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Fax uniform validated

hi there my name is brody o'donnell and my student number is 214083414 and this is my week 3 video assignment so starting off the question here um we have a bunch of givens and um assumptions to be made so starting off we're given that um this object is a droplet so right off the bat we're going to assume that this is a sphere and we're given the diameter we're given t initial t and t surrounding our t infinity and we're given the convection coefficient the conduction coefficient and the heat capacity coefficient and we're also given the density of the material so other assumptions to be made here so as i mentioned in the question we can neglect radiation we're also going to assume constant thermal properties so in terms of the convection the conduction coefficient and the heat capacity and um we're going to assume uh that heat convection or convection heat transfer is uniform and we're also going to assume one-dimensional conduction through the object so starting off the governing equations uh first of all we have be its number for determining lump system and lc to go with and then we have the lump system equations here so first is the temperature equation along with um b to go with the temperature equation and the um the heat capacity or the uh sorry the heat value equation so right here and then we also have all these spatial variation equations um so in this case we're dealing with a sphere so let's put the sheet here so we're using all of the approximate one term solution equations for a sphere plus the q max equation here so those are all written down plus we also have importantly we have fourier's number to actually determine if we can use um the one term approximation equation so starting with the analysis so let's look at question a um okay so question a says assuming a lumped system uh determine the temperature of the droplet after four seconds okay so since we're using lump system we're going to use our lump system equations to determine the temperature okay so just gonna sub in a couple values here solve for the surface area solve for the volume of the sphere and then solve for our b value and then once we get the b value we can sub back into the main equation and we can solve for our temperature at 4 seconds which ends up being 462.5 kelvin so then moving on to b we're determining the the energy lost from zero to four seconds and for that we're going to use our heat equation here for lump system so first just determining the mass which is just going to be volume times density and subbing in there we can determine our mass value down here and then moving forward we can sub in our mass into our q equation and our other values here and we can solve for 15.646 kilojoules of energy lost from zero to four seconds okay so now on to c evaluate the validity of the uniform temperature assumption okay so in this case the uniform temperature assumption is only valid if the be it number is less than or equal to 0.1 so now we're going to solve for the be it number solve for lc which is just volume over area or surface area and so we determined that the biet number is equal to 0.333 which is greater than 0.1 so this means that the system is it's invalid to use a lumped system analysis because be it number is too large okay so now on to d so our answer for c was invalid so that means we have to continue on to d and determine the temperature at the surface of the sphere at t equals four and determine the total energy lost okay so first thing we're going to do is we're going to calculate fourier's number to determine if we can use the one term approximation so we can only use it if fourier's number is greater than 0.2 so going through and basically just subbing in all of our values here we determine that 40's number is equal to 0.667 which is greater than 0.2 so we're good to use the one term approximation okay so part one of d is to determine the temperature at t equals four seconds so we're just going to use this first equation right here saphir and determine the temperatures also so the first thing to do is we have to go to the table 4-2 and in this case we have to recalculate recalculate b its number for this for these tables for a sphere so we did that and our b a number ends up being 1.0 and in this case we're dealing with sphere so we're going to use 1.5708 and 1.2732 for our lambda and our a1 values okay so that's everything i just explained there here's b it number and then our lambda and our a value and next thing we can do is we can actually just sub into this sphere equation and we're going to start off with the right hand side of the equation so we can calculate our theta value so we just sub and solve here and then once we have our theta value we can use that to easily calculate the temperature value at the surface so i'm once again subbing and solving and we find out that our temperature is equal to 487.6 kelvin at the surface so this is interesting as we calculated the temperature at part a and um let's just go back to that so we calculated to be 462 but now we're calculating it to be 487. so that's pretty big difference and um which explains that the lum system was not valid and we needed to use the one term approximation so moving on to part two we have to determine the energy lost from zero to four seconds so we're going to use our our our heat equation here and so starting off we have to calculate our theta naught value for a sphere which is this equation right here and basically just subbing and solving we already have all these values found and calculate that and then we also go through and we have to calculate our q max value once again subbing and solving q max equal to 18.09 kilojoules and then we can finally sub into our main equation here and determine that the energy lost is equal to 14.6 kilojoules from 0 to 4 seconds so once again we compare this to what we found in part a and which was 15.6 so you know that's about that's one kilojoule off um from the lumped system analysis so once again that proves why we need to use the one term approximation thank you