in heat transfer problem sometimes we want to determine how long it will take to cool down a subject or uh we want to determine the temperature of a subject after a certain time um so we cannot assume that the heat trans problem is not uh is a steady state problem um so temperature in this case uh is function of time as well as the position um but if the temperature is a function of time as well as positions the heat transfer problem is uh very complicated so there are several method to tackle uh this problems so first lumped cleance method and the second is exact solution uh this is basically the solving the heat division equations uh we learned in week one uh so there will be um two uh second order uh in space and the false order in time um and actually this is beyond the scope of this course so we are going to we are not going to solve uh the Trent problem analytical and the third method is use uh uh is to use a numerical method so uh after this video I'm going to talk about uh solving the Trent conduction problem using numerical method uh in this video we are going to focus on lump compon method so the let's take a look at the concept of lump capacitance method uh this is an idealization of the heat conduction problem so we assume that the object have a uniform temperature so typically uh temperature uh may be a function of position so in rectangular coordinate temperature may be function of X Y and Z position and it's also a function of time uh to simp implify this uh heat transfer problem so we may assume that temperature is only function of time so we have an object and the temperature at the surface and also the inside of the medium may have a uniform temperature or have a very negligibly small temperature difference between the surface of the object and at the temperature uh inside the object then we assume that it has a uniform temperature so temperature is not function of position but it only function of time this greatly simplify the heat transfer problem at the small uh expanse of um accuracy so for example copper ball from oven so copper ball has very high thermoconductivity so he conducts very well uh through the solid medium so assuming that the copper bow is very small with a high thermoconductivity so it may have very small temperature difference between the outer surface of the copper and the center very center of the copper ball so we assume that it has a uniform temperature um temperature is only function of time so in this case we may assume the system is a lumped system so we use the lump capacitor method to predict the temperature of the object at certain time T so now we want to determine the relationship uh between the temperature of the object and the time so con a solid body that is initially at a specially uniform temperature ti so assuming this this is our object temperature TI initially ti so this is a lump system has a uniform temperature at any instant time T So initially or time equal zero temperature of the object is TI it is sudden sub molds in a fleet at temperature T Infinity so Fleet temperature is T Infinity so there's the temperature difference between the surrounding Fleet and the solid object uh there will be heat transfer between the object and the surrounding Fleet so take the energy balance uh relations the rate of change of internal energy if equals to rate of convective heat transfer on the surface of the body so if the heat removed from the suface solid suface to the surrounding then the energy lost the rate of change of internal energy energy lost by the solid object equals to the rate of convective heat transfer uh from the solid surface to the surrounding fluid so energy of the solid object may be expressed using specific kit so solid for solid incompressible substance CP or CV or we can just write C for incompressible substance CP equal CV or we just write it C specific hit the definition of specific heit is the energy required to raise the temperature of a substance per unit mass by a degree Celsius or degree Kelvin so it has the unit of energy per unit Mass de cius right or kelvin uh so this is the definition of specific kit so if you multiply by Row V Row V is uh Mass right multiply by mass and multiply by DT temperature so this term had a unit of watt right Jew per second on the right hand side this is the rate of convective heat transfer from the surface to the surrounding flet it also have the unit of what um to integrate uh this equations uh we awesome introduce Theta Theta is T minus t Infinity here T is the temperature of the solid object which is a function of time it's only function of time it's not function of position since this is a lump system T is function of time T Infinity is a constant it's the surrounding flid temperature so D Theta D seta equals DT since T Infinity is constant uh so we now we see D set and set and DT so we now we are able to integrate this equations uh with respect to set and time at Time Zero uh Theta temperature is t i minus t infinity or Theta I a certain time T temperature is T minus t Infinity again T is function of time T minus t Infinity so integrated from Theta I initial temperature to the final temperature at time T Theta so we got this relations by integrating uh this this expression so set over seta I or seta is defined as T minus t Infinity IDE by set I or T minus t Infinity T initial temperature of the solid object T Infinity surrounding fle temperature so this is temperature over temperature so it's a dimensionist number dimensionist temperature on the right hand side we also we should also have dimensionist number right since left hand side is dimensionist right hand side must be also dimensionist so exponential of minus H as as is the what is as as coming from the convection heat transfer heat transfer so as s is the surface area heat transfer area by convection is the surface area that touches the surrounding flid so a heat transfer area over density row times C specific heit times V volume of the solid object um so this term is also dimensionist right so this term is Dimension is what is the unit of time time is second right so this term must be 1/ 2 and we this quantity is defined as so this is the reciprocal of this values reciprocal of this value is defined as time constant to to time constant which has the unit of second right since this is the reciprocal of this the unit is second so density time specific heat time volume divide by convection heat trans coefficient times surface area so you you you can do the unit uh analysis and the unit of this term should be uh second so this is defined as time constant so this expression may be Rewritten as this dimensional is temperature equals exponential my negative T time over time constant to so time second time constant also has the unit of second so second over second become dimensionless so left hand side and right hand side are both uh both dimensionless so this is a very important equations uh in this topic uh so this is the final result and we may use this Expressions to determine the temperature of a body temperature of an object at time t so here T Infinity is a constant surrounding F temperature TI is the initial temperature of the object solid object T is the time uh to is this property row density specific heat volume H convection heat transfer coefficient a s is a heat transfer area so here time and uh sorry temperature and time are the only variable in this equation so at certain time T if time is given we can determine the temperature at the given time or we may be able to use this Expressions to determine the time at the given temperature if the temperature of the body is given we can use this equation to solve it for time T So note that this expression may be used for uh the time or for the temperature so uh when you so let's talk more about the time constant to uh when time equals to when in this expression when time equals to it becomes -1 become -1 exponential netive 1 is 368 so this this dimensionist temperature becomes 368 368 and you can plot this how the temperature various of an object so y AIS is dimensional is temper temperature xais time so Dimension is temperature so at time equals z temperature at time equal Z is TI so T minus t Infinity over T minus t Infinity so initially dimensional is temperature become one so initially It Starts Here dimension temper dimensionist temperature is one as time increases this Dimension is temperature decreases and when time equals time constant this value Dimension is temperature become 368 right and as time further increases Dimension is temperature approaches zero okay so actually this graph may be also um drawn on T temperature and time um graph so temperature maybe initial temperature and let's say this is the final temperature T Infinity so initial temperature of the object is greater than the final temperature but then graph May Decay as this or if the initial temperature is lower than the final temperature so if the object is cold surrounding flee temperature is hot then the graph may look like this okay so this is um so outer this is uh heating so object is heated this is cooling object is being cooled and uh both graph uh may be Express in dimensionless form and actually dimensionist temperature always start from one and it approaches zero no matter uh What uh condition it has so either cooling or heating Dimension temperature approaches zero um so this time constants uh is a property of the characteristic of a object so it has the unit of second if the time constant is small then we can say that uh the object responds to a change of a temperature quickly if the time constant is large then we can say that the object slowly to response to temperature change okay so you can also find the time constant of a temperature measuring devices like like a thermocouples or thermometer or RTD or biometric thermometer so you can determine the time constant of the temperature measuring device so if you want to have a fast uh response to a temperature change you you must use a temperature measuring device with a small time constant uh if the um Trent response doesn't really matter uh in your temperature measurement then you may use a large time constant temperature measuring device um so when the time uh again uh when the time becomes time constant Dimension is time Dimension is temperature is point 3 6 8 okay uh this can be interpreted as you know so we can express this expression uh solve it for T temperature so it for T and from this expression we can say that at time constant to time constant to is Def def as the time required for the medium for the object to reach 63.2% of the instant temperature change from initial to the final temperature 63.2% of the temperature change so for example um if the initial temperature so let's say temp initial temperature 10° C final temperature is 100° C then uh so time constant is defined as the you know 63.2% of the instant temperature change so 60.3 of 632 * uh 10 - 100 + 100 this is the temperature at one time constant right so we can determine the time constant of um object based on this relations and the definition of time constant can be also uh expressed as this um so in practice iCal uh heat transfer problems we want to justify whether it is valid to use the lump capacitor System since this is a simplification uh this is an approximation method so we need to know whether it is reasonable to assume that uh the system has a uniform temperature or you know we may not be able to you know assume that the system does uh system has a uniform temperature across the uh system so we use a bon number a bon number is a dimensionless number Beyond number which is defined as H * l/ k h convection heat trans coefficient l l is the characteristic length or volume of the object over a a is the surface area volume over a so volume has the unit of cubic meter a has the unit of square meter right so the unit of characteristic length is meter uh it's called characteristic length K is the thermoconductivity of the fleet so actually it's not the conduct thermal conductivity of the fleet but it's the thermal conductivity of the solid so you need to be careful K is the thermal conductivity of the solid not the fluid um so p number can be Rewritten as this so multiply temperature difference in the denominator in the numerator so H * delta T which is relevant to convection at the surface of the body in the denominator we have K time Delta t/ L this is relevant to conduction within the body so definition of Bon number is the ratio of convection over conduction this expression may be Rewritten as l/ k a over 1 / ha so we are already familiar with this expression l/ Ka is the conduction resistance 1 / ha is convection resistance so it can be defined as the conduction ratio of conduction resistance over convection resistance so p number will be small if if the conduction resistance is small which means heat conducts well within the body beond number is large if the conduction resistance is large while beon number is large when the convection at the surface is large beon number is small when the conduction within the body is large right so in general uh beond if the beond number is less than or equal to .1 we assume that the lump capacitance system is applicable so the smaller the B number the more likely uh the system uh can be assumed as a lumped Capac system so based on the definition of beon number small body small body means L is small right L is small so Bon number is small high K high thermoconductivity of the solid medium or beond small beond number and low H low be number right so small body high K low H uh most likely satisfy the Criterion for lump system so small cable with the low H low H when do you have low H in air cooled in air or cooled in water cooled in air so air has much less what smaller specific H and density so H for air is much smaller than H4 water so if the object is cooled or heated it in air more likely beond number the small will be smaller so more likely we can assume we can apply the lump capacitance system uh compare the Slender body or round body slender body has a smaller L compared to round body so smaller be a number for slender body more likely uh we can apply the lump cap Sun System compare the cool by fan or cooled without a fan so naturally cooled so H value for no fan will be smaller right so smaller AG smaller p number more likely Lum capacitor system will be applicable so let's take a look at this example uh we have a metal plate with thermoconductivity density specific hit temperature initial temperature uh we submerged uh the plate into the fluid at 800° C so metal plate low temperature the fleet high temperature so it's heating the plate H value uh is 200 wat per M Square Kelvin uh so if we assume so we want to um predict the temperature of the plate after 2 minutes so if we assume that the plate has a uniform temperatures and we know the temperature and time relations then you know the analysis will be very simple if we assume that it has a uniform temperature but if we cannot apply the lump capacitor model uh then the analysis will be a lot more complicated so first things uh to do is to see uh if it's reasonable to assume this is a lump system so to determine the lump system uh we need to find Bon number so if the Bon number is less than 0.1 the lump capacitor model will be applicable so assuming that it has a constant properties uniform H convection heat trans coefficient so Bon number is h l/ k h 200 wat per M Square Kelvin L characteristic length so L is is V volume over surface area so here um surface area actually this is um it has a six sides right six side top bottom and four sides but uh actually the height of this plate is very small 1 cm so compared to top and bottom surfaces this the side four sides are relatively small so we neglect the four side uh surface area so volume is L time a let's say a is this area L * a is the Sur uh volume and a surface area is top and bottom so 2 a so a a cancels out l 1 cm or 10 mm 2ide by 2 so 5 mm that is the characteristic length beond number H L over K thermoconductivity of the fleet or of the plate so again thermal conductivity of the solid K is the thermoconductivity of the solid so maybe in the uh you know problems you also know the thermoconductivity of the fleet you know don't get confused K thermoconductivity of the solid not the fluid so K be careful of using the correct K so calculate HL over K which is 00556 this is dimensionist number make sure you unify the dimensions 00556 which is much smaller than1 there's no doubt that we can assume lump capacitance system uh so we want to predict the temperature of the object of the matter after 2 minutes so we use the [Music] temperature temperature and time relations right temperature and time so we write it for temperature which is function of time okay so t surrounding fle temperature 800° C TI initial temperature of the plate exponential negative time 120 seconds 2 minutes or 120 seconds over to what is to time constant time constant again time constant row dens time specific time v/ a or characteristic length L over H so density 2800 specific hit 800 characteristic length 005 mide by H 200 right so 12 second IDE by time constant also has second so temperature after 120 seconds can be calculated 289 de C so initially temperature of the plate is 20° C surrounding flee temperature 800° C after 2 minutes temperature Rises and become 689 DE cius after 2 2 minutes so you can determine the temperature of the plate at any time well uh this question maybe you know made to determine the time and at you know time at certain you know temperature of the place maybe 100° C so it may be asked to determine the time how much time will it take for the metal to reach at 100° C or you know whatever the temperature so we just use this relations to determine the time at a given temperature well we use this expression to determine the temperature at a specific time all right uh this is all for lump cap capacitance model thank you