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HYDRAULIC 1 CVE 303 BASICS OF FLUID FLOW Types of flow – The type of flow depends on the manner in which the particles unite, the particles group themselves in a variety of ways e.g. regular or irregular. The type of flow is identified by the Reynolds number. vL R Flow of an IDEAL FLUID – No Viscousity Flow of a REAL FLUID – Viscousity included Laminar (low velocity, motion in layers) & Turbulent Flow ( high velocity, chaotic motion) Steady flow (conditions at any point remain constant, may differ from point to point) Uniform Flow ( velocity is the same at any given point in the fluid) Flow lines : Path lines, Stream lines and Streak lines Straight line streamlines of moving particle- One dimensional flow (Flow in pipes) Curve – Two dimensional flow (Flow over a spillway) Streamlines represented in space – Three dimensional (flow in a river bed) Motion of Fluid Particles Lagrangian method Eulerian method Equation of continuity of a liquid flow: If an incompressible liquid us is continuously flowing through a pipe or a channel (whose cross-sectional area may or may not be constant) the quantity of liquid passing per second is the same at all sections. Q1 Q2 Q3 a1v1 a2 v2 a3v3 Questions Water is flowing through a tapered pipe having end diameter of 150m and 50m respectively. Find the discharge at the larger end and velocity head at the smaller end if the velocity of water at the larger end is 2m/s. A circular pipe of 250 mm diameter carries an oil of specific gravity 0.8 at the rate of 120 litres/s and under a pressure of 20kPa. Calculate the total energy in meters at a point which is 3m above the datum line. A horizontal pipe 100 m long uniformly tapers from 300 mm diameter to 200 mm diameter. What is the pressure head at the smaller end, if the pressure at the larger end is 100 KPa and the pipe is discharging 50 litres of water per second. (5 marks BERNOULLI’S EQUATION AND ITS APPLICATIONS Introduction Energy of liquid in motion Potential: Energy a liquid posses by virtue of its position Kinetic: Energy posessed by a liquid by virtue of its motion Pressure head of a liquid particle in motion Total Energy : This is the sum of a liquids Potential, Kinetics and Pressure energy v2 p E Z 2g w Total Head of a liquid in motion 2 v p H Z 2g w BERNOULLI’S EQUATION (CONTD.) Definition: For a perfect incompressible fluid, flowing in a continuous stream, the total energy of a particle remains the same, while the particles moves from one point to another. Limitations of Bernoulli’s equation Practical application of Bernoulli’s equation Venturimeter ( Convergent cone, throat, divergent cone) Discharge through a venturimeter ca1a2 Q c.a2 .v2 2 gh a1 a2 2 2 Inclined venturimeter Orificemeter Pitot tube Questions A pipe 500 m long has a slope of 1 in 100 and tapers from 1 m diameter at the higher end to 0.5 m at the lower end. The quantity of water flowing is 900 l/sec. If the pressure at the higher end is 70KPa. Find the pressure at the lower end. A pipe AB branches into two pipes C and D. The pipe has a diameter of 0.45 m at A, 0.3 m at B, 0.2 m at C and 0.15 m at D. Find the discharge at A if the velocity of water at A is 2 m/s also find the velocities at B and D if the velocity at C is 4 m/s. A venturimeter with a 150 mm diameter at inlet and 100 mm at throat is laid with its axis horizontal and is used for measuring the flow of oil specific gravity 0.9. The oil mercury differential manometer shows a guage difference of 200 mm. Assume coefficient of meter as 0.98. Calculate the discharge in litres per minute A venturimeter has 400 mm diameter at the main and 150 mm at the throat. If the difference of pressure is 250 mm of mercury and the metre coefficient is 0.97, calculate the discharge of oil (specific gravity = 0.75) through the venturimeter OPEN CHANNEL FLOW General Definition: An open channel is a passage through which the water flows under the force of gravity and atmospheric pressure e.g. canals, sewers, aqauduct. It is also known as free- surface flow or gravity flow. The flow here is not due to pressure as in the case of pipe flow. Most open channel flow are turbulent flow and froude number is the relevant parameter here Types of open channel flow Natural streams and rivers Artificial canals, flumes Sewers, tunnels, partially full pipelines Gutters Applications of artificial channels Water power development Irrigation Water supply Drainage Flood control Reynolds no. for pipes IR pipe VD Reynolds no. for open channels IRchannel VRn ; Rn A P Wetted perimeter: Length of cross-sectional border in contact with water. Length where friction acts. Slopes in open channel flow OPEN CHANNEL FLOW (CONTD.) Uniform flow through open channels Chezy’s formula V C Rn So Values of Chezy’s constant in the formula for discharge in open channel Mannings formula 1 2 3 12 V Rn S o n Basin’s formula Kutter’s formula Geometric properties of cross-sections Trapezoidal channels Triangular channels Rectangular channels Exercises A trapezoidal channel 3.5 m wide at the bottom has side and bed slopes of 1:1 and 1:1000 respectively. Using manning’s formula, find the discharge through the channel, if the depth of water is 0.5 m. Take N = 0.03. A non symmetrical trapezoidal cross-section with b = 1.2 m, m1 = 2 and m2 = 1, So = 0.00009, Q = 120 m3/s and y = 0.6 m, find the manning’s roughness factor n. Unnatural stream’s cross-section can be approximated by a parabolic shape with B = 4m and a depth y = 2 m. If the stream is laid on a slope of 0.001 m/m and n = 0.028, determine the discharge. SPECIFIC ENERGY Introduction v2 q3 E y y 2g 2 gy 2 Specific energy diagram Critical conditions E = Emin i.e. dE 0 dy Critical depth q2 yc 3 g Critical velocity q vc gyc yc Subcritical: y1 > yc tranquil, upper stage flow V < Vc Supercritical: y1 < yc rapid, lower stage flow V >Vc Froude’s number v F gy Critical depth in non rectangular channels Occurrence of critical flow Critical energy 2 yc Ec 2 Emin 3 3 qmax gyc 3 Occurrence of Critical Flow Change from mild to steep slope in a channel Entrance from a reservoir into a steep slope channel Free fall from mild slope channel Free fall from steep slope channel Humps and Contractions Questions A rectangular channel 3.25 m wide discharges 2600 litres of water per second. What is the critical depth and critical velocity? A channel of rectangular section 8 m wide is discharging water at the rate of 12 m3/s with an average velocity of 1.2 m/s. Find the type of flow? A rectangular channel 3 m wide carries 4 m3/s of water in subcritical uniform flow at a depth of 1.2 m, a frictionless hump is to be installed across the bed. Find the critical hump height? HYDRAULIC JUMP Introduction : Local non-uniform flow phenomenon – supercritical flow going into subcritical. The shooting flow is an unstable type of flow and does not continue on the downstream side, the flow transform itself to the streaming flow by increasing its depth. The rise in water level which occurs during the transformation of the unstable shooting flow to the stable streaming flow is called hydraulic jump. Depth relations Energy losses hij y2 y1 3 4 y1 y2 v2 v2 H1 H 2 d1 1 d 2 2 2g 2g Types of jumps Stilling basins Questions A horizontal rectangular channel of constant breadth has a sluice opening from the bed upwards. When the sluice is partially opened water issues at 6 m/s with a depth of 600 mm. Determine the loss of head per KN of water. A discharge of 1000 l/s flows along a rectangular channel 1.5 m wide. If a standing wave is to be formed at a point where the upstream depth is 180 mm. What would be the rise in water level? Water flows at the rate of 1 m3/s along a channel of rectangular section of 1.6 m width. If a standing wave occurs at a point where upstream depth is 250 mm. Find the rise in water level after the hydraulic jump. Also find the loss of head in the standing wave.

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