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Hydraulic analysis of complex piping systems (updated)

1. Given: Pipe characteristics (D, L, e), fluid properties (ν), flow conditions (Q or V) 2. Calculate Reynold's number (Re) using the given flow parameters 3. Determine friction factor (f) from Moody diagram or equations based on Re and relative roughness (e/D) 4. Use Darcy-Weisbach equation to calculate head loss (hf) or solve for unknown parameter (Q or V)

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Hydraulic analysis of complex piping systems (Losses in pipes, pipes in parallel and pipes in series) 1 Hydraulics Dr. Mohsin Siddique
Steady Flow Through Pipes 2 Laminar Flow: flow in layers Re<2000 (pipe flow) Turbulent Flow: flow layers mixing with each other Re >4000 (pipe flow)
Steady Flow Through Pipes 3 Reynold’s Number(R or Re): It is ratio of inertial forces (Fi) to viscous forces (Fv) of flowing fluid For laminar flow: Re<=2000 For transitional flow: 2000<Re<4000 For Turbulent flow: Re>= 4000 . . Re . . . . . . . . Velocity Volume Mass Velocity Fi Time Time Fv Shear Stress Area Shear Stress Area Q V AV V AV V VL VL du VA A A dy L ρ ρ ρ ρ ρ τ µ υµ µ = = = = = = = = Where ; V is avg. velocity of flow in pipe ν is kinematic viscosity L is characteristic/representative linear dimension of pipe. It is diameter of pipe (circular conduits) or hydraulic radius (non-circular conduits). νµ ρ VDVD Re == Note: For non-circular section, we need to use hydraulic radius (Rh) instead of diameter (D) for the linear dimension (L). Values of critical Reynolds no.
Steady Flow Through Pipes 4 Hydraulic Radius (Rh) or Hydraulic Diameter: It is the ratio of area of flow to wetted perimeter of a channel or pipe. P A perimeterwetted Area Rh == For Circular Pipe ( )( ) h h RD D D D P A R 4 4 4/ 2 = === π π For Rectangular pipe DB BD P A Rh 2+ == B D Note: hydraulic Radius gives us indication for most economical section. More the Rh more economical will be the section. νν h e VRVD R 4 == By replacing D with Rh, Reynolds’ number formulae can be used for non-circular sections as well.
Head Loss in Pipes 5 Total Head Loss=Major Losses+ Minor Losses Major Loss: Due to pipe friction Minor Loss: Due to pipe fittings, bents and valves etc
Head Loss in Pipes due to Friction 6 The head loss due to friction in a given length of pipe is proportional to mean velocity of flow (V) as long as the flow in laminar. i.e., But with increasing velocity, as the flow become turbulent the head loss also varies and become proportion toVn Where n ranges from 1.75 to 2 Log-log plot for flow in uniform pipe (n=2.0 for rough wall pipe; n=1.75 for smooth wall pipe) VH f ∝ n f VH ∝
Frictional Head Loss in Conduits of Constant Cross-Section 7 Consider stead flow in a conduit of uniform cross-section A.The pressure at section 1 & 2 are P1 & P2 respectively.The distance between the sections is L. For equilibrium in stead flow, ∑ == 0maF Figure: Schematic diagram of conduit 0sin 21 =−−− APPLWAP oτα P= perimeter of conduit = Avg. shear stress between pipe boundary and liquid oτ 012 21 =−      − −− PL L zz ALAPAP oτγ αsin12 =      − L zz
Frictional Head Loss in Conduits of Constant cross-section 8 012 21 =−      − −− PL L zz ALAPAP oτγ Dividing the equation by Aγ ( ) 012 21 =−−−− A PL zz PP o γ τ γγ fL o hh A PLP z P z ===      +−      + γ τ γγ 2 2 1 1 Therefore, head loss due to friction, hf, can be written as; h oo f R L A PL h γ τ γ τ == Lh g v z P g v z P +++=++ 22 2 2 2 2 2 1 1 1 γγ Remember !! For pipe flow For stead flow in pipe of uniform diameter v1=v2 Lhz P z P =      +−      + 2 2 1 1 γγ This is general equation and can be applied to any shape conduit having either Laminar or turbulent flow. P A Rh =Q
Determining Shear Stress 9 For smooth-walled pipes/conduits, the average shear stress at the wall is Using Rayleigh's Theorem of dimensional analysis, the above relation can be written as; Rewriting above equation in terms of dimension (FLT), we get ( ),,,, VRf ho ρµτ = ( )                           = ncb a T L L FT L FT LK L F 4 2 22 ( )ncba ho VRk ... ρµτ = LlengthR L F areaforce h o == == 2 /τ ( ) 22 4233 /. //// / − == === = FTLmsN LFTLaFLM TLV µ ρ( ) ( ) ( ) ( )( )ncba TLLFTFTLLKFL /4222 −−− =
Determining Shear Stress 10 According to dimensional homogeneity, the dimension must be equal on either side of the equation, i.e., ( ) ( ) ( ) ( )( )ncba TLLFTFTLLKFL /4222 −−− = )(20: )(422: )(1: iiincbT iincbaL icbF →−+= →+−−=− →+= Solving three equations, we get 1;2;2 −=−=−= ncnbna Substituting values back in above equation ( ) ( ) 2 2 122 ...... V VR kVRkVRk n hnnnn h ncba ho ρ µ ρ ρµρµτ − −−−       === ( ) 22 VRk n eo ρτ − = Setting = we get 2 2 V Cfo ρτ = Where, Cf is the coefficient of friction ( ) 2−n eRk 2/fC (Re)fCf =
Determining Shear Stress 11 Now substituting the equation of avg. shear stress in equation of head loss, For circular pipe flows, Rh=D/4 Where, f is a friction factor. i.e., The above equation is known as pipe friction equation and as Darcy- Weisbach equation and is used for calculation of pipe-friction head loss for circular pipes flowing full (laminar or turbulent) 2/2 VCfo ρτ = h o f R L h γ τ = h f h f f gR LVC R LVC h 22 22 == γ ρ g V D L fh g V D L C Dg LVC h f f f f 2 2 4 42 4 2 22 = == fCf 4=Q
Friction Factor for Laminar and Turbulent Flows in Circular Pipes 12 Smooth and Rough Pipe A smooth pipe is the one which behaves as smooth pipe i.e. (frictionless). It is quite possible that a pipe may behave as smooth at a certain flow condition while at some other condition it may behave rough. Mathematically; Smooth Pipe Rough Pipe Transitional mode Turbulent flow near boundary = = v e δ Roughness height Thickness of viscous sub-layer Smooth pipe Rough pipe f D fV v Re 14.1414.14 == ν δ ve δ< ve δ> ve δ< ve δ14> vv e δδ 14≤≤
Friction Factor for Laminar and Turbulent Flows in Circular Pipes 13 For laminar flow For turbulent flow Re 64 =f2000Re < 51.2 Re log2 1 f f = 9.6 Re log8.1 1 = f 25.0 Re 316.0 =f 7/1 max       = or y u u Def / 7.3 log2 1 =         +−= f De f Re 51.2 7.3 / log2 1         +      −= Re 9.6 7.3 / log8.1 1 11.1 De f Colebrook Eq. for turbulent flow in all pipes Halaand Eq. For turbulent flow in all pipes Von-karman Eq. for fully rough flow Blacius Eq. for smooth pipe flow Seventh-root law 5 10Re3000 ≤≤ From Nikuradse experiments Colebrook Eq. for smooth pipe flow for smooth pipe flow 4000Re >
Friction Factor for Laminar and Turbulent Flows in Circular Pipes 14 The Moody chart or Moody diagram is a graph in non- dimensional form that relates the Darcy-Weisbach friction factor, Reynolds number and relative roughness for fully developed flow in a circular pipe. The Moody chart is universally valid for all steady, fully developed, incompressible pipe flows.
Friction Factor for Laminar and Turbulent Flows in Circular Pipes 15 For laminar flow For non-laminar flow eR f 64 =         +−= f De f Re 51.2 7.3 / log2 1 Colebrook eq.
Friction Factor for Laminar and Turbulent Flows in Circular Pipes 16 The friction factor can be determined by its Reynolds number (Re) and the Relative roughness (e/D) of the Pipe.( where: e = absolute roughness and D = diameter of pipe)
17 Absolute roughness
Problem Types 18 Type 1: Determine f and hf, Type 2: Determine Q Type 3: Determine D
Problem 19 Find friction factor for the following pipe e=0.002 ft D=1ft KinematicViscosity, ν=14.1x10-6ft2/s Velocity of flow,V=0.141ft/s Solution: e/D=0.002/1=0.002 R=VD/ ν =1x0.141/(14.1x10-6)=10000 From Moody’s Diagram; f=0.034 ___________ Re 51.2 7.3 / log2 1 =         +−= f f De f
Problem-Type 1 20 Pipe dia, D= 3 inch & L=100m Re=50,000 ʋ=1.059x10-5ft2/s (a): Laminar flow: f=64/Re=64/50,000=0.00128 ft gD fLV HLf 0357.0 )12/3)(2.32(2 )12.2)(100(00128.0 2 22 === sftV VVD /12.2 10059.1 )12/3( 50000Re 5 =⇒ × =⇒= − ν
Problem-Type 1 21 Pipe dia, D= 3 inch & L=100m Re=50,000 ʋ=1.059x10-5ft2/s (b): Turbulent flow in smooth pipe: i.e.: e=0 0209.0 50000 51.2 7.3 0 log2 Re 51.2 7.3 / log2 1 =         +−=         +−= f ff De f ft gD fLV HLf 582.0 )12/3)(2.32(2 )12.2)(100(0209.0 2 22 ===
Problem-Type 1 22 Pipe dia, D= 3 inch & L=100m Re=50,000 ʋ=1.059x10-5ft2/s (c): Turbulent flow in rough pipe: i.e.: e/D=0.05 0720.0 50000 51.2 7.3 05.0 log2 Re 51.2 7.3 / log2 1 =         +−=         +−= f ff De f ft gD fLV HLf 01.2 )12/3)(2.32(2 )12.2)(100(0720.0 2 22 ===
Problem-Type 1 23 hL=? memDmL 0005.0;25.0;1000 === smsmQ /10306.1;/051.0 263 − ×== ν ( ) ( ) 002.025.0/0005.0/ 10210306.1/25.0039.1/ 56 == ×=××== − De VDR ν 0.0245f DiagramsMoody'From = m gD fLV hL 39.5 2 2 == smAQV /039.1/ ==Q
Problem-Type 2 24 gDfLVhL 2/2 =
Problem-Type 2 25 For laminar flow For non-laminar flow eR f 64 =         +−= f De f Re 51.2 7.3 / log2 1 Colebrook eq.
Problem-Type 3 26
Problem 27 gD flV h f De f Lf 2 Re 51.2 7.3 / log2 1 2 =         +−=
Problem 28
Problem 29
MINOR LOSSES 30 Each type of loss can be quantified using a loss coefficient (K). Losses are proportional to velocity of flow and geometry of device. Where, Hm is minor loss and K is minor loss coefficient.The value of K is typically provided for various types/devices NOTE: If L > 1000D minor losses become significantly less than that of major losses and hence can be neglected. g V KHm 2 2 =
Categories of Minor Losses 31 These can be categorized as 1. Head loss due to contraction in pipe 1.1 Sudden Contraction 1.2 Gradual Contraction 2. Entrance loss 3. Head loss due to enlargement of pipe 3.1 Sudden Enlargement 3.2 Gradual Enlargement 4. Exit loss 5. Head loss due to pipe fittings 6. Head loss due to bends and elbows
Minor Losses 32 Head loss due to contraction of pipe (Sudden contraction) A sudden contraction (Figure) in pipe usually causes a marked drop in pressure in the pipe because of both the increase in velocity and the loss of energy of turbulence. g V KH cm 2 2 2 = Head loss due to sudden contraction is Where, kc is sudden contraction coefficient and it value depends up ratio of D2/D1 and velocity (V2) in smaller pipe
Minor Losses 33 Head loss due to contraction of pipe (Gradual Contraction) Head loss from pipe contraction may be greatly reduced by introducing a gradual pipe transition known as a confusor as shown Figure. g V KH cm 2 ' 2 2 = Head loss due to gradual contraction is Where, kc ’ is gradual contraction coefficient and it value depends up ratio of D2/D1 and velocity (V2) in smaller pipe
Minor Losses 34 Entrance loss The general equation for an entrance head loss is also expressed in terms of velocity head of the pipe: The approximate values for the entrance loss coefficient (Ke) for different entrance conditions are given below g V KH em 2 2 =
Minor Losses 35 head loss due to enlargement of pipe (Sudden Enlargement) The behavior of the energy grade line and the hydraulic grade line in the vicinity of a sudden pipe expansion is shown in Figure The magnitude of the head loss may be expressed as ( ) g VV Hm 2 2 21 − =
Minor Losses 36 head loss due to enlargement of pipe (Gradual Enlargement) The head loss resulting from pipe expansions may be greatly reduced by introducing a gradual pipe transition known as a diffusor The magnitude of the head loss may be expressed as ( ) g VV KH em 2 2 21 − = The values of Ke’ vary with the diffuser angle (α).
Minor Losses 37 Exit Loss A submerged pipe discharging into a large reservoir (Figure ) is a special case of head loss from expansion. ( ) g V KH dm 2 2 = Exit (discharge) head loss is expressed as where the exit (discharge) loss coefficient Kd=1.0.
Minor Losses 38 Head loss due to fittings valves Fittings are installed in pipelines to control flow.As with other losses in pipes, the head loss through fittings may also be expressed in terms of velocity head in the pipe: g V KH fm 2 2 =
Minor Losses 39 Head loss due to bends The head loss produced at a bend was found to be dependent of the ratio the radius of curvature of the bend (R) to the diameter of the pipe (D).The loss of head due to a bend may be expressed in terms of the velocity head as For smooth pipe bend of 900, the values of Kb for various values of R/D are listed in following table. g V KH bm 2 2 =
Minor Losses 40
Numerical Problems 41
Numerical Problems 42
Pipes in Series 43 If a pipeline is made up of lengths of different diameters, as shown in figure, conditions must satisfy the continuity and energy equations; If Q is given, the problem is straight forward. Total head loss can be estimated by adding the contributions from various sections as; νπ VD R gD LV fh D Q V L === , 2 , 4 2 2 Friction coefficient f, can be constant values given in problem or variable estimated from material and flow properties. i.e. e/D,V, R and f If head loss is given and discharge is required then either of the two methods namely equivalent velocity head method or equivalent length method many be used.;
Pipes in Series 44 (1). Equivalent-velocity-head method: In this method a more accurate approach of DarcyWeisbach is applied to estimate Q. ...321 +++= LLLL hhhh Above equation considering minor losses becomes as; ... 222 2 3 3 3 3 2 2 2 2 2 2 1 1 1 1 +      ++      ++      += ∑∑∑ g V K D L f g V K D L f g V K D L fhL Now using continuity equation, we can write 3 2 31 2 12 2 21 2 1 3 2 32 2 21 2 13 2 32 2 21 2 1221 , 444 VDVDVDVD VDVDVDVDVDVDQQQ == ==⇒==⇒== πππ From above, all velocities can be expressed in terms of chosen velocity (V). (1). Equivalent velocity head method (2). Equivalent length method If Q is not given:
Pipes in Series 45 (2). Equivalent length method: In this method, all pipes are expressed in terms of equivalent lengths to one given pipe size By equivalent length is meant a length, Le, of pipe of a certain diameter, De, and friction factor, fe, which for the same flow will give same head loss as the pipe under consideration of length, L, diameter, D, and friction factor, f 5 2 2 2/ 2/       == D De fe f L gVe gV D De fe f LLe
Pipes in Series 46 Problem: Suppose in the Figure the pipes 1, 2, 3 are 300m of 30cm diameter, 150m of 20cm diameter and 250m of 25cm diameter, respectively, of new cast iron and are conveying 15oC water. If h=10m, find the rate of flow from A to B. Figure
47 From moody diagram, we
Pipes in Series 48 5 2 2 2/ 2/       == D De fe f L gVe gV D De fe f LLe
Pipes in Parallel 49 In the case of flow through two or more parallel pipes, as shown in fig, the continuity and energy equations establish the following relations which must be satisfied. According to Darcy-Weisbach equation, the head loss including minor losses can be written as; g V K D L fhL 2 2       += ∑       + = ∑K D L f gh V L2
Pipes in Parallel 50 According to continuity equation L L hC K D L f gh AAVQ =       + == ∑ 2 Where, C is constant for given pipe. Now according to governing equation for pipe flow, we can write as; ( ) L LLL hCCCQ hChChCQ QQQQ ... .. ... 321 221 321 +++= +++= +++= Above equations enable us to find a first estimate of hL and the distribution of flows and velocities in pipes. Using these, we can next make improvements to the values of f and if necessary repeat them until we finally make a correct determination of hL and distribution of flows.
Pipes in Parallel 51 Find the flow rates in each pipe ?
Pipes in Parallel 52 Energy eq. Head loss Eq. for parallel pipes
Pipes in Parallel 53
Problem 54 Suppose in the Figure the pipes 1, 2, 3 are 150m of 80mm diameter, 60m of 50mm diameter and 120m of 60mm diameter, respectively, of new wrough iron pipe (e=0.000046m). If hL=6m, find the rate of flow from A to B. Take kinematic viscosity as 4.26x10-5m2/s Answer: 1.5L/s
Problem 55
Problem 56 In fig. pipe 1 is 500ft of 2-in, pipe 2 is 350ft of 3-in, and pipe 3 is 750ft of 4- in diameter, all of smooth brass (e=0.0005ft). Crude oil (s=0.855 and kinematic viscosity of 7.6x10-5ft2/s) is flow at 0.7 cfs. Find the head loss from A to B and the flow in each pipe. Answer:
Problem 57
58
Thank you Questions…. 59

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Hydraulic analysis of complex piping systems (updated)

  • 1.
    Hydraulic analysis ofcomplex piping systems (Losses in pipes, pipes in parallel and pipes in series) 1 Hydraulics Dr. Mohsin Siddique
  • 2.
    Steady Flow ThroughPipes 2 Laminar Flow: flow in layers Re<2000 (pipe flow) Turbulent Flow: flow layers mixing with each other Re >4000 (pipe flow)
  • 3.
    Steady Flow ThroughPipes 3 Reynold’s Number(R or Re): It is ratio of inertial forces (Fi) to viscous forces (Fv) of flowing fluid For laminar flow: Re<=2000 For transitional flow: 2000<Re<4000 For Turbulent flow: Re>= 4000 . . Re . . . . . . . . Velocity Volume Mass Velocity Fi Time Time Fv Shear Stress Area Shear Stress Area Q V AV V AV V VL VL du VA A A dy L ρ ρ ρ ρ ρ τ µ υµ µ = = = = = = = = Where ; V is avg. velocity of flow in pipe ν is kinematic viscosity L is characteristic/representative linear dimension of pipe. It is diameter of pipe (circular conduits) or hydraulic radius (non-circular conduits). νµ ρ VDVD Re == Note: For non-circular section, we need to use hydraulic radius (Rh) instead of diameter (D) for the linear dimension (L). Values of critical Reynolds no.
  • 4.
    Steady Flow ThroughPipes 4 Hydraulic Radius (Rh) or Hydraulic Diameter: It is the ratio of area of flow to wetted perimeter of a channel or pipe. P A perimeterwetted Area Rh == For Circular Pipe ( )( ) h h RD D D D P A R 4 4 4/ 2 = === π π For Rectangular pipe DB BD P A Rh 2+ == B D Note: hydraulic Radius gives us indication for most economical section. More the Rh more economical will be the section. νν h e VRVD R 4 == By replacing D with Rh, Reynolds’ number formulae can be used for non-circular sections as well.
  • 5.
    Head Loss inPipes 5 Total Head Loss=Major Losses+ Minor Losses Major Loss: Due to pipe friction Minor Loss: Due to pipe fittings, bents and valves etc
  • 6.
    Head Loss inPipes due to Friction 6 The head loss due to friction in a given length of pipe is proportional to mean velocity of flow (V) as long as the flow in laminar. i.e., But with increasing velocity, as the flow become turbulent the head loss also varies and become proportion toVn Where n ranges from 1.75 to 2 Log-log plot for flow in uniform pipe (n=2.0 for rough wall pipe; n=1.75 for smooth wall pipe) VH f ∝ n f VH ∝
  • 7.
    Frictional Head Lossin Conduits of Constant Cross-Section 7 Consider stead flow in a conduit of uniform cross-section A.The pressure at section 1 & 2 are P1 & P2 respectively.The distance between the sections is L. For equilibrium in stead flow, ∑ == 0maF Figure: Schematic diagram of conduit 0sin 21 =−−− APPLWAP oτα P= perimeter of conduit = Avg. shear stress between pipe boundary and liquid oτ 012 21 =−      − −− PL L zz ALAPAP oτγ αsin12 =      − L zz
  • 8.
    Frictional Head Lossin Conduits of Constant cross-section 8 012 21 =−      − −− PL L zz ALAPAP oτγ Dividing the equation by Aγ ( ) 012 21 =−−−− A PL zz PP o γ τ γγ fL o hh A PLP z P z ===      +−      + γ τ γγ 2 2 1 1 Therefore, head loss due to friction, hf, can be written as; h oo f R L A PL h γ τ γ τ == Lh g v z P g v z P +++=++ 22 2 2 2 2 2 1 1 1 γγ Remember !! For pipe flow For stead flow in pipe of uniform diameter v1=v2 Lhz P z P =      +−      + 2 2 1 1 γγ This is general equation and can be applied to any shape conduit having either Laminar or turbulent flow. P A Rh =Q
  • 9.
    Determining Shear Stress 9 Forsmooth-walled pipes/conduits, the average shear stress at the wall is Using Rayleigh's Theorem of dimensional analysis, the above relation can be written as; Rewriting above equation in terms of dimension (FLT), we get ( ),,,, VRf ho ρµτ = ( )                           = ncb a T L L FT L FT LK L F 4 2 22 ( )ncba ho VRk ... ρµτ = LlengthR L F areaforce h o == == 2 /τ ( ) 22 4233 /. //// / − == === = FTLmsN LFTLaFLM TLV µ ρ( ) ( ) ( ) ( )( )ncba TLLFTFTLLKFL /4222 −−− =
  • 10.
    Determining Shear Stress 10 Accordingto dimensional homogeneity, the dimension must be equal on either side of the equation, i.e., ( ) ( ) ( ) ( )( )ncba TLLFTFTLLKFL /4222 −−− = )(20: )(422: )(1: iiincbT iincbaL icbF →−+= →+−−=− →+= Solving three equations, we get 1;2;2 −=−=−= ncnbna Substituting values back in above equation ( ) ( ) 2 2 122 ...... V VR kVRkVRk n hnnnn h ncba ho ρ µ ρ ρµρµτ − −−−       === ( ) 22 VRk n eo ρτ − = Setting = we get 2 2 V Cfo ρτ = Where, Cf is the coefficient of friction ( ) 2−n eRk 2/fC (Re)fCf =
  • 11.
    Determining Shear Stress 11 Nowsubstituting the equation of avg. shear stress in equation of head loss, For circular pipe flows, Rh=D/4 Where, f is a friction factor. i.e., The above equation is known as pipe friction equation and as Darcy- Weisbach equation and is used for calculation of pipe-friction head loss for circular pipes flowing full (laminar or turbulent) 2/2 VCfo ρτ = h o f R L h γ τ = h f h f f gR LVC R LVC h 22 22 == γ ρ g V D L fh g V D L C Dg LVC h f f f f 2 2 4 42 4 2 22 = == fCf 4=Q
  • 12.
    Friction Factor forLaminar and Turbulent Flows in Circular Pipes 12 Smooth and Rough Pipe A smooth pipe is the one which behaves as smooth pipe i.e. (frictionless). It is quite possible that a pipe may behave as smooth at a certain flow condition while at some other condition it may behave rough. Mathematically; Smooth Pipe Rough Pipe Transitional mode Turbulent flow near boundary = = v e δ Roughness height Thickness of viscous sub-layer Smooth pipe Rough pipe f D fV v Re 14.1414.14 == ν δ ve δ< ve δ> ve δ< ve δ14> vv e δδ 14≤≤
  • 13.
    Friction Factor forLaminar and Turbulent Flows in Circular Pipes 13 For laminar flow For turbulent flow Re 64 =f2000Re < 51.2 Re log2 1 f f = 9.6 Re log8.1 1 = f 25.0 Re 316.0 =f 7/1 max       = or y u u Def / 7.3 log2 1 =         +−= f De f Re 51.2 7.3 / log2 1         +      −= Re 9.6 7.3 / log8.1 1 11.1 De f Colebrook Eq. for turbulent flow in all pipes Halaand Eq. For turbulent flow in all pipes Von-karman Eq. for fully rough flow Blacius Eq. for smooth pipe flow Seventh-root law 5 10Re3000 ≤≤ From Nikuradse experiments Colebrook Eq. for smooth pipe flow for smooth pipe flow 4000Re >
  • 14.
    Friction Factor forLaminar and Turbulent Flows in Circular Pipes 14 The Moody chart or Moody diagram is a graph in non- dimensional form that relates the Darcy-Weisbach friction factor, Reynolds number and relative roughness for fully developed flow in a circular pipe. The Moody chart is universally valid for all steady, fully developed, incompressible pipe flows.
  • 15.
    Friction Factor forLaminar and Turbulent Flows in Circular Pipes 15 For laminar flow For non-laminar flow eR f 64 =         +−= f De f Re 51.2 7.3 / log2 1 Colebrook eq.
  • 16.
    Friction Factor forLaminar and Turbulent Flows in Circular Pipes 16 The friction factor can be determined by its Reynolds number (Re) and the Relative roughness (e/D) of the Pipe.( where: e = absolute roughness and D = diameter of pipe)
  • 17.
  • 18.
    Problem Types 18 Type 1:Determine f and hf, Type 2: Determine Q Type 3: Determine D
  • 19.
    Problem 19 Find friction factorfor the following pipe e=0.002 ft D=1ft KinematicViscosity, ν=14.1x10-6ft2/s Velocity of flow,V=0.141ft/s Solution: e/D=0.002/1=0.002 R=VD/ ν =1x0.141/(14.1x10-6)=10000 From Moody’s Diagram; f=0.034 ___________ Re 51.2 7.3 / log2 1 =         +−= f f De f
  • 20.
    Problem-Type 1 20 Pipe dia,D= 3 inch & L=100m Re=50,000 ʋ=1.059x10-5ft2/s (a): Laminar flow: f=64/Re=64/50,000=0.00128 ft gD fLV HLf 0357.0 )12/3)(2.32(2 )12.2)(100(00128.0 2 22 === sftV VVD /12.2 10059.1 )12/3( 50000Re 5 =⇒ × =⇒= − ν
  • 21.
    Problem-Type 1 21 Pipe dia,D= 3 inch & L=100m Re=50,000 ʋ=1.059x10-5ft2/s (b): Turbulent flow in smooth pipe: i.e.: e=0 0209.0 50000 51.2 7.3 0 log2 Re 51.2 7.3 / log2 1 =         +−=         +−= f ff De f ft gD fLV HLf 582.0 )12/3)(2.32(2 )12.2)(100(0209.0 2 22 ===
  • 22.
    Problem-Type 1 22 Pipe dia,D= 3 inch & L=100m Re=50,000 ʋ=1.059x10-5ft2/s (c): Turbulent flow in rough pipe: i.e.: e/D=0.05 0720.0 50000 51.2 7.3 05.0 log2 Re 51.2 7.3 / log2 1 =         +−=         +−= f ff De f ft gD fLV HLf 01.2 )12/3)(2.32(2 )12.2)(100(0720.0 2 22 ===
  • 23.
    Problem-Type 1 23 hL=? memDmL 0005.0;25.0;1000=== smsmQ /10306.1;/051.0 263 − ×== ν ( ) ( ) 002.025.0/0005.0/ 10210306.1/25.0039.1/ 56 == ×=××== − De VDR ν 0.0245f DiagramsMoody'From = m gD fLV hL 39.5 2 2 == smAQV /039.1/ ==Q
  • 24.
  • 25.
    Problem-Type 2 25 For laminarflow For non-laminar flow eR f 64 =         +−= f De f Re 51.2 7.3 / log2 1 Colebrook eq.
  • 26.
  • 27.
  • 28.
  • 29.
  • 30.
    MINOR LOSSES 30 Each typeof loss can be quantified using a loss coefficient (K). Losses are proportional to velocity of flow and geometry of device. Where, Hm is minor loss and K is minor loss coefficient.The value of K is typically provided for various types/devices NOTE: If L > 1000D minor losses become significantly less than that of major losses and hence can be neglected. g V KHm 2 2 =
  • 31.
    Categories of MinorLosses 31 These can be categorized as 1. Head loss due to contraction in pipe 1.1 Sudden Contraction 1.2 Gradual Contraction 2. Entrance loss 3. Head loss due to enlargement of pipe 3.1 Sudden Enlargement 3.2 Gradual Enlargement 4. Exit loss 5. Head loss due to pipe fittings 6. Head loss due to bends and elbows
  • 32.
    Minor Losses 32 Head lossdue to contraction of pipe (Sudden contraction) A sudden contraction (Figure) in pipe usually causes a marked drop in pressure in the pipe because of both the increase in velocity and the loss of energy of turbulence. g V KH cm 2 2 2 = Head loss due to sudden contraction is Where, kc is sudden contraction coefficient and it value depends up ratio of D2/D1 and velocity (V2) in smaller pipe
  • 33.
    Minor Losses 33 Head lossdue to contraction of pipe (Gradual Contraction) Head loss from pipe contraction may be greatly reduced by introducing a gradual pipe transition known as a confusor as shown Figure. g V KH cm 2 ' 2 2 = Head loss due to gradual contraction is Where, kc ’ is gradual contraction coefficient and it value depends up ratio of D2/D1 and velocity (V2) in smaller pipe
  • 34.
    Minor Losses 34 Entrance loss Thegeneral equation for an entrance head loss is also expressed in terms of velocity head of the pipe: The approximate values for the entrance loss coefficient (Ke) for different entrance conditions are given below g V KH em 2 2 =
  • 35.
    Minor Losses 35 head lossdue to enlargement of pipe (Sudden Enlargement) The behavior of the energy grade line and the hydraulic grade line in the vicinity of a sudden pipe expansion is shown in Figure The magnitude of the head loss may be expressed as ( ) g VV Hm 2 2 21 − =
  • 36.
    Minor Losses 36 head lossdue to enlargement of pipe (Gradual Enlargement) The head loss resulting from pipe expansions may be greatly reduced by introducing a gradual pipe transition known as a diffusor The magnitude of the head loss may be expressed as ( ) g VV KH em 2 2 21 − = The values of Ke’ vary with the diffuser angle (α).
  • 37.
    Minor Losses 37 Exit Loss Asubmerged pipe discharging into a large reservoir (Figure ) is a special case of head loss from expansion. ( ) g V KH dm 2 2 = Exit (discharge) head loss is expressed as where the exit (discharge) loss coefficient Kd=1.0.
  • 38.
    Minor Losses 38 Head lossdue to fittings valves Fittings are installed in pipelines to control flow.As with other losses in pipes, the head loss through fittings may also be expressed in terms of velocity head in the pipe: g V KH fm 2 2 =
  • 39.
    Minor Losses 39 Head lossdue to bends The head loss produced at a bend was found to be dependent of the ratio the radius of curvature of the bend (R) to the diameter of the pipe (D).The loss of head due to a bend may be expressed in terms of the velocity head as For smooth pipe bend of 900, the values of Kb for various values of R/D are listed in following table. g V KH bm 2 2 =
  • 40.
  • 41.
  • 42.
  • 43.
    Pipes in Series 43 Ifa pipeline is made up of lengths of different diameters, as shown in figure, conditions must satisfy the continuity and energy equations; If Q is given, the problem is straight forward. Total head loss can be estimated by adding the contributions from various sections as; νπ VD R gD LV fh D Q V L === , 2 , 4 2 2 Friction coefficient f, can be constant values given in problem or variable estimated from material and flow properties. i.e. e/D,V, R and f If head loss is given and discharge is required then either of the two methods namely equivalent velocity head method or equivalent length method many be used.;
  • 44.
    Pipes in Series 44 (1).Equivalent-velocity-head method: In this method a more accurate approach of DarcyWeisbach is applied to estimate Q. ...321 +++= LLLL hhhh Above equation considering minor losses becomes as; ... 222 2 3 3 3 3 2 2 2 2 2 2 1 1 1 1 +      ++      ++      += ∑∑∑ g V K D L f g V K D L f g V K D L fhL Now using continuity equation, we can write 3 2 31 2 12 2 21 2 1 3 2 32 2 21 2 13 2 32 2 21 2 1221 , 444 VDVDVDVD VDVDVDVDVDVDQQQ == ==⇒==⇒== πππ From above, all velocities can be expressed in terms of chosen velocity (V). (1). Equivalent velocity head method (2). Equivalent length method If Q is not given:
  • 45.
    Pipes in Series 45 (2).Equivalent length method: In this method, all pipes are expressed in terms of equivalent lengths to one given pipe size By equivalent length is meant a length, Le, of pipe of a certain diameter, De, and friction factor, fe, which for the same flow will give same head loss as the pipe under consideration of length, L, diameter, D, and friction factor, f 5 2 2 2/ 2/       == D De fe f L gVe gV D De fe f LLe
  • 46.
    Pipes in Series 46 Problem:Suppose in the Figure the pipes 1, 2, 3 are 300m of 30cm diameter, 150m of 20cm diameter and 250m of 25cm diameter, respectively, of new cast iron and are conveying 15oC water. If h=10m, find the rate of flow from A to B. Figure
  • 47.
  • 48.
  • 49.
    Pipes in Parallel 49 Inthe case of flow through two or more parallel pipes, as shown in fig, the continuity and energy equations establish the following relations which must be satisfied. According to Darcy-Weisbach equation, the head loss including minor losses can be written as; g V K D L fhL 2 2       += ∑       + = ∑K D L f gh V L2
  • 50.
    Pipes in Parallel 50 Accordingto continuity equation L L hC K D L f gh AAVQ =       + == ∑ 2 Where, C is constant for given pipe. Now according to governing equation for pipe flow, we can write as; ( ) L LLL hCCCQ hChChCQ QQQQ ... .. ... 321 221 321 +++= +++= +++= Above equations enable us to find a first estimate of hL and the distribution of flows and velocities in pipes. Using these, we can next make improvements to the values of f and if necessary repeat them until we finally make a correct determination of hL and distribution of flows.
  • 51.
    Pipes in Parallel 51 Findthe flow rates in each pipe ?
  • 52.
    Pipes in Parallel 52 Energyeq. Head loss Eq. for parallel pipes
  • 53.
  • 54.
    Problem 54 Suppose in theFigure the pipes 1, 2, 3 are 150m of 80mm diameter, 60m of 50mm diameter and 120m of 60mm diameter, respectively, of new wrough iron pipe (e=0.000046m). If hL=6m, find the rate of flow from A to B. Take kinematic viscosity as 4.26x10-5m2/s Answer: 1.5L/s
  • 55.
  • 56.
    Problem 56 In fig. pipe1 is 500ft of 2-in, pipe 2 is 350ft of 3-in, and pipe 3 is 750ft of 4- in diameter, all of smooth brass (e=0.0005ft). Crude oil (s=0.855 and kinematic viscosity of 7.6x10-5ft2/s) is flow at 0.7 cfs. Find the head loss from A to B and the flow in each pipe. Answer:
  • 57.
  • 58.
  • 59.