Fluid Mechanics CE 256 PDF

CE 256: Fluid Mechanics Introduction CE 256: Fluid Mechanics Lectures Laboratory work Assessment: 70% Examination 30% Continuous assessment Continuous Assessment: Mid-Sem Examination Lab work (To be defended) Attendance in class Assignments Many others...

CE 256: Fluid Mechanics Introduction CE 256: Fluid Mechanics Lectures Laboratory work Assessment: 70% Examination 30% Continuous assessment Continuous Assessment: Mid-Sem Examination Lab work (To be defended) Attendance in class Assignments Many others FLUID DYNAMICS Fluid Dynamics is a sub-discipline of Fluid Mechanics that deals with fluid flow. Sub-disciplines of fluid dynamics: Aerodynamics (Study of air and other gasses in motion. Hydrodynamics (Study of liquids in motion) Conclusion: Study and analysis of fluid in motion.  Continuity and Conservation of Matter Basic Terms to Note M = Mass flow rate, [M/T] Q = Volume flow rate, [L / T] 3 W = Weight flow rate, N/s Mass flow rate Mass of fluid Mass flow rate ( M ) = Time taken to collect that mass of fluid Example: An empty tank weighs 10.0kg. After 20 minutes of collecting water, the tank weighed 100.0kg. Determine the Mass flow rate.  100.0 − 10.0 Mass flow rate = M = = 7.5  10 -2 kg / s 20  60 Question The Mass flow rate of a fluid from a pipeline is 8.5 kg/s. Determine the time it will take to fill a container with 105 kg of the fluid. mass Time = mass flow rate 105 =  12.4 s 8.5  Volume flow rate (Discharge) Volume of fluid flowing per unit time. Volume of fluid Discharge = Q = Time taken to collect that volume of fluid Weight flow rate Weight of fluid flowing per unit time. Weight of fluid W= Time taken to collect that weight of fluid Note the following relations. M = Q W = Q Example If it takes 10 min to fill a reservoir of dimension 500 cm by 1000 cm by 200 cm from a pipe. What is the discharge of the pipe in m3/s. Volume 5 10  2 Discharge = = Time 10  60 100 = = 0.17 m / s 3 600 If the density of a fluid is 1000 kg/m3 and the mass flow rate ( M ) = 0.72kg / s. Determine the Discharge? Mass flow rate ( M ) = Q Hence M 0.72 Q= = = 7.0  10 m / s -4 3  1000  Discharge and mean velocity ▪ Flow through a pipe x u Pipe um umax Velocity in the pipe is not constant across the pipe cross section. Velocity is zero at the walls and increases to a maximum at the centre, then decreases symmetrically to the other wall. Variation of velocity across the section is known as the velocity profile or distribution. Area, A x umt x Area, A Pipe Cylinder of fluid Let Cross sectional area of a pipe at point X be A Mean velocity at section X to be um During time t, a cylinder of fluid will pass section x-x. Volume of fluid (cylinder) is given as Aumt. Discharge (Q) = Volume per unit time Volume A x um x t Q= = The idea, of mean velocity Time t multiplied by the area gives the discharge, applies to all Q = Aum or Q = Au situations - not just pipe flow If Cross-sectional area, A of a pipe is 1.2 x 10-3m2 Discharge, Q is 4.8 l/s, Then the Mean velocity, um, of flow is Q 4.8x10 -3 um = = = 4.0 m/s A 1.2x10 -3  Continuity Principle of conservation of mass: Matter cannot be created or destroyed - (it is simply changed into a different form of matter). Applying this principle to fixed volumes (control volumes, surfaces, fixed region) Gambia  In General, for any control volume, Mass entering  Mass leaving  Change of mass in the control = + per unit time  per unit time  volume per unit time   For steady flow, no change in mass in the control volume, Change of mass in the control   =0 volume per unit time  Applying the principle of conservation of mass to analyse flowing fluids. Consider a streamtube (a pipe). ❖ No fluid flows across the boundary made by the streamlines. ❖ Mass enters the streamtube at point 1 and only leaves through point 2 of the streamtube. 2 A2 u2 1 A1 u1 This means for steady state, mass entering per  Mass leaving per   =  unit time at end 1  unit time at end 2  1Q1 = 2Q2 1 A1u1 =  2 A2 u2 For the existence of steady flow, 1 A1u1 =  2 A2 u2 = constant  For steady flow, quantity of fluid passing any section (point) will remain constant (the same). Ø = 100mm Ø = 50mm 1 2 No fluid is added/removed between points 1 & 2 M1 = M 2 or 1A1u1 = 2 A2u2 For incompressible fluid (density constant), 1 =  2 = const. A1u1 = A2u2 = Q1 = Q2 = constant Continuity equation  Example: Fluid with the following parameters. T = 70o C, u1 = 8 m/s, Ø= Ø=  70 = 9.59kN/m ; 70 = 978kg/m 3 3 100mm 50mm 1 2 Find u2, Q2 , W2 , M 2 Soln. Using Continuity Equation A1 A1u1 = A2u2 ; u2 = ( )u1 A2 D 2  100 2  50 2 A1 = 1 = = 7854 mm 2 A2 = = 1963 mm 2 4 4 4 Hence 7854 u2 =  8m/s = 32 m/s 1963 Discharge  2  1m 2  Q1 = Q2 = A1u1 = A2u2 = (1963m ) (32m/s) = 0.063m 3 /s ( )  103 mm 2    Weight flow rate W = Q;  70 = 9.59kN/m 3 ;  W = (9.59kN/m )(0.063m /s) = 0.60kN/s 3 3 Mass flow rate M = Q;  70 = 978kg/m 3 ;  M = (978kg/m 3 )(0.063m 3 /s ) = 616kg/s Some applications of the continuity equation ❖A pipe which expands or diverges ❖Determination of velocities in pipes coming from a junction. Q1 = Q2 + Q3 A1u1 = A2 u2 + A3u3 Exercise If pipe 1 diameter = 150mm, mean velocity 4.2m/s, pipe 2 diameter 100mm takes 30% of total discharge and pipe 3 diameter 60mm. What are the values of discharge and mean velocity in each pipe?