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Chapter 3 x Integral Relations for a Control Volume The integral term ³ wV dV ds | 1 h is very small and will be neglected, and p1 p2.. Then wt dt 12 ª 2gh º V1 | « ¬ D 1 »¼ h ³ ho dh h1/2 293 , where D (D/d)4; but also V1 ³ dt, or: h o (b) the tank is empty when [] dh , separate and integrate: dt 2 1/2 t ª 2g º « ¬ D 1 »¼ 1/2 ª 1/2 ­ g ½ º «ho ® ¾ t» , D ¯ 2(D 1) ¿ ¼» ¬« 0 in (a) above, or tfinal 3.180 The large tank of incompressible liquid in Fig. P3.180 is at rest when, at t 0, the valve is opened to the atmosphere. Assuming h | constant (negligible velocities and accelerations in the tank), use the unsteady frictionless Bernoulli equation to derive and solve a differential equation for V(t) in the pipe. §D· ¨ ¸ ©d¹ 4 Ans. (a) [2(D 1)g/ho]1/2. Ans. (b) Fig. P3.180 Solution: Write unsteady Bernoulli from 1 to 2: 2 ³ 1 wV V22 V12 gz 2 | gz1 , where p1 ds wt 2 2 p2 , V1 | 0, z 2 | 0, and z1 dV dV L, so the diff. eqn. is 2L V2 dt dt This first-order ordinary differential equation has an exact solution for V The integral approximately equals V §V t· Vfinal tanh ¨ final ¸ , where Vfinal © 2L ¹ 2gh h const 2gh 0 at t 0: Ans. 3.181 Modify Prob. 3.180 as follows. Let the top of the tank be enclosed and under constant gage pressure po. Repeat the analysis to find V(t) in the pipe.