) 2 ] 2. Covered in this lecture: 1. Minor losses 2. Equivalent Lengths 3. Noncircular pipes 4. Multiple pipe systems - series Minor Losses

Transcription

1 Covered in this lecture: 1. Minor losses 2. Equivalent Lengths 3. Noncircular pipes 4. Multiple pipe systems - series Minor Losses S Losses due to entrances, expansions, bends, valves, fittings, etc. S A int necessarily minor, can be significant S Caused by turbulence, flow separation S Denoted by h m or h lm = K V2 where K (or K e, K c, K b, K E, etc.) is determined experimentally for each fitting. Sudden Expansion in which h e = K V2 2 1 = [1 - (D 1) D 2 ] V2 1 2 K = [1 - ( D 1 D 2 ) 2 ] 2 1

2 Sudden Contraction S The process of converting pressure to velocity (shown 1 to 0) is very efficient, the vena contracta is point 0 S The process of converting velocity back to pressure to is not as efficient (shown 0 to 2) V V2 S Head loss is expressed in terms of V 2 h c = K c V 2 2 S Values are available in Tables S Entrance loss K e from reservoir varies (0.78 reentrant, 0.5 for square-edged, to 0.04 for well rounded) S Exit loss to a reservoir is always K E = 1.0 2

3 Equivalent Length The head loss through a component is expressed in terms of a length of pipe L e needed for the same head loss, so h m = f L e V 2 D Now if K = S? i s in the section, and we know f and D, we can write L e = KD f Example: If all the minor loss add up to K = 20 in 5000 ft of line of 12 inch diameter line, and if f = 0.020, the equivalent length of the minor losses L e = KD 20 (12 in 1 ft/12 in ) = = 1000 ft f 0.02 So 5000 ft ft = 6000 ft of pipe without the minor losses would have the same amount of head loss as the pipe section with the minor losses. Noncircular Pipes The D used in our equations is actually the hydraulic diameter, D h = 4A P where A is the area of the flow, and P is the wetted perimeter of the pipe section. For a circular pipe D h = 4A P = 4p D2 4 2pD/2 = D 3

4 Example: What are the Reynolds number,relative roughness, and head loss for 1000 ft for a 8 in by 1 ft rectangular cast iron pipe (e = ft carrying water (? = 1 x 10-5 ft 2 /s) at 9 ft/s? D h = 4A P = 4(8/12 ft) 1 ft = 0.80 ft 2 (8/12 ft) + 2(1 ft) Using D h as D Re = VD? = (9 ft/s)(0.80 ft) ft 2 = 720, 000 /s Also use D h for D in the calculation of relative roughness Á ft = = D 0.80 ft From Moody, f = 0.022, so hf = f L D V 2 h f = ft 0.80 ft (9 ft/s) 2 = 34.6 ft ft/s2 4

5 Example: If steel pipe with D=15 cm, Find i) the discharge through the pipe in the flowing figure for H = 10 m, and ii) determine the head loss H for Q = 60 l/s. 1 H =? Globe Valve Water at 20 deg C Square-edged entrance 12 m Standard Elbows 2 Q =? 30 m 60 m Part i) Writing the energy equation between points 1 and 2, using 2 as the datum (z = 0) and including all losses V 2 2 L = m V 2 2 H = V f 102 m 0.15 m V V V2 2 velocity head at 1 or pressure head at 1 elevation head at 2 pressure head at 2 entrance loss friction loss elbow loss valve loss exit loss Guess f = 0.022, so H = V2 2 ( f ) 10 = V2 2 ( f ) V = 2.63 m/s 5

6 So for? = 1.01 µm 2 /s, e/d = , then Moody gives (2.63 m/s )(0.15m) Re = 1.01 µm 2 = 391, 000 /s For which f = Repeating the procedure gives V 2 = 2.60 m/s, Re = 380,000 and f = The discharge is Q = V 2 A 2 = (2.60 m/s)( p 4 )(0.15 m)2 = m 3 /s = 45.9 l/s Part ii) Q is known, solution is straightforward V 2 = Q A = 0.06 m3 /s = 3.40 m/s Re = 505, 000 f = (p /4)(0.15 m) 2 and H = V2 2 ( f ) (3.4 m/s)2 H = ( ) = m m/s2 Using equivalent lengths Using the technique of equivalent lengths, the value of f is approximated, say f = The sum of minor losses is 13.3, and the kinetic energy at point 2 is considered a minor loss, so L e = KD = = 90.7 m f So the total pipe length is 30 m + 12 m + 60 m m = m. Solving for Part i) of the problem V m = f L + L e D = f m (V 2 m/s) m m/s 2 If f = 0.022, V 2 = 2.63 m/s, Re = 391,000, so from Moody new f =

7 If f = 0.023, V 2 = 2.58, so Q = 45.6 l/s. S Normally it is not necessary to use the new f to improve L e, unless your first guess if f is very poor. 7

8 Multiple pipe systems When fluid flows from one pipe to another of a different diameter, it is considered as connected in series. A 1 2 H B K e Not to scale In general two type of problems S H is desired for a Q, or S Q is desired for a given H Writing the energy equation from A to B including all minor losses pressure head at A pressure head at B V 2 1 H = K e velocity head at A elevation head at B velocity head at B entrance loss + f 1 L 1 D 1 V friction loss Pipe 1 V K Exp Sudden Expansion loss Pipe 1-2 L + f 2 V D 2 friction loss Pipe 2 V K Exit Expansion Exit loss Pipe 2 8

9 Using the continuity equation V 1 D 1 = V 2 D 2 to eliminate V 2 from the above equation, along with our diameter calculation for K exp = [1 - (D 1 /D 2 ) 2 ] 2 and setting K exit = 1, we get H = V2 1 [K e + f 1 L 1 D 1 + [1 - ( D 1 D 2 ) 2 ] 2 + f 2 L 2 D 2 ( D 1 D 2 ) 4 + ( D 1 D 2 ) 4 ] For pipes of known lengths and sizes, we can reduce this to H = V2 1 (C 1 + C 2 f 1 + C 3 f 2 ) Where C 1, C 2, C 3 are known. S With Q given, Re is computed and f read from Moody S With H given, V 1, f 1, f 2 are unknown. Assume values, solve for V, compute estimate of Re, solve for f s, etc. Example: Suppose for our Figure, Ke = 0.5 L 1 = 1000 ft D 1 = 2 ft e 1 = ft L 2 = 800 ft D 2 = 3 ft e 2 = ft? = ft 2 /s, H = 20 ft We get H = V2 1 [0.5 + f 1 ( ) + (1 - (2 3 )2 ) 2 + f 2 ( )(2 3 )4 + ( 2 3 )4 ] 20 = V2 1 ( f f 2 ) 9

10 S Now for e 1 /D 1 = , and e 2 /D 2 = , From the Flat part (complete turbulent) of Moody f 1 = f 2 = Solving for V 1, V 1 = 9.49 ft/s, and so V 2 = 4.21 ft/s, for which 9.49 ft/s 2 ft Re 1 = = 1, 898, ft 4.21 ft/s 3 ft Re 2 = = 1, 263, ft S With these values, from Moody f 1 = f 2 = So, V 1 = 9.46 ft/s, so Using Equivalent Pipes Q = 9.46 ft/s p 4 (2 ft)2 = 29.8 cfs S Equivalent lengths can be used since two pipe systems are said to be equivalent when the same head loss produces the same discharge in both systems Writing the head loss for the first pipe h f 1 = f 1 L 1 D 1 Q 2 1 (D 2 1 p /4)2 = f 1 L 1 D 5 1 8Q 2 1 p 2 g 10

11 Writing the head loss for the second pipe h f 2 = f 2 L 8Q D 5 p 2 2 g S For equvilent pipe, flow is equal, an the head loss is the same then: For the two to be equivalent pipes h f 1 = h f 2 Q 1 = Q 2 Setting h f1 = h f2, and simplifying f 1 L 1 D 5 = f 2 L 2 D Solving for L 2 gives L 2 = L 1 f 1 f 2 ( D 2 D 1 ) 5 S This is the length of a second pipe to be equivalent to that of the first pipe Example: Replace 300 m of 25 cm pipe with an equivalent length of 15 cm pipe S Approximate values of f 1 and f 2, assuming complete turbulence If for this example f 1 = and f 2 = 0.018, then L 2 = 300 m cm ( cm )5 = 25.9 m S For the assumed conditions, 25.9 m of 15 cm pipe has the same head loss as 300 m of 25 cm pipe. 11

12 Example: Solve the previous reservoir example using equivalent pipes S First express minor losses in terms of equivalent lengths, using f 1 and f 2 as fully turbulent For pipe 1 For pipe 2 entrance expansion K 1 = [1 - (2/3) 2 ] 2 = L e1 = K 1 D 1 f 1 = = 65 ft K 2 = 1 L e2 = K 2 D 2 = 1 3 = 200 ft f So we now have 1000 ft + 65 ft = 1065 ft of 2 ft diam. pipe 800 ft ft = 1000 ft of 3 ft diam. pipe Expressing the 3 ft diam. pipe in terms of 2 ft diam. pipe L e = L 2 = L 1 f 1 f 2 ( D 2 D 1 ) 5 Le = 1000 ft ft ( ft )5 = 79 ft By adding this to the 2 ft pipe, the problem is now equivalent to 1065 ft + 79 ft = 1144 ft of 2 ft diam. pipe So if e = 0.05 ft, and H = 20 ft, With f = 0.025, V = 9.5 ft/s, Re = 1,900,000. Converges, so Q =9.5p = 29.9 ft 3 /s 12

13 H = f 1144 ft 2 ft V 2 13