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Unformatted text preview: An illustration of a mass balance with a deforming control volume has already been given in Example 3.2. The controlvolume mass relations, Eq. (3.20) or (3.21), are fundamental to all fluidflow analyses. They involve only velocity and density. Vector directions are of no consequence except to determine the normal velocity at the surface and hence whether the flow is in or out. Although your specific analysis may concern forces or moments or MAE 101A: make sure that mass is balanced as Mechanics energy, you must always Introductory Fluid part of the analysis; otherwise the results will be unrealistic and probably 2 Midterm rotten. We shall see in the examples which follow how mass conservation is constantly checked in performing an analysis 02/22/10 of other fluid properties. In Newton's law, Eq. (3.2), the property being differentiated is the linear momentum .4 The Linear Momentum Problem 1 mV. Therefore our Write down the B mV and ofdB/dm V, and application Theorem (15 points) dummy variable is general form the Reynolds Transport Equation for conservation of momentum and describethe linearmomentum relationof each term. of the Reynolds transport theorem gives in words the meaning for a deformable control volume d (mV)syst dt 1. F d dt
CV V d CS V (Vr n) dA (3.35) The following points concerning this relation should be strongly emphasized: The term V is the fluid velocity relative to an inertial (nonaccelerating) coordinate system; otherwise Newton's law must be modified to include noninertial The LHS of the equation is the (see the end of thisof linear momentum mV of the system. relativeacceleration terms rate of change section). 2. The linearThe term F is of the system is conserved. on the controlvolume matemomentum the vector sum of all forces acting rial considered as a free body; i.e., it includes surface forces on all fluids and the forces F the sum of ALL forces acting on the controlvolume material considered as a free body, that is surface forces (for instance, pressure forces) as well as body forces (such as gravity). next term in the rate of change of the linear momentum within the control volume (storage of linear momentum within the control volume) finally, the last term of the equation is the flux of linear momentum, entering and leaving the control volume throughout the control surfaces. Problem 2 (35 points) The open cylindrical tank in the figure contains water at 20o C and is being filled through sections 1 and 3. Assuming incompressible flow, apply the mass conservation equation in integral form and: (a) First, derive an analytic expression for the waterlevel change dh/dt in terms of arbitrary volume flows (Q1 , Q2 , Q3 ) and tank diameter d. for a control volume enclosing the tank, as shown in figure, apply the continuity equation in integral form, assuming: unsteady flow incompressible uniform velocity on each inlet and outlet d dt dV +
CV CS v n dA = 0 1 orem to find an expression mass Msalt within the tank. tube of radius R and length The fluid entered the tube w/3. As the fluid exits the halpy profiles are approxicpTw 1 2 r2 R2 flow. First derive an analytic expression for the waterlevel change dh/dt in terms of arbitrary volume flows (Q1, Q2, Q3) and tank diameter d. Then, if the water level h is constant, determine the exit velocity V2 for the given data V1 3 m/s and Q3 0.01 m3/s. n3
1 3 Q3 = 0.01 m 3/s comment on their physical lux of enthalpy through the CV
h 2 iform concentration C by introducing fresh air at rea Ai on one wall and exty V0 through a duct A0 on ession for the instantaneous thin the room. hrough a closed tank, as in 6 cm and the volume flow 5 cm and the average vewhat is (a) Q3 in m3/h and n1
V1 D1 = 5 cm V2 n2
Water D2 = 7 cm d P3.14 2 Figure 1: Problem steadily through the P3.15 Water, assumed incompressible, flows 2 round pipe in Fig. P3.15. The entrance velocity is constant, u U0, and the exit velocity approximates turbulent flow, u umax(1 r/R)1/7. Determine the ratio U0 /umax for this since flow. Q r = R=
CS v n dA = vA hence
P3.15 d2 dh r + Q2  Q1  Q3 = 0 4 dt
U0 u(r) 3 x = 0 dh dt t 40 kg/s through the noz(b) m and D2 5 cm, compute Then, if the water level h is constant, determine the exit velocity V2 for the given P3.16 An incompressible fluid flows past an impermeable flat data sec(a) section 1 and (b) V1 = 3 m/s and Q3 = 0.01 m3P3.16, with a uniform inlet profile u U0 plate, as in Fig. /s. and a cubic polynomial then: if h is constant, then dh/dt = 0, andexit profile Q2 = Q1 + Q3 = 0.01 + finally obtaining: V2 = 4.13 m/s (0.05)2 (3.0) = + (0.07)2 V2 4 4 = Q2 + Q1 + Q3 x = L d2 /4 2 Problem 3 (50 points) A pipe of diameter d = 1 m carries water of density = 1000 kg/m3 at a volume flow rate Q = 5 m3 /s, with a pressure at this inlet of 200 kP a. The pipe splits into two smaller pipes, both of diameter d2 = d3 = 0.5 m, shown in the figure. The flow at both outlets is at a static pressure of 139.26 kP a and the flow rate Q splits symmetrically. Find the reaction forces at the junction Fx and Fy exerted by the flow, following the steps below: (a) Apply the mass conservation equation in integral form to the appropriate control volume. Continuity equation in integral form is d dt Under the assumptions: steady state incompressible uniform velocity on each inlet and outlet control volume and normal to each control surface as shown in figure and knowing that: (v n)1 = V1 , and: Q=
CS1 dV +
CV CS v n dA = 0 (v n)2 = V2 , (v n)3 = V3 v n dA = V1 A1 the mass conservation equation states: V1 A1 + V2 A2 + V3 A3 = 0, (b) From part (a) find the velocities V2 and V3 . and since Q2 = Q3 = Q/2 Q2 = V2 d2 (d/2)2 2 = V2 = Q/2; 4 4 85 8Q = = 12.73 m/s 2 d 12 Q3 = V 3 d2 (d/2)2 3 = V3 = Q/2; 4 4 or Q2 + Q3 = Q V2 = V3 = and V1 = V3 /2 = 6.37 m/s 3 (c) Express the horizontal momentum conservation equation in integral form for the appropriate control volume in terms of Fx , , V1 , A, V3 , A3 , p1 and p3 . momentum conservation equation: d dt v dV +
CV CS v(v n) dA = 
CS pn = F 
CS pgauge n, now, projecting on the horizontal direction, d dt v ex dV +
CV CS v ex (v n) dA = 
CS pn ex = Fx 
CS pgauge n ex , using the abovementioned assumptions, and noticing that: (v n)1 = V1 , (v ex )1 = V1 , (n ex )1 = 1, where = 45o the horizontal component of the momentum conservation equation leads to: V12 A1 + V32 cos A3 = Fx + p1 A1  p3 cos A3 (d) From (c) calculate the reaction force Fx and indicate its direction. Fx = V12 A1 + V32 cos A3  p1 A1 + p3 cos A3 Fx = 147.1 kN to the lef t (v n)2 = V2 , (v ex )2 = 0, (n ex )2 = 0, (v n)3 = V3 (v ex )3 = V3 cos (n ex )3 = cos (e) Express the vertical momentum conservation equation in integral form for the appropriate control volume in terms of Fy , , V2 , A2 , V3 , A3 , p2 and p3 . now, projecting on the vertical direction, d dt v ey dV +
CV CS v ey (v n) dA = 
CS pn ey = Fx 
CS pgauge n ey , using the abovementioned assumptions, and noticing that: (v n)1 = V1 , (v n)2 = V2 , 4 (v n)3 = V3 (v ey )1 = 0, (n ex )1 = 0, where = 45o (v ey )2 = V2 , (n ex )2 = 1, (v ey )3 = V3 sin (n ey )3 =  sin the horizontal component of the momentum conservation equation leads to: V22 A2  V32 sin A3 = Fy  p2 A2 + p3 sin A3 (f) From (e) calculate the reaction force Fy and indicate its direction. Fy = V22 A2  V32 sin A3 + p2 A2  p3 sin A3 Fy = 17.33 kN upwards n3 CV V2 d2 n1 V1 d3 n2
Figure 2: Problem 3 V3 5 ...
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 Winter '10
 Marsden
 Fluid Dynamics

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