MAE 101A Homework 3 Solution /5/018
Munon 3.6: What preure gradient along the treamline, /d, i required to accelerate water upward in a vertical pipe at a rate of 30 ft/? What i the anwer if the flow i downward? Preure gradient, /d a = 30 ft teady 1 z V g = V ρ a = V V + V t θ = ±90 γ = 6.4 lbf ft 3 ρ = 1.94 lug ft 3 Equation 1 Equation Invicid Incompreible fluid Up (θ = 90 ) Down (θ = 90 ) Conider Eq. 1, which if you recall from the derivation (ee pg 3) i eentially F = ma in the treamline direction. 1 ρ z V g = V Since p = p() the partial derivative / = /d. From geometry it can be een z/ = in θ. For teady flow, acceleration in the treamline direction reult in a change in peed a decribed by a = V V. d = γ in θ ρa Calculate the preure gradient to accelerate the flow upward (θ = 90 ) at a = 30 ft/ d d = γ in θ ρa lbf lug ft 1 lbf = (6.4 ft3) in(90 ) (1.94 ) (30 ft 3 ) [ d = 10.6 lbf ft 3 1 lug ft ] Calculate the preure gradient to accelerate the flow downward (θ = 90 ) at a = 30 ft/ d d = γ in θ ρa lbf lug ft 1 lbf = (6.4 ft3) in( 90 ) (1.94 ) (30 ft 3 ) [ d = 4. lbf ft 3 1 lug ft ] Dicuion Preure and gravity force act to accelerate the fluid. When the flow i directed upward gravity oppoe fluid motion reulting in greater preure drop; when directed downward gravity contribute to fluid motion and the preure gradient i much maller in magnitude. Since g > a the preure gradient i actually poitive. 1
Munon 3.1: Water flow around the vertical two-dimenional bend with circular treamline and contant velocity a hown in Fig. P3.1. If the preure i 40 kpa at point (1), determine the preure at point () and (3). Aume that the velocity profile i uniform a indicated. Preure at () & (3) V = 10 m/ Steady p and p 3 p 1 = 40 kpa Invicid R 1 = 6 m Incompreible fluid ρ = 1000 kh/m 3 Uniform velocity profile γ = 9.81 kn/m 3 θ = 0 g z 1 = V n ρ n R g z + 1 = V n ρ n R (Where n i poitive toward center of curvature) (Where n i poitive outward from center of curvature) Equation 3a Equation 3b a n = V /R Equation 4 Conider Eq. 3a on the left and Eq. 3b on the right. Thee equation decribe the phyic of F n = ma n with n directed toward and away from the center of curvature repectively (ee pg 3). Since the phyic are the ame the olution will be the ame uing either equation. Equation 3a g z 1 = V n ρ n R Equation 3b g z + 1 = V n ρ n R Since p = p(n) the partial derivative / n = /dn. From geometry it can be een, z n z/ n = co θ z n dn = (ρg co θ + ρ V R ) Separation of variable can be ued to integrate, = (ρg co θ + ρ V R ) dn dn z/ n = co θ = ρg co θ + ρ V R V = (ρg co θ + ρ ) dn R Note that n = f(r) and thi relationhip need to be accounted for in the integration, n = R 1 R and dn = dr n = R 1 + R and dn = dr = (ρg co θ + ρ V R ) dr p R = p (ρ V + γ co θ) dr 1 R 1 R V = (ρg co θ + ρ ) dr R p p 1 = (ρv ln R + γr co θ) (ρv ln R 1 + γr 1 co θ) p = p 1 + ρv ln R R 1 + γ co θ (R R 1 ) At R = R = 5 m and θ = 0 At R = R 3 = 4 m and θ = 0 p = 40 kpa + (10 3 kg m 3) (10 m ) ln ( 5 ) + (9810 N 6 m3) (5 m 6 m) p = 11.958 kpa p 3 = 40 kpa + (10 3 kg m 3) (10 m ) ln ( 4 ) + (9810 N 6 m3) (4 m 6 m) p 3 = 0.167 kpa Dicuion: Note on problem 3.6, an imbalance of preure and gravity force reult in fluid acceleration in the treamline direction. Here, the curvature in treamline caue a normal acceleration and reult in a preure variation in the normal direction. In the treamline direction, ince a = 0, preure and gravity force are balanced o a hydrotatic preure variation occur in the direction.
Derivation - Dynamic in Streamline Direction Conider a force balance in the treamline direction applied to a fluid element, F = ma Force include preure on each urface (expreed uing a Taylor erie expanion) and the component of weight in the treamline direction. (p 1 δ) δa (p + 1 δ) δa δw = δm a z δδa δm g = δm a Divide by the volume of fluid element δ = δδa z γ = ρa 1 z g = a ρ Equation 1 In general V = V(t, ) and a = V Thu, for teady flow: 1 ρ t g z + V V = V V Derivation - Dynamic in Normal Direction Conider a force balance in the normal direction (pointing toward the center of curvature) F n = ma n Force include preure on each urface (expreed uing a Taylor erie expanion) and the component of weight in the normal direction. (p 1 δn) δa (p + 1 δn) δa δw n n n = δm a n z V δnδa δmg = δm n n R Divide by the volume of fluid element δ = δnδa z V γ = ρ n n R 1 ρ n z g = V n R (Equation 3a) Alternatively the normal direction could be flipped (pointing away from the center of curvature) F n = ma n (p 1 δn) δa (p + 1 δn) δa + δw n n n = δm a n z V δnδa δm g = δm n n R Notice the ign remain the ame on the preure and weight term. The direction of thee force are incorporated in the partial derivative. The fluid i till accelerated toward the center of curvature o a n = V /R. Divide by the volume of fluid element δ = δnδa z V + ρg = ρ n n R 1 ρ n z + g = V n R (Equation 3b) n 3
Munon 3.4: What i the minimum height for an oil (SG = 0.75) manometer to meaure airplane peed up to 30 m/ at altitude up to 1500 m? The manometer i connected to a Pitot-tatic tube a hown in Fig. P3.4.t Height, h V = 30 m/ Steady z = 1500 m Incompreible SG = 0.75 Invicid Static manometer fluid γ m γ a p 1 + 1 ρv 1 + γz 1 = p + 1 ρv + γz = contant p p 1 = γ m h SG = ρ ρ w = γ γ w Auming the air i incompreible and invicid we can apply the Bernoullli Equation to air at location 1 &. p 1 + 1 ρ av 1 + γ a z 1 = p + 1 ρ av + γ a z = contant Location i the tagnation tate where the fluid i brought to ret (V = 0) a it enter the pitot tube. p 1 + 1 ρv 1 = p p p 1 = 1 ρv 1 Since air i light, SG 10 3 compared to SG = 0.75, the weight of air in the manometer i negligible. The preure at the fluid interface of the manometer matche the preure at the repective inlet of the pitot-tube. Fluid tatic applied to the incompreible manometer fluid give, p p 1 = γ m h γ m h = 1 ρ av 1 h = ρ av 1 γ m h = ρ av 1 SGγ w h = (1.06 kg/m3 )(30 m/) (0.75)(9810 N m 3) h = 6.48 cm Dicuion Variou preure definition were introduced in thi problem. Static, dynamic, and tagnation preure are all commonly referenced in fluid mechanic o it i important to undertand and remember how each i defined. If you place the palm of your hand a flow, uch a in front of a fan, you will feel the tagnation preure the preure aociated with the fluid at ret. The back of your hand would feel the tatic preure, which i the thermodynamic preure of the fluid. The difference between the two i the dynamic preure. You mut exert a force/area equal to the dynamic preure to keep your hand teady. 4
Munon 3.56: Find the water ma flow rate at the nozzle outlet O hown in Fig. P3.56, and calculate the maximum height to which the water tream will rie. The water denity i 1.9 lug/ft 3, and the flow i invicid. Ma flow rate, m z 1 = 0 Steady Height of jet, h 3 z = 1 ft + 6ft = 15 ft Incompreible d = 1 in = 0.0833 ft Invicid along treamline A = 0 ft 3 Large tank ρ = 1.9 lug/ft 3 p 1 + 1 ρv 1 + γz 1 = p + 1 ρv + γz = contant m = ρav = ρq, Q = AV Since the flow i teady, incompreible, and invicid, apply B.E along a treamline: p 1 + 1 ρv 1 + γz 1 = p + 1 ρv + γz Location 1 and are both expoed to atmopheric preure o p 1 = p = p atm. Alo ince A 1 A we can aume V 1 0. γz 1 = 1 ρv + γz PE at the urface i converted to KE at nozzle exit V = g(z 1 z ) = (3. ft ) (15 ft) = 31.1 ft/ m = ρav = π 4 ρd V = ( π lug ) (1.9 ) (0.0833 4 ft 3 ft) (31.1 ft lug ) = 0.31 m = 0.31 lug p 1 + 1 ρv 1 + ρgz 1 = p 3 + 1 ρv 3 + ρgz 3 ρgz 1 = ρgz 3 z 1 = z 3 = 0 ft KE at the nozzle exit () i converted back to PE at (3) Dicuion No loe occur o the tream of fluid will rie to the ame elevation a the tank urface. In reality loe do occur. Lo ource include friction along the tank wall, nozzle inlet, pipe wall, corner bend, and nozzle outlet. Thee loe reduce the mechanical energy available to the flow reulting in a horter tream out of the nozzle. 5
Munon 3.69: Water i iphoned from the tank hown in Fig. P3.69. Determine the flowrate from the tank and the preure at point (1), (), and (3) if vicou effect are negligible. Flow rate, Q z 0 = 0 Steady Preure at (1), (), and (3) z 1 = 8 ft Incompreible fluid z = 8 ft Invicid z 3 = 0 Large tank z 4 = 3 ft p + 1 ρv + γz = contant m = ρav = ρq, Q = AV Apply BE along a treamline from urface (0) to the iphon outlet (4) p 0 + 1 ρv 0 + ρgz 0 = p 4 + 1 ρv 4 + ρgz 4 Location (0) and (4) are both expoed to atmopheric preure o p 0 = p 4 = p atm. Alo ince A 0 A 4 we can aume V 0 0. V 4 = g(z 0 z 4 ) = (3. ft ) (3 ft) = 13.9 ft/ Q 4 = A 4 V 4 = π 4 ( 1 ft) (13.9 ft ) Q 4 = 0.303 ft3 Apply B.E. from urface (0) to the bottom of the tank (1). Note for mot of the tank, except near the iphon inlet, V 0 o one would expect hydrotatic preure variation. p 0 + 1 ρv 0 + γz 0 = p 1 + 1 ρv 1 + γz 1 0 = p 1 + ρgz 1 p 1 = ρgz 1 p 1 = (6.4 lbf ft3) ( 8 ft) p 1 = 499 lbf ft From conervation of ma, m 4 = m 3 = m. Since ρ and the area of the tube are contant V 4 = V 3 = V Apply BE from 0 to Apply BE from 0 to 3 p 0 + 1 ρv 0 + γz 0 = p + 1 ρv + γz p 0 + 1 ρv 0 + γz 0 = p 3 + 1 ρv 3 + γz 3 0 = p + 1 ρv + γz 0 = p 3 + 1 ρv 3 + γz 3 p = 1 ρv γz p 3 = 1 ρv 3 γz 3 = 1 ρv 3 p = 1 kg (1.94 m 3) (13.9 ft ) (6.4 lbf ft3) ( 8 ft) p 3 = 1 kg (1.94 m 3) (13.9 ft ) p = 311.8 lbf/ft p 3 = 187.4 lbf/ft Dicuion: The velocity of the fluid in the tube reduce the preure, p 3 < p 0, cauing a preure difference allowing the fluid to flow and be iphoned from the tank. 6
Munon 3.76: The pecific gravity of the manometer fluid hown in Fig. P3.76 i 1.07. Determine the volume flowrate, Q, if the flow i invicid and incompreible and the flowing fluid i (a) water, (b) gaoline, or (c) air at tandard condition. Volume flowrate, Q SG = 1.07 Steady h 1 = h = 0 Incompreible fluid h 3 = 0.053 m Invicid h 4 = 0.051 m D = 0.09 m g = 9.81 m/ p 1 + 1 ρv 1 + γz 1 = p + 1 ρv + γz = contant p = p 1 + γ(h h 1 ) m = ρav = ρq, Q = AV Since the flow i teady, incompreible, and invicid the B.E. can be applied between (1) and () p 1 + 1 ρv 1 γh 1 = p + 1 ρv γh A the flow impact the tagnation inlet of the pitot tube the fluid i brought to ret V = 0. p 1 + 1 ρv 1 = p V 1 = g (p p 1 ) γ Fluid tatic can be ued to relate the preure p 1 and p. p 1 + γh 4 + γ m (h 3 h 4 ) γh 3 = p p p 1 = γ(h 3 h 4 ) + γ m (h 3 h 4 ) p p 1 = (γ m γ)(h 3 h 4 ) V 1 = g (γ m γ)(h 3 h 4 ) γ Q = AV = π 4 D g (γ m γ)(h 3 h 4 ) γ or Q = π 4 D g (SGγ w γ)(h 3 h 4 ) γ Fluid γ [N/m 3 ] Q [m 3 /] Water 9810 0.00105 Gaoline 6670 0.0030 Air 1 0.11779 Dicuion A the denity (or equivalently γ) decreae the volumetric flow rate, Q, increae. One explanation i that the dynamic preure, 1 ρv, i eentially what the manometer i meauring. Since the manometer reading i contant in each cae the dynamic preure mut alo remain contant. A denity decreae velocity mut increae to keep the dynamic preure contant reulting in the invere relationhip between γ and Q. 7
Munon 3.81: Air flow through the device hown in Fig. P3.81. If the flowrate i large enough, the preure within the contriction will be low enough to draw the water up into the tube. Determine the flowrate, Q, and the preure needed at ection (1) to draw the water into ection (). Neglect compreibility and vicou effect. Volume flowrate, Q D 1 = D 3 = 0.05 m Steady Preure P 1 D = 0.05 m Incompreible fluid h = 0.3 m Invicid p 1 + 1 ρv 1 + γz 1 = p + 1 ρv + γz = contant p p 1 = γ m h m = ρav = ρq Apply Bernoulli from location to 3 where the flow exit a a free jet p + 1 ρ av + γ a z = p 3 + 1 ρ av 3 + γ a z 3 The two location at the ame elevation o z = z 3 and location (3) i expoed to the atmophere o p 3 = p atm p + 1 ρ av = p atm + 1 ρ av 3 From continuity, m = m 3 ρ a A V = ρa 3 V 3 V = A 3 A V 3 = ( D 3 D ) V 3 = 4V 3 Subtitute into Bernoulli, p + 8ρ a V 3 = p atm + 1 ρ av 3 V 3 = (p atm p ) 15ρ a From hydrotatic the preure at location i p = p atm γ w h V 3 = (p atm p atm +γ w h) 15ρ a = γ wh 15ρ a V 3 = γ wh = (9810 N/m3 )(0.3 m) = 18. m 15ρ a 15(1.184 kg/m 3 ) Q 3 = A 3 V 3 = π 4 D 3 V 3 = π 4 (0.05 m) (18. m ) Q = Q 3 = 0.0357 m3 From conervation of ma, m 1 = m 3. Since ρ 1 = ρ 3 and A 1 = A 3 the velocity V 1 = V 3. p 1 + 1 ρ av 1 + γ a z 1 = p 3 + 1 ρ av 3 + γ a z 3 p 1 = p 3 = p atm Dicuion The preure at the inlet matche the preure at the outlet becaue there are no loe. Generally the lo of mechanical energy would require p 1 > p 3 to upport the flow. 8
H3.1 (Ch 4): The velocity field repreenting flow in a corner i given by, V = Axi Ayj where A = 1. a. Determine the equation for the treamline in thi flow. Explain the relevance of A. b. Plot everal treamline in the firt quadrant including the one that pae through the point (x, y) = (0,0) and the one that pae through the point (x, y) = (,). c. Show that the parametric equation for fluid particle motion are given by x p = c 1 e At and y p = c e At. d. Determine the equation for the pathline of the particle located at the point (x, y) = (,) at the intant t = 0. e. How doe thi pathline compare with the treamline through the ame point. Explain. Streamline Eq. V = Axi Ayj Fluid kinematic Pathline Eq. A = 1 dx dy = u v u p = dx p dt v p = dy p dt Streamline Pathline Pathline a. Determine the equation for the treamline in thi flow. Explain the relevance of A. Streamline are everywhere tangential to the velocity field a een to the right dx dy = u v dx = Ax dy Ay dx x = dy y ln(x) = ln(y) + c a ln(x) + ln(y) = c a ln(xy) = c a xy = e c a = C The parameter A act to cale the velocity field. Since it cale both u and v equally it doe not change the direction of the velocity vector. The treamline i tangent to the velocity field o A drop out and you ee the ame treamline for any A. However, it hould be noted that A doe affect the magnitude of velocity. A fluid element travering a treamline will move fater or lower depending on A. y = C x b. Plot everal treamline in the firt quadrant including the one that pae through the point (x, y) = (0,0) and the one that pae through the point (x, y) = (,). Each value of C repreent a different treamline. You can elect a coordinate pair and olve for the treamline paing through that point. At (x, y) = (0,0), C = 0, and xy = 0 At (x, y) = (,), C = 4, and y = 4/x Notice the treamline are tangent to velocity field. 9 C = 0
c. Show that the parametric equation for fluid particle motion are given by x p = c 1 e At and y p = c e At. u p = dx p dt dx p = Axdt v p = dy p dt dy p = Aydt dx p x = Adt ln x = At + c x x = exp(at + c x ) x = c 1 exp(at) dy p y = Adt ln y = At + c y y = exp( At + c y ) y = c exp( At) d. Determine the equation for the pathline of the particle located at the point (x, y) = (,) at the intant t = 0. x = c 1 exp(at) y = c 1 exp( At) t = 1 A ln ( x c 1 ) t = 1 A ln ( y c ) 1 A ln ( x c 1 ) = 1 A ln ( y c ) ln ( x c 1 ) = ln ( y c ) ln ( x c 1 ) = ln ( c y ) y = c c 1 x = C x At (x, y) = (,), C = 4 y = 4 x e. How doe thi pathline compare with the treamline through the ame point. Explain. Flow i teady o treamline, pathline, and treakline are the ame. 10