Integral Relations for a Control Volume Part 7 (Solutions)

3.151 Water flows through a circular nozzle, exits into the air as a jet, and strikes a plate. The force required to hold the plate steady is 70 N. Assuming frictionless onedimensional flow, estimate (a) the velocities at sections (1) and (2); (b) the mercury manometer reading h. Solution: (a) First examine the momentum of the jet striking the plate, Fig. P3.151 ! in uin = −ρ A2 V22 ! F = F = −m !π " 70 N = −(998) # $ (0.032 )(V22 ) V2 = 9.96 m/s %4 & VA Then V1 = 2 2 = A1 Ans. (a) !π " (9.96) # $ (0.032 ) %4& or V1 = 0.9 m/s π (0.12 ) 4 Ans. (a) (b) Applying Bernoulli, p2 − p1 = 1 1 ρ V22 − V12 = (998)(9.962 − 0.92 ) = 49,100 Pa 2 2 ( ) And from our manometry principles, h= 49,100 ∆p = ≈ 0.4 m Ans. (b) ρg (133,100 − 9790) 3.152 A free liquid jet, as in Fig. P3.152, has constant ambient pressure and small losses; hence from Bernoulli’s equation 2 z + V /(2g) is constant along the jet. For the fire nozzle in the figure, what are (a) the minimum and (b) the maximum values of θ for which the water jet will clear the corner of the building? For which case will the jet velocity be higher when it strikes the roof of the building? Fig. P3.152 Solution: The two extreme cases are when the jet just touches the corner A of the building. For these two cases, Bernoulli’s equation requires that V12 + 2gz1 = (100)2 + 2g(0) = VA2 + 2gzA = VA2 + 2(32.2)(50), or: VA = 82.3 ft s The jet moves like a frictionless particle as in elementary particle dynamics: Vertical motion: z = V1 sinθ t − 1 2 gt ; Horizontal motion: x = V1 cosθ t 2 Eliminate “t” between these two and apply the result to point A: zA = 50 = xA tan θ − gx2A (32.2)(40)2 ; clean up and rearrange: = 40 tan θ − 2V21 cos2 θ 2(100)2 cos2 θ 2 tan θ = 1.25 + 0.0644 sec θ , solve for θ = 85.94° Ans. (a) and 55.40° Ans. (b) Path (b) is shown in the figure, where the jet just grazes the corner A and goes over the top of the roof. Path (a) goes nearly straight up, to z = 155 ft, then falls down to pt. A. 3.153 For the container of Fig. P3.153 use Bernoulli’s equation to derive a formula for the distance X where the free jet leaving horizontally will strike the floor, as a function of h and H. For what ratio h/H will X be maximum? Sketch the three trajectories for h/H = 0.4, 0.5, and 0.6. Solution: The velocity out the hole and the time to fall from hole to ground are given by Vo = 2g(H − h) t fall = 2h/g Then the distance travelled horizontally is X = Vot fall = 2 h(H − h) Ans. Fig. P3.153 Maximum X occurs at h = H/2, or Xmax = H. When h = 0.25H or 0.75H, the jet travels out to H = 0.866H. These three trajectories are shown in the sketch on the previous page. 3.154 In Fig. P3.154 the exit nozzle is horizontal. If losses are negligible, what should the water level h cm be for the free jet to just clear the wall? Solution: The fall distance is 0.3 m = (1/2)gt2, hence t = [2(0.3)/g]. The exit velocity is V = √(2gh). Then x = 0.4 m = Vo t = (2gh)1/2[2(0.3)/g]1/2 = 2(0.3h)1/ 2 , or: h ≈ 0.133 m Ans. Fig. P3.154 3.155 Bernoulli’s 1738 treatise Hydrodynamica contains many excellent sketches of flow patterns. One, however, redrawn here as Fig. P3.155, seems physically misleading. What is wrong with the drawing? Solution: If friction is neglected and the exit pipe is fully open, then pressure in the closed “piezometer” tube would be atmospheric and the fluid would not rise at all in the tube. The open jet coming from the hole in the tube would have V ≈ √(2gh) and would rise up to nearly the same height as the water in the tank. 3.156 A blimp cruises at 75 mi/h through sea-level standard air. A differential pressure transducer connected between the nose and the side of the blimp registers 950 Pa. Estimate (a) the absolute pressure at the nose and (b) the absolute velocity of the air near the blimp side. Fig. P3.155 Solution: Assume that the nose reads “stagnation” pressure and the side reads the local side pressure and senses a local velocity not equal to the blimp speed. The nose velocity is zero. Then Bernoulli’s equation, assuming ∆z = 0, yields 1 1 1! kg " 2 2 2 ρVnose = pside + ρVside , or : pnose − pside = 950 Pa = # 1.225 3 $ Vside 2 2 2% m & m = 88 mi/hr Ans. (b) solve for Vside ≈ 39.4 s Now relate Bernoulli to the ambient upstream conditions, Vblimp = 75 mi/hr = 33.5 m/s: pnose + 1 1.225 2 2 pnose = patm + ρ Vvehicle = 101350 + (33.5) = 102, 000 Pa 2 2 Ans. (a) 3.157 The manometer fluid in Fig. P3.157 is mercury. Estimate the volume flow in the tube if the flowing fluid is (a) gasoline and (b) nitrogen, at 20°C and 1 atm. Solution: For gasoline (a) take ρ = 1.32 slug/ft3. For nitrogen (b), R ≈ 297 J/kg ⋅ °C and Fig. P3.157 ρ = p/RT = (101350)/[(297)(293)] ≈ 1.165 kg/m 3 = 0.00226 slug/ft3. For mercury, take ρ ≈ 26.34 slug/ft3. The pitot tube (2) reads stagnation pressure, and the wall hole (1) reads static pressure. Thus Bernoulli’s relation becomes, with ∆z = 0, 1 p1 + ρ V12 = p2 , or V1 = 2(p2 − p1 )/ρ 2 The pressure difference is found from the manometer reading, for each fluid in turn: (a) Gasoline: ∆ p = (ρ Hg − ρ )γη = (26.34 − 1.32)(32.2)(1/12 ft) ≈ 67.1 lbf/ft 2 2 V1 = [2(67.1)/1.32]1/2 = 10.1 ft ft 3 ! π "! 3 " , Q = V1 A1 = (10.1) # $# $ = 0.495 s s % 4 &% 12 & Ans. (a) (b) N2 : ∆ p = (ρHg − ρ )gh = (26.34 − 0.00226)(32.2)(1/12) ≈ 70.7 lbf/ft2 2 V1 = [2(70.7)/0.00226]1/2 = 250 ft3 ft ! π "! 3 " , Q = V1 A1 = (250)# $# $ ≈ 12.3 s s % 4 &% 12 & Ans. (b) 3.158 In Fig. P3.158 the flowing fluid is CO2 at 20°C. Neglect losses. If p1 = 170 kPa and the manometer fluid is Meriam red oil (SG = 0.827), estimate (a) p2 and (b) the gas flow rate in m3/h. Solution: Estimate the CO2 density as ρ = 3 p/RT = (170000)/[189(293)] ≈ 3.07 kg/m . The manometer reading gives the downstream pressure: Fig. P3.158 p1 − p 2 = ( ρ oil − ρ CO2 )gh = [0.827(998) − 3.07](9.81)(0.08) ≈ 645 Pa Hence p 2 = 170,000 − 645 ≈ 169400 Pa Ans. (a) Now use Bernoulli to find V2, assuming p1 ≈ stagnation pressure (V1 = 0): p1 + or: V2 = 1 1 ρ (0)2 ≈ p 2 + ρ V22 , 2 2 2(p 1 − p 2) 2(645) m = ≈ 20.5 3.07 s ρ 2 3 Then Q = V2 A2 = (20.5)(π /4)(0.06) = 0.058 m /s ≈ 209 m3 hr Ans. (b) 3.159 Our D = 0.625-in-diameter hose is too short, and it is 125 ft from the d = 0.375-in-diameter nozzle exit to the garden. If losses are neglected, what is the minimum gage pressure required, inside the hose, to reach the garden? Solution: Assume that the water jet from the hose approximates the trajectory of a frictionless particle. Then ∆x = 125 ft can be translated to the jet velocity needed: 2 ∆ xmax = 125 ft = Vjet g 2 = Vjet 32.2 , solve for V jet = 63.44 ft/s Then write Bernoulli’s equation and continuity from outside to inside of the nozzle: 2 4 ft $ ! # 0.375 $ " ρ 2 ! # d $ 4 " 1.94 slug/ ft3 # Vjet %1 − ' ( & = 63.4 %1 − ' & ' ( ( 2 2 s * %+ ) 0.625 * ,& ) +% ) D * ,& lbf = 3400 2 Ans. ft pinside − patm = 3.160 The air-cushion vehicle in Fig. P3.160 brings in sea-level standard air through a fan and discharges it at high velocity through an annular skirt of 3-cm clearance. If the vehicle weighs 50 kN, estimate (a) the required airflow rate and (b) the fan power in kW. Solution: The air inside at section 1 is nearly stagnant (V ≈ 0) and supports the weight and also drives the flow out of the interior into the atmosphere: p1 ≈ po1 : po1 − patm = Fig. P3.160 weight 50,000 N 1 2 1 2 = = ρ Vexit = (1.205)Vexit ≈ 1768 Pa 2 area 2 2 π(3 m) Solve for Vexit ≈ 54.2 m/s, whence Qe = Ae Ve = π (6)(0.03)(54.2) = 30.6 m3 s Then the power required by the fan is P = Qe∆p = (30.6)(1768) ≈ 54000 W Ans. 3.161 A necked-down section in a pipe flow, called a venturi, develops a low throat pressure which can aspirate fluid upward from a reservoir, as in Fig. P3.161. Using Bernoulli’s equation with no losses, derive an expression for the velocity V1 which is just sufficient to bring reservoir fluid into the throat. Fig. P3.161 Solution: Water will begin to aspirate into the throat when pa − p1 = ρ gh. Hence: 1 1 Volume flow: V1 = V2 (D2 /D1 ) 2 ; Bernoulli (∆ z = 0): p1 + ρ V12 ≈ patm + ρ V22 2 2 Solve for pa − p1 = D 2gh ρ 4 2 (α −1)V2 ≥ ρ gh, α = 2 , or: V2 ≥ α 4 −1 2 D1 Similarly, V1, min = α 2V2, min = 2gh 1 − (D1 /D2 )4 Ans. Ans. 3.162 Suppose you are designing a 3 × 6-ft air-hockey table, with 1/16-inch-diameter holes spaced every inch in a rectangular pattern (2592 holes total), the required jet speed from each hole is 50 ft/s. You must select an appropriate blower. Estimate the volumetric flow rate (in ft3/min) and pressure rise (in psi) required. Hint: Assume the air is stagnant in the large manifold under the table surface, and neglect frictional losses. Solution: Assume an air density of about sea-level, 0.00235 slug/ft3. Apply Bernoulli’s equation through any single hole, as in the figure: p1 + ∆prequired = p1 − pa = ρ 2 ρ 2 V = pa + V jet , or: 2 1 2 lbf 0.00235 lbf ρ 2 (50)2 = 2.94 2 = 0.0204 2 V jet = 2 2 ft in Ans. The total volume flow required is 2 ft " π ! 1/16 " ! Q = VA1− hole(# of holes) = # 50 $ # ft $ (2592 holes) s & 4 % 12 % & ft3 ft3 Ans. = 2.76 = 166 s min It wasn’t asked, but the power required would be P = Q ∆p = (2.76 ft3/s)(2.94 lbf/ft2) = 8.1 ft·lbf/s, or about 11 watts. 3.163 The liquid in Fig. P3.163 is kerosine at 20°C. Estimate the flow rate from the tank for (a) no losses and (b) pipe losses hf ≈ 4.5V 2/(2g). 3 Solution: For kerosine let γ = 50.3 lbf/ft . Let (1) be the surface and (2) the exit jet: Fig. P3.163 V12 V22 2 V2 p1 p + +z = 2 + + z 2 + h f , with z 2 = 0 and V1 ≈ 0, h f = K γ 2g 1 γ 2g 2g Solve for V22 2g (1 + K) = z1 + p1 − p2 (20 − 14.7)(144) = 5+ ≈ 20.2 ft γ 50.3 We are asked to compute two cases (a) no losses; and (b) substantial losses, K ≈ 4.5: 2(32.2)(20.2)" (a) K = 0: V2 = !' ( 1+ 0 ) * 1/2 2 = 36.0 π 1 ft ft 3 , Q = 36.0 #% $& ≈ 0.197 s s 4 + 12 , Ans. (a) 2 (b) K = 4.5: V2 = π 1 2(32.2)(23.0) ft ft 3 = 16.4 , Q =16.4 #! $" ≈ 0.089 s 1+ 4.5 s 4 % 12 & 3.164 An open water jet exits from a nozzle into sea-level air, as shown, and strikes a stagnation tube. If the centerline pressure at section (1) is 110 kPa and losses are neglected, estimate (a) the mass flow in kg/s; and (b) the height H of the fluid in the tube. Solution: Ans. (b) Fig. P3.164 Writing Bernoulli and continuity between pipe and jet yields jet velocity: 4 4 ρ 2 ! # D jet $ " 998 2 ! # 4 $ " V jet %1 − ' V jet %1 − ' ( & , ( & = 110000 −101350 = 2 2 %+ ) D1 * &, %+ ) 12 * &, m solve Vjet = 4.19 s π kg 2 Then the mass flow is m! = ρ A jetV jet = 998 (0.04) (4.19) = 5.25 Ans. (a) 4 s p1 − p a = (b) The water in the stagnation tube will rise above the jet surface by an amount equal to the stagnation pressure head of the jet: H = R jet + V 2jet 2g = 0.02 m + (4.19)2 = 0.02 + 0.89 = 0.91 m 2(9.81) Ans. (b) 3.165 A venturi meter, shown in Fig. P3.165, is a carefully designed constriction whose pressure difference is a measure of the flow rate in a pipe. Using Bernoulli’s equation for steady incompressible flow with no losses, show that the flow rate Q is related to the manometer reading h by Q= A2 1 − (D2 /D1 ) 4 2 gh( ρM − ρ ) ρ Fig. P3.165 where ρ M is the density of the manometer fluid. Solution: and 2: First establish that the manometer reads the pressure difference between 1 p1 − p2 = ( ρ M − ρ )gh (1) Then write incompressible Bernoulli’s equation and continuity between (1) and (2): ( ∆z = 0): 2 2 p1 V 1 p 2 V 2 + ≈ + ρ 2 ρ 2 and V2 = V1(D 1/D 2) 2 , Q = A 1V1 = A 2V2 Eliminate V2 and (p1 − p2 ) from (1) above: Q = A 2 2gh( ρM − ρ)/ ρ 1 − (D2 /D1 )4 A ns. 3.166 A wind tunnel draws in sea-level standard air from the room and accelerates it into a 1-m by 1-m test section. A pressure transducer in the test section wall measures ∆p = 45 mm water between inside and outside. Estimate (a) the test section velocity in mi/hr; and (b) the absolute pressure at the nose of the model. Solution: (a) First apply Bernoulli from the atmosphere (1) to (2), using the known ∆p: 3 p a − p 2 = 45 mm H 2O = 441 Pa; ρ a = 1.225 kg/m ; p 1 + Since V1 ≈ 0 and p1 = pa , we obtain V2 = ρ 2 ρ 2 V1 ≈ p2 + V 2 2 2 2 ∆p 2(441) m mi = = 26.8 = 60 ρ 1.225 s hr Ans. (a) (b) Bernoulli from 1 to 3: both velocities = 0, so pnose = pa ≈ 101350 Pa. Ans. (b) 3.167 In Fig. P3.167 the fluid is gasoline at 20°C at a weight flux of 120 N/s. Assuming no losses, estimate the gage pressure at section 1. 3 Solution: For gasoline, ρ = 680 kg/m . Compute the velocities from the given flow Fig. P3.167 rate: ! W 120 N/s m3 Q= , = = 0.018 s ρ g 680(9.81) V1 = 0.018 m 0.018 m = 3.58 ; V2 = = 9.16 s s π (0.04) 2 π (0.025) 2 Now apply Bernoulli between 1 and 2: 2 V22 p1 V 1 p p (3.58)2 0(gage) (9.16)2 + + gz1 ≈ 2 + + gz 2 , or: 1 + +0 ≈ + + 9.81(12) 2 2 680 2 ρ 2 ρ ρ Solve for p1 ≈ 104, 000 Pa (gage) Ans. 3.168 In Fig. P3.168 both fluids are at 20°C. If V1 = 1.7 ft/s and losses are neglected, what should the manometer reading h ft be? Solution: By continuity, establish V2: V2 = V1 (D1 /D2 )2 = 1.7(3/1)2 = 15.3 ft s Now apply Bernoulli between 1 and 2 to establish the pressure at section 2: p1 + ρ 2 ρ V1 + ρ gz1 = p2 + V22 + ρ gz2 , 2 2 Fig. P3.168 or: p1 + (1.94/2)(1.7)2 + 0 ≈ 0 + (1.94/2)(15.3)2 + (62.4)(10), p1 = 848 psf This is gage pressure. Now the manometer reads gage pressure, so p1 − pa = 848 lbf ft2 = ( ρmerc − ρwater )gh = (846 − 62.4)h, solve for h ≈ 1.08 ft Ans . 3.169 Once it has been started by sufficient suction, the siphon in Fig. P3.169 will run continuously as long as reservoir fluid is available. Using Bernoulli’s equation with no losses, show (a) that the exit velocity V2 depends only upon gravity and the distance H and (b) that the lowest (vacuum) pressure occurs at point 3 and depends on the distance L + H. Solution: Fig. P3.169 Write Bernoulli from 1 to 2: 2 2 V22 p1 V 1 p V2 pa p + + z1 ≈ 2 + + z2 , or: + 0 + z1 ≈ a + +z γ 2g γ 2g γ γ 2g 2 Solve for V2 = Vexit = 2g(z1 − z2 ) = 2gH Ans. Since the velocity is constant throughout the tube, at any point C inside the tube, p C + γ z C ≈ p a + γ z 2 , or, at point 3: p C, min = p a − γ (z 3 − z 2 ) = pa − ρg(L + H) Ans. 3.170 If losses are neglected in Fig. P3.170, for what water level h will the flow begin to form vapor cavities at the throat of the nozzle? Solution: Applying Bernoulli from (a) to (2) gives Torricelli’s relation: V2 = √(2gh). Also, V1 = V2 (D2 /D1 )2 = V2 (8/5)2 = 2.56V2 Fig. P3.170 Vapor bubbles form when p1 reaches the vapor pressure at 30°C, pvap ≈ 4242 Pa (from 3 Table A.5), while ρ ≈ 996 kg/m at 30°C (Table A.1). Apply Bernoulli between 1 and 2: 2 V22 p1 V 1 p 4242 (2.56V2 ) 2 100000 V 22 + + gz 1 ≈ 2 + + gz 2, or: + +0 ≈ + +0 ρ ρ 2 2 996 2 996 2 Solve for V 22 = 34.62 = 2gh, or h = 34.62/[2(9.81)] ≈ 1.76 m A ns. 3.171 For the 40°C water flow in Fig. P3.171, estimate the volume flow through the pipe, assuming no losses; then explain what is wrong with this seemingly innocent question. If the actual flow rate is Q = 3 40 m /h, compute (a) the head loss in ft and (b) the constriction diameter D which causes cavitation, assuming that the throat divides the head loss equally and that changing the constriction causes no additional losses. Solution: Fig. P3.171 Apply Bernoulli between 1 and 2: 2 V2 p1 V 1 p + + z1 ≈ 2 + 2 + z2 , or: 0 + 0 + 25 ≈ 0 + 0 + 10, or: 25 = 10 ?? γ 2g γ 2g This is madness, what happened? The answer is that this problem cannot be free of losses. There is a 15-m loss as the pipe-exit jet dissipates into the downstream reservoir. Ans. (a) (b) Examining analysis (a) shows that the head loss is 15 meters. For water at 40°C, the 3 vapor pressure is 7375 Pa (Table A.5), and the density is 992 kg/m (Table A.1). Now write Bernoulli between (1) and (3), assuming a head loss of 15/2 = 7.5 m: 2 V23 p1 V1 p g Q 40/3600 0.0141 + + gz1 = 3 + + gz 3 + hf,total , where V3 = = = ρ 2 ρ 2 2 A 3 (π /4)D2 D2 Thus 101350 7375 (0.0141/D2 )2 + 0 + 9.81(25) ≈ + + 0 + (9.81)(7.5) 992 992 2 Solve for D 4 ≈ 3.75E−7 m 4, or D ≈ 0.0248 m ≈ 25 mm Ans. This corresponds to V3 ≈ 23 m/s. 3.172 The 35°C water flow of Fig. P3.172 discharges to sea-level standard atmosphere. Neglecting losses, for what nozzle diameter D will cavitation begin to occur? To avoid cavitation, should you increase or decrease D from this critical value? Fig. P3.172 Solution: At 35°C the vapor pressure of water is approximately 5600 Pa (Table A.5). Bernoulli from the surface to point 3 gives the Torricelli result V3 = √(2gh) = √2(32.2)(6) ≈ 19.66 ft/s. We can ignore section 2 and write Bernoulli from (1) to (3), with p1 = pvap and ∆z = 0: 2 2 2 2 p 1 V1 p 2 V 2 117 V 1 2116 V 3 , or: , + ≈ + + ≈ + ρ 2 ρ 2 1.93 2 1.93 2 ! D " but also V1 = V3 # $ % 1/12 & Eliminate V1 and introduce V3 = 19.66 2 ft to obtain D4 = 3.07E−4, D ≈ 0.132 ft A ns. s To avoid cavitation, we would keep D < 0.132 ft, which will keep p1 > pvapor.