Page 282 - Analysis, Synthesis and Design of Chemical Processes, Third Edition
P. 282

to obtain this limiting value.


                    Example  9.21  illustrates  the  use  of  each  of  the  above  formulas  to  calculate  the  yearly  depreciation
                    allowances.


                    Example 9.21


                    The  fixed  capital  investment  (excluding  the  cost  of  land)  of  a  new  project  is  estimated  to  be  $150.0

                    million,  and  the  salvage  value  of  the  plant  is  $10.0  million.  Assuming  a  seven-year  equipment  life,
                    estimate the yearly depreciation allowances using the following:
                          a.   The straight-line method
                          b.   The sum of the years digits method
                          c.   The double declining balance method
                                                             6
                                                                                6
                                We have FCI  = $150 × 10 , S = $10.0 × 10 , and n = 7 years.
                                             L
                                Sample calculations for year 2 give the following:
                                For straight-line depreciation, using Equation (9.22),
                                                              6
                                                6
                                d  = ($150 × 10  – $10 × 10 )/7 = $20 × 10      6
                                2
                                For SOYD depreciation, using Equation (9.23),
                                                                           6
                                                             6
                                d  = (7 + 1 – 2) ($150 × 10  – $10 × 10 )/28 = $30 × 10       6
                                2
                                For double declining balance depreciation, using Equation (9.24),
                                                       6
                                                                        6
                                d  = (2/7) ($150 × 10  – $42.86 × 10 ) = $30.6 × 10       6
                                2
                    A summary of all the calculations is given in Table E9.21 and presented graphically in Figure E9.21.


                    Table E9.21 Calculations and Results for Example 9.21: The Depreciation of Capital Investment for
                                                                     7
                    a New Chemical Plant (All Values in $10 )
   277   278   279   280   281   282   283   284   285   286   287