Consider fully developed conditions in a circular tube with...

Consider fully developed conditions in a circular tube with constant surface temperature Ts < Tm. Determine whether a small- or large- diameter tube is more effective in minimizing heat loss from the flowing fluid characterized by a mass flow rate of . Consider laminar conditions. Note: For laminar flow q' = 3.66 ? k (Ts - Tm). Select one: a. The heat loss rate is independent of the tube diameter. b. The heat loss rate is larger for large diameters tubes. c. The heat loss rate is larger for small diameters tubes. d. Non of them. e. All of them.

Transcript

00:01 Hello student according to this question so here the luminar flow okay luminar flow this is given by what this is given by an u and this is equals to hd divided by k okay but we know that new cell number newseld number is what is constant okay so we know this thing that this is constant right so product of product of h and d is also constant is constant is constant right so where we consider the formula for q which is equal to q equals to mcp delta t okay so we put the value so what is for this we get h multiplied by pi dl multiplied by l mt d okay so here what is that l m t d so l mt d will be equals to log mean temperature log mean temperature difference okay this is the meaning of the lmtd okay now this is q equals to hd q equals to what hd multiplied by pi l multiplied by l mpd okay so this is the expression for the q now we consider for the turbulent flow okay for the turbulent flow okay we are considering for the turbulent flow right so n u will be equals to hd divided by k and this is equals to f multiplied by r e power 0 .8 okay so what is this equals to so we consider the mass flow rate this is equals to what row a and v okay so what is that value equals to this is equal to row multiplied by pi divided by four and d square multiplied by v okay so from here what is this value equals to so for the constant okay here for the constant mv f multiplied by d power minus two okay and multiplied by the r e and this is equals to row v d divided by m u right so from here m multiplied by d by d by d will be equals to f d multiplied by d power minus one okay so from here n u will be equals to hd divided by k okay what is that value of n u this is f multiplied by by r e and 0 .8 okay and this is equal to f multiplied by d power minus 0 .8 so from here what is the value of h so h will be equals to d power minus 1 .8 right what is the value of h a so h a equals to f multiplied by d power minus 0 .8 so from here the heat transfer okay heat transfer is equal to heat transfer is equal to f multiplied by d power minus 0 point 8...

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