exam1

# Tb 2 tb 2 a0 a12 1 2 g h0 391s solution to mt2

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Unformatted text preview: t The drain time TB is such that h(t=TB) = 0: € € 2 2 h0 = 2 g /[ A0 A12 − 1] .TB => € 2 TB = 2 [ A0 A12 − 1] / 2 g h0 ≈ 39.1s Solution to MT2 - Problem #3 (Using velocity not flow rate): (a) The CV is the part of fluid inside the pipe, then cut off at inlet (2) and at outlet (2) Using Conservation of Mass Equation, we can relate the velocities at the inlet (1) and outlet (2): v1 A1 = v 2 A2 ⇒ v 2 A1 D12 1 = = 2= ≅2 v1 A2 D2 0.7 2 (1) Using Bernoulli’s Equation with no loss, we can relate the pressures at the inlet(1) and outlet(2): 2 ρv12 v2 ⇒ p2 = 1 − + p1 2 v1 v12 p1 v2 2 p2 += + 2ρ 2 ρ € This gives the pressure ratio: € 2 p2 ρv12 v 2 ρv12 1 3ρv12 = 1+ 1− = 1+ 1 − ≅ 1− p1 2 p1 v1 2 p1 0.7 4 2 p1 The pressure-velocity “parameter” (2) 2 p1 ρv1 shows up naturally, as in the “bent pipe example” in class. € (b) Steady flow, no time derivative term, RTT for momentum in the x-direction: ( ) 2 € −v1ρv1 A1 = p1 A1 + Fx , ⇒ − v1 ρ + p1 A1 = Fx , or Note that the velocities are all related to Q1 € v1 = Fx p = −1 + 12 , (Note < 0.) ρv 21 A1 ρv 1 4 Q1 = 0.49v 2 πD12 (c) The equation for conservation of linear momentum applied in the y-direction: p = 1 + 2 2...
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## This note was uploaded on 01/24/2014 for the course ME 106 taught by Professor Morris during the Spring '08 term at Berkeley.

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