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Because of the above drawbacks, it is generally recommended to
            avoid using heating as a treatment process if at all possible. Otherwise,
            some of the benefits of heating may be realized with the minimum amount
            of heating.


            5.4.2  Methods of Heating Oil Emulsions
            The fuel used to supply heat in oil-treating operations is practically natural
            gas. Under some special conditions, crude oil may be used.
                 Heaters are generally of two basic types:
                 1. Direct heaters, in which oil is passed through a coil exposed to
                    the hot flue gases of the burned fuel or to introduce the emulsion
                    into a vessel heated using a fire tube heater.
                 2. Indirect heaters, in which heat is transferred from the hot flue
                    gases to the emulsion via water as a transfer medium. The
                    emulsion passes through tubes immersed in a hot water bath.
                 In general, the amount of free water in the oil emulsion will be a
            factor in determining which method is to be used. If free water is found to
            be 1–2%, then use an indirect heater. If the free-water content is more
            enough to hold a level around the fire tube, then use a direct heater. Both
            types are shown in Figure 10.


            5.4.3  Heat Requirement
            The amount of heat transferred to, or gained by, a fluid is generally
            given by

                 q ¼ mc T     Btu=h                                       ð4Þ
            where q is the rate of heat transferred/gained (Btu/h), m is the mass flow

            rate of fluid (lb/h), c is the specific heat of fluid (Btu/lb- F), and  T is the

            increase in temperature due to heat transferred ( F).
                 The mass flow rate, m, could be calculated from the volumetric flow
            rate, Q, that is normally given in BPD (bbl/day) as follows:
                                                        !
                                              3         3
                        bbl   1   day       ft        ft   lb
                 m ¼ Q                5:61      62:4
                        day  24    h       bbl        lb   h
            Therefore,
                 m ¼ 14:59
Q ffi 15
Q lb=h                                  ð5Þ

            where 
 is the specific gravity of the liquid.






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