Execise sheet 8 (Modeling lage- and small-scale flows) last edited June 18, 2018 These lectue notes ae based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. Except othewise indicated, we assume that fluids ae Newtonian, and that: ρ wate = 1 000 kg m 3 ; p atm. = 1 ba; ρ atm. = 1,225 kg m 3 ; T atm. = 11,3 C; µ atm. = 1,5 10 5 N s m 2 ; д = 9,81 m s 2. Ai is modeled as a pefect gas (R ai = 287 J K 1 kg 1 ; γ ai = 1,4; c pai = 1 005 J kg 1 K 1 ). In a highly-viscous (ceeping) steady flow, the dag F D exeted on a spheical body of diamete D at by flow at velocity U is quantified as: F D sphee = 3π µu D (8/40) 8.1 Volcanic ash fom the Eyjafjallajökull Çengel & al. [16] E10.2 In 2010, a volcano with a complicated name and unpedictable mood decided to gound the entie Euopean ailine industy fo five days. We conside a micoscopic ash paticle eleased at vey high altitude ( 50 C, 0,55 ba, 1,474 10 5 N s m 2 ). We model it as a sphee with 50 µm diamete. The density of volcanic ash is 1 240 kg m 3. 1. What is the teminal velocity of the paticle? 2. Will this teminal velocity incease o decease as the paticle pogesses towads the gound? (give a bief, qualitative answe) 8.2 Wate dop Çengel & al. [16] 10-21 A ainy day povides yet anothe oppotunity fo exploing fluid dynamics (fig. 8.12). A wate dop with diamete 42,4 µm is falling though ai at 25 C and 1 ba. 1. Which teminal velocity will it each? 2. Which velocity will it each once its diamete will have doubled? 183
Figue 8.12 A sketched diagam showing the geomety of wate dops of vaious sizes in fee fall. When thei diamete is lowe than 2 mm, wate dops ae appoximately spheical (B). As they gow beyond this size, thei shape changes and they eventually beak-up (C-E). They neve display the classical shape displayed in A, which is caused only by suface tension effects when they dip fom solid sufaces. Figue CC-by-sa by Ryan Wilson 8.3 Hanga oof based on White [13] P8.54 Cetain flows in which both compessibility and viscosity effects ae negligible can be descibed using the potential flow assumption (the hypothesis that the flow is eveywhee iotational). If we compute the two-dimensional lamina steady fluid flow aound a cylinde pofile, we obtain the velocities in pola coodinates as: v = 1 ψ θ = U cos θ v θ = ψ = U sin θ ) (1 R2 2 ) (1 + R2 whee the oigin ( = 0) is at the cente of the cylinde pofile; θ is measued elative to the fee-steam velocity vecto; U is the incoming fee-steam velocity; and R is the (fixed) cylinde adius. Based on this model, in this execise, we study the flow ove a hanga oof. 2 (8/25) (8/26) Wind with a nealy-unifom velocity U = 100 km h 1 is blowing acoss a 50 m-long hanga with a semi-cylindical geomety, as shown in fig. 8.13. The adius of the hanga is R = 20 m. 1. If the pessue inside the hanga is the same as the pessue of the faaway atmosphee, and if the wind closely follows the hanga oof geomety (without any flow sepaation), what is the total lift foce on the hanga? (hint: we accept that sin 3 x dx = 1 3 cos3 x cos x + k). 184
Figue 8.13 A semi-cylindical hanga oof. Wind with unifom velocity U flows pependicula to the cylinde axis. Figue CC-0 o.c. 2. At which position on the oof could we dill a hole to negate the aeodynamic lift foce? 3. Popose two easons why the aeodynamic foce measued in pactice on the hanga oof may be lowe than calculated with this model. 8.4 Cabling of the Wight Flye deived fom Munson & al. [18] 9.106 The Wight Flye I, the fist poweed and contolled aicaft in histoy, was subjected to multiple types of dag. We have aleady studied viscous fiction on its thin wings in execise 7.4. The data in figue 8.14 povides the oppotunity to quantify dag due to pessue. A netwok of metal cables with diamete 1,27 mm ciss-cossed the aicaft in ode to povide stuctual igidity. The cables wee positioned pependiculaly to the ai flow, which came at 40 km h 1. The total cable length was appoximately 60 m. What was the dag geneated by the cables? 185
Figue 8.14 Expeimental measuements of the dag coefficient applying to a cylinde and to a sphee as a function of the diamete-based Reynolds numbe [Re] D, shown togethe with schematic depictions of the flow aound the cylinde. By convention, the dag coefficient C D C F D F D 1 2 ρsu 2 (eq. 6/2 p.130) compaes the dag foce F D with the fontal aea S. Both figues fom Munson & al.[18] 186
8.5 Ping pong ball Munson & al. [18] E9.16 A seies of expeiments is conducted in a wind tunnel on a lage cast ion ball with a smooth suface; the esults ae shown in fig. 8.15. These measuement data ae used to pedict the behavio of a ping pong ball. Table tennis egulations constain the mass of the ball to 2,7 g and its diamete to 40 mm. 1. Is it possible fo a ball thown at a speed of 50 km h 1 to have a pefectly hoizontal tajectoy? 2. If so, what would be its deceleation? 3. How would the dag and lift applying on the ball evolve if the ai viscosity was pogessively deceased to zeo? Figue 8.15 Expeimental measuements of the lift and dag coefficients applying on a otating sphee in an steady unifom flow. Figue fom Munson & al.[18] 187
8.6 Lift on a symmetical object non-examinable Biefly explain (e.g. with answes 30 wods o less) how lift can be geneated on a sphee o a cylinde, with diffeential contol bounday laye contol; with the effect of otation. Daw a few steamlines in a two-dimensional sketch of the phenomenon. 8.7 Ai flow ove a wing pofile Non-examinable. Fom Munson & al. [18] 9.109 The chaacteistics of a thin, flat-bottomed aifoil ae examined by a goup of students in a wind tunnel. The fist investigations focus on the bounday laye, and the eseach goup evaluate the bounday laye thickness and make sue that it is fully attached. Once this is done, the goup poceeds with speed measuements all aound the aifoil. Measuements of the longitudinal speed u just above the bounday laye on the top suface ae tabulated below: x/c (%) y/c (%) u/u 0 0 0 2,5 3,72 0,971 5 5,3 1,232 7,5 6,48 1,273 10 7,43 1,271 20 9,92 1,276 30 11,14 1,295 40 10,49 1,307 50 10,45 1,308 60 9,11 1,195 70 6,46 1,065 80 3,62 0,945 90 1,26 0,856 100 0 0,807 On the bottom suface, the speed is measued as being constant (u = U ) to within expeimental eo. What is the lift coefficient on the aifoil? 188
8.8 Flow field of a tonado Fom Çengel & al. [16] E9-5, E9-14 & E10-3 The continuity and Navie-Stokes equations fo incompessible flow witten in cylindical coodinates ae as follows: ρ v t ρ ρ v + v + v θ = ρд p + µ [ vθ t v θ v2 θ + v z ( v ) [ 1 v θ + v + v θ = ρд θ 1 [ vz t p θ + µ v z + v + v θ = ρд z p z + µ 1 v + 1 v z v θ θ + v ] v θ v θ + v z z [ ( 1 v ) θ ] v θ + v v z z z ( v ) z [ 1 v 2 + 1 2 2 v ( θ ) 2 2 2 v θ θ + 2 v ( z) 2 v θ 2 + 1 2 2 v θ ( θ ) 2 + 2 2 v θ + 2 v θ ( z) 2 + 1 2 ] v z 2 ( θ ) 2 + 2 v z ( z) 2 ] ] (8/42) (8/43) (8/44) v θ θ + v z z = 0 (8/45) In this execise, we ae inteested in solving the pessue field in a simplified desciption of a tonado. Fo this, we conside only a hoizontal laye of the flow, and we conside that all popeties ae independent of the altitude z and of the time t. We stat by modeling the tonado as a votex impating an angula velocity such that: v θ = Γ 2π in which Γ is the ciculation (measued in s 1 ) and emains constant eveywhee. (8/46) 1. What fom must the adial velocity v take in ode to satisfy continuity? Among all the possibilities fo v, we choose the simplest in ou study, so that: v = 0 (8/47) all thoughout the tonado flow field. 2. What is the pessue field thoughout the hoizontal laye of the tonado? The flow field descibed above becomes unphysical in the vey cente of the tonado votex. Indeed, ou model fo v θ is typical of an iotational votex, which, like all iotational flows, is constucted unde the pemise that the flow is inviscid. Howeve, in the cente of the votex, we ae confonted with high velocity gadients ove vey small distances: viscous effects can no longe be neglected and ou model beaks down. It is obseved that in most such votices, a votex coe egion foms that otates just like a solid cylindical body. This flow is otational and its govening equations esult in a ealistic pessue distibution. 189
3. What is the pessue field within the otational coe of the tonado? Make a simple sketch showing the pessue distibution as a function of adius thoughout the entie tonado flow field. 4. [An opening question fo cuious students] What is the basic mechanism of votex stetching? How would it modify the flow field descibed hee? 190
Answes 8.1 1) At teminal velocity, the weight of the sphee equals the dag. This allows us to D obtain U = дρ 2 sphee 18µ = 0,1146 m s 1 : unbeaably slow when you ae stuck in an aipot! With U, check that the Reynolds numbe indeed coesponds to ceeping flow: [Re] D = 0,334. 8.2 Same as pevious execise: U 1 = 4,578 10 2 m s 1 and U 2 = 0,183 m s 1, with Reynolds numbes of 0,113 and 0,906 espectively (thus ceeping flow hypothesis valid). 8.3 1) expess oof pessue as a function of θ using eq. 8/29 p.175 on eq. 8/26, then integate the vetical component of foce due to pessue: F L oof = 1,575 MN. 2) θ F=0 = 54,7 8.4 A simple eading gives F D = 6,9 N, W = 76 W. 8.5 Yes a eading gives ω = 83 ev/s. 191
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