Heat Transfer from Extended Surface (Fins)

we are going to analyze the heat transfer from extended surface extended suface or fins so uh so far we have studied heat conduction through solid medium so heat is conducted through solid medium and in many applications uh we should have um any ways to remove the heat from the solid surface to the surrounding uh by convection so according to Newton's low of cooling here it shows a Newton's low of cooling Q the rate of heat transfer by convection is proportional to H H is a convection heat transfer coefficient as s is the heat transfer area so in convection heat transfer area is the surface of the solid object that touches the surrounding fluid and temperature difference Q is also proportional to the temperature difference TS is the temperature of the solid suface T Infinity is the temperature of the fluid so to enhance the heat transfer from the solid suface to the surrounding fluid um we may uh increase H convection heat transfer coefficient or as heat transfer area or temperature difference between the solid surface and the free stream temperature uh typically we cannot change the surrounding uh fluid temperature T Infinity right so we may uh extract the heat from the salid surface to the surrounding air uh we cannot change the temperature of air and the purpose of uh using extended surface is is to minimize the decrease the temperature of the solid suface so TS most cases uh is fixed we cannot change TS H convection heat trans coefficient uh is dependent on many different parameters including fluid velocity flow motion and the suface conditions like roughness of the suface uh so the most effective way to increase H value is to increase the fluid velocity by using an external means such as a fan or pump so using pump or fan uh we can increase the fluid velocity but usually it's uh expensive to generate uh forc convection another uh way to increase increase the heat transfer by convection is to increase the heat transfer area this is more economical way to increase the heat transfer by convection so we are going to study uh about the inan heat transfer by convection by increasing the surface area as um we are going to in uh this is called a fins so attaching extended surface or fin to increase the surface area um for enhancing the heat transfer from the solid surface to the surrounding Fleet here are examples of fins so at the back side of the refrigerator you see um uh like a extended uh suface and the you see the coil uh the refrigerant uh uh flow along the coils and the fins are attached to uh increase the surface area so refrigerant uh is hot uh in the condenser and it reject heat to the surrounding and the heat transfer uh is enhanced by increasing the surface area and you see this heat sink uh us typically it's installed on on top of CPU or GPU uh to efficiently extract the heat uh from the electronic devices and this is a car radiator Al the coolant flow through the channels and it has a you you can see the fin small fins attached to the channels and to enhance the uh heat transfer area between the channel and the surrounding air and other types of heat SN it actually um used in combination with the fan so fan also generate um false convection and you see this is a motor and it has a plate like um Fin and another types of heat sink uh this is used uh in combination with a heat pipe and plate uh fins so heat pipe is uh very efficient device to uh remove the heat move the heat from one place to another so now uh let's see how we analyze the heat transfer across the fin and it's called fin equation so uh this is our solid suface so we have um solid suface and the temperature at the surface is TS and temperature of the solid suface and the fin base this is fin base which is also TS um and you see a single cylindrical fin attached to the solid suface so heat is conducted through the fin okay heat is conducted through the fin and let's say this actual direction is X AIS uh heat is conducted in X Direction assuming this is a onedimensional heat conduction which means you slice the uh cylinder cylindrical Fin and it'll be like a circular shape and at any point on this circular in this circle the temperature is uniform so there's no heat transfer uh in radial Direction but there's only heat transfer in axal Direction X X Direction so we can assume that this is a onedimensional heat conduction uh let's take a small volume element uh with with uh the distance Delta X so this small uh cylindrical volume element um has a parameter P parameter is P or if the radius of the cylinder is R then the p is 2 pi r and 2 pi r * Delta X is the surface area that touches the surrounding Fleet so there will be heat transfer uh from by convection from the outer surface of the F to the surrounding flid right in AAL Direction there theat conduction uh and from the solid surface outer surface of the cylinder to the surrounding fluid uh there's a heat convection so we can take the energy balance equation on the volume elements under stady States one dimensional heat conduction no heat generation so rate of heat transfer uh by conduction at X so on the left surface the distance from the thin base to the left surface is X so QX that's going in so assuming that she transfer in X Direction increasing X Direction uh that's going in and the rate of heat transfer leaving the control volume rate of heat conduction at X Plus Del Delta X QX plus Delta X also there's a heat transfer from the volume element to the surrounding fluid by convection rate of heat convection from the volume element equal zero on the right hand side if there's any change in the energy in the volume elements uh we have some T here but since this is under Ste State there's no change of the energy content within the volume element uh with time so that's why right hand side uh we have zero so using this uh energy balance equations and uh F LW of heat conduction and Newton's law of cooling we can uh get uh this relations and take the limits as Delta X approaches zero uh using the definition of d derivative uh it can be simpli simplified as this assuming that thermal thermoconductivity K is constant AC AC is the cross-sectional area P Pi R square uh assuming that this is this is a uniform cross-section uh fin so a does not change with X so K * a is also constant so you can divide this equations with k a c okay and to solve this equations conveniently we introduce a excess temperature which is defined as TX temperature at position x minus t Infinity T Infinity is the fluid temperature so this is a this is called excess temperature and then the equation uh can be written as this Theta is TX minus t Infinity m is defined as this uh this is simply coming from the previous an balance equation so this is a linear homogeneous second order equation with constant coefficient and the solution of this uh defent heat deficient equation is shown here and C1 C2 are arbitrary constants m is defined here so we are not going to solve this uh we will just use the final result uh this one has two integration constants so we need a two boundary conditions right this is a second order in space so we need to have a two boundary conditions to determine C1 and C2 so the first boundary condition is the temperature at the fin base temperature at the fin base is TB right TB temperature at the fin base TB is same as the solid suface temperature TS uh since we introduced seta B seta excess temperature uh at temperature at the fin base TB minus t infinity or seta B and there are four different cases uh we are going to look at four different cases at the uh thermal boundary condition at the fin tip so first convection heat transfer second adiabetic condition third prescrib temperature fors uh infin it fin so when you look at the fin so this is fin base base this is tip so here's are some area so we are talking about this area uh at the fin tip there may be heat convection at the fin tip so from this surface area to the surrounding there may be heat convection or we may assume this is insulated insulated so there's no heat transfer or we know the temperature at the fin tip specific temperature at the fin tip say 50° C so the temperature the fin tip is given or infinitely long the fin is infinitely long imagine that uh th based temperature 100° C if it's infinitely long temperature uh decreases uh as X increases as the distance increases eventually the temperature at the fin tip approaches the surrounding uh FL temperature so this is infinite fin uh boundary condition so we can express uh these thermal boundary conditions in mathematical form that is called boundary condition so if uh there's a convection at the fin tip H * Theta H time Theta Theta is the temperature difference between the fin tip and the surrounding right this is based on Newton's low of cooling H times temperature difference uh we didn't so this is a flux right we didn't multiply it by a so this is flux on the right hand side we also have a flux K * temperature gradient K * D Theta over DX at the fin tip are diabetic condition if it's adiabetic what is zero Q is zero at the fin tip or k * DT / DX is zero right so we can say since K is constant DT / DX is zero in this case we defined Theta so D Theta over DX is zero so since you know the T Infinity is constant so D Theta equals DT prescribed temperature uh since TL is given uh we can denote it as seta L infinitely long since the temperature at the fin uh tip same as the surrounding free temperature T Infinity minus t Infinity becom zero so these are the four different um boundary conditions and okay again we are not going to solve it but uh our textbook shows the you know uh the solution uh at each uh tip conditions so first case convective heat transfer what we are interested in uh in the fin is the temperature distribution seta over seta B this is a dimension list temperature distribution right temperature over temperature so Dimension is temperature distribution or uh we also interested in the total rate of heat transfer uh through the fin from the fin to the surrounding flet so here we see a large M large m is also defined here um and it's actually expressed it expressed in hyperbolic cosine s functions and case b are diabetic conditions temperature distributions and the rate of heat transfer prescribed temperature also shown um temperature distributions and the rate of heat transfer Infinity long fin temperature distributions and the rate of heat transfer so you just need to know how to use uh this table at each boundary condition um so let's see how the temperature profile look like uh in each boundary condition so first convection heat transfer at the fin tip the temperature at the fin base is the largest and temperature decreases and there will be so at the fin tip this is the at the fin tip at the fin tip um TL and there will be so this is the so you see T Infinity is the surrounding fluid temperature this is a short fin so short fin the fin tip temperature make may be still greater than the surrounding Fleet temperature so here you need to understand that um you need to understand so the larger the temperature difference so you see the temperature difference between the fin and the surround uh Fin and the surrounding Fleet surrounding Fleet temp temperature is fixed T Infinity so this is the this Arrow actually means he temperature difference between this point and the surrounding let's say this is T Infinity this is TS and temperature actually temperature of the surface de temp tempature of the fin surface decreases which means the temperature difference between the fin and the surrounding become smaller and smaller right and where does it have the greatest heat transfer which point does it have the greatest heat transfer between the fin and the surrounding at the fin base right so compare the temperature difference between the fin and the surrounding this is the largest this is the smallest the larger the temperature difference the greater the heat transfer so at the fin tip there's small temperature uh gradient between the fin and the surrounding so there's a smaller heat transfer at the fin base there's large temperature difference between the fin and the surrounding there so there's a greater uh Heat transfer so adiabetic fin tip if it's ad diabetic Q equal 0 so DT / DX is zero or D Theta over DX is zero so you take the slope tangent at the fin tip the slope temperature slope temperature gradient must be zero prescribe temperature temperature at the fin tip TL is given and not necessarily you the temperature Pro temperature um gradient at the fin tip is not necessarily zero okay so likely it's not zero infinitely long fin so temperature decreases the fin is very long so this also indicate the length fin is very long and eventually the fin tip temperature is uh close to the surrounding F temperature so here there's no temperature difference between the fin tip and the surrounding so then means there's no heat transfer there are no heat transfer there may be very small heat transfer a little more heat transfer a more heat transfer so actually it's not efficient to make the fin this long right so it's inefficient to make the fin uh to be very long so you need to maybe you can have a small fin this lengths and instead of having one long fin you may choose to have multiple fins to be more efficient uh there are several uh performance metrics uh to evaluate fins so first metric is a fin efficiency second metric is fin Effectiveness or we can also use Q to evaluate the fin the total rate of heat transfer from the fin so first fin efficiency so fin efficiency is defined as actual heat transfer weight from the fin divide by ideal heat transfer from the fin actual versus ideal qf actual qmax ideal so which case so we have a temperature distribution here along the fin 100 to you know it has a uniform temperature at 100° C another case temperature decreases 100 to 10 surrounding free temperature 10° C which is more efficient I idea so actually we are comparing the the heat transfer rate between the fin and the surrounding between the fin and the surrounding so the surrounding temperature is 10° C so compare the temperature difference between the fin and the surrounding so compare the temperature difference between the fin and the surrounding so compare this with 10 so actually if it's ideal ideal fin which has uni from temperature the temperature gradient maintained high right temperature difference is 90 at any point along the fin actual case temperature decreases maybe due to the low thermoconductivity of the fin temperature difference between the fin and the surrounding decreases 90 80 70 60 50 so there's less and less heat transfer As you move further away from the fin bits right so actual fin has a lower heat transfer rate from the fin to the surrounding ideal heat transfer ideal fin has a uniform heat uh temperature distribution across the fin and it maintain large temperature gradient thus it inance the it uh provide the maximum uh heat transfer it from the fin to the surrounding okay so this is fin efficiency efficiency is smaller than one right since it's comparing actual versus ideal so fin efficiency is smaller than one fin Effectiveness Epsilon uh numerator is still the same actual heat transfer R from the fin in the denominator Q with no fin so actual heat transfer you know the he transfer from the fin surface to the surrounding and he transfer rate from the surface area so assuming that there is no Fin and he transfer we are comparing the heat transfer uh from the fin Surface versus this area assuming that you know instead of having a fin we you know we just calculate the heat transfer from the same crosssectional area as the fin okay so we are comparing large versus small large versus small so fin Effectiveness must be greater than one right okay so fin effectivess must be greater than one fin efficiency is smaller than one um so now let's consider like what shape of fin would be more efficient so we see uh case D case D actually um we want to we want to know the qf the rad of transfer from the fin to the surrounding and since a b c and they are you know in they include hyperbolic um s cosine so it's complicated and let's just look at uh case D so m m is here so we are going to compare um we are going to see we are going to see M so fin Effectiveness can be expressed as this so uh here the higher the K the higher the effectiveness and you can also see p is the parameter of the fin if it's circular fin is 2 pi R AC is pi r squ so high p over AC um this p and AC cross-sectional area are dependent on the shape of the fin so to have high p over AC we should have thin plate like fin plate fin or Slender body fin okay so the shape of the fin should be flat or Slender instead of having cylinderical Fin and um Effectiveness uh is inversely proportional to H which means it'll be more effective uh to use the fin in involving low H value so H value for gas uh is smaller than a value in liquid so it's more effective to use uh fin in gas side which has lower H value than uh the fin in liquid side okay uh also lastly I want to show you the efficiency of common fin shape so in our textbook table 3.5 shows different shapes of Fin and um so by the shape of the fin uh we can calculate the efficiency of the fin from this efficiency we can actually get the actual heat transfer from uh different shapes of fin also one thing that I didn't mention uh t 3.4 uh this is for uniform cross-section so it can be used for any shape of fin as long as it has the uniform crosssection

we are going to analyze the heat transfer from extended surface extended suface or fins so uh so far we have studied heat conduction through solid medium so heat is conducted through solid medium and in many applications uh we should have um any ways to remove the heat from the solid surface to the surrounding uh by convection so according to Newton&#39;s low of cooling here it shows a Newton&#39;s low of cooling Q the rate of heat transfer by convection is proportional to H H is a convection heat transfer coefficient as s is the heat transfer area so in convection heat transfer area is the surface of the solid object that touches the surrounding fluid and temperature difference Q is also proportional to the temperature difference TS is the temperature of the solid suface T Infinity is the temperature of the fluid so to enhance the heat transfer from the solid suface to the surrounding fluid um we may uh increase H convection heat transfer coefficient or as heat transfer area or temperature difference between the solid surface and the free stream temperature uh typically we cannot change the surrounding uh fluid temperature T Infinity right so we may uh extract the heat from the salid surface to the surrounding air uh we cannot change the temperature of air and the purpose of uh using extended surface is is to minimize the decrease the temperature of the solid suface so TS most cases uh is fixed we cannot change TS H convection heat trans coefficient uh is dependent on many different parameters including fluid velocity flow motion and the suface conditions like roughness of the suface uh so the most effective way to increase H value is to increase the fluid velocity by using an external means such as a fan or pump so using pump or fan uh we can increase the fluid velocity but usually it&#39;s uh expensive to generate uh forc convection another uh way to increase increase the heat transfer by convection is to increase the heat transfer area this is more economical way to increase the heat transfer by convection so we are going to study uh about the inan heat transfer by convection by increasing the surface area as um we are going to in uh this is called a fins so attaching extended surface or fin to increase the surface area um for enhancing the heat transfer from the solid surface to the surrounding Fleet here are examples of fins so at the back side of the refrigerator you see um uh like a extended uh suface and the you see the coil uh the refrigerant uh uh flow along the coils and the fins are attached to uh increase the surface area so refrigerant uh is hot uh in the condenser and it reject heat to the surrounding and the heat transfer uh is enhanced by increasing the surface area and you see this heat sink uh us typically it&#39;s installed on on top of CPU or GPU uh to efficiently extract the heat uh from the electronic devices and this is a car radiator Al the coolant flow through the channels and it has a you you can see the fin small fins attached to the channels and to enhance the uh heat transfer area between the channel and the surrounding air and other types of heat SN it actually um used in combination with the fan so fan also generate um false convection and you see this is a motor and it has a plate like um Fin and another types of heat sink uh this is used uh in combination with a heat pipe and plate uh fins so heat pipe is uh very efficient device to uh remove the heat move the heat from one place to another so now uh let&#39;s see how we analyze the heat transfer across the fin and it&#39;s called fin equation so uh this is our solid suface so we have um solid suface and the temperature at the surface is TS and temperature of the solid suface and the fin base this is fin base which is also TS um and you see a single cylindrical fin attached to the solid suface so heat is conducted through the fin okay heat is conducted through the fin and let&#39;s say this actual direction is X AIS uh heat is conducted in X Direction assuming this is a onedimensional heat conduction which means you slice the uh cylinder cylindrical Fin and it&#39;ll be like a circular shape and at any point on this circular in this circle the temperature is uniform so there&#39;s no heat transfer uh in radial Direction but there&#39;s only heat transfer in axal Direction X X Direction so we can assume that this is a onedimensional heat conduction uh let&#39;s take a small volume element uh with with uh the distance Delta X so this small uh cylindrical volume element um has a parameter P parameter is P or if the radius of the cylinder is R then the p is 2 pi r and 2 pi r * Delta X is the surface area that touches the surrounding Fleet so there will be heat transfer uh from by convection from the outer surface of the F to the surrounding flid right in AAL Direction there theat conduction uh and from the solid surface outer surface of the cylinder to the surrounding fluid uh there&#39;s a heat convection so we can take the energy balance equation on the volume elements under stady States one dimensional heat conduction no heat generation so rate of heat transfer uh by conduction at X so on the left surface the distance from the thin base to the left surface is X so QX that&#39;s going in so assuming that she transfer in X Direction increasing X Direction uh that&#39;s going in and the rate of heat transfer leaving the control volume rate of heat conduction at X Plus Del Delta X QX plus Delta X also there&#39;s a heat transfer from the volume element to the surrounding fluid by convection rate of heat convection from the volume element equal zero on the right hand side if there&#39;s any change in the energy in the volume elements uh we have some T here but since this is under Ste State there&#39;s no change of the energy content within the volume element uh with time so that&#39;s why right hand side uh we have zero so using this uh energy balance equations and uh F LW of heat conduction and Newton&#39;s law of cooling we can uh get uh this relations and take the limits as Delta X approaches zero uh using the definition of d derivative uh it can be simpli simplified as this assuming that thermal thermoconductivity K is constant AC AC is the cross-sectional area P Pi R square uh assuming that this is this is a uniform cross-section uh fin so a does not change with X so K * a is also constant so you can divide this equations with k a c okay and to solve this equations conveniently we introduce a excess temperature which is defined as TX temperature at position x minus t Infinity T Infinity is the fluid temperature so this is a this is called excess temperature and then the equation uh can be written as this Theta is TX minus t Infinity m is defined as this uh this is simply coming from the previous an balance equation so this is a linear homogeneous second order equation with constant coefficient and the solution of this uh defent heat deficient equation is shown here and C1 C2 are arbitrary constants m is defined here so we are not going to solve this uh we will just use the final result uh this one has two integration constants so we need a two boundary conditions right this is a second order in space so we need to have a two boundary conditions to determine C1 and C2 so the first boundary condition is the temperature at the fin base temperature at the fin base is TB right TB temperature at the fin base TB is same as the solid suface temperature TS uh since we introduced seta B seta excess temperature uh at temperature at the fin base TB minus t infinity or seta B and there are four different cases uh we are going to look at four different cases at the uh thermal boundary condition at the fin tip so first convection heat transfer second adiabetic condition third prescrib temperature fors uh infin it fin so when you look at the fin so this is fin base base this is tip so here&#39;s are some area so we are talking about this area uh at the fin tip there may be heat convection at the fin tip so from this surface area to the surrounding there may be heat convection or we may assume this is insulated insulated so there&#39;s no heat transfer or we know the temperature at the fin tip specific temperature at the fin tip say 50° C so the temperature the fin tip is given or infinitely long the fin is infinitely long imagine that uh th based temperature 100° C if it&#39;s infinitely long temperature uh decreases uh as X increases as the distance increases eventually the temperature at the fin tip approaches the surrounding uh FL temperature so this is infinite fin uh boundary condition so we can express uh these thermal boundary conditions in mathematical form that is called boundary condition so if uh there&#39;s a convection at the fin tip H * Theta H time Theta Theta is the temperature difference between the fin tip and the surrounding right this is based on Newton&#39;s low of cooling H times temperature difference uh we didn&#39;t so this is a flux right we didn&#39;t multiply it by a so this is flux on the right hand side we also have a flux K * temperature gradient K * D Theta over DX at the fin tip are diabetic condition if it&#39;s adiabetic what is zero Q is zero at the fin tip or k * DT / DX is zero right so we can say since K is constant DT / DX is zero in this case we defined Theta so D Theta over DX is zero so since you know the T Infinity is constant so D Theta equals DT prescribed temperature uh since TL is given uh we can denote it as seta L infinitely long since the temperature at the fin uh tip same as the surrounding free temperature T Infinity minus t Infinity becom zero so these are the four different um boundary conditions and okay again we are not going to solve it but uh our textbook shows the you know uh the solution uh at each uh tip conditions so first case convective heat transfer what we are interested in uh in the fin is the temperature distribution seta over seta B this is a dimension list temperature distribution right temperature over temperature so Dimension is temperature distribution or uh we also interested in the total rate of heat transfer uh through the fin from the fin to the surrounding flet so here we see a large M large m is also defined here um and it&#39;s actually expressed it expressed in hyperbolic cosine s functions and case b are diabetic conditions temperature distributions and the rate of heat transfer prescribed temperature also shown um temperature distributions and the rate of heat transfer Infinity long fin temperature distributions and the rate of heat transfer so you just need to know how to use uh this table at each boundary condition um so let&#39;s see how the temperature profile look like uh in each boundary condition so first convection heat transfer at the fin tip the temperature at the fin base is the largest and temperature decreases and there will be so at the fin tip this is the at the fin tip at the fin tip um TL and there will be so this is the so you see T Infinity is the surrounding fluid temperature this is a short fin so short fin the fin tip temperature make may be still greater than the surrounding Fleet temperature so here you need to understand that um you need to understand so the larger the temperature difference so you see the temperature difference between the fin and the surround uh Fin and the surrounding Fleet surrounding Fleet temp temperature is fixed T Infinity so this is the this Arrow actually means he temperature difference between this point and the surrounding let&#39;s say this is T Infinity this is TS and temperature actually temperature of the surface de temp tempature of the fin surface decreases which means the temperature difference between the fin and the surrounding become smaller and smaller right and where does it have the greatest heat transfer which point does it have the greatest heat transfer between the fin and the surrounding at the fin base right so compare the temperature difference between the fin and the surrounding this is the largest this is the smallest the larger the temperature difference the greater the heat transfer so at the fin tip there&#39;s small temperature uh gradient between the fin and the surrounding so there&#39;s a smaller heat transfer at the fin base there&#39;s large temperature difference between the fin and the surrounding there so there&#39;s a greater uh Heat transfer so adiabetic fin tip if it&#39;s ad diabetic Q equal 0 so DT / DX is zero or D Theta over DX is zero so you take the slope tangent at the fin tip the slope temperature slope temperature gradient must be zero prescribe temperature temperature at the fin tip TL is given and not necessarily you the temperature Pro temperature um gradient at the fin tip is not necessarily zero okay so likely it&#39;s not zero infinitely long fin so temperature decreases the fin is very long so this also indicate the length fin is very long and eventually the fin tip temperature is uh close to the surrounding F temperature so here there&#39;s no temperature difference between the fin tip and the surrounding so then means there&#39;s no heat transfer there are no heat transfer there may be very small heat transfer a little more heat transfer a more heat transfer so actually it&#39;s not efficient to make the fin this long right so it&#39;s inefficient to make the fin uh to be very long so you need to maybe you can have a small fin this lengths and instead of having one long fin you may choose to have multiple fins to be more efficient uh there are several uh performance metrics uh to evaluate fins so first metric is a fin efficiency second metric is fin Effectiveness or we can also use Q to evaluate the fin the total rate of heat transfer from the fin so first fin efficiency so fin efficiency is defined as actual heat transfer weight from the fin divide by ideal heat transfer from the fin actual versus ideal qf actual qmax ideal so which case so we have a temperature distribution here along the fin 100 to you know it has a uniform temperature at 100° C another case temperature decreases 100 to 10 surrounding free temperature 10° C which is more efficient I idea so actually we are comparing the the heat transfer rate between the fin and the surrounding between the fin and the surrounding so the surrounding temperature is 10° C so compare the temperature difference between the fin and the surrounding so compare the temperature difference between the fin and the surrounding so compare this with 10 so actually if it&#39;s ideal ideal fin which has uni from temperature the temperature gradient maintained high right temperature difference is 90 at any point along the fin actual case temperature decreases maybe due to the low thermoconductivity of the fin temperature difference between the fin and the surrounding decreases 90 80 70 60 50 so there&#39;s less and less heat transfer As you move further away from the fin bits right so actual fin has a lower heat transfer rate from the fin to the surrounding ideal heat transfer ideal fin has a uniform heat uh temperature distribution across the fin and it maintain large temperature gradient thus it inance the it uh provide the maximum uh heat transfer it from the fin to the surrounding okay so this is fin efficiency efficiency is smaller than one right since it&#39;s comparing actual versus ideal so fin efficiency is smaller than one fin Effectiveness Epsilon uh numerator is still the same actual heat transfer R from the fin in the denominator Q with no fin so actual heat transfer you know the he transfer from the fin surface to the surrounding and he transfer rate from the surface area so assuming that there is no Fin and he transfer we are comparing the heat transfer uh from the fin Surface versus this area assuming that you know instead of having a fin we you know we just calculate the heat transfer from the same crosssectional area as the fin okay so we are comparing large versus small large versus small so fin Effectiveness must be greater than one right okay so fin effectivess must be greater than one fin efficiency is smaller than one um so now let&#39;s consider like what shape of fin would be more efficient so we see uh case D case D actually um we want to we want to know the qf the rad of transfer from the fin to the surrounding and since a b c and they are you know in they include hyperbolic um s cosine so it&#39;s complicated and let&#39;s just look at uh case D so m m is here so we are going to compare um we are going to see we are going to see M so fin Effectiveness can be expressed as this so uh here the higher the K the higher the effectiveness and you can also see p is the parameter of the fin if it&#39;s circular fin is 2 pi R AC is pi r squ so high p over AC um this p and AC cross-sectional area are dependent on the shape of the fin so to have high p over AC we should have thin plate like fin plate fin or Slender body fin okay so the shape of the fin should be flat or Slender instead of having cylinderical Fin and um Effectiveness uh is inversely proportional to H which means it&#39;ll be more effective uh to use the fin in involving low H value so H value for gas uh is smaller than a value in liquid so it&#39;s more effective to use uh fin in gas side which has lower H value than uh the fin in liquid side okay uh also lastly I want to show you the efficiency of common fin shape so in our textbook table 3.5 shows different shapes of Fin and um so by the shape of the fin uh we can calculate the efficiency of the fin from this efficiency we can actually get the actual heat transfer from uh different shapes of fin also one thing that I didn&#39;t mention uh t 3.4 uh this is for uniform cross-section so it can be used for any shape of fin as long as it has the uniform crosssection

Transcript for:Heat Transfer from Extended Surface (Fins)

Transcript for:
Heat Transfer from Extended Surface (Fins)