44 | 0 3 U n i f o r m O p e n C h a n n e l 13. Explain the wet perimeter and hydraulic radius for the drainage of water tank as shown in Figure 3.13. [Session 1: 2023/2024 – Sec. B: Q1(a)] Answer: The wetted perimeter is the perimeter of the cross-sectional area that is ‘wet’ (contact with the flowing water at any section) Formula for wet perimeter; ?? = ?? + 2?? Hydraulic radius can be defined as the ratio between the area of cross-sectional flow and the wetted perimeter. Formula for hydraulic radius; ?? = ?? ?? = ???? ?? + 2?? 14. Water is transported in a rectangular channel cast iron has a bottom width of 10.0 m at velocity 3.0 m/s. The terrain of the bottom channel drops 1.5 : 1000 length. Determine the minimum height of the channel under uniform-flow condition. Given n = 0.013. [Session 1: 2023/2024 – Sec. B: Q1(b)] Answer: Width, B = 10.0 m Velocity, V = 3.0 m/s Manning’s coefficient, n = 0.013 Bed slope, 1.5: 1000 ??0 = 1.5 1000 ??0 = 0.0015 Figure 3.13 B = 10.0 m y b y
45 | 0 3 U n i f o r m O p e n C h a n n e l Area, ?? : ?? = ???? ?? = 10.0 × ?? ?? = 10?? Wetted Perimeter, ?? : ?? = ?? + 2?? ?? = 10.0 + 2?? Hydraulic Radius, ?? : ?? = ?? ?? ?? = 10?? 10 + 2?? Value of depth using Manning Equation, ?? : ?? = ?? 2 3 × ??0 1 2 ?? 3.0 = [ 10?? 10 + 2?? ] 2 3 (0.0015) 1 2 0.013 0.039 = [ 10?? 10 + 2?? ] 2 3 × 0.0387 (1.008) 3 2 = [ 10?? 10 + 2?? ] 1.0117 = [ 10?? 10 + 2?? ] 1.0117(10 + 2y) = 10?? 10.117 + 2.0234y = 10?? 10.117 = 10?? − 2.0234y ?? = 10.117 7.977 ?? = ??. ????????
46 | 0 3 U n i f o r m O p e n C h a n n e l 15. A trapezoidal channel with a bottom width of 0.8 m, trapezoidal angle of 50°, built on bottom slope of 0.007 as shown in Figure 3.15. If the flow depth is 0.52 m and n = 0.016, determine the flow rate of water through the channel. [Session 1: 2023/2024 – Sec. B: Q1(c)] Answer: Depth, y = 0.52 m Width, B = 0.8 m Manning’s coefficient, n = 0.016 Bed slope, ??0 = 0.007 ?????????????????????? ??????????, ?? = 50° ; ?? = 90ᵒ − 50ᵒ = 40° ?? = 40° Value of ?? : ?????? ??° = ???????????????? ???????????????? ?????? ??° = ?? ?? ?????? 40° = ?? 0.52 ?? = 0.52 × ?????? 40° ?? = 0.4363 ?? Figure 3.15 ?? 40° ?? 1 ?? = 0.52 m ?? ?? ?? = 0.52 m ?? T 50° Z y = 0.52 m 1 B = 0.8 m
47 | 0 3 U n i f o r m O p e n C h a n n e l Value of ?? : 1 ?? = ?? ?? 1 ?? = 0.52 0.436 0.52 ?? ?? = 0.4363 ?? = 0.4363 0.52 ?? = 0.839 Area, ?? : ?? = ???? + ???? 2 ?? = [0.8(0.52) + 0.839(0.52) 2 ] ?? = 0.643 ??2 Wetted Perimeter, ?? : ?? = ?? + 2??√1 + ?? 2 ?? = 0.8 + [2(0.52)√1 + (0.839) 2] ?? = 2.158 ?? Hydraulic Radius, ?? : ?? = ?? ?? ?? = 0.643 2.158 ?? = 0.298 ?? Value of discharge using Manning Equation, ?? : ?? = ?? × ?? 2 3 × ??0 1 2 ?? ?? = 0.643 × (0.298) 2 3 × (0.007) 1 2 0.016 ?? = ??. ?? ???? /?? Another option: ?????? ??° = ???????????????? ???????????????? ?????? 40° = ?? 1 ?? = 1 × ?????? 40° ?? = 0.839
48 | 0 3 U n i f o r m O p e n C h a n n e l 16. With the aid of a diagram, explain uniform flow in open channel for a drainage system in Civil Engineering. [Session 1: 2023/2024 – Sec. B: Q2(a)] Answer: Flow in a channel is said to be uniform if the depth, slope, cross-section and velocity remain constant over a given length of the channel. Uniform flow conditions are commonly encountered in practice in long straight sections of channels with constant slope, constant roughness, and constant cross section as shown in figure below. (Amat Sairin Demun, 2017) 17. Water is transported at a rate of 10.0 m3/s in uniform flow in open channel that surface is asphalt lined (n = 0.016). The bottom slope is 0.002. If the z value is 0.5, determine the dimensions of the best cross-section if the shape of channel is trapezoidal. [Session 1: 2023/2024 – Sec. B: Q2(b)] Answer: Discharge, Q = 10.0 m3/s Manning’s coefficient, n = 0.016 Side slope, z = 0.5 Bed slope, ??0 = 0.002 Width for best cross section, ?? : ?? + 2???? = 2??√1 + ?? 2 ?? + 2(0.5)?? = 2??√1 + (0.5) 2 ?? + ?? = 2.236?? ?? = 2.236?? − ?? ?? = 1.236?? Uniform flow y constant Slope, ??0 = constant v = velocity 1 z z = 0.5 y B
49 | 0 3 U n i f o r m O p e n C h a n n e l Area, ?? : ?? = ???? + ???? 2 ?? = [1.236??(??) + 0.5(??) 2 ] ?? = 1.736?? 2 Hydraulic Radius for best cross section, ?? : ?? = ?? ?? = 0.5?? ???????? ?????????????????? ???? ?????? ?????????????? ?????????? ?????????????? ???????????????? ∶ ?? = ?? × ?? 2 3 × ??0 1 2 ?? 10.0 = 1.736?? 2 × (0.5??) 2 3 × (0.002) 1 2 0.016 10(0.016) = 1.736?? 2 × 0.63?? 2 3 × 0.045 0.16 = 0.049?? 8 3 ?? 8 3 = 0.16 0.049 ?? = 3.265 3 8 ?? = ??. ?????? ?? Width, ?? : ?? = 1.236?? ?? = 1.236 × 1.559 ?? = ??. ?????? ??
50 | 0 3 U n i f o r m O p e n C h a n n e l 18. Water flowing in a V – shaped open channel with a bottom slope of 0.0015 and n = 0.0013 as in Figure 3.18. Determine the depth of the channel if this channel needs to deliver water at flow rate 4.0 m3/s. [Session 1: 2023/2024 – Sec. B: Q2(c)] Answer: Discharge, Q = 4.0 m3/s Manning’s coefficient, n = 0.0013 Bed slope, ??0 = 0.0015 ???????????????? ??????????, ??ᵒ = 60° ; ??ᵒ = 90ᵒ − 60ᵒ = 30° ?????? 30° = 0.577 ?????? 30° = 0.866 Area, ?? : ?? = ?? 2 × ?????? ??ᵒ ?? = ?? 2 × ?????? 30ᵒ ?? = ?? 2 × 0.577 ?? = 0.577?? 2 Wetted Perimeter, ?? : ?? = 2 × ?? ?????? ??° ?? = 2 × ?? ?????? 30° ?? = 2 × ?? 0.866 ?? = 2.309?? Figure 3.18 ??° ?? ??ᵒ = 60° ?? ??˚ ?? ?? ?? ??ᵒ = 60° 60°
51 | 0 3 U n i f o r m O p e n C h a n n e l Hydraulic Radius, ?? : ?? = ?? ?? ?? = 0.577?? 2 2.309?? ?? = 0.25?? Value of depth using Manning Equation, ?? : ?? = ?? × ?? 2 3 × ???? 1 2 ?? 4.0 = 0.577?? 2 × (0.25??) 2 3 × (0.0015) 1 2 0.0013 4.0 = 6.822?? 8 3 ?? 8 3 = 4 6.822 ?? = (0.586) 3 8 ?? = ??. ?????? ?? 19. With the aid of diagram, explain uniform flow in open channel [Session 2: 2023/2024 – Sec. B: Q1(a)] Answer Flow in an open channel to be uniform if the depth, slope, cross-section and velocity remain constant over the given length of the channel. The slope of the energy gradient (S), the water surface slope (Sw) and the channel bed slope (So) are equal; that is S = Sw = So EGL Datum Water level ??2 ??2 ?? 2 2?? ⁄
52 | 0 3 U n i f o r m O p e n C h a n n e l 20. Based on Figure 3.20 below, calculate the best hydraulic cross-section for rectangular channel to convey 12.5 m3/s discharge with Manning coefficient of 0.03 and bed slope of the channel is 0.0005 [Session 2: 2023/2024 – Sec. B: Q1(b)] Answer Discharge, Q = 12.5 m3/s Manning’s coefficient, n = 0.03 Bed slope, ??0 = 0.0005 Width for most effective cross section, ?? : ?? = 2?? Area, ?? : ?? = ???? ?? = 2?? × ?? ?? = 2?? 2 Hydraulic Radius for most effective cross section, ?? : ?? = ?? ?? ?? = 2?? 2 4?? = 0.5?? ?? = 0.5?? B = 2y y Figure 3.20 b y
53 | 0 3 U n i f o r m O p e n C h a n n e l ???????? ?????????????????? ???? ?????? ?????????????? ?????????? ?????????????? ???????????????? ∶ ?? = ?? × ?? 2 3 × ??0 1 2 ?? 12.5 = (2?? 2 ) (0.5??) 2 3 [0.0005] 1 2 0.03 12.5 = 0.939?? 8 3 ?? 8 3 = 12.5 0.939 ?? = [ 12.5 0.939] 3 8 ?? = (0.0133) 3 8 ?? = ??. ???? ?? Width, ?? : ?? = 2?? ?? = 2 × 2.64 ?? = ??. ???? ?? 21. An open trapezoidal channel with side slope 1:2, bed slope 1:4000 and the water depth is 3.26 m. By using Manning coefficient, n = 0.058, calculate discharge in m3/s as show in Figure 3.21. [Session 2: 2023/2024 – Sec. B: Q1(c)] Answer Depth, y = 3.26 m Width, B = 5.55 m Manning’s coefficient, n = 0.058 Side slope 1:2, so z = 2.0 Figure 3.21 1 2 z = 2 y B 1 2 3.26 m 5.55 m
54 | 0 3 U n i f o r m O p e n C h a n n e l Bed slope, 1: 4000, ??0 = 1 4000 ??0 = 0.25 × 10−3 = 0.0025 Area, ?? : ?? = ???? + ???? 2 ?? = [5.55(3.26) + 2.0(3.26) 2 ] ?? = 39.348 ??2 Wetted Perimeter, ?? : ?? = ?? + 2??√1 + ?? 2 ?? = 5.55 + [2(3.26)√1 + (2.0) 2] ?? = 20.129 ?? Hydraulic Radius, ?? : ?? = ?? ?? ?? = 39.348 20.129 ?? = 1.955 ?? Value of discharge using Manning Equation, ?? : ?? = ?? × ?? 2 3 × ??0 1 2 ?? ?? = (39.348)(1.955) 2 3 [0.0025] 1 2 0.058 ?? = ????. ?????? ???? ?? ⁄
55 | 0 3 U n i f o r m O p e n C h a n n e l 22. Explain the term of hydraulic gradient [Session 2: 2023/2024 – Sec. B: Q2(a)] Answer • Hydraulic gradient or bed slope is the slope at the bottom or base of the channel • Movement of fluid flow is also influenced by the bed slope • The slope of the hydraulic grade line • This is the slope of the water surface in an open channel, the slope of the water surface of the groundwater table, or the slope of the water pressure for pipe under pressure 23. A cement-lined rectangular channel of 5 m wide carries water at the rate of 30 m3/s as shown in Figure 3.23. Calculate the value of Manning’s coefficient if the bed slope to maintain a depth of 1.5 m is 1/725. [Session 2: 2023/2024 – Sec. B: Q2(b)] Answer Depth, y = 1.5 m Width, B = 5.0 m Discharge, Q = 30.0 m3/s Bed slope, 1: 725, ??0 = 1 725 ??0 = 1.379 × 10−3 = 0.001379 Area, ?? : ?? = ???? ?? = [5.0 × 1.5] ?? = 7.5 ??2 B = 5.0 m y = 1.5 m Figure 3.23 5.0 m 1.5 m
56 | 0 3 U n i f o r m O p e n C h a n n e l Wetted Perimeter, ?? : ?? = ?? + 2?? ?? = 5.0 + 2(1.5) ?? = 8.0 ?? Hydraulic Radius, ?? : ?? = ?? ?? ?? = 7.5 8.0 ?? = 0.938 ?? Manning Coefficient, ?? : ?? = ?? × ?? 2 3 × ??0 1 2 ?? 20.0 = (7.5) × (0.938) 2 3 × ( 1 725) 1 2 ?? ?? = 0.267 30.0 ?? = ??. ?????? × ????−?? = ??. ?????? 24. One of the advantages of an open trapezoidal channel is it has maximum discharge. A Trapezoidal channel has a 1V:2H side slope and bed slope 1 in 2000. The area of the section is 48 m2. Determine the dimension of the channel for Figure 3.24 and the discharge of the channels for the most economical section if Manning coefficient, n = 0.017. [Session 2: 2023/2024 – Sec. B: Q2(c)] Answer Area, A = 48.0 m2 Manning’s coefficient, n = 0.017 Side slope, 1V:2H, z = 2.0 Figure 3.24 1 2 z = 2 y B Z y 1 b
57 | 0 3 U n i f o r m O p e n C h a n n e l Bed slope, 1: 2000, ??0 = 1 2000 ??0 = 5.0 × 10−4 = 0.0005 Width for best cross section, ?? : ?? + 2???? = 2??√1 + ?? 2 ?? + 2(2.0)?? = 2??√1 + (2.0) 2 ?? + 4?? = 4.472?? ?? = 4.472?? − 4?? ?? = 0.472?? ???????? ?????????????????? ???? ?????? ?????????????? ?????????? ???????? ???????????????? ∶ ?? = ???? + ???? 2 48.0 = [0.472??(??) + 2(??) 2 ] 48.0 = 2.472 ?? 2 ?? = ( 48 2.472) 1 2 ?? = ??. ?????? ?? Width, ?? : ?? = 0.472?? ?? = 0.472 × 4.407 ?? = ??. ???? ?? Hydraulic Radius for best cross section, ?? : ?? = ?? ?? = 0.5?? ?? = 0.5 × (4.407) ?? = 2.203 ?? ?????? ?????????????????? ?????? ?????? ???????? ???????????????????? ?????????????? ?????????? ?????????????? ???????????????? ∶ ?? = ?? × ?? 2 3 × ??0 1 2 ?? ?? = (48) (2.203) 2 3 (0.0005) 2 3 0.017 ?? = ??????. ?????????? ?? ⁄
04: NON-UNIFORM OPEN CHANNEL
59 | 0 4 N o n - U n i f o r m O p e n C h a n n e l 04 : NON-UNIFORM OPEN CHANNEL 1. With the aid of a diagram, describe the definition of specific energy. [Session 1: 2022/2023 – Sec. B: Q3(a)] Answer: The specific energy also can be defined as the total energy, is measured relative to the channel bottom and is composed of potential energy due to the depth of the fluid plus kinetic energy due to its velocity (Robert, 2006). The specific energy, Es of a flowing liquid is given as: ???? = ?? + ?? 2 2?? Where: y = Depth of flow (m) V = Average velocity of flow (m/s) g = Gravity acceleration (m2/s) 2. A flow rate of water through an open rectangular channel is 35m3/s. Given that width of channel and depth of water are 12m and 1.2m. Determine : (i) Specific Energy (ii) Froud Number (iii) Types of flow [Session 1: 2022/2023 – Sec. B: Q3(b)] Answer: Flow rate, Q = 35.0 m3/s Gravity acceleration, g = 9.81 m2/s y ?? 2 2?? Z Datum Specific Energy Line Free Water Surface Channel Bottom y =1.2 m b =12.0 m
60 | 0 4 N o n - U n i f o r m O p e n C h a n n e l i. Specific Energy, ES Area, A : ?? = ?? ?? ?? ?? = 12.0 ?? 1.2 ?? = 14.4 ??2 Velocity, V : ?? = ?? ?? ?? ?? = ?? ?? ?? = 35.0 14.4 ?? = 2.431 ??/?? Specific Energy, ???? : ???? = ?? + ?? 2 2?? ???? = 1.2 + 2.3412 2 ?? 9.81 ???? = ??. ?????? ?? ii. Froude Number, ???? ∶ ???? = ?? √(????) ???? = 2.341 √(9.81??1.2) ???? = ??. ?????? iii. Type of flow : Froude Number, Fr = 0.709 < 1.0, therefore flow of water is subcritical flow.
61 | 0 4 N o n - U n i f o r m O p e n C h a n n e l 3. An open rectangular channel with 7m wide flowing water at rate of 21m3/s. Given that Manning’s roughness coefficient is 0.012. Determine : (i) Critical depth (ii) Critical velocity (iii) Minimum specific energy (iv) Hydraulic gradient [Session 1: 2022/2023 – Sec. B: Q3(c)] Answer: Flow rate, Q = 21.0 m3/s Gravity acceleration, g = 9.81 m2/s Manning roughness coefficient, n = 0.012 i. Critical depth, ???? : ???? = √[ ??2 ?? 2?? ] 3 ???? = √[ (21.0) 2 (7.0) 2 ?? (9.81) ] 3 ???? = ??. ?????? ?? ii. Critical velocity, ???? : ???? = √(?????? ) ???? = √(9.81 ?? 0.972) ???? = ??. ?????? ??/?? iii. Minimum specific energy, ???? ?????? : ???? ?????? = 3 2 ???? ???? ?????? = 3 2 ?? 0.972 ???? ?????? = ??. ?????? ?? y b = 7.0 m
62 | 0 4 N o n - U n i f o r m O p e n C h a n n e l iv. Hydraulic gradient, So Area, A : ?? = ?? ?? ???? ?? = 7.0 ?? 0.972 ?? = 6.804 ??2 Wet Perimeter, P : ?? = ?? + 2???? ?? = 7.0 + (2 ?? 0.972) ?? = 8.944 ?? Hydraulic Radius, R : ?? = ?? ?? ?? = 6.804 8.944 ?? = 0.761 ?? Hydraulic Gradient using Manning Equation, ???? : ?? = 1 ?? ?? ?? ?? ?? 2 3 ?? ???? 1 2 21.0 = 1 0.012 ?? 6.804 ?? 0.761 2 3 ?? ???? 1 2 21.0 = 472.614 ?? ???? 1 2 ???? 1 2 = 0.044 ???? = ??. ?????? 4. Describe FIVE (5) uses of hydraulic jump. [Session 1: 2022/2023 – Sec. B: Q4(a)] Answer: 5 Uses of hydraulic jump: i. To dissipate energy ii. For aeration iii. For chemical diffusion iv. To increase flow level v. Reduce uplift pressure
63 | 0 4 N o n - U n i f o r m O p e n C h a n n e l 5. A hydraulic jump occurs in a 3.2m wide rectangular channel, flow depth before the jumps 0.72m. Discharge in the channel is 13.5m3/s. Determine : (i) Flow depth after the jump (ii) Type of jump (iii) Energy loss and power due to jump [Session 1: 2022/2023 – Sec. B: Q4(b)] Answer: Discharge, Q = 13.5 m3/s Density, ρ = 1000 kg/m3 Gravity acceleration, g = 9.81 m2/s i. Flow depth after the jump, y2 Velocity before the jump, ??1 : ?? = ?? ?? ?? ?? = ?? ?? ??1 = ?? ?? ?? ??1 ??1 = 13.5 3.2 ?? 0.72 ??1 = 5.859 ??/?? Froude Number before the jump, ????1 : ????1 = ??1 √(????1 ) ????1 = 5.859 √(9.81??0.72) ????1 = 2.205 y1 = 0.72 m b = 3.2 m
64 | 0 4 N o n - U n i f o r m O p e n C h a n n e l Flow depth after the jump, ??2 : ??2 = ??1 2 [ √1 + 8????1 2 − 1] ??2 = 0.72 2 [ √1 + 8(2.205) 2 − 1] ???? = ??. ?????? ?? ii. Type of hydraulic jump Froude Number, ????1 = 2.205 < 2.5, therefore hydraulic jump is weak jump. iii. Energy loss, EL and power, PL due to jump Energy loss, ???? : ???? = (??2 − ??1) 3 4??1??2 ???? = ( 1.914 − 0.72) 3 4 ?? 0.72 ?? 1.914 ???? = ??. ?????? ?? Power loss, ???? : ???? = ?????????? ???? = 1000 ?? 9.81 ?? 13.5 ?? 0.309 ???? = ???? ??????. ?????? ???????? 6. Water flows at the rate of 1m3/s along a channel of rectangular section of 1.6m width. If a standing wave occurs at a point where upstream depth is 250mm. Calculate: (i) The increase value in water level after the hydraulic jump. (ii) Loss head of water. [Session 1: 2022/2023 – Sec. B: Q4(c)] Answer: Water flows, Q = 1.0 m3/s Density, ρ = 1000 kg/m3 Gravity acceleration, g = 9.81 m2/s y1 = 0.25 m b = 1.6 m
65 | 0 4 N o n - U n i f o r m O p e n C h a n n e l i. The increase value in water level after the hydraulic jump, Δy Velocity before the jump (upstream), V1 : ?? = ?? ?? ?? ?? = ?? ?? ??1 = ?? ?? ?? ??1 ??1 = 1.0 1.6 ?? 0.25 ??1 = 2.5 ??/?? Froude Number before the jump (upstream), Fr1 : ????1 = ??1 √(????1 ) ????1 = 2.5 √(9.81??0.25) ????1 = 1.596 Flow depth after the jump (downstream), y2 : ??2 = ??1 2 [ √1 + 8????1 2 − 1] ??2 = 0.25 2 [ √1 + 8(1.596) 2 − 1] ??2 = 0.453 ?? The increase value in water level after the hydraulic jump, Δy : Δy = ??2 − ??1 Δy = 0.453 − 0.25 ???? = ??. ?????? ?? ii. Loss head of water, EL ???? = (??2 − ??1) 3 4??1??2 ???? = (0.453 − 0.25) 3 4 ?? 0.25 ?? 0.453 ???? = ??. ?????? ??
66 | 0 4 N o n - U n i f o r m O p e n C h a n n e l 7. Explain supercritical flow. [Session 2: 2022/2023 – Sec. B: Q3(a)] Answer: i. Supercritical flow is dominated by inertial forces and behaves as rapid or instable flow. ii. Supercritical flow transitions to subcritical through a hydraulic jump which is represents a high energy loss with erosive potential. iii. When the actual depth is less than critical depth, it is classified as supercritical. iv. Supercritical flow has a Froude Number greater than one. (Fr > 1.0) 8. Water flows at the rate of 7.3m3/s through an open channel of a rectangular section with 3.5m of width. If a wave occurs at a point where the upstream depth is 500mm, determine the height of the hydraulic jump. [Session 2: 2022/2023 – Sec. B: Q3(b)] Answer: Water flows, Q = 7.3 m3/s Density, ρ = 1000 kg/m3 Gravity acceleration, g = 9.81 m2/s Velocity before the jump (upstream), ??1 : ?? = ?? ?? ?? ?? = ?? ?? ??1 = ?? ?? ?? ??1 ??1 = 7.3 3.5 ?? 0.5 ??1 = 4.171 ??/?? y1 = 0.5 m b = 3.5 m
67 | 0 4 N o n - U n i f o r m O p e n C h a n n e l Froude Number before the jump (upstream), ????1 : ????1 = ??1 √(????1 ) ????1 = 4.171 √(9.81??0.5) ????1 = 1.883 Flow depth after the jump (downstream), ??2 : ??2 = ??1 2 [ √1 + 8????1 2 − 1] ??2 = 0.5 2 [ √1 + 8(1.883) 2 − 1] ??2 = 1.105 ?? Height of the hydraulic jump, Δy : Δy = ??2 − ??1 Δy = 1.105 − 0.5 ???? = ??. ?????? ?? 9. Water flowing in an open channel with a flow rate per unit width of 12m3/s/m and an upstream depth of 1.5m. If the flow produces a hydraulic jump, calculate the depth after the jump, velocity after the jump and energy loss. [Session 2: 2022/2023 – Sec. B: Q3(c)] Answer: Flow rate, q = 12.0 m3/s/m Density, ρ = 1000 kg/m3 Gravity acceleration, g = 9.81 m2/s y1 = 1.5 m b
68 | 0 4 N o n - U n i f o r m O p e n C h a n n e l i. The depth after the jump (downstream), y2 Velocity before the jump (upstream), ??1 : ?? = ?? ?? ?? ?? = ?? ?? ??1 = ?? ?? ?? ??1 ; ?? = ?? ?? ??1 = ?? ??1 ??1 = 12.0 1.5 ??1 = 8.0 ??/?? Froude Number before the jump (upstream), ????1 : ????1 = ??1 √(????1 ) ????1 = 8.0 √(9.81??1.5) ????1 = 2.085 Flow depth after the jump (downstream), ??2 : ??2 = ??1 2 [ √1 + 8????1 2 − 1] ??2 = 1.5 2 [ √1 + 8(2.085) 2 − 1] ??2 = 3.736 ?? ii. Velocity after the jump (downstream), ???? ∶ ??2 = ?? ??2 ??2 = 12.0 3.736 ???? = ??. ?????? ??/??
69 | 0 4 N o n - U n i f o r m O p e n C h a n n e l iii. Energy loss, ???? ∶ ???? = (??2 − ??1) 3 4??1??2 ???? = (3.736 − 1.5) 3 4 ?? 1.5 ?? 3.736 ???? = ??. ?????? ?? 10. With an aid of a diagram, explain specific energy in open channel. [Session 2: 2022/2023 – Sec. B: Q4(a)] Answer: The relationship between the energy, discharge and depth of flow in an open channel gives the specific energy in the system (Suleyman & Badronnisa, 2001). The specific energy of a liquid in an open channel is the total energy when the bottom of the channel is used as datum and is expressed as summation of flow depth and velocity head (Hamidah et al., 2023). The specific energy, Es of a flowing liquid is given as: ???? = ?? + ?? 2 2?? Where: y = Depth of flow (m) V = Average velocity of flow (m/s) g = Gravity acceleration (m2/s) y ?? 2 2?? Z Datum Specific Energy Line Free Water Surface Channel Bottom
70 | 0 4 N o n - U n i f o r m O p e n C h a n n e l 11. A rectangular channel carrying supercritical stream is having an energy loss of 0.8m in the jump. Calculate the sequent depth before jump and after jump if the inlet Froude Number is 1.78. [Session 2: 2022/2023 – Sec. B: Q4(b)] Answer: Energy Loss, EL = 0.8 m Inlet Froud Number, Fr1 = 1.78 Flow depth after the jump (downstream), y2 : ??2 = ??1 2 [ √1 + 8????1 2 − 1] ??2 = ??1 2 [ √1 + 8(1.78) 2 − 1] ??2 = 2.066??1 Flow depth before the jump (upstream), y1 : ???? = (??2 − ??1) 3 4??1??2 0.8 = (2.066??1 − ??1) 3 4 ?? ??1?? 2.066??1 0.8 = (1.066??1) 3 8.264??1 2 0.8 = 0.147??1 ???? = ??. ?????? ?? Flow depth after the jump (downstream), y2 : ??2 = 2.066??1 ??2 = 2.066 ?? 5.442 ???? = ????. ?????? ?? y1 y2
71 | 0 4 N o n - U n i f o r m O p e n C h a n n e l 12. A rectangular open channel with a width of 5.0m is carrying water at 11000 liter/s. If the velocity of the water is 2.4m/s, calculate specific energy of the flowing water, critical depth, critical velocity and type of flow. [Session 2: 2022/2023 – Sec. B: Q4(c)] Answer: Velocity, V = 2.4 m/s Gravity acceleration, g = 9.81 m2/s Flow Rate, Q : ?? = 11000 Ɩ/?? ?? = 11000 Ɩ ?? ?? 1 ??3 1000 Ɩ ?? = 11.0 ??3 /s i. Specific Energy, ES Area, A : ?? = ?? ?? ?? ?? = ?? ?? ?? = 11.0 2.4 ?? = 4.583 ??2 Flow depth, y : ?? = ?? ?? ?? ?? = ?? ?? ?? = 4.583 5 ?? = 0.917 ?? y b = 5.0 m
72 | 0 4 N o n - U n i f o r m O p e n C h a n n e l Specific Energy, ???? : ???? = ?? + ?? 2 2?? ???? = 0.917 + 2.4 2 2 ?? 9.81 ???? = ??. ?????? ?? ii. Critical depth, ???? ∶ ???? = √[ ??2 ?? 2?? ] 3 ???? = √[ (11.0) 2 (5.0) 2 ?? (9.81) ] 3 ???? = ??. ???? ?? iii. Critical velocity, ???? : ???? = √(?????? ) ???? = √(9.81 ?? 0.79) ???? = ??. ?????? ??/?? iv. Type of flow : ???? = ?? √(????) ???? = 2.4 √(9.81??0.917) ???? = 0.8 Froude Number, ???? = 0.8 < 1.0, therefore flow of water is subcritical flow.
73 | 0 4 N o n - U n i f o r m O p e n C h a n n e l 13. During a steady flow in an open channel, the flow rate is constant and graph depth, y versus specific energy, Es is plotted as shown in Figure 4.13. Show the position of subcritical flow, super critical flow and critical depth of graph in Figure 4.13 below. [Session 1: 2023/2024 – Sec. B: Q3(a)] Answer: 14. A rectangular channel with width of 250.0cm delivered water with a specific energy head of 1.5m. Determine the flow rate of the water if the flow of water is critical. [Session 1: 2023/2024 – Sec. B: Q3(b)] Answer: Specific energy, Es = 1.5 m Water flow is critical, Fr = 1.0 Figure 4.13 Es y yc Esmin Subcritical flow, Fr < 1.0 Supercritical flow, Fr > 1.0 Critical depth, Fr = 1.0 y b = 2.5 m Es y yc Esmin
74 | 0 4 N o n - U n i f o r m O p e n C h a n n e l Froud Number, ???? : ???? = ?? √(????) 1.0 = ?? √(9.81 ?? ??) ?? = 1.0 ?? √(9.81 ?? ??) ?? 2 = 9.81?? Depth of water flow, ?? : ???? = ?? + ?? 2 2?? 1.5 = ?? + 9.81?? 2 ?? 9.81 1.5 = ?? + 0.5?? ?? = 1.0 ?? Area, A : ?? = ?? ?? ?? ?? = 2.5 ?? 1.0 ?? = 2.5 ??2 Velocity, V : ?? 2 = 9.81?? ?? 2 = 9.81 ?? 1.0 ?? = √9.81 ?? = 3.132 ??/?? Flow rate, Q : ?? = ?? ?? ?? ?? = 2.5 ?? 3.132 ?? = ??. ???? ???? /??
75 | 0 4 N o n - U n i f o r m O p e n C h a n n e l 15. A rectangular channel with a 1.5m wide is to be built to transport water at the flow rate of 10.0m3/s. If the velocity of water through that channel is 2.6m/s, calculate the specific energy, critical depth, critical velocity and minimum specific energy. [Session 1: 2023/2024 – Sec. B: Q3(c)] Answer: Flow rate, Q = 10.0 m3/s Velocity, V = 2.6 m/s Gravity acceleration, g = 9.81 m2/s i. Specific Energy, ES Area, A : ?? = ?? ?? ?? ?? = ?? ?? ?? = 10.0 2.6 ?? = 3.846 ??2 Flow depth, y : ?? = ?? ?? ?? ?? = ?? ?? ?? = 3.846 1.5 ?? = 2.564 ?? Specific Energy, ???? : ???? = ?? + ?? 2 2?? ???? = 2.564 + 2.6 2 2 ?? 9.81 ???? = ??. ?????? ?? y b = 1.5 m
76 | 0 4 N o n - U n i f o r m O p e n C h a n n e l ii. Critical depth, ???? ∶ ???? = √[ ??2 ?? 2?? ] 3 ???? = √[ (10.0) 2 (1.5) 2 ?? (9.81) ] 3 ???? = ??. ?????? ?? iii. Critical velocity, ???? ∶ ???? = √(?????? ) ???? = √(9.81 ?? 1.655) ???? = ??. ?????? ??/?? v. Minimum specific energy, Es min : ???? ?????? = 3 2 ???? ???? ?????? = 3 2 ?? 1.655 ???? ?????? = ??. ?????? ?? 16. With an aid of a diagram, show the phenomena of hydraulics jump which is frequently observed in open channel flow such as rivers and spillways. [Session 1: 2023/2024 – Sec. B: Q4(a)] Answer: the phenomena of hydraulics jump Supercritical flow, Fr>1.0 Subcritical flow, Fr<1.0 Critical Flow, Fr=1.0 y1 y2
77 | 0 4 N o n - U n i f o r m O p e n C h a n n e l 17. A water flow in a 10.0m wide channel at velocity of 1.7m/s and a flow depth of 0.5m. The water then undergoes a hydraulic jump and the flow depth after jump is measured to be 4.0m. Determine the power loss during the jump. [Session 1: 2023/2024 – Sec. B: Q4(b)] Answer: Wide of channel, b = 10.0 m Velocity, V1 = 1.7 m/s Gravity acceleration, g = 9.81 m2/s Flow rate, Q : ?? = ?? ?? ?? ?? = (?? ?? ??1) ?? ??1 ?? = (10.0 ?? 0.5) ?? 1.7 ?? = 8.5 ??3 /?? Energy loss, ???? : ???? = (??2 − ??1) 3 4??1??2 ???? = (4.0 − 0.5) 3 4 ?? 4.0 ?? 0.5 ???? = 5.359 ?? Power loss, ???? : ???? = ?????????? ???? = 1000 ?? 9.81 ?? 8.5 ?? 5.359 ???? = ?????? ??????. ?????? ???????? y1 = 0.5 m y2 = 4.0 m
78 | 0 4 N o n - U n i f o r m O p e n C h a n n e l 18. Water is discharged from a sluice gate into a 3.0m wide rectangular horizontal channel has undergone a hydraulic jump. The flow depth and velocity at upstream before the jump are 1.2m and 9.0m/s respectively. Determine flow depth at downstream, Froude number after jump and height of hydraulic jump. [Session 1: 2023/2024 – Sec. B: Q4(c)] Answer: Wide of channel, b = 3.0 m Velocity before the jump (upstream), V1 = 9.0 m/s Gravity acceleration, g = 9.81 m2/s Froude Number before the jump (upstream), ????1 : ????1 = ??1 √(????1 ) ????1 = 9.0 √(9.81??1.2) ????1 = 2.623 Flow depth after the jump (downstream), ???? : ??2 = ??1 2 [ √1 + 8????1 2 − 1] ??2 = 1.2 2 [ √1 + 8(2.623) 2 − 1] ???? = ??. ?????? ?? Flow rate, Q : ?? = ?? ?? ?? ?? = (?? ?? ??1) ?? ??1 ?? = (3.0 ?? 1.2) ?? 9.0 ?? = 32.4 ??3 /?? y1 = 1.2 m y2 upstream downstream
79 | 0 4 N o n - U n i f o r m O p e n C h a n n e l Velocity after the jump (downstream), ??2 : ?? = ?? ?? ?? ?? = ?? ?? ??2 = ?? ?? ?? ??2 ??2 = 32.4 3.0 ?? 3.892 ??2 = 2.775 ??/?? Froude Number after the jump (downstream), ?????? : ????2 = ??2 √(????2 ) ????2 = 2.775 √(9.81??3.892) ?????? = ??. ?????? Height of the hydraulic jump, ???? : Δy = ??2 − ??1 Δy = 3.892 − 1.2 ???? = ??. ?????? ?? 19. The velocity of flow at critical depth is known as critical flow. Identify types of flow based on the critical depth flow condition with the aid of a diagram. [Session 2: 2023/2024 – Sec. B: Q3(a)] Answer: Supercritical flow, y < yc Subcritical flow, y > yc Critical Flow, y = yc y1 y2 yc
80 | 0 4 N o n - U n i f o r m O p e n C h a n n e l 20. A flow rate of water through an open rectangular channel is 30 m3/s. Given the width of channel and depth of water are 10m and 1m. Determine specific energy, Froude Number and types of flow. [Session 2: 2023/2024 – Sec. B: Q3(b)] Answer: Flow rate, Q = 30.0 m3/s Gravity acceleration, g = 9.81 m2/s i. Specific Energy, ES Area, A : ?? = ?? ?? ?? ?? = 10.0 ?? 1.0 ?? = 10.0 ??2 Velocity, V : ?? = ?? ?? ?? ?? = ?? ?? ?? = 30.0 10.0 ?? = 3.0 ??/?? Specific Energy, ???? : ???? = ?? + ?? 2 2?? ???? = 1.0 + 3.0 2 2 ?? 9.81 ???? = ??. ?????? ?? y =1.0 m b =10.0 m
81 | 0 4 N o n - U n i f o r m O p e n C h a n n e l ii. Froude Number, Fr ???? = ?? √(????) ???? = 3.0 √(9.81??1.0) ???? = ??. ?????? iii. Type of flow Froude Number, Fr = 0.958 < 1.0, therefore flow of water is subcritical flow. 21. An open rectangular channel with 6m wide flowing water at rate of 18m3/s. Given the Manning’s roughness coefficient is 0.012. Determine critical depth, critical velocity, minimum specific energy and hydraulic gradient at critical depth. [Session 2: 2023/2024 – Sec. B: Q3(c)] Answer: Flow rate, Q = 18.0 m3/s Gravity acceleration, g = 9.81 m2/s Manning roughness coefficient, n = 0.012 i. Critical depth, yc ???? = √[ ??2 ?? 2?? ] 3 ???? = √[ (18.0) 2 (6.0) 2 ?? (9.81) ] 3 ???? = ??. ?????? ?? ii. Critical velocity, Vc ???? = √(?????? ) ???? = √(9.81 ?? 0.972) ???? = ??. ?????? ??/?? y b = 6.0 m
82 | 0 4 N o n - U n i f o r m O p e n C h a n n e l iii. Minimum specific energy, Es min ???? ?????? = 3 2 ???? ???? ?????? = 3 2 ?? 0.972 ???? ?????? = ??. ?????? ?? iv. Hydraulic gradient, So Area, A : ?? = ?? ?? ???? ?? = 6.0 ?? 0.972 ?? = 5.832 ??2 Wet Perimeter, P : ?? = ?? + 2???? ?? = 6.0 + (2 ?? 0.972) ?? = 7.944 ?? Hydraulic Radius, R : ?? = ?? ?? ?? = 5.832 7.944 ?? = 0.734 ?? Hydraulic Gradient using Manning Equation, So : ?? = 1 ?? ?? ?? ?? ?? 2 3 ?? ???? 1 2 18.0 = 1 0.012 ?? 5.832 ?? 0.734 2 3 ?? ???? 1 2 18.0 = 395.458 ?? ???? 1 2 ???? 1 2 = 0.046 ???? = ??. ??????
83 | 0 4 N o n - U n i f o r m O p e n C h a n n e l 22. The critical depth and critical velocity are parameters for non-uniform flow. Identify the terms of the two parameters. [Session 2: 2023/2024 – Sec. B: Q4(a)] Answer: i. Critical depth, ???? ∶ The depth of flow corresponding to minimum specific energy is called critical depth and donated by ???? . Thus, the critical depth is given by: ???? = √[ ??2 ?? 2?? ] 3 ii. Critical velocity, ???? : The velocity of flow at critical depth is known as critical flow and donated by Vc. Thus, the critical velocity is given by: ???? = √(?????? ) 23. The hydraulic jump occurs in a 3.1m wide rectangular channel, flow depth before the jump 0.62m. Discharge in the channel is 12.5m3/s. Determine flow depth after the jump, type of jump, energy loss and power due to jump. [Session 2: 2023/2024 – Sec. B: Q4(b)] Answer: Wide of channel, b = 3.1 m Discharge, Q = 12.5 m3/s Gravity acceleration, g = 9.81 m2/s Velocity, ??1 : ?? = ?? ?? ?? ?? = ?? ?? y1 = 0.62 m y2
84 | 0 4 N o n - U n i f o r m O p e n C h a n n e l ??1 = ?? ?? ?? ??1 ??1 = 12.5 3.1 ?? 0.62 ??1 = 6.504 ??/?? Froude Number, Fr : ????1 = ??1 √(????1 ) ????1 = 6.504 √(9.81??0.62) ?????? = ??. ?????? i. Flow depth after the jump (downstream), y2 : ??2 = ??1 2 [ √1 + 8????1 2 − 1] ??2 = 0.62 2 [ √1 + 8(2.637) 2 − 1] ??2 = 2.023 ii. Type of hydraulic jump : ?????? = ??. ???? ∴ The type of hydraulic jump is oscillating jump. iii. Energy loss, EL : ???? = (??2 − ??1) 3 4??1??2 ???? = (2.023 − 0.62) 3 4 ?? 2.023 ?? 0.62 ???? = ??. ???? ?? iv. Power loss, PL : ???? = ?????????? ???? = 1000 ?? 9.81 ?? 12.5 ?? 0.55 ???? = ???? ??????. ???? ????????
85 | 0 4 N o n - U n i f o r m O p e n C h a n n e l 24. Water flow at the rate of 1m3/s along a channel of rectangular section of 1.5m width. If a hydraulic jump occurs at a point where upstream depth is 350mm. Calculate the increase in water level after the hydraulic jump and energy loss of water. [Session 2: 2023/2024 – Sec. B: Q4(c)] Answer: Wide of channel, b = 1.5 m Flow rate, Q = 1.0 m3/s Gravity acceleration, g = 9.81 m2/s Velocity, ??1 : ?? = ?? ?? ?? ?? = ?? ?? ??1 = ?? ?? ?? ??1 ??1 = 1.0 1.5 ?? 0.35 ??1 = 1.905 ??/?? Froude Number, ????1 : ????1 = ??1 √(????1 ) ????1 = 1.905 √(9.81??0.35) ?????? = ??. ?????? Flow depth after the jump (downstream), y2 : ??2 = ??1 2 [ √1 + 8????1 2 − 1] ??2 = 0.35 2 [ √1 + 8(1.028) 2 − 1] ??2 = 0.363 m y1 = 0.35 m y2 = 4.0 m yc
86 | 0 4 N o n - U n i f o r m O p e n C h a n n e l The increase value in water level after the hydraulic jump, ???? : Δy = ??2 − ??1 Δy = 0.363 − 0.35 ???? = ??. ?????? ?? Energy loss, ???? : ???? = (??2 − ??1) 3 4??1??2 ???? = (0.363 − 0.35) 3 4 ?? 0.363 ?? 0.35 ???? = ??. ?????? ?? ????−?? ??
REFERENCE
88 | R e f e r e n c e REFERENCE Amat Sairin Demun (2017). Hydraulics Engineering: Problems and Solutions. University Teknologi Malaysia. Britannica, The Editors of Encyclopedia. Archimedes’ Principle. Encyclopedia Britannica, 2 Jul. 2024, https://www.britannica.com/science/Archimedes-principle. Accessed 4 July 2024. Hamidah Zakaria, Mohd Yuzha Usoff & Nor Aziah Fatma Abdul Ayah @ Abdul Aziz. (2023). A Textbook of Hydraulics. Dungun Terengganu: Politeknik Sultan Mizan Zainal Abidin. Final Examination Session I:2022/2023 DCC50222: Hydraulics. Bahagian Peperiksaan & Penilaian, Jabatan Pendidikan Politeknik & Kolej Komuniti, Kementerian Pendidikan Tinggi Malaysia. Final Examination Session II:2022/2023 DCC50222: Hydraulics. Bahagian Peperiksaan & Penilaian, Jabatan Pendidikan Politeknik & Kolej Komuniti, Kementerian Pendidikan Tinggi Malaysia. Final Examination Session I: 2023/2024 DCC50222: Hydraulics. Bahagian Peperiksaan & Penilaian, Jabatan Pendidikan Politeknik & Kolej Komuniti, Kementerian Pendidikan Tinggi Malaysia. Final Examination Session II:2023/2024 DCC50222: Hydraulics. Bahagian Peperiksaan & Penilaian, Jabatan Pendidikan Politeknik & Kolej Komuniti, Kementerian Pendidikan Tinggi Malaysia. Les Hamill (2011). Understanding Hydraulics 3rd Edition. Palgrave Macmillan. Robert L. M. (2006). Applied Fluid Mechanics (6th ed.). Pearson Prentice Hall. Suleyman A, Muyibi & Badronnisa Yusuf. (2001). Hydraulics for Engineers and Technologists. Universiti Putra Malaysia.
PAST YEAR QUESTION HYDRAULICS for Polytechnic Rahayu Hayat is a Senior Lecturer at the Department of Civil Engineering, Politeknik Sultan Idris Shah since 2022. She was awarded Bachelor of Civil Engineering (1999) from Universiti Teknologi Malaysia (UTM). She tought the Politeknik students almost 22 years. Rozilaili Mustapa is a Senior Lecturer at the Department of Civil Engineering, Politeknik Sultan Idris Shah since 2003. She was awarded Bachelor Sciences of Civil Engineering (2001) from Universiti Teknologi Malaysia (UTM) and a Master of Technical and Vocational Education (2003) from KUiTTHO. She tought Hydraulics to the Politeknik students almost 21 years. Seti Suhadaini Mohammed is a Senior Lecturer at the Department of Civil Engineering, Politeknik Sultan Idris Shah since 2004. She was awarded Bachelor of Civil Engineering (2002) from Universiti Sains Malaysia (USM) and a Master of Technical and Vocational Education (2003) from KUiTTHO. She tought the Politeknik students almost 21 years. Politeknik Sultan Idris Shah Sungai Lang 45100 Sungai Air Tawar Selangor Darul Ehsan No. Tel : 03 32806200 Faks : 03 32806400 Website : http://psis.mypolycc.edu.my