1OPEN CHANNEL FLOW234Question – What is the most obvious difference between pipe flow and open channel flow????????????? (in terms of flow conditions and energy situation) Typical open channel shapes – Figure 14.156Types of open channel flows – Steady flow – when discharge (Q) does not change with time. Uniform flow – when depth of fluid does not change for a selected length or section of the channel Uniform steady flow – when discharge does not change with time and depth remains constant for a selected section - cross section should remain unchanged – referred to as a prismatic channel Varied steady flow – when depth changes but discharge remains the same (how can this happen?) Varied unsteady flow – when both depth and discharge change along a channel length of interest. Rapidly varying flow – depth change is rapid Gradually varying flow – depth change is gradual7 Figure 14.3 Section 1 – rapidly varying flow Section 2 – gradually varying flow Section 3 – hydraulic jump Section 4 – weir and waterfall Section 5 – gradually varying Section 6 – hydraulic drop due to change in channel slope8Hydraulic radius of open channel flow A parameter that is used often Ratio of flow cross sectional area (A) and wetted perimeter (WP) R = A/ WP Hydraulic radius R for various channel shapes – Figure 14.1910Kinds (types) of open channel flow Reynolds number for pipe flow – υvDNR= Reynolds number for channel flow – υvRNR= For pipe flow – NR < 2000 – laminar NR > 4000 – turbulent For channel flow – NR < 500 – laminar NR > 2000 – turbulent11 Another “number” for channel flow! Froude Number [NF] (gravity versus inertial forces) hFgyvN = Where yh is referred to as the hydraulic depth and given as – yh = A/T where A is the area and T is the top width of the channel NF = 1.0 or when v = (gy)1/2 - critical flow NF < 1.0 – subcritical flow NF > 1.0 – super critical flow A combination of both the numbers is used to describe channel flow conditions.12Uniform steady flow and Manning’s Equation When discharge remains the same and depth does not change then we have uniform steady flow. In this condition – The surface of water is parallel to the bed of the channel Or S = Sw Where S is the slope of the channel The slope of the channel can be expressed as – - An angle = 1 degrees - As percent = 1% - Or as fraction = 0.01 or 1 in 10013 Velocity of flow (v) in a channel can be computed numerous empirical equations – One of them is Mannings equation – 2/13/20.1SRnv = This the SI units form of the equation with v (meters/sec) and R (meters). Where n is the Manning’s coefficient (dimensionless) – values developed through experimentation Possible n values for various channel surfaces – Table 14.11415 In English units the Manning’s equation form is – 2/13/249.1SRnv = Where v is in feet/sec and the R value is in feet. If velocity is known, the discharge (Q) can then be computed as – Q = A*v 2/13/20.1SARnQ = Where Q is in m3/s For uniform flow, Q is referred to as Normal discharge16The above equation can also be re-arranged such that – 2/13/2SnQAR = The left hand term is simply based on channel geometry.17Problem 14.2 Determine normal discharge for a 200 mm inside diameter common clay drainage tile running half-full if the slope drops 1 m over 1000 m. S = 1/1000 = 0.001 A = (1/2) * (π D2/4) = 0.5*π*(0.2)2/4 = 0.0157 m2 WP = (1/2) * (π D) = 0.5*π*0.2 = 0.3141 m R = 0.05 m From Table 14.1 n for clay tile = 0.013 Substitute these values in the equation –182/13/20.1SARnQ = And we get 2/13/2)001.0()05.0(*0157.0*013.00.1=Q Q = 5.18 x 10-3 m3/s19Problem 14.3 Calculate slope of channel below If Q = 50 ft3/s Formed unfinished concrete channel Equation that you will use 2/13/20.1SARnQ = Or203/22/149.1 ARQnS = Compute A = 12 ft2 WP = 9.66 ft R = A/WP = 12/9.66 = 1.24 ft Manning’s n for concrete channel = 0.017 Substitute And S = 0.00169 Drop 1.69 ft for every 1000 ft.21Problem 14.4 Design rectangular channel in formed unfinished concrete Q = 5.75 m3/s S = 1.2% Normal depth = ½ of the width of the channel Since we have to design the channel – the equation that should be used – 2/13/2SnQAR = RHS is known. RHS = 0.017*5.75/(0.012)1/2 = 0.0892 Now we know that y = b/2 Express Area and the hydraulic radius in terms of b. A = by = b2/222 WP = b+ 2y = 2b R = A/WP = b/4 Therefore, LHS = AR2/3 = b2/2 * (b/4) 2/3 = RHS = 0.892 B = 1.76 m y = 1.76/2 m23Problem 14.5 In the problem above the final width was set at = 2m and the maximum Q = 12 m3/s; find the normal depth for this maximum discharge. OK again, 2/13/2SnQAR = RHS = 0.017*12/(0.012)1/2 = 1.86 B = 2m A = 2y WP = 2+2y R = 2y/(2+2y) Therefore LHS =2486.122223/2=⎟⎟⎠⎞⎜⎜⎝⎛+ yyy Cannot solve this directly, will have to do trial and error. Set up a Table and compare y (m) A (m2) WP (m) R (m) R2/3 AR2/3 Required change in y 2.0 4.0 6.0 0.667 0.763 3.05 Make y lower 1.5 3.0 5.0 0.600 0.711 2.13 Make y lower 1.35 2.7 4.7 0.574 0.691 1.86 OK25Conveyance and most efficient channel shapes Look at the RHS of the equation 2/13/20.1SARnQ = Other than the S term, all other terms are related to channel cross section and its features. These terms together are referred to as the Conveyance (K) of the channel 3/20.1ARnK = OR 2/1SKQ =26 K is maximum when WP is the least for a give area  this is also the most efficient cross section for conveying flow For circular section – half full flow is the most efficient For other shapes – see Table 14.3 from the text.2728 Compound Sections When channel shape changes with flow depth – typical in natural stream sections during flooding During floods – water spills over the flood plain You need to know Q at various depths or vice-versa – so that you can design channels or determine channel safety for various flood magnitudes29 Cazenovia Creek in Buffalo during “normal” flow conditions30 Cazenovia Creek during flood!31Problem 14.21E Figure 14.21 – natural channel with levees Channel – earth with grass cover, n = 0.04 S = 0.00015 Determine normal Q for depth = 3 and 6 ft.3233Assignment # 9 - 14.3E - 14.9M - 14.10M - 14.14M34Compound section – More realistic situation – channel roughness