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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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