Hydraulic Engineering lecture notes

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1 CE2253-APPLIED HYDRAULIC ENGINEERING (FOR IV – SEMESTER) UNIT - I OPEN CHANNEL FLOW 1 2 UNIT – I OPEN CHANNEL FLOW OPEN CHANNEL FLOW - TYPES AND REGIMES OF FLOW - VELOCITY DISTRIBUTION IN OPEN CHANNEL – WIDE OPEN CHANNEL – SPECIFIC ENERGY – CRITICAL FLOW AND ITS COMPUTATION. S.NO 2 MARKS PAGE NO 1. Define open channel 5 2. What are the classifications of flow in an open channel? 5 3. Define steady flow and unsteady flow 5 4. Define Uniform flow and Non-Uniform flow. 6 5. What is rapidly varied flow? 6 6. What is gradually varied flow? 6 7. What is laminar and turbulent flow? 7 8. What is TRANSITION state? 7 9. Give a brief note on Sub-critical, Critical, Super critical flow. 7 Give the formula relating to velocity and discharge in 10. 7 chezy’s formula Give the BAZIN, GANGUILLET-KUTTER, MANNINGS 11. 8 formulas for chezy’s constant. 12. Give the formula for total energy. 8 13. Define specific energy. 8 14. How do you calculate specific energy? 8 15. What is specific energy curve? 9 16. Define critical depth. 9 17. What is critical velocity? 9 Represent minimum specific energy in terms of critical 18. 9 depth. 19. What is critical flow? 10 20. Define streaming flow. What it is otherwise called as? 10 21. When does a super critical flow occur? 10 22. What is alternate depth? 10 23. Comparison between open channel flow and pipe flow 10 24. What are the types of channels? 11 25. What are regimes of flow? 11 26. What is wide open channel? 11 What do you mean by velocity distribution in an open 27. 11 channel? 2 3 28. What are the applications of specific energy? 12 What are the factors affecting Manning’s roughness 29. 12 coefficient? What is top width, wetted area, wetted perimeter, Hydraulic 30. 12 mean radius, hydraulic depth? S.NO 16 MARKS PAGE NO Find the diameter of a circular sewer pipe which is laid at a slope of 1 in 8000 and carries a discharge of 800 1 13 litres/second.when flowing half full. Take the value of Manning’s N=0.02 i) Find the specific energy of flowing water through a rectangular channel of width 5m when the discharge is 10 3 m /s and depth of water is 3m. 2 14 (ii) Find the critical depth and critical velocity of water flowing through a rectangular channel of width 5m, when 3 discharge is 15m /s. The discharge of water through a rectangular channel of 3 width 8m,is 15 m /s when depth of flow of water is 1.2 3 m.Calculate:Specific energy of flowing water, Critical 14 depth and critical velocity, Value of minimum specific energy. i) The specific energy for a 5m wide rectangular channel is to be 4Nm/N.If the rate of flow of water through the 4 15 3 channel is 20 m /s.Determine the alternate depth of flow. ii) Derive the froude number. Condition for maximum discharge for given value of 5 16 specific energy. i) Find the velocity of flow and rate of flow of water through a rectangular channel of 6m wide and 3m deep, when it is running full. The channel is having bed slope as 1 in 2000. 6 Take chezy’s constant C = 55. 18 ii) Find the slope of the bed of a rectangular channel 5m when depth of water is 2m and rate of flow is given as 20 m³/s. Take chezy’s constant, C = 50. A flow of water of 100 lts/sec flows down in rectangular flume of width 600mm and having adjustable bottom 7 slope. If chezy’s constant C is 56, find the bottom slope 19 necessary for uniform flow with a depth of flow of 300mm. Also find the conveyance K of the flume. Find the discharge through a trapezoidal channel of width 8 20 8m and side slope of 1 horizontal to 3 vertical. The depth 3 4 of flow of water is 2.4m and value of Chezy’s constant, C = 50. The slope of the bed of the channel is given 1 in 4000. Find the bed slope of trapezoidal channel of bed width 6m, depth of water 3m and side slope of 3 horizontal to 4 9 20 vertical, when the discharge through the channel is 30 m³/s. Take Chezy’s Constant, C= 70 i) Find the discharge of water through the channel shown in the fig. Take the value of Chezy’s constant = 60 and slope of the bed as 1 in 2000. 10 21 ii) Find the rate of flow of water through a V- Shaped channel as shown in the fig. Take the value of C = 55 and slope of the bed 1 in 2000 11 With neat sketches give the computation of critical flow. 23 i) Derive the equation for minimum specific energy in terms of critical depth 12 24 ii) How do you obtain the specific energy curve explain briefly? 13 Derive the equation for critical depth. 26 i) Give the application of specific energy and discharge curve. ii) Discharge curve: iii) Uniform flow occurs at a depth of 1.5 m in a long 14 rectangular channel 3m wide and laid at a slope of 27 0.0009.If Manning’s N = 0.015.Calculate (1) max height of jump on the flow to produce the critical depth. (2) The width of the contraction which will produce critical depth without increasing the upstream depth of flow 4 5 APPLIED HYDRAULIC ENGINEERING UNIT – I OPEN CHANNEL FLOW OPEN CHANNEL FLOW - TYPES AND REGIMES OF FLOW - VELOCITY DISTRIBUTION IN OPEN CHANNEL – WIDE OPEN CHANNEL – SPECIFIC ENERGY – CRITICAL FLOW AND ITS COMPUTATION. Two Marks Questions and Answers 1. Define open channel. A liquid flowing at atmospheric pressure through a passage is known as flow in open channels. The flow of water through pipes at atmospheric pressure or when the level of water in the pipe is below the top of the pipe, is also classified as open channel flow. 2. What are the classifications of flow in an open channel? 1. Steady and unsteady flow 2. Uniform flow and non-uniform flow 3. Laminar flow and turbulent flow 4. Sub-critical, critical, and super critical flow 3. Define steady flow and unsteady flow. Steady Flow If the flow characteristics such as depth of flow, velocity of flow, rate of flow at any point in open channel flow do not change with respect to time, the flow is said to be steady flow. ∂v ∂Q ∂y =0 , =0 or = 0 ∂t ∂t ∂t Unsteady Flow 5 6 If at any point in open channel flow, the velocity of flow, depth of flow or rate of flow at any point in open channel flow changes with respect to time, the flow is said to be steady flow. ∂v ∂Q ∂y ≠ 0 , or ≠ 0 or ≠ 0 ∂t ∂t ∂t 4. Define Uniform flow and Non-Uniform flow. Uniform flow If for a given length of the channel, the velocity of flow, depth of flow, slope of the channel and cross-section remain constant, the flow is said to be uniform. ∂v ∂y =0 , =0 for uniform flow ∂S ∂S Non – uniform flow If for a given length of the channel, the velocity of flow, depth of flow, slope of the channel and cross-section do not remain constant, the flow is said to be non - uniform flow. ∂v ∂y ≠ 0 , ≠ 0 for non-uniform flow ∂S ∂S 5. What is rapidly varied flow? It is defined as that flow in which depth of flow changes abruptly over a small length of the channel. 6. What is gradually varied flow? If the depth of flow in a channel changes gradually over a long length of the channel, the flow is said to be gradually varied flow. 7. What is laminar and turbulent flow? 6 7 Laminar flow The flow in open channel is said to be laminar if the Reynolds number (R ) is e than 500 or 600 ρVR Reynolds number = μ Turbulent flow If the Reynolds number is more than 2000, the flow is said to be turbulent in open channel flow. 8. What is TRANSITION state? If the R lies between 500 to 2000, the flow is considered to be in transition e state. 9. Give a brief note on Sub-critical, Critical, Super critical flow. Sub critical flow: The flow in open channel is said to be sub-critical if the Froude number is less than 1. V F = e gD Critical Flow: The flow in open channel is said to be critical if the Froude number is 1. Super critical flow: The flow in open channel is said to be super critical if the Froude number is greater than 10. Give the formula relating to velocity and discharge in chezy’s formula. Velocity V = C mi Discharge Q = A C mi 7 8 11. Give the BAZIN, GANGUILLET-KUTTER, MANNINGS formulas for chezy’s constant. a) Bazin formula 157.6 C = K 1.81+ m b) ganguillet-kutter formula 0.00155 1 23+ + i N C = 0.00155 N 1+(23+ ) i m c) Manning’s formula 1 1 6 C = m N 12. Give the formula for total energy. 2 V TOTAL ENERGY = z + h + 2g 8 9 13. Define specific energy. It is defined as energy per unit weight of the liquid with respect to the bottom of the channel. 14. How do you calculate specific energy? 2 V E = h + 2g h = depth of liquid V = Mean velocity of flow g= Acceleration due to gravity. 15. What is specific energy curve? It is defined as the curve which shows variation of specific energy with depth of flow 16. Define critical depth. It is defined as the depth of flow of water at which the specific energy is minimum. The depth of flow of water at C is known as critical depth. 1 2 3 q  h =   c g   17. What is critical velocity? The velocity of flow at critical depth is known as critical velocity. V = gh c c Where, V = Critical velocity c h Critical depth c= g= acceleration due to gravity 18. Represent minimum specific energy in terms of critical depth. 9 10 3h c E min = 2 Where, h = critical depth c E = minimum specific energy min 19. What is critical flow? It is defined as that flow at which the specific energy is minimum or the flow corresponding to critical depth is defined as critical flow. V c V = gh (or) = 0 c c gh c 20. Define streaming flow. What it is otherwise called as? When the depth of flow in a channel is greater than the critical depth (h ), the flow c is said to be sub-critical flow (or) streaming flow (or) tranquil flow.F 1.0. e 21. When does a super critical flow occur? When the depth of flow in a channel is less than the critical depth (h ) ,the flow is c said to be super-critical flow(or) shooting flow(or) the torrential flow.(F 1.0) e 22. What is alternate depth? In the specific energy curve, the point C corresponds to the minimum specific energy and the depth of flow at C is called critical depth. The depths corresponding to points G & H are called alternate depth. 23. Comparison between open channel flow and pipe flow S.No Aspects Open channel Pipe flow flow 1 Causes of flow Gravity flow Hydraulic pressures 2 Geometric of May have any Generally round in sections shape cross sections 3 Surface Varies widely with Depending upon roughness depth of flow the material of the 10 11 pipe 4 Piezometric head Z + Y = HGL Z + P/W = HGL Coincides with the does not coincide water surface with water surface. 5 Velocity Maximum velocity Maximum at distribution occurs at a little center of pipe and distance below the at pipe wall V = 0 water surface. 6 Law Obeys frouds law Obeys Reynolds inertia force law inertia /gravity force. force/viscous force. 24. What are the types of channels? The types are: i) Natural surface: It has irregular sections of varying shapes. Ex. Rivers, streams etc. ii) A channel without any cover at top is known as open channel .Ex. Irrigation channels iii) Prismatic channel: A channel with constant beds slope and the same cross section along its length. iv) Experimental: The cross section of the channel is proportional to any power of depth of flow in channel. Ex;Rectangular,Triangular v) Non Exponential: Trapezoidal and circular channel are non exponential channels. 25. What are regimes of flow? Regimes of flow are the result of joint influence of viscosity and gravity. The four common stages of flow, viz, laminar, turbulent, sub critical and super critical flows. TYPES: 1) Sub critical laminar 2) Super critical laminar 3) Sub critical turbulent 4) Super critical turbulent. 26) What is wide open channel? If the width of the channel is equal to or more than ten times the depth of flow it may be called as wide open channel. For experimental or analytical purposes the flow in 11 12 general region wide open channel may be considered as a same as flow in the rectangular channel of infinite width. 27) What do you mean by velocity distribution in an open channel? The non uniform distribution of velocity in an open channel is due to 1) Presence of free surface. 2) Frictional resistance along the channel boundary 3) The velocity distribution in a channel is measured with help of Pitot tube. 28) What are the applications of specific energy? The application of specific energy is: 1) Analysis of flow through channel transmission. 2) Flow over raised channel bottom 3) Flow through sluice gate openings. 29) What are the factors affecting Manning’s roughness coefficient? The factors affecting Manning’s roughness coefficient: 1) surface roughness 2) Channel irregularities 3) Silting and scouring 4) Obstruction 5) Size and shape of channel 6) Seasonal changes. 7) Suspended material and bed load. 8) Canal alignment 30) What is top width, wetted area, wetted perimeter, Hydraulic mean radius, hydraulic depth? Top width is the width of the channel section at the free surface. (The width of the liquid surface exposure to atmospheric pressure).Wetted area is the cross sectional area 12 13 of flow section in the channel. Wetted perimeter the cross sectional area of flow section of the channel. Hydraulic mean radius (R): R= A/P Hydraulic depth (D): D = A/T 16 Marks Questions and Answers 1. Find the diameter of a circular sewer pipe which is laid at a slope of 1 in 8000 and carries a discharge of 800 litres/second.when flowing half full. Take the value of Manning’s N=0.02 Given: Slope of pipe, I = 1/8000 3 Discharge Q= 800 lts/s =0.8 m /s Manning’s, N=0.02 Dia of sewer pipe= D Depth of Flow, d= D/2 π 1 2 Area of flow A = D X 4 2 2 πD = 8 πD Wetted perimeter= 2 2 πD 8 Hydraulic mean depth, m= A/P = = D/4 πD 2 13 14 2 πD 1   1/6 Q= AC mi = X X mi  m  8 N   2 πD 1/6 1/2 0.8= x 1/0.02 m x m x i 8 2 πD 1 1   1 π 2 2/3 0.8= x 1/0.02 m + x x 1/0.02 x m x 0.01118.   = D 8 6 2   8000 8 0.2195 2 2/3 8/3 = 0.2195 x D x (D/4) = D 2/3 4 0.8 8/3 D = =9.1848 0.0871 3/8 D= (9.1848) = 2.296m 2) i) Find the specific energy of flowing water through a rectangular channel of 3 width 5m when the discharge is 10 m /s and depth of water is 3m. Width b=5m 3 Q= 10 m /s h= 3m V= Q/area=10/b x h = 10 /5 x3 = 2/3 2 E= h + V /2g 2 (2/3) E= 3 + = 3.0226m 2x9.81 (ii) Find the critical depth and critical velocity of water flowing through a 3 rectangular channel of width 5m, when discharge is 15m /s. b= 5m 3 Q= 15 m /s 2 Discharge per unit, q= Q/b=15/5 = 3 m /s 14 15 2 1/3 2 1/3 Critical depth h = (q / g) = (3 /9.81) = 0.972 m c Critical velocity Vc = 9.81x0.972 = gxh c = 3.088 m/s 3 3) The discharge of water through a rectangular channel of width 8m,is 15 m /s when depth of flow of water is 1.2 m.Calculate: 1) Specific energy of the flowing water 2) Critical depth and critical velocity 3) Value of minimum specific energy Given: 3 Q = 15 m /s b = 8m h = 1.2 m Discharge per unit width= Q/b=15/8 Velocity of flow, V=Q/area=15/bx h=15/8 x1.2 = 1.5625 m/s 2 2 V (1.5625) (i) Specific energy E = h + =1.2+ =1.324m 2g 8x9.81 1 1 2 3 2 3     q 1.875 (ii) Critical depth h =   =   = 0.71 m c     g 9.81     (iii) Critical velocity Vc= gxh = 9.81x0.71 = 2.639 m/s. c (iv) Minimum Specific energy (E ): min 3h 3x0.71 c E = = =1.065m min 2 2 4) i) The specific energy for a 5m wide rectangular channel is to be 4Nm/N.If the 3 rate of flow of water through the channel is 20 m /s.Determine the alternate depth of flow. b = 5m 2 V E = 4Nm/N = 4m E = h + V= Discharge/area =Q/b x h= 20 /5 x h 2g 15 16 3 Q = 20 m /s E= 4 2 V 2 E = h + = h+ (4/h) x 1/2g 2g 8 = h + 2 gxh 8 0.8155 4 = h + = h + 2 2 9.81xh h 2 3 3 2 4h = h + 0.8155 (or) h – 4 h + 0.8155 =0 This is a cubic equation solving by trial (or) error h = 3.93 m and 0.48 ii)Derive the froude number. The square root of ratio of inertia force of flowing liquid to gravity force F i F = e F g 2 F = ρAV = mass acceleration i F = M g = ρADg g D = wetted area / Top width of channel = A/T 2 2 ρAV V V F = = = e ρADg Dg Dg 5) Condition for maximum discharge for given value of specific energy. The specific energy at any section of channel is given 16 17 2 V Q Q E = y + ; V = = 2g A by 2 Q E = y + 2 2 2b y g 2 Q E – y = 2 2 2b y g 2 2 2 Q = (E-Y) 2 b y g 2 2 3 = 2b g Ey – y 2 3 Q = b 2g(Ey − y …………………………….. (1) 2 3 For discharge Q to be maximum to expression Ey – y should be maximum Hence differentiate (1) with respect to y and equate it to zero 1 dQ b   2 3 2 2 = 2g Ey − y 2g 2Ey −3y   dy 2     b 1 2 =  2g(2Ey −3y ) 2 3 2  2g(Ey − y )    dQ = 0 dy 2 bg(2Ey−3y ) = 0 3) 2g(Ey− y 2 2E bgy – 3 bgy = 0 E = 3/2 y y = 2/3 E Maximum Discharge: 17 18 3 3 3 Q = b 2g( y − y ) 2 3 Q = b gy 4 8 2 3 Q = b 2g(E E − E 9 27 3/2 Q = 1.705 b E max The specific energy is minimum when it is equal to 3/2 times value of depth of critical flow. Here the equation (2) represents the specific energy and is equal to 3/2 times the depth of flow. Therefore the equation (2) represents the specific energy and y is the critical depth. Hence the condition for maximum discharge for given value of specific energy is that the depth of flow should be critical. 6) i) Find the velocity of flow and rate of flow of water through a rectangular channel of 6m wide and 3m deep, when it is running full. The channel is having bed slope as 1 in 2000. Take chezy’s constant C = 55. Given: Width of rectangle channel, b = 6m. Depth d = 3m Solution: Area = b × d = 6 × 3 = 18m² Bed Slope, i = 1 in 2000 = 1/2000 Chezy’s constant C = 55 Perimeter P = b+2d = 6 + 2 × 3 = 12m Hydraulic mean depth, m = A/P = 18/12 = 1.5m V = C√1.5 × 1/2000 = 1.506 m/s Q = V × Area = 1.506 × 18 = 27.108m³/s. ii) Find the slope of the bed of a rectangular channel 5m when depth of water is 2m and rate of flow is given as 20 m³/s. Take chezy’s constant, C = 50. Given: Width of channel b = 5m. Depth of water d = 2m Rate of flow Q = 20 m³/s. C = 50 Bed Slope = i Solution: Q = AC √mi 18 19 A = Area = b × d = 5 × 2 = 10 m² m = A/P = 10/ (b+2d) = 10 / (5+2×2) = 10 /(5+4) = 10/9 m 20.0 = 10 × 50 × √ 10/9 × i √ 10/9 × i = 20/500 = 2/50 10i/9 = 4/2500 i = 4/2500 × 9/10 = 36/25000 = 1/25000/36 = 1/694.44 Therefore Bed slope is 1 in 694.44 7) A flow of water of 100 lts/sec flows down in rectangular flume of width 600mm and having adjustable bottom slope. If chezy’s constant C is 56, find the bottom slope necessary for uniform flow with a depth of flow of 300mm. Also find the conveyance K of the flume. Given: Discharge Q = 100 lts/s = 100/1000 = 0.1 m³/s. b = 600mm = 0.6m d = 300mm = 0.3m A = b×d = 0.6 × 0.3 = 0.18m² C = 56 Slope of bed = i Solution: Hydraulic mean depth, m = A/P = 0.18/(b+2d) = 0.18/(0.6+2 × 0.3) = 0.18/1.2 = 0.15m Q = AC √mi 0.1 = 0.18 × 56 √ 0.15×i √ 0.15×i = 0.1/(0.18 × 56) Squaring on both sides 0.15i = (0.10/0.18 × 56) ² = 0.00098418 i = 0.00098418 / 0.15 = 0.0006512 = 1/1/0.0006512 = 1/1524 Therefore Slope of the bed is 1 in 1524. Conveyance K of the channel Q = AC√mi = Q = K√i Where, K = AC√m Conveyance of channel section K = AC√m = 0.18 × 56 ×√0.15 = 3.9039 m³/s. 8) Find the discharge through a trapezoidal channel of width 8m and side slope of 1 horizontal to 3 vertical. The depth of flow of water is 2.4m and value of Chezy’s constant, C = 50. The slope of the bed of the channel is given 1 in 4000. Given: Width b = 8m Side Slope = 1 horizontal to 3 vertical Depth d = 2.4m Chezy’s constant C = 50, Bed Slope I = 1/4000 19 20 Depth CF = 2.4 Solution: Horizontal dts BE = 2.4 × 1/3 = 0.8m Therefore Top Width of the channel, CD = AB + 2 × BE = 8.0 + 2×0.8 = 9.6m Therefore Area of trapezoidal channel, ABCD is given as, A = (AB + CD) × CE/2 = (8+9.6) × 2.4/2 = 17.6 × 1.2 = 21.12m² Wetted Perimeter, P = AB + BC + AD = AB = 2BC BC = √BE² + CE² = √ (0.8)² + (2.4)² = 2.529m P = 8 + 2 × 2.529 = 13.058m Hydraulic mean depth m = A/P = 12.12/13.058 = 1.617m Q = AC√mi = 21.12 × 50 √1.617×1/4000 = 21.23 m³/s. 9) Find the bed slope of trapezoidal channel of bed width 6m, depth of water 3m and side slope of 3 horizontal to 4 vertical, when the discharge through the channel is 30 m³/s. Take Chezy’s Constant, C= 70 Given: Bed Width, b = 6.0m Depth of flow, d = 3.0m Side Slope = 3 Horizontal to 4 vertical Discharge Q = 30 m³/s Depth of flow CE = 3m Chezy’s Constant = 70 CE = 3m BE = 3 × ¾ = 9/4 = 2.25m Therefore Top Width, CD = AB + 2 × BE = 6.0 + 2 × 2.25 = 10.50m Wetted Primeter, P = AD + AB + BC = AB + 2ABC (…BC = AD) = AB + 2 √ (BE² + CE²) = 6.0 + 2√ (2.25) ² + (3) ² = 13.5m A = Area of trapezoidal ABCD = (AB+CD) × CE /2 = (6+10.50)/2 × 3 = 24.75m² Hydraulic mean depth, m = A/P = 24.75/13.50 = 1.833 Q = AC√mi 30.0 = 24.75 × 70 √1.833×i = 2345.6√ i i = (30/2345.6) ² = 1/(2345.6/30) ² = 1/6133 i = 1/6133 10) i) Find the discharge of water through the channel shown in the fig. Take the value of Chezy’s constant = 60 and slope of the bed as 1 in 2000. Given: Chezy’s Constant C = 60 Bed Slope, i = 1/2000 20

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