The pressure integral yields j s p pa n ds j p1

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Unformatted text preview: θ) 100Vj 24 4 Outlet d 20 2 Shower head π π ρV 2 d2 sin θ = ρ U 2 d2 sin θ 16 j 4 where we again make use of the fact that Vj = 2U . The pressure integral yields −j · S (p − pa ) n dS = −j · (p1 − pa ) π2 π d i − j · (p2 − pa ) d2 (−i) = 0 4 4 Thus, the vertical momentum equation simplifies to Ry = π ρ U 2 d2 sin θ 4 Bernoulli’s Equation: All of the conditions required for Bernoulli’s equation to hold prevail for this flow. So, applying the equation between the supply-pipe’s inlet and outlet, we have 1 1 p1 + ρ U 2 = p2 + ρ 2 2 1 U 2 2 =⇒ 3 p 1 − p2 = − ρ U 2 8 792 CHAPTER 6. CONTROL-VOLUME METHOD So, the horizontal force is Rx = π2 3 d − ρ U2 4 8 + π ρ U 2 d2 4 3 − cos θ 4 = 8 3 πρ U 2 d2 1 − cos θ 32 3 Therefore, the force required to hold the pipes and shower head in place is R= 3 8 1 πρ U 2 d2 1 − cos θ i + πρ U 2 d2 sin θ j 32 3 4 (c) In the case for which Rx = 0, we have 1− 8 cos θ = 0 3 =⇒ cos θ = Therefore, the angle θ is θ = cos−1 3 8 = 68o 3 8...
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