Pipe Flow April 8 and 15, 2008

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

Piping System What We Want to Do • Determine losses from friction forces in Fundamentals of Fluid Mechanics, 5/E by Bruce straight pipes and joints/valves Munson, Donald Young, and 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 equation 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 D for circular cross sections Driving force is pressure is gravity 5 Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, 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 layers of fluid results for laminar flows have fluctuations

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

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

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

– Laminar flow: l e = 0.06Re D • Look at arbitrary element, with length ℓ, l e = 4.4Re1 6 – Turbulent flow: D and radius r, in fully developed flow

– Turbulent flow rule of thumb ℓe ≈ 10D • What are forces on this element?

Fully Developed Flow Extend Relation to Wall No change in momentum Flow Direction

F = πr 2 p −πr 2 ()p − Δp −τ 2πr = 0 ∑ x 1 1 l • Have Δp= 2τℓ/r for any r: 0 < r < R = D/2 2τ • Pressure drop is due • For wall r = R = D/2 and τ = τ = wall Δp = l w 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

2 • Can get ⎡ ⎛ r ⎞ ⎤ exact u = uc ⎢1− ⎜ ⎟ ⎥ ⎢ ⎝ R ⎠ ⎥ equation for ⎣ ⎦ R R pressure • Laminar shear drop stress profile 128μlQ found from Δp = uc du πD4 τ = μ • Laminar 2 ⎡ ⎛ r ⎞ ⎤ dr du 2r 8μuc Fundamentals of Fluid Mechanics, 5/E by Fundamentals of Fluid Mechanics, 5/E by τ = μ = μuc = r Bruce Munson, Donald Young, and Theodore velocity u = uc ⎢1− ⎜ ⎟ ⎥ Bruce Munson, Donald Young, and Theodore 2 2 Okiishi. Copyright © 2005 by John Wiley & R Okiishi. Copyright © 2005 by John Wiley & dr R D Sons, Inc. All rights reserved. profile ⎣⎢ ⎝ ⎠ ⎦⎥ Sons, Inc. All rights reserved. 17 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 ⎡ 2 ⎤ mean velocity, V 2 ⎛ r ⎞ Q = VA = VπR = udA = u2πrdr = uc ⎢1− ⎜ ⎟ ⎥2πrdr –Flow = m & V = ρVAV = ρV2(πR2) ∫ ∫ ∫ ⎢ ⎝ R ⎠ ⎥ Momentum A 0 0 ⎣ ⎦ 2 2 3 2 –FlowKE = m& V /2 = ρVAV /2 = ρV (πR )/2 RR R ⎡ r3 ⎤ ⎡r 2 r 4 ⎤ R2 • Accurate representation uses profile Q = 2πu ⎢ rdr − dr⎥ = 2πu ⎢ − ⎥ = 2πu c ∫∫2 c 2 c 2 ⎢ R ⎥ ⎢ 2 4R ⎥ 4 R ⎡ 2 ⎤ ⎣00 ⎦ ⎣ ⎦0 ⎛ r ⎞ 4 2 Flow = ρudAu = ρ⎢u ⎜1− ⎟⎥ 2πrdr = ρV A Momentum c ⎜ 2 ⎟ ∫ ∫ ⎢ R ⎥ 3 R2 2Q 2VA 2VπR2 A 0 ⎣ ⎝ ⎠⎦ 3 Q = πuc ⇒ uc = = = = 2V 2 R ⎡ 2 ⎤ 3 2 R2 R2 R2 u 1 ⎛ r ⎞ V π π π FlowKE = ρudA = ρ⎢uc ⎜1− ⎟⎥ 2πrdr = 2ρA ∫ 2 2 ∫ ⎜ R2 ⎟ 2 A 0 ⎣⎢ ⎝ ⎠⎦⎥ Centerline uc is twice the mean velocity, V 19 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 u = u(t)dt u’ = velocity • Turbulent flows: random fluctuations exchange T ∫ eddies of high momentum from the center with t0 fluctuation = u – u low momentum flow from near-wall regions

Momentum Exchange Turbulence Regions/Profiles turbulent eddy viscosity, η du τ = ()μ + η

Fundamentals dr of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi. • Very thin viscous sublayer next to wall Laminar flow – Turbulent flow – Copyright © 2005 by John random eddies have Wiley & Sons, – 0.13% of R = 3 in for H20 at u = 5 ft/s structure Inc. All rights molecular motion 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 Profile Effect of Velocity Profile Munson, Donald Young, and Theodore Okiishi. Copyright © 2005 by John 1 n Wiley & Sons, Inc. All u ⎛ r ⎞ • Analysis similar to one used for laminar rights reserved. = ⎜1− ⎟ Vc ⎝ R ⎠ flow profile Turbulent – Determine momentum and kinetic energy flow for mean velocity velocity – Correction factor multiplies average V profiles 2 3 4 results to give integrated u and u values n = 6: Re = 1.5x10 ; Vc/V = 1.264 with n a n = 8: Re = 4x105;V/V = 1.195 c function of n Re Momentum KE n = 10: Re = 3x106;V/V = 1.155 c Reynolds 6 1.5x104 1.027 1.077 Laminar: Vc/V = 2 V = Q/A number 8 4x105 1.016 1.046 6 25 10 3x10 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 Mechanics, 5/E by Bruce • Typical roughness values for different Munson, Donald Young, pressure and Theodore Okiishi. materials expressed as roughness 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 V 2 p V 2 z + 2 + 2 = z + 1 + 1 + h − h find ε 2 γ 2g 1 γ 2g s L • Previous applications allowed us to compute the head loss from all other Fundamentals of Fluid 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) p V 2 p V 2 z + 2 + 2 = z + 1 + 1 + h − h 2 γ 2g 1 γ 2g s L p p Δp h = 1 − 2 = • Will use friction factor f for Δp in such L γ γ γ flows, but we are really getting hL •Calculated Δpfor 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 = = f ⎜ , ⎟ with empirical fit to equations for 1 ⎜ ⎟ l ρV 2 ⎝ μ D ⎠ turbulent flows 2 D γh ρV 2 • Results expressed as Moody diagram f = L ⇒ h = f l 1 L l ρV 2 D 2g 2 D 33 34

• Colebrook equation 1 ⎛ ε D 2.51 ⎞ = −2.0log ⎜ + ⎟ 10⎜ ⎟ (turbulent) f ⎝ 3.7 Re f ⎠

1.11 • Haaland equation 1 ⎛ 6.9 ⎛ ε D ⎞ ⎞ ≈ −1.8log ⎜ + ⎜ ⎟ ⎟ 10⎜ ⎟ (turbulent) f ⎝ Re ⎝ 3.7 ⎠ ⎠

Fundamentals of • Laminar Fluid Mechanics, 5/E 128μ Q 256μ π 2 by Bruce Munson, l V D Donald Young, and Δp 4 3 4 64 64 Theodore Okiishi. f = = πD = πD = = Copyright © 2005 by 2 John Wiley & Sons, 1 l 2 1 l 2 ρV ρVD Re Inc. All rights ρV ρV reserved. 2 D 2 D μ

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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 ρV 2 Q Q 16 Q2 flow, mass flow, or velocity) Δp = f l V = = ⇒ V 2 = π 2 4 –Get A = πD2/4 D 2 A D2 π D 4 – Get V = Q/A or V = m & /ρA if not given V 2 2 2 ρ ρ 16 Q 8 f ρQ 8 f m – Find ρ and μ for fluid at given T and P Δp = f l V 2 = f l = l = l & D 2 D 2 π2 D4 π2 D5 π2 ρD5 – 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 -4 – Similar to D dependence in laminar flow – Compute Δp= f (ℓ/D) ρV2/2

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Sample Problem Sample Problem Solution ft π π D = ()6.065 in = 0.5054 ft A = D2 = ()0.5054 ft 2 = 0.2006 ft 2 • You have been asked to size a pump 12 in 4 4 for an airport fuel delivery system. JP-4 5 slug m 16.61 ft 3 -5 V = & = s = fuel (ρ = 1.50 slug/ft , μ = 1.2x10 1.50 slug ρA ()0.2006 ft3 s slug/ft·s) has to travel 0.5 mi through ft3 commercial steel, schedule 40 pipe with 1.50 slug 16.61 ft ()0.5054 ft a nominal 6 in diameter. The flow rate ρVD ft3 s Re = = =1.05x106 > 4100 is 5 slug/s. What is the head loss? μ 1.2x10−5 slug • Schedule 40 pipe: OD = 6.625 in; ft ⋅ s Since Re > thickness = 0.280 in; ID = 6.065 in 4,100, flow is turbulent 39 40

Sample Problem Solution II Sample Problem Solution III 2 2 2 ε 0.00015 ft ε = 0.00015 for commercial ρV 0.0156 0.5 mi 5280 ft 1.50 slug 1lb f ⋅ s ⎛16.61 ft ⎞ = = 0.000297 ΔP = f l = ⎜ ⎟ D 0.5054 ft steel (Table 8.1, page 433) D 2 2 0.5054 ft mi ft3 slug ⋅ ft ⎝ s ⎠ 16876 lb 117.2 lb Find f from Moody diagram (page 434) ΔP = f = f =117.2 psi 2 2 6 ft in f (Re =1.05x10 ,ε D = 0.000297) = 0.0155 • For shaft head to overcome this lead loss W&shaft Check f value with Colebrook equation ΔP ΔP mΔP net in = h ≥ h = = ⇒ W& ≥ & ⎛ ⎞ s L shaft 1 ⎜ ε D 2.51 ⎟ ⎛ 0.000297 2.51 ⎞ m& g γ ρg net in ρ = −2.0log10 + = −2.0log10⎜ + ⎟ f ⎜ 3.7 Re f ⎟ ⎝ 3.7 1.05x106 0.0155 ⎠ ⎝ ⎠ 5 slug 16876 lb f 1 1 2 = 8.005 ⇒ f = = 0.0156 m& ΔP s ft hp ⋅ s f 8.0052 W&shaft ≥ = =102 hp net in ρ 1.50 slug 550 ft ⋅lbf Use f = 0.0156 41 ft3 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 40 pipe should be used? – Get V = Q/A or V = m& /ρA if not given V – Compute Re = ρVD/μ and ε/D • Compute required Δp – Find f from diagram or equation 1.50 slug 550 ft ⋅lb ρW& 200 hp f • Laminar f = 64/Re; Colebrook for turbulent shaft 3 () net in ft hp ⋅ s 33000 lb f – Compute Δp = f (ℓ/D) ρV2/2 Δprequired = = = calculated m& 5 slug ft 2 – Iterate on D until Δpcalculated = Δprequired s 43 44

Iterative Solution Iterative Problem Solution ft π π D = ()5 in = 0.4167 ft A = D2 = ()0.4167 ft 2 = 0.1364 ft 2 • The calculation we just did for D = 6.065 12 in 4 4 in gave Δp = 16876 psf an error of 5 slug m 24.45 ft V = & = s = 16876 psf – 33000 psf = –16124 psf 1.50 slug ρA ()0.1364 ft3 s ft3 Count Dguess (in) Δpcomputed (psf) Error (psf) 1.50 slug 24.45 ft 1 6.065 16876 –16124 ()0.1364 ft ρVD ft3 s Re = = =1.27x106 > 4100 • Take second guess of D = 5 in and μ 1.2x10−5 slug repeat calculations done previously to ft ⋅ s Since Re > 4,100, flow is find Δpcomputed turbulent 45 46

Iterative Problem Solution II Iterative Problem Solution III ε 0.00015 ft ε = 0.00015 for commercial 2 2 2 = = 0.00036 l ρV 0.0160 0.5 mi 5280 ft 1.50 slug 1lb f ⋅ s ⎛ 24.45 ft ⎞ steel (Table 8.1, page 433) ΔP = f = ⎜ ⎟ D 0.4167 ft D 2 2 0.4167 ft mi ft3 slug ⋅ ft ⎝ s ⎠

Find f from Moody diagram (page 434) 45564 lb 316.4 lb ΔP = f = f = 316.4 psi f (Re =1.27x106,ε D = 0.000297) = 0.0159 ft 2 in2 Check f value with Colebrook equation • We now have two iterations ⎛ ⎞ 1 ⎜ ε D 2.51 ⎟ ⎛ 0.00036 2.51 ⎞ Count D (in) Δp (psf) Error (psf) = −2.0log10 + = −2.0log10⎜ + ⎟ guess computed f ⎜ 3.7 Re f ⎟ ⎝ 3.7 1.27x106 0.0159 ⎠ ⎝ ⎠ 1 6.065 16876 –16124 1 1 = 7.894 ⇒ f = = 0.0160 f 7.8942 2 5 45564 12564 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 Count Dguess (in) Δpcomputed (psf) Error (psf) Di − Di−1 Di − Di−1 Di+1 = Di + ()ei+1 − ei = Di − ei e − e ei − ei−1 1 6.065 16876 –16124 i i−1 0 D − D 5 − 6.065 2 5 45564 12564 D = D − e i i−1 = 5 −12564 = 5.466 i+1 i i 3 5.466 28780 –4219 ei − ei−1 12564 − ()−16124 4 5.349 32176 –823 Count D (in) Δp (psf) Error (psf) guess computed 5 5.321 33072 72 1 6.065 16876 –16124 6 5.323 32999 –1 2 5 45564 12564

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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 Δp = f ℓ/D ρV2/2 Schedule 40) calculated – Iterate on V until Δp = Δp – Only choice is 6 in (nominal) calculated required – 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 2 – Fanning factor = τw / (ρV /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 Δp = K V = Pipe velocity Minor Losses EntranceL L 2 Losses • Fittings in pipe systems modeled as

loss coefficients, KL Fundamentals of Fluid Sharp edged: 2 2 2 Mechanics, 5/E by Bruce Reentrant: V ΔpL V ρV Munson, Donald Young, K = 0.5 h = K ⇒ = K ⇒ Δp = K and Theodore Okiishi. K = 0.8 L L L L L L Copyright © 2005 by John L 2g ρg 2g 2 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: Well rounded: 55 KL = 0.2 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 by John Wiley & r/D = 0 is 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 – 2 A1/A2)

59 60 Slightly rounded Well rounded

ME 390 – Fluid Mechanics 10 Pipe flow April 8 and 15, 2008

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Problem with Minor Losses Problem Solution π 2 π 2 A = D = ()0.05250 m ρ = ()SG ρref = • 4 kg/s of oil with SG = 0.82 and μ = 0.05 4 4 2 999 kg 819.2 kg kg·m/s is pumped from one tank to = 0.002165 m2 (0.82) = s s another. The line of 2-in Schedule-40 4 kg pipe has a total length of 40 m, with two m 2.26 m V = & = s = gate valves and six elbows (regular ρA 819.2 kg 3 s ()0.002165 m flanged 90o). The entrance is rounded m3 with an r/D ratio of 0.1. 819.2 kg 2.26 m ()0.05250 m • Find pressure loss with both valves open ρVD 3 s Re = = m =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 laminar 64

Problem Solution II Problem Solution III

2 Find Δpmajor directly from laminar flow equation Δpminor = (Loss coefficient sum) times ρV /2 m 4 kg m3 0.04883 m3 Q = & = = Could also ρ s 819.2 kg s use f = 64/Re 2 ρV 2 3.07 819 kg ⎛ 2.26 m ⎞ 1 N ⋅ s2 6,397 N 3 Δp = K = = 0.05 N ⋅ s 0.004883 m minor ()∑ L 3 ⎜ ⎟ 2 ()128 ()40 m 2 2 m ⎝ s ⎠ kg ⋅m m 128μlQ m2 s 52369 N Δpmajor = = = = 52.369 kPa πD4 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, K = 0.15; 6 elbows, K = 6(0.3) = 1.8. Δptotal = = 58.8 kPa L L 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, D = 4A/P 4A h D = – A is cross-sectional area for flow h P – P is wetted perimeter ρVD Re = – For a circular pipe where A = πD2/4 and P h μ = πD, D = 4(πD2/4) / (πD) = D h ρV 2 • For turbulent flows use Moody diagram ΔP = f l Dh 2 with D replaced by Dh in Re, f, and ε/D 2C μ Q • For laminar flows, f = C/Re (both based ΔP = l π D4 on Dh) – see next slide for C values h 67 Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, 68 and Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

Problem Solution

3 • An 10-in-square, commercial steel air 4A 4L2 125 ft min o D = = = L Q min 60 s 3 ft conditioning duct contains air at 80 F h P 4L V = = = A ft 2 s and atmospheric pressure and has a =10 in = 0.8333 ft 10 in 2 () 2 flow rate of 125 ft3/min. Find the 144 in 3 ft ()0.8333 ft pressure drop per unit duct length VD Turbulent flow Re = h = s =1.78x105 o h −4 2 • Property data at 80 F (Table B.3) ρ = ν 1.69x10 ft for Reh > 4100 0.002286 slug/ft3; ν = 1.69x10-4 ft2/s s ε 0.00015 ft ε = 0.00015 for commercial • Solution: find Reh to see if flow is = = 0.00018 laminar or turbulent then find f and Δp D 0.8333 ft steel (Table 8.1, page 433)

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Solution II Recommended Air Velocity Find f from Moody diagram (page 434) Air Velocity Air Ducts f (Re =1.27x106,ε D = 0.00018) = 0.0172 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 1 ⎛ ε D 2.51 ⎞ ⎛ 0.00018 2.51 ⎞ = −2.0log ⎜ + ⎟ = −2.0log ⎜ + ⎟ 10⎜ ⎟ 10⎜ 5 ⎟ Warm air for house heating 0.8 - 1.0 2.6 - 3.3 f ⎝ 3.7 Re f ⎠ ⎝ 3.7 1.78x10 0.0172 ⎠ Vacuum cleaning pipe 8 - 15 26 - 49 1 1 = 7.611 ⇒ f = = 0.0173 Use f = 0.0173 f 7.6112 Compressed air pipe 20 - 30 66 - 98

2 2 2 −5 Ventilation ducts (hospitals) 1.8 - 4 5.9 - 13 ΔP 1 ρV 0.0173 1 0.00229 slug 1lb f ⋅ s ⎛ 3 ft ⎞ 1.78x10 lb f = f = = 3 ⎜ ⎟ 3 Ventilation ducts (offices) 2.0 - 4.5 6.5 - 15 l D 2 2 0.8333 ft ft slug ⋅ ft ⎝ s ⎠ ft

71 72 http://www.engineeringtoolbox.com/flow-velocity-air-ducts-d_388.html

ME 390 – Fluid Mechanics 12