 opher Sunil Lattain Kuilkarani, instructor of mechanical engineering, valchan institute of technology, today I am going to deliver a video session, on steady state heat conduction. At the end of this session, you will be able to calculate, the temperature distribution and heat flow rate, through composite structures. The contents of this video session are, heat transfer through composite structures, the concept of overall heat transfer coefficient as we are aware that most of the practical applications involve composites structures or composite walls The composite structure means the structure which is made up of material having various conductivity For example if we see the walls of refrigerator walls of cold storage walls of furnace these are made up of various material in this case to find out the heat transfer through such composite structure we require to use electrical analogy method if the first problem we will take off heat transfer through composite wall for the purpose let us consider one composite wall which is made up of two different material having thermal conductivity k1 and k2 and having the thickness in the direction of heat flow for material one the thickness is b1 and for material two the thickness is b2 the temperatures on the outer side and inner surface of the wall are T1 and T3 and interface temperature between the two materials is T2 let us assume that this particular wall is also having a convective environment on inner side of this particular wall we are assuming that the hot flue gases are there and heat transfer coefficient is HI whereas on outer side the ambient air is present with heat transfer coefficient of HO now to write down the heat flow rate equation for this composite wall we can develop the electrical circuit which is now if we consider this problem in this problem we are having two conductive resistance because we are having two material and on left hand side there will be a convective resistance by using electrical analogy method we know that q is equal to HA delta T if we rearrange that q will be equal to 1 upon delta T upon 1 upon HA so convective resistance is given by 1 upon HA H into A therefore I can write the first resistance which is there between the temperatures of TG and T1 as 1 upon HA into A the conductive resistance for first material I can write it as B1 upon K1A for second material I can write the conductive resistance as B2 upon K2A and on outer side the heat transfer from the furnace outer surface to the surrounding air will be happening by convection for which the temperature of surrounding air is PA and heat transfer coefficient is HO so I can write here the convective resistance as 1 upon HO into A since in case of plane wall the area of heat flow if we see here it will be constant in the normal to the direction of heat flow so here the area of heat flow we can assume to be throughout uniform which is A so I can write the equation for heat flow rate as q is equal to now I will take the temperature difference at the two ends that is temperature of hot gases minus temperature of ambient air divided by summation of all these resistances that is 1 upon HA into A plus B1 upon K1A plus B2 upon K2A plus 1 upon HO into A since area of heat flow is constant we can take it common from the denominator and I can take it as 1 upon HA to the left hand side and by rearranging the equation as q is equal to TG minus TA into A divided by 1 upon HI plus B1 upon K1 plus B2 upon K2 plus 1 upon HO so by using this electrical analogy method we have converted the steady state heat conduction through composite wall into electric circuit and using the electrical analogy method we have obtained very easily the equation for heat flow rate this way we can use the electrical analogy to solve any number of layers if they are present we have to take into account that how many materials are there accordingly we have to add the conductive thermal resistances in the circuit if they are arranged in series we have to take the resistances in series if they are arranged in parallel we have to take the using the law of electrical energy parallel resistances we can find as 1 upon R equivalent is equal to 1 upon R1 plus 1 upon R2 so we have to take equivalent resistances if they are parallel for series we can directly as R equivalent will be equal to R1 plus R2 now let us see the concept of overall heat transfer coefficient now usually the heat transfer by combined modes of conduction and convection are defined by using the term overall heat transfer coefficient and equation the heat flow rate is expressed in terms of overall heat transfer coefficient as q is equal to ua into delta t now if we give the equation number 1 to these and already we have just now derived the equation for heat flow rate through composite wall as q is equal to tg minus ta into a divided by summation of various resistances like 1 upon hi plus b1 by k1 plus b2 by k2 plus 1 upon ho so if we compare the equations 1 and 2 by making the comparison on the right hand side we will get that 1 upon ua into a will be equal to 1 upon summation of all resistances so we can write it as u into a is equal to if we cancel the a we can write directly u is equal to 1 upon summation r therefore we will be getting u is equal to 1 upon hi plus b1 by k1 plus b2 by k2 plus 1 upon ho so for a for wall cylinder or spear we can use the concept of overall heat transfer coefficient to find out the heat flow rate when it is occurring by combined modes of conduction and convection now let us see how to find out the heat flow rate through composite cylinder which is made up of two materials again we are assuming that these cylinders are coaxial and length of cylinder is l and it is also exposed to the convective environment because we have assumed that the inner radius cylinder radius of inner cylinder is r1 and outer radius is r2 whereas inner radius of outer cylinder is r2 and outer radius of outer cylinder layer is r3 and assuming that the materials are having thermal conduct with k1 and k2 let the hot gas or hot steam is flowing through inside this particular hollow cylinder for which the heat transfer coefficient is h i and on the outside we are having the ambient at temperature ta and heat transfer coefficient as to now again by using a electrical analogy method we can write the electrical circuit for this composite cylinder problem as it is having again two conductive resistances for two materials having conductivities k1 and k2 and two convective resistance on outer surface so we can write the convective resistance as first convective resistance as one upon h i into a i conductive resistance for first layer of the material ln of r2 by r1 upon 2 pi k1l ln of r3 by r2 upon 2 pi k2l and one upon h o into a o the major difference between plain wall and cylinder is that in case of cylinder the area of heat flow does not remain constant so we can write as the area a i for inner cylinder will be equal to 2 pi r1l whereas the outer surface area of the cylinder through which the heat will be launched by convection will be given by 2 pi r3 into l so by putting this equation this a i and a o values in the above resistances I can write the equation for heat flow rate as q is equal to tg minus ta divided by summation r summation of all thermal resistances which I can further write it as q is equal to tg minus ta divided by one upon h i into inner surface area of the inner sphere that is 2 pi r1l plus the conductive resistance for plain cylinder we know that ln of r2 by r1 divided by 2 pi k1l similarly I can write the conductive resistance for second sphere l ln of r3 by r2 divided by 2 pi k2l plus convective resistance on the outer surface that is one upon h o into 2 pi r3l if we take 2 pi l as the common term from the denominator we can write this equation as q is equal to 2 pi l tg minus ta divided by 1 upon h i into r1 plus ln of r2 by r1 divided by k1 plus ln of r3 by r2 divided by k2 plus 1 upon h o into r3 so this equation gives us the heat flow rate for composite cylinder which is passed by pkna tata magroil publishing company limited new deli and heat and mass transfer by R.K. Rajput S.San and company limited new deli thank you