Problem 4.93[Difficulty: 3]Rxyx CSV2V1p2p1Given:Data on adiabatic flow of airFind:Force of air on pipeSolution:Basic equation: Continuity, and momentum flux in x direction, plus ideal gas equationpρR⋅T⋅=Assumptions: 1) Steady flow 2) Ideal gas CV 3) Uniform flowFrom continuityρ1−V1⋅A1⋅ρ2V2⋅A2⋅+0=ρ1V1⋅A⋅ρ2V2⋅A⋅=ρ1V1⋅ρ2V2⋅=For x momentumRxp1A⋅+p2A⋅−V1ρ1−V1⋅A⋅()⋅V2ρ2V2⋅A⋅⋅+=ρ1V1⋅A⋅V2V1−⋅=Rxp2p1−A⋅ρ1V1⋅A⋅V2V1−⋅+=For the airρ1P1RairT1⋅=ρ1200101+103×Nm2⋅kg K⋅286.9 N⋅m⋅×160273+K⋅×=ρ13.15kg
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.