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Unformatted text preview: MECH 411 Introduction to Fluid Mechanics Jan 18, 2005 American University of Beirut, Fall 20042005 Handout # PS6 Problem Set 6 Solution Textbook: 6.15, 6.62, 6.91, 6.103, 6.118 Required Sections: 6.1, 6.2, 6.3, 6.4, 6.6 (Effect of Rough Walls and Moody Chart only), 6.7, 6.8 (no turbulent flow solution), 6.9, 6.10 (series and parallel only.) 1 Problem 6.15 d = 8 mm L = 30 cm 1 2 p 1 = p a p 2 = p a 3000 Pa d = 8 mm L = 30 cm 1 2 p 1 = p a p 2 = p a 3000 Pa Figure 1: Problem 6.15 (a) ρ = 1200 kg/m 3 , μ = 6 kg/m.s, d = 0 . 008 m, L = 0 . 3 m. We assume that the flow is straw is operating between between (1) atmospheric pressure p a = 101 . 4 kPa and (2) lung pressure p l = p a − 3 kPa. Conservation of energy between (1) and (2) Δ p ρg + Δ z = h f (1) where Δ p = p 1 − p 2 = p a − p l , Δ z = z 1 − z 2 = − L (we assume the student is drinking the straw in a vertical position.), then h f = p a − p l ρg − L (2) The head loss is given by h f = f L d V 2 2 g = f L d 8 Q 2 π 2 d 4 g (3) where Q = V πd 2 4 is volume flow rate. The friction factor, assuming Laminar flow is, f = 64 Re = 64 ν V d = 16 νπd Q (4) Combining Eqs. (3) and (4), we get Q = πρgd 4 128 μL h f = πρgd 4 128 μL parenleftBigg p a − p l ρg − L parenrightBigg 2 For...
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