 mey shall Mr Sunil Dutthay and Kulkarni go to deliver a video session on steady-state heat conduction 5. In this video session, we are going to discuss the concept of critical radius of insulation. At the end of this session, you'll be able to analyse the critical radius of insulation for cylinder and spear and you'll be able to explain its physical significance. also you will be able to derive the equation for critical radius of insulation for cylinder and sphere the contents of this video session are critical radius of insulation for cylinder and sphere and physical significance of critical radius of insulation Now, when we talk about the critical radius of thickness of insulation as we are aware that the insulation is used to reduce the heat loss by conduction and convection so it is very natural to expect that more insulation we put the lesser will be the heat loss from the surface but it is not always true that addition of insulation will reduce the heat loss in case of plane surfaces it happens that whenever we put more and more insulation the heat loss from the surface or geometry will decrease however in case of cylindrical and spherical surfaces it is observed that when we put the insulation insulation additional insulation the initially the heat flow rate increases and after certain radius of insulation or thickness of insulation it decreases so the radius or thickness of insulation at which the heat loss is maximum is called as the critical radius of insulation for understanding the critical radius of insulation concept we will consider one cylinder which is made up of an insulating material for this cylinder the inner radius of cylinder is R1 which can be outer radius of conducting tube and R2 is outer radius of insulation let us consider T1 and T2 are inner and outer surfaces of this insulating cylinder TA is the surrounding air temperature and HO is the heat transfer coefficient now when we consider this insulation cylinder we can using an electrical analogy method we can draw the electrical circuit for this particular geometry in which you can recall that whenever we are having any conductive layer and convective environment just recall that for conductive resistance of cylinder is given by ln of R2 by R1 upon 2 pi kL whereas the convective resistance which is given as 1 upon HO into A where here the outer radius outer surface area of this cylinder is 1 upon 2 pi R2L into HO so we are having in the electrical circuit two resistances one is called as the conductive resistance and other is called as the convective resistance this conductive resistance or resistance due to insulation is nothing but a material resistance which is expressed as ln of R2 by R1 upon 2 pi kL whereas the convective resistance can be considered as a surface resistance which is equal to 1 upon 2 pi R2L into HO now if we observe when we put more and more insulation from the equation of this particular material resistance as the R2 will increase with the addition of insulation outer radius will increase thus material resistance will increase whereas when we put more and more insulation with the addition of insulation on the outer surface of cylinder this particular surface resistance decreases because it is in the denominator now if we observe the variation of these resistances then it is found that initially when we put the insulation on the cylindrical surface on the outer surface of cylinder this particular material resistance increases slowly up to the certain radius and afterward it increases at a higher rate so rate of increase of this material resistance or conductive resistance initially is gradual and after certain radius that is critical radius it increases at a faster rate whereas the surface resistance relatively decreases at a faster rate up to critical radius after critical radius it decreases slowly so effect of these two is that the equivalent resistance which is the equivalent of material resistance and surface resistance it decreases initially up to the critical radius and after critical radius it goes on increasing so net effect on the heat flow rate through the insulating surface is that up to critical radius of insulation the heat flow rate goes on increasing at a critical radius where the equivalent resistance is minimum as we can see from this figure at this critical radius the heat flow rate from the surface is maximum and this radius or thickness of insulation we call it as either critical radius of insulation or critical thickness of insulation now when the insulation is further added beyond critical radius we find that this particular heat flow rate will go on decreasing if we observe this concept of critical radius of insulation can be used with advantage under two circumstances firstly in case of steam carrying pipes our aim is that we want to reduce the heat loss from the steam pipe in that case we should always keep the outer radius of insulation that is R2 greater than critical radius of insulation so that the heat loss from the steam carrying pipe can be reduced whereas in case of electrical conductor the heat is generated due to flow of current which is given by qg equal to i square r where i is the electrical current in ampere side r is the electrical resistance in ohm now in case of electrical conductor whatever heat is generated due to passage of current it should be dissipated as fast as possible to increase the life of conductor and to maintain the temperature of conductor steady in that case the care is to be taken that we should keep the outer radius of insulation equal to critical radius of insulation so that the heat loss from the conductor will be maximum now let us derive the equation for critical radius of insulation for cylinder now as we have seen in the earlier figure considering the same geometry we can derive the critical radius of insulation for cylinder the equation for heat flow rate for hollow cylinder is given by ti minus ta divided by ln of R2 by R1 upon 2 pi kL plus 1 upon 2 pi R2L into HO now this first is the material resistance as we have discussed earlier and this is the surface resistance that is convector resistance which can be modified as taking 2 pi L common in the denominator and we can take it into the numerator so we can write the equation for q equal to 2 pi L ti minus ta upon ln of R2 by R1 upon k plus 1 upon R2 into HO now as we are aware that the heat loss from the surface is maximum at critical radius so in order to obtain the equation for critical radius we have to differentiate this equation with respect to R2 assuming the inner surface temperature ambient air temperature length of the cylinder inner radius R1 and thermal conductivity of the material as constant when we differentiate this equation with respect to R2 and equating to 0 now we will get the equation for critical radius however as the numerator in this equation is constant we will differentiate the only denominator and equate it to 0 we can write as d by dr2 of ln of R2 by R1 upon k plus 1 upon R2 into HO equal to 0 so we know that the differentiation of ln of R2 by k will be equal to 1 upon R2 into k minus the differentiation of ln of R1 by k as R1 and k we are considering constant it will be 0 so we can write directly ln of R2 by k minus the differentiation of 1 upon R2 into HO equal to 0 is given by minus 1 upon R2 square into HO and equate to 0 so by rearranging the terms we get that R2 equal to k upon H0 so R2 equal to k upon H0 where k is the thermal conductivity of the material and HO is the outer convective coefficient convection coefficient on the outer surface this equation gives us the value of critical radius of insulation for cylinder which we can write it as k upon H0 equal to RC similarly now we can derive the equation of critical radius of insulation for sphere for spherical geometry let us consider one hollow sphere which is again made up of an insulating material having thermal conductivity k and let us assume that the inner radius of this sphere is R1 outer radius is R2 thermal conductivity of material is k the surrounding air temperature be ta and the convective coefficient on outer surface is HO so you can recall that by electrical analogy method if you recall q equal to ti minus to divided by summation of resistances now in this case also we are having two resistances in the path of law of heat first resistance is the insulating resistance due to the insulating material plus outer resistance on the outer surface that is convective resistance so q is equal to ti minus to now if you recall that the conductive resistance for spherical surface is given by R2 minus R1 upon 4 pi k R1 R2 plus convective resistance is 1 upon HO into 4 pi R2 square which we can rearrange as taking 4 pi common in the denominator and we can take it into numerator by dividing into numerator and denominator by 4 pi we will get q is equal to 4 pi ti minus to upon R2 minus R1 upon k R1 R2 plus 1 upon HO into R2 square now in order to obtain the equation for critical radius of insulation as we have done for cylinder similarly as the numerator is constant we will differentiate the denominator considering k R1 ti to as constant if we differentiate the denominator I can write it as d by dr2 of simplifying the first term as R2 minus R1 upon k R1 R2 L 1 upon k R1 minus 1 upon k R2 plus HO into R2 square equal to 0 so on differentiation since the first term is constant its derivative is 0 minus 1 upon k R2 derivative is 1 upon 2 k R2 square and if I take the derivative of 1 upon HO R2 square which we get as 1 upon HO R2 cube so after rearranging we are getting the equation as 1 upon 2 k R2 square equal to 1 upon HO into R2 cube therefore by rearranging this term we will get R2 equal to 2 k upon HO so this is nothing but the equation for critical radius of insulation for sphere now as we have discussed this concept of critical radius of insulation one should ensure that when we are applying the insulation on the surface if the pipe radius radius of the pipe is lesser than the critical radius then applying the insulation on the surface will always reduce the heat loss however if the outer radius of insulation is greater than critical radius then when we put the insulation the heat loss will always decrease however if the outer radius of pipe is less then up to critical radius the heat loss will increase and after critical radius it will decrease so this is the critical radius of insulation concept the references for this video session are heat transfer by P.K. Nag Tata McGrawill Publishing Company Ltd heat and mass transfer by R.K. Rajput thank you