hydraulics quicknotes Are You ready? - n u r u l zamri na ahmadsecond edition :2023
Published by: Politeknik Sultan Azlan Shah Behrang Stesyen, Behrang 35950 Perak. Tel: 05-4544431 Faks: 05-4544993 Email: http://www.psas.edu.my First published 2022 Second published 2023 All rights reserved. No part of this publication may be reproduced stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without permission of Sultan Azlan Shah Polytechnics. PERPUSTAKAAN NEGARA MALAYSIA Hydraulics Quick Notes
SYNOPSIS HYDRAULICS QUICK NOTES is specially written for Civil Engineering Diploma students taking hydraulics course at Polytechnic Malaysia. It contains all the topics stipulated in revised Hydraulics DCC50222 syllabus. Hydraulics covers the application in hydrostatic and hydrodynamic fluids. It involves discussion on hydrostatic concept and basic equations of stability and buoyancy. It also emphasizes on the application of constituents of pump and open channel flow concept appropriately in solving hydraulic problem. All hydraulics concepts are presented clearly in simple way for easy understanding. it is hoped that this book will help Civil Engineering Diploma students master the basic concept of hydraulics as well as prepare them to work in the field of civil engineering. Nurulzamrina binti Ahmad
Nurulzamrina binti Ahmad is a lecturer of Civil Engineering Department at Politeknik Sultan Azlan Shah. She graduated from Kolej Universiti Tun Hussein Onn (KUiTTHO) in 2005 with a Bachelor of Civil Engineering with honour and Degree of Master of Technical and Vocational Education from Universiti Tun Hussein Onn Malaysia (UTHM) in 2007. She has more than 15 years of teaching experience. Hello! I am Nurulzamrina binti Ahmad I'll be your hydraulics lecturer. I love teaching because I got to help my students understand the world and each other. And I'd love to get to know you all my lovely students, can you introduce yourselves. I hope that this e-book will help Civil Engineering Diploma students master the basic concept of hydraulics as well as prepare them to work in the filed of civil engineering.
1. Hydrostatic Force 2. Buoyancy & Floatation 3. Pump 4. Uniform Open Channel 5. Non-uniform Open Channel -Mulakan sesuatu dengan niat yang baikTable of content DCC50222 HYDRAULICS Pages .....1 .....14 .....24 .....36 .....49
CHAPTER 1 HYDROSTATIC FORCE PLO 1 : apply knowledge of applied mathematics, applied science, engineering fundamentals and an engineeringspecialization as specified in DK1 to DK4 respectively to wide practical procedures and practices PLO 2 : identify and analyze well- defined engineering problems reaching substantiated conclusions using codified method of analysis specific to their field of activity (DK1 to DK4) 1
F=PxA 2. Pressure increases in direct proportion to the depth of water. The formula of hydrostatic forces (F) : 4. Pressure at any point in the water acts in all directions at the same magnitude. 3. Pressure in a continuous volume of water is the same at all points that are at the same depth or elevation. principles of hydrostatic pressure: 1. Pressure depends only on the depth of water above the point in question (not on the water surface area). Force = pressure x area 2
Force, F = PxA Pressure, P = Force/Area P = F/A P = pgh unit: P : N/m @ Pascal (Pa) F : Newton (N) A : m p : kg/m (density) g : m/s (gravity) h : m (height) 2 3 2 2 3
Hydrostatic forces acting on submerged surfaces: a. VERTICAL PLANE b. INCLINED SURFACE c. CURVED SURFACE Hydrostaticforces Important Information: 4
Recall back!! These are the typical examples for the area moment of inertia. 1 2 3 How to find the centroid of simple composite shapes? 5
A . VERT I C A L Resultant force Center of pressure height hp = center of pressure height from fluid surface hc = centroid height from fluid surface A = active area of object submerged plane surfaces 6
B . I N C L I N E D S U R F A C E 7
Resultant Force, Centre of pressure height, 1. 2. angled triangular gate? how to analyze? hc? A? angle? 8
CC.. CCUURRVVEEDD SSUURRFFAACCEE Fv 1. F 2. 3. 4. H 9
Formula Match Direction of FH resultant force Force on the fluid element due to the weight of water above curve Force on the fluid element due to horizontal hydrostatic forces on curve V Volume of the water above curve Centroid height from fluid surface to curve FRThe force that counters all other forces 10
example 1.1 1. 2. 11
example 1.2 1. 2. 12 12
example 1.3 13
Buoyancy & Floatation Chapter 2 PLO 1 : apply knowledge of applied mathematics, applied science, engineering fundamentals and an engineering specialization as specified in DK1to DK4 respectively to wide practical procedures and practices PLO 2 : identify and analyze well- defined engineering problems reaching substantiated conclusions using codified method of analysis specific to their field of activity (DK1to DK4) "Teachers can open the door, but you must enter it yourself" Titanic 14
What do fluids do? A fluid is any substance that can flow and alter its shape. Fluids Why do some objects sink and others floats? Pebbles sink Newton's Law Buoyancy Upward force exerted by a fluid which acts on a fully submerged or floating body. Density Mass per unit volume. Cruise ships float 15
ARCHIMEDES' PRINCIPLE a submerged object will experience a buoyant force equal to the weight of the displaced fluid. an object will float if it is less dense than the fluid. steel is dense but air is not. When an object is completely or partially immersed in a fluid, the fluid exerts an upward force on the object equal to the weight of the fluid displaced by the object When object of 5kg is gradually immersed in the water, the weight of an object decreased to 3kg. Subtracting these two scales readings, 5kg and 3kg, results in differences is 2kg. This is called buoyant force that exerted on the object and it same as the weight of the water of an object displaced. It is important to note that the buoyant force does not depend on the weight or shape of the submerged object, only on the weight of the displaced fluid. 16
floating - sinking a floating object displaces fluid based on its mass a sinking object displaces fluid based on its volume the volume of the water increases by precisely the volume of the person the volume of the water increases by precisely the volume of the steel bar Floating object Sinking object Weight of water displaced = Weight of object 17
1.Stable equilibrium STABILITY OF FLOATING BODIES 2.Unstable equilibrium If a body floating in a liquid does not return back to its original position and heels farther away when given a small angular displacement, then the body is said to be in unstable equilibrium. W and F give an OVERTURNING MOMENT that rotates the body even further in the tilted direction. Metacentre, M, lies below the centre of gravity, G, then the body is unstable. The metacentric height, GM, is negative (GM < 0). If a body floating in a liquid returns back to its original position, when given a small angular displacement, then the body is said to be in stable equilibrium. W and F give a RESTORING MOMENT that rotates the body back to its untilted position. Metacentre, M, lies above the centre of gravity, G, then the body is stable. The metacentric height, GM, is positive (GM = z - z > 0). Floating object is acted on by the forces of gravity (G) and forces of buoyancy (B). When the body is tilted the centre of buoyancy moves to a new position, B', because the shape of the displaced volume changes. A new point, M, called the METACENTRE. This is the point where a vertical line drawn upwards from the new centre of buoyancy, B', of the tilted body intersects the line of symmetry of the body. The buoyancy force, FB , now acts through B'. B B 3.Neutral equilibrium If a body floating in a liquid occupies a new position and remains at rest in this new position, when given a small angular displacement, then the body is said to be in neutral equilibrium. The metacentric height, GM, is zero (G=M). M G 18
METACENTRE It is defined as the point about which a body starts oscillating when the body is tiled by a small angle. It is the point at which the line of action of the force of buoyancy will meet the normal axis of the body when the body is given small angular displacement. Metacentric height: Is the distance between the centre of gravity of floating body and the metacentre. LEARN SOMETHING NEW EVERYDAY Differences of stability between... 1.floating bodies 2.fully submerged bodies When the body undergoes an angular displacement about a horizontal axis, the shape of the immersed volume changes and so the centre of buoyancy moves relative to the body. As a result of above observation stable equilibrium can be achieved, under certain condition, even when G is above B Consider a submerge body in equilibrium whose centre of gravity is located below the centre of buoyancy. If the body is tilted slightly in any direction, the buoyant force and the weight always produce a restoring couple trying to return the body to its original position. 19
STEP TO CALCULATE GM d? H? G? O? B? M? H = height of object d = depth of immersion Ixx = second moment of area Vd = volume of displace liquid Calculate the dept of immersion, d Calculate the height of Metacentre, GM 1. 2. GM = MB - BG MB = Ixx/Vd BG = OG - OB BG = (H/2) - (d/2) OG = H/2 , OB = d/2 GM = + (positive) [type of equilibrium is stable] GM = - (negative) [type of equilibrium is unstable] GM = 0 [type of equilibrium is neutral] "action is the fundamental key to success” . o 20
A rectangular wooden block 100 cm x 50 cm x 40 cm is left floating in the water. If it is observed that a quarter of the volume of the block of wood lies above the water surface, calculate the mass of the wooden block. Idea Name: Date: example 2.1 "success is a process, not an event" Question Solution Solution 1.Draw the object mass,m = p x g x Vd g m = 1000 x (1x0.5x0.3) m = 150 kg Question Explain the differences between centre of pressure, centre of gravity and centre of buoyancy. point of application of hydrostatic force point of application of the weight of the body point of application of the weight of displace liquid 21
A block of wood 4 m long, 2 m wide and 1 m deep is floating horizontally in the water. If the density of wood is 700 kg/m3, calculate the: a) mass of wooden block b) volume of water displaced c) depth of immersion d) position of center of buoyancy Name: Date: example 2.2 "success doesn't come to you, you go to it" Question Solution Solution Solution 1.Draw the object Density, density of object = mass of object volume of object given; density of wooden block = 700kg/m Vo = 4x2x1 = 8m therefore; a) mass of wooden block = 700 x 8 (density x volume) = 5600kg b) volume of water displaced = mass of wooden block density of water = 5600/1000 = 5.6m c) depth of immersion, d volume of water displaced = b x l x d 5.6 = 4 x 2 x d d = 0.7m d) centre of buoyancy, OB OB = d / 2 = 0.7 / 2 = 0.35m 3 3 3 22
Determine the metacentric height of a vehicle to transport pontoons that across a strait of sea water with a density of 1150 kg/m3. The pontoon measuring is 27 m in long, 19 m in wide and 9 m in high and the weight of the pontoon is 500 tones. Name: Date: example 2.3 "success is the sum of small efforts, repeated day in and day out " Question Solution Solution Solution 1.Draw the object mass of object, mo = 500 x 1000 = 500000kg 1150 x 9.81 x (27x19xd) = 500000 x 9.81 d = 0.85m Ixx = bd3/12 = 15432.75m Vd = 27x19x0.85 = 436.05m BM = Ixx/Vd = 15432.75 / 436.05 = 35.392m BG = OG - OB = (9/2) - (0.85/2) = 4.075m GM = MB - BG = 35.392 - 4.075 = 31.317m (stable equlibrium) 3 4 23
Pump Today a reader, tomorrow a leader PLO 1 : apply knowledge of appl ied mathematics, appl ied science, engineering fundamentals and an engineering special ization as specified in DK1 to DK4 respectively to wide practical procedures and practices PLO 2 : identify and analyze wel l- defined engineering problems reaching substantiated conclusions using codified method of analysis specific to their field of activity (DK1 to DK4) 24
Introduction Pump: The hydraulic machines – convert to the mechanical energy into hydraulic energy. The hydraulic energy is in the form of pressure. Centrifugal pump A centrifugal pump is a rotating machine in which flow and pressure are generated dynamically. The energy changes occur by virtue of two main parts of the pump, the impeller and the volute or casing. The function of the casing is to collect the liquid discharged by the impeller and to convert some of the kinetic (velocity) energy into pressure energy. Dynamic (Centrifugal) Pumps – Energy is continuously added to increase the fluid velocities within the machine. Positive Displacement Pumps – Energy is periodically added by application of force to one or more movable boundaries of enclosed, fluidcontaining volumes. 1. 2. Classification of Pumps The main components of a centrifugal pump is: a. Impeller (with blades) b. Casing (to keep the impeller) c. Suction pipe (reservoir-pump) d. Delivery pipe (pump-delivery point) e. Valve (to control) f. Motor (to rotate the impeller shaft) Axial Flow: ▸The flow through impeller is parallel to shaft axis low head and very high discharge. The axial flow type pumps are used for the application of medium head and high discharge. Radial Flow: The impeller discharges fluid at right angles to the shaft axis. In this centrifugal pump in which the pressure is developed wholly by centrifugal force. The radial type pumps are used for the application of high head and low discharge. Mixed Flow: ▸The flow direction is partly axial and partly radial. Hence has a result the flow is diagonal. The mixed flow type pumps are used for the application of medium head and high discharge. 25
In a reciprocating pump, a piston or plunger moves up and down. During the suction stroke, the pump cylinder fills with fresh liquid, and the discharge stroke displaces it through a check valve into the discharge line. Reciprocating pumps can develop very high pressures. Plunger, piston and diaphragm pumps are under these type of pumps. RReecciipprrooccaattiinngg ppuummppss Comparison between Centrifugal Pumps & Reciprocating Pumps. The discharge is continuous and smooth. It can handle large quantity of liquid. It can be used for lifting highly viscous liquids. It is used for large discharge through smaller heads. Centrifugal Pumps The discharge is fluctuating and pulsating. It handle small quantity fluid only. It is used only for lifting pure water or less viscous liquids. It is meant for small discharge and high heads. Reciprocating Pumps 26
axial flow radial flow mixed flow Impeller vane shapes Centrifugal Pump the roots of education are biter, but the fruit is sweet Various types of impeller vane shapes are used in centrifugal pump engineering. They include the conventional axial, mixed flow and radial vane forms. enclosed vane semi-open vane open vane 27
Head & Efficienccy System head, Hsystem Static head, Hs Efficiency, n Suction head, hs Delivery head, hd The sum of suction head and delivery head is known as static head. Hs = hs + hd The vertical height of the centre line of the centrifugal pump above the water surface in the tank or pump from which water is to be lifted. This height is also called suction lift. The vertical distance between the centre line of the pump and the water surface in the tank to which water is delivered is known as delivery head. is defined as the head against which a centrifugal pump has work. It is given by the following expression: hL = friction head loss in suction and delivery pipe f = frictional coefficient l = pipe length Q = flowrate d = diameter of pipe 28
Pumps System Characteristic Graph Operational points Optimal Point Pumps System Pumps installed in series to provide equal pressure between two points. The distance between the two pumps is usually limited to a few meters. Pumps in Serial - Head Added Pumps in Parallel - Flow Rate Added The pump is installed in two or more of the resulting flow will be routed to one channel, it is considered in parallel. ▸Q series = QA = QB ▸H series = HA + HB ▸P series = PA + PB ▸The efficiency: ▸H Parallel = HA = HB ▸Q Parallel = QA + QB ▸P Parallel = PA + PB ▸The efficiency: 29
CAVITATION THANK YOU! “There is no elevator to success. You have to take the stairs.” If the incoming liquid is at a pressure with insufficient margin above its vapour pressure, then vapour cavities or bubbles appear along the impeller vanes just behind the inlet edges. Liquid entering the impeller eye turns and is split into separate streams by the leading edges of the impeller vanes, an action which locally drops the pressure below that in the inlet pipe to the pump. cavitation has three undesirable effects! 1) The collapsing cavitation bubbles can erode the vane surface, especially when pumping water-based liquids. 2) Noise and vibration are increased, with possible shortened seal and bearing life. 3) The cavity areas will initially partially choke the impeller passages and reduce the pump performance. In extreme cases, total loss of pump developed head occurs. 30
I N S P I R A T I O N centrifugal pump cavitation 31
Example 3.1 A centrifugal pump has a performance data as shown below, when it was operating at a required speed of 1800 revolution/second. Determine the head, flow rate and efficiency at a optimum point. Solution Question Graph of head and efficiency vs Q 32
Example 3.2 Based on the data given below, determine the head and efficiency from the graph drawn. Solution Question Efficiency Graph of head and efficiency vs Q Q series = QA = QB H series = HA + HB 33
Table below shows the data of a centrifugal pump working at a speed of 720 rpm. This pump is used to deliver water 16m high. The total length of pipeline is 2600m and using 500mm diameter pipe. By assuming there is no minor energy loss in the pipeline and coefficient friction, ƒ is 0.0025. a .Plot the graphs of pump and system characteristics b. Determine the flow rate, efficiency and head developed for this pump from the graph c .Find the power of pump at the operating point d If a similar pump is fitted in series with the existing pump, get a flow rate and power required for this new system Example 3.3 Solution Question 34 36 36 32 28 23 10 From the question, Hs = 16m f = 0.0025 l = 2600m d = 0.5m 34
b) from the graph, at operating point then value are: efficiency = 62% head = 18m Q = 0.2m3/s c) Power output, Po = pghQ = 1000 x 9,81 x 18 x 0.2 = 35.316kW d) new system (2H-fitted in series) efficiency = 83% head = 25.5m Q = 0.38m3/s 35
uniform open channel 36
Introduction Flow in open channels is defined as the flow of a liquid with a free surface. A free surface is a surface having constant pressure such as atmospheric pressure. 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. The flow of water, in an open channel, is not due to any pressure as in the case of pipe flow. But it is due to the slope of the bed of the channel. (referred as free-surface flow or gravity flow) Note: Program Learning Outcome (PLO) PLO 1 : apply knowledge of applied mathematics, applied science, engineering fundamentals and an engineering specialization as specified in DK1 to DK4 respectively to wide practical procedures and practices PLO 2 : identify and analyze well- defined engineering problems reaching substantiated conclusions using codified method of analysis specific to their field of activity (DK1 to DK4) PLO 9 : function effectively as an individual, and as a member in diverse technical teams. Open Channel 37
two piezometers are placed at section 1 and 2. the water level maintained by the pressure in the pipe at elevations represented by the hydraulics grade line or hydraulic gradient. the pressure exerted by the water in each section of pipe is shown in the tube by the height y of a column of water above the centre line of the pipe. the total energy of the flow of the section is the sum of elevation z of the centre line, the piezometric head y and the velocity head the energy is represented in figure as the energy grade line or energy gradient. the loss of energy represented by physical condition-is usually round. flow driven by-pressure work. Pipe flow: The flow of water in a conduit may be either open channel flow or pipe flow. the two kinds of are similar in many ways but differ in one important respect. Open channel flow must have a free surface, whereas pipe flow has none. A free surface is subject to atmospheric pressure. in pipe flow there exist no direct atmospheric flow but hydraulic pressure only. Differences between pipe flow and open channel flow Pipe flow Open channel flow this is simplified by assuming parallel flow with a uniform velocity distribution and that the slope of channel is small. flow conditions are complicated by the position of the free surface which will change with time and space. And also by the fact that depth of flow, the discharge, and the slopes of the channel bottom and the free surface are all dependent. physical condition-it can be any shape. flow driven by-gravity (potential energy) Open channel flow: 38
Type of Channel irregular shape nor prismatic i.e: river, hillsides, rivulets, tidal estuaries materials of construction can vary widely (mainly of earth) the surface roughness will often change with time distance and even elevation. Natural Channel regular shape (man made - irregular cross-section and prismatic channel) i.e: drains, culverts, sewer, tunnels, irrigation canals, navigation canals, spillways. Artificial Channel rectangular parabolic triangular trapezoidal circular 39
irrigation type of channel navigation sewer hillside tidal estuaries spillway 40
Uniform Flow And Non-Uniform Flow While traveling along a river, you may notice a sequence of pools and rapids. These are the areas of different flow conditions. To classify types of flow, we examine two flow conditions: the uniformity of the flow within the stream and the steadiness of the flow over time. Flow conditions often change within the same river. Certain sections of the river, or reaches, often transition from one type of flow condition to another and back again. UNIFORM FLOW: The water depth, flow area, discharge and the velocity distribution at all section throughout unchanged. The energy gradient line, the water surface line and the channel bottom line must be parallel to each other. The slope of these parallel lines are the same. Uniform flow in an open channel must satisfy the following conditions: 1. 2. 3. NON-UNIFORM FLOW: Non- Uniform flow for reaches of channel where energy line , water surface and bed slope are not parallel. If at a given instant, the velocity is not the same at every point, the flow is non-uniform. This can be caused by: 1) differences in depth of channel 2) differences in width of channel 3) differences in the nature of bed 4) differences in slope of channel 5) obstruction in the direction of flow 41
Hydraulic Gradient, Wet Perimeter and Hydraulic Radius 1 Hydraulic radius is the term used to describe the shape of a channel. It is the ratio between the length of the wetted perimeter and the cross-section area. The higher the hydraulic radius the lower the amount of water in contact with the bed and banks which means there is less friction and water can move at a higher velocity. 2 Hydraulic gradient is defined as the hydraulic head across the soil divided by the length of the flow path through the soil. 3 Top width, T, is the width of the channel section at the free surface. Water Area, A, is the cross-sectional area of the flow normal to the direction of flow. Wetted Perimeter, P Hydraulic Radius, R = A/P Hydraulic Depth, D = A/T The difference in the wetted perimeter of a river between the upper and lower course of a river. 4 Which river is the most efficient? This river has a higher wetted perimeter in comparison to its volume, which increases friction and reduces velocity (upper course) This river has a smaller wetted perimeter in comparison with its volume, because it has smooth banks, friction is reduced and this allows velocity to increase (middle/lower course) 42
DETERMINE THE HYDRAULIC RADIUS Calculate the hydraulic radius Stream A and Stream B Stream ___________ is the most efficient because the value of hydraulic radius higher than Stream ___________ HYDRAULIC RADIUS = 1.87 Find the hydraulic radius, R for figure below. Answer: Area, A = 24 m2 Wetted perimeter, P = 15.7 m Hydraulic radius, R = 1.53 m it always seems impossible until it is done Open channel flow section geometries 43
This equation was developed in 1889 by the Irish engineer, Robert Manning. In addition to being empirical, the Manning Equation is a dimensional equation, so the units must be specified for a given constant in the equation. Manning's equation is a widely used empirical equation that relates several uniform open channel flow parameters. For commonly used units the Manning Equation and the units for its parameters are as follows: Channel Surface Coefficient, n Asbestos cement 0.011 Brass 0.011 Brick 0.015 Cast-iron, new 0.012 Concrete, steel forms 0.011 Concrete, wooden forms 0.015 Concrete, centrifugally spun 0.013 Copper 0.011 Corrugated metal 0.022 Galvanized iron 0.016 Lead 0.011 Plastic 0.009 Steel - Coal-tar enamel 0.010 Steel - New unlined 0.011 Steel - Riveted 0.019 Wood stave 0.012 Most economical section of channels Manning's Equation Q=AV Ais the cross-sectional area of flow perpendicular to the flow direction in m2 S is the bottom slope of the channel in m/m (dimensionless). n is a dimensionless empirical constant called the Manning Roughness coefficient R is the hydraulic radius = A/P. Q is the volumetric flow rate passing through the channel reach in m3/s V is the velocity passing through in m/s The best hydraulic (the most efficient) cross-section for a given Q, n, and S is the one with a minimum excavation and minimum lining cross-section. A = A min and P = P min. The minimum cross-sectional area and the minimum lining area will reduce construction expenses and therefore that crosssection is economically the most efficient one. The cross-section with the minimum wetted perimeter is the best hydraulic cross section within the crosssections with the same area since lining and maintenance expenses will reduce substantially. Channel of most economical section can also be defined as the one in which the discharge is maximum for a given cross-sectional area, or R maximum, or P minimum. 44