Question 1: Figure 1: Schematic

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Jacob Lawrence Bailey
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1 Question : θ Figure : Schemtic Consider chnnel of height with rectngulr cross section s shown in the sketch. A hinged plnk of length L < nd t n ngle θ is locted t the center of the chnnel. You my ssume tht L is smll compred to the length perpendiculr to the sketch.) The plnk hs mss per unit width, m. A constnt volume flux per unit width, Q, of n inviscid, incompressible fluid is pplied t the inlet of the chnnel. Find ll equilibrium vlues for θ include BOTH stble nd unstble equilibri!) To simplify your clcultion, you my ssume tht we hve chosen the mss, m, nd volume flux, Q, such tht for t lest one equilibrium stte, θ is smll. Solution: 4 points totl, for ech equilibrium nd ech for stbility.) Before we do ny clcultions we cn see by inspection tht for inviscid flow, there re two trivil equilibrium positions. If the plnk is verticlly ligned, the pressures on the front nd bck blnce since the flow is symmetric) nd the weight of the plnk is supported by the hinge:

2 Stremlines FBD Thus there is n equilibrium t θ = π/ nd t θ = 3π/. Note, neither of these re physiclly relizble since, in rel fluid tht hs some viscosity, seprtion t the plnk ends will brek the front/bck symmetry. To find the stbility of there two solutions, we perturb wy from equilibrium nd inspect the resulting torques. For the θ = 3π/ solution: Stremlines FBD Considering the torque bout the hinge, both grvity nd pressure ct to restore the plnk to its equilibrium position so θ = 3π/ is STABLE. Drwing similr picture for π/, we find tht the pressure is stbilizing nd grvity is destbilizing so we need to do clcultion to figure out which one wins we will come bck to this fter the next section). 5 points totl: 3 points for mss conservtion, 3 points for conservtion of momentum Bernoulli), 6 points for force nd torque blnce on the plte, points to put it ll together nd find θ eq i.e. integrte!) nd point for stbility.) Solving this problem exctly is nontrivil since techniclly, we would need to

3 solve ψ = with no flux boundry conditions t the plnk nd t the top nd bottom chnnel wlls. To pproximte the equilibrium conditions, we need to estimte the torque on the plnk. Since there re severl resonble ssumptions you cn mke t this point, credit ws given if you ) mde resonble set of ssumptions ) pplied conservtion of mss, conservtion of momentum in this cse Bernoulli) correctly, nd 3) pplied force nd torque blnce on the plnk in wy tht is consistent with your resonble ssumptions. Note tht symmetry is broken in this problem becuse, if the plnk is tilted θ =, it is no longer centered in the chnnel. Even if you ignored this broken symmetry nd ssumed your plnk is in n infinite flow, there is still net torque on the plnk due to the flow since the stgntion points on the front nd bck of the plnk re not ligned see the perturbtion picture bove) so the torque rising from pressure cn counter blnce the torque due to grvity. For smll θ, one resonble pproximtion to mke is tht roughly hlf of the flux goes bove the plnk nd hlf below. Thus, by conservtion of mss, From the FBD digrm: Q = vt x)y T x) = v B x)y B x). ) we cn write down verticl force blnce: L [ ] mg F hinge = p B x ) p T x ) cos θ dx ) nd torque blnce round the center of the plnk: L [ ] L p B x ) p T x ) x L/ F hinge cos θ =. 3) Strictly speking this is not exctly correct unless there is second plnk, s indicted in light grey in Figure, preventing flow bove centerline from circulting below the plnk. 3

4 Following two stremline, one tht psses just bove the plnk nd one just below, we cn use Bernoulli to find p B nd p T : Hence p T x) = p B x) ρv in ρvin = p T x) ρv B x) 4) [ ] p B x) p T x) = ρ v T x) v B x) 5) Using conservtion of mss nd y T x) = x tn θ nd y B x) = x tn θ we find v T x) v B x) = Q 4 y y For smll θ, tn θ θ nd θ) n nθ note this is lso true for n < ). So fter little bit of lgebr we find: Inserting this in eqution ) Similrly, from eqution 3) Hence mg F hinge = T B 6) v T x) v B x) Q θ x 7) 3 L ρ Q θ 3 x dx = Q L ρ 4 3 θ. 8) F hinge = ρq L 3 θ. 9) θ eq 3mg3 ρq L ) Actully this clcultion is even esier if you tke the torque blnce bout the hinge... then you don t need to pply force blnce since you don t need F hinge in the torque blnce eqution.) The lift in this problem is supplied becuse the flow bove the plnk must speed up s it goes through the nrrowing gp. Thus, this equilibrium is UNSTABLE since, if θ is incresed, the flow bove the plnk must go even fster, further lowering the pressure, mplifying the perturbtion. Likewise, if θ is decresed, the flow bove slows down, gin mplifying the perturbtion. Now we cn lso ddress the stbility of the verticl equilibrium position. We hve chosen the weight of the plnk nd the flow rte of the fluid s.t. θ eq. Thus, if we consider position θ > θ eq, we know tht the torques rising from the flow will bet the torques due to grvity. Hence, the verticl plnk is STABLE. point.) Comment on other equilibri positions using symmetry rguments these my vry depending on wht ssumptions you mde erlier). 4

.5 Fll 5 Quiz Solutions. Lift force on circulr disk Photo removed for copyright resons. See film "Pressure Fields nd Fluid Accelertion" t http://web.mit.edu/fluids/www/shpiro/ncfmf.html Inflow: Q =!

5 .5 Fll 5 Quiz Solutions. Lift force on circulr disk Photo removed for copyright resons. See film "Pressure Fields nd Fluid Accelertion" t Inflow: Q =! V = constnt Model s D rdil) outflow of fluid; constnt density, inviscid stedy flow " v = # v r v! v " f r ' = # ' ) Point P is stgntion point with mximum pressure: henceforth we ignore the smll hydrosttic term). pr = ) = p stg = p!v!gh ) In the outer region r > ) we hve pure rdil flow; no centripetl ccelertion, no ccelertion cross!p stremlines; hence!n =!p!" =. Pressure field! function of!. In the inner flow the stremlines re curved; there is pressure grdient norml to stremlines since pressure decreses towrds center of curvture; we therefore expect: p s < p Q < p P ; Stgntion pressure is mximum vlue in system. Hence, since p s is smll we my ctully get seprtion here nd recirculting region nlogous to the ven contrct in jet; viscous effects my ply role here).! b) Velocity Field: V r = C r r! " # C r r " The second condition rises becuse for r > mss blnce gives:! V = Q =!hv r or v r = V hr! C = V hence h r = C = V Equting velocities t rdius gives! C = V h h c) Pressure Field!!!!" Applying Bernoulli Eqution long stremline from PQ ignoring!gz term)

6 .5 Fll 5 Quiz Solutions or by substituting for v r p T = p!v = p r! v r r) in ech region r <, r > ) we find: "# p r)! p = ) "V #! r 4h r ' * "V #! 4 4h r r, In fct on deeper considertion, you cn see tht this entire nlysis is relly only vlid or plusible) for h!, otherwise the proposed velocity field nd pressure field do not even mke sense. In outer region r > ) we expect v r! s r! nd hence tht p r t r! ". In the inner region, the stgntion pressure p stg = p!v!gz << ) ) is the lrgest pressure nd! s r! nd pproches p! = p then velocity decreses nd pressure decreses However, if h! then we cnnot reconcile these two vritions in pr)! The nlysis is wrong becuse of seprtion of the strem ner the shrp corner nd the resulting formtion of vortices. The role of viscosity is to smooth the shrp grdients in velocity close to r nd djust the pressure field so tht for given rdius plte, the pressure field smoothly recovers to p t. d) Totl Force on Disk Clerly the pressure decreses qudrticlly with r in the inner region nd the gge pressure t r = becomes negtive if:! 4h < or h!! h i.e. ) There is then region where pressure cting on disk is upwrds. This lift my overcome the weight of the disk s V increses. Note tht the result requires use of gge pressure becuse p cts on both sides). The pressure force downwrds on the disk is given by with element of re { da =!rdr} F P =! p r) "rdr = "#V! r r 3 4h ' dr! "#V r 4 ' ) 4h r * dr

7 .5 Fll 5 Quiz Solutions or finlly =!"V r # r 4 ' 6h )!"V r # 4 4h ln r ' ) =!"V # 4 ' 6h )!"V * 4 -, 4h. / ln F P =!"V # 4 ' 8h )!"V * 4 -, 4h. / ln ) ) 3) Note tht i) since <, term 3 is negtive ii) First term is lwys ve iii) Term chnges sign when: 4 8h! i.e. h! 4 8 = For Given Conditions h =. cm, =. cm, = 5 cm; the criticl rdius is therefore: r * = h =.cm < ; hence pressure does go negtive in inner region. For lift we require F P W =!velift force from sub tmospheric pressure ve grvity force grms) =! ir =.kg / m 3 =. " #3 g / cm 3!. " #3)V 5 # 4 ' 8 ". )!. " #3.36V!.57V =! or V = 8.84 m/s * 4 V -, 4 ".. / ln 5 = # e) Unstedy flow: the Bernoulli eqution pplied between points P where the stgntion pressure is given by eq. nd point gives #! "v r r,t) ds "t ' p)!v ) * ' p stg ) = P The only thing tht is vrying in this problem is the gp ht). We use conservtion of mss with cylindricl control volume of rdius r nd height ht). We cn then derive n expression for the chnge in velocity with time s follows; dm sys dt = = dm cv dt # nda! v " v c which gives in ech seprte region): A = d!r h) "!r V!rh!v r for r nd = d!r h for dt dt "! V!rh!v r r 3

8 .5 Fll 5 Quiz Solutions Vr ht)! nd hence v r r,t) = V ' rht)! r dh ht) dt r dh ht) dt Note tht!h = dh dt < so the velocity is incresed, nd lso note tht if h! = then these expressions for the velocities reduce to expressions in prt b). We substitute these expressions into the integrl term, nd brek it down into two prts integrted long stremline ds = dr). The unstedy Bernoulli eqution then becomes: 6 8 7! V # " h! h ' ) h! h " h!! *. 5rdr, h / ! V r r " r # # h "! h ' ) h! h " h!! - *., h / r 3 5dr p!v 3 4 " p = 45 Assuming tht p = p stg = p!v nd v = V ht)!!h ht) we obtin: "!h h! h!! # ' h! ) V ln,!h * -. h / V 4 h! h!h 4 34! V = so the gp seprtion decreses with time in rther complex nonliner wy! This eqution cn be solved s n Initil Vlue Problem with conditions ht = ) = h ; ht! = ) = ). 4

Math 8 Winter 2015 Applications of Integration

Math 8 Winter 2015 Applications of Integration Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl