 వావర్నిలియకి సినౕంలూవాసిలొతి ఆరాం ఆరింప్ సిమింుమన్యని మావ్లికికారాంపిమా పార్విమేకిద్లీతాల్టి. చాణలల్త్లిటాపా. కోసితా ವನೆವದ ಹೊದ ಮೈರೋ ಹ್ರೋ ಮೋಚ್ಟ್ ಪೈ � zen ಗಳಾವ್ಅ ಹೊದ೛ ಆವ೏ ದೈ ಸಿ call of skinny బల్ ఆప్ ఏమ౒ినాని మైర్ంర్, ఎరిసిలు స౉OOD 1 music నియాడťళదవధి, టనాడünüz నిటనానబ moinsత౿ selfieisions పరినానదరచి ఘరొష్ంల . the heat loss always decreases whereas in case of cylindrical and spherical surfaces it is not always true with the addition of heat insulation the heat loss from cylindrical and spherical surfaces under circumstances they may increase so let us see why this happens for that purpose we will consider one cylinder which is made up of insulating material let us assume that the inner radius of this cylinder is R1 outer radius of this cylinder is R2 the inner and outer surfaces are T1 and T2 and TA is the ambient temperature to which the heat will be lost from this insulating surface by convection and HO is the outside air heat transfer coefficient now if we consider the heat transfer by convection then there are going to be two resistances first resistance will be the material resistance due to the insulating material which we already know that by using the electrical analogy method the thermal resistance for cylindrical geometry is ln of R2 by R1 divided by 2 pi k where R1 is the inner radius of insulating cylinder or outer radius of tube surface and R2 is the outer radius of insulating resistance whereas another resistance which will come into picture will be 1 upon 2 pi R2L into HO which is known as the surface resistance or convective resistance now we can see that when we put the more and more insulation when we add the insulation at a more thickness with the outer radius will increase and with the increase of R2 the material resistance that is R insulation will increase whereas the surface resistance or outside resistance due to convection will decrease now how this will vary let us see initially initially with if we assume that the inner radius is R1 when we go on increasing the thickness of insulation up to certain radius R2 that is called as the critical radius of insulation when the R2 will increase then it is found that the resistance which is offered due to the material initially it increases at a slower rate and then after RC it increases at a faster rate and so this is the material resistance or conductive resistance whereas the surface resistance which is offered due to the convective it increases at a faster rate initially and then it decreases slowly so this is called as the surface resistance so net effect if we see so net effect is that initially because the surface resistance decreases at a faster rate compared to material resistance the net heat flow rate will initially increase because the equivalent resistance or total thermal resistance offered due to surface and material resistance initially it decreases and then it increases so if we compare the heat loss from the insulated surface then it is found that the heat loss up to critical radius it goes on increasing and at a critical radius this heat loss is maximum and after critical radius the heat loss will decrease so concept of critical radius which we can apply with the advantage under two different circumstances for stream carrying pipes our aim is that we want to always reduce the heat loss so while the care is taken that the thickness of insulation or radius of insulation after putting insulation is kept always outer radius is kept always kept greater than critical radius that addition of insulation will reduce the heat loss however in case of electrical conductor the use of insulation is beneficial in two ways if you put an insulation which is equal to having the outer radius equal to critical radius then the heat dissipation rate will be maximum in case of electrical conductor the heat is generated due to the flow of current in ampere and R is the electrical resistance so this heat generated has to be dissipated to the surrounding at a faster rate if you dissipate this then the current carrying capacity of electrical conductor will be more therefore in case of electrical conductor we are providing insulation which will be having the outer radius outer radius equal to critical radius and it is also beneficial from point of view of preventing the electrical shock so from safety point of view now let us see how to derive the equation for critical radius of insulation for cylinder for cylinder we can write the equation of heat flow rate equal to Ti minus Ta divided by material resistance plus surface resistance which we can express 2 pi L into Ti minus Ta upon ln of R2 by R1 divided by K plus 1 upon R2 into HU now as we know that when the outer radius of insulation is equal to critical radius so differentiating this equation with respect to R2 and equating to 0 assuming the R1 K HO Ti Ta and length of cylinder as constant so because the numerator in this equation is a constant term we can differentiate only the denominator and we can get the relation for critical radius of insulation for cylinder so I can write it as d by dr2 of ln of R2 by R1 divided by K plus 1 upon R2 into HO equal to 0 so if I differentiate this denominator term with respect to R2 I will get ln of R2 minus upon K minus ln of R1 by but the logarithm of ln of R2 by K will be 1 upon R2 into K minus ln of R1 by K will be a constant term so this will be 0 the differentiation will be 0 minus 1 upon R2 into HO the differentiation of 1 upon R2 will be minus 1 upon R2 square into HO will be equal to 0 so if we rearrange this term we will get that R2 into K equal to 1 upon R2 square into HO therefore we will get R2 equal to critical radius of insulation equal to K upon H0 so this is an expression for the critical radius of insulation for cylindrical surfaces similarly we can derive the equation for critical radius of insulation for sphere so if we have done for cylinder we can write the equation for Q as Q is equal to Ti minus T upon R2 minus R1 divided by 4 pi K R1 R2 upon 1 upon HO into 4 pi R2 square so which we can take common term 4 pi from the denominator and shift it to the numerator we get the equation so if we write the differentiation term dQ by dr2 equal to 0 for cylinder by differentiating denominator we can get the equation as we can write this equation as dQ by dr2 equal to d by dr2 of the denominator term since numerator is constant I can write it as 1 upon K R1 minus R1 upon it will be 1 upon K R2 R1 will get cancelled plus 1 upon HO into R2 square will be equal to 0 so if we differentiate this term we get that since first term is constant derivative will be 0 so minus 1 upon K into R2 square the derivative of 1 upon R2 square minus 1 upon HO into R2 and this will be equal to 0 so this term will become now if we rearrange this term the equation will become this 1 upon HO into R2 equal to 1 upon K into R2 square whereas this particular derivative was minus 1 upon 2 into HO into R2 so if we take this then after rearranging this sum is of course a positive term so we will get the R2 is equal to 2K upon HO so R2 is equal to 2K upon HO is the equation for critical radius of insulation for cylinder the references which I have referred are heat transfer by PK now and heat and mass transfer by Rajpur thank you