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1、ProblemsandSolutionstoChemicalEngineeringPrinciples化工原理教研组编Chapter1FluidMechanics1.Theflamegasfromburningtheheavyoilisconstitutedof8.5%CO2，7.5O2，76%N2，8%H2O(involume).Whenthetemperatureandpressureare500℃and1atm,respectively,calculatethedensityofthemixedgas.Solut
2、ion:ThemolecularweightofthegaseousmixtureMnisMn=My+My+My+My=44×0.085+32×0.075+28×0.76+18×0.08=28.86kg/kmolUnder500℃,1atm,thedensityofthegaseousmixtureisρ==×=0.455kg/m³2.Thereadingofvacuumgaugeintheequipmentis100mmHg,trytocalculatetheabsolutepressureandthegaugepr
3、essure,respectively.Giventhattheatmosphericpressureinthisareais740mmHg.Solution:TheabsolutepressureintheequipmentisequaltothatatmospherepressureminusvacuumP(absolute)=740―100=640mmHg=640×=8.53×10N/m²Thegaugepressure=-vacuum=-100mmHg=-（100×）=―1.33×10N/m²orthegaug
4、epressure=-（100×1.33×10）=―1.33×10N/m²3.Asshowninthefigure,thereservoirholdstheoilwhosedensityis960kg/m³.Theoillevelis9.6mhigherthanthebottomofthereservoir.Thepressureabovetheoillevelisatmosphericpressure.Thereisaroundhole(Φ760mm)atthelowerhalfofthesidewall,thece
5、nterofwhichis800mmfromthebottomofthereservoir.Thehand-holedoorisfixedbysteelbolts(14mm).Iftheworkingstressoftheboltsis400kgf/cm2,howmanyboltsshouldbeneeded?Solution:Supposethestaticpressureoftheliquidon0-0levelplaneisp,thenpistheaveragepressureoftheliquidactingo
6、nthecover.Accordingtothebasichydrostaticsequationp=p+ρghTheatmospherepressurewhichactsontheouterflankofthecoverispa,thenpressuredifferencebetweentheinnerflankandouterflankisΔp=p―p=p+ρgh―ρghΔp=960×9.81(9.6 ―0.8)=8.29×10N/m²ThestaticpressurewhichactsonthecoverisΔp
7、×=8.29×10N/m2ThepressureoneveryscrewistheNumberofscrew=3.76×10=6.234.Therearetwodifferentialpressuremeterfixedonthefluidbedreactor,asshowninthefigure.ItismeasuredthatthereadingareR1=400mm，R2=500mm,respectively.TrytocalculatethepressuresofpointsAandB.Solution:The
8、reisagaseousmixtureintheU-differentialpressuremeter.Supposearethedensitiesofgas,waterandmercury,respectively,thenthepressuredifferencecouldbeignoredfor《.ThenAccording
