Problem 4.95[Difficulty: 3]RxyxCS V2V1p2p1V3ρ12Given:Data on heated flow of gasFind:Force of gas on pipeSolution:Basic equation: Continuity, and momentum flux in x directionpρR⋅T⋅=Assumptions: 1) Steady flow 2) Uniform flowFrom continuityρ1−V1⋅A1⋅ρ2V2⋅A2⋅+m3+0=V2V1ρ1ρ2⋅m3ρ2A⋅−=where m3= 20 kg/s is the mass leavingthrough the walls (the software does notallow a dot)V2170ms⋅62.75×20kgs⋅m32.75 kg⋅×10.15 m2⋅×−=V2322ms=For x momentumRxp1A⋅+p2A⋅−V1ρ1−V1⋅A⋅
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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.