# Chapter 10. Open- Channel Flow

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1 Updated: Sept Created by Dr. İsmail HALTAŞ Created: Sept Chapter 10 Open- Channel Flow based on Fundamentals of Fluid Mechanics 6th EdiAon By Munson 2009* *some of the Figures and Tables in these lecture slides are adopted from this reference for educaaonal purposes only

2 Learning ObjecAves ASer compleang this chapter, you should be able to: discuss the general characterisacs of open- channel flow. use a specific energy diagram. apply appropriate equaaons to analyze open- channel flow with uniform depth. calculate key properaes of a hydraulic jump. determine flow rates based on open- channel flow- measuring devices. 2

3 IntroducAon Natural and Man- made Open Channel Photos 3

4 IntroducAon Open- Channel flow is a flow which has a free surface, i.e. a surface on which the pressure is equal to local atmospheric pressure. The main driving force is the weight of the fluid, gravity forces the fluid flow to downhill. P atm P atm Free Surface 4

5 IntroducAon The difference between open- channel flow and the pipe flow is in the fundamental mechanism that drives the flow. Open Channel Flow Only Gravity drives Pipe Flow Pressure (and in some cases also Gravity) drives If the pipe is not full, it is not possible to maintain the pressure difference, p 1 p 2. So it is actually open channel flow! 5

6 ClassificaAon of Open- Channel Flow The flow is called steady if the depth at a given locaaon does not change with Ame. If the depth at a given locaaon changes with Ame, flow is called unsteady. dy dt = 0 dy Steady dt 0 Unsteady An open- channel flow is classified as uniform flow (UF) if the depth of flow does not vary along the channel. Conversely it is nonuniform flow or varied flow if the depth varies with distance. dy dx = 0 dy Uniform dx 0 Varied 6

7 ClassificaAon of Nonuniform Flow Nonuniform flows are further classified as rapidly varying flow (RVF) if the flow depth changes considerably over a relaavely short distance. Gradually varying flows (GVF) are those in which the flow depth changes slowly with distance along the channel. dy dx 1 Rapidly Varying Flow dy <<1 Gradually Varying Flow dx 7

8 ClassificaAon of Nonuniform Flow 8

9 Laminar or Turbulent Flow Open- Channel flow may be laminar, transiaonal, or turbulent. Which type of flow occurs depends on the Reynolds number, Re. Re < 500 Laminar Re > 1250 Turbulent Re = ρvr H otherwise TransiAonal µ V : (cross- secaonal) average velocity of the fluid R H : hydraulic radius of the channel cross- secaon ρ : fluid density μ : dynamic viscosity of the fluid Reynolds number indicates the effect of viscosity on the flow. Viscous effect is not as important in Open- Channel flow compared to Pipe flow! 9

10 Hydraulic Radius, R H R H = A P A : Flow Area P : Weoed Perimeter D H : Hydraulic Diameter D H = 4R H Examples of R h for common geometries shown in Figure at the les. 10

11 ProperAes of Water Temperature Viscosity, μ Temperature Density, ρ [ C] x 10-3 [Pa s] [ C] [kg/m 3 ] [Pa s] = [N s/m 2 ] = [kg/(m s)] 11

12 Wave Speed 12

13 Wave Speed ConAnuity Eq. ConAnuity EquaAon: Q in = Q out cyb = ( c +δv )( y +δy )b ( c = y +δy )δv Small amplitude waves with δy << y δy c = y δv δy 13

14 Wave Speed Momentum Eq. Momentum EquaAon: F = ( mv ) out ( mv ) in 1 2 γ y2 b 1 2 γ ( y +δy ) 2 b = ρbcy"# ( c +δv ) c\$ % Where; Mass flow rate m in = m out = ρbcy HydrostaAc Forces F 1 = 1 2 γ y2 b F 2 = 1 2 γ ( y +δy ) 2 b 14

15 Wave Speed Momentum Eq. Momentum EquaAon: 1 2 γ y2 b 1 2 γ ( y +δy ) 2 b = ρbcy"# ( c +δv ) c\$ % Small amplitude waves with δy << y δv δy = g c 15

16 Wave Speed ConAnuity EquaAon Momentum EquaAon δv c = y δv + δy = g δy c c = gy 16

17 Froude Number Open- Channel flow may be subcriacal, criacal, or supercriacal. depending on the Froude number, Fr. Fr = V gy Fr < 1 subcriacal Fr = 1 criacal Fr > 1 supercriacal V V V V < gy V = gy V > gy 17

18 Froude Number Open- Channel flow may be subcriacal, criacal, or supercriacal. depending on the Froude number, Fr. Fr > 1 supercriacal V > gy Fr = 1 criacal V = gy Fr < 1 subcriacal V < gy 18

19 Example Problem 1 If the velocity of flow is given as a funcaon of flow depth as follows. For what range of water depth will a surface wave on the river be able to travel upstream? V = 3.36y 2/3 (SI Unit System) 19

20 SoluAon 1 If the velocity of flow is given as a funcaon of flow depth as follows. For what range of water depth will a surface wave on the river be able to travel upstream? V = 3.36y 2/3 (SI Unit System) In order for the wave to travel upstream flow should be subcriacal... V < gy 3.36y 2 3 < y 1 2 g y 1 6 < 9.81 y < m

21 Energy ConsideraAons Average booom slope for Mississippi River is S 0 =

22 Energy ConsideraAons Energy equaaon: p 1 γ + V 2 1 2g + z = p 2 1 γ + V 2 2 2g + z + h 2 L z 1 z 2 = S 0 pressure is hydrostaac so; p 1 γ = y p 2 1 γ = y 2 h L = S f fricaon slope 22

23 Energy ConsideraAons Energy equaaon: p 1 γ + V 2 1 2g + z = p 2 1 γ + V 2 2 2g + z + h 2 L y 1 + V 2 1 2g = y 2 + V 2 2 2g + ( S f S 0 ) Specific Energy, E: E = y + V 2 2g E 1 = E 2 + ( S f S 0 ) 23

24 Specific Energy Specific Energy, E: Consider a prismaac channel whose cross- secaonal shape is a rectangle of E = y + V 2 width b, the specific energy can be 2g wrioen in terms of flowrate per unit width, q (where Q = const.) q = Q b = Vyb b = Vy V = q E = y + q2 y 2gy 2 " % y 3 Ey 2 + \$ q2 ' = 0 # 2g & SoluAon of the equaaon; y sub, y sup, and y neg 24

25 Specific Energy Specific Energy, E: E = y + V 2 2g y neg : no physical meaning y sub and y sup : alternate depths # & lim E y ( y) = lim y % y + q2 ( = y \$ 2gy 2 ' Asymptotically approaches to y = E lim E y 0 ( y) = 25

26 Specific Energy E min is reached at criacal depth y c E(y) = y + q2 2gy 2 To find the minimum of E(y) de q2 =1 dy gy = " % y c = \$ q2 ' E g min = 3y c # & 2 V c = q = gy c (Fr = 1) y c So criacal condiaon occurs at E min 26

27 Specific Energy E min is reached at criacal depth y c, So criacal condiaon occurs at E min subcriacal flow supercriacal flow 27

28 Example Problem 2 Water flows up a 15 cm tall ramp in a constant width rectangular channel at a rate q = 0.6 m 2 /s. The upstream depth is 70 cm and viscous effects are negligible. Determine the elevaaon of the water surface downstream of the ramp. q = 0.6 m 2 /s y 2 y 1 = 70 cm z 2 = 15 cm (2) Ramp z 1 = 0 (1) profile view 28

29 SoluAon 2 Water flows up a 15 cm tall ramp in a constant width rectangular channel at a rate q = 0.6 m 2 /s. The upstream depth is 70 cm and viscous effects are negligible. Determine the elevaaon of the water surface downstream of the ramp. Viscous effects are neglected, h L = 0. Then the energy equaaon; y 1 + V 2 1 2g + z 1 = y 2 + V 2 2 2g + z q = 0.6 m 2 /s 2 y 1 = 0.7 m z 1 = 0 m 0.74 = y 2 + q2 V 1 = q / y 1 = 0.86 m/s 2gy z 2 = 0.15 m y 2 = 0.52 y 2 = y 2 =

30 SoluAon 2 cont. y 2 = 0.52 y 2 = y 1 = 0.7 m y 2 = y 2 = 0.52 m 0.15 m Which one to select?? Fr 1 = V 1 = 0.33 <1 y c = 0.33 m gy 1 q = 0.6 m subcriacal 2 /s flow at (1) E min = 0.5 m E (2) (1) 1 = 0.74 m E (m) 30 y (m)

31 SoluAon 2 cont. E 1 + z 1 = E 2 + z 2 y 1 = 0.7 m E 1 = E y 2 = 0.52 m E 1 = 0.74m E 2 = 0.59m 0.15 m 1 3! \$ y c = # q2 & = 0.33m " g % y c = 0.33 m E min = 3y c 2 = 0.5m q = 0.6 m 2 /s subcriacal flow at (2) therefore; E min = 0.5 m E (2) (1) 1 = 0.74 m y 2 = 0.52 m E (m) Water Elev = z 2 + y 2 = 0.67 m 31 y (m)

32 Example Problem 2b Water flows up a 15 cm tall ramp in a constant width rectangular channel at a rate q = 0.6 m 2 /s. The upstream depth is 70 cm and viscous effects are negligible. Determine the elevaaon of the water surface downstream of the ramp If there is a bump on the ramp with a crest elevaaon at 24 cm. q = 0.6 m 2 /s (p) z p = 24 cm y 2 y 1 = 70 cm Bump z 2 = 15 cm (2) Ramp z 1 = 0 (1) profile view 32

33 SoluAon 2b Water flows up a 15 cm tall ramp in a constant width rectangular channel at a rate q = 0.6 m 2 /s. The upstream depth is 70 cm and viscous effects are negligible. Determine the elevaaon of the water surface downstream of the ramp If there is a bump on the ramp with a crest elevaaon at 24 cm. E 1 + z 1 = E p + z p z p = 0.24 m E E 1 = E p = 0.74 m E min = 0.5 m E p = 0.5m E p = 0.5 = E min = 0.5 criacal flow at (p) 33

34 SoluAon 2b cont. criacal flow at (p) y 1 = 0.7 m E p + z p = E 2 + z 2 y 2 = 0.52 m E 2 = E min E 2 = 0.59m y c = 0.33 m supercriacal flow at (2) therefore; q = 0.6 m 2 /s E y 2 = m min = 0.5 m E (2) (1) 1 = 0.74 m (p) E (m) Water Elev = z 2 + y 2 = m 34 y (m)

35 Example Problem 2c Water flows up a 15 cm tall ramp in a constant width rectangular channel at a rate q = 0.8 m 2 /s. The upstream depth is 25 cm and viscous effects are negligible. Determine the elevaaon of the water surface downstream of the ramp. y 2 q = 0.8 m 2 /s z 2 = 15 cm y 1 = 25 cm (2) Ramp z 1 = 0 (1) profile view 35

36 SoluAon 2c Water flows up a 15 cm tall ramp in a constant width rectangular channel at a rate q = 0.8 m 2 /s. The upstream depth is 25 cm and viscous effects are negligible. Determine the elevaaon of the water surface downstream of the ramp. y 1 + V 2 1 2g + z 1 = y 2 + V 2 2 2g + z q = 0.8 m 2 /s 2 y 1 = 0.25 m z 1 = 0 m 0.77 = y 2 + q2 V 1 = q / y 1 = 3.2 m/s 2gy z 2 = 0.15 m y 2 = 0.48 y 2 = 0.35 y 2 =

37 SoluAon 2c cont. y 2 = 0.48 y 2 = 0.35 q = 0.8 m 2 /s y 2 = Which one to select?? Fr 1 = V 1 = 2.04 >1 y c = 0.4 m gy m y 2 = 0.35 m supercriacal y flow at (1) 1 = 0.25 m E min = 0.6 m E (2) (1) 1 = 0.77 m supercriacal flow at (2) E (m) y 2 = 0.35 m 37 y (m)

38 Channel TransiAons and Chocking Cases Cases I- A: Upward Step SubcriAcal Flow (constant width) E 1 = E 2 + Δz q y 2 y 1 Δz (2) (1) profile view y 1 y 2 Δz y c 38 E 2 E 1 y (m) E (m)

39 Channel TransiAons and Chocking Cases Cases I- B: Upward Step SupercriAcal Flow (constant width) E 1 = E 2 + Δz y 2 y q 1 Δz (2) (1) profile view Δz y c y 2 y 1 39 E 2 E 1 y (m) E (m)

40 Channel TransiAons and Chocking Cases Cases II- A: Downward Step SubcriAcal Flow (constant width) E 1 + Δz = E 2 y q 1 y 2 (1) Δz profile view (2) y 2 y 1 Δz y c 40 E 1 E 2 y (m) E (m)

41 Channel TransiAons and Chocking Cases Cases II- B: Downward Step SupercriAcal Flow (constant width) E 1 + Δz = E 2 q y 1 y 2 (1) Δz profile view (2) Δz y c y 1 y 2 41 E 1 E 2 y (m) E (m)

42 Channel TransiAons and Chocking Cases Cases III: Channel Expansion (constant bed elevaaon) q 1 = Q/b 1 q 1 q 2 b q 2 = Q/b 2 1 b 2 q 2 (1) top view y (2) 2 q 1 y 1 subcriacal y 2 y c 1 y 1 y c 2 y c 2 y c 1 y 1 supercriacal y 2 y 1 y 2 E 1 = E 2 (1) profile view (2) E (m) 42 y (m)

43 Channel TransiAons and Chocking Cases Cases IV: Channel ContracAon (constant bed elevaaon) q 1 = Q/b 1 b 1 q 1 q 2 b 2 q 2 = Q/b 2 q 1 (2) (1) top view y 1 q 2 y 2 subcriacal y 1 y c y c y y c 1 y c 2 y 2 supercriacal y 1 y y 1 2 E 1 = E 2 (1) profile view (2) E (m) 43 y (m)

44 Channel TransiAons and Chocking Cases Cases V: Choking: Upward Step (constant width) E 1 = E min + Δz max y 1 q y c y 2 =y 1 Δz max (1) profile view (2) y 1 Δz max y c 44 E 1 = E 2 y (m) E (m)

45 Channel TransiAons and Chocking Cases Cases VI: Choking: ContracAon q 1 = Q/b 1 q 2 = Q/b 2 E b q 1 1 q 2 1 = E 2 = E 3 = E min b min b 3 = b 1 q 1 (2) (1) top view (3) q 2 subcriacal y 1 y c y 1 y c y 3 (1) (2) (3) E profile view 1 = E 2 = E 3 = E min E (m) 45 y (m)

46 Gradually Varied Flow dy Gradually Varying Flow dx <<1 H = V 2 Total energy at secaon (1) 2g + y + z Energy equaaon between secaon (1) & (2) H 1 = H 2 + h L dh Slope of energy line dx = dh L dx = S f dz Slope of channel booom dx = S 0 46

47 Gradually Varied Flow dh dx = d! V 2 dx 2g + y + z \$ # & = V dv " % g dx + dy dx + S 0 S f = V dv g dx + dy dx + S 0 S f S 0 = V dv g dx + dy dx For a given flowrate per unit width q, in a rectangular channel of constant width b, V = q/y dv dx = q dy y 2 dx = V dy y dx 47

48 Gradually Varied Flow S f S 0 = V dv g dx + dy dv + dx dx = q dy y 2 dx = V dy y dx S f S 0 = V " g V dy % \$ '+ dy # y dx & dx S f S 0 = dy " dx 1 V 2 % \$ ' = dy ( 1 ) Fr2 # gy & dx dy dx = S Rate of change of fluid depth dy/dx, depends f S 0 on the local channel booom slope S 0, slope of 1 Fr 2 the energy line S f, and the Froude number Fr. 48

49 Gradually Varied Flow dy dx = S f S 0 1 Fr 2 If S f = S 0 then dy/dx = 0 Uniform Depth Flow! The constant flow depth in uniform flow is called normal depth, and denoted as y n. 49

50 ClassificaAon of Surface Profiles for GVF ClassificaAon of water surface profiles as a funcaon of Fr, S 0, S f, and iniaal flow depth y. Bed slope S 0 is classified as - Steep : y n < y c - CriAcal : y n = y c - Mild : y n > y c - Horizontal : S 0 = 0 - Adverse : S 0 < 0 IniAal depth y is classified as 1 : y > y n 2 : y n > y > y c 3 : y c > y 50

51 ClassificaAon of Surface Profiles for GVF S 0 > S c 51 Mild (M) Steep (S)

52 ClassificaAon of Surface Profiles for GVF S 0 = S c 52 CriAcal (C) Horizontal (H) Adverse (A)

53 Example Surface Profiles 53

54 Example Surface Profiles cont. 54

55 Uniform Depth Flow dy Uniform Depth Flow dx = 0 S 0 = S f Flow depth (and flow velocity) is constant along the channel and it is called normal depth y n. Uniform flow develops in long straight prismaac channels with no booom slope changes. Uniform depth is maintained as long as the slope, cross- secaon, and surface roughness of the channel remain unchanged. During uniform flow, the terminal velocity is reached, and the head loss equals to the channel bed elevaaon drop. 55

56 Uniform Depth Flow : Chezy Formula dy Uniform Depth Flow dx = 0 S 0 = S f V 2 Darcy- Weisbach Formula h L = f R H 8g V 2 h L = S f = f S 0 = S f R H 8g 1 V 2 S 0 = f V = 8g S 0 R H C = 8g Chezy R H 8g f f Coefficient V = C S 0 R H : Chezy Formula 56

57 Uniform Depth Flow : Manning EquaAon V = 1 n R h S 0 : Manning EquaAon n: Manning Coefficient (SI Unit System) 57

58 Example Problem 3 Water flows uniformly in the canal of trapezoidal cross secaon shown in the Figure. The canal booom slopes at S 0 = The canal is lined with new finished concrete. Determine the flow rate and the Froude Number for this flow. y = 2 m 45 o 45 o w = 4 m 58

59 SoluAon 3 Water flows uniformly in the canal of trapezoidal cross secaon shown in the Figure. The canal booom slopes at S 0 = The canal is lined with new finished concrete. Determine the flow rate and the Froude Number for this flow. y = 2 m A = 12 m 2 w = 4 m P = 9.66 m y = 2 m Θ = 45 o R H = 1.24 m 45 o 45 o S 0 = n = w = 4 m V = 1 n R h S 2 0 =1.21m / s Q = VA =14.46cms Fr = V gy = subcriacal flow! 59

60 Example Problem 4 Water flows uniformly in the excavated weedy earth channel of trapezoidal cross secaon shown in the Figure. The channel booom slopes at S 0 = The flow rate Q = 10 cms. Determine the flow depth and the Froude Number for this flow. y 45 o 45 o w = 4 m 60

61 SoluAon 4 Water flows uniformly in the excavated weedy earth channel of trapezoidal cross secaon shown in the Figure. The channel booom slopes at S 0 = The flow rate Q = 10 cms. Determine the flow depth and the Froude Number for this flow. w = 4 m n = 0.03 y Θ = 45 o S 0 = o 45 o Q = 10 cms w = 4 m A = 2y + y 2 Q = 1 n AR h S 2 0 = 1 2y + y ( 2y + y2) 2 2 3! \$ # & P = 4 + 2y 2 " 4 + 2y 2 % 2y + y2 R H = 10 = 1 2y + y ( 2y + y2) 2 2 3! \$ # & y 2 " 4 + 2y 2 % 61

62 SoluAon 4 cont. Water flows uniformly in the excavated weedy earth channel of trapezoidal cross secaon shown in the Figure. The channel booom slopes at S 0 = The flow rate Q = 10 cms. Determine the flow depth and the Froude Number for this flow. w = 4 m n = 0.03 y Θ = 45 o S 0 = o 45 o Q = 10 cms w = 4 m 10 = 1 2y + y ( 2y + y2) 2 2 3! \$ # & by trial and error methods y n = m 0.03 " 4 + 2y 2 % Fr =

63 Flow Urniversity irve is called apidly Zvirve aried ﬂow Z(irve RVF) if the ﬂ ow Udniversity epth hzas au niversity large Uoniversity irve Ud niversity change ver a szhort istance. dy 1 dx Rapidly varied is observed ﬂUow niversity at suluice gates, wniversity eirs, niversity U a iun niversity waterfalls, brupt changes cross secaons. Signiﬁcant 3D and transient eﬀects are o bserved i.e. irve backﬂow, szeparaaon etc. Flow over weir 63 Rapidly Varied Flow

64 Rapidly Varied Flow Hydraulic Jump When water at high velocity discharges into a zone of lower velocity, a rather abrupt rise occurs in the water surface. dy dx = V 2 V 1 The rapidly flowing water is abruptly slowed and increases in height, converang some of the flow's iniaal kineac energy into an increase in potenaal energy, with some energy irreversibly lost through turbulence to heat. 64

65 Rapidly Varied Flow Hydraulic Jump Hydraulic jump is a transiaon from supercriacal flow to subcriacal flow. dy dx = 65

66 Hydraulic Jump ignore the wall shear Momentum equaaon in x direcaon: stress F 1 F 2 = ρq(v 2 V 1 ) F 1 = γ y 2 1 b 2 F 2 = γ y 2 2 b 2 2 y 1 Q = V 1 y 1 b 2 y = V y 1 1 ( g V 2 V 1 ) 66

67 Hydraulic Jump y b ConAnuity equaaon in x direcaon: V 1 y 1 b = V 2 y 2 b = Q assume rectangular V 2 = V secaon 1y 1 y 2 67

68 Hydraulic Jump ConAnuity EquaAon + Momentum EquaAon 2 V 1 y 1 b = V 2 y 2 b = Q y 1 2 y = V y 1 1 ( g V V 2 1) 2 y 1 2 y = V y " 1 1 V 1 y 1 % \$ V 1 ' = V 2 y 1 1 ( y 1 y 2 ) g # y 2 & gy 2 2 " y 2 % " \$ ' + y % 2 \$ ' 2Fr 2 1 = 0 # y 1 & # y 1 & y 2 depth raao across the hydraulic = 1 ( y jump Fr )

69 Hydraulic Jump Energy Loss Energy equaaon in x direcaon: y 1 + V 2 1 2g = y + V g + h L h L = y 1 y 2 + V 2 V 2 head loss across the hydraulic 1 2 jump 2g 69

70 Hydraulic Jump Energy dissipaaon OSen, hydraulic jumps are avoided because they dissipate valuable energy However, in some cases, the energy must be dissipated so that it doesn t cause damage h L h A measure of performance of = L E 1 y 1 + V 2 a hydraulic jump is its fracaon 1 of energy dissipaaon, or 2g energy dissipaaon raao 70

71 Example Problem 5 Water on the horizontal apron of the 30 meter wide spillway shown in the Figure has a depth of 0.18 meter and velocity of 5 m/s. Determine the downstream depth aser the jump, Froude numbers before and aser the jump and the head loss due to viscous effect within the jump. b = 30 m V 2 y 2 V 1 = 5 m/s y 1 = 0.18 m (1) (2) 71

72 SoluAon 5 Water on the horizontal apron of the 30 meter wide spillway shown in the Figure has a depth of 0.18 meter and velocity of 5 m/s. Determine the downstream depth and velocity aser the jump, Froude numbers before and aser the jump and the head loss due to viscous effect within the jump. b = 30 m Fr 1 = V 1 = 3.8 supercriacal flow! y 1 = 0.18 m gy 1 V 1 = 5 m/s y 2 = 1 ( y Fr ) 2 1 y 2 = y ( Fr ) 2 = 0.87 m 1 V 2 = V 1y 1 =1.03 m/s h L = y 1 y 2 + V 2 V = 0.53 m y 2 2g 72

73 Hydraulic Jump ClassificaAon 73

74 Flow Control and Measurement Flow rate in pipes and ducts is controlled by various kinds of valves. In Open Channel flows, flow rate is controlled by paraally blocking the channel. These structures can also be used to measure the flow rate. a) Weir b) Underflow gate 74

75 Broad- Crested Weir Nearly uniform criacal flow is achieved at the top of the broad- crested weir, y = y c! Q = C wb b g 2 3 H \$ 3 2 # & " % C wb = H P 1 2! \$ w # & " 2 + H P w % 75

76 Sharp- Crested Weir SubcriAcal flow becomes criacal as it approaches to the weir. Flow discharges as a supercriacal flow stream that resembles a free jet. 2 Q = C wrec 3 b 2gH Q = C wtri 15 tan! θ \$ # & 2gH 5 2 " 2 %! C wrec = H \$ # & " P w % C w

77 Underflow Gates SubcriAcal flow becomes criacal as it approaches to the weir. Flow discharges as a supercriacal flow stream that resembles a free jet. q = C d a 2gy 1 q : flow rate per unit width C d

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