Integral Relations for a Control Volume Part 1 (Solutions)

Integral Relations for a Control Volume 3.1 Discuss Newton’s second law (the linear momentum relation) in these three forms: % F = ma %F = d ( mV) dt %F = " d! # ! Vρ dυ $ # $ dt system & ' Solution: These questions are just to get the students thinking about the basic laws of mechanics. They are valid and equivalent for constant-mass systems, and we can make use of all of them in certain fluids problems, e.g. the #1 form for small elements, #2 form for rocket propulsion, but the #3 form is control-volume related and thus the most popular in this chapter. 3.2 Consider the angular-momentum relation in the form # MO = " d ! $ ! ( r × V) ρ dυ % dt $ system %' & What does r mean in this relation? Is this relation valid in both solid and fluid mechanics? Is it related to the linear-momentum equation (Prob. 3.1)? In what manner? Solution: These questions are just to get the students thinking about angular momentum versus linear momentum. One might forget that r is the position vector from the moment-center O to the elements ρ dυ where momentum is being summed. Perhaps rO is a better notation. 3.3 For steady laminar flow through a long tube (see Prob. 1.12), the axial velocity 2 2 distribution is given by u = C(R − r ), where R is the tube outer radius and C is a constant. Integrate u(r) to find the total volume flow Q through the tube. Solution: The area element for this axisymmetric flow is dA = 2π r dr. From Eq. (3.7), R 2 2 Q = ! u dA = ! C (R −r )2π r dr = 0 π CR 4 2 Ans. 3.4 Discuss whether the following flows are steady or unsteady: (a) flow near an automobile moving at 55 m/h; (b) flow of the wind past a water tower; (c) flow in a pipe as the downstream valve is opened at a uniform rate; (d) river flow over a spillway of a dam; and (e) flow in the ocean beneath a series of uniform propagating surface waves. Solution: (a) steady (except for vortex shedding) in a frame fixed to the auto. (b) steady (except for vortex shedding) in a frame fixed to the water tower. (c) unsteady by its very nature (accelerating flow). (d) steady except for fluctuating turbulence. (e) Uniform periodic waves are steady when viewed in a frame fixed to the waves. 3.5 A theory proposed by S. I. Pai in 1953 gives the following velocity values u(r) for turbulent (high-Reynolds number) airflow in a 4-cm-diameter tube: r, cm u, m/s 0 6.00 0.25 5.97 0.5 5.88 0.75 5.72 1.0 5.51 1.25 5.23 1.5 4.89 1.75 4.43 2.0 0.00 Comment on these data vis-a-vis laminar flow, Prof. 3.3. Estimate, as best you can, the total volume flow Q through the tube, in m3/s. Solution: The data can be plotted in the figure below. As seen in the figure, the flat (turbulent) velocities do not resemble the parabolic laminarflow profile of Prob. 3.3. (The discontinuity at r = 1.75 cm is an artifact—we need more data for 1.75 < r < 2.0 cm.) The volume flow, Q = ! u(2π r)dr, can be estimated by a numerical quadrature formula such as Simpson’s rule. Here there are nine data points: ! ∆r " Q = 2π (r1u1 + 4r2u2 + 2r3u3 + 4r4u4 + 2r5u5 + 4r6u6 + 2r7u7 + 4r8u8 + r9u9 ) # $ % 3 & For the given data, Q ≈ 0.0059 m 3 /s Ans. 3.6 When a gravity-driven liquid jet issues from a slot in a tank, as in Fig. P3.6, an approximation for the exit velocity distribution is u ≈ 2g(h − z), where h is the depth of the jet centerline. Near the slot, the jet is horizontal, two-dimensional, and of thickness 2L, as shown. Find a general expression for the total volume flow Q issuing from the slot; then take the limit of your result if L ! h. Solution: Fig. P3.6 Let the slot width be b into the paper. Then the volume flow from Eq. (3.7) is Q = ! u dA = +L ! −L [2g(h − z)]1/2 b dz = 2b √ (2g)[(h + L)3/2 − (h − L)3/2 ] Ans. 3 In the limit of L ! h, this formula reduces to Q ≈ (2Lb) (2gh) Ans. 3.7 In Chap. 8 a theory gives the velocities for flow past a cylinder: vr = U cos θ(1 − R 2 /r 2 ) vθ = −U sinθ (1 + R2 /r 2 ) Compute the volume flow Q passing through surface CC in the figure. Fig. P3.7 Solution: This problem is quite laborious and illustrates the difficulty of working with polar velocity components passing through a cartesian (plane) surface. From Eq. (3.7), Q = ! ( V ⋅ n) dA = +R ! (v r cosθ −vθ sin θ )b dy, where y = r sin θ −R We have to integrate over surface CC, where r varies from 2R to 2.24R and θ varies from 153.4° to 206.6°. Inserting the velocity components as given we obtain +R Q= ! [U cos 2θ (1 − R 2 /r 2 ) + Usin 2θ (1 + R 2 /r 2 )]bdy |x=−2R = 1.6URb A ns. −R The integration is messy. In Chaps. 4&8 we find the result easily using the stream function. 3.8 Three pipes steadily deliver water at 20°C to a large exit pipe in Fig. P3.8. The velocity V2 = 5 m/s, and the exit flow rate Q4 = 120 m3/h. Find (a) V1; (b) V3; and (c) V4 if it is known that increasing Q3 by 20% would increase Q4 by 10%. Solution: (a) For steady flow we have Q1 + Q2 + Q3 = Q4, or Fig. P3.8 V1 A1 + V2 A2 + V3 A3 = V4 A4 (1) Since 0.2Q3 = 0.1Q4, and Q4 = (120 m3/h)(h/3600 s) = 0.0333 m3/s, V3 = Q4 (0.0333 m3 /s) = = 5.89 m/s π 2 A3 (0.062 ) 2 Ans. (b) Substituting into (1), !π V1 # %4 " !π 2 $ (0.04 ) + (5) # & %4 " !π 2 $ (0.05 ) + (5.89) # & %4 " 2 $ (0.06 ) = 0.0333 & V1 = 5.45 m/s From mass conservation, Q4 = V4A4 (0.0333 m3 /s) = V4 ( π )(0.062 )/4 V4 = 5.24 m/s Ans. (c) Ans. (a) 3.9 A laboratory test tank contains seawater of salinity S and density ρ. Water enters the tank at conditions (S1, ρ1, A1, V1) and is assumed to mix immediately in the tank. Tank water leaves through an outlet A2 at velocity V2. If salt is a “conservative” property (neither created nor destroyed), use the Reynolds transport theorem to find an expression for the rate of change of salt mass Msalt within the tank. Solution: By definition, salinity S = ρsalt/ ρ . Since salt is a “conservative” substance (not consumed or created in this problem), the appropriate control volume relation is ! " dMsalt |system = d ## ! ρ s dυ $$ + Sm! 2 − S1m! 1 = 0 dt dt % CV & or: dMs | = S ρ A V − Sρ A2 V2 dt CV 1 1 1 1 Ans. 3.10 Water flowing through an 8-cm-diameter pipe enters a porous section, as in Fig. P3.10, which allows a uniform radial velocity vw through the wall surfaces for a distance of 1.2 m. If the entrance average velocity V1 is 12 m/s, find the exit velocity V2 if (a) vw = 15 cm/s out of the pipe walls; (b) vw = 10 cm/s into the pipe. (c) What value of vw will make V2 = 9 m/s? Fig. P3.10 Solution: (a) For a suction velocity of vw = 0.15 m/s, and a cylindrical suction surface area, Aw = 2π (0.04)(1.2) = 0.3016 m2 Q1 = Qw + Q2 2 (12)( π )(0.08 )/4 = (0.15)(0.3016) + V2 ( π )(0.082 )/4 V2 = 3 m/s A ns. (a) (b) For a smaller wall velocity, vw = 0.10 m/s, (12)(π )(0.082 )/4 = (0.10)(0.3016) + V2 (π )(0.082 )/4 V2 = 6 m/s A ns. (b) (c) Setting the outflow V2 to 9 m/s, the wall suction velocity is, (12)( π )(0.08 2)/4 = (v w )(0.3016) + (9)( π )(0.08 2)/4 v w = 0.05 m/s = 5 cm/s out 3.11 A room contains dust at uniform concentration C = ρdust/ρ. It is to be cleaned by introducing fresh air at an inlet section Ai, Vi and exhausting the room air through an outlet section. Find an expression for the rate of change of dust mass in the room. Solution: This problem is very similar to Prob. 3.9 on the previous page, except that here Ci = 0 (dustfree air). Refer to the figure in Prob. 3.9. The dust mass relation is ! " dMdust |system = 0 = d ## ! ρdust dυ $$ + Cout m! out − Cin m! in, dt dt % CV & dMdust or, since Cin = 0, we obtain |CV = −C ρA oVo A ns. dt To complete the analysis, we would need to make an overall fluid mass balance. 3.12 The pipe flow in Fig. P3.12 fills a cylindrical tank as shown. At time t = 0, the water depth in the tank is 30 cm. Estimate the time required to fill the remainder of the tank. Fig. P3.12 Solution: For a control volume enclosing the tank and the portion of the pipe below the tank, d ! ρ dv " + m! out − m! in = 0 $ dt # ! dh + (ρ AV )out − (ρ AV )in = 0 dt dh 4 ! " # π$ = 998 % & (0.122 )(2.5 − 1.9) ( = 0.0153 m/ s, ) 4* dt 998(π )(0.752 ) '+ , ρπ R 2 ∆t = 0.7/0.0153 = 46 s Ans. 3.13 Water at 20°C flows steadily at 40 kg/s through the nozzle in Fig. P3.13. If D1 = 18 cm and D2 = 5 cm, compute the average velocity, in m/s, at (a) section 1 and (b) section 2. Fig. P3.13 Solution: The flow is incompressible, hence the volume flow Q = (40 kg/s)/(998 kg/m3) = 0.0401 m3/s. The average velocities are computed from Q and the two diameters: V1 = Q 0.0401 = = 1.58 m/s A1 (π /4)(0.18)2 Ans. (a) V2 = Q 0.0401 = = 20.4 m/s A2 (π /4)(0.05)2 Ans. (b) 3.14 The open tank in the figure contains water at 20°C. For incompressible flow, (a) derive an analytic expression for dh/dt in terms of (Q1, Q2, Q3). (b) If h is constant, determine V2 for the given data if V1 = 3 m/s and Q3 = 0.01 m3/s. Solution: For a control volume enclosing the tank, " π d 2 dh d ! + ρ (Q2 − Q1 − Q3 ), # ! ρ dυ $$ + ρ (Q2 − Q1 − Q3 ) = ρ # dt % CV 4 dt & solve dh Q1 + Q3 − Q2 = dt (π d2 /4) Ans. (a) If h is constant, then Q2 = Q1 + Q3 = 0.01 + π π (0.05)2 (3.0) = 0.0159 = (0.07)2 V2 , 4 4 solve V2 = 4.13 m/s Ans. (b) 3.15 Water flows steadily through the round pipe in the figure. The entrance velocity is Vo. The exit velocity approximates turbulent flow, u = umax(1 − r/R)1/7. Determine the ratio Uo/umax for this incompressible flow. Solution: Inlet and outlet flow must balance: R Q1 = Q2 , or: 1/7 R r" 49π 2 ! 2 ! U o 2π r dr = ! u max #% 1− R $& 2π r dr , or: U oπ R = u max 60 R 0 0 Cancel and rearrange for this assumed incompressible pipe flow: 49 Uo = Ans. umax 60 3.16 An incompressible fluid flows past an impermeable flat plate, as in Fig. P3.16, with a uniform inlet profile u = Uo and a cubic polynomial exit profile ! 3η − η 3 " y u ≈ Uo # $ where η = Fig. P3.16 δ % 2 & Compute the volume flow Q across the top surface of the control volume. Solution: For the given control volume and incompressible flow, we obtain δ δ ! 3y y 3 " 0 = Q top + Q right − Q left = Q + ! U o # − $ b dy − ! U ob dy % 2δ 2δ 3 & 0 0 3 5 = Q + U o bδ − U o bδ , solve for Q = U ob δ A ns. 8 8 3.17 Incompressible steady flow in the inlet between parallel plates in Fig. P3.17 is uniform, u = Uo = 8 cm/s, while downstream the flow develops into the parabolic laminar profile u = az(zo − z), where a is a constant. If zo = 4 cm and the fluid is SAE 30 oil at 20°C, what is the value of umax in cm/s? Fig. P3.17 Solution: Let b be the plate width into the paper. Let the control volume enclose the inlet and outlet. The walls are solid, so no flow through the wall. For incompressible flow, zo 0 = Qout − Q in = zo ! az(zo − z)b dz − ! 0 Uo b dz = abzo3/6 − Uo bzo = 0, or: a = 6Uo /zo2 0 Thus continuity forces the constant a to have a particular value. Meanwhile, a is also related to the maximum velocity, which occurs at the center of the parabolic profile: z " !z "! 2 2 2 At z = z o/2: u = u max = a # o $ # z o − o $ = az o/4 = (6U o/z o )(z o/4) 2 & % 2 &% 3 3 cm A ns. or: umax = Uo = (8 cm/s) = 12 s 2 2 Note that the result is independent of zo or of the particular fluid, which is SAE 30 oil. 3.18 An incompressible fluid flows steadily through the rectangular duct in the figure. The exit velocity profile is given by u ≈ umax (1 – y2/b2)(1 – z2/h2). (a) Does this profile satisfy the correct boundary conditions for viscous fluid flow? (b) Find an analytical expression for the volume flow Q at the exit. (c) If the inlet flow is 300 ft3/min, estimate umax in m/s. Solution: (a) The fluid should not slip at any of the duct surfaces, which are defined by y = ±b and z = ±h. From our formula, we see u ≡ 0 at all duct surfaces, OK. Ans. (a) (b) The exit volume flow Q is defined by the integral of u over the exit plane area: Q = ! ! u dA = 16bhu max = 9 +h +b ! 2 y "! 2 z " −h −b % &% & ! 4b " ! 4h " $# $ 3 &% 3 & ! ! umax #1 − b2 $ #1 − h2 $ dy dz = umax #% Ans. (b) 3 3 (c) Given Q = 300 ft /min = 0.1416 m /s, we need duct dimensions, also (sorry). Let us take b = h = 10 cm. Then the maximum exit velocity is Q = 0.1416 m3 16 = (0.1 m)(0.1 m)umax , solve for u max = 7.96 m/s s 9 Ans. (c) 3.19 Water from a storm drain flows over an outfall onto a porous bed which absorbs the water at a uniform vertical velocity of 8 mm/s, as shown in Fig. P3.19. The system is 5 m deep into the paper. Find the length L of bed which will completely absorb the storm water. Fig. P3.19 Solution: For the bed to completely absorb the water, the flow rate over the outfall must equal that into the porous bed, Q1 = Q PB; or (2 m/s)(0.2 m)(5 m) = (0.008 m/s)(5 m)L L ≈ 50 m A ns. 3.20 Oil (SG-0.91) enters the thrust bearing at 250 N/hr and exits radially through the narrow clearance between thrust plates. Compute (a) the outlet volume flow in mL/s, and (b) the average outlet velocity in cm/s. Solution: The specific weight of the oil is (0.91)(9790) = 8909 N/m3. Then Q2 = Q1 = 250/3600 N/s 3 8909 N/m = 7.8 × 10 −6 Fig. P3.20 m3 mL = 7.8 s s Ans. (a) But also Q2 = V2 π (0.1 m)(0.002 m) = 7.8× 10− 6 , solve for V2 = 1.24 cm s Ans. (b) 3.21 A dehumidifier brings in saturated wet air (100 percent relative humidity) at 30°C and 1 atm, through an inlet of 8-cm diameter and average velocity 3 m/s. After some of the water vapor condenses and is drained off at the bottom, the somewhat drier air leaves at approximately 30°C, 1 atm, and 50 percent relative humidity. For steady operation, estimate the amount of water drained off in kg/h. (This problem is idealized from a real dehumidifier.) Solution: Recall that “relative humidity” is the ratio of water-vapor density present in the air to the saturated vapor density at the same temperature. From Tables A-4 and A-5 for water at 30°C, Rgas = 461 J/(kg ·°K) and pvap = 4242 Pa. Then, at 100% humidity, ρvapor = p vap RT = 4242 = 0.0304 kg/m 3 (461)(303) 2 3 Meanwhile, the inlet volume flow of wet air is Qin = AV = (π/4)(0.08 m) (3 m/s) = 0.0151 m /s. It follows that the mass of water vapor entering is ρvapQin = 4.58E−4 kg/s. If the air leaves at 50% relative humidity, exactly half of this is drained away: ! drained away = (1/2)(4.58E −4)(3600) ≈ 0.82 m 3.22 The converging-diverging nozzle shown in Fig. P3.22 expands and accelerates dry air to supersonic speeds at the exit, where p2 = 8 kPa and T2 = 240 K. At the throat, p1 = 284 kPa, T1 = 665 K, and V1 = 517 m/s. For steady compressible flow of an ideal gas, estimate (a) the mass flow in kg/h, (b) the velocity V2, and (c) the Mach number Ma2. Solution: kg hr Ans. Fig. P3.22 The mass flow is given by the throat conditions: 284000 kg " π m$ kg # = 0.0604 ! = ρ1A1V1 = '! m (0.01 m) 2 %517 3( s &* s ) + (287)(665) m , 4 Ans. (a) For steady flow, this must equal the mass flow at the exit: 0.0604 kg m ! 8000 " π 2 = ρ2 A2 V2 = # $ (0.025) V2 , or V2 ≈ 1060 s s % 287(240) & 4 Ans. (b) Recall from Eq. (1.39) that the speed of sound of an ideal gas = (kRT)1/2. Then Mach number at exit: Ma = V2 /a 2 = 1060 ≈ 3.41 Ans. (c) [1.4(287)(240)]1/2 The hypodermic needle in the figure contains a liquid (SG = 1.05). If the serum is 3 (a) if leakage , how should the plunger be to be injected steadily at in the plunger clearance is ; and (b) if is 1 w? Solution: Q=6 (a) For incompressible flow, the volume flow is the same at piston and exit: cm3 in3 in π 2 = 0.366 = A1V1 = (0.75 in) V1, solve Vpiston = 0.83 s 4 s s Ans. (a) (b) If there is 10% leakage, the piston must deliver both needle flow and leakage: A1V1 = Qneedle + Qclearance = 6 + 0.1(6) = 6.6 V1 = 0.91 in s cm 3 in 3 π = 0.403 = (0.75)2 V1 , s s 4 Ans. (b) 3.24 Water enters the bottom of the cone in the figure at a uniformly increasing average velocity V = Kt. If d is very small, derive an analytic formula for the water surface rise h(t), assuming h = 0 at t = 0. Solution: For a control volume around the cone, the mass relation becomes d dt ( ! ρ dυ )− m! in = 0= Integrate: ρ Solve for π d! π " ρ (h tanθ )2 h$ − ρ d 2Kt dt #% 3 4 & π 3 2 π h tan θ = ρ d2 Kt2 3 8 !3 " h(t) = # Kt 2d 2 cot 2θ $ %8 & 1/3 Ans. 3.25 As will be discussed in Chaps. 7 and 8, the flow of a stream Uo past a blunt flat plate creates a broad low-velocity wake behind the plate. A simple model is given in Fig. P3.25, with only half of the flow shown due to symmetry. The velocity profile behind the plate is idealized as “dead air” (near-zero velocity) behind the plate, plus a higher velocity, decaying vertically above the wake according to the variation u ≈ Uo + ∆Ue−z/L, where L is the plate height and z = 0 is the top of the wake. Find ∆U as a function of stream speed Uo. Fig. P3.25 Solution: For a control volume enclosing the upper half of the plate and the section where the exponential profile applies, extending upward to a large distance H such that exp(–H/L) ≈ 0, we must have inlet and outlet volume flows the same: H Qin = − H L" ! ) dz, or: Uo # H + $ = U o H + ∆ UL 2& % 1 Cancel Uo H and solve for ∆ U ≈ Uo Ans . 2 ! Uo dz = Qout = !0 L/2 (Uo + ∆ U e −z/L 3.26 A thin layer of liquid, draining from an inclined plane, as in the figure, will have 2 2 a laminar velocity profile u = Uo(2y/h − y /h ), where Uo is the surface velocity. If the plane has width b into the paper, (a) determine the volume rate of flow of the film. (b) Suppose that h = 0.5 in and the flow rate per foot of channel width is 1.25 gal/min. Estimate Uo in ft/s. Solution: (a) The total volume flow is computed by integration over the flow area: h ! 2y y 2 " 2 Q = ! Vn dA = ! Uo # − 2 $ b dy = Uo bh 3 h h & % 0 Ans. (a) (b) Evaluate the above expression for the given data: Q = 1.25 gal ft 3 2 2 ! 0.5 " ft$ , = 0.002785 = Uo bh = Uo (1.0 ft )# s min 3 3 % 12 & ft Ans. (b) solve for U o = 0.10 s