HW_4_Soln

HW_4_Soln - ME 309 Fall 2008 Section 2 (Merkle) Homework #...

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ME 309 Fall 2008 Section 2 (Merkle) Homework # 4 Due Fri. 5 September 2008 Problem 4-1. Water enters a garden-hose nozzle with an average velocity of 1.8 m/s . The diameter at the nozzle entrance is 15 mm . The nozzle is open such that the effective flow area at the exit is an annulus with outer and inner diameters of 6 mm and 5 mm , respectively. What is the average water velocity at the exit annulus? SOLUTION: Known : V in , D in , D out,eff Find: V out Sketch : Assumptions: Use conservation of mass Fluid is incompressible Constant and uniform fluid properties Steady State Solution : Apply conservation of mass principle: dt dM m m m CV gen out in = + - Third and fourth terms are zero: 0 = - out in m m out out out in in in A u A u ρ = properties are constant—density is constant ( ) 2 2 2 005 . 0 006 . 0 * / 015 . 0 * 8 . 1 / - = = π in in out A u u = 36.82 m/s Problem 4-2 . Consider a cylindrical tank with a diameter D of 1.0 m and a height H of 1.5 m as shown on the sketch. Water ( ρ = 1000 kg/m 3 ) enters the tank at the bottom through a tube with diameter d of 5 mm at a constant average velocity of 4 m/s. Initially (time t = 0), the tank is half full as shown; water continues to enter until the tank begins to overflow at time t full . Selecting the water inside the tank at any instant as the control volume of interest, write or Sketch of Cylindrical Tank
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develop a formal explicit expression of mass conservation for this control volume. Your result should be in the form of an ordinary differential equation. Solve your differential equation for t full and determine a numerical value using the data given. SOLUTION: Known : Cylindrical tank, D, H, ρ , d , u in Find : t full Sketch : Above Assumptions : Constant inlet velocity Constant density No mass generation Solution : Conservation of mass gives: dt dM m m m CV gen out in = + - simplifications (no outflow; no generation) give: dt dM m CV in = Incorporate variables from problem: dt dh D h D dt d u d in 4 4 4 2 2 2 π = = Simplifying: in u D d dt dh 2 2 = Integrate from h/2 to h: ( ) full in in t u D d t t u D d h h 2 2 0 2 2 2 = - = - Solve for t full : m h s s d u D h t in full 5 2 7500 005 . 0 * 4 1 2 5 . 1 2 2 2 2 2
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HW_4_Soln - ME 309 Fall 2008 Section 2 (Merkle) Homework #...

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