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Integral budgets: mass and momentum Lecture 7 Mecânica de Fluidos Ambiental 2015/2016.

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Presentation on theme: "Integral budgets: mass and momentum Lecture 7 Mecânica de Fluidos Ambiental 2015/2016."— Presentation transcript:

1 Integral budgets: mass and momentum Lecture 7 Mecânica de Fluidos Ambiental 2015/2016

2 Motivation Statics problems basically require only the density of the fluid and knowledge of the free surface position Most flow problems require the analysis of an arbitrary state of variable fluid motion defined by the geometry, boundary conditions and the law of mechanics Three basic approaches to the analysis of arbitrary flow problems: 1. Control volume, or large-scale, analysis (Chap. 3). 2. Differential, or small-scale, analysis (Chap. 4). 3. Experimental, or dimensional, analysis (Chap. 5). Mecânica de Fluidos Ambiental 2015/2016

3 Motivation In analyzing fluid motion, we might take one of two paths: (1) seeking to describe the detailed flow pattern at every point (x, y, z) in the field (“differential” approach). (2) working with a finite region, making a balance of flow in versus flow out, and determining gross flow effects such as the force or torque on a body or the total energy exchange (“control volume” method). Mecânica de Fluidos Ambiental 2015/2016

4 Aim: Provide tools to perform simple integral calculations E.g. Compute the force exerted by a water jet on a plate, as in the figure Mecânica de Fluidos Ambiental 2015/2016

5 System versus control volumes All the laws of mechanics are written for a system System is defined as an arbitrary quantity of mass of fixed identity Everything external to this system is denoted by the term surroundings (“vizinhanças”), and the system is separated from its surroundings by its boundaries (”fronteiras”) The laws of mechanics then state what happens when there is an interaction between the system and its surroundings In order to convert a system analysis into a control- volume analysis we must convert our mathematics to apply to a specific region rather than to individual masses (Reynolsd transport theorem). Mecânica de Fluidos Ambiental 2015/2016

6 Laws of mechanics Mecânica de Fluidos Ambiental 2015/2016

7 Laws of mechanics Mecânica de Fluidos Ambiental 2015/2016

8 System versus control volumes In analyzing a control volume, we convert the system laws to apply to a specific region, which the system may occupy for only an instant The basic laws are reformulated to apply to this local region called a control volume All we need to know is the flow field in this region, and often simple assumptions will be accurate enough (such as uniform inlet and/or outlet flows) Mecânica de Fluidos Ambiental 2015/2016

9 Volume and Mass Rate of Flow Mecânica de Fluidos Ambiental 2015/2016

10 Volume and Mass Rate of Flow Mecânica de Fluidos Ambiental 2015/2016

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12 What is production - destruction Depends on the property considered. In case of organisms it is growth minus respiration/excretion In case of momentum it is the result of forces In case of kinetic energy it is the result of work supplied …. Mecânica de Fluidos Ambiental 2015/2016

13 Reynolds transport theorem The changing rate of a property inside a control volume occupied by the fluid is equal to the changing rate inside the material system located inside the control plus what is flowing in, minus what is flowing out. Mecânica de Fluidos Ambiental 2015/2016

14 Demonstration of Reynolds theorem Mecânica de Fluidos Ambiental 2015/2016 SYS 2 SYS 1 SYS 3 Let’s consider a conduct and 3 portions fluid (systems), SYS 1, SYS2 and SYS 3 that are moving. Let’s consider a space control volume (not moving -fixed) that at time “t” is completely filled by the fluid SYS 2 (coincident with SYS 2). CV SYS 2 SYS 1 SYS 3 CV Time = t Time = t+∆t Between time= t and time =(t+∆t) inside the control volume properties can change because some fluid flew in (SYS1) and other fluid flew out (SYS2) and the properties of those systems have changed in time. At time t+  t the fluid SYS 2 has moved slightly to the right and fluid SYS 1 has also moved to the right.

15 Demonstration of Reynolds theorem Mecânica de Fluidos Ambiental 2015/2016 SYS 2 SYS 1 SYS 3 CV SYS 2 SYS 1 SYS 3 CV

16 Demonstration of Reynolds theorem Mecânica de Fluidos Ambiental 2015/2016

17 Demonstration of Reynolds theorem Mecânica de Fluidos Ambiental 2015/2016

18 Reynolds transport theorem – physical interpretation 1-time rate of change of an arbitraryextensive parameter of a system (rate of change of mass, momentum...) 2 –rate of change of B within the control volume as the fluid flows through it 3- net flowrate of the parameter B across the entire control surface (may be negative, zero or positive depending on the particular situation) Mecânica de Fluidos Ambiental 2015/2016 Or: 132

19 If the Volume is infinitesimal Becomes: But: And thus: Dividing by the volume (  x 1  x 2  x 3 ): Mecânica de Fluidos Ambiental 2015/2016

20 Reynolds transport theorem Mecânica de Fluidos Ambiental 2015/2016

21 Momentum conservation {the rate of accumulation of a property inside a control volume} = {what flows in – what flows out} + {production – destruction} Production is the result of the forces applied in the sense of the flow and destruction of the forces in the opposite sense. Mecânica de Fluidos Ambiental 2015/2016

22 Momentum conservation Forces applied to a volume of fluid (momentum production/consumption) Mecânica de Fluidos Ambiental 2015/2016 Forças superficie – pressão (força normal) e o atrito (força tangencial) Forças mássicas – peso e força electromagnética

23 Integral momentum budget A taxa de variação da quantidade de movimento pode então ser calculada através do balanço da quantidade de movimento que entra/sai e das forças aplicadas ao volume de controlo: Se o escoamento for estacionário: Mecânica de Fluidos Ambiental 2015/2016

24 Integral momentum budget Mecânica de Fluidos Ambiental 2015/2016 (Gradient Theorem)

25 Integral Momentum Budget Where F i is the summation of pressure forces other than those acting at inlet and outlet plus the summation of the friction forces. This is an algebraic equation applicable if: the flow is stationary and incompressible the velocity and pressure are uniform at each inlet and outlet Mecânica de Fluidos Ambiental 2015/2016

26 Example 1 Calculate the force exerted by the fluid over the deflector neglecting friction V=2 m/s and jet radius is 2 cm and theta is 45 . Mecânica de Fluidos Ambiental 2015/2016 D=2CmA=0.000315m2 V=2 m/s Q=0.000629m3/s Theta45º1.26N 0.88971N N Fx=-0.37N Fy=0.88971N F=0.963015N We have a flow with an inlet and an outlet. Velocity has a component at the inlet and two at the outlet Pressure is atmospheric at inlet and outlets and thus the velocity modulus remains constant. We have to compute budgets along both directions x and y.

27 Example 2 The Jet is hitting the surface perpendicularly (V j =3m/s), but the surface is moving (Vc=1m/s). D=10 cm. Calculate the force and power supplied. Mecânica de Fluidos Ambiental 2015/2016 If the control volume is moving, fluxes depend on the flow velocity relative to the control volume:

28 Mecânica de Fluidos Ambiental 2015/2016 Relative or absolute reference Discharge must be computed using relative velocity. Transported velocity can be the relative or the absolute velocity. Usually the relative velocity is more intuitive.

29 Mecânica de Fluidos Ambiental 2015/2016 Example 2Vc=1m/s D=10cm Vj=3m/sA=0.007864m2 VRj=2Q=0.015728m3/s Theta9031.46N 0N 0Symmetrical Fx=-31.46N Fy=0N F=31.456N Symmetrical

30 Summary The integral (algebraic) momentum equation describes the momentum conservation principle (Newton law) assuming simplified solutions for momentum flux calculations. It is useful when flow is stationary and incompressible. Becomes more useful when associated to the Bernoulli Energy Conservation Equation. Mecânica de Fluidos Ambiental 2015/2016


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