Pipeline Engineering

Pipeline Engineering - PIPELINE ENGINEERING FLUID FLOW...

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PIPELINE ENGINEERING FLUID FLOW Mechanical Energy Balance gz v d p V WF o ∆∆ ++ =− 2 2 (1-1) potential energy expansion work Kinetic energy Work added/ Sum of friction change change subtracted by losses compressors or pumps/expanders Note that the balance is per unit mass. In differential form F W VdV vdp gdz o δ = + + (1-2) Rewrite as follows ( ) dp g dz V dV F W o − ⋅ + ρδ (1-3) Divide by dL ( L is the length of pipe) dp dL g dz dL V dV dL F L W L Tot o + + ρρ ρ (1-4) or: dp dL dp dL dp dL dp dL Tot elev accel frict = + + (1-5) ( δ δ W L o is usually ignored, as the equation applies to a section of pipe) The above equation is an alternative way of writing the mechanical energy balance. It is not a different equation. The differential form of the potential energy change is
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Natural Gas Basic Engineering Copyright: Miguel Bagajewicz. No reproduction allowed without consent 2 dZ φ d L g dZ dL g = sin φ (1-6) Friction losses: We use the Fanning or Darcy-Weisbach equation (Often called Darcy equation) δ F Vf D dL = 2 2 (1-7) an equation that applies for single phase fluids, only (two phase fluids are treated separately). The friction factor, in turn, is obtained from the Moody Diagram below. Figure 1-1: Moody Diagram Friction factor equations. (Much needed in the era of computers and excel) Laminar Flow f = 16 Re (1-8)
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Natural Gas Basic Engineering Copyright: Miguel Bagajewicz. No reproduction allowed without consent 3 Turbulent Flow f a = 0 046 . Re (1-9) smooth pipes: a=0. Iron or steel pipes a=0.16 Turbulent Flow 1 2 37 251 10 f D f =− + log . . Re ε (Colebrook eqn) (1-10) Equivalent length of valves and fittings: Pressure drop for valves and fittings is accounted for as equivalent length of pipe. Typical values can be obtained from the following Table. Table 1-1: Equivalent lengths for various fittings. Fitting e L D 45 O elbows 15 90 O elbows, std radius 32 90 O elbows, medium radius 26 90 O elbows, long sweep 20 90 O square elbows 60 180 O close return bends 75 180 O medium radius return bends 50 Tee (used as elbow, entering run) 60 Tee (used as elbow, entering branch) 90 Gate Valve (open ) 7 Globe Valve (open ) 300 Angle Valve (open) 170 Pressure Drop Calculations Piping is known. Need pressure drop. (Pump or compressor is not present.) Incompressible Flow a) Isothermal ( ρ is constant) Tot dp dZ dV dF =- g +V + dL dL dL dL ρ ⎛⎞ ⎜⎟ ⎝⎠ (1-11) for a fixed φ V constant dV = 0
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Natural Gas Basic Engineering Copyright: Miguel Bagajewicz. No reproduction allowed without consent 4 2 2 L FV f D δ ⎛⎞ =⋅ ⎜⎟ ⎝⎠ (1-12) ∆∆ pg Z V f L D F =− + ⋅ ⋅ + ρ 2 2 (1-13) b) Nonisothermal It will not have a big error if you use ρ (T average ), v(T average ) Exercise 1-1: Consider the flow of liquid water (@ 20 o C) through a 200 m, 3” pipe, with an elevation change of 5 m. What is the pressure drop?
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This note was uploaded on 09/05/2011 for the course CHE 5480 taught by Professor Staff during the Spring '11 term at OKCU.

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Pipeline Engineering - PIPELINE ENGINEERING FLUID FLOW...

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