Chapter 5 Mass, Momentum, and Energy Equations

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 Chater 5 Mass, Momentum, and Energy Equations Flow Rate and Conservation of Mass. cross-sectional area oriented normal to velocity vector (simle case where ) U constant: Q volume flux U [m/s m m 3 /s] U constant: Q Ud Similarly the mass flux m ρud. general case Q m nd cosθd ρ ( n) d

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 average velocity: Q Examle: t low velocities the flow through a long circular tube, i.e. ie, has a arabolic velocity distribution (actually araboloid of revolution). i.e., centerline velocity u u max r R a) find Q and Q nd ud π R ud u(r)rdθdr 0 0 u(r)rdr π R 0 d πrdr π u u(r) and not θ d θ 0 π

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 3 Q π R r u max rdr 0 R u max u maxπ R Continuity Equation RTT can be used to obtain an integral relationshi exressing conservation of mass by defining the extensive roerty B M such that β. B M mass β db/dm General Form of Continuity Equation dm or 0 d ρd + ρ d C ρ d d ρd C net rate of outflow of mass across rate of decrease of mass within C Simlifications: d. Steady flow: ρd 0 C

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 4. constant over discrete d (flow sections): ρ d ρ 3. Incomressible fluid (ρ constant) d d d conservation of volume C 4. Steady One-Dimensional Flow in a Conduit: ρ 0 ρ + ρ 0 for ρ constant Q Q Some useful definitions: Mass flux m ρ d olume flux Q d verage elocity Q / verage Density ρ ρd Note: m ρq unless ρ constant

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 5 Examle ρ d 0 d ρd C i.e., -ρ - ρ + ρ 3 3 3 + ρ *Steady flow *,,3 50 fs *@ varies linearly from zero at wall to max at ie center *find m 4, Q 4, max 0 *water, ρ w.94 slug/ft 3 4 4 d 4 0 ρ const..94 lb-s /ft 4.94 slug/ft 3 m 4 ρ 4d4 ρ( + 3 ) 3 50f/s.94 π 50 ( +.5 ) 44 4.45 slugs/s m 4

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 6 Q 4 m 4 ρ. 75 ft 3 /s 4 Q 4 4 d 4 velocity rofile r o r r π max 0 0 o 4 4 (θ) rdθdr d 4 π π ro max 0 o max r r ro r r r 0 o rdr dr 4 Q 3 3 πr o πr o max max π max r r 0 0 3 r 3r o r o 0 πmaxro 3 Q4 max. 86fs πro 3 3 πr o max

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 7 Momentum Equation RTT with B M and β d [ F S + FB ] ρd + ρ R d C velocity referenced to an inertial frame (non-accelerating) R velocity referenced to control volume F S surface forces + reaction forces (due to ressure and viscous normal and shear stresses) F B body force (due to gravity) lications of the Momentum Equation Initial Setu and Signs. Jet deflected by a late or a vane. Flow through a nozzle 3. Forces on bends 4. Problems involving non-uniform velocity distribution 5. Motion of a rocket 6. Force on rectangular sluice gate 7. Water hammer Derivation of the Basic Equation dbsys d Recall RTT: βρd + βρr d C General form for moving but non-accelerating reference frame R velocity relative to S absolute velocity Subscrit not shown in text but imlied! i.e., referenced to C Let, B M linear momentum β must be referenced to inertial reference frame

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 8 d(m) F Newton s nd law where d ρd + C ρ R d Must be relative to a non-accelerating inertial reference frame ΣF vector sum of all forces acting on the control volume including both surface and body forces ΣF S + ΣF B ΣF S sum of all external surface forces acting at the, i.e., ressure forces, forces transmitted through solids, shear forces, etc. ΣF B sum of all external body forces, i.e., gravity force ΣF x + R x ΣF y -W + R y free body diagram R resultant force on fluid in C due to w and τ w i.e., reaction force on fluid Imortant Features (to be remembered) ) ector equation to get comonent in any direction must use dot roduct x equation d F x ρud + ρur d C Carefully define coordinate system with forces ositive in ositive direction of coordinate axes

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 9 y equation d F y ρvd + ρvr d C z equation d F z ρwd + ρwr d C ) Carefully define control volume and be sure to include all external body and surface faces acting on it. For examle, (R x,r y ) reaction force on fluid (R x,r y ) reaction force on nozzle 3) elocity must be referenced to a non-accelerating inertial reference frame. Sometimes it is advantageous to use a moving (at constant velocity) reference frame. Note R s is always relative to. i.e., in these cases used for B also referenced to C (i.e., R )

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 0 4) Steady vs. Unsteady Flow d Steady flow ρd 0 C 5) Uniform vs. Nonuniform Flow ρ R d change in flow of momentum across Σρ R uniform flow across 6) F res nd fd f nds f constant, f 0 0 for constant and for a closed surface i.e., always use gage ressure 7) Pressure condition at a jet exit S at an exit into the atmoshere jet ressure must be a lication of the Momentum Equation. Jet deflected by a late or vane Consider a jet of water turned through a horizontal angle

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 C and are for jet so that F x and F y are vane reactions forces on fluid d x-equation: F x Fx ρud + ρu d F ρu steady flow x ρ ( ) + ( ) x ρ x continuity equation: F x ρq( x x ) ρ ρ ρq for y-equation: F F ρv y y F y ρ y ( ) + ρ y ( ) ρq( y y ) for above geometry only where: x x - cosθ y - sinθ y 0 note: F x and F y are force on fluid - F x and -F y are force on vane due to fluid If the vane is moving with velocity v, then it is convenient to choose C moving with the vane

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 i.e., R - v and used for B also moving with vane x-equation: F ρu d x R F x ρ x [-( v ) ] + ρ x [-( v ) ] Continuity: 0 ρ d R i.e., ρ(- v ) ρ(- v ) ρ(- v ) Q rel F x ρ(- v )[ x x ] Q rel on fluid x ( v ) x x ( v ) x Power -F x v i.e., 0 for v 0 F y ρq rel ( y y ). Flow through a nozzle Consider a nozzle at the end of a ie (or hose). What force is required to hold the nozzle in lace? C nozzle and fluid (R x, R y ) force required to hold nozzle in lace

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 3 ssume either the ie velocity or ressure is known. Then, the unknown (velocity or ressure) and the exit velocity can be obtained from combined use of the continuity and Bernoulli equations. Bernoulli: + z z + ρ ρ γz + ρ + γz + ρ Continuity: Q D d D + ρ d 4 Say known: 4 ρ ( D ) d To obtain the reaction force R x aly momentum equation in x- direction 0 / F x d uρd + C ρu ρu d steady flow and uniform flow over R x + ρ (- ) + ρ ( ) ρq( - )

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 4 R x ρq( - ) - To obtain the reaction force R y aly momentum equation in y- direction F ρv since no flow in y-direction y 0 R y W f W N 0 i.e., R y W f + W N Numerical Examle: Oil with S.85 flows in ie under ressure of 00 si. Pie diameter is 3 and nozzle ti diameter is S γ ρ.65 g 4.59 ft/s D/d 3 3.3 ft/s Q π R x 4.48 706.86 569 lbf 4 R z 0 lbf.76 ft 3 /s This is force on nozzle 3. Forces on Bends Consider the flow through a bend in a ie. The flow is considered steady and uniform across the inlet and outlet sections. Of rimary concern is the force required to hold the bend in lace, i.e., the reaction forces R x and R y which can be determined by alication of the momentum equation.

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 5 Continuity: 0 ρ ρ + ρ i.e., Q constant x-momentum: F x ρu cosθ + R x ρ ρ Q x ( ) + ρx ( ) ( ) y-momentum: ρv F y sin θ + R y w f w b ρy y x x ρ Q( ) y R x, R y reaction force on bend i.e., force required to hold bend in lace ( ) + ρ ( ) y 4. Problems involving Nonuniform elocity Distribution See text. 5 6

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 6 5. Force on a rectangular sluice gate The force on the fluid due to the gate is calculated from the x- momentum equation: ρu F F x ( ) + ρ ( ) + FGW Fvisc F ρ GW F F + ρq( ) Fvisc y y γ y b γ y b + ρq( ) F + F GW bγ ( y y ) + ρq( ) ρq b y Moment of Momentum Equation See text. 9 y usually can be neglected Q yb Q y b

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 7 Energy Equations Derivation of the Energy Equation The First Law of Thermodynamics The difference between the heat added to a system and the work done by a system deends only on the initial and final states of the system; that is, deends only on the change in energy E: rincile of conservation of energy E Q W E change in energy Q heat added to the system W work done by the system E E u + E k + E total energy of the system otential energy kinetic energy Internal energy due to molecular motion The differential form of the first law of thermodynamics exresses the rate of change of E with resect to time de Q W rate of work being done by system rate of heat transfer to system

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 8 Energy Equation for Fluid Flow The energy equation for fluid flow is derived from Reynolds transort theorem with B system E total energy of the system (extensive roerty) β E/mass e energy er unit mass (intensive roerty) û + e k + e de d Cρed + ρe d d Q W ρ( uˆ+ e ) ( ˆ k + e d + ρ u+ ek + e) d C This can be ut in a more useable form by noting the following: e e k TotalKEof mass with velocity M / mass M E γ z gz (for E due to gravity only) M ρ d Q W ρ gz uˆ d ρ gz uˆ d + + + + + C Cs rate of work rate of change flux of energy done by system of energy in C out of C (ie, across ) rate of heat transfer to sysem

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 9 Rate of Work Comonents: W W s + W f For convenience of analysis, work is divided into shaft work W s and flow work W f W f net work done on the surroundings as a result of normal and tangential stresses acting at the control surfaces W f ressure + W f shear W s any other work transferred to the surroundings usually in the form of a shaft which either takes energy out of the system (turbine) or uts energy into the system (um) Flow work due to ressure forces W f (for system) Note: here uniform over C System at time t + t System at time t Work force distance at W t rate of work W (on surroundings) neg. sign since ressure force on surrounding fluid acts in a direction oosite to the motion of the system boundary at W t W

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 0 In general, W f for more than one control surface and not necessarily uniform over : W f d ρ d ρ W W + W f f fshear Basic form of energy equation Q W s W fshear ρ d ρ d ρ + gz + uˆ d + ρ + gz + uˆ d C d Q W ˆ s W fshear gz u d ρ + + C + ρ + gz+ uˆ + d ρ henthaly Usually this term can be eliminated by roer choice of C, i.e. normal to flow lines. lso, at fixed boundaries the velocity is zero (no sli condition) and no shear stress flow work is done. Not included or discussed in text!

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 Simlified Forms of the Energy Equation Energy Equation for Steady One-Dimensional Pie Flow Consider flow through the ie system as shown Energy Equation (steady flow) Q W ˆ s ρ + gz+ + u d ρ Q ρ W s + gz u d + + + ρ ρ + + + ρ 3 ˆ ρ 3 gz uˆ ρ d *lthough the velocity varies across the flow sections the streamlines are assumed to be straight and arallel; consequently, there is no acceleration normal to the streamlines and the ressure is hydrostatically distributed, i.e., /ρ +gz constant.

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 *Furthermore, the internal energy u can be considered as constant across the flow sections, i.e. T constant. These quantities can then be taken outside the integral sign to yield 3 Q W ˆ s + + gz+ u ρ d ρ d ρ + 3 + gz ˆ + u ρ d + ρ d ρ Recall that So that Q d ρ d ρ m mass flow rate 3 ρ Define: ρ d α α m K.E. flux K.E. flux for constant across ie 3 i.e., α d kinetic energy correction factor Q W + + gz ˆ ˆ + u+ α m + gz + u + α m ρ ρ m ( ) 3 ˆ α ˆ α Q W + gz u gz u ρ + + + ρ + + + Nnote that: α if is constant across the flow section α > if is nonuniform laminar flow α turbulent flow α.05 may be used

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 3 Shaft Work Shaft work is usually the result of a turbine or a um in the flow system. When a fluid asses through a turbine, the fluid is doing shaft work on the surroundings; on the other hand, a um does work on the fluid W s W t W where W t and W are work magnitudes of ower time Using this result in the energy equation and deviding by g results in W W ˆ ˆ t u u Q + + z+ α + + z + α + mg γ mg γ g mg mechanical art thermal art Note: each term has dimensions of length Define the following: h W mg W ρqg W γq h t W t mg h L uˆ uˆ Q g mg headloss

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 4 Head Loss In a general fluid system a certain amount of mechanical energy is converted to thermal energy due to viscous action. This effect results in an increase in the fluid internal energy. lso, some heat will be generated through energy dissiation and be lost (i.e. -Q ). Therefore the term from nd law h L uˆ uˆ Q g gm > 0 reresents a loss in mechanical energy due to viscous stresses Note that adding Q to system will not make h L 0 since this also increases u. It can be shown from nd law of thermodynamics that h L > 0. Dro over and understand that in energy equation refers to average velocity. Using the above definitions in the energy equation results in (steady -D incomressible flow) γ + α + z + h + α + z + h t + h L g γ g form of energy equation used for this course!

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 5 Comarison of Energy Equation and Bernoulli Equation ly energy equation to a stream tube without any shaft work Infinitesimal stream tube α α Energy eq : γ + + z + + z + h L g γ g If h L 0 (i.e., µ 0) we get Bernoulli equation and conservation of mechanical energy along a streamline Therefore, energy equation for steady -D ie flow can be interreted as a modified Bernoulli equation to include viscous effects (h L ) and shaft work (h or h t ) Summary of the Energy Equation The energy equation is derived from RTT with B E total energy of the system β e E/M energy er unit mass

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 6 û + +gz internal KE PE de d ρed + C ρe d Q W heat add work done from st Law of Thermodynamics W W + W + W s Neglected in text resentation v shaft work done on or by system (um or turbine) ressure work done on iscous stress work on W d ρ C ( ρ) d W s W Q W t t W + W ˆ d ρed + C e u+ + gz ρ ( e + e) d For steady -D ie flow (one inlet and one outlet): ) Streamlines are straight and arallel /ρ +gz constant across

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 7 ) T constant u constant across 3 3) define α d KE correction factor γ ρ 3 ρ d α α mechanical energy 3 m + α + z + h + α + z + h t + h L g γ g Thermal energy h h t W W t mg mg has Note: each term units of length is average velocity (vector droed) and corrected by α uˆ ˆ u Q hl g mg head loss > 0 reresents loss in mechanical energy due to viscosity

EGL EGL + h L for h h t 0 57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 8 Concet of Hydraulic and Energy Grade Lines + α + z + h + α + z + h t + h L γ g γ Define HGL + z γ EGL + z + α γ g g HGL corresonds to ressure ta measurement + z EGL corresonds to stagnation tube measurement + z EGL HGL if 0 oint-by-oint alication is grahically dislayed h L f L D g i.e., linear variation in L for D,, and f constant f friction factor f f(re) ressure ta: γ stagnation tube: γ h + α g h h height of fluid in ta/tube EGL + h EGL + h t + h L EGL EGL + h h t h L abrut change due to h or h t f L D g

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 9 Helful hints for drawing HGL and EGL. EGL HGL + α /g HGL for 0 L.&3. h L f in ie means EGL and HGL will sloe D g downward, excet for abrut changes due to h t or h + z + + z + γ g γ g HGL EGL - h L h L for abrut exansion g + h L

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 30 4. 0 HGL z 5. for L h L f constant L D g EGL/HGL sloe downward 6. for change in D change in i.e., linearly increased for increasing L with sloe f D g i.e. πd πd 4 4 D D change in distance between HGL & EGL and sloe change due to change in h L

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 3 7. If HGL < z then /γ < 0 i.e., cavitation ossible condition for cavitation: va N 000 m N gage ressure va,g atm atm 00,000 m va, g γ 0m 980 N/m 3

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 3

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 33

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 34 lication of the Energy, Momentum, and Continuity Equations in Combination In general, when solving fluid mechanics roblems, one should use all available equations in order to derive as much information as ossible about the flow. For examle, consistent with the aroximation of the energy equation we can also aly the momentum and continuity equations Energy: + α γ + z + h + α + z + h t + h L g γ g Momentum: F s ρ ρ ρq Continuity: Q constant ( ) one inlet and one outlet ρ constant

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 35 brut Exansion Consider the flow from a small ie to a larger ie. Would like to know h L h L (, ). nalytic solution to exact roblem is extremely difficult due to the occurrence of flow searations and turbulence. However, if the assumtion is made that the ressure in the searation region remains aroximately constant and at the value at the oint of searation, i.e,, an aroximate solution for h L is ossible: ly Energy Eq from - (α α ) + z + + z + + h L γ g γ g Momentum eq. For C shown (shear stress neglected) Wsin α ρu Fs ρ ( ) + ρ ( ) z γ ρ ρ L L W sin α next divide momentum equation by γ

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 36 γ ( ) γ γ g g g z z from energy equation L g g h g g + + L g g h + L g h [ ] L g h If, L h g continutity eq.

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 37 Forces on Transitions Examle 7-6 Q.707 m 3 /s head loss. g (emirical equation) Fluid water 50 kpa D 30 cm D 0 cm F x? First aly momentum theorem ρu F x F x + ρ ( ) + ρ ( ) F x ρq( ) + force required to hold transition in lace

57:00 Mechanics of Fluids and Transort Processes Chater 5 Professor Fred Stern Fall 006 38 The only unknown in this equation is, which can be obtained from the energy equation. γ + + + h L g γ g note: z z and α γ + h L dro in ressure g g g Fx ρq( ) + γ + h L g (note: if 0 same as nozzle) In this equation, Q/ 0 m/s Q/.5 m/s h L..58m g continuity i.e. > F x 8.5 kn is negative x direction to hold transition in lace