ES06 Fluid Mechani UNIT D: Flow Field Analysis ROAD MAP... D-0: Reynolds Transport Theorem D-1: Conservation of Mass D-: Conservation of Momentum D-3: Conservation of Energy ES06 Fluid Mechani Unit D-0: List of Subjects Eulerian and Lagrangian Streamline, Streakline, and Pathline System v.s. Control Volume Reynolds Transport Theorem Types of Control Volume
Page 1 of 1 Unit D-0 Eulerian and Lagrangian Eulerian and Lagrangian: Different Approaches in Flow Field Analysis There are two general approaches in analyzing fluid mechani problems: Eulerian and Lagrangian methods Eulerian method: uses the field representation concept Lagrangian method: involves tracing individual fluid particles as they move and determining how the fluid properties associated with these particles change as a function of time the fluid particles are tagged or identified, and their properties determined as they move Eulerian method = INTEGRATION of properties into the FIELD of flow (this field is often called, the CONTROL VOLUME). Lagrangian method = DIFFERENTIATION of properties onto each PARTICLE of flow (the collection of all particles of flow is often called, the SYSTEM).
Page of 1 Streamline, Streakline, and Pathline Unit D-0 dy dx v u Flow Visualization: 3 Different Methods in Flow Field Representations Streamline: a line that is everywhere tangent to the velocity field (mathematical concept) For -D flow, streamline can be written as: dy v dx u Streakline: a line consists of all particles in a flow that have previously passed through a common point (experimental concept) Pathline: a line of traced particle as it flows from one point to another (Lagrangian concept) Textbook (Munson, Young, and Okiishi), page 156 Pathline = path of individual particle (Lagrangian) Streakline = streak (line or locus of multiple particles path, originated from a common location (Eulerian, as this is common in experimental approach) Streamline = is a line defined under specific conditions only (steady flow) For steady flow: pathline = streakline = streamline
Page 3 of 1 Unit D-0 System v.s. Control Volume (1) System = Lagrangian Approach Control Volume = Eulerian Approach B mb to move and interact with its surroundings System approach: a system is a collection of matter of fixed identity (always the same atoms or fluid particles), which may move, flow, and interact with its surroundings (Lagrangian concept) Control volume approach: a control volume is a volume in space (a geometric entity, independent of mass) through which fluid may flow (Eulerian concept) All of the laws governing the motion of a fluid are stated in their basic form in terms of a system approach (not a control volume approach) laws of fluid motion Let B represent any fluid parameter (velocity, acceleration, mass, momentum, etc.) and b represent the amount of that parameter per unit mass, then: B mb Extensive Property Mass of the portion of fluid of interest Intensive Property
Page 4 of 1 Unit D-0 System v.s. Control Volume () First, consider a system ( sys ): For infinitesimal fluid particles of size and mass, the amount of an extensive property that a system possesses at a given instant is: Most of the laws governing fluid motion involve the time rate of change of an extensive property of a fluid system: Next, consider a control volume ( cv ): In terms of control volume (cv) and control surface (), the time rate of change of an extensive property within a control volume is:
Page 5 of 1 EXAMPLE 4.7 Textbook (Munson, Young, and Okiishi), page 171 Unit D-0 Time Rate of Change for a System and a Control Volume Let B m (mass as an extensive property) and b 1 d dv dbsys dm sys sys (1) dt dt dt This represents the time rate of change of system mass ( system mass here is essentially the mass of all fire extinguisher fluid particles). As the release valve is open ( t 0), the system (fluid particles) simply move from one place (inside of the tank) to another (outside of the tank). Therefore, the system mass remains constant, or: dbsys 0 dt d dv dbcv dmcv cv () dt dt dt This represents the time rate of change of control volume mass ( control volume mass here is essentially the mass contained within the fixed control volume: means inside of the fire extinguisher). As the release valve is opened ( t 0), the mass within the fixed control volume (inside of the fire extinguisher) decreases. Therefore: db cv 0 dt
Page 6 of 1 Unit D-0 Reynolds Transport Theorem (1) The system at time : SYS = CV B t) B ( t) sys( cv The system at time B sys t t t : SYS = CV I + II ( t t) Bcv( t t) BI ( t t) BII( t t) The change in the amount of B in the system in the time interval t is given by: Bsys ( t) Bcv( t) Bsys Bsys( t t) Bsys( t) t t Bcv( t t) BI ( t t) BII( t t) Bsys( t) t Bcv( t t) Bcv( t) BI ( t t) BII( t t) t t t
Page 7 of 1 Unit D-0 Reynolds Transport Theorem () B cv( cv t t) B ( t) t In the limit ( t 0 ), this term is seen to be the time rate of change of the amount of B within the control volume: BI ( t t) t This term represents the rate at which the parameter B flows IN to the control volume across the inflow control surface The amount of B within region I (inflow) is its amount per unit volume, times the volume: hence, In limit ( t 0 ):
Page 8 of 1 Unit D-0 Reynolds Transport Theorem (3) B II ( t t) t This term represents the rate at which the parameter B flows OUT of the control volume across the outflow control surface The amount of B within region II (outflow) is its amount per unit volume, times the volume: hence, In limit ( t 0 ): Reynolds Transport Theorem (for a flow of a variable cross section pipe) Taking the limit of the LHS of the equation: Bsys DBsys lim (substantial, or total, change of system property) t0 t Dt Taking the limit of the RHS of the equation: Bcv t t Bcv t Bcv lim (change of property within the control volume) t0 t t BI or II t t & lim Bin or Bout (property s flux, in or out) t0 t DBsys Bcv Bout Bin (Reynolds Transport Theorem) Dt t
Page 9 of 1 Unit D-0 Reynolds Transport Theorem (4) DB Dt sys B t cv B out B in B t cv A V b AV b 1 1 1 1 The time rate of change in the amount of B in the system and that for the control volume is: Bsys( t t) Bsys( t) DBsys Bcv lim B out B in t0 t Dt t DB Dt sys B t cv B out B in B t cv A V b AV b 1 1 1 1 Total Rate of Change of B of a System Time Rate of Change of B within a Control Volume Flow Rate of B across a Control Surface Generalized form in Reynolds Transport Theorem (FLUX NOTATION) The surface dividing the control volume from surroundings is defined as control surface Let us define a unit vector nˆ, which defines the direction of outward normal at each control surface Let us define an angle, which defines the angle between the velocity vector and outward pointing normal to the surface (unit vector nˆ )
Page 10 of 1 Unit D-0 Reynolds Transport Theorem (5) Using the net flux notation, the Reynolds transport theorem can be written as: Note that: The general form of Reynolds transport theorem can therefore be written as: (for a fixed, non-deforming control volume) The physical interpretation of the material derivative v.s. Reynolds transport theorem: D( ) ( ) ( V )( ) Dt t System (or Lagrangian) Approach Time Rate of Change Unsteady Effects Convective Effects Control Volume (or Eulerian) Approach
Page 11 of 1 EXAMPLE 4.8 Textbook (Munson, Young, and Okiishi), page 174 Unit D-0 Use of the Reynolds Transport Theorem Let B m (mass as an extensive property) and b 1 The Reynolds transport theorem can be written as: Dmsys mcv mcv mout min out AoutVout in AinVin Dt t t dv m cv cv where, t dt in AV in in 0 Hence, dv Dmsys m cv cv mout out AoutV out Dt t dt Conservation of mass is the time rate of change of system mass is zero, Dmsys or, 0 Dt Therefore, dv Dmsys m dv cv cv cv mout out AoutV out 0 or, outaoutvout Dt t dt dt
Page 1 of 1 Unit D-0 Types of Control Volume (a) fluid flows through a pipe (fixed in space) (b) rectangular control volume surrounding the jet engine (either fixed or moving) (c) deforming control volume (either fixed or moving) Selection of Control Volume for Flow Analysis Selection of control volume is arbitrary. You can choose any control volume. However, the poor selection of control volume will complicate the analysis (still can draw conclusions, yet amount of mathematical work involved will be dramatically different). Often, special type of control volumes are employed, in order to simplify analysis: o DEFORMING control volume: the control volume with one (or more) control surface(s) have certain velocity (V ). Unit D-1 o MOVING control volume: the control volume, which is moving with the velocity (V cv ). Unit D-
ES06 Fluid Mechani UNIT D: Flow Field Analysis ROAD MAP... D-0: Reynolds Transport Theorem D-1: Conservation of Mass D-: Conservation of Momentum D-3: Conservation of Energy ES06 Fluid Mechani Unit D-1: List of Subjects Conservation of Mass Continuity Equation Deforming Control Volume
Page 1 of 6 Unit D-1 Conservation of Mass Time rate of change of the mass of the coincident system = Time rate of change of the mass of the contents of the coincident control volume + Net rate of flow of mass through the control surface For a system and a fixed, non-deforming cv (control volume) that are coincident at an instant of time (t), the Reynolds transport theorem with B = mass and b = 1 provides: The conservation of mass is simply, time rate of change of the system mass = 0 DM sys 0 where, Dt (Continuity Equation)
Page of 6 Unit D-1 Continuity Equation (1) Steady/Unsteady Flow () Mass/Volume Flowrate (3) Sign Convention For a steady flow: When all of the differential quantities are summed over the entire control surface: Representative (or average) velocity Mass flowrate: The average velocity: If the velocity is considered uniformly distributed (called one-dimensional flow): V V The dot product V nˆ is + for the flow out of the control volume and for the flow into the control volume Therefore, the mass flowrate is + for the flow out of the control volume and for the flow into the control volume: m m 0 out in If the steady flow is also incompressible, the net amount of volume flowrate through the control surface is also zero: Qout Qin 0 For unsteady flow: + if the mass of the contents of the control volume is increasing, and if decreasing
Page 3 of 6 Class Example Problem Related Subjects... Continuity Equation Unit D-1 Starting from the continuity equation: dv ˆ da 0 t V n cv Since the flow is steady, dv 0 t, the equation becomes: V n ˆ da 0 cv Let us apply this equation for single inflow, section (1), and outflow, section (): 1V 1A1 V A 0, or m1 m 0 Therefore, V A 1V 1A1 The velocity (average value) at section () can be found as: p 1 A RT1 A p T 77 kpa 40 K V 1 1 1 1 V1 V1 V1 05 m/s = 314.13 m/s A p A p T1 45 kpa 68 K RT
Page 4 of 6 EXAMPLE 5.5 Textbook (Munson, Young, and Okiishi), page 198 Unit D-1 Conservation of Mass Unsteady Flow Starting from the continuity equation: dv ˆ da 0 t V n cv As a fixed control volume is defined (the bathtub), the flow within the control volume is NOT steady (it will change the mass over time). airdvair mair waterdvwater mwater 0 t t air water volume volume Let us consider only the flow of water (since we are interested in water depth): waterdvwater mwater 0 t water volume Note that: waterdvwater water h ft5 ft 1.5 ft h A j water volume Therefore, water h ft5 ft 1.5 ft h A j mwater t h or, water 10 ft Aj mwater waterq t h Qwater Q Hence, water (since A j 10 ft ) t 10 ft Aj 10 ft The time rate of change of the water depth is: 3 1 ft 9 gal/min 7.48 gal h Qwater 1 in = 1.44 in/min t 10 ft 10 ft 1 ft
Page 5 of 6 Unit D-1 Deforming Control Volume V CS Occasionally, a deforming control volume can simplify the solution of a problem A deforming control volume involves changing volume size and control surface movement The continuity equation can be written in terms of relative velocity: V W V c s Velocity of the control surface as seen by the a fixed observer
Page 6 of 6 EXAMPLE 5.9 Textbook (Munson, Young, and Okiishi), page 04 Unit D-1 Conservation of Mass Deforming Control Volume If the control volume is defined as a deforming control volume (contains only water), the continuity equation can be written as: dv ˆ da 0 t Wn cv The first term: h dv h ft5 ft 10 ft t t t cv The second term: ˆ h W n da Vj Aj t h h Therefore, 10 ft Vj Aj 0 t t h h h VA j j Qwater Q or, 10 ft Vj A => j water t t t 10 ft Aj 10 ft Aj 10 ft The time rate of change of the water depth is: 3 1 ft 9 gal/min 7.48 gal h Qwater 1 in = 1.44 in/min t 10 ft 10 ft 1 ft
ES06 Fluid Mechani UNIT D: Flow Field Analysis ROAD MAP... D-0: Reynolds Transport Theorem D-1: Conservation of Mass D-: Conservation of Momentum D-3: Conservation of Energy ES06 Fluid Mechani Unit D-: List of Subjects Conservation of Momentum Momentum Equation Moving Control Volume
Page 1 of 7 Unit D- Conservation of Momentum Time rate of change of the linear momentum of the System Time rate of change of = the linear momentum + of the contents of the control volume Net rate of flow of linear momentum through the control surface For a system and a fixed, non-deforming cv that are coincident at an instant of time (t), the Reynolds transport theorem with B = momentum (mass times velocity) and b = V provides: Newton s second law of motion states: Time rate of change of the linear Sum of external forces momentum of the system = acting on the system The conservation of momentum is simply, time rate of change of the linear momentum of the system = sum of external forces acting on the system where, (Linear Momentum Equation)
Page of 7 Unit D- Momentum Equation (1) Coordinate System () External Forces (3) Sign Convention When the flow is uniformly distributed over a section of the control surface (1-D flow) where flow into or out of the control volume occurs, the integral operations are simplified Linear momentum is directional The flow of positive or negative linear momentum into a control volume can be determined by the dot product: V nˆ Time rate of change of the linear momentum of the contents of a non-deforming control volume is zero for steady flow The forces due to atmospheric pressure acting on the control surface may need consideration If the control surface is selected so that it is perpendicular to the flow where fluid enters or exits the control volume, the surface force exerted at these locations by fluid outside the control volume on fluid inside will be due to pressure The external forces have an algebraic sign, positive if the force is in the assigned positive coordinate direction and negative otherwise Only external forces acting on the contents of the control volume are considered in the linear momentum equation The force required to anchor an object will generally exist in response to surface pressure and/or shear forces acting on the control surface
Page 3 of 7 EXAMPLE 5.10 Textbook (Munson, Young, and Okiishi), page 195 Unit D- Linear Momentum Change in Flow Direction F Az Starting from the momentum equation: dv ˆ da t V V V n F cv For steady flow: V V n ˆ da F x-direction component: u Vn ˆ da F x => u1 V nˆ da(1) (1) u V n ˆ da () () Fx (1) () z-direction component: w Vn ˆ da F z => w1 V nˆ da(1) (1) w V n ˆ da () () Fz (1) () At each control surface: (1): u1 V1, w1 0, Vn ˆ V (1) 1 (): u V cos, w V sin, Vn ˆ V () Note that from continuity, Q1 Q => AV 1 1 AV => since A1 A A, V1 V V Hence, x momentum equation becomes: FAx V V A V cos VA V A V cos A V A1 cos 3 10 ft/s 1.94 slugs/ft 0.06 ft 1 cos = 11.64 1 cos lb Also, z momentum equation becomes: FAz V sin VA V sin A 10 ft/s 1.94 slugs/ft 3 sin 0.06 ft = 11.64sin lb
Page 4 of 7 Class Example Problem Related Subjects... Momentum Equation Unit D- y ˆn 1 (1) (3) ˆn 3 () x Starting from the momentum equation: dv ˆ da t V V V n F cv For steady flow: V V n ˆ da F x-direction component: uvn ˆ da Fx => u1 V nˆ da(1) (1) u V nˆ da() () u3 V n ˆ da (3) (3) Fx (1) () (3) y-direction component: vvn ˆ da Fy => v1 V nˆ da(1) (1) v V nˆ da () () v3 V n ˆ da (3) (3) Fy (1) () (3) At each control surface: (1): u1 V, v1 0, Vn ˆ (1) V (): u Vcos30, v Vsin30, Vn ˆ () V (3): u3 V, v3 0, Vn ˆ (3) V Hence, x and y momentum equation becomes: V V A1 V cos30 VA V VA 3 FH => V A 1 V cos30 A V A3 FH (eqn. 1) V sin30 VA FV => V sin30 A FV (eqn. ) Also, from continuity: Q1 Q Q3 => AV 1 AV A3V => A1 A A3 (eqn. 3) Combining eqns. 1,, & 3 yields: A 3 3 1 A cos30 A A A A cos30 A A cos30 1 3 cos30 1 FH FV = 0.68 A sin30 A sin30 A sin30 sin30 ˆn
Page 5 of 7 Class Example Problem Related Subjects... Momentum Equation Unit D- y ˆn 1 (1) FR x () ˆn Starting from the momentum equation: dv ˆ da t V V V n F cv For steady flow: V V n ˆ da F x-direction component: u Vn ˆ da F x => u1 V nˆ da(1) (1) u V n ˆ da () () Fx (1) () At each control surface: (1): u1 V1, Vn ˆ V (1) 1 (): u V sin0, Vn ˆ V () Hence, x momentum equation becomes: 1 V1 V 1A1 V sin 0 V A FAx FRx FAx water h1 A1 cos0 1 => V 1 A1 V sin 0 A FAx waterh1 A1 cos0 (eqn. 1) A1 h1 Also, from continuity: Q1 Q => AV 1 1 AV => V V1 V1 A h (eqn. ) Combining eqns. 1 & yields (note that in terms of per unit depth, A1 h1 & A h): 1 h 1 FAx waterh1 cos0 V1 h1 V1 sin 0 h h 1 3 3 4 ft 6.4 lb/ft 4 ft cos 0 10 ft/s 1.94 slugs/ft 4 ft 10 ft/s sin 0 1.94 slugs/ft 3 1 ft 1 ft = 183.46 lb
Page 6 of 7 Unit D- Moving Control Volume For a system and an inertial, non-deforming, and moving control volume that are both coincident at an instant of time, the Reynolds transport theorem can be written in terms of relative velocity: V W V cv Velocity of the control volume as seen by the a fixed observer Hence, the linear momentum equation for an inertial, moving, and non-deforming control volume that involves steady (instantaneous or time-average) flow is:
Page 7 of 7 EXAMPLE 5.17 Textbook (Munson, Young, and Okiishi), page 19 Unit D- Linear Momentum Moving Control Volume z R z R x x For a moving control volume: W W n ˆ da F x-direction component: W ˆ xwn da F x => W1 W 1 A1 W cos 45 W A Rx z-direction component: W ˆ zwn da F z => W sin 45 W A Rz (ignoring the weight of water) Page From continuity, Q 1 Q => AW 1 1 AW => since 8 of 8 A1 A A, W1 W W where, W1 W W V1 V0 100 ft/s 0 ft/s 80 ft/s EXAMPLE 5.17 Hence, x momentum equation becomes: 3 Rx W A W cos45 A W A 1 cos45 80 ft/s 1.94 slugs/ft 0.006 ft 1 cos45 = 1.8 lb Also, z momentum equation becomes: Linear Momentum Moving Control Volume 3 Rz W sin 45 A 80 ft/s sin 45 1.94 slugs/ft 0.006 ft = 5.68 lb Magnitude of the force, exerted by the water is: F R 1.8 5.68 = 57.0 lb Direction of the force is: 1 R z 15.68 tan tan R x 1.8 = 67.5 F 67.5 R 5.68 lb 1.8 lb Textbook (Munson, Young, and Okiishi), page 19
ES06 Fluid Mechani UNIT D: Flow Field Analysis ROAD MAP... D-0: Reynolds Transport Theorem D-1: Conservation of Mass D-: Conservation of Momentum D-3: Conservation of Energy ES06 Fluid Mechani Unit D-3: List of Subjects Conservation of Energy Energy Equation
Page 1 of 5 Unit D-3 Conservation of Energy (1) Time rate of increase of the total stored energy of the system Time rate of increase = of the total stored + energy of the contents of the control volume Net rate of flow of the total stored energy out of the control volume through the control surface For a system and a fixed, non-deforming cv that are coincident at an instant of time (t), the Reynolds transport theorem with B = energy and b = e gives: The first law of thermodynami for a system is: Time rate of Net time rate of Net time rate of increase of the total = energy addition by + energy addition by stored energy of the heat transfer into the work transfer into system system the system Total stored energy per unit mass for each particle is: = Internal Energy + Kinetic Energy + Potential Energy D V V e dv u gz dv u gz ˆ da Qnet Wnet Dt t V n sys cv in in
Page of 5 Unit D-3 Conservation of Energy () For a control volume that is fixed and non-deforming, the energy equation becomes: W W W net shaft normal in net in stress Work transfer can also occur at the control surface when a force associated with fluid normal stress acts over a distance in case of the simple pipe flow: (Flow Work) Textbook (Munson, Young, and Okiishi), page 31 Including the flow work, the energy equation can be written as: W net in Flow Work Combining the flow work into the total energy:
Page 3 of 5 Unit D-3 Energy Equation (1) (1) Steady Flow () 1-D Flow (3) Single Stream (Pipe) (4) Enthalpy Simplification of the Energy Equation (1) Steady flow () 1-D flow (3) Pipe flow (single in/out flow) For a steady flow or steady in mean (cyclical) flow, the energy equation can be simplified as: Enthalpy If the properties within parenthesis are all assumed to be uniformly distributed over the flow cross-sectional areas involved (1-D flow), the integration becomes simple and provides the following equation: m out m in If there is only one stream entering and leaving the control volume, then: min mout m
Page 4 of 5 Unit D-3 Energy Equation () For steady & 1-D flow with single in & out: Simplification of the Energy Equation (continued) (4) Incompressible flow (density = constant) (5) Constant density Dividing by the mass flowrate and rearranging: W w m loss of useful or available energy Therefore, the energy equation is: Multiplying by fluid density () and dividing by the specific weight (): The equation involves energy per unit weight which has the unit of length called, head Shaft Head: Head Loss: NOTE: Shaft head could be positive or negative, depending on the device: Pump Head (energy is added IN to the system): hs hp Turbine Head (energy is taken OUT from the system): hs ht
Page 5 of 5 Class Example Problem Related Subjects... Energy Equation Unit D-3 z pout Vout pin Vin Starting from the energy equation: zout zin hs hl g g Note that: pout 0 (gage pressure) Vout 0 (velocity becomes zero, by reaching 60 ft elevation) zout 60 ft 1 in pin 10 psi 10 1, 440 lb/ft (gage pressure) 1 ft 3 Q 1.5 ft /s V in 17.19 ft/s A in 1 ft 4 in 4 1 in zin 0 hl 0 (head loss is ignored) Therefore, pin Vin 1,440 lb/ft 17.19 ft/s hs hp zout 60 ft 3.33 ft 3 g 6.4 lb/ft 3. ft/s 3 3 1 hp Wshaft Qh s 6.4 lb/ft 1.5 ft /s3.33 ft = 5.5 hp net in 550 ft lb/s