Page 288 - Modelling in Transport Phenomena A Conceptual Approach
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268   CHAPTER 8.  STEADY MICROSCOPIC BALANCES WITHOUT GEN.


            around 3 W/ m2. K.  Therefore, the maximum value of  the critical radius is approx-
            imately  3.3cm,  and  in most practical  applications,  this will  not pose  a problem.
            Therefore,  critical  radius  of  insulation  is of  importance  only for small  diameter
            wires  or tubes.



            Example 8.8  Steam (fluid A) flows in a pipe  of  R1 = 30cm  and  R2  = 45cm.
            The pipe  is surrounded  by fluid B.  Calculate  the overall heat  transfer coefficients
            and sketch the representative temperature profiles for the following  cases:

            a) (hA) =10W/m2.K; (hB)=5000W/m2.K k=2000W/m2.K
            b) (hA) = 5000 W/ m2. K; (hB) = 8000 W/m2. K; k = 0.02 W/ m2. K
            c) (hA) = 5000W/m2.K; (hB) = 10W/m2.K; k = 2000W/m2.K

            Solution

            a) Note that the dominant  resistance to heat transfer .is that of fluid A. Therefore,
            one  expects the largest temperature  drop in this region.  Hence  Eqs.  (8.2-42) and
            (8.2-43) give  the overall heat transfer coeficients (1s
                    uA = (h)               =
                                 -1
                                    = (h~) 10W/m2.K

                                    -1
                    UB = (-) R2       = (hA) (2) o(3o) =6.67W/m2.K
                                                   =
                            (hA )R1                     45
             The expected  temperature profile for this me is shown below.












            b) In this case  the  dominant  resistance  to heat  transfer is that of  the pipe  wall.
             The overall heat transfer coeficients are

                                  k       -     0.02
                       VA =                              = 0.16 W/ m2. K
                             R1 ln(R2IR1) - (0.3) 1n(45/30)
                                  k              0.02
                       UB =               -               = 0.11 W/ m2. K
                             R2 ln(RzlR1) - (0.45) ln(45/30)
             The expected temperature profile for this case is shown below:
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