MECH 314 Problems 8, Spring 2014

MECH 411 Introduction to Fluid Mechanics Jan 18, 2005 American University of Beirut, Fall 2004-2005 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 p1 = pa p2 = pa - 3000 Pa d = 8 mm L = 30 cm 1 2 p1 = pa p2 = pa - 3000 Pa Figure 1: Problem 6.15 (a) ρ = 1200 kg/m3, 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 pa = 101.4 kPa and (2) lung pressure pl = pa − 3 kPa. Conservation of energy between (1) and (2) ∆p ρg + ∆z = hf (1) where ∆p = p1 − p2 = pa − pl, ∆z = z1 − z2 = −L (we assume the student is drinking the straw in a vertical position.), then hf = pa − pl ρg − L (2) The head loss is given by hf = f L d V 2 2g = f L d 8Q2 π2d4g (3) where Q = V πd2 4 is volume flow rate. The friction factor, assuming Laminar flow is, f = 64 Re = 64ν V d = 16νπd Q (4) Combining Eqs. (3) and (4), we get Q = πρgd4 128µLhf = πρgd4 128µL �pa − pl ρg − L � 2 For L = 30 cm, the volume flow rate is Q = −0.029 cm3/sec; the student would be unable to drink the milk shake. (b) For L = 15 cm, the volume flow rate is Q = 0.138 cm3/sec. We can easily verify that Re< 2000; the flow is laminar as we assumed. 3 Problem 6.62 L = 2000 ft, z2 − z1 = 120 ft, Q = 3 ft3/s. Pipe: Cast iron d = 6 in, pump efficiency 0.75. Power of pump in hp? From Table 6.1 Page 367, the surface roughness for cast iron is ǫ = 0.00085 ft. The energy equation between (1) and (2) is ∆p ρg + ∆z + ∆(V 2) 2g = hf + � hm − hp (5) where ∆() = ()1 − ()2, hf is the friction head loss, and hp is the head delivered by the pump. The minor losses � hm are neglected in this problem. Referring to Figure P6.62 in book, we assume (1) and (2) correspond to locations on the surfaces of the two tanks so that p1 = p2 = pa. Then hp = hf − (z1 − z2) With hf = f(L/d) (V 2/2g), we get hp = f L d V 2 2g − (z1 − z2) The average velocity is V = 4Q/πd2 = 15.28 ft/s. For water µ = 2.09E − 5 slug/ft.s and ρ = 1.94 slug/ft3. Then Red = V d/ν = 7.09 × 105. The flow is turbulent. The friction factorcan be obtained from (a) Moody Chart Fig. 6.13 page 366, (b) Eq. 6.48 page 365, or (c) Eq. 6.49 page 366. We shall use Eq. 6.49 because it is easier to use 1 f 1/2 ≃ −1.8log   6.9 Red + �ǫ/d 3.7 �1.11  (6) leading to f = 0.02274136153, hf = 330 ft. Then hp = hf −(z1−z2) = 330+120 = 450 ft. The power of the pump is therefore P = ρQghp ξ = 204 hp where ξ = 0.75 is the pump efficiency. 4 Problem 6.91 L = 0.6 m, V = 2 m/s. Channel: isoceles cross section: a = 2 cm, β = 80o. SAE Oil at 20 C: ρ = 870 kg/m3, µ = 0.104 kg/m.s from From Table A.3 page 811. The Perimeter of the cross section is P = 2a(1 + sin(β/2)) = 0.06571 m. The cross section area is A = a2 sin(β/2) cos(β/2) = 0.000197 m2. The hydraulic diameter is dh = 4A/P = 0.012 m. For Reynolds number is Re= V dh/ν = 200.58: Laminar flow. For Laminar flow in triangular duct with θ = β/2 = 40o, From Table 6.4 page 383, the friction factor f = 52.9/Re = 0.2637. The friction head loss is hf = f(L/dh) (V 2/2g) = 2.6933 m. The pressure drop is obtained from the energy equation ∆p ρg + ∆z = hf With ∆z = 0, then the pressure drop is ∆p = ρghf = 22.96 kPa. 5 Problem 6.103 From Table 6.1 Page 367, the surface roughness for cast iron is ǫ = 0.00085 ft. Energy equation from (1) at surface to (2) at surface according to Figure P6.103 page 436 ∆p ρg + ∆z + ∆(V 2) 2g = hf + � hm − hp + ht (7) For ∆z = z1 − z2 = 45 ft, pump head hp = 0, turbine head ht = 0, and p1 = p2 = pa. Also V1 ≃ V2 = 0 (recall that (1) and (2) are at surfaces of tanks and not correspond to flow inside pipe) then ∆z = hf + � hm = V 2 a 2g faLa da + V 2 b 2g fbLb db + V 2 a Kent 2g + V 2 a Kexp 2g + V 2 b Kexit 2g = V 2 a 2g �faLa da + d4 a d4 b fbLb db + Kent + Kexp + d4 a d4 b Kexit � (8) The subscripts a and b refer to the two sections of the pipe. The minor losses are sharp entrance, sudden expansion and sharp exit: Figure 6.22 page 390 for sharp entrance (sudden contraction with D = ∞): d/D = 0 , we get Kent = 0.4. Figure 6.22 page 390 for sudden expansion: d/D = 1/2, we get Kexp = 0.5. Figure 6.22 page 390 for sharp exit (sudden expansion with D = ∞) : d/D = 0, we get Kexp = 1. Then V 2 a 2g (240fa + 7.5fb + 0.9625) = 45 (9) We assume trubulent flow. (We will validate later). Initial guess: fully rough flow, Moody Chart: so for ǫ/da = 0.0102, fa = 0.038 and ǫ/db = 0.00051, fb = 0.0165. Eq. (9): Va = 16.85 ft/s, then Rea = Vada/ν = 1.303E5, Reb = Vbdb/ν = 65171. Update from Eq. (6) fa = fa(Rea, ǫ/da) = 0.03866, fb = 0.03188. Eq. (9): Va = 16.629 ft/s, then Rea = 1.286E5, Reb = 64314. Update from Eq. (6) fa = 0.03866, fb = 0.03189. The converged solution is Va = 16.628 ft/s, Q = 0.09 ft3/s. 6