in this video we are going to solve heat diffusion equations to determine the temperature distributions uh through the medium uh it shows a procedure how to approach the uh heat conduction problems uh first obtain the differential equation or heat deficient equation in its simplest form so you need to make appropriate assumptions to simplify the heat diffusion equation and specify two boundary conditions uh heat diffusion equations uh is a second order uh in space so we need to have uh two boundary conditions to determine the two integration constants obtain the general solution of the differential equation uh with the arbitrary integration constants and finally determine the integration constants by applying the boundary conditions let's take a look at the example consider a large plane wall of l. 6 cm that is the distance between the two surfaces as you can see in this scheme itic L is the distance between the two surfaces a uh 160 square cm this is the heat transfer area so if you look at this uh in three dimensions this world maybe look like this and a is uh this area so that is normal to the heat flow normal to the heat uh flow Direction K thermal conductivity 60 watt per met de C left side of the wall is subjected to uniform heat flux qou prime notation Q with dou Prime denotes heat flux the rate of heat transfer per unit area that's why it has the unit of watt per met squared right watt is the rate of heat transfer per area me squared the right side of surface temperature is 112° C determine the temperature of the left surface uh in this schematic T1 is the temperature on the left surface T2 is the temperature on the right surface so T1 is unknown T2 is 112° C uh if there's a temperature gradient temperature difference between T1 and T2 between the two surfaces heat flow from higher temperature to lower temperature and uh it's always good idea to draw schematic uh with all the informations uh that are given uh in the problem statement to better understand the question so uh Q Prime heat flux uh on the left surface and right surface temperature remain constant at 112° C so it's also a good idea to draw the direction of the flow uh here actually the temperature on the right surface it remain a constant temperature and uh there's a heat flux on the left surface so it's reasonable to say that a hit flow from left to right uh so for some questions you may not sure you you may not be sure that uh the actual direction of the heat flow then you can still assume the direction of the heat flow and if your assumption is not correct uh you may end up getting a negative Q that just tell you that the actual direction of the heat flow is the opposite to the assumed Direction so Q here we uh assume that uh the heat flow from left to right and we are starting with the heat diffusion equation so here uh it has a I wrote down a general heat diffusion equation and first thing what we need to do is to simplify uh the heat deficient equations and specify two boundary condition so simplify this equation uh with the appropriate assumption so first uh I I'm going to assume that this is at steady state condition uh which means temperature at a given point given location does not change with time temperature does not change with the time onedimensional heat conduction um heat flow only in One Direction in this case uh if I assume this is x axis hit flow on X AIS uh this also means that if you slice the wall you slice the wall uh perpendicular to x axis you may end up having this surface and at any point on the surface the temperature is uniform if the temper if there's no temperature gradient there's no heat flow the driving force of heat conduction is the temperature difference so there's no heat flow heat transfer uh across the this uh green uh plate there is only temperature gradient in xais so heat flow only in xaxis so this is one dimensional heat conduction problem constant k K typically thermoconductivity typically varies with temperature uh but we are going to assume uh K is constant K is 60 wat per met de C so sometimes you will see K in watt per meter Kelvin instead of De cius so you know um Kelvin is 273.15 plus de C uh so how do you convert 60 watt per met de C to Watt per meter Kelvin so actually it's still 60 watt per met kin the same uh because this is wat per de cus wat per de CSUS this is wat per Kelvin so you see the change in degrees celsius same as change in celvin right so what per degree celsius you know there's no uh magnitude change when it converted to Watt per Kelvin last assumption no heat generation so you know I introduced uh three different cases when we need to consider the heat generation but uh this war is a plain wall uh no heat generation so using these assumptions uh let me simplify the heat division equation steady state so wisom drops out temperature does not vary with uh time onedimensional heat conduction temperature is only function of X Conant K this K is constant we can take this k out of the derivative since K is a constant no heat generation Q do drops out so the simplified heat division equation becomes this and we also need to specify two boundary conditions so boundary conditions so we have two boundaries left surface right surface on the left surface uniform heat flux so uniform heat flux goes into the surface so Q qou Prime that goes into the left surface and it is conducted to the plain W so negative netive K DT over DX at x = z so this is uh one uh boundary condition and the second boundary condition uh the right surface temperature is given so T 2 112° C so we simplify the heat division equations and identified two boundary conditions so what is the next step obtain the general solution of the differential equations so we are going to obtain General solutions by integrating the heat division equation so since this is a second order in space we integrate it once DT / DX become C1 here C1 is arbitrary constant [Music] and integrate it one more time C1 x + C2 C2 is also arbitrary constant so this is a tempure profile we will uh get so here C1 and C2 are unknown for now and we are going to apply boundary conditions to determine C1 and C2 so first boundary condition uh at xal Z right on the left surface so here we have a x coordinate uh on the left Sur at the left surface uh that is xal z so this is how we set up the coordinate system at x = 0 um Q so using this uh boundary condition false boundary condition qw Prime so negative NE negative become positive K DT over DX so here D2 DX is some constant C1 * C1 = 0 so C1 = Q dou Prime over K now we apply the second boundary conditions at the right surface x equal l l is the distance between the two SES uh so we apply this in to this equations so temperature L TL or T2 equal C1 l+ C2 so C1 we already determin C1 so C2 is cative C1 l+ T2 is 112 right so we can write temporature since we determined C1 and C2 temper qou Prime over k [Music] x C2 qou Prime k l + 112° [Music] C uh we can take Q Prime over k l - x + 112 and plug in uh actual values qou Prime 50 kilowatt per M squared 50,000 wat per square M and K 60 wat per M de C or kelvin L it's 6 cm or6 m and on 12° C and this is 8 33.3 .06 112° [Music] C so it ask to determine the temperature at the left surface which is so x equals 0 a 33 time6 + 112 117° cus okay uh it isn't ask to determine the rate of heat transfer through uh through the plane wall but we can still get the rate of heat transfer through uh the plane one using since qw Prime is given qw Prime is Q over a since so Q is Q Prime time a q Prime is given a is given so we can get Q right or Q is alsoa K a DT / DX so K 60 a a is 160 cm squared so you need to convert this to me squar DT over DX temperature gradient DT temperature uh vary from 17 uh so we got 117 on the left surface minus 112 So 5° difference and the distance is 6 cm right so 6 cm actually you know we what we did is this is X1 X2 X2 - X1 uh is positive and T2 - T1 is actually negative so actually -5° C so negative negative become positive so this also give us Q so check uh so Q Prime 50 kilowatt per M squar time area 160 and check if these two values are the same they they must be the same uh it's just I just showed you the two different way to determine the rate of heat transfer through the plane wall okay um next I'm going to explain about the heat division equations in three dimensions so typically heat May flow in three uh you know in three dimensions uh so this what we see here is the um rectangles it's a small volume element and we can apply uh the energy balance uh to this small volume element and get this heat division equations and as you can see uh so this is X the heat conduction in X directions heat conduction in y direction and heat conduction in Z Direction so this is the heat defent equation in cian coordinates or in rectangular coordinates uh in previous video we only showed the energy balance in X Direction only but if the heat flow if there's a temperature gradient in Y and Z direction we need to add these additional terms okay and Q do uh heat Generations on the right hand side that is the change rate of change of the energy content uh in the control volume in the volume element um the heat diffusion equation may be expressed in cylindrical coordinate so if we are interested in the heat transfer uh in through a pipe so maybe uh the inner surface temperature of the pipe is different from the outer surface of the temperature then the there may be heat transfer um across the pipe so you know we what we are interested in is the temperature at the specific location of the in the medium uh it is easier to use cylindrical coordinate to represent the position in a pipe so if the shape of the medium is in cylindrical shape then uh it's easier to use heat division equation in cylindrical coordinate instead of using cian the heat deficient equation in rectangular coordinate so this equations shows the heat diffusion equations in cylind cylindrical coordinates so what you see T is a function of r distance from the center line uh through the cylinder V and Z that is the angular direction of the temperature so uh temperature is the position of the location uh was expressed in terms of r v and Z okay so this equation is in also three dimensions and it can be also simplified uh with the appropriate assumption so we are going to take a look at the example shortly sometimes we may want to analyze the heat temperature distribution in spherical medium uh then it's easier to uh use the heat division equations in cylindrical coordinate and again temperature is expressed in r v and seta also this equation may be simplified uh depending on the the Assumption uh we make so you don't need to this this heat defis equation in cylindrical coordinate and spherical coordinates are also derived derived based on the energy balance relation uh and I'm going to show you an example uh of a heat defion equation in cylindrical coordinate so consider a hot water hpe of length l so and inner radius R1 outer radius R2 thermoconductivity k k is uniform across the medium in a surface temperature T1 so at R1 temperature of the inner surface is T1 and outer surface radius R2 temperature is T2 determine the temperature distribution and Q the rate of heat transfer through the pipe so again I drew a schematic and also it include all the informations given in the problem statement and here um Q is in the rate of heat transfer uh is in radial Direction and we also Al uh use the make an appropriate assumptions to simplify the heat division equations so heat division equation that we are going to use is this since we have a cylinder we are going to use this heat diffusion equations to get the temperature distributions and Q uh this is steady state which means temperature does not vary with time onedimensional heat conduction so here you need to be careful about the direction of the heat flow heat does not flow in actual Direction heat flow in radial Direction this is correct heat flow in radial Direction heat does not flow in actual Direction uh since what we have is the temperature between the inner surface and the outer surface inner surface in know surface this is the inner surface outer surface this is the outer surface so there will be a temp so if you see the cross-section of this pipe um so maybe we can say the pipe contains hot fluid so inner surface temperature is high outer surface temperature is low so there's a temperature gradient between the two surfaces and heat flow in radial Direction radial Direction and this may be AUM as a onedimensional heat conduction although heat flow in you know all directions uh the temperature at a given R temperature at given R let's say this is R so at a given R temperature is uniform at a given R so on this ring at this ring temperature is uniform so we can assume temperature is only function of r so this is one this may be assume as a onedimensional heat conduction constant K so K uh remain constant although there's a temperature gradient within the medium no heat generation so we start from this heat diffusion equation in three dimensions and St States one dimensional heat conduction temperature is f only function of r this drops out this drops out no generation so we can write the simplified heat diffusion equation this is the simplified heat diffusion equation so we can multiply r on both sides of the equations so this can be drops out since on the right hand side 0 * R is zero uh and there's no K right I actually so K was here and you can divide uh the equation by K so K also drops out since on the right hand side we only have zero and you are 10 tempted to remove r as well right so divide the equation by R so you want to remove R but we cannot divide the equation by R since this R is within the derivative R is not a constant so we cannot take R out of out of the derivative nor we can not uh divide the equation by R so this is the simplified uh heat diffusion equations with uh these four uh assumption and we need to identify two boundary conditions so first boundary conditions uh temperature at the inner surface is T1 second boundary condition temp temp at the outer surface T2 we have simplified heat division equations with the two boundary conditions now we are going to integrate the heat diffusion equations this is also second order in space so we need to integrate it twice and apply the boundary conditions to determine the integration constant so integrating the differential equations once what we get so we integrate it with respect to R so what we get is Dr R DT over Dr R on the right hand side we have C1 C1 is also arbitrary constant and divide the equation by R and we integrate it again with respect to R uh so on the left hand side t on the right hand side so C1 C1 is constants now integration of one/ R is is natural log R right and we got another integration constant C2 C2 is also arbitrary constant so to determine C1 and C2 uh we use boundary conditions so apply boundary condition so at R1 temperature T1 so T1 at R1 this is the first boundary condition second boundary condition at R2 temperature is T2 so solve these two equations two unknown two equations so we can solve it for C1 and C2 so we get C1 and C2 and plug in these two integration constants back to temperature equation and it can be simplified as this so R here R without subscript is a variable R1 is the uh radius uh inner surface radius R2 is the outer surface radius so only variable here is R so temperature is function of r temperature is function of r so using this equations uh we can get the temperature across the pipe at any position R uh this question also ask to determine Q the rate of heat transfer through the wall so through the pipe world so Q is K a DT / Dr R so here F low heat conduction in cylindrical coordinate in one Dimensions can be written as this DT over Dr R and what is a what is a heat transfer area so what is a so you see this you see this uh Blue Area like a cylinder cylinder surface area so this is the heat transfer area at R so again heat transfer area in F low of heat conduction is normal to the heat flow you see the direction of the heat flow in radial directions so the heat transfer area is the outer surface area of the cylinder which is 2 pi r l so this is the heat transfer area which is function of r and DT d r DT d r can be found from here we already know C1 so the r can be written as this well we can simplify this equations [Music] okay so now I have a question um so Q is the rate of heat transfer um how does q vary Q vary with r q is not function of r right it's not function you cannot find R here R1 R2 are constant we cannot find R in this equation so Q is constant Q is actually the total rate of heat transfer uh across the pipe Q remain constant how about Q Prime Q Prime is the rate of heat transfer per unit area rate of heat transfer per unit area is Q over what is area area is the heat transfer area which is 2 pi r l so qou Prime Q is constant Q Prime is inversely proportional to R so why why Q Prime heat flux is inversely proportional to R so s so this is r as R increases how about a a increases right as R increases area heat transfer area also increases that's why heat flux is inversely proportional to R okay okay so next question uh is about uh heat diffusion equation in spherical coordinate so this question is very similar to um the the the example that we just uh went over so I will just quickly go over uh the equ uh the problem so sometimes we want to determine the temperature distributions across the spherical uh system so this is the sh division equation in spherical system and with the appropriate assumption St States no heat Generations one dimensional heat conduction and we also want to determine temperature distribution and Q the rate of heat transfer so simplified uh equation is 1 / r² so divide the equation by K and we can also multiply both side by r² so this is the simplified equations the boundary conditions same as the previous one the inner surface so this is uh like a spherical container and inner surface temperature outer surface temperatures are given and also the heat flow uh in radial Direction and also integrate the heat diffusion equations twice with respect to R so what we have so r² DT / Dr R becomes some constant and integrated again t is 1 / r² so integration of 1 / r² is 1 / r 1 / R and we need a two boundary conditions so again boundary we already got the boundary conditions apply the boundary conditions so first one at R1 temperature T1 so T1 at R1 T2 at [Music] R2 so two equations two unknown we can solve it for C1 and C2 so again TR R temperature distribution is only function of r and Q is K a DT over d r so temperature is only function of r one dimensional heat conduction in radial Direction so what is heat transfer area in SPH spherical uh medium 4 * < R 2 and DT over Dr R DT over Dr R is here so we got C1 so DT over Dr r c1/ r² or so Q is this uh it's not function of r either so Q is constant total rate of heat transfer constant but let's see how Q Prime change Q over area Pi r² so Q constant and Q Prime um as R increases as R increases qou Prime his flux decreases quadratically okay so this uh is all uh I have uh in chapter 2