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Appendix A   Combustion Fundamentals           277




                 Metric conversions for these units are as follows: ft   0.304 8   m; lbf   4.448   N;
             and lb   0.453 6   kg.
                 Other quantities, such as velocity, area, and pressure, described in terms of the
             basic quantities, are known as secondary quantities.

             2.1.1 Mass and Force
             Sir Isaac Newton’s famous second law of motion states: “The acceleration produced
             by a particular force acting on a body is directly proportional to the magnitude of the
             force and inversely proportional to the mass of the body.”

                                     Force 
 Mass   Acceleration

             Where
                       force   N (lbf),
                       mass   kg (lb), and
                                                      2
                                                                   2
                 acceleration   g   local gravity 9.807 m/s (32.174 ft/sec ).
                 It is important to note the choice of the word “proportional” over the more com-
             monly used (but incorrect) equal. To express Newton’s law as an equality, it is neces-
             sary to multiply the right-hand side (RHS) by 1/g , where g   universal gravitation
                                                         c        c
                                          2
             constant   32.174 (ft-lb)/(lbf-sec ) (1 lbf   4.448 N).
                 From the foregoing, the units of g/g are lbf/lb (lbf/lb   9.806 65   N/kg).
                                                   c
             Whenever pounds of force must be converted to pounds of mass, it must be multi-
             plied by (g/g ) (note that ft-lb   1.356   N • m).
                         c
                 To illustrate, calculate how much force you would exert on a spring bathroom
             scale standing on the moon, which has approximately one-sixth of the earth’s gravity:

                                               2
                                                                      2
                      lbf   lb   ([32.174/6] ft/sec )/[(32.174 (ft-lb)/(lbf-sec )]   lb/6
                 Therefore, a mass of 200 lb, would require a force of 200/6   33.33 lbf. One of the
             most common situations that requires use of the ratio (g/g ) is the determination of
                                                                  c
             pressure, which is typically represented to the force exerted over a specific area by a
             column of liquid of known height and density:

                           Pressure (lbf/sq ft)   Height (ft)   Density (lb/cu ft)

                 By multiplying the RHS of the equation by (g/g ), we get
                                                           c
                                                                               2
                                                         2
                    lbf/sq ft   ft   (lb/cu ft)   (32.174 ft/sec )/[(32.174 ft-lb)/(lbf-sec )]
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