1. LAMINAR & TURBULENT FLOWSTypes of flows in pipes and how they influence energy losses inpipes.Laminar flow:Where the fluid moves slowly in layers in a pipe, without muchmixing among the layers.Typically occurs when the velocity is low or the fluid is veryviscous. 1
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3. Turbulent flowOpposite of laminar, where considerable mixing occurs,velocities are high.Laminar and Turbulent flows can be characterized andquantified using Reynolds Number established by Osborne Reynoldsand is given as – 3
4. V Dρ VD NR = = η νWhere NR – Reynolds numberV – velocity of flow (m/s)D – diameter of pipe (m)ρ – density of water (kg/m3)ή – dynamic viscosity (kg/m.s)ν – kinematic viscosity (m2/s)when – units are considered – NR is dimensionless.NOTE – Reynolds number directly proportional to velocity& inversely proportional to viscosity!NR < 2000 – laminar flowNR > 4000 – Turbulent flow 4
5. For 2000 < NR < 4000 – transition region or critical region –flow can either be laminar of turbulent – difficult to pin downexactly.More viscous fluid will tend to have laminar flow or lowerReynolds number.Reynolds numbers for some real-life examples – • Blood flow in brain ~ 100 • Blood flow in aorta ~ 1000 • Typical pitch in major league baseball ~ 200000 • Blue whale swimming ~ 300000000 5
6. Problem 8.1:Determine the Reynolds number for –Glycerin at 25CPipe at D = 150 mm = 0.15 mVelocity = V = 3.6 m/sρ = 1258 kg/m3ή = 0.96 Pa.s or (kg/m.s) V DρNR = η = 3.6 * 0.15 * 1258 / 0.96 = 708Nr < 2000; therefore flow is laminar! 6
7. Friction losses in PipesVary with laminar or turbulent flowEnergy equation can be given as –p1/γ + z1 + v12/2g + hA – hR – hL = p2/γ + z2 + v22/2gwhere hA, hR, hL are the heads associated with addition,removal and friction loss in pipes, respectively.note that the terms are added on the LHS of the equation!The head loss in pipes = hL can be expressed as – hL = f * (L/D) * v2/2g- Darcy’s equation for energy loss (GENERAL FORM)Wheref – friction factorL – length of pipeD – diameter of pipev – velocity of flowHow do these factors affect losses??????????? 7
8. Similarly,Another equation was developed to compute hL under Laminarflow conditions only 32 η L V hL = γ D2 - called the Hagen-Poiseuille equationIf you equate Darcy’s equation and Hagen-Poiseuille equationThen we can find the friction factor f 64η 64 f = = V D ρ NRThus the friction factor is a function of Reynold’s number! 8
9. Example Problem 8.4Determine energy loss if –Glycerin at 25CL = 30mD = 150 mm = 0.15 mV = 4.0 m/sDensity = 1258 k/m3Dynamic viscosity = 0.96 Pa.sFirst determine the Reynolds number to determine if it islaminar flow V DρNR = η= 4.0 * 0.15 * 1258 / 0.96 = 786 < 2000 - Laminar flowTherefore Darcy’s equationhL = f * (L/D) * (v2/2g) = 64/NR * (L/D) * (v2/2g) = 13.2 m 9
10. Friction losses for Turbulent FlowFor Laminar flow we got a nice equation to compute the frictionfactor – dependent ONLY on Reynolds number!In case of Turbulent flow – friction factor computed basedon inside roughness of the pipe and Reynolds number!Roughness is expressed as - Relative roughness = D/ε(not the best way to express it!) • More rough pipe – low D/ε • Less rough pipe – high D/εFriction factor for Turbulent flow ~ Relative roughness &Reynolds numberRoughness values for various pipes – Table 8.2 10
11. Some Roughness values from Table 8.2Material Roughness ε (m)Plastic 3.0 x 10-7Steel 4.6 x 10-5Galvanized iron 1.5 x 10-4Concrete 1.2 x 10-4Question - **Why doesn’t ε affect loss in laminar flowconditions ????????? 11
12. The Moody DiagramDeveloped to provide the friction factor for turbulent flow forvarious values of Relative roughness and Reynold’s number!Curves generated by experimental data. 12
13. Key points about the Moody Diagram – 1. In the laminar zone – f decreases as Nr increases! 2. f = 64/Nr. 3. transition zone – uncertainty – not possible to predict - 4. Beyond 4000, for a given Nr, as the relative roughness term D/ε increases (less rough), friction factor decreases 5. For given relative roughness, friction factor decreases with increasing Reynolds number till the zone of complete turbulence 6. Within the zone of complete turbulence – Reynolds number has no affect. 7. As relative roughness increases (less rough) – the boundary of the zone of complete turbulence shifts (increases)If you know the – relative roughness, Reynolds number –you can compute the friction factor from the MoodyDiagram 13
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15. Swamee and Jain converted the graph into an equation!For the turbulent portion -(less than 1% error) 0.25 f = 2 ⎡ ⎛ 1 ⎞ 5.74 ⎤ ⎢ log ⎜ ⎜ 3.7 ( D / ε ) ⎟ + N 0.9 ⎥ ⎟ ⎢ ⎝ ⎣ ⎠ R ⎥ ⎦ 15
16. Example problem 8.5:Determine friction factor if water at 160 F is flowing at 30ft/s in an uncoated ductile iron pipe having an insidediameter of 1.0 inch.Determine Reynolds number: V Dρ VDNR = = η νD = 1/12 = 0.0833 ft; V = 30 ft/s; ν = 4.38 x 10-6 ft2/sNR = 5.70 x 105So its turbulent flow, we will have to use Moody diagramDetermine relative roughness –ε for ductile iron pipe from table 8.2 in text = 8 x 10-4 ftD/ ε = 0.0833/8 x 10-4= 104, use 100 for moody diagramUsing those values and going into Moody’s diagram, we get – 16
17. f = 0.0038. 17
18. Hazen-William formulaAlternate formula to compute head loss due to frictionApplicable for – - Water - D – larger than 2 inches and less than 6ft - Velocity should not exceed 10 ft/sThe formula is unit-specificFor SI units – 1.852 ⎡ Q ⎤hL = L ⎢ ⎥ ⎣ 0.85 A C h R 0.63 ⎦WhereQ = 0.85 A C h R 0.63 s 0.54Where R – hydraulic radius = area/perimeter of pipe.Ch – Hazen Williams Coefficientand s – slope gradient of the pipe. 18
19. Head Additions and Losses due to Pumps and Motors(CHAPTER 7)Bernoulli’s Energy equation can be given as –p1/γ + z1 + v12/2g + hA – hR – hL = p2/γ + z2 + v22/2gwhere hA, hR, hL are the heads associated with addition, removaland friction loss in pipes, respectively.also known as the GENERAL ENERGY EQUATIONrepresents the system in reality (not idealized conditions) 19
20. hA – head/energy additions due to pumps.hR – head/energy removal due to motors.hL – head losses due to friction and losses due to fittings, bends,valves.MAJOR LOSSES – motors, friction lossMINOR LOSSES – valves, fittings, and bendsTypicallyhL = K * v2/2gthat’s what we saw before with respect to Darcy’s Equation. 20
21. Example Problem 7.1: Finding TOTAL head loss due tofriction, valves, bends etc. –Given –Water moving in the systemQ = 1.20 ft3/sCalculate TOTAL head loss due to valves, bends, and pipefriction.How should we proceed??????????? 21
22. Apply the General Energy Equation for pts 1 and 2! p1/γ + z1 + v12/2g - hL = p2/γ + z2 + v22/2gcancel out the terms – p1/γ + z1 + v12/2g - hL = p2/γ + z2 + v22/2gand we get – hL = (z1- z2) - v22/2g(z1- z2) = 25 ftHow do we find v2???Area of 3 inch dia pipe = 0.0491 ft2v2 = 1.20 / 0.0491 = 24.4 ft/shL = 25 – (24.4)2/2*32.2 = 15.75 ft 22
23. Example Problem 7.2: - Now we will introduce a Pump anddetermine the energy added by a pump! • Q = 0.014 m3/s • Fluid – Oil at Sg = 0.86; γ = 0.86 * 9.81 = 8.44 kN/m3 • hL = head loss due to friction, fittings, etc. = 1.86 Nm/NEnergy added by Pump???????????????????? 23
24. Apply the General Energy Equation for pts. A and B! pA/γ + zA + vA2/2g + hA - hL = pB/γ + zB + vB2/2gRearrange the equation!hA = hL + (pB – pA) / γ + (zB – zA) + (vB2 - vA2 ) / 2g(pB – pA) / γ = (296-(-28))/8.44 = 38.4 m(zB – zA) = 1.0 mHow will you find velocities?????Area at A = 0.004768 m2Area at B = 0.002168 m2vA = 2.94 m/svB = 6.46 m/splug all terms back in the original equation –hA = hL + (pB – pA) / γ + (zB – zA) + (vB2 - vA2 ) / 2g 24
25. and hA = 1.86 + 38.4 + 1.0 + 1.69 = 42.9 m or 42.9 N.m/N 25
26. Power supplied by PumpsPower of the pump = energy being transferredPA = hA * W (where W is the weight flow rate)= hA * γ *QSI units of power = watt (W) = 1.0 N.m/N or 1.0 JouleUS units of Power = lb-ft/s1 horsepower = 1 hp = 550 lb-ft/s1 hp = 745.7 WattEfficiency of the Pump= (Power delivered to the fluid/ Power supplied to the pump)= (PA/PI) 26
27. Example Problem 7.3: Determine the efficiency of the PumpPI = 3.85 hpQ = 500 gal/min of oil = 1.11 ft3/sγ of oil = 56.0 lb/ft3What is PA????And what is efficiency (em) ????So how do you proceed???????????????????? 27
28. Apply the General Energy Equation for pts 1 and 2! p1/γ + z1 + v12/2g + hA = p2/γ + z2 + v22/2gRearrange the equation! hA = (p2 – p1) / γ + (z2 – z1) + (v22 - v12 ) / 2glet’s solve for each termγm = 13.54*62.4 = 844.9 lb/ft3pressure difference using the manometer –p1 + γo y + 20.4*γm – 20.4*γo - γo y = p2(p2 – p1) / γo = (γm*20.4 / γo) – 20.4= (844/56 -1) * 20.4 = 24.0 ftElevation difference is zero!Find out the velocity head term??? 28
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