot SD and the third one so you have to evaluate what is the rate of energy stored in the system at the given instant where the temperature can be expressed in this form and the last part is what is the time rate of temperature change at x equals 0 x equals 0.25 and x equals 0.5 meters so once again the problem I described the problem once again where we have a wall in which some amount of heat is generated some amount of heat comes into the system and it lives out of the system and as a result of imbalance of these three quantities the energy stored in the wall will also change however the temperature at a given instant of time can be expressed in the polynomial form of a plus BX plus CX square the relevant quantities for example el ABC are provided as well as the amount of heat generation and the area through which the wall is the area of the wall that is also provided so this area cross sectional area is 10 meter square the density thermal conductivity and specific heat of the wall material is also provided what you have to find out is what is the heat transfer rate of heat transfer entering the wall that is at x equals 0 living the wall which is at x equals L and what is the rate of energy stored at this for this system at that that instant of time and time rate of temperature change so del T by Del time at these three locations so how do you solve the problem as I mentioned the first thing that you need to identify is when you look at x equals zero plane the amount of energy which enters the the the wall as a result of which the temperature profile is provided so if I take my this as my control surface then Q dot in must be equal to minus K a DT DX at X equal zero at this location the amount of heat that comes in is going to be taken up by conduction on this side and the conductive heat flow is ka DT DX at x equals zero so by equating Q in with the conductive heat flux and knowing temperature as a function of time I can find out what is DT DX and evaluate what is DT DX at x equals zero and therefore find out what is what is the value of what is the value of Q in so my Q in would simply be minus K ay DT DX at x equals zero and the temperature profile is known to me similarly my Q dot out would be a Q dot out would be equals minus K a DT DX at x equals L so the T as a function of X is known to me so I can find out what is DT DX what is DT DX at x equals L the value of K and the value of a unknown to me so I'd be able to find out what is Q dot out so these two it should be able to evaluate now what is e dot store how do I write a dot store so from my energy equation e dot in plus y dot generated - E dot out is equal to e dot store that's your that sugar that's your energy conservation equation so my Y dot in is nothing but Q dot in my a dot out is nothing but my Q out what is e dot j the value of e dot g is is provided as a dot G equals where the Q dot is provided as thousand watt per meter cube so when everything is expressed on a per unit volume basis my Y dot in is known a dot G is known a dot out is known so I should be able to find out what is a dot what is a dot store and I leave this part for you to find out for you to think and solve and what I can tell you is that this del T by Del temperature is you will see that it's not a function of X value of temperature change with time that you determine it will not be a function of X so the time rate of temperature change at x equals 0.25 or 0.5 all are going to be equal what I am what I am going to do is I will give you the values the answers to this problem and you can try it on your own so Q dot in is going to be hundred twenty kilowatts the Q dot out is going to be 160 kilowatts and the energy stored in the system is this part is going to be equal to minus 30 kilowatt so that means the the system is losing this much of at at that instant of time and you would say that Delta divided that temperature change with time the value of this would be minus four point six nine into ten to the power minus four degree centigrade per second so the answer to part the the first one the in is 120 the out is 160 the stored is minus state third minus 30 and the time rate of temperature change is going to be minus four point 7 into 10 to the power minus 4 degree centigrade per degree centigrade per second in the the way this this part is going to be solved I will simply write it once again the governing equation for this case would be K by Rho CP become del 2 T by Del X square plus Q dot by Rho CP is equal to Del temperature by Delta so this is the form of the governing equation which you would obtain by writing the energy equation and cancelling the terms since T is a function of X and time so all the terms that that will those are Delta T del Y square and delta T del Z square will will be will be off will be set to 0 temperature as a function of X is known to you so you should be able to find out what is delta T del X square the values of K Rho CP Q dot are known to you so if only unknown here is Dell temperature by Delta and looking at the expression of TX T as a function of X you'd see that delta T by Del X square is going to be a constant it's simply going to be 2 C so delta T by Del X square is simply going to be 2 C so the entire left hand side which is time rate of change of temperature is going to be a constant so it's immaterial at which value of x you are measuring the time rate of temperature change it not going to be a function of X and when you plug in the values you would see that the time rate of change of temperature would be minus 4 point 7 into 10 to the power minus 4 degree centigrade per second so this completes our today's lecture where we have seen what could be the relevant boundary conditions and we have we have discussed a problem where a direct application of force law and the conservation of energy can be used to arrive at numbers for the heat which comes into the system going out of the system and time rate of change of temperature and so on so I believe if you go through this your concept of conductive heat transfer must be clear right now and the use of use of the use of the heat diffusion equation in the next classes we will see how this heat diffusion equation can be modified to obtain special cases solutions for special cases and what form would it take for not only for Cartesian coordinate systems but also for cylindrical and spherical systems