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Pipe flow April 8 and 15, 2008 Outline Pipe Flow • Laminar and turbulent flows • Developing and fully-developed flows Larry Caretto • Laminar and turbulent velocity profiles: Mechanical Engineering 390 effects on momentum and energy Fluid Mechanics • Calculating head losses in pipes – Major losses from pipe only April 8 and 15, 2008 – Minor losses from fittings, valves, etc. • Noncircular ducts 2 Piping System What We Want to Do • Determine losses from friction forces in Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and straight pipes and joints/valves Theodore Okiishi Copyright © 2005 by John Wiley – Will be expressed as head loss or & Sons, Inc. All rights reserved. “pressure drop” hL = ΔP/γ • Will show that this is head loss in energy equa- tion if variables other than pressure change • System consists of – Losses in straight pipes are called “major” losses – Straight pipes – Losses in fittings, joints, valves, etc. are – Joints and valves called “minor” losses – Inlets and outlets – Minor losses may be greater than major – Work input/output 3 losses in some cases 4 Pipe Cross Section The Pipes are Full • Most pipes have circular cross section • Consider only flows where the fluid to provide stress resistance completely fills the pipe • Main exception is air conditioning ducts • Partially filled pipes are considered • Consider round pipes first then extend under open-channel flow analysis to non-circular cross sections – Extension based on using same equations as for circular pipe by defining hydraulic diameter = 4 (area) / (perimeter), which is Driving force Driving force D for circular cross sections is pressure is gravity Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, 5 6 Donald Young, and Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. ME 390 – Fluid Mechanics 1 Pipe flow April 8 and 15, 2008 Laminar vs. Turbulent Flow Laminar vs. Turbulent Flow II • Most flows of engineering interest are turbulent – Analysis relies mainly on experimentation guided by dimensional analysis – Even advanced computer models, called computational fluid dynamics (CFD) rely on “turbulence models” that have large degree • Laminar flows of empiricism have smooth • Can get some (very limited) analytical • Turbulent flows results for laminar flows layers of fluid have fluctuations Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and 7 8 Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Laminar vs. Turbulent Flow III Flow Development • Condition of flow as laminar or turbulent depends on Reynolds number • For pipe flows – Re = ρVD/μ < 2100 is laminar – Re = ρVD/μ > 4000 is turbulent – 2100 < Re < 4000 is transition flow • Other flow geometries have different characteristics in Re = ρVLc/μ and different values of Re for laminar and turbulent flow limits 9 Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and 10 Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Developing Flows Developing Flows II • Entrance regions and bends create • Entrance regions and bends create changing flow patters with different changing flow patters with different head losses head losses • Once flow is “fully developed” the head • Once flow is “fully developed” the head loss is proportional to the distance loss is proportional to the distance • Entrance pressure drop is complex – Complete entrance region treated under minor losses – Will not treat partial entrance region here 11 Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and 12 Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. ME 390 – Fluid Mechanics 2 Pipe flow April 8 and 15, 2008 Developing Flows III Fluid Element in Pipe Flow • After development region, pressure drop (head loss) is proportional to pipe length • Equations for entrance region length, ℓe le – Laminar flow: = 0.06 Re D le • Look at arbitrary element, with length ℓ, – Turbulent flow: = 4.4 Re1 6 and radius r, in fully developed flow D – Turbulent flow rule of thumb ℓe ≈ 10D • What are forces on this element? 13 Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and 14 Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Fully Developed Flow Extend Relation to Wall No change in momentum Flow Direction ∑F = πr 2 p1 − πr 2 ( p1 − Δp ) − τ 2πrl = 0 • Have Δp = 2τℓ/r for any r: 0 < r < R = D/2 x 2τl • Pressure drop is due • For wall r = R = D/2 and τ = τw = wall Δp = r to viscous stresses shear stress: Δp = 2τwℓ/R = 4τwℓ/D Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and 15 Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and 16 Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Fully Developed Laminar Flow Fully Developed Laminar Flow II • Can get ⎡ ⎛ r ⎞2 ⎤ exact u = uc ⎢1 − ⎜ ⎟ ⎥ equation for ⎢ ⎝R⎠ ⎥ ⎣ ⎦ R R pressure • Laminar shear drop stress profile 128μlQ found from Δp = uc du πD 4 τ=μ • Laminar ⎡ ⎛ r ⎞2 ⎤ dr du 2r 8μu Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore velocity u = uc ⎢1 − ⎜ ⎟ ⎥ Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore τ=μ = μuc 2 = 2c r dr R D ⎢ ⎝R⎠ ⎥ Okiishi. Copyright © 2005 by John Wiley & Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. profile ⎣ 17 ⎦ Sons, Inc. All rights reserved. 18 ME 390 – Fluid Mechanics 3 Pipe flow April 8 and 15, 2008 Fully Developed Laminar Flow III Effect of Velocity Profile • What is centerline velocity, uc? • Momentum and kinetic energy flow for ⎡ ⎛r⎞ R R 2⎤ mean velocity, V ∫ ∫ Q = VA = VπR 2 = udA = u 2πrdr = uc ⎢1 − ⎜ ⎟ ⎥ 2πrdr ⎢ ⎝R⎠ ⎥ ∫ & – FlowMomentum = m V = ρVAV = ρV2(πR2) A 0 0 ⎣ ⎦ & V2/2 = ρVAV2/2 = ρV3(πR2)/2 – FlowKE = m ⎡R R 3 r ⎤ ⎡r2 r4 ⎤ R R2 • Accurate representation uses profile ⎢0 ∫ Q = 2πuc ⎢ rdr − ∫ dr ⎥ = 2πuc ⎢ − 2 ⎥ = 2πuc R2 ⎥ ⎢ 2 4R ⎥0 ⎣ ⎦ 4 R ⎡ ⎛ r 2 ⎞⎤ 2 ⎣ ⎦ 4 ∫ ∫ FlowMomentum = ρudAu = ρ⎢uc ⎜1 − 2 ⎟⎥ 2πrdr = ρV 2 A 0 ⎜ R ⎟ 0 ⎢ ⎝ ⎠⎥ ⎣ ⎦ 3 R2 2Q 2VA 2VπR 2 A Q = πuc ⇒ uc = = = = 2V 3 u 2 1 ⎡ ⎛ r 2 ⎞⎤ R πR 2 πR 2 πR 2 V3 2 ∫ FlowKE = ρudA = ∫ ρ⎢uc ⎜1 − 2 ⎟⎥ 2πrdr = 2ρA 2 2 ⎢ ⎝ ⎜ R ⎟ ⎠⎥ 2 Centerline uc is twice the mean velocity, V 19 A 0 ⎣ ⎦ 20 Turbulent Flow Turbulent Flow Quantities Velocities at one point • For laminar and turbulent flows, the as a function of time velocity at the wall is zero – This is called the no-slip condition – Momentum is maximum in the center of the flow and zero at the wall u(t) = instantaneous • Laminar flows: momentum transport from wall t0 +T velocity to center is by viscosity, τ = μdu/dr 1 • Turbulent flows: random fluctuations exchange u= T ∫ u(t )dt u’ = velocity eddies of high momentum from the center with t0 fluctuation = u – u low momentum flow from near-wall regions 21 Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and 22 Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Momentum Exchange Turbulence Regions/Profiles turbulent eddy viscosity, η τ = (μ + η) du Fundamentals dr of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and • Very thin viscous sublayer next to wall Theodore Okiishi. Laminar flow – Turbulent flow – Copyright © – 0.13% of R = 3 in for H20 at u = 5 ft/s 2005 by John random eddies have Wiley & Sons, Inc. All rights molecular motion structure reserved. • Flat velocity profile in center of flow Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and 23 Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and 24 Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. ME 390 – Fluid Mechanics 4 Pipe flow April 8 and 15, 2008 Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, Profile Effect of Velocity Profile and Theodore Okiishi. 1n Copyright © 2005 by John Wiley & Sons, Inc. All u ⎛ r⎞ • Analysis similar to one used for laminar rights reserved. = ⎜1 − ⎟ Vc ⎝ R ⎠ flow profile – Determine momentum and kinetic energy Turbulent flow for mean velocity velocity – Correction factor multiplies average V profiles results to give integrated u2 and u3 values n = 6: Re = 1.5x104; Vc/V = 1.264 with n a n = 8: Re = 4x105; Vc/V = 1.195 function of n Re Momentum KE n = 10: Re = 3x106; Vc/V = 1.155 6 1.5x104 1.027 1.077 Laminar: Vc/V = 2 V = Q/A Reynolds number 8 4x105 1.016 1.046 25 10 3x106 1.011 1.031 26 Pipe Pipe Roughness roughness effects in • Effect of rough walls on pressure drop viscous may depend on surface roughness of sublayer pipe affects Fundamentals of Fluid • Typical roughness values for different Mechanics, 5/E by Bruce Munson, Donald Young, pressure materials expressed as roughness and Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All drop in length, ε, with units of feet or meters rights reserved. turbulent • Only turbulent flows depend on flow roughness length, laminar flows do not No effect on laminar flow 27 28 Use this Energy Equation table (p • Energy equation between inlet (1) and 433 of outlet (2) text) to p V2 p V2 z2 + 2 + 2 = z1 + 1 + 1 + hs − hL find ε γ 2g γ 2g • Previous applications allowed us to Fundamentals of Fluid compute the head loss from all other Mechanics, 5/E by Bruce Munson, Donald Young, and data in this equation Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights – Call this the measured head loss reserved. • We can compute it, but we have no way of knowing its cause 29 30 ME 390 – Fluid Mechanics 5 Pipe flow April 8 and 15, 2008 Pressure Drop/Head Loss Pressure Drop/Head Loss • We now seek a design calculation for hL • We now seek a design calculation for hL • Use level pipe (z1 = z2) with constant • Use level pipe (z1 = z2) with constant area (V1 = V2) and no shaft head (hs = 0) area (V1 = V2) and no shaft head (hs = 0) p2 V22 p V2 z2 + + = z1 + 1 + 1 + hs − hL γ 2g γ 2g p1 p2 Δp • Will use friction factor f for Δp in such hL = − = γ γ γ flows, but we are really getting hL • Calculated Δp for z1 = z2, V1 = V2, and hs – Extend to more general flows later p1 p2 Δp = 0 gives hL for more general flows hL = − = 31 γ γ γ 32 Head Loss in Pipes How do we get f? • Dimensional analysis shows that • Have said that f = f(Re, ε/D) dimensionless pressure drop, Δp/ρV2, is • What is form of this relationship? a function of Reynolds number, ρVD/μ, • For laminar flow we will later show that the ℓ/D ratio and relative roughness, ε/D f = 64/Re • Expressed in terms of friction factor, f • Relationship determined experimentally Δp ⎛ ρVD ε ⎞ f = = ⎜ μ ,D⎟ f⎜ ⎟ with empirical fit to equations for 1 l ⎝ ⎠ ρV 2 turbulent flows 2D f = γhL ⇒ hL = f l ρV 2 • Results expressed as Moody diagram 1 l D 2g ρV 2 33 34 2D Moody Diagram Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi. Moody Diagram Equations Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. • Colebrook equation 1 ⎛ε D 2.51 ⎞ = −2.0 log10 ⎜ + ⎟ (turbulent) ⎜ 3.7 Re f ⎟ f ⎝ ⎠ • Haaland equation 1 ⎛ 6.9 ⎛ ε D ⎞1.11 ⎞ ≈ −1.8 log10 ⎜ +⎜ ⎟ ⎟ (turbulent) f ⎜ Re ⎝ 3.7 ⎠ ⎟ ⎝ ⎠ Fundamentals of Fluid Mechanics, 5/E • Laminar 128μlQ 256μ π 2 by Bruce Munson, V D Donald Young, and Δp πD 4 = πD 3 4 64 64 Theodore Okiishi. f = = = = Copyright © 2005 by 1 l 1 l ρV 2 ρVD Re ρV ρV John Wiley & Sons, 2 2 Inc. All rights reserved. 2D 2D μ 35 36 ME 390 – Fluid Mechanics 6 Pipe flow April 8 and 15, 2008 Wholly Turbulent Flows Pressure Drop Problems • Large Reynolds numbers: f independent • Find the pressure drop given fluid data, of Re depends only on ε/D pipe dimensions, ε, and flow (volume l ρV 2 Q Q 16 Q 2 flow, mass flow, or velocity) Δp = f V= = ⇒ V2 = D 2 A π D2 π2 D 4 – Get A = πD2/4 4 – Get V = Q/A or V = m/ρA if not given V & l ρ 2 l ρ 16 Q 2 8 fl ρQ 2 8 fl m 2 & – Find ρ and μ for fluid at given T and P Δp = f V = f = 2 = 2 D2 D2π D2 4 π D 5 π ρD 5 – Compute Re = ρVD/μ and ε/D – Find f from diagram or equation • Pressure drop varies as D-5 • Laminar f = 64/Re; Colebrook for turbulent – Similar to D-4 dependence in laminar flow – Compute Δp = f (ℓ/D) ρV2/2 37 38 Sample Problem Sample Problem Solution π 2 π D = (6.065 in ) D = (0.5054 ft )2 = 0.2006 ft 2 ft • You have been asked to size a pump = 0.5054 ft A= 12 in 4 4 for an airport fuel delivery system. JP-4 5 slug m& 16.61 ft fuel (ρ = 1.50 slug/ft3, μ = 1.2x10-5 V= = s = slug/ft·s) has to travel 0.5 mi through ft 3 ( ρA 1.50 slug 0.2006 ft 3 s ) commercial steel, schedule 40 pipe with a nominal 6 in diameter. The flow rate 1.50 slug 16.61 ft (0.5054 ft ) ρVD ft 3 s Re = = = 1.05 x106 > 4100 is 5 slug/s. What is the head loss? μ 1.2 x10 −5 slug • Schedule 40 pipe: OD = 6.625 in; ft ⋅ s Since Re > thickness = 0.280 in; ID = 6.065 in 4,100, flow is 39 turbulent 40 Sample Problem Solution II Sample Problem Solution III ε 0.00015 ft l ρV 2 0.0156 0.5 mi 5280 ft 1.50 slug 1 lb f ⋅ s ⎛ 16.61 ft ⎞ 2 2 ε = 0.00015 for commercial = = 0.000297 ΔP = f = ⎜ ⎟ D 0.5054 ft steel (Table 8.1, page 433) D 2 2 0.5054 ft mi ft 3 slug ⋅ ft ⎝ s ⎠ Find f from Moody diagram (page 434) ΔP = 16876 lb f = 117.2 lb f = 117.2 psi ft 2 in 2 f (Re = 1.05 x106 , ε D = 0.000297) = 0.0155 • For shaft head to overcome this lead loss & Wshaft Check f value with Colebrook equation net in ΔP ΔP mΔP & = hs ≥ hL = = & ⇒ Wshaft ≥ ⎛ε D ⎞ γ ρg ρ ⎟ = −2.0 log10 ⎛ 0.000297 + ⎞ 1 2.51 2.51 & mg = −2.0 log10 ⎜ + ⎜ ⎜ 3.7 ⎟ ⎟ net in ⎜ 3.7 Re f ⎟ ⎝ 1.05 x106 0.0155 ⎠ f ⎝ ⎠ 5 slug 16876 lb f 1 1 mΔP & s ft 2 hp ⋅ s = 8.005 ⇒ f = = 0.0156 & Wshaft ≥ = = 102 hp f 8.0052 ρ 1.50 slug 550 ft ⋅ lb f net in Use f = 0.0156 41 ft 3 42 ME 390 – Fluid Mechanics 7 Pipe flow April 8 and 15, 2008 Pressure Drop Problems II A Harder Problem • Find the diameter for a given pressure • You have a 200 hp pump to deliver 5 drop given fluid data, ε, and flow slug/s of JP-4 fuel (ρ = 1.50 slug/ft3, μ = (volume flow, mass flow, or velocity) 1.2x10-5 slug/ft·s) over 0.5 mi. What – Find ρ and μ for fluid at given T and P diameter of commercial steel, schedule – Guess D; get A = πD2/4 & – Get V = Q/A or V = m/ρA if not given V 40 pipe should be used? – Compute Re = ρVD/μ and ε/D • Compute required Δp – Find f from diagram or equation 550 ft ⋅ lb f • Laminar f = 64/Re; Colebrook for turbulent & ρWshaft 1.50 slug (200 hp ) net in ft 3 hp ⋅ s 33000 lb f – Compute Δpcalculated = f (ℓ/D) ρV2/2 Δprequired = = = & m 5 slug ft 2 – Iterate on D until Δpcalculated = Δprequired s 43 44 Iterative Solution Iterative Problem Solution π 2 π D = (5 in ) D = (0.4167 ft )2 = 0.1364 ft 2 ft • The calculation we just did for D = 6.065 = 0.4167 ft A= 12 in 4 4 in gave Δp = 16876 psf an error of 5 slug m& s 24.45 ft 16876 psf – 33000 psf = –16124 psf V= = = ft 3 ( ρA 1.50 slug 0.1364 ft 3 s ) Count Dguess (in) Δpcomputed (psf) Error (psf) 1 6.065 16876 –16124 1.50 slug 24.45 ft (0.1364 ft ) ρVD ft 3 s Re = = = 1.27 x106 > 4100 • Take second guess of D = 5 in and μ 1.2 x10 −5 slug repeat calculations done previously to ft ⋅ s Since Re > find Δpcomputed 4,100, flow is 45 turbulent 46 Iterative Problem Solution II Iterative Problem Solution III ε 0.00015 ft ε = 0.00015 for commercial l ρV 2 0.0160 0.5 mi 5280 ft 1.50 slug 1 lb f ⋅ s ⎛ 24.45 ft ⎞ 2 2 = = 0.00036 ΔP = f = ⎜ ⎟ D 0.4167 ft steel (Table 8.1, page 433) slug ⋅ ft ⎝ D 2 2 0.4167 ft mi ft 3 s ⎠ Find f from Moody diagram (page 434) 45564 lb f 316.4 lb f ΔP = = = 316.4 psi f (Re = 1.27 x10 , ε D = 0.000297) = 0.0159 6 ft 2 in 2 Check f value with Colebrook equation • We now have two iterations ⎛ε D ⎞ 1 = −2.0 log10 ⎜ + 2.51 ⎟ = −2.0 log10 ⎛ 0.00036 + ⎜ 2.51 ⎞ ⎟ Count Dguess (in) Δpcomputed (psf) Error (psf) ⎜ 3.7 Re f ⎟ ⎜ 3.7 ⎟ f ⎝ ⎠ ⎝ 1.27 x106 0.0159 ⎠ 1 6.065 16876 –16124 1 1 = 7.894 ⇒ f = = 0.0160 2 5 45564 12564 f 7.8942 Use f = 0.0160 47 48 ME 390 – Fluid Mechanics 8 Pipe flow April 8 and 15, 2008 Iterative Problem Solution IV Iterative Problem Solution V • Use linear interpolation to get new guess, • Continue iterations until error is “small” Di+1 that sets error ei+1 to zero Di − Di −1 D −D Count Dguess (in) Δpcomputed (psf) Error (psf) Di +1 = Di + (ei +1 − ei ) = Di − ei ei − e i −1 ei − ei −1 i i −1 1 6.065 16876 –16124 0 Di − Di −1 5 − 6.065 2 5 45564 12564 Di +1 = Di − ei = 5 − 12564 = 5.466 ei − ei −1 12564 − (− 16124 ) 3 5.466 28780 –4219 4 5.349 32176 –823 Count Dguess (in) Δpcomputed (psf) Error (psf) 5 5.321 33072 72 1 6.065 16876 –16124 6 5.323 32999 –1 2 5 45564 12564 49 50 Iterations and Reality Pressure Drop Problems III • Commercial pipe and tubing only comes • Find the flow rate for a given pressure in fixed sizes drop given fluid data, ε, and diameter – Get A = πD2/4 – Extra iterations not needed once the minimally acceptable commercial size is – Find ρ and μ for fluid at given T and P found – Guess V – Compute Re = ρVD/μ and ε/D – In this case available nominal diameters – Find f from diagram or equation are 5 in and 6 in with actual inside • Laminar f = 64/Re; Colebrook for turbulent diameters of 5.047 in and 6.065 in (for – Compute Δpcalculated = f ℓ/D ρV2/2 Schedule 40) – Iterate on V until Δpcalculated = Δprequired – Only choice is 6 in (nominal) & – Compute Q or m as desired 51 52 Different Friction Factors Minor Losses • The friction factor definition we are • Determine pressure drop (head loss) in using here is the common one a variety of flow passages – Called the Darcy friction factor if – Entrance into a piping system clarification is needed – Exit from a piping system • Another definition, called the Fanning – Expansion in a piping system friction factor is sometimes seen – Contraction in a piping system – Fanning factor = τw / (ρV2/2) – Valves of various types (with different • From the relationship that τw = DΔp/4ℓ we get opening fractions) the result that the Fanning factor is one fourth – Fittings (elbows, tees, bends, unions) of the Darcy factor 53 54 ME 390 – Fluid Mechanics 9 Pipe flow April 8 and 15, 2008 ρV 2 V = Pipe velocity ΔpL = K L Minor Losses Entrance Losses 2 • Fittings in pipe systems modeled as loss coefficients, KL Fundamentals of Fluid Reentrant: Sharp edged: V2 Δp L V2 ρV 2 Mechanics, 5/E by Bruce hL = K L ⇒ = KL ⇒ Δp L = K L Munson, Donald Young, and Theodore Okiishi. KL = 0.8 KL = 0.5 2g ρg 2g 2 Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. • KL depends on geometry and Re r – For flows dominated by inertia effects KL is a function of geometry only D • Alternative process, not given here, uses equivalent length for minor losses Slightly rounded: 55 KL = 0.2 Well rounded: 56 KL = f(r/D) Rounded Inlet KL Full KE loss Slightly cannot be rounded KL recovered in = 0.2 for r/D sharp-edged = 0.055 entrance Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi. Copyright © 2005 r/D = 0 is by John Wiley & Sons, Inc. All square inlet rights reserved. Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and 57 58 Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. New Area Sudden contraction (left) Reentrant KL = 1 for all Sharp edged Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, exit flows and Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All For sudden rights reserved. expansion (right) KL = ( 1 – A1/A2)2 59 60 Slightly rounded Well rounded ME 390 – Fluid Mechanics 10 Pipe flow April 8 and 15, 2008 Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. 61 62 Problem with Minor Losses Problem Solution π π A = D 2 = (0.05250 m )2 ρ = (SG )ρ ref = • 4 kg/s of oil with SG = 0.82 and μ = 0.05 4 4 999 kg 819.2 kg kg·m/s2 is pumped from one tank to = 0.002165 m 2 (0.82) = s s another. The line of 2-in Schedule-40 4 kg pipe has a total length of 40 m, with two V= & m = s = 2.26 m gate valves and six elbows (regular flanged 90o). The entrance is rounded ρA 819.2 kg 0.002165 m3 m3 ( s ) with an r/D ratio of 0.1. • Find pressure loss with both valves open ρVD 819.2 kg 2.26 m (0.05250 m ) Re = = m3 s = 1940 < 2100 • 2-in schedule 40 pipe has OD = 2.375 in μ 0.050 kg and thickness = 0.154 in, so ID = 2.067 m⋅s Since Re < in = 0.05250 m 2,100, flow is 63 64 laminar Problem Solution II Problem Solution III Find Δpmajor directly from laminar flow equation Δpminor = (Loss coefficient sum) times ρV2/2 3 3 & m 4 kg m 0.04883 m Could also Q= = = ρ s 819.2 kg s use f = 64/Re (∑ K L ) ρV 2 2 3.07 819 kg ⎛ 2.26 m ⎞ 1 N ⋅ s 2 6,397 N Δpminor = = ⎜ ⎟ = (128) 0.05 2 ⋅ s (40 m ) 0.004883 m 52369 N 3 N 2 2 m3 ⎝ s ⎠ kg ⋅ m m2 128μlQ m s Δpmajor = = = = 52.369 kPa πD 4 π(0.05250 m )4 m2 52,369 N 6,397 N Δptotal = Δpmajor + Δpminor = + Minor losses coefficients: rounded entrance (r/D m2 m2 = 0.1), KL = 0.12; exit, KL = 1; fully open gate 58,766 N valve, KL = 0.15; 6 elbows, KL = 6(0.3) = 1.8. Δptotal = = 58.8 kPa m2 Total KL = 0.12 + 1 + 0.15 + 1.8 = 3.07 65 66 ME 390 – Fluid Mechanics 11 Pipe flow April 8 and 15, 2008 Noncircular Ducts • Define hydraulic diameter, Dh = 4A/P 4A Dh = – A is cross-sectional area for flow P – P is wetted perimeter ρVD Re h = – For a circular pipe where A = πD2/4 and P μ = πD, Dh = 4(πD2/4) / (πD) = D l ρV 2 • For turbulent flows use Moody diagram ΔP = f Dh 2 with D replaced by Dh in Re, f, and ε/D • For laminar flows, f = C/Re (both based 2C μlQ ΔP = on Dh) – see next slide for C values π Dh4 Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, 67 68 and Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Problem Solution • An 10-in-square, commercial steel air 125 ft 3 min 4 A 4 L2 Dh = = =L Q min 60 s 3 ft conditioning duct contains air at 80oF P 4L V= = 2 = and atmospheric pressure and has a = 10 in = 0.8333 ft A (10 in ) ft 2 2 s 144 in flow rate of 125 ft3/min. Find the pressure drop per unit duct length VDh 3 ft (0.8333 ft ) Turbulent flow Re h = = s = 1.78 x105 • Property data at 80oF (Table B.3) ρ = ν 1.69 x10 −4 ft 2 for Reh > 4100 0.002286 slug/ft3; ν = 1.69x10-4 ft2/s s • Solution: find Reh to see if flow is ε 0.00015 ft ε = 0.00015 for commercial = = 0.00018 steel (Table 8.1, page 433) laminar or turbulent then find f and Δp D 0.8333 ft 69 70 Solution II Recommended Air Velocity Find f from Moody diagram (page 434) Air Velocity Air Ducts f (Re = 1.27 x10 , ε D = 0.00018) = 0.0172 6 m/s ft/s Combustion air ducts 12 - 20 40 - 66 Check f value with Colebrook equation Air inlet to boiler room 1-3 3.3 - 9.8 ⎛ε D ⎞ 1 = −2.0 log10 ⎜ + 2.51 ⎟ = −2.0 log10 ⎛ 0.00018 + ⎜ 2.51 ⎞ ⎟ Warm air for house heating 0.8 - 1.0 2.6 - 3.3 ⎜ 3.7 Re f ⎟ ⎜ 3.7 ⎟ f ⎝ ⎠ ⎝ 1.78 x105 0.0172 ⎠ 1 1 Vacuum cleaning pipe 8 - 15 26 - 49 = 7.611 ⇒ f = = 0.0173 Use f = 0.0173 f 7.6112 Compressed air pipe 20 - 30 66 - 98 −5 Ventilation ducts (hospitals) 1.8 - 4 5.9 - 13 0.00229 slug 1 lb f ⋅ s ⎛ 3 ft ⎞ 1.78 x10 lb f 2 2 ΔP 1 ρV 2 0.0173 1 = f = ⎜ ⎟ = Ventilation ducts (offices) 2.0 - 4.5 6.5 - 15 l D 2 2 0.8333 ft ft 3 slug ⋅ ft ⎝ s ⎠ ft 3 71 72 http://www.engineeringtoolbox.com/flow-velocity-air-ducts-d_388.html ME 390 – Fluid Mechanics 12

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