Pipe Flow

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