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Chapter 5 Mass, Momentum, and Energy Equations

57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 1. Chapter 5 mass , Momentum, and Energy Equations Flow Rate and Conservation of mass 1. cross-sectional area oriented normal to velocity vector (simple case where V A). U = constant: Q = volume flux = UA [m/s m2 = m3/s]. U constant: Q = UdA. A. = UdA. Similarly the mass flux = m A. 2. general case Q = V ndA. CS. = V cos dA. CS. = (V n )dA. m CS. 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 2. Q. average velocity: V =. A. Example: At low velocities the flow through a long circular tube, pipe, has a parabolic velocity distribution (actually paraboloid of revolution). r 2 . u = u max 1 . R .. , centerline velocity a) find Q and V. Q = V ndA = udA. A A. 2 R. udA = u (r ) rd dr A 0 0. R. = 2 u (r ) rdr 0. dA = 2 rdr 2.

Chapter 5 Mass, Momentum, and Energy Equations Flow Rate and Conservation of Mass 1. cross-sectional area oriented normal to velocity vector (simple case where V ⊥ A) U = constant: Q = volume flux = UA [m/s × m2 = m3/s] U ≠ constant: Q = ∫ A UdA Similarly the mass flux = =∫ρ A m UdA 2. general case = ∫ θ = ∫ ⋅ CS CS V cos dA Q ...

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Transcription of Chapter 5 Mass, Momentum, and Energy Equations

1 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 1. Chapter 5 mass , Momentum, and Energy Equations Flow Rate and Conservation of mass 1. cross-sectional area oriented normal to velocity vector (simple case where V A). U = constant: Q = volume flux = UA [m/s m2 = m3/s]. U constant: Q = UdA. A. = UdA. Similarly the mass flux = m A. 2. general case Q = V ndA. CS. = V cos dA. CS. = (V n )dA. m CS. 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 2. Q. average velocity: V =. A. Example: At low velocities the flow through a long circular tube, pipe, has a parabolic velocity distribution (actually paraboloid of revolution). r 2 . u = u max 1 . R .. , centerline velocity a) find Q and V. Q = V ndA = udA. A A. 2 R. udA = u (r ) rd dr A 0 0. R. = 2 u (r ) rdr 0. dA = 2 rdr 2.

2 U = u(r) and not d = 2 . 0. 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 3. R r 2 1. Q = 2 u max 1 rdr = u max R 2. R 2. 0 . 1. V = u max 2. Continuity Equation RTT can be used to obtain an integral relationship expressing conservation of mass by defining the extensive property B = M. such that = 1. B = M = mass = dB/dM = 1. General Form of Continuity Equation dM d =0= dV + V dA. dt dt CV CS. or d V dA = dV. CS dt CV. net rate of outflow rate of decrease of of mass across CS mass within CV. Simplifications: d 1. Steady flow: dV = 0. dt CV. 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 4. 2. V = constant over discrete dA (flow sections): V dA = V A. CS CS. 3. Incompressible fluid ( = constant). d V dA = . CS.. dt CV. dV conservation of volume 4. Steady One-Dimensional Flow in a Conduit: V A = 0.

3 CS. 1V1A1 + 2V2A2 = 0. for = constant Q1 = Q2. Some useful definitions: mass flux = V dA. m A. Volume flux Q = V dA. A. Average Velocity V = Q/A. 1. Average Density = dA. A. Q unless = constant Note: m 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 5. Example *Steady flow *V1,2,3 = 50 fps *@ u V varies linearly from zero at wall to Vmax at pipe center 4 , Q4, Vmax *find m 0 *water, w = slug/ft3. d V dA = 0 = dV. CS dt CV 4. m , - 1V1A1 - 2V2A2 + 3V3A3 + V4 dA 4 = 0. A4. = const. = lb-s2 /ft4 = slug/ft3. 4 = V4 dA 4 = V(A1 + A2 A3). m V1=V2=V3=V=50f/s =. 144.. (. 50 12 + 2 2 2. 4. ). = slugs/s 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 6. 4 = .75 ft3/s Q4 = m = V4 dA 4. A4. velocity profile . ro 2 . Q4 = Vmax 1 r rd dr 0 0 ro . dA4. V4 V4( ).. ro r.

4 = 2 Vmax 1 rdr 0 ro 1 2. r V. Q 3 o max V4 = =. ro . r2 A ro2. = 2 Vmax r dr 1. 0 ro = Vmax 3. r2 r0. r3. ro . = 2 Vmax . 2 0. 3ro 0 .. 1 1 1. = 2 Vmax ro 2 = ro 2 Vmax 2 3 3. Q4. Vmax = = fps 1 2. ro 3. 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 7. Momentum Equation RTT with B = MV and = V. [FS + F B ] =. d VdV + V V R dA. dt CV CS. V = velocity referenced to an inertial frame (non-accelerating). VR = velocity referenced to control volume FS = surface forces + reaction forces (due to pressure and viscous normal and shear stresses). FB = body force (due to gravity). Applications of the Momentum Equation Initial Setup and Signs 1. Jet deflected by a plate or a vane 2. 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.

5 Water hammer Derivation of the Basic Equation General form for dBsys d moving but Recall RTT: = dV + V R dA non-accelerating dt dt CV CS reference frame VR=velocity relative to CS=V VS=absolute velocity CS. Subscript not shown in text but implied! , referenced to CV. Let, B = MV = linear momentum V must be referenced to =V inertial reference frame 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 8. d(M V) d = F = V dV + V V R dA. dt dt CV CS. Must be relative to a non-accelerating Newton's 2nd law inertial reference frame where F = vector sum of all forces acting on the control volume including both surface and body forces = FS + FB. FS = sum of all external surface forces acting at the CS, , pressure forces, forces transmitted through solids, shear forces, etc. FB = sum of all external body forces, , gravity force Fx = p1A1 p2A2 + Rx Fy = -W + Ry R = resultant force on fluid in CV due to pw and w free body diagram , reaction force on fluid Important Features (to be remembered).

6 1) Vector equation to get component in any direction must use dot product Carefully define coordinate x equation system with forces positive in d Fx = udV + u V R dA positive direction of dt CV CS coordinate axes 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 9. y equation d Fy = vdV + v V R dA. dt CV CS. z equation d Fz = wdV + w V R dA. dt CV CS. 2) Carefully define control volume and be sure to include all external body and surface faces acting on it. For example, (Rx,Ry) = reaction force on fluid (Rx,Ry) = reaction force on nozzle 3) Velocity V must be referenced to a non-accelerating inertial reference frame. Sometimes it is advantageous to use a moving (at constant velocity) reference frame. Note VR = V. Vs is always relative to CS. , in these cases V used for B also referenced to CV. ( , V = VR).

7 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 10. 4) Steady vs. Unsteady Flow d Steady flow VdV = 0. dt CV. 5) Uniform vs. Nonuniform Flow V V R dA = change in flow of momentum across CS. CS. = V VR A uniform flow across A. 6) Fpres = p ndA fdV = f nds V S. f = constant, f = 0. = 0 for p = constant and for a closed surface , always use gage pressure 7) Pressure condition at a jet exit at an exit into the atmosphere jet pressure must be pa Application of the Momentum Equation 1. Jet deflected by a plate or vane Consider a jet of water turned through a horizontal angle 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 11. CV and CS are for jet so that Fx and Fy are vane reactions forces on fluid d x-equation: Fx = Fx = udV + u V dA. dt CS. Fx = u V A steady flow CS.

8 = V1x ( V1A1 ) + V2 x (V2 A 2 ). continuity equation: A1V1 = A2V2 = Q for A1 = A2. V1 = V2. Fx = Q(V2x V1x). y-equation: Fy = Fy = v V A. CS. Fy = V1y( A1V1) + V2y( A2V2). = Q(V2y V1y). for above geometry only where: V1x = V1 V2x = -V2cos V2y = -V2sin V1y = 0. note: Fx and Fy are force on fluid - Fx and -Fy are force on vane due to fluid If the vane is moving with velocity Vv, then it is convenient to choose CV moving with the vane 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 12. , VR = V - Vv and V used for B also moving with vane x-equation: Fx = u V R dA. CS. Fx = V1x[-(V Vv)1A1] + V2x[-(V Vv)2A2]. Continuity: 0 = V R dA. , (V-Vv)1A1 = (V-Vv)2A2 = (V-Vv)A. Qrel Fx = (V-Vv)A[V2x V1x]. Qrel on fluid V2x = (V Vv)2x V1x = (V Vv)1x Power = -FxVv , = 0 for Vv = 0. Fy = Qrel(V2y V1y). 2. Flow through a nozzle Consider a nozzle at the end of a pipe (or hose).

9 What force is required to hold the nozzle in place? CV = nozzle and fluid (Rx, Ry) =. force required to hold nozzle in place 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 13. Assume either the pipe velocity or pressure is known. Then, the unknown (velocity or pressure) and the exit velocity V2 can be obtained from combined use of the continuity and Bernoulli Equations . 1 1. Bernoulli: p1 + z1 + V12 = p 2 + z 2 + V22 z1=z2. 2 2. 1 1. p1 + V12 = V22. 2 2. Continuity: A1V1 = A2V2 = Q. 2. A1 D . V2 = V1 = V1. A2 d . 1 2 D . 4. p1 + V1 1 = 0. 2 d .. 1/ 2.. 2p1 . Say p1 known: V1 = 4 . 1 d .. D.. ( ). To obtain the reaction force Rx apply momentum equation in x- direction d Fx = u dV + u V dA. dt CV CS. = u V A steady flow and uniform CS flow over CS. Rx + p1A1 p2A2 = V1(-V1A1) + V2(V2A2). = Q(V2 - V1).

10 57:020 Mechanics of Fluids and Transport Processes Chapter 5. Professor Fred Stern Fall 2006 14. Rx = Q(V2 - V1) - p1A1. To obtain the reaction force Ry apply momentum equation in y- direction Fy = vV A = 0 since no flow in y-direction CS. Ry Wf WN = 0 , Ry = Wf + WN. Numerical Example: Oil with S = .85 flows in pipe under pressure of 100 psi. Pipe diameter is 3 and nozzle tip diameter is 1 S . = = g V1 = ft/s D/d = 3. V2 = ft/s 2. 1 . Q = V2. Rx = = 569 lbf 4 12 . Rz = 10 lbf = .716 ft3/s This is force on nozzle 3. Forces on Bends Consider the flow through a bend in a pipe. The flow is considered steady and uniform across the inlet and outlet sections. Of primary concern is the force required to hold the bend in place, , the reaction forces Rx and Ry which can be determined by application of the momentum equation. 57:020 Mechanics of Fluids and Transport Processes Chapter 5.


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