Chapter 4

Chapter 4 - 69 iViAE 31.61: Fluid Mechanics Notes by Paavo...

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Unformatted text preview: 69 iViAE 31.61: Fluid Mechanics Notes by Paavo Sepri LECTURE 10 Chagter 5 5.0 APPLICATION OF A CONTROL VOLUME TO FLUID MOTION Here, we apply the Reynolds Transport Theorem to three basic principles of Physics: (1) Conservation of Mass. (2) Newton's 2'“Cl Law of Motion (Linear Momentum Equation). (3) First Law of Thermodynamics (Energy Equation). 5.1 CONSERVATION OF MASS Recall: We discussed two ways to View fluid motion: (1) Lagrangian: Follow the fluid element as it moves in space. (2) Euierian: Observe how a fluid moves into and out of a volume that is fixed in space. I (Jim/Q. TunneJQ-a / ,— fl / / K r“ ""9- VH’) (r) W (n we) mm wi‘lfi w CEZZQ 7%“ _,__.,. —-—-e- (Z) // /'r / fr/f/f/f/ (2) VF. ( my, dati'omr? Here, in order to discuss conservation of mass, we first consider the Lagrangian approach: Follow a volume of fluid bounded by surface, S(z‘), such that no mass crosses the boundary as this volume, V(r) , moves in space. Define: Ma “1de = total mass within V0) (5.1) V0) Then, in the Lagrangian view, the conservation of mass principle states that: iii dt %% pail/3c _=. 0 (by construction) (5-2) Efired m) @ Although this statement is self—evident, we do not know how VG) changes in time, h and we do not know how p(t,i) varies within V(t) . Therefore, we seek an alternative View, which is easier for computational purposes. Apply the Reynolds Transport Theorem (Eq. (4.12)): .91. I New; -=~ “JED—Rde + Hpmmds = 0 (5.3) 5” Wm} V at S Re-write as foltows: V3, 2 "" “91(fi"7)rdsr " H1011 (fi'fi)11'dSII (S‘Ba) V S; 51: Interpret this equation through the following diagram: Here, the surface integral over S, represents the mass influx rate into the fixed volume, and the surface integral over S,, represents the mass out-flux rate from the volume. The net mass flux rate across the complete boundary, S , is comprised of the sum of these two integrals. @ 5.1.1 SIMPLE IDEALIZATION OF THE MASS CONSERVATION PRINCIPLE Pour water into a bucket that has a hole in it: Rage“ "—\ min, Rewconsider Eq.(5.3a): fl e-Hmfz-mdw Jim-W511 V SI S” This may be re-written with different symbols as follows: 51% = min W moat : pvl‘nAin m pvoutAaut Three cases are possible: (1) If 1521.5 n’zom , then the water level, Mr), decreases with increasing time. (2) If rial." > firm, then the water level, Mr), increases with increasing time. (3) If mm: m then the water level, ha), remains constant with increasing time. our ' The last case would be called a steady flow (except for surface ripples or internal turbulence in the fluid). @ 5.1.2 ANOTHER EXAMPLE OF THE MASS CONSERVATION PRINCIPLE Consider oil being poured into a conical funnel: (B if 63% 1001'! p water Given: 2 0.8 and mm = 0.12kg/s at some instant of time. Find: (1) The average flow speed at the funnel exit. (2) The volume flow rate through the funnel at steady state. (3) If m,” suddenly stops 6.62., no more input), at what rate will ha) change at: this instant? SOLUTION: Part (1): mm 3 pvam w your m 111:; w 0.12kg/ s 2 p out 3 2 .4 m 800k / 1m 10 ( g M m>[ 0,] Calculate to conclude: v : 0_4775m/S 0 [If Part (2): Assume steady state, then: n‘z. a n‘: m our The volume flow rate, Q, is defined by: 0.12kg/S w=15x104m3fs 215063711? /S g m Part (3): If the inflow is suddenly stopped, then Eq.(5.4) becomes: = —mout Since 1112?.” m 0 I Now the volume of a cone is given by the formula: V 2—251”? tan2(6) Here: 1101‘) The previous equation becomes: d 77 — mh3tan2 (9 Jmm, 961113 ( ) on Take the derivative: dh 71.")7121'Lan2 9 ———=——r§1 p ( ) our Solve for the time rate of change of the oil level: g3 M m cit plfi‘z ~0.12kg / s (800kg/s)(7564cm2)[ out ,_,. Here: rmhtama) and tan(45“)=1 2 10“ XL cm2 J m —0.7460cm / S MAE 3161: Fluid Mechanics Notes by Paavo Sepri LECTURE 11 Chagter 5 5.2 INTEGRAL FORM OF THE MOMENTUM EQUATION As with the conservation of mass principle (Lagrangian View), we follow a fluid element of fixed mass as it moves in space. ? Wt) Here: M : m.de E constant m) Appiy Newton's 2‘“1 Law of Motion to M : d ‘ _ W(Mv) = F dz‘ Or, in integrai form: fl mm] (as) VG) @ There are two types of forces that are applied to M : (1) Surface Force: Molecules from outside of the bounding surface, S (t), collide with molecules on the inside. Such a force would be called a short-range force. A £= M +£% (Pg “£4. Sham" 55.9.65 :4 <3" 5-“— “F cowPoweui: “'5 1: (2) Long—range Force: In our case this wouid be gravity. The presence of the earth produces a downward force on every massive body. Other examples might be externally applied magnetic or electric fieids on susceptibie fluids. 35.";234“ we On the larger fluid element, the total forces are described by integrals: m HfidS Where: fs = force per unit area. (5.6) so) P Surf:er FEW 2 HI}?st Where: f3 = force per unit volume. (5.7) W) As in the case of mass conservation, we use the Reynoids Transport Theorem {Eq.(4.12)] to transform to the Eulerian cowardinate frame: 3; m deV. ~2- fll—aigfldm + lime-mg = mgdw mars (5.3) {‘ r) V S V0} SO) 5.2.1 EXAMPLE #1 OF THE MOMENTUM EQUATION Consider a stationary curved pipe, through which a fluid flows at a steady rate. What is the relation between force on the pipe and the fluid flow? A. M I “3 f , YLI W s'uu" ace cross»- Sector/L x)? r, " Q‘il‘ 3 n r \\\6’ Elk \( A A Ii. “‘2 n3 “'3 First, apply conservation of mass [Eq.(5.3)]: g; p0,?)de a [$61K + flpmws =0 V{:“c,r) 6 Since the flow is steady, we a O , and the corresponding volume integral vanishes. at There are three surface integrals to be considered: J.ij (73$)! dSI + “1011 (fi'vh dSII 4" flpm (mi—5)!!! 6113171 2 0 SI 5;: 511; By considering averaged properties aiong the three surfaces, these may be simplified as foliows: —pllei + pzvaz + 9 : 0 Note that the third integral vanishes identically for either of two cases: (1) V} a 0 (for viscous flow - the fluid sticks to the surface), or (2) 173%,, a 0 (for inviscid flow — the flow is tangential to the surface). Next, assuming incompressible flow ( p1 = p2 ), the conservation of mass equation results in the statement that mass flow rate is constant and uniform throughout the pipe cross-sections: pviAi w pvaz E m (1) Next, apply the momentum equation {Eq.(5.8)]: F mflgjdififlpfifirmdsw mirde “ids V 5 V0) so) Again, since the flow is steady, the first integral vanishes, and there are three surface integrals to be considered: 13* H Haw-MS; + j [ pram-m c181; + flame-ems,” 3! SH SH: As before, these integrals may be approximated as foilows: "plvifiiAi '5' [7215172142 + 0 x F The third integral vanishes for the same reasons as given for the conservation of mass. Note that the above is a vector equation (it may be decomposed into three parts; by symmetry the z-component vanishes identically). Use Eq.(1) here to obtain the approximation for the momentum equation: we; w“1.71) =F (2) Next, we consider the forces applied to the pipe: .4. A F: [New Jijids ; f3=“ng VG) Si!) Thus, “Ifng 2 —ijg}'dI/' = mMgf' (gravitational force on the fluid in the pipe) V0) V0) The net force applied to the three surfaces is: [[13,de + Hinds” + fldSm m F; St S :1 5:1; Assume that the flows into surface 1 and out of surface 2 are uniform and paraliei to their respective directions. We consider pressure only. Then: “LOIS; 21011411T Note: 1721 2—1? and j"; m~pifii S; Also: is” 0’8” = p2A2[-—f 005(6) + fsinmfl 511 Call: Hinds}; = Fwall' SH: Combine with Eq.(2): rm, w 171) = wng + F + p2A2[wfcos(t9) e jsinwn + p14? (3) wall Split this vector equation into its two components: A i : 17";[122 003(6) ~14] m FWJr - 192442 cos(8)+ p114! (3a) j : —n‘w2 si1'1(£9) :2 ~Mg + Fwy + pZA2 sin(6) (‘31:) Here we have expressions for the force components on the pipe in terms of pipe geometry and flow speeds and pressures. Also note: The force on the pipe exerted by the internal flow is equal in magnitude but opposite in direction to the force on the fluid as exerted by the pipe (action and reaction). 5.2.2 EXAMPLE #2 OF THE MOMENTUM EQUATION Given a jet engine in steady operation as shown, find the thrust generated: are; we; e comrrem We 4% Begin with the conservation of mass equation, as before: “91711-41 + pzuzAz +7731: m 0 (1) Next, consider the x-component of the momentum equation: fl gigfilde + Jim—um + flew/12 = F; (2) Since the flow is steady, the first integral vanishes, and the rest may be approximated as foliows: “plqul +102“§A2 : Fx (3) Combine Eqs. (1) and (3): F; 2 ~p1u12/11 «1— [92245112 2 p1u1A1[(1.01)u2 wul] since: plug/12 w (1.011095%!1 Apply the given numerical values: -m,‘ :__ £13511 fl‘ 2 f5 W [3: lbf Thrust~ 5; (0.0805fl3)(300 )(10.8fi‘ {(1009008) 3008] 32.2 (15172)“) S2 S Conclude: Thrust =(—4933lbf)f ‘ ...
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Chapter 4 - 69 iViAE 31.61: Fluid Mechanics Notes by Paavo...

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