Chapter 6 Losses due to Fluid Friction 1
Objectives ä To measure the pressure drop in the straight section of smooth, rough, and packed pipes as a function of flow rate. ä To correlate this in terms of the friction factor and Reynolds number. ä To compare results with available theories and correlations. ä To determine the influence of pipe fittings on pressure drop ä To show the relation between flow area, pressure drop and loss as a function of flow rate for Venturi meter and Orifice meter.
Losses due to Friction Mechanical energy equation between locations 1 and in the absence of shaft work: P1 P V1 V ( ) ( ) ( z1 z) g g g g For flow in a horizontal pipe and no diameter change (V1=V), then : F P 1 P Hagen-Poiseuille Law-E5.10 F Q x 18 4 D o OR Or F/g= h Loss = 18μQL/(πρgD 4 ) and F/g F g hloss Because of (5.4) : F 4x D (6.13) (6.14) h L w where Thus, the shear stress at the wall is responsible for the losses due to friction F u =Friction head/unit mass P1 P g F in J/kg 3 Q m
Losses due to Friction Total head loss, h L (=F/g), is regarded as the sum of major losses and minor losses h L major, due to frictional effects in fully developed flow in constant area tubes, h L minor, resulting from entrance, fitting, area changes, and so on. 4
Losses due to Friction/The Friction Factor In order to determine an expression for the losses due to friction we must resort to experimentation. F L V D By introducing the friction factor, f: where L=length of the pipe, D=diameter of the pipe, V=velocity, F f L V D. (6.15) F P 1 P where f ( L / F D)( V / ) or f P( D / L) V / f is called Darcy friction factor 5
Friction Factor now later ( P gz V ) P work P The Darcy friction factor f is defined as f P( D / L) V / We know the wall shear stress f V w / and L 0 τ w P D 4 (E6-1) L, major P L The friction factor is the ratio between wall shear stress and flow inertial force. L P P L, min or 6
( P gz Major Loss and Friction Factor V ) P work P L 0 P With the introduction of friction factor, we can calculate major loss by L P L, major P L, min or P L, Major f L D V (E6-) Friction factor Pipe geometry factor Dynamic pressure Therefore, our job now is find the friction factor for various flows. 7
Friction Factor - Laminar Pipe Flow For a pipe with a length of L, the pressure gradient is constant, the pressure drop based on Hagen-Poiseuille Law, P 3VL / D Divided both sides by the dynamic pressure V / and L/D f P( D / L) 3VL / D D 64 V / V / L VD 64 Re We have f 64 Re 8
Major Loss and Friction Factor The mechanical energy equation can be written in terms of heads : P1 P V1 V ( ) ( ) ( z1 z) g g g g W shaft mg h Loss (6.17) L V f L, D g h Major Knowledge of the friction factor allows us to estimate the loss term in the energy equation. This head loss can also be calculated using H-B equation 9
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Example 1 losses in 11
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Example 13
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Case : Turbulent Flow When fluid flow at higher flow rates, the streamlines are not steady, not straight and the flow is not laminar. Generally, the flow field will vary in both space and time with fluctuations that comprise "turbulence When the flow is turbulent the velocity and pressure fluctuate very rapidly. The velocity components at a point in a turbulent flow field fluctuate about a mean value. Time-averaged velocity profile can be expressed in terms of the power law equation, n =7 is a good approximation. u V max r 1 R 1/ n 15
Friction Factor-Turbulent Pipe Flow For a laminar flow, the friction factor can be analytically derived. It is impossible to do so for a turbulent flow so that we can only obtain the friction factor from empirical results. For turbulent flow, it is impossible to analytically derive the friction factor f, which can ONLY be obtained from experimental data. In addition, most pipes, except glass tubing, have rough surfaces. The pipe surface roughness is quantified by a dimensionless number, relative pipe roughness (ε / D ), where ε is pipe roughness and D is pipe diameter. For laminar pipe flow, the flow is dominated by viscous effects hence surface roughness is not a consideration. However, for turbulent flow, the surface roughness may protrude beyond the laminar sublayer and affect the flow to a certain degree. Therefore, the friction factor f can be generally written as a function of Reynolds number and pipe relative roughness There are several theoretical models available for the prediction of shear stresses in turbulent flow. 16
Surface Roughness Additional dimensionless group /D need to be characterize Thus more than one curve on friction factor- Reynolds number plot Fanning diagram or Moody diagram Depending on the laminar region. If, at the lowest Reynolds numbers, the laminar portion corresponds to f =16/Re Fanning Chart (or f = 64/Re Moody chart) 17
Friction Factor and Pipe Roughness Most pipes, except glass tubing, have rough surfaces - Pipe surface roughness, - Relative pipe roughness, / D The surface roughness may affect the friction factor. Generally, we have f (Re, ) D 18
Pipe Surface Roughness 19
Friction Factor of Turbulent Flow If the surface protrusions are within the viscous layer, the pipe is hydraulically smooth; f 0.316 Re 1/ 4 If the surface protrusions extend into the buffer layer, f is a function of both Re and /D; f (Re, D) For large protrusions into the turbulent core, f is only a function of /D. f ( D) 0
Friction Factor for Smooth, Transition, f P L Smooth pipe, Re>3000 and Rough Turbulent flow D U Rough pipe, [ (D/ε)/(Re ƒ) <0.01] Transition function for both smooth and rough pipe Or 1 f 1 f 1 4.0 * logre* f 0.4 f 0.079Re 0.5 f 4.0 * log D.8 4.0 * log D D/.8 4.0 * log 4.67 Re f 1 1
Fanning Diagram 1 f 4.0 * log D D/.8 4.0 * log 4.67 Re f 1 1 f 4.0 * log D.8 f =16/Re
Friction Factor The Moody Chart 3
Example: Comparison of Laminar or Turbulent pressure Drop Air under standard conditions flows through a 4.0-mmdiameter drawn tubing with an average velocity of V = 50 m/s. For such conditions the flow would normally be turbulent. However, if precautions are taken to eliminate disturbances to the flow (the entrance to the tube is very smooth, the air is dust free, the tube does not vibrate, etc.), it may be possible to maintain laminar flow. (a) Determine the pressure drop in a 0.1-m section of the tube if the flow is laminar. (b) Repeat the calculations if the flow is turbulent. Straight and horizontal pipe and same diameters give same velocity: Z 1 =Z =0 V 1 =V and thus p / g h Loss 4
Solution 1/ Under standard temperature and pressure conditions =1.3kg/m 3, μ=1.7910-5 Ns/m The Reynolds number R e VD /... 13,700 Turbulent flow If the flow were laminar and using Darcy friction f=64/re=` =0.00467 p f 1 V... 0. 179kPa D If the flow were laminar and using Fanning friction 1 p 4 f V... 0. 179kPa D F=16/Re=` =0.001167 5
Solution / If the flow were turbulent From Moody chart f=φ(re, smooth pipe) =0.08 p f 1 V... 1. 076kPa D From Fanning chart f=φ(re, smooth pipe) =0.007 1 p 4 f V... 1. 076kPa D 6
Example Straight and horizontal pipe and same diameters give same velocity: Z 1 =Z =0 V 1 =V and thus p gh Loss pump pressure 7
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Example 0.04 m 30
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f 0.079Re 0.5 for fanning 33
Example: Determine Head Loss Crude oil at 140 F with γ=53.7 lb/ft 3 and μ= 810-5 lb s/ft (about four times the viscosity of water) is pumped across Alaska through the Alaska pipeline, a 799-mile-along, 4-ft-diameter steel pipe, at a maximum rate of Q =.4 million barrel/day = 117ft 3 /s, or V=Q/A=9.31 ft/s. Determine the horsepower needed for the pumps that drive this large system. 34
Solution 1/ The energy equation between points (1) and () p 1 V1 p V z1 hp z hl g g h P is the head provided to the oil by the pump. Assume that z 1 =z, p 1 =p =V 1 =V =0 (large, open tank) Minor losses are negligible because of the large lengthto-diameter ratio of the relatively straight, uninterrupted pipe. V hl hp f... 17700 ft D g f=0.014 from chart ε/d=(0.00015ft)/(4ft), Re=.. 35
Solution / The actual power supplied to the fluid power=q P =Qρgh unit of power (SI): Watt =N.m/s=J/s Power 1hp gqh P... 0000hp 550 ftlb / s 36
Minor Losses 1/5 Most pipe systems consist of considerably more than straight pipes. These additional components (valves, bends, tees, and the like) add to the overall head loss of the system. Such losses are termed MINOR LOSS. The flow pattern through a valve 37
90 0 Bend 90 0 Elbow 45 0 Elbow Tees-line flow Tees-branch flow 38
Valves Globe valve Angle valve Gate valve Ball valve 39
Minor Losses /5 The theoretical analysis to predict the details of flow pattern (through these additional components) is not, as yet, possible. The head loss information for essentially all components is given in dimensionless form and based on experimental data. The most common method used to determine these head losses or pressure drops is to specify the loss coefficient, K L 40
Minor Losses 3/5 K L h V L min or / g p 1 V p Minor losses are sometimes given in terms of an equivalent length eq K L h 1 L V min or eq K K L L D f V g The actual value of K L is strongly dependent on the geometry of the component considered. It may also dependent on the fluid properties. That is f eq D V g K (geometry,re) L 41
Minor Losses 4/5 For many practical applications the Reynolds number is large enough so that the flow through the component is dominated by inertial effects, with viscous effects being of secondary importance. In a flow that is dominated by inertia effects rather than viscous effects, it is usually found that pressure drops and head losses correlate directly with the dynamic pressure. This is the reason why the friction factor for very large Reynolds number, fully developed pipe flow is independent of the Reynolds number. 4
Minor Losses 5/5 This is true for flow through pipe components. Thus, in most cases of practical interest the loss coefficients for components are a function of geometry only, K (geometry ) L 43
Minor Losses Coefficient For Example Entrance flow Entrance flow condition and loss coefficient (a) re-entrant, K L = 0.8 (b) sharp-edged, K L = 0.5 (c) slightly rounded, K L = 0. (d) well-rounded, K L = 0.04 44
Summary of Minor Losses Major losses: Associated with the friction in the straight portions of the pipes Minor losses: Due to additional components (pipe fittings, valves, bends, tees etc.) and to changes in flow area (contractions or expansions) Method 1: We try to express the head loss due to minor losses in terms of a loss coefficient, K L : F g minor losses K L V g Values of K L can be found in the literature (for example, see Table next ppt, for losses due to pipe components. 45
Minor Losses Component Elbows K L Regular 90, flanged 0.3 Regular 90, threaded 1.5 Long radius 90, flanged 0. Long radius 90, threaded 0.7 Long radius 45, flanged 0. Regular 45, threaded 0.4 180 return bends 180 return bend, threaded 0. 180 return bend, flanged 1.5 Tees Line flow, flanged 0. Line flow, threaded 0.9 Branch flow, flanged 1.0 Branch flow, threaded.0 Component K L Union, threaded 0.8 Valves Globe, fully open 10 Angle, fully open Gate, fully open 0.15 Gate, ¼ closed 0.6 Gate, ½ closed.1 Gate, ¾ closed 17 Ball valve, fully open 0.05 Ball valve, 1/3 closed 5.5 Ball valve, /3 closed 10 46
47 Minor Losses The mechanical energy equation can be written: g V K g V D L f mg W z z g V g V g P g P L shaft ) ( ) ( ) ( 1 1 1 (6.)
48 Minor Losses Method : Using the concept of equivalent length, which is the equivalent length of pipe which would have the same friction effect as the fitting. Values of equivalent length (L/D) can be found in the literature g V D L f mg W z z g V g V g P g P equiv shaft ) ( ) ( ) (. 1 1 1 (6.3)
Example: Determine Pressure Drop Water at 60 F flows from the basement to the second floor through the 0.75-in. diameter copper pipe (a drawn tubing) at a rate of Q = 1.0 gal/min (= 0.067 ft 3 /s) and exits through a faucet of diameter 0.50 in. as shown in Figure Determine the pressure at point (1) if: (a) all losses are neglected, (b) the only losses included are major losses, or (c) all losses are included. 49
V 1 Solution 1/4 Q A 1.34 10... 8.70ft / s 5 lb s / ft The energy equation 1.94slug / ft Re VD / 45000 3 The flow is turbulent p1 g V1 p V z1 z g g g p h L z 1 (V V ) h 1 1 L z 1 V 0,z Q / A 0ft,p 0(free... 19.6ft / s jet) Head loss is different for each of the three cases. 50
Solution /4 (a) If all losses are neglected (h L =0) 1 p1 z (V V1 )... 1547lb /ft 10.7psi 0.000005 (b) If the only losses included are the major losses, the head loss is h V1 f L D g / D 810 5 chart Re 45000 f = 0.015 1 ( 60 ft) V1 p1 z ( V V1 ) 4f... 306lb / ft 1. 3psi D 51
Solution 3/4 (c) If major and minor losses are included p 1 z 1 ( V V 1 ) f D V1 g K L V p 1 1.3psi 1.3psi K L V (1.94slugs / ft 3 ) (8.70 ft / s) [10 4(1.5) ] p 1 1.3psi 9.17psi 30.5psi 5
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Noncircular Ducts The empirical correlations for pipe flow may be used for computations involving noncircular ducts, provided their cross sections are not too exaggerated. The correlation for turbulent pipe flow are extended for use with noncircular geometries by introducing the hydraulic diameter, D h, defined as D h 4A P Where A is crosssectional area, and P is wetted perimeter. 55
Noncircular Ducts 3/4 The friction factor can be written as f=c/re h, where the constant C depends on the particular shape of the duct, and Re h is the Reynolds number based on the hydraulic diameter. Re V Dh The hydraulic diameter is also used in the definition of the friction factor, h, and the relative L f ( / Dh)(V / g) roughness /D h. 56
Define hydraulic diameter, D h : Non circular conduits D h 4 x (cross sectional area) wetted perimeter Then use Reynolds number based on hydraulic diameter, D h : Re V Dh 57
58 Non circular conduits Pipe of circular cross-section D D 4 D 4 D h Annulus (inside diameter D 1, outside D ) 1 1 1 h D D D D 4 D 4 D 4 D Rectangular conduit (area ab) b a ab b a (ab) 4 D h
Example Solved before 59
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Multiple Pipe Systems Pipes in series Q 1 Q Q 3 F L AB F L 1 F L F L 3 61
Multiple Pipe Systems Parallel pipes Q F L Q 1 F Q F 1 L L3 Q 3 6