Page 249 - Chemical Process Equipment - Selection and Design
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8.12.  INSULATION  OF  EQUIPMENT  219
           EXAMPLE  8,15--(continued)                               From (Eq. (4),
                in  the  radiant  section.  Repeat  the  calculation  with  an
                estimate of  60(E6)                                    T, = 920°F.
                   HI =:  248 + (77.14 - 60)(E6)/195,394 = 335.7,   28-31.
                    TI = 542,                                          LMTD = 735.6
                     T, = 100 + 0.5(542 + 900) = 821,
                    T, = 1550.5,                                    mean gas film temp is
                     F = 0.16498,
                                                                       4 = 0.5(400 + 547 + 735.6) = 841.3.
                    QR = 3190(0.6498)(28,727) = 59.55(E6).
                                                                    Since G = 0.325 lb/(sec)(sqft),
                Interpolating,
                            ?    8,   T,                               V, = 5.6 Btu/(hr)(sqft)("F)  [(Eq. (1091,
                   ~Bss"PII.3Kl           Qcalsd  Q/A
                    57.86   565  832.5 1553.7  59.43                         17'64(E6)  4282 sqft,
                    60.00   642  821   1550.5  59.55                       = 735.60 =
                 Interpolation  [547   1551.2  59.50  99001
                                                                          4282   = 64.1 bare tubes
                                                                       1.7344(38.5)
           26-27.
                   Q,,,,   = (7'7.14 - 59.50)(E6)                   or 16 rows of 4 tubes each. Spacing the same as of  the shield
                        = 17.54(E6).                                tubes.
                                                                       Beyond  the  first  two  rows,  extended  surfaces can  be
                Fraction lost in stack gas                          installed.
                   Qs/Q, = 1 - 0.02 - 0.75 = 0.23.                     Total rows = 2 + 1412 = 9.




           bank, and is called the equivalent cold plane area. Evaluation of the   is found with the trial calculation described in Steps 22-23  of  Table
           exchange factox  F is explained in item 9 of  Table 8.16. It depends   8.17 and may utilize the computer program of  Table 8.18.
           on the emissivity of  the gas and the ratio  of  refractory area A,,, to
           the equivalent cold plane area aA,.  In turn, A,,, =A -A,,  where   8.12. INSULATION OF  EQUIPMENT
           A is the area of  the inside walls, roof, and floor that are covered by
           refractory.                                        Equipment  at  high  or  low  temperatures  is  insulated  to  conserve
              In the convection zone of  the heater,  some heat  also is trans-   energy,  to  keep  process  conditions from fluctuating with  ambient
           fenred  by  direct  radiation  and  reflection.  The  several  contribu-   conditions, and to protect personnel who have occasion to approach
           tions to overall heat transfer specifically in the convection zone of fired   the  equipment.  A  measure  of  protection  of  the  equipment  metal
           heaters were coirrelated by Monrad [Ind. Eng. Chern. 24,505 (1932)].   against atmospheric corrosion also may be a benefit. Application of
           The combined effects are approximated by  item  10 of  Table 8.16,   insulation is  a  skilled trade.  Its  cost  runs  to  8-9%  of  purchased
           which is adequate for estimating purposes. The relation depends on   equipment cost.
           the temperature of  the gas film which is taken to be the sum of  the   In figuring heat transfer between equipment and surroundings,
           average  process  temperature  and  one-half  of  the  log  mean   it  is  adequate  to  take  account  of  the  resistances  of  only  the
           temperature difference between process and flue gas over the entire   insulation and  the  outside film.  Coefficients of  natural  convection
           tube bank. The temperature of  the gas entering the convection zone   are  in  Table  8.9  and  properties  of  insulating materials  at  several





              EXAMPLE 8.16                                                  1 + 0.0722vm
                                                                          -
              Applieaticaln of the Wilson-Lobo-Hottel  Equation           -      1.8327
           In the  case  of  Example  8.15,  25%  excess  air  was  employed,
           corresponding  to  19.0 lb/air/lb  fuel,  the  heat  release  was   :.  Q,  = 95.82(106) Btu/hr,
           Qf  = 102.86(10')  Btu/hr, and cuA,  = 3036. The effect will be found
           of  changing  the  excess  air  to  10%  (16.721b  air/lb  fuel)  on  the   which is the heat release with 10% excess air.
           amount  of  fuel  to  be  fired  while  maintaining  the  same  heat
           absorption.
              Ratioing Eq. (8.45) to yield the ratio of the releases at the two   With 25% excess air,  Q/Q, = 111.8327 = 0.5456,
           conditions,                                            With 10% excess air,  Q/Q, = 0.5456(102.86/95.82) = 0.5857,
                 Qf2   .=   1 + (16.72/4200)a,/3036           which  shows that  approximately 7%  more  of  the  released  heat  is
              102.86(106:)  1 + (19.O/4200)'d102.86(1O6)/3036   absorbed when the excess air is cut from 25% down io 10%.
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