Page 208 - Excel for Scientists and Engineers: Numerical Methods
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CHAPTER 8                ROOTS OF EQUATIONS                          185



               Problems

                   Answers to the following problems are found in the folder "Ch. 08 (Roots of
                   Equations)" in the "Problems & Solutions" folder on the CD.


                1.  A circuit consisting of a source, a resistor  and  a  load, has  a current  i that
                   oscillates as a function of time t according to the following:

                                            7T                    7T
                                   i = 2.5 sin(-)e-2.5'  + 2.5 sin(2.5t - -)
                                            4                     4
                   Find the first time after t = 0 when the current reaches zero.)


               2.  In pipe flow problems the relationship
                                            aD3 + bD + c = 0
                   is encountered.  Solve for D, if a = 700, b = -2.9,  c = -300.


                3.  When the sparingly soluble salt BaC03 is dissolved  in water, the following
                   simultaneous equilibria apply:
                   BaC03 e Ba2++ C03'-                     Ksp = [Ba2'][C032-] = 5.1 x  lo-'

                   C0:-   + H20 + HCO< + OH-       Kb = [OH-][ HCO<]/[ C032-] = 2.1 x  lo4
                   Employing  mass-  and  charge-balance  equations,  the  following relationship
                   can be obtained for a saturated solution of BaC03 in water:
                                   [Ba2+I2  - JKbKsp [Ba2+]1'2 - Ksp = 0

                   Find the concentration of free Ba2'  in the saturated solution.


                4.  A  solution  of  0.10  M  nitric  acid  (HNO3)  is  saturated  with  silver  acetate
                   (AgAc), a sparingly soluble salt.  The system of mass-  and charge-balance
                   equations describing the system is
                    NO^-] = 0.1 o                                         (mass balance)

                   [Ag']  = S                                             (mass balance)
                   [Ac-] + [HAc] = S                                      (mass balance)
                   [Ag']  + [W] = [Ac-] + [NOS-]                         (charge balance)
                   [Ag'][Ac-]  = 4.0 x                                               KSp
                   [H'][Ac-]/[HAc]  =1.8 x                                           KO
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