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 M you 101A: make sure that mass is balanced as Mechanics energy,AEmust 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 ofdthe Reynoldsapplication Theorem (15 points) dummy variable is general form B/dm V, and Transport quation 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 Vd 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 property being diﬀerentiated is mend of this section). relativeacceleration terms (see the V , the linear momentum. 2. The term F is the vector sum of all forces acting on the controlvolume mate• the forces rialthe sum of ALL forces i.e., it includes surface forces on all fluids and considered as F considered as a free body; acting on the controlvolume material 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 mass within the control volume. • ﬁnally, the last term of the equation is the ﬂux of linear momentum, entering and leaving the control volume. 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 since ρ dV +
CV CS ρv · n dA = 0 1 rem to find an expression ass Msalt within the tank. ube of radius R and length he fluid entered the tube /3. As the fluid exits the alpy 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 omment on their physical ux of enthalpy through the form concentration C y introducing fresh air at ea Ai on one wall and exy V0 through a duct A0 on ssion for the instantaneous hin the room. rough a closed tank, as in cm and the volume flow 5 cm and the average vehat is (a) Q3 in m3/h and CV
h 2 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 flow. v · n dA = vA Q=
CS r=R πd2 dh ρ + ρQ2 − ρQ1 − ρQ3 = 0 4 dt r hence
U0 dh x=0 u(r) dt = −Q2 + Q1 + Q3 πd2 /4 x=L P3.15 (b) Then, if the water level h is constant, determine the exit velocity V2 for the given 40 kg/s through the noz3 data V1 = and D2 5 cm, compute 3 m/s and Q3 = 0.01 m /s. fluid flows past an impermeable flat P3.16 An incompressible (a) section 1 and (b) if h is constant, then dh/dt = 0, and then: secplate, as in Fig. P3.16, with a uniform inlet profile u U0 and a cubic polynomial exit profile π π Q2 = Q1 + Q3 = 0.01 + (0.05)2 (3.0) = + (0.07)2 V2 4 4 ﬁnally obtaining: V2 = 4.13 m/s 2 Problem 3 (50 points) A pipe of diameter d = 1 m carries water of density ρ = 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 ρ dV +
CV CS ρv · n dA = 0 (v · n)2 = V2 , (v · n)3 = V3 v · n dA = V1 A1 the mass conservation equation states: −ρV1 A1 + ρV2 A2 + ρV3 A3 = 0, (b) From part (a) ﬁnd the velocities V2 and V3 . and since Q2 = Q3 = Q/2 Q2 = V2 πd2 π (d/2)2 2 = V2 = Q/2; 4 4 8·5 8Q = = 12.73 m/s 2 πd π · 12 Q3 = V 3 πd2 π (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 , ρ, V1 , A, V3 , A3 , p1 and p3 . momentum conservation equation: d dt ρv dV +
CV CS ρv (v · n) dA = −
CS pn = F −
CS pgauge n, now, projecting on the horizontal direction, d dt ρv · ex dV +
CV CS ρv · 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 θ = 45o the horizontal component of the momentum conservation equation leads to: −ρV12 A1 + ρV32 cos θA3 = Fx + p1 A1 − p3 cos θA3 (d) From (c) calculate the reaction force Fx and indicate its direction. Fx = −ρV12 A1 + ρV32 cos θA3 − p1 A1 + p3 cos θA3 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 θ (n · ex )3 = cos θ (e) Express the vertical momentum conservation equation in integral form for the appropriate control volume in terms of Fy , ρ, V2 , A2 , V3 , A3 , p2 and p3 . now, projecting on the vertical direction, d dt ρv · ey dV +
CV CS ρv · 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 θ = 45o (v · ey )2 = V2 , (n · ex )2 = 1, (v · ey )3 = −V3 sin θ (n · ey )3 = − sin θ the horizontal component of the momentum conservation equation leads to: ρV22 A2 − ρV32 sin θA3 = Fy − p2 A2 + p3 sin θA3 (f) From (e) calculate the reaction force Fy and indicate its direction. Fy = ρV22 A2 − ρV32 sin θA3 + p2 A2 − p3 sin θA3 Fy = 17.33 kN upwards n3 CV V2 d2 n1 V1 d3 n2
Figure 2: Problem 3 V3 5 ...
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 Spring '08
 Sakar

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