Page 428 - Handbook of Energy Engineering Calculations
P. 428

the  flow  rate  of  flue  gas,  w   =  450,000  lb/h  (56.7  kg/s);  flue-gas  average
                                                   g
               specific heat, c  = 0.25 Btu/(lb/°F) [1.05 kg/(kg · K)]; overall coefficient of
                                  p
                                                                             2
                                                                                                    2
               heat transfer of the boiler tubes, U = 13 Btu/(h · ft  · °F) [73.8 W/(m  · K)];
                                                                            2
                                                                                       2
               area of the boiler tubes exposed to heat, A = 900 ft  (83.6 m ); other values as
               before.
                  Then, Q = 450,000 × 0.25 × (2000 – t ) = 13 × 900 × (2000 – t )/{2.3 ×
                                                                                                  2
                                                                  2
               log   [(1455.4)/(t   –  544.6)]}.  or,  log   [(1455.4)/(t   –  544.6)]  =  13  ×
                                                                                  2
                                                                 10
                   10
                                     2
               900/(2.3 × 450.000 × 0.25) = 0.0452. The antilog of 0.0452 = 1.11, hence,
               [(1455.4/(t  – 544.7)] = 1.11, and t  = (1455.4/1.11) + 544.6 = 1850°F (1280
                                                          2
                            2
               K).
               4. Find the heat of vaporization of the water
               From the Steam Tables, the heat of vaporization of the water at 1000 lb/in                   2

               (6900 kPa), h  = 649.5 Btu/lb (1511 kJ/kg).
                                fg

               5. Compute the steam-generating rate of the boiler tube bank

               Heat absorbed by the water = heat transferred by the flue gas, or Q = w  × h                fg
                                                                                                     s
               = w  × c  × (t  – t ), where the mass of steam generated is w  in lb/h (kg/s);
                                1
                                      2
                                                                                          s
                          p
                    g
               other values as before. Then, w  × 649.5 = 450,000 × 0.25 × (2000 – 1850) =
                                                     s
                                                                                                 6
                            6
               16.9  ×  10   Btu/h  (4950  kJ/s)  (4953  kW).  Thus,  w   =  16.9  ×  10 /649.5  =
                                                                               s
               26,000 lb/h (200 kg/s).
               Related  Calculations.  The  Steam  Tables  appear  in  Thermodynamic
               Properties  of  Water  Including  Vapor,  Liquid,  and  Solid  Phases,  1969,
               Keenan,  et  al.,  John  Wiley  &  Sons,  Inc.  Use  later  versions  of  such  tables
               whenever available, as required.



               ENERGY DESIGN ANALYSIS OF SHELL-AND-TUBE HEAT
               EXCHANGERS




               Determine  the  heat  transferred,  shellside  outlet  temperature,  surface  area,

               maximum number of tubes, and tubeside pressure drop for a liquid-to-liquid
               shell-and-tube heat exchanger such as that in Fig.  10,  when  the  conditions
   423   424   425   426   427   428   429   430   431   432   433