in this video we are going to start chapter 3 uh the objectives of chapter 3 is to learn to use the analogy between thermal resistance and electrical resistance and to analyze the heat transfer from extended surfaces or they are also called fins so let's start with theral resistance concept so chapter three is still about conduction uh take a look at this uh plain wall and it shows the temperature distributions within the wall and observe how the temperature varies within the wall so on the left surface uh temperature 20° C so at any point on the left surface temperature remain constant and the on the mid plane the temperature remain at 11° C on the right surface of the W temperature remain at 3° C so driving force of heat transfer is the temperature gradient so there's a is no temperature gradient um in on the plane uh if you look at the planes uh which is uh perpendicular to X AIS there is no temperature gradient and this is x-axis so in y AIS and in Z axis there is no temperature gradient uh which which means there's no heat transfer in Y and Z Direction and there is a significant temperature gradient in X Direction 20 on the left surface mid plane 11 de C and 3° C on the right surface since there is a significant temperature variation in X Direction heat will flow from higher temperature locations to a lower temperature location so heat flow in x Direction uh if the temperature uh does not change with time so at any given point temperature remain the same it's called steady state condition so in heat transfer uh whenever we say it's a steady state that means temperature at a given location uh does not change with time temperature still vary with location but temperature at a given location does not change with time onedimensional heat conduction uh since heat flow only in X Direction uh this is onedimensional heat conduction so we are going to take analogy between electrical resistance and thermal resistance so we already know ohms law so electric current I uh is proportional to Delta V voltage difference potential difference and inversely proportional to uh electrical resistance uh this is ohms low and electrons flow through the circuit and uh you know the heat conduction uh may be expressed as this this is f law of heat conduction Q can be Rewritten as q = k a DT / DX or um so we have negative sign since DT / DX is negative or we can write it as k a delta T over L right so here um I I uh didn't use I didn't add a negative sign since delta T is large temperature minus smaller temperature so which is positive and L is also positive so I didn't add negative sign and this can be Rewritten as delta T over L over k a so l/ k a is inversely proportional to Q so l/ k a is called thermal resistance which is you know uh so thermal resistance corresponds to electrical resistance so compare these two equations i electric current Q the rate of heat transfer so electric current is proportional to voltage difference uh Q the rate of heat transfer is proportional to delta T temperature difference so delta T is the driving force of heat transfer voltage difference potential difference difference is the driving force of the electric current and electric current is inversely proportional to Electric electrical resistance Q is inversely proportional to theral resistance which is L over k a l the distance between the two surfaces or distance where the heat transfer occurs K thermoconductivity of the medium a heat transfer area so in the thermal resistance l n a are dependent on the geometry of the medium K depends on the material property so ohm slow may be expressed in the electric circuit is this so uh here is two points indicates the the uh the potential V1 and V2 and here's a electrical resistance and electric current flow through the resistance in the same manner uh we can express the heat conduction through the resistance thermal resistance so here is the two points where the different uh with the different temperatures and thermal resistance by conduction which is l/ Ka and heat flow through the resistance by T1 temperature difference over R thermal resistance uh we can also apply this to convection and radiation so convection you know we use the Newton's law of cooling Q equal ha time temperature difference so T Infinity here is the free stream temperature TS is the suface temperature H convection heat transfer coefficient a is the heat transfer area so a typically is the surface area that touches the fluid so convection always occur between the fluid and the solid surface so a is the heat transfer area or the area surface area that touches the fluid so uh Newton slow of cooling may be expressed as this temperature difference over 1/ ha so 1/ ha may be regarded as thermal resistance by convection so thermal resistance by conduction is L over k a Thal resistance by convection is 1/ ha a and it can be also expressed in thermal circuit so temperature T Infinity TS and thermal resistance 1 / ha by convection and heat flow by convection so electric current corresponds to rate of heat transfer electric resistance corresponds to Thermal resistance voltage difference drives the electric current or temperature difference drives the heat flow heat transfer and this analogy uh may be used only for steady state heat conduction one diens heat conduction and no heat generation okay so you need to remember this uh thermal resistance conent may be used under these three conditions uh the reason why we use this analogy is to simplify the heat conduction problem and solve it uh conveniently uh let's take a look at um single layer uh heat transfer problem single layer heat transfer and this plain wall is surrounded by Fleet so on the left hand side there's a fleet with a temperature T infinity1 and H1 convection heat trans coefficient and we have a solid wall with a thermoconductivity k and the length is L and heat transfer area is a on the right hand side uh there's another fluid with a temperature T Infinity 2 and convection heat trans coefficient H2 so assuming this is a onedimensional heat conduction steady state and the wall does not produce any heat no heat generation okay so we apply the thermal resistance concept to this heat transfer problem uh so we can draw a thermal resistance Network this is called thermal resistance Network or thermal resistance circuit uh we have temperature T Infinity 1 T1 T2 T Infinity 2 so T Infinity 1 T1 T2 and T Infinity 2 these temperatures are connected through thermal resistance so T Infinity one between t Infinity 1 and T1 there is a thermal resistance by convection between T1 and T2 is they are connected by thermal resistance conduction thermal resistance AR Ka between T2 and T Infinity 2 they are connected through convection thermal resistance one over ha and note that they are connected in series they connected in series okay so actually when they are connected in series so we need to note that um you know q1 so here q1 is the rate of heat transfer by convection Q2 is the rate of heat conduction heat transfer by conduction Q3 is the rate of heat transfer by convection so here uh we can say that under St State conditions one dimensional heat conduction the heat generation the rate of heat convection into the world Q convection into the world equals rate of heat conduction through the wall equals rate of heat convection from the wall right this occurs under state state one dimensional heat transfer and no heat generation conditions uh so heat conduction on the left hand side of the wall may be expressed as this Newton slope of cooling H1 a T infinity1 minus T1 so here uh it can be Rewritten as this uh so 1 / ha is the thermal resistance by convection uh you may wonder why we took T Infinity 1 minus T1 instead of T1 minus t Infinity 1 so our assumption is that you know the heat transfer Ur from higher temperature to lower temperature from left to right he transfer into the wall um so we assume you know temper shift flow always higher to lower temperature so assume T Infinity one is greater than T1 higher temperature to lower temperature from left to right okay Q conduction for is low conduction we don't have negative sign since T1 is greater than T2 positive L length is positive so we don't need to add negative sign here also from left to right T1 is greater than T2 right heat flow from left to right so that's why T1 minus T2 is not T2 minus T1 and uh we can uh express it uh using the thermal resistant concept T1 minus T2 temperature difference over thermal resistance by conduction on the right hand side uh again heat convection um rewrite it uh with the thermal resistance concept one over ha so again here heat flow from left to right T2 is greater than T infinity2 that's why we have T2 minus t Infinity 2 he flow out of the wall since T2 is greater than t infinity2 so in actual problem there's a possibility that we don't know the actual heat uh flow directions uh then we can just assume and in this uh in this problem we just assume that heat flow left to right so T Infinity one is the largest followed by T1 T2 and T Infinity 2 is the smallest temperature so since the rate of heat transfer through the wall uh by convection equals rate of heat transfer by conduction and same as the rate of heat convection from uh con rate of heat convection uh from the wall so we can write it as this so Newton's low of cooling on the left hand side F low of heat conduction and Newton's slope of cooling on the right hand side they are all the same and this is the summation of the uh denominate uh the numerator and summation of the denominator so if if they are the same we can say 1 + 2 + 4 also 2 + 4 + 8 are also the same summation of the numerator over summation of the denominator still the same so here um the summation of so in the denominator this is the summation of the each thermal resistances thermal resistance by convection thermal resistance by conduction and Thal resistance by convection so summation of all the Thal resistance is called total resistance total resistance so when the thermal resistances are connected in series the total Thal resistance is simply the summation of all resistances when they are connected in series so this is same as the electric resist electrical resistance is in you know electric circuit right so you just uh to get the total uh electrical resistance you just add up all the resistances that are connected in series so same for thermal resistance Network uh so multi-layer world so if the medium is in medium is composite two layers so let's see how we treat this multi-layer thermal resistances in multi-layer uh again uh we regard this as a steady stage onedimensional heat transfer and no heat generation uh thermal resistance network is simply adding additional resistance so this is almost the same as the previous example you just simply add additional uh thermal resistance again the rate of heat transfer by convection same as rate of heat conduction through m material layer one first layer and rate of heat conduction through the second layer and the rate of heit convection uh in the second Fleet they are all the same okay and total thermal resistance is just simply uh you know some summation of all the resistances so we can write this and Q so here so actually I express it in different way uh so this is not different way so what is this this is the rate of heat transfer um uh through the left fluid left fluid to the wall right so temperature difference T infinity1 T infinity1 minus T1 so this is the heat transfer by convection which is equal to T Infinity 1 minus T2 so look at this T Infinity 1 minus T2 this is the rate of heat transfer between t Infinity 1 and T2 two so we can take the rate of heat transfer this through this two resistances between two temperatures so now we add the Thal resistances between the two temperature T Infinity 1 and T2 there a heit thermal resistance by convection and thermal resistance by condu uction right we can also write it as this T Infinity 1 and T3 so T infinity1 and T3 what thermal resistance is uh exist thermal resistance by convection thermal resistance by conduction and thermal resistance by conduction so there are three thermal resistances are connected in series so that's why we see these three Thal resistances or we can write it t Infinity minus t t Infinity 1 minus t Infinity 2 over R toal so between t Infinity 1 and T Infinity 2 we have four resistances th resistance by convection th resistance by conduction th resistance by conduction and thermal resistance by convection so R total okay so this is between t Infinity one and T Infinity 2 so the reason why I express it in different form is you know some sometimes maybe you want to determine T2 sometimes you want to determine T3 and all the remaining terms are known right so basically using this um Expressions you can acquire any temperature at the interface T1 T2 T3 if T Infinity one and T infinity2 are known we can get any temperature at the interface so this example um actually we want to determine the inter the temperature at the interface T1 T2 T3 and T4 and we use the thermal resistance uh Network concept uh to determine the temperature at the interface so this is a double uh paint glass glass um and you know in between the glass uh Del a air or is filled with uh you know uh low ther thermoconductivity uh gas or it's under vacuum so to minimize the heat transfer between the two uh space so on the left surface uh left side uh we have a fleet with a temperature 20° C and convection heat trans coefficient 10 W per M Square de C again uh you can make uh you can write it as 10 watt per square meter Kelvin right it's the same thing uh and the thermoconductivity of glass is given and also thinness of the glass is given 4 mm and the gap between the two glass is uh 10 mm filled with air air the thermal conductivity of air is 026 wat per met kin or wat per meter de Celsius and another glass pan and the area of the glass pain is 1.2 M squared this is the heat transfer area since heat assuming this is a onedimensional heat conduction in from left to right so area is perpendicular to the heat flow Direction uh let's see how to determine T1 uh let's see how to determine T1 first so assuming this is a onedimensional heat conduction steady stat and constant K so kg and K A remain constant and doat generation so we can draw a ther thermal uh resistance Network so thermal resist we can draw a thermal resistant Network so let me move this to the side and so there's a uh let me first draw a point so this is T Infinity 2 this is T Infinity 1 and this is T1 T2 T3 T4 so so each uh temperature are connected through res Sumer resistance so we have 1 2 3 4 five five resistances and so false resistance is through by uh convection so H1 a this is the thermal resistance through Q convection one let's say this is q q convection one the second resistance is by conduction so L1 over k G A and let's say this is Q conduction one and next between T2 and T3 another uh conduction so this is air uh since H is not given right for air and thermoconductivity is given so assuming that there's no fluid MO motion and there's only pure conduction through air so this there's another conduction two and thermal resistance by conduction is K L2 over k a a k a is the thermal resistance of air and another thermal uh resistance by conduction so L3 kga and on the right hand side of the glass pane there's a heat convection so Thal resistance by convection one over H2 a okay so this is the thermal resistance Network and so actually so let's determine the values of each thermal resistance one over H1 a so one over H1 is given 10 watt per met squ de cus a uh is 1.2 2 m Square given so Thal resistance by conduction H sorry convection and L1 over kga this is the thermal resistance by conduction through glass pane so 04 [Music] M kg 78 wat per M de C area 0427 de C For What and L2 through air point1 m02 six thermoconductivity of air times area 3205 and L3 um same as uh thermal resistance through the second uh glass pane is same as the thermal resistance through the false glass pane and thermal resistance by convection on the right hand side of the fluid one/ H2 H2 is different from H1 02083 so what is our total so summation of all resistances which is 433 2° C per so summation of 1 2 3 4 5 resistances so this is the total resistances and let's first get Q so Q so Q in general same as Q convection one same as Q conduction one same as Q conduction 2 Q conduction 3 Q convection two on the steady state so you know same the same rate of heat transfer uh occurs through each thermal contact thermal resistances so Q may be written as so here T1 T2 T3 T4 are unknown T Infinity one and T Infinity 2 are known right T Infinity 1 and T Infinity 2 are known so we can actually get Q using T Infinity 1 and T Infinity 2 and the resistan between the two temperature is all total so we can write 20° cus -10° C R total is 4332 De C per wat so Q is 69.2 watt through the first flid through the first glass pane through air through second glass pane through the second uh Fleet they all remain the same Q remain the same okay and we also want to determine um T1 so we can express uh Q in terms of T1 we can express Q in terms of T1 since Q uh Q equals q = q convection 1 right qal Q convection 1 Q is with just determined and Q convection one what is Q convection one Q convection one t Infinity 1 minus T1 over thermal contact resistance by convection 1 over H1 a so the only unknown is T1 so uh T Infinity 1 minus T1 over 1/ H1 a so the only unknown is T1 we can solve it for T1 right every all the remaining uh parameters are known so we can solve it for T1 which is t infinity1 - q/ H1 a so 20° cus [Music] minus 69.2 watt times um 8 0 33 we o 33 H1 H1 is point sorry it's a 10 10 watt per M squ de C time 1.2 M Square so we got um Al actually need to divide it H1 Q H1 a you need to divide it so divided by this one and T1 one is acquired 14.2 de C so uh T infinity1 is 20° C and on the very right hand side temperature negative 10° C under steady state conditions you know heat flow so since this is greater than the outer surface I mean on the right hand side FID temperature we know that flow from left to right so T1 must be smaller than T Infinity one right so that heat flow to the right so T1 should be smaller than 20° cus and T1 is 14.2 which is smaller than T Infinity one so this is reasonable and T2 how do we get t two so we need to also get Expressions we need to somehow Express uh Q in terms of T2 so here actually there are different ways to express T2 so one thing you could do is um one thing is to get the temporature difference between t Infinity one and T2 right and there are two resistances one through convection and the other one through false glass pain then that term equals to Q then we can determine T2 that is the only unknown or you can we can also establish Q equals the heat transfer through the false glass pane and since T1 was already determined right 14.2 de C we can also solve it for T2 so there are different ways to get T2 and T2 I will just give you the answer for T2 13.9 de C which is smaller than T1 so it still makes sense right since heat flow to the right uh in the right directions uh T2 should be smaller than T1 so in the same manner you can get T3 and T4 okay again uh this can be only applied uh to steady state one dimensional heat conduction uh without heat generation