Integral Relations for a Control Volume Part 8 (Solutions)

3.173 The horizontal wye fitting in Fig. P3.173 splits the 20°C water flow rate 3 2 equally, if Q1 = 5 ft /s and p1 = 25 lbf/in (gage) and losses are neglected, estimate (a) p2, (b) p3, and (c) the vector force required to keep the wye in place. Solution: First calculate the velocities: Fig. P3.173 Q 5.0 ft 2.5 ft ft = = 25.46 ; V2 = = 50.93 , V3 = 28.65 V1 = 2 A1 (π /4)(6/12) 2 s s s π ( /4)(3/12) Then apply Bernoulli from 1 to 2 and then again from 1 to 3, assuming ∆z ≈ 0: p2 = p1 + 1.94 ρ 2 V 1 − V22 = 25(144) + [(25.46)2 − (50.93)2 ] ≈ 1713 psfg Ans. (a) 2 2 p3 = p1 + 1.94 ρ 2 (V − V23 ) = 25(144) + [(25.46)2 − (28.65)2 ] ≈ 3433 psfg 2 1 2 ( ) Ans. (b) (c) to compute the support force R (see figure above), put a CV around the entire wye: ! Fx = R x + p 1A 1 − p 2A 2sin30° − p 3A 3sin50 ° = ρQ 2V2 sin30° + ρQ 3V3sin50 ° − ρQ 1V1 = R x + 707 − 42 − 229 = 124 + 106 − 247, or: R x = −453 lbf (to left) Ans. (c) ! Fy = R y − p 2 A 2 cos 30 ° + p 3 A 3 cos 50° = ρQ 2V2 cos 30° + ρQ 3 ( − V3 ) cos 50° = R y − 73 + 193 = 214 − 89, or: R y ≈ + 5 lbf (up) Ans. (c) 3.174 In Fig. P3.174 the piston drives water at 20°C. Neglecting losses, estimate the exit velocity V2 ft/s. If D2 is further constricted, what is the maximum possible value of V2? Solution: piston: Fig. P3.174 Find p 1 from a freebody of the ! Fx = F + p a A1 − p1A1, or: p1 − pa = 10.0 lbf (π /4)(8/12) 2 ≈ 28.65 lbf ft 2 Now apply continuity and Bernoulli from 1 to 2: V1A1 = V2 A2 , or V1 = 2 2 1 p V 1 pa V 2 ≈ + V2 ; 1 + 4 ρ 2 ρ 2 2 Introduce p1 − pa and substitute for V1 to obtain V2 = V2 = 5.61 ft s 2(28.65) , 1.94(1 − 1/16) Ans. If we reduce section 2 to a pinhole, V2 will drop off slowly until V1 vanishes: Severely constricted section 2: V2 = 2(28.65) ft ≈ 5.43 s 1.94(1 − 0) 3.175 If the approach velocity is not too high, a hump in the bottom of a water channel causes a dip ∆h in the water level, which can serve as a flow measurement. If, as shown in Fig. P3.175, ∆h = 10 cm when the bump is 30 cm high, what is the volume flow Q1 per unit width, assuming no losses? In general, is ∆h proportional to Q1? Solution: Ans. Fig. P3.175 Apply continuity and Bernoulli between 1 and 2: V1 h1 = V2 h2 ; V21 V 22 2g∆ h + h1 ≈ + h2 + H, solve V21 ≈ 2 2 Ans . 2g 2g (h1 /h2 ) − 1 We see that ∆h is proportional to the square of V1 (or Q1), not the first power. For the given numerical data, we may compute the approach velocity: h 2 = 2.0 − 0.3 − 0.1 = 1.6 m; V1 = 2(9.81)(0.1) m = 1.87 s [(2.0/1.6)2 −1] whence Q1 = V1 h1 = (1.87)(2.0) ≈ 3.74 m3 s⋅ m Ans . 3.176 In the spillway flow of Fig. P3.176, the flow is assumed uniform and hydrostatic at sections 1 and 2. If losses are neglected, compute (a) V2 and (b) the force per unit width of the water on the spillway. For mass conservation, Solution: V2 = V1h1 /h 2 = 5.0 V1 = 7.14V1 0.7 Fig. P3.176 (a) Now apply Bernoulli from 1 to 2: 2 V22 V21 p1 V 1 p (7.14V1 ) 2 + + h1 ≈ 2 + + h 2 ; or: 0 + + 5.0 ≈ 0 + + 0.7 2g 2g γ 2g γ 2g Solve for V21 = 2(9.81)(5.0 − 0.7) m m , or V1 = 1.30 , V2 = 7.14V1 = 9.28 s s [(7.14)2 − 1] Ans. (a) (b) To find the force on the spillway (F ←), put a CV around sections 1 and 2 to obtain γ γ ! 2 − V1), or, using the given data, ! F x = −F + h 21 − h 22 = m(V 2 2 1 N F = (9790)[(5.0) 2 −(0.7) 2] − 998[(1.30)(5.0)](9.28 −1.30) ≈ 68300 2 m Ans. (b) 3.177 For the water-channel flow of Fig. P3.177, h1 = 1.5 m, H = 4 m, and V1 = 3 m/s. Neglecting losses and assuming uniform flow at sections 1 and 2, find the downstream depth h2, and show that two realistic solutions are possible. Solution: Combine continuity and Bernoulli between 1 and 2: V2 = V1 Fig. P3.177 2 V22 V12 h 1 3(1.5) V1 (4.5/h 2) 2 ; = + h1 + H ≈ + h2 = + 1.5+ 4 ≈ + h2 h2 h2 2g 2g 2(9.81) 2(9.81) Combine into a cubic equation: h32 − 5.959 h22 + 1.032 = 0. The three roots are: h2 = − 0.403 m (impossible); h2 = +5.93 m (subcritical); h2 = + 0.432 m (supercritical) Ans. 3.178 For the water channel flow of Fig. P3.178, h1 = 0.45 ft, H = 2.2 ft, and V1 = 16 ft/s. Neglecting losses and assuming uniform flow at sections 1 and 2, find the downstream depth h2. Show that two realistic solutions are possible. Solution: The analysis is quite similar to Prob. 3.177 - continuity + Bernoulli: V2 = V1 Fig. P3.178 h1 16(0.45) V12 V2 V12 (7.2/h2 ) 2 ; = + h1 = 2 + h 2 + H = + 0.45 = + h 2 + 2.2 h2 h2 2g 2g 2(32.2) 2(32.2) Combine into a cubic equation: h23 − 2.225 h22 + 0.805 = 0 . The three roots are: h2 = −0.540 ft (impossible); h2 = +2.03 ft (subcritical); h 2 = + 0.735 ft (supercritical) Ans. 3.179 A cylindrical tank of diameter D contains liquid to an initial height ho. At time t = 0 a small stopper of diameter d is removed from the bottom. Using Bernoulli’s equation with no losses, derive (a) a differential equation for the free-surface height h(t) during draining and (b) an expression for the time to to drain the entire tank. Solution: 2 !1 Write continuity and the unsteady Bernoulli relation from 1 to 2: 2 p V2 p V2 A ∂V ! D" ds + 2 + 2 + gz 2 = 1 + 1 + gz 1; Continuity: V 2 = V1 1 = V1 # $ 2 2 A2 ∂t ρ ρ %d& The integral term ! dV ∂V ds ≈ 1 h is very small and will be neglected, and p1 = p2.. Then dt ∂t 12 dh ! 2gh " 4 V1 ≈ # , where α = (D/d) ; but also V1 = − , separate and integrate: dt % α − 1&$ h ! ho dh ! 2g " = −' 1/2 h - α − 1(. 1/2 t !o 2 1/2 4 ! g $ " # % D& dt, or: h = ' h1/2 * t( , α = + , o −) 1 2(α − 1) 2 (. / d0 '- Ans. (a) 1/2 (b) the tank is empty when [] = 0 in (a) above, or tfinal = [2(α − 1)g/ho] . Ans. (b) 3.180 The large tank of incompressible liquid in Fig. P3.180 is at rest when, at t = 0, the valve is opened to the atmosphere. Assuming h ≈ constant (negligible velocities and accelerations in the tank), use the unsteady frictionless Bernoulli equation to derive and solve a differential equation for V(t) in the pipe. Solution: 2 ! 1 Fig. P3.180 Write unsteady Bernoulli from 1 to 2: V2 V2 ∂V ds + 2 + gz 2 ≈ 1 + gz 1, where p 1 = p 2, V1 ≈ 0, z 2 ≈ 0, and z 1 = h = const 2 2 ∂t dV dV + V2 = 2gh L, so the diff. eqn. is 2L dt dt This first-order ordinary differential equation has an exact solution for V = 0 at t = 0: The integral approximately equals ! V t" V = Vfinal tanh # final $ , where Vfinal = % 2L & 2gh A ns. 3.181 Modify Prob. 3.180 as follows. Let the top of the tank be enclosed and under constant gage pressure po. Repeat the analysis to find V(t) in the pipe. Solution: The analysis is the same as Prob. 3.180, except that we now have a (constant) surface-pressure term at point 1 which contributes to Vfinal: 2 !1 2 2 2 V p V dV V p ∂V ds + 2 + gz 2 ≈ o + 1 + gz1 = L+ = o + gh, with V = 0 at t = 0. 2 2 dt 2 ∂t ρ ρ The solution is: 2po !V t" V = Vfinal tanh # final $, where Vfinal = + 2gh ρ % 2L & Ans. 3.182 The incompressible-flow form of Bernoulli’s relation, Eq. (3.77), is accurate only for Mach numbers less than about 0.3. At higher speeds, variable density must be accounted for. The most common assumption for compressible fluids is isentropic flow of an ideal gas, k or p = Cρ , where k = cp/cv. Substitute this relation into Eq. (3.75), integrate, and eliminate the constant C. Compare your compressible result with Eq. (3.77) and comment. Solution: We are to integrate the differential Bernoulli relation with variable density: p = Cρ k , so dp = kCρ k− 1dρ , k = c p /c v Substitute this into the Bernoulli relation: dp kCρ k −1 dρ + V dV + g dz = + V dV + g dz = 0 ρ ρ Integrate: ! kCρ k−2 dρ + ! V dV + ! g dz = ! 0 = constant k–1 The first integral equals kCρ /(k − 1) = kp/[ ρ(k − 1)] from the isentropic relation. Thus the compressible isentropic Bernoulli relation can be written in the form kp V2 + + gz = constant (k− 1)ρ 2 Ans. It looks quite different from the incompressible relation, which only has “p/ρ.” It becomes more clear when we make the ideal-gas substitution p/ ρ = RT and cp = kR/(k − 1). Then we obtain the equivalent of the adiabatic, no-shaft-work energy equation: c pT + V2 + gz = constant 2 Ans. 3.183 The pump in Fig. P3.183 draws gasoline at 20°C from a reservoir. Pumps are in big trouble if the liquid vaporizes (cavitates) before it enters the pump. (a) Neglecting losses and assuming a flow rate of 65 gal/min, find the limitations on (x, y, z) for avoiding cavitation. (b) If pipefriction losses are included, what additional limitations might be important? Solution: (a) From Table A.3, ρ = 680 kg/ 3 m and pv = 5.51E+4. z2 − z1 = y + z = y+ z = p1 − p2 ( pa + ρgy ) − pv = ρg ρg (100,000 − 55,100) + y z = 6.73 m (680)(9.81) Fig. P3.183 Thus make length z appreciably less than 6.73 (25% less), or z < 5 m. Ans. (a) (b) Total pipe length (x + y + z) restricted by friction losses. Ans. (b) 3.184 For the system of Prob. 3.183, let the pump exhaust gasoline at 65 gal/min to the atmosphere through a 3-cm-diameter opening, with no cavitation, when x = 3 m, y = 2 2.5 m, and z = 2 m. If the friction head loss is hloss ≈ 3.7(V /2g), where V is the average velocity in the pipe, estimate the horsepower required to be delivered by the pump. Solution: Since power is a function of hp, Bernoulli is required. Thus calculate the velocity, ! m 3 /s " (65 gal/min)# 6.3083E−5 gal/min$& % Q V= = = 5.8 m/s π A (0.032 ) 4 The pump head may then be found, Vj 2 p1 p + z1 = 2 + z2 + h f − h p + 2g γ γ 100,000 + (680)(9.81)(2.5) 100,000 3.7(5.8 2) (5.8 2) − 2.5 = +2 + −h p + (680)(9.81) (680)(9.81) 2(9.81) 2(9.81) h p = 10.05 m P = γ Qh p = (680)(9.81)(0.0041)(10.05) P = 275 W = 0.37 hp Ans. 3.185 Water at 20°C flows through a 3 vertical tapered pipe at 163 m /h. The entrance diameter is 12 cm, and the pipe diameter reduces by 3 mm for every 2 meter rise in elevation. For frictionless flow, if the entrance pressure is 400 kPa, at what elevation will the fluid pressure be 100 kPa? Solution: Bernouilli’s relation applies, p1 Q2 p Q22 + z1 + 1 2 = 2 + z2 + γ γ 2gA 1 2gA 22 (1) d 2 = d1 − 0.0015(z 2 − z1) (2) Where, 3 3 Also, Q1 = Q2 = Q = (163 m /h)(h/3600s) = 0.0453 m /s; γ = 9790; z1 = 0.0; p1 = 400,000; and p2 = 100,000. Using EES software to solve equations (1) and (2) simultaneously, the final height is found to be z ≈ 27.2 m. The pipe diameter at this elevation is d2 = 0.079 m = 7.9 cm. FUNDAMENTALS OF ENGINEERING EXAM PROBLEMS: Answers FE3.1 In Fig. FE3.1 water exits from a nozzle into atmospheric pressure of 101 kPa. If the flow rate is 160 gal/min, what is the average velocity at section 1? (a) 2.6 m/s (b) 0.81 m/s (c) 93 m/s (d) 23 m/s (e) 1.62 m/s FE3.2 In Fig. FE3.1 water exits from a nozzle into atmospheric pressure of 101 kPa. If the flow rate is 160 gal/min and friction is neglected, what is the gage pressure at section 1? (a) 1.4 kPa (b) 32 kPa (c) 43 kPa (d) 22 kPa (e) 123 kPa FE3.3 In Fig. FE3.1 water exits from a nozzle into atmospheric pressure of 101 kPa. If the exit velocity is V2 = 8 m/s and friction is neglected, what is the axial flange force required to keep the nozzle attached to pipe 1? (a) 11 N (b) 36 N (c) 83 N (d) 123 N (e) 110 N FE3.4 In Fig. FE3.1 water exits from a nozzle into atmospheric pressure of 101 kPa. If the manometer fluid has a specific gravity of 1.6 and h = 66 cm, with friction neglected, what is the average velocity at section 2? (a) 4.55 m/s (b) 2.4 m/s (c) 2.8 m/s (d) 5.55 m/s (e) 3.4 m/s FE3.5 A jet of water 3 cm in diameter strikes normal to a plate as in Fig. FE3.5. If the force required to hold the plate is 23 N, what is the jet velocity? (a) 2.85 m/s (b) 5.7 m/s (c) 8.1 m/s (d) 4.0 m/s (e) 23 m/s FE3.6 A fireboat pump delivers water to a vertical nozzle with a 3:1 diameter ratio, as in Fig. FE3.6. If friction is neglected and the flow rate is 500 gal/min, how high will the outlet water jet rise? (a) 2.0 m (b) 9.8 m (c) 32 m (d) 64 m (e) 98 m FE3.7 A fireboat pump delivers water to a vertical nozzle with a 3:1 diameter ratio, as in Fig. FE3.6. If friction is neglected and the pump increases the pressure at section 1 to 51 kPa (gage), what will be the resulting flow rate? (a) 187 gal/min (b) 199 gal/min (c) 214 gal/min (d) 359 gal/min (e) 141 gal/min FE3.8 A fireboat pump delivers water to a vertical nozzle with a 3:1 diameter ratio, as in Fig. FE3.6. If duct and nozzle friction are neglected and the pump provides 12.3 feet of head to the flow, what will be the outlet flow rate? (a) 85 gal/min (b) 120 gal/min (c) 154 gal/min (d) 217 gal/min (e) 285 gal/min FE3.9 Water flowing in a smooth 6-cm-diameter pipe enters a venturi contraction with a throat diameter of 3 cm. Upstream pressure is 120 kPa. If cavitation occurs in the throat at a flow rate of 155 gal/min, what is the estimated fluid vapor pressure, assuming ideal frictionless flow? (a) 6 kPa (b) 12 kPa (c) 24 kPa (d) 31 kPa (e) 52 kPa FE3.10 Water flowing in a smooth 6-cm-diameter pipe enters a venturi contraction with a throat diameter of 4 cm. Upstream pressure is 120 kPa. If the pressure in the throat is 50 kPa, what is the flow rate, assuming ideal frictionless flow? (a) 7.5 gal/min (b) 236 gal/min (c) 263 gal/min (d) 745 gal/min (e) 1053 gal/min Fig. FE3.1 Fig. FE3.5 Fig. FE3.6 COMPREHENSIVE PROBLEMS C3.1 In a certain industrial process, oil of density ρ flows through the inclined pipe in the figure. A U-tube manometer with fluid density ρm, measures the pressure difference between points 1 and 2, as shown. The flow is steady, so that fluids in the U-tube are stationary. (a) Find an analytic expression for p1 − p2 in terms of system parameters. (b) Discuss the conditions on h necessary for there to be no flow in the pipe. (c) What about flow up, from 1 to 2? (d) What about flow down, from 2 to 1? Solution: (a) Start at 1 and work your way around the U-tube to point 2: or: p1 + ρgs + ρgh − ρm gh − ρgs − ρg ∆z = p2 , p1 − p2 = ρ g∆ z + ( ρm − ρ ) gh where ∆z = z2 − z1 Ans. (a) (b) If there is no flow, the pressure is entirely hydrostatic, therefore ∆p = ρg and, since ρm ≠ ρ, it follows from Ans. (a) above that h = 0 Ans. (b) (c) If h is positive (as in the figure above), p1 is greater than it would be for no flow, because of head losses in the pipe. Thus, if h > 0, flow is up from 1 to 2. Ans. (c) (d) If h is negative, p1 is less than it would be for no flow, because the head losses act against hydrostatics. Thus, if h < 0, flow is down from 2 to 1. Ans. (d) Note that h is a direct measure of flow, regardless of the angle θ of the pipe. 3 C3.2 A rigid tank of volume υ = 1.0 m is initially filled with air at 20°C and po = 100 kPa. At time t = 0, a vacuum pump is turned on and evacuates air at a constant volume flow rate Q = 80 L/min (regardless of the pressure). Assume an ideal gas and an isothermal process. (a) Set up a differential equation for this flow. (b) Solve this equation for t as a function of (υ , Q, p, po). (c) Compute the time in minutes to pump the tank down to p = 20 kPa. [Hint: Your answer should lie between 15 and 25 minutes.] Solution: becomes The control volume encloses the tank, as shown. The CV mass flow relation d dt ( ! ρdυ) + ! m! out − ! m! in = 0 Assuming that ρ is constant throughout the tank, the integral equals ρυ, and we obtain ! ρ" dρ Q Qt = − ! dt , yielding ln # $ = − ρ υ υ % ρo & Where ρo is the initial density. But, for an isothermal ideal gas, ρ/ ρo = p/po. Thus the time required to pump the tank down to pressure p is given by υ dρ + ρQ = 0, or : dt ! υ ! p" ln # $ Ans. (a, b) Q % po & 3 (c) For our particular numbers, noting Q = 80 L/min = 0.080 m /min, the time to pump a 3 1 m tank down from 100 to 20 kPa is t=− t= − 1.0 m3 ! 20 " ln # $ = 20.1 min 0.08 m3 /min % 100& Ans. (c) C3.3 Suppose the same steady water jet as in Prob. 3.40 (jet velocity 8 m/s and jet diameter 10 cm) impinges instead on a cup cavity as shown in the figure. The water is turned 180° and exits, due to friction, at lower velocity, Ve = 4 m/s. (Looking from the left, the exit jet is a circular annulus of outer radius R and thickness h, flowing toward the viewer.) The cup has a radius of curvature of 25 cm. Find (a) the thickness h of the exit jet, and (b) the force F required to Fig. C3.3 hold the cupped object in place. (c) Compare part (b) to Prob. 3.40, where F = 500 N, and give a physical explanation as to why F has changed. Solution: For a steady-flow control volume enclosing the block and cutting through the jets, we obtain ΣQin = ΣQout, or: Vj V j D 2j π 2 Dj = Veπ [ R2 − ( R − h)2 ], or: h = R − R2 − 4 Ve 4 Ans. (a) For our particular numbers, h = 0.25 − (0.25)2 − 8 (0.1) 2 = 0.25 − 0.2398 = 0.0102 m = 1.02 cm 4 4 Ans. (a) (b) Use the momentum relation, assuming no net pressure force except for F: ! Fx = −F = m! jet ( −Ve ) − m! jet (V j ), or: F = ρV j π 2 D (V + Ve ) Ans. (b) 4 j j For our particular numbers: F = 998(8) π (0.1) 2(8 + 4) = 752 N to the left 4 Ans. (b) (c) The answer to Prob. 3.40 was 502 N. We get 50% more because we turned through 180°, not 90°. Ans. (c) C3.4 The air flow beneath an air hockey puck is very complex, especially since the air jets from the table impinge on the puck at various points asymmetrically. A reasonable approximation is that, at any given time, the gage pressure on the bottom of the puck is halfway between zero (atmospheric) and the 2 stagnation pressure of the impinging jets, po = 1/2 ρVjet . (a) Find the velocity Vjet required to support a puck of weight W and diameter d, with air density ρ as a parameter. (b) For W = 0.05 lbf and d = 2.5 inches, estimate the required jet velocity in ft/s. Solution: bottom: (a) The puck has atmospheric pressure on the top and slightly higher on the ( punder − pa ) A puck = W = 1! ρ 4 W "π 0 + V 2jet $ d2 , Solve for V jet = 2 #% 2 d πρ &4 Ans. (a) For our particular numbers, W = 0.05 lbf and d = 2.5 inches, we assume sea-level air, 3 ρ = 0.00237 slug/ft , and obtain V jet = 4 0.05 lbf = 50 ft/s (2.5/12 ft) π (0.00237 slug/ft 3 ) Ans. (b) C3.5 Neglecting friction sometimes leads to odd results. You are asked to analyze and discuss the following example in Fig. C3.5. A fan blows air vertically through a duct from section 1 to section 2, as shown. Assume constant air density ρ. Neglecting frictional losses, find a relation between the required fan head hp and the flow rate and the elevation change. Then explain what may be an unexpected result. Solution: Neglecting frictional losses, hf = 0, and Bernoulli becomes, p1 V12 p V2 + +z1 = 2 +z 2 + 2 −h p ρ g 2g ρg 2g Fig. C3.5 p1 V12 p + ρg (z1 − z2 ) V22 + +z = 2 + + z2 − h p ρg 2g 1 ρg 2g Since the fan draws from and exhausts to atmosphere, V1 = V2 ≈ 0. Solving for hp, h p = ρ g (z1 − z 2 ) + ρ gz 2 − ρ gz 1 = 0 Ans. Without friction, and with V1 = V2, the energy equation predicts that hp = 0! Because the air has insignificant weight, as compared to a heavier fluid such as water, the power input required to lift the air is also negligible.