PAST YEAR QUESTION HYDRAULICS for Polytechnic SETI SUHADAINI BINTI MOHAMMED ROZILAILI BINTI MUSTAPA RAHAYU BINTI HAYAT
First Edition, 2024 Seti Suhadaini binti Mohammed / Rozilaili binti Mustapa / Rahayu binti Hayat Department of Civil Engineering, Politeknik Sultan Idris Shah, Sabak Bernam, Selangor
POLITEKNIK SULTAN IDRIS SHAH POLYTECHNIC AND COMMUNITY COLLEGE EDUCATION DEPARTMENT MINISTRY OF HIGHER EDUCATION COPYRIGHT RESERVED Publication 2024 No part of this book may be reproduced in any form and by any means including electronic, mechanical, photocopying, recording and so on without the written permission of the Author and Publisher of Sultan Idris Shah Polytechnic eISBN 978-629-7742-052 Published by; Politeknik Sultan Idris Shah Sg. Lang, 45100 Sg. Air Tawar, Selangor Darul Ehsan No. Tel : 03-3280 6200 No. Fax : 03-3280 6400 Laman web: https://psis.mypolycc.edu.my
i PREFACE This book, Hydraulics Past Year Questions for Polytechnic, is a carefully curated resource designed to aid students in their revision and preparation for final examinations in hydraulic engineering. Understanding the essential concepts of hydraulics is crucial for any aspiring engineer, particularly in polytechnic courses where practical application and problem-solving skills are paramount. This book focuses on providing students with past examination questions to solidify their knowledge and give them a clear idea of what to expect in their assessments. The content covers fundamental topics in hydraulics, including Hydrostatics Forces on Surfaces, Buoyancy and Floatation, Uniform Open Channel Flow, and Non-Uniform Open Channel Flow. Each section presents questions that challenge students to apply core principles to a variety of scenarios, helping them understand the theoretical foundations while gaining practical problem-solving experience. With real questions from previous final exams, this collection serves as a focused tool for effective and targeted revision. Every question in this book is accompanied by detailed solutions, allowing students to self-assess and clarify any misunderstandings. By working through these past questions, students can improve their technical accuracy, build confidence, and hone the analytical skills necessary for tackling complex hydraulic problems. It is our hope that Hydraulics Past Year Questions for Polytechnic will serve as an invaluable resource for students, providing clarity and preparation as they work towards their academic and professional goals in the field of hydraulics. Seti Suhadaini, Rozilaili, Rahayu JKA, PSIS 2024
ii CONTENT Hydraulic Past Year Question for Polytechnic No Topic Page Preface i Content ii Synopsis iii 01 Hydrostatic Forces on Surfaces 1 02 Buoyancy and Floatation 18 03 Uniform Open Channel Flow 32 04 Non-Uniform Open Channel Flow 58 Reference 87
iii SYNOPSIS About This Book Hydraulics Past Year Questions for Polytechnic is an essential study guide designed to support polytechnic students in mastering key concepts and preparing for their hydraulics examinations. This book compiles a selection of past final exam questions, giving students a focused approach to revising and reinforcing their understanding of hydraulics. It is organized around core topics including Hydrostatic Forces on Surfaces, Buoyancy and Floatation, Uniform Open Channel Flow, and Non-Uniform Open Channel Flow. Each topic includes real examination questions with detailed solutions, offering students insights into the types of questions they may encounter, and the methods needed to solve them effectively. With this targeted revision guide, students can develop their problem-solving skills, deepen their knowledge, and build confidence in tackling hydraulics questions. Ideal for students preparing for final exams, Hydraulics Past Year Questions for Polytechnic serves as a practical and reliable resource for achieving academic success in hydraulic engineering.
01: HYDROSTATIC FORCE ON SURFACE
2 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e 01 : HYDROSTATIC FORCE ON SURFACE 1. When a surface is submerged in a fluid, force developed on the surface due to the fluid. Illustrate the position of hydrostatic force (F), centroid of object (G) and center of pressure (P) acting on a plane surface submerged in a fluid. [Session 1: 2022/2023 – Sec. A: Q1(a)] Answer: 2. A triangular plate of 1.66m base and 2.0m height is immersed vertically in liquid as shown in Figure 1.2 with specific weight of 10kN/m3. Calculate the total hydraulic force on the plate (F) and the depth of center of pressure (hp). [Session 1: 2022/2023 – Sec. A: Q1(b)] Answer: i. Total Hydrostatic Force, FR Specific Weight, ɣ : ɣ = ???? ɣ = 10 ????/??3 = 10??103 ??/m3 Figure 1.2 G P F FR hcp hcg 1.66 m 2.0 m
3 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e Depth of center of object, hcg : ℎ???? = 1 3 ?? 2.0 ℎ???? = 0.667 ?? Area for triangular plate, A : ?? = 1 2 ?? ?? ?? ?? ?? = 1 2 ?? 1.66 ?? 2.0 ?? = 1.66 ??2 Total Hydrostatic Force, FR : ???? = ????ℎ?????? ???? = ɣℎ?????? ???? = 10??103 ?? 0.667 ?? 1.66 ???? = ???? ??????. ?? ?? ii. Depth of center of pressure, hcp Moment of Inertia, Ic : ???? = ???? 3 36 ???? = 1.66 ?? 2.0 3 36 ???? = 0.369 ??4 Inclined Angle, θ = 90ᵒ Depth of center of pressure, hcp: ℎ???? = ???? ??????2 ?? ??ℎ???? + ℎ???? ℎ???? = 0.369 ?? ??????2 90˚ 1.66 ?? 0.667 + 0.667 ?????? = ??. ?? ??
4 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e 3. Calculate the horizontal and vertical force exerted by the fluid on the curved vane AB as shown in Figure 1.3. Given the fluid density 800kg/m3, vane length 6.7m and radius 15m. Answer: i. Horizontal Force, FH Density, ρ = 800 kg/m3 Gravitational Acceleration, g = 9.81 m/s2 Length, L = 6.7 m Radius, r = 15.0 m Depth of center of object, hcg : ℎ???? = ?? + ?? 2 ℎ???? = 8.0 + 15.0 2 ℎ???? = 15.5 ?? Area, A : ?? = ?? ?? ?? ?? = 15.0 ?? 6.7 ?? = 100.5 ??2 8.0 m 15.0 m A B [Session 1: 2022/2023 – Sec. A: Q1(c)] Figure 1.3
5 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e Horizontal Force, FH : ???? = ????ℎ?????? ???? = 800 ?? 9.81 ?? 15.5 ?? 100.5 ???? = 12 225 222 ?? ???? = ???? ??????. ?????? ???? ii. Vertical Force, FV Volume, V : = − Volume of Cuboid, V1 : ??1 = ?? ?? ?? ?? ?? ??1 = 15.0 ?? (15.0 + 8.0) ?? 6.7 ??1 = 2311.5 ??3 Volume of Quarter Cylinder, V2 : ??2 = ?? ?? 2 4 ?? ?? ??2 = ?? ?? 15.0 2 4 ?? 6.7 ??2 = 1183.988 ??3 Volume, V : ?? = ??1 − ??2 ?? = 2311.5 − 1183.988 ?? = 1127.512 ??3 Vertical Force, FV : ???? = ?????? ???? = 800 ?? 9.81 ?? 1127.512 ???? = 8 848 714.176 ?? ???? = ?? ??????. ?????? ???? V V1 V2
6 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e 4. Describe the centre of gravity and centre of pressure in hydrostatic. [Session 2: 2022/2023 – Sec. A: Q1(a)] Answer: i. Centre of Gravity, G Centre of gravity, G is defined as the centre of the point where their body acts (Hamidah et al., 2023). It also can be defined as a point which only the gravitational force acts. ii. Centre of Pressure, P Centre of Pressure is defined as the point of application of the resultant force on the surface of body or object (Hamidah et al., 2023). 5. Identify the vertical force exerted by the fluid on the curved vane BC as shown in Figure 1.5. Given the fluid density of 900kg/m3, vane length of 2.0m and radius of 4.0m. Session 2: 2022/2023 – Sec. A: Q1(b)] Answer: Vertical Force, FV Density, ρ = 900 kg/m3 Gravitational Acceleration, g = 9.81 m/s2 Length, L = 2.0 m Radius, r = 4.0 m Height, H = 8.0 m Figure 1.5 4.0 m 8.0 m A B C
7 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e Volume, V : = + Volume of Cuboid, V1 : ??1 = ?? ?? ?? ?? ?? ??1 = 4.0 ?? 8.0 ?? 2.0 ??1 = 64.0 ??3 Volume of Quarter Cylinder, V2 : ??2 = ?? ?? 2 4 ?? ?? ??2 = ?? ?? 4.0 2 4 ?? 2.0 ??2 = 25.133 ??3 Volume, V : ?? = ??1 + ??2 ?? = 64.0 + 25.133 ?? = 89.133 ??3 Vertical Force, FV : ???? = ?????? ???? = 900 ?? 9.81 ?? 89.133 ???? = 786 955.257 ?? ???? = ??????. ?????? ???? V V1 V2
8 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e 6. A triangular plate of 1.0m base and 2.0m height is immersed in liquid, as shown in Figure 1.6, with specific gravity of 0.8. Calculate the total hydrostatic force on the plate (FR) and location of the centre of pressure (hp). [Session 2: 2022/2023 – Sec. A: Q1(c)] Answer: i. Total Hydrostatic Force, FR Density, ρ = 800 kg/m3 Gravitational Acceleration, g = 9.81 m/s2 Centroid of triangular plate, S : ?? = 1 3 ?? 2.0 ?? = 0.667 ?? Depth of centre of object, hcg : ?????? ??˚ = ???????????????? ???????????????????? ?????? 40˚ = ?? ?? ?? = ?? ?? ?????? 40˚ ?? = 0.667 ?? ?????? 40˚ ?? = 0.429 ?? ℎ???? = ?? + ?? ℎ???? = 4.0 + 0.429 ℎ???? = 4.429 ?? Figure 1.6 40˚ S = 0.667 m y x=4.0 m 2.0 m 40˚ 1.0 m S y hcg 4.0 m 2.0 m 40˚ 1.0 m
9 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e Area for triangular plat, A : ?? = 1 2 ?? ?? ?? ?? ?? = 1 2 ?? 1.0 ?? 2.0 ?? = 1.0 ??2 Total Hydrostatic Force, FR : ???? = ????ℎ?????? ???? = 800 ?? 9.81 ?? 4.429 ?? 1.0 ???? = ???? ??????. ?????? ?? ii. Depth of centre of pressure, hcp Moment of Inertia, Ic : ???? = ???? 3 36 ???? = 1.0 ?? 2.0 3 36 ???? = 0.222 ??4 Inclined Angle, θ = 40ᵒ Depth of centre of pressure, hcp: ℎ???? = ???? ??????2 ?? ??ℎ???? + ℎ???? ℎ???? = 0.222 ?? ??????2 40˚ 1.0 ?? 4.429 + 4.429 ?????? = ??. ???? ??
10 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e 7. Centroid mean centre of point where its body weight acts. Using the related diagram, illustrate TWO (2) geometric of the centroid of plate. [Session 1: 2023/2024 – Sec. A: Q1(a)] Answer: i. Centroid for square plate ii. Centroid for rectangular plate iii. Centroid for triangular plate iv. Centroid for circle plate G B ?? 2 G B H ?? 2 B ?? 3 H 2?? 3 D D ?? 2
11 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e 8. A plate as shown in Figure 1.8 with 2.0m diameter submerged vertically into water. Distance from the fluid surface to the centre of gravity is 5.0m. Identify the total pressure on the plate. [Session 1: 2023/2024 – Sec. A: Q1(b)] Answer: Total Hydrostatic Force, FR Density of water, ρ = 1000 kg/m3 Gravitational Acceleration, g = 9.81 m/s2 Depth of centre of object, hcg = 5.0 m Area for circle plate, A : ?? = ?? ?? 2 4 ?? = ?? ?? 2.0 2 4 ?? = 3.142 ??2 Total Hydrostatic Force, FR : ???? = ????ℎ?????? ???? = 1000 ?? 9.81 ?? 5.0 ?? 3.142 ???? = ?????? ??????. ?? ?? Figure 1.8 5.0 m
12 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e 9. Figure 1.9 shows a triangular plate of 1.25m base and 2.75m height was immersed in water. The distance from the upper and edge of the plate to the water surface is 1.0m and 2.0m. Determine the hydrostatic force on the plate and position of pressure. [Session 1: 2023/2024 – Sec. A: Q1(c)] Answer: i. Total Hydrostatic Force, FR Density of water, ρ = 1000 kg/m3 Gravitational Acceleration, g = 9.81 m/s2 Depth of centre of object, hcg : ℎ???? = 1 3 ?? (??2 − ??1 ) + ??1 ℎ???? = 1 3 ?? (2.0 − 1.0) + 1.0 ℎ???? = 1.333 ?? Area for triangular plate, A : ?? = 1 2 ?? ?? ?? ?? ?? = 1 2 ?? 1.25 ?? 2.75 ?? = 1.719 ??2 Total Hydrostatic Force, FR : ???? = ????ℎ?????? ???? = 1000 ?? 9.81 ?? 1.333 ?? 1.719 ???? = ???? ??????. ?????? ?? Figure 1.9 ℎ???? 1.25 m 2.0 m ??˚ 1.0 m 2.75 m 1.25 m 2.0 m Ɵ 1.0 m 2.75 m
13 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e ii. Depth of centre of pressure, hcp Moment of Inertia, Ic : ???? = ???? 3 36 ???? = 1.25 ?? 2.753 36 ???? = 0.722 ??4 Inclined Angle, ??˚ : ?????? ??˚ = ???????????????? ???????????????????? ?????? ??˚ = 1.0 2.75 ?????? ??˚ = 0.364 ??˚ = ??????−1 0.364 ??˚ = 21.324ᵒ Depth of centre of pressure, hcp: ℎ???? = ???? ??????2 ?? ??ℎ???? + ℎ???? ℎ???? = 0.722 ?? ??????2 21.324˚ 1.719 ?? 1.333 + 1.333 ?????? = ??. ?????? ?? 10. In a rectangular tank contains water. When a surface is submerged in water, force action on the surface, bottom and ends of the tank. Illustrate the pressure on the bottom and the ends of the tank. [Session 2: 2023/2024 – Sec. A: Q1(a)] Answer: ??˚ 2.75 m 1.0 m Pressure on tank bottom Pressure on the ends of tank free surface, ρ= 0 Pressure on the ends of tank
14 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e 11. A rectangular plate in Figure 1.11 with a width of 2m and a depth of 3.5m is held vertically in water at 3.35m below the surface of the free water. Identify the hydrostatic force at one surface of the plate and the depth of center pressure. Given the specific weight of water is 9.81 kN/m3. [Session 2: 2023/2024 – Sec. A: Q1(b)] Answer: i. Total Hydrostatic Force, FR Specific Weight, ɣ : ɣ = ???? ɣ = 9.81 ????/??3 = 9.81??103 ??/m3 Depth of centre of object, ℎ???? : ℎ???? = 1 2 ?? 3.5 + 3.35 ℎ???? = 5.1 ?? Area for rectangular plate, A : ?? = ?? ?? ?? ?? = 2.0 ?? 3.5 ?? = 7.0 ??2 Total Hydrostatic Force, ???? : ???? = ????ℎ?????? ???? = ɣℎ?????? ???? = 9.81??103 ?? 5.1 ?? 7 ???? = ?????? ?????? ?? Figure 1.11 2 m 3.35 m 3.5 m water surface
15 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e ii. Depth of centre of pressure, hcp Moment of Inertia, ???? : ???? = ???? 3 12 ???? = 2 ?? 3.5 3 12 ???? = 7.146 ??4 Inclined Angle, ??˚ = 90ᵒ Depth of centre of pressure, ℎ???? : ℎ???? = ???? ??????2 ?? ??ℎ???? + ℎ???? ℎ???? = 7.146 ?? ??????2 90˚ 7.0 ?? 5.1 + 5.1 ?????? = ??. ?? ?? 12. Given the Specific Gravity of fluid is 1.2 to the curved blade as shown in Figure 1.12. The curved blade radius is 2.1 m and 5.2 m long. The upper edge of the curved blade from the fluid surface is 3.8 m. Determine the magnitude and direction of the resultant force. [Session 2: 2023/2024 – Sec. A: Q1(c)] Answer: Horizontal Force, FH Density, ???????????? : ???????????? = ??. ?? ?? ???????????? ???????????? = 1.2 ?? 1000 ???????????? = 1200 kg/m3 Figure 1.12 2.1 m 3.8 m + Curved Blade
16 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e Gravitational Acceleration, g = 9.81 m/s2 Length, L = 5.2 m Radius, r = 2.1 m Depth of centre of object, ℎ???? : ℎ???? = ?? + ?? 2 ℎ???? = 3.8 + 2.1 2 ℎ???? = 4.85 ?? Area, A : ?? = ?? ?? ?? ?? = 2.1 ?? 5.2 ?? = 10.92 ??2 Horizontal Force, ???? : ???? = ????ℎ?????? ???? = 1200 ?? 9.81 ?? 4.85 ?? 10.92 ???? = ?????? ??????. ?????? ?? Vertical Force, FV Volume, V : = + Volume of Cuboid, ??1 : ??1 = ?? ?? ?? ?? ?? ??1 = 2.1 ?? 3.8 ?? 5.2 ??1 = 41.496 ??3 V V1 V2
17 | 0 1 H y d r o s t a t i c F o r c e o n S u r f a c e Volume of Quarter Cylinder, ??2 : ??2 = ?? ?? 2 4 ?? ?? ??2 = ?? ?? 2.1 2 4 ?? 5.2 ??2 = 18.011 ??3 Volume, ?? : ?? = ??1 + ??2 ?? = 41.496 + 18.011 ?? = 59.507 ??3 Vertical Force, ????: ???? = ?????? ???? = 1200 ?? 9.81 ?? 59.507 ???? = ????????????. ?????? ?? Magnitude of Resultant Force, ???? : ???? = √(????) 2 + (????) 2 ???? = √(623 468.664) 2 + (700516.404) 2 ???? = ?????? ??????. ?????? ?? Direction of Resultant Force, ?? : ?????? ?? = ???? ???? ?? = ??????−1 ( ???? ???? ) ?? = ??????−1 ( 700516.404 623 468.664 ) ?? = 48.331ᵒ
02: BUOYANCY AND FLOATATION
19 | 0 2 B u o y a n c y a n d F l o a t a t i o n F M B G F G M B 02 : BUOYANCY AND FLOATATION 1. There are 3 types of equilibrium of floating bodies namely stable, unstable and neutral. Describe TWO (2) of them with a related diagram. [Session 1: 2022/2023 – Sec. A: Q2(a)] Answer: Stable Equilibrium If the point M is above G, the floating body will be in stable equilibrium. MG is positive (+ve). Unstable Equilibrium If the point M is below G, the floating body will be in unstable equilibrium. MG is negative (-ve). Neutral Equilibrium If the point M is at the G, the floating body will be in neutral equilibrium. MG is positive (+ve). W W B W M G
20 | 0 2 B u o y a n c y a n d F l o a t a t i o n 2. A wooden block of length 3.0 m, width 1.25 m, depth 0.75 m and its float-water specific weight is 6.4 kN/m3 Calculate: i. Volume of water displaced (m3), ???? ii. Position of center of Buoyancy (m), OB [Session 1: 2022/2023 – Sec. A: Q2(b)] Answer: Length, L = 3.0 m Width, w = 1.25 m Height, H = 0.75 m Specific Weight, ???? = 6.4 kN/m3 i. Volume of water displaced (m3), ???? : ?????????? = ?????????????? ??????????, ???? ?????????? = ???????????????????? ?????????? ?????????? = ?????????? ???????? = ?????????? 6.4 ?? 103 ?? [3 ?? 1.25 ?? 0.75] = 1000 ?? 9.81 ?? ???? ???? = ??. ?????? ???? ii. Position of center of Buoyancy (m), ???? : ???????????? ???? ?????????? ??????????????????, ???? = ?? ?? ?? ?? ℎ ????????ℎ?? ???? ???????????????? ????????, ℎ: ℎ = ???? ?? ?? ?? ℎ = 1.835 3 ?? 1.25 ℎ = 0.489 ?? ???????????? ???? ????????????????,???? ∶ ???? = ℎ 2 ???? = 0.489 2 ???? = ??. ?????? ??
21 | 0 2 B u o y a n c y a n d F l o a t a t i o n 3. A pontoon has mass of 70 tones and size of 8 m width, 20 m long and 3 m height. The pontoon is in the sea with a density of 1025 kg/m3. Calculate metacentric if the pontoon is loaded with 1000 kg gravel stone. [Session 1: 2022/2023 – Sec. A: Q2(c)] Answer: Length, L = 20.0 m Width, w = 8.0 m Height, H = 3.0 m Density of seawater, ???? = 1025 kg/m3 ???????????????? = 70 ?????????? ???????????????? = 70 x 1000 = 70,000 kg ???????????? ???? ???????????????? ????????,??: ?????????? = ?????????????? ??????????, ???? ????????????????+???????????? = ???????????????????? ?????????? ???? = ?????????? (70,000 + 1,000) ?? 9.81 = 1025 ?? 9.81 ?? (8 ?? 20 ?? ℎ) 696, 510 = 1, 608, 840 ℎ ℎ = 0.433 ?? ???????????????? ?????????????? ?????? ???????????? ???? ?????????????? (????)?????? ?????? ???????????? ???? ???????????????? (????), ????: ???? = ???? − ???? ???? = ?? 2 − ℎ 2 ???? = 3.0 2 − 0.433 2 ???? = 1.284 ?? Moment of inertia, ???? ∶ ???? = ???? 3 12 @ ?? ?? ?? 3 12 ???? = 20.0 ?? 8.0 3 12 ???? = 853.33 ??4 ???????????? ???? ?????????? ??????????????????, ???? ∶ ???? = ?? ?? ?? ?? ℎ ???? = 20.0 ?? 8.0 ?? 0.433 ???? = 69.28 ??3
22 | 0 2 B u o y a n c y a n d F l o a t a t i o n ???????????????? ?????????????? ?????? ???????????? ???? ???????????????? ?????? ?????? ???????? ???????????? ???? ?? ???????????????? ????????, ????: ???? = ???? ???? ???? = 853.33 69.28 ???? = 12.32 ?? ?????????????????????? ????????????, ????: ???? = ???? − ???? ???? = 12.32 − 1.284 ???? = ????. ???????? ?? MG value is positive (+ve), So, the pontoon is in stable condition. 4. Describe the Archimedes principle for buoyancy force using appropriate diagram. [Session 2: 2022/2023 – Sec. A: Q(a)] Answer: Archimedes, stating that anybody completely or partially submerged in a fluid (gas or liquid) at rest is acted upon by an upward, or buoyant, force, the magnitude of which is equal to the weight of the fluid displaced by the body. (Source: https://www.britannica.com/science/Archimedes-principle)
23 | 0 2 B u o y a n c y a n d F l o a t a t i o n 5. A block of wood with specific gravity = 0.7 is partially submerged in water. The dimension of the wood is 50.0 cm x 30.0 cm x 20.0 cm as shown in Figure 2.5. Estimate the height of the block that is above the water. [Session 2: 2022/2023 – Sec. A: Q2(b)] Answer: Length, L = 50.0 cm = 0.5 m Width, w = 30.0 cm = 0.3 m Height, H = 20.0 cm = 0.2 m Specific Gravity, ??. ?? = 0.7 Density, ?????????? : ?????????? = ??. ?? ?? ???????????? ?????????? = 0.7 ?? 1000 ?????????? = 700 kg/m3 Height of block immersed in water, ?? : ?????????? = ?????????????? ??????????, ???? ?????????? ?????????? = ???????????????????? ?????????? ???? ?????? = ???? ?????? 700 ?? 9.81 ?? (0.5 ?? 0.3 ?? ??0.2) = 1000 ?? 9.81 ?? (0.5 ?? 0.3 ?? ℎ) 206.01 = 1471.5 ℎ ℎ = 206.01 1471.5 ℎ = 0.14 ??3 Height of the block that is above the water, ???? ∶ ℎ?? = ?? − ℎ ℎ?? = 0.2 − 0.14 ℎ?? = 0.06 ?? Figure 2.5 20.0 cm 50.0 cm
24 | 0 2 B u o y a n c y a n d F l o a t a t i o n 6. A solid buoy cylinder with a specific gravity of 0.8 is used in seawater with 1.25 specific gravity. The buoy cylinder with a 3.0 m diameter and 3.0 m height is floating upright. Calculate its meta centric height. [Session 2: 2022/2023 – Sec. A: Q2(c)] Answer: Height, H = 3.0 m Diameter, D = 3.0 m Specific Gravity of Cylinder, ??. ?????????? = 0.8 Specific Gravity of Seawater, ??. ???????????? = 1.25 Density of body, ?????????? : ?????????? = ??. ?? ?? ???????????? ?????????? = 0.8 ?? 1000 ?????????? = 800 kg/m3 Density of seawater, ???????????? : ???????????? = ??. ?? ?? ???????????? ???????????? = 1.25 ?? 1000 ???????????? = 1250 kg/m3 ???????????? ???? ????????????????, ???? ∶ ?????????? = ???? 2 4 ?? ?? ?????????? = ??(3.0) 2 4 ?? 3.0 ?????????? = 21.206 ??3 ???????????? ???? ?????????????????? ??????????, ????: ???? = ???? 2 4 ?? ?? ???? = ??(3) 2 4 ?? ℎ ???? = 7.069ℎ
25 | 0 2 B u o y a n c y a n d F l o a t a t i o n ???????????? ???? ???????????????? ????????,??: ?????????? = ?????????????? ??????????, ???? ?????????? = ???????????????????? ?????????? ???? ?????? = ???? ?????? 800 ?? 9.81 ?? 21.206 = 1250 ?? 9.81 ?? 7.069ℎ 166424.688 = 866683.613ℎ ℎ = 166424.688 866683.613 ℎ = 1.920 ?? ???????????????? ?????????????? ?????? ???????????? ???? ?????????????? (????)?????? ?????? ???????????? ???? ???????????????? (????), ????: ???? = ???? − ???? ???? = ?? 2 − ℎ 2 ???? = 3 2 − 1.92 2 ???? = 0.54 ?? Moment of inertia, ???? ∶ ???? = ???? 4 64 ???? = ??(3.0) 4 64 ???? = 3.976 ??4 ???????????????? ?????????????? ?????? ???????????? ???? ???????????????? ?????? ?????? ???????? ???????????? ???? ?? ???????????????? ????????, ????: ???? = ???? ???? ???? = 3.976 13.572 ???? = 0.293 ?? ?????????????????????? ????????????, ????: ???? = ???? − ???? ???? = 0.293 − 0.54 ???? = −??. ?????? ?? MG value is negative (− ????), So, the cylinder is in unstable condition. ???????????? ???? ?????????? ??????????????????, ???? ∶ ???? = 7.069ℎ ???? = 7.069 ?? 1.92 ???? = 13.572 ??3
26 | 0 2 B u o y a n c y a n d F l o a t a t i o n 7. With an aid of diagram, show the distance between centre gravity of body (OG) and the centre of buoyancy (OB) from the bottom of floating body. [Session 1: 2023/2024 – Sec. A: Q2(a)] Answer: 8. A solid cylinder of diameter 2.55 m has a height of 4.5 m. Identity the volume of the cylinder when it is floating in water vertically. The specific gravity of cylinder = 0.6. [Session 1: 2023/2024 – Sec. A: Q2(b)] Answer: Height, H = 4.5 m Diameter, D = 2.55 m Specific Gravity of Cylinder, ??. ?????????? = 0.6 Density of Water, ???? = 1000 ????/??3 ?????????????? ???? ?????????? ????????????????: ?????????? = ??. ?? ?? ???????????? ?????????? = 0.6 ?? 1000 ?????????? = 600 ????/??3 ???????????? ???? ?????????? ????????????????, ????: ?????????? = ???? 2 4 ?? ?? ?????????? = ??(2.55) 2 4 ?? 4.5 ?????????? = 22.982 ??3 W G H h B O Centre gravity of body,???? = ?? 2 Centre of buoyancy,???? = ℎ 2
27 | 0 2 B u o y a n c y a n d F l o a t a t i o n ???????????? ???? ?????????? ??????????????????, ????: ?????????? = ?????????????? ??????????, ???? ?????????? = ???????????????????? ?????????? ?????????? = ?????????? 600 ?? 9.81 ?? 22.982 = 1000 ?? 9.81 ?? ???? ???? = 1352272.052 9810 ???? = ??????. ?????? ???? 9. A wooden block with the specific weight of 7.36kN/m3 floats in seawater. If the size of block is 1.5m(length) x 0.8m (width) x 0.5m height, determine the meta centric height. (Density of water, ρsw = 1025kg/m3) [Session 1: 2023/2024 – Sec. A: Q2(c)] Answer: Length, L = 1.5 m Width, w = 0.8 m Height, H = 0.5 m Specific Weight, ???? = 7.36 ????/??3 Density of seawater, ???????????? = 1025 ????/??3 ???????????? ???? ???????????? ??????????, ????: ???? = ?? ?? ?? ?? ?? ???? = 1.5 ?? 0.8 ?? 0.5 ???? = 0.6 ??3 ???????????? ???? ?????????? ??????????????????, ????: ???? = ?? ?? ?? ?? ℎ ???? = 1.5 ?? 0.8 ?? ℎ ???? = 1.2ℎ
28 | 0 2 B u o y a n c y a n d F l o a t a t i o n ???????????? ???? ???????????????? ????????,??: ?????????? = ?????????????? ??????????, ???? ?????????? = ???????????????????? ?????????? ???? ?????? = ???? ?????? ???????? = ?????????? 7.36 ?? 103?? 0.6 = 1025 ?? 9.81 ?? 1.2ℎ ℎ = 4416 12066.3 ℎ = 0.366 ?? ???????????????? ?????????????? ?????? ???????????? ???? ?????????????? (????)?????? ?????? ???????????? ???? ???????????????? (????), ????: ???? = ???? − ???? ???? = ?? 2 − ℎ 2 ???? = 0.5 2 − 0.366 2 ???? = 0.067 ?? Moment of inertia, ???? ∶ ???? = ???? 3 12 @ ?? ?? ?? 3 12 ???? = 1.5 ?? 0.8 3 12 ???? = 0.064 ??4 ???????????????? ?????????????? ?????? ???????????? ???? ???????????????? ?????? ?????? ???????? ???????????? ???? ?? ???????????????? ????????, ????: ???? = ???? ???? ???? = 0.064 0.439 ???? = 0.146 ?? ?????????????????????? ????????????, ????: ???? = ???? − ???? ???? = 0.146 − 0.067 ???? = ??. ?????? ?? MG value is negative (+ ????), So, the wooden block is in stable condition. ???????????? ???? ?????????? ??????????????????, ???? ∶ ???? = 1.2ℎ ???? = 1.2 ?? 0.366 ???? = 0.439 ??3
29 | 0 2 B u o y a n c y a n d F l o a t a t i o n 10. Stability of floating body occurs when the body undergoes an angular displacement at a horizontal axis. The shape of the immersed volume changes and the center of buoyancy moves relatively with the body. Explain the neutral equilibrium of floating body with the aid of a diagram. [Session 2: 2023/2024 – Sec. A: Q2(a)] Answer: If a body, when given a small angular displacement, occupies a new position and remains at rest in this new position, it is said to possess a neutral equilibrium. For Neutral equilibrium, the position of metacenter M coincides with G. Where: GM = 0, (no rotation moment) therefore, object position would be neutral. 11. A cylinder object with a diameter of 2.0 m and a height of 1.5 m has a weight in the air at 3500 N. The object is immersed in water and has a weight at 2000 N. If the object is in equilibrium, identify the volume of water displaced and the height of the immersed object. [Session 2: 2023/2024 – Sec. A: Q2(b)] Answer: Height, H = 1.5 m Diameter, D = 2.0 m Weight of cylinder in the air, ???? = 3500 ?? Weight of cylinder in the water, ???? = 2000 ?? ???????? ?????????? ???? ??ℎ?? ????????????????, ??: ?? = ????− ???? ?? = 3500 − 2000 ?? = 1500 ?? B W M G
30 | 0 2 B u o y a n c y a n d F l o a t a t i o n ???????????? ???? ?????????? ??????????????????, ????: ?????????? = ?????????????? ??????????, ???? ?????????? = ???????????????????? ?????????? ?? = ???? ?????? 1500 = 1000 ?? 9.81 ?? ???? ???? = 4416 12066.3 = 0.153 ??3 ???????????? ???? ???????????????? ????????,??: ???? = ???? 2 4 ?? ℎ 0.153 = ??(2) 2 4 ?? ℎ ?? = ??. ?????? ?? 12. A wooden cube sized 0.7 m floating in the water as shown in Figure 2.12. Determine the meta centric height and equilibrium of the wooden cube. [Session 2: 2023/2024 – Sec. A: Q2(c)] Answer: Length, L = 0.7 m Width, w = 0.7 m Height of body, H = 0.7 m Height of immersed body, h = 0.4 m Density of water, ???????????? = 1000 ????/??3 Figure 2.12 0.7 m 0.4 m
31 | 0 2 B u o y a n c y a n d F l o a t a t i o n ???????????????? ?????????????? ?????? ???????????? ???? ?????????????? (????)?????? ?????? ???????????? ???? ???????????????? (????), ????: ???? = ???? − ???? ???? = ?? 2 − ℎ 2 ???? = 0.7 2 − 0.4 2 ???? = 0.15 ?? Moment of inertia, ???? ∶ ???? = ?? ?? ?? 3 12 ???? = 0.7 ?? 0.7 3 12 ???? = 0.02 ??4 ???????????????? ?????????????? ?????? ???????????? ???? ???????????????? ?????? ?????? ???????? ???????????? ???? ?? ???????????????? ????????, ????: ???? = ???? ???? ???? = 0.020 0.196 ???? = 0.102 ?? ?????????????????????? ????????????, ????: ???? = ???? − ???? ???? = 0.102 − 0.15 ???? = −??. ?????? ?? MG value is negative (− ????), So, the equilibrium of wooden cube is in unstable condition. ???????????? ???? ?????????? ??????????????????, ???? ∶ ???? = ?? ?? ?? ?? ℎ ???? = 0.7 ?? 0.7 ?? 0.4 ???? = 0.196 ??3
03: UNIFORM OPEN CHANNEL
33 | 0 3 U n i f o r m O p e n C h a n n e l 03 : UNIFORM OPEN CHANNEL 1. Explain uniform flow in open channel. [Session 1: 2022/2023 – Sec. B: Q1(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. A uniform flow may be either steady or unsteady, depending upon whether the discharge varies with time; unsteady uniform flow is rare in practice. (Les Hamill, 2011) 2. A trapezoidal channel carries water at a rate equal to 10 m3/s, Manning’s coefficient, n is 0.01 and bed slope is 0.0035. The side slope of this channel, Z is 2:1. Determine the width (B) of this channel if the allowable velocity of flow is 2.5 m/s. [Session 1: 2022/2023 – Sec. B: Q1(b)] Answer: Flow rate, Q = 10.0 m3/s Velocity. V = 2.5 m/s Manning’s coefficient, n = 0.01 Bed slope, ??0 = 0.0035 Side slope 2:1, so z = ½ = 0.5 Area, ?? : ?? = ?? × ?? ?? = ?? ?? ?? = 10.0 2.5 ?? = 4.0 ??2 Width equation, ?? : ?? = ???? + ???? 2 4.0 = ???? + 0.5?? 2 ?? = 4.0 − 0.5?? 2 ?? 2 1 z = 1/2 y B
34 | 0 3 U n i f o r m O p e n C h a n n e l Wetted Perimeter equation, ?? : ?? = ?? + 2??√1 + ?? 2 ?? = 4.0 − 0.5?? 2 ?? + 2??√1 + 0.5 2 ?? = 4.0 ?? − 0.5?? + 2.24?? ?? = 4.0 ?? + 1.736?? Hydraulic Radius Equation, ?? : ?? = ?? ?? ?? = 4.0 4.0 ?? + 1.736?? Value of depth using Manning Equation, ?? : ?? = ?? × ?? 2 3 × ??0 1 2 ?? 10 = 4.0 [ 4.0 4.0 ?? + 1.736?? ] 2 3 [0.0035] 1 2 0.01 [ 4.0 4.0 ?? + 1.736?? ] 2 3 = 0.4226 [ 4.0 4.0 ?? + 1.736?? ] = [0.4226] 3 2 4.0 = 0.2747 [ 4.0 ?? + 1.736??] 4.0 = 1.0988 ?? + 0.4769?? 4?? = 1.0988 + 0.4769?? 2 0.4769?? 2 − 4?? + 1.0988 = 0 ?? = ??. ?????? ??
35 | 0 3 U n i f o r m O p e n C h a n n e l Value of Width, ?? : ?? = 4.0 − 0.5?? 2 ?? ?? = 4.0 − 0.5(0.284) 2 0.284 ?? = ????. ?????? ?? 3. An open trapezium channel with side slope 1:2, bottom slope 1:4000 and the depth is 1.25 m. Using Manning’s coefficient, n = 0.025, calculate discharge in m3/s as shown in Figure 3.3. [Session 1: 2022/2023 – Sec. B: Q1(c)] Answer: Depth, y = 1.25 m Width, B = 3.0 m Manning’s coefficient, n = 0.025 Bed slope, 1: 4000, ??0 = 0.00025 Side slope 1:2, so z = 2.0 Area, ?? : ?? = ???? + ???? 2 ?? = [3.0(1.25) + 2.0(1.25) 2 ] ?? = 6.875 ??2 Wetted Perimeter, ?? : ?? = ?? + 2??√1 + ?? 2 ?? = 3.0 + 2(1.25)√1 + (2.0) 2 ?? = 8.59 ?? Figure 3.3 1 2 z = 2 y B 2 1 1.25 m 3.0 m
36 | 0 3 U n i f o r m O p e n C h a n n e l Hydraulic Radius, ?? : ?? = ?? ?? ?? = 6.875 8.59 ?? = 0.8 ?? Value of discharge using Manning Equation, ?? : ?? = ?? × ?? 2 3 × ??0 1 2 ?? ?? = (6.875) (0.8) 2 3 [ 1 4000] 1 2 0.025 ?? = ??. ?????? ???? ?? ⁄ 4. Explain hydraulic gradient. [Session 1: 2022/2023 – Sec. B: Q2(a)] Answer: 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 pipes under pressure (Les Hamill, 2011) 5. A cement-lined rectangular channel 6.0 m wide carries water at the rate of 20 m3/s. Calculate the value of Manning’s coefficient, if the slope required to maintain a depth of 1.5 m is 1/625. [Session 1: 2022/2023 – Sec. B: Q2(b)] Answer: Depth, y = 1.5 m Width, B = 6.0 m Discharge, Q = 20.0 m3/s Bed slope, 1: 625, ??0 = 1 625 ??0 = 1.6 × 10−3 Area, ?? : ?? = ???? ?? = [6.0 × 1.5] ?? = 9.0 ??2 B = 6.0 m y = 1.5 m
37 | 0 3 U n i f o r m O p e n C h a n n e l Wetted Perimeter, ?? : ?? = ?? + 2?? ?? = 6.0 + 2(1.5) ?? = 9.0 ?? Hydraulic Radius, ?? : ?? = ?? ?? ?? = 9.0 9.0 ?? = 1.0 ?? Manning Coefficient, ?? : ?? = ?? × ?? 2 3 × ??0 1 2 ?? 20.0 = (9.0) (1.0) 2 3 (1.6 × 10−3 ) 1 2 ?? ?? = 0.36 20.0 ?? = ??. ?????? 6. A trapezoidal channel is required to flow the water of 20 m3/s at the minimum cross section. Determine the best cross section for the channel if the base gradient is 1:1200. Take the side gradient is 2 verticals: 3 horizontal and n = 0.014. [Session 1: 2022/2023 – Sec. B: Q2(c)] Answer: Discharge, Q = 20.0 m3/s Manning’s coefficient, n = 0.014 Side slope 2:3, so z = 3/2 = 1.5 Bed slope, 1: 1200, ??0 = 1 1200 ??0 = 0.00083 2 3 z = 3/2 y B
38 | 0 3 U n i f o r m O p e n C h a n n e l Width for best cross section, ?? : ?? + 2???? = 2??√1 + ?? 2 ?? + 2(1.5)?? = 2??√1 + (1.5) 2 ?? = 3.606?? – 3y ?? = 0.606?? Area, ?? : ?? = ???? + ???? 2 ?? = [0.606??(??) + 1.5(??) 2 ] ?? = 2.106?? 2 Hydraulic Radius for best cross section, ?? : ?? = ?? ?? = 0.5?? Value of depth using Manning Equation, ?? : ?? = ?? × ?? 2 3 × ??0 1 2 ?? 20.0 = (2.106?? 2 ) (0.5??) 2 3 (0.00083) 1 2 0.014 20.0 = 2.736?? 8 3 ?? = [ 20 2.736] 3 8 ?? = ??. ?????? ?? Width, ?? : ?? = 0.606?? ?? = 0.606 × 2.108 ?? = ??. ?????? ??
39 | 0 3 U n i f o r m O p e n C h a n n e l 7. Describe uniform flow and non-uniform flow in an open channel. [Session 2: 2022/2023 – Sec. B: Q1(a)] Answer: Uniform Flow: the velocity, depth and cross-sectional area of flow are constant along the channel. Non-uniform flow: the velocity, depth and cross-sectional area of flow are different section along the channel. (Les Hamill, 2011) 8. Water flows through an open rectangular channel with a width base of 6.0 m and bed slope of 1 in 1000. If the Manning’s coefficient for the channel is 0.013 and a steady flow depth is 400 cm along the channel, calculate the velocity of flow. [Session 2: 2022/2023 – Sec. B: Q1(b)] Answer: Depth, y = 400 cm = 4.0 m Width, B = 6.0 m Manning’s coefficient, n = 0.013 Bed slope, 1: 1000, ??0 = 1 1000 ??0 = 1.0 × 10−3 = 0.001 Area, ?? : ?? = ?? × ?? ?? = [6.0 × 4.0] ?? = 24.0 ??2 Wetted Perimeter, ?? : ?? = ?? + 2?? ?? = 6.0 + 2(4.0) ?? = 14.0 ?? B = 6.0 m y = 4.0 m
40 | 0 3 U n i f o r m O p e n C h a n n e l Hydraulic Radius, ?? : ?? = ?? ?? ?? = 24.0 14.0 ?? = 1.714 ?? Value of velocity using Manning Equation, ?? : ?? = ?? 2 3 × ??0 1 2 ?? ?? = (1.714) 2 3 (0.001) 1 2 0.013 ?? = ??. ?????? ??/?? 9. Water flows uniformly through an open trapezium channel at 1.0 m depth. The width base of the channel is 5.0 m, and the side slope is 1V: 2H on both sides. Calculate the water discharge if a bed slope of channel is 1 in 1000 and the Manning’s coefficient, n = 0.045 [Session 2: 2022/2023 – Sec. B: Q1(b)] Answer: Depth, y = 1.0 m Width, B = 5.0 m Manning’s coefficient, n = 0.045 Side slope 1:2, so z = 2.0 Bed slope, 1: 1000, ??0 = 1 1000 ??0 = 1.0 × 10−3 = 0.001 Area, ?? : ?? = ???? + ???? 2 ?? = [5.0(1.0) + 2.0(1.0) 2 ] ?? = 7.0 ??2 Wetted Perimeter, ?? : ?? = ?? + 2??√1 + ?? 2 ?? = 5.0 + [2(1.0)√1 + (2.0) 2] ?? = 9.472 ?? 1 2 z = 2 y B
41 | 0 3 U n i f o r m O p e n C h a n n e l Hydraulic Radius, ?? : ?? = ?? ?? ?? = 7.0 9.472 ?? = 0.739 ?? Value of discharge using Manning Equation, ?? : ?? = ?? × ?? 2 3 × ??0 1 2 ?? ?? = (7.0)(0.739) 2 3 [0.001] 1 2 0.045 ?? = ??. ?????? ???? ?? ⁄ 10. Describe the wetted perimeter and bed slope. [Session 2: 2022/2023 – Sec. B: Q2(a)] Answer: Wetted perimeter, P is the length of channel perimeter that is wetted or covered by flowing water. Bed slope, ?? @ ???? is the slope at the bottom or base of the channel. Also known as hydraulic gradient. 11. Water flows through half of circular channel with a diameter of 1.5 m. Calculate the bed slope required if the flow rate of water is 0.83 m3/s and the Manning’s coefficient of roughness is 0.01. [Session 2: 2022/2023 – Sec. B: Q2(b)] Answer: Diameter, D = 1.5 m Discharge, Q = 0.83 m3/s Manning’s coefficient, n = 0.01 Depth, ??: ?? = ?? 2 ?? = 1.5 2 ?? = 0.75 ??
42 | 0 3 U n i f o r m O p e n C h a n n e l Area, ?? : ?? = 1 2 × ???? 2 4 ?? = 1 2 × ??(1.5) 2 4 ?? = 0.884 ??2 Wetted Perimeter, ?? : ?? = ???? 2 ?? = ??(1.5) 2 ?? = 2.356 ?? Hydraulic Radius, ?? : ?? = ?? ?? ?? = 0.884 2.356 ?? = 0.375 ?? Value of bed slope using Manning Equation, ???? : ?? = ?? × ?? 2 3 × ??0 1 2 ?? 0.83 = (0.884)(0.375) 2 3 (??0 ) 1 2 0.01 0.83 = 45.970 ??0 1 2 ??0 1 2 = 0.83 45.970 ??0 = 0.0182 ???? = ??. ???????????? = ??. ???? × ????−??
43 | 0 3 U n i f o r m O p e n C h a n n e l 12. A rectangular glazed brick channel has to carry a flow rate of 0.42 m3/s of water. If the bed slope is 0.0005 and the Manning’s coefficient is 0.013, calculate the most effective crosssection channel. [Session 2: 2022/2023 – Sec. B: Q2(c)] Answer: Discharge, Q = 0.42 m3/s Manning’s coefficient, n = 0.013 Bed slope, ??0 = 0.0005 Width for most effective cross section, ?? : ?? = 2?? Area, ?? : ?? = ???? ?? = 2?? × ?? = 2?? 2 Hydraulic Radius for most effective cross section, ?? : ?? = ?? ?? = 0.5?? ???????? ?????????????????? ???? ?????? ?????????????? ?????????? ?????????????? ???????????????? ∶ ?? = ?? × ?? 2 3 × ??0 1 2 ?? 0.42 = (2?? 2 ) (0.5??) 2 3 [0.0005] 1 2 0.013 5.46 × 10−3 = (2?? 2 ) × 0.63?? 2 3 × 0.22 5.46 × 10−3 = 0.028?? 8 3 ?? = (0.195) 3 8 ?? = ??. ?????? ?? Width, ?? : ?? = 2?? ?? = 2 × 0.541 ?? = ??. ?????? ?? B = 2 y y