fluids-1 - FLUID FLOW Mechanical Energy Balance dp V 2 gZ =...

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FLUID FLOW Mechanical Energy Balance potential expansion kinetic work added/ sumof energy work energy change subtracted by friction change pumps or losses compressors Notethat the balanceis per unit mass. In differential form: - = + + F W V dp Z g o 2 2 ρ ( 29 o W F dV V dZ g dp δ δ ρ + - - - = ChE 4253 - Design I
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FLUID FLOW Mechanical Energy Balance Divideby dL, (L is thelength of thepipe) or: is usually ignored, as theequation applies to a pipe section. Theaboveequation is an alternativeway of writing themechanical energy balance. It is not a different equation L W L F dL dV V dL dZ g dL dp o Tot δ δ ρ δ δ ρ ρ ρ - + + - = frict accel elev Tot dL dp dL dp dL dp dL dp + + = L W o δ δ ChE 4253 - Design I
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Mechanical Energy Balance Potential energy change: Friction Losses: Fanning equation: This equation applies to singlephasefluids. Thefriction factor is obtained fromthe“Moody Diagram” (seePT&W page487). dZ φ dL φ sin g dL dZ g = dL D f V dF 2 2 = ChE 4253 - Design I
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Mechanical Energy Balance Friction factor equations. (Useful for computers and Excel) Laminar Flow Smooth pipes: a = 0.2 Iron or steel pipes: a = 0.16 Colebrook equation for turbulent flow. Equivalent length of valves and fittings . Pressuredrop for valves and fittings is accounted for as equivalent length of pipe. SeePT&W for a tablecontaining thesevalues (page490). Re 16 = f a f Re 046 . 0 = + - = f D f Re 51 . 2 7 . 3 log 2 1 10 ε ChE 4253 - Design I
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Mechanical Energy Balance - Fluid Flow Scenario I Need pressuredrop in known pipes (pump or compressor is not present.) Incompressible Flow a) Isothermal ( ρ is constant) for a fixed ρ V constant dV = 0 Integral form: b) Nonisothermal It will not havea big error if you use ρ (T average ) + + - = L F dL dV V dL dZ g dL dp Tot δ δ ρ + + + - = F D L L f V Z g p e 2 2 ρ ChE 4253 - Design I
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Mechanical Energy Balance - Fluid Flow Compressible Flow a) Relatively small changein T (known)
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