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# The x momentum equation is s u u n ds i s p pa

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Unformatted text preview: the reaction force, Rx . The x-momentum equation is S ρu (u · n) dS = −i · S (p − pa ) n dS + Rx 6.88. CHAPTER 6, PROBLEM 88 791 First, the net flux of x momentum out of the control volume is S ρu (u · n) dS π 1 = (−ρ U ) −U d2 + − ρ U 4 2 Inlet 1 π2 π U d + (−ρVj cos θ) 100Vj 24 4 Outlet d 20 2 Shower head 3 1 πρ U 2 d2 − πρVj2 d2 cos θ 16 16 = Using the result of Part (a) to eliminate Vj , S ρu (u · n) dS = π ρ U 2 d2 4 3 − cos θ 4 Next, the pressure differs from its atmospheric value, pa , only at the inlet and outlet, wherefore −i · S π2 π π d i − i · (p2 − pa ) d2 (−i) = d2 (p2 − p1 ) 4 4 4 (p − pa ) n dS = −i · (p1 − pa ) So, the x-momentum equation simplifies to Rx = π2 π d (p1 − p2 ) + ρ U 2 d2 4 4 3 − cos θ 4 y -Momentum Principle: Denoting the y component of the reaction force by Ry , the momentum principle in the y direction is S ρv (u · n) dS = −j · S (p − pa ) n dS + Ry The net flux of y momentum out of the control volume is S ρv (u · n) dS π = (0) −U d2 + (0) 4 Inlet = 1 π2 π U d + (ρVj sin...
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