Conservation of Mass - Clarkson University

Applying the definition given in Equation (12) yieldsRQ 1 2V = av2πrV ( r) dr rV ( r)drπR = πR = R2 2 20 0R∫ ∫ (13)Two example problems are considered next. The first example involves using the steady versionof the equation of conservation of mass to make calculations involving a natural gas pipeline(source: Fluid Mechanics for Chemical Engineers by Noel de Nevers). The second exampleshows how to use the unsteady version of the equation of conservation of mass to calculate therate of change of the height of liquid in a tank.Example 1 -- Natural Gas PipelineD1 = 2.00 ftD2 = 2.50V1 = 33.7 ft / sV2= ?ft1 2p1 = 820 psiap2 = 450 psiaT1 = 70 FT2 = 58 FGas Constant R = 10.73g( f/ )2 3lb in • ftMolecular Weight = 16• R( lb mole)lblb moleFind m, V2 2This is a long pipeline, which is shown schematically. The natural gas flows through a section ofcircular pipe of diameter 2.00 feet for some distance, and then through a larger section of circularpipe of diameter 2.50 feet. The pressure and temperature at locations 1 and 2 are specified. Weneed to find the velocity at location 2, and the mass flow rate at location 2.6