ME 309 Fluid Mechanics Fall 2006 Solutions to Exam3. (ME309_Fa2006_soln3 Solutions to Exam 3)

Transcription

1 Fll 6 Solutions to Exm3 (ME39_F6_soln3 Solutions to Exm 3)

2 Fll 6. ( pts totl) Unidirectionl Flow in Tringulr Duct (A Multiple-Choice Problem) We revisit n old friend, the duct with n equilterl-tringle cross section. The coordinte system is plced so tht the flow is in the x direction (coming out of the pge), one of the vertices of the tringle is t the origin of the yz plne, nd the ltitude (length ) is coincident with the y xis. z y Circle the letter (, b, c, or d) next to the correct nswer to the following question nd drw the typicl differentil segment (OK to drw this in the tringle in the figure bove) corresponding to tht nswer. Question: the expression for the flow rte Q in the duct is given by which of the following?. y/ 3 y / 3 dzdy b. / / dzdy c. y/ 3 y / 3 dydz d. None of the bove. The typicl differentil segment hs been drwn in the figure (the shded trpezoid).

3 Fll 6. (3 + = 4 pts totl) Drg on Prchute To estimte the drg on prchute, consider it s hemisphere of rdius R with the open end fcing the oncoming flow strem (drg coefficient C D =.4, bsed on the frontl projected re).. Clculte the terminl speed (in m/s) ssuming tht the prchute jump occurs on erth, the density is tht of stndrd ir, the prchute rdius R =.5 meters nd the combined mss M of the skydiver nd the prchute is kg. At terminl velocity (i.e. no ccelertion) force blnce on the skydiver-prchute combined body gives: W( = Mg) = FD From the definition of the drg coefficient, we hve F = C ρv π R. D D Solve for V nd then V, using dt for ρ ir from the formul sheet nd M, C D nd R s given bove. V Mg [kg] 9.8[m/s ] = = = 3 ρc πr.3[kg/m ].4 π.5 [ m ] D 7.56 [m/s] b. A novice skydiver would like to trin for the impct t lnding by jumping (without prchute) from tower of moderte height. How tll should this tower be to reproduce the pproprite impct velocity from prt ()? Check your nswer with common sense, i.e., would you be willing to jump without prchute from tht height? Initilly, objects flling from rest ccelerte under the ction of grvity with negligible drg (the V is too smll) so from elementry Newtonin mechnics/physics we hve: ( gt) V V = gt, H = gt = g = g. H V (7.56m) = = =.9 [m] g 9.8[m/s ] (This does not result in fether-soft lnding, but it is fesible height for jump without prchute).

4 Fll 6 3. (+++++ = 5 pts totl) The Momentum Integrl Eqution for Boundry Lyer Flow Consider the control volume (CV) for boundry lyer flow of n incompressible fluid s shown in the figure below. Your chllenge in this problem is to pply mss nd momentum blnce on the pproprite elements of the CV to derive expressions for the terms tht come together to form the momentum integrl eqution. U(x) in the free strem y/δ(x y b c δ(x) x d x+ u/u(x Throughout this problem, ssume unit width w = in the dimension coming out of the pge nd use liner pproximtion for the velocity profile in the boundry lyer, s shown in the figure on the right. u = U( x) η where η y/ δ ( x) =,. By pplying Newton s Lw of Viscosity on the ssumed velocity profile, determine the sher stress τ w cting on the fluid due to the surfce (the plte) nd give this expression s function of µ, ρ, U, du/, δ, nd dδ/ (or subset of these terms). (wll sher stress) u y U( x) τ w = µ = µ U( x) = µ y y δ( x) δ( x) b. Use the Bernoulli eqution in the free strem to relte the pressure grdient dp/ to µ, ρ, U, du/, δ, nd dδ/ (or subset of these terms). Applying Bernoulli in the free strem gives the reltion: dp du p+ ρu = constnt, nd tking the x-derivtive of tht reltion we get + ρu =. Therefore, the pressure grdient is: dp = ρu du. c. Determine the pressure force cting on the surfce b nd on the surfce cd s function of µ, ρ, U, du/, δ, nd dδ/ (or subset of these terms). The two expressions cnnot be expressed s function of the bove vribles; (digression: when the pressure forces re combined from surfces b, cd nd bc mny terms cncel nd the net pressure force from the three surfces my be expressed in terms of the required set of vribles). Going bck to the exm, for the two terms in question, the best we cn do is s written in the box below.

5 Fll 6 YOUR ANSWERS (two expressions): Pressure force on b: pδ w= pδ dp d du d Pressure force on cd: ( p )( δ δ + δ + ) w = ( p ρu )( δ + ) d. Apply mss blnce to the CV nd determine the expression for the mss flow rte crossing in through the surfce bc in terms of µ, ρ, U, du/, δ, nd dδ/ (or subset of these terms). Use m = ( m + m ) nd find expressions for the two mss flow rtes on the right side of the eqution. bc b cd m udy w b = δ δ y ρ = ρu dy = ρuδ dot product VdA< i (negtive) for flow in to CV. m cd δ =+ δ ρ udy w d = ρuδ ρuδ ρu x = + δ + x+ So combining these two mss flow rtes gives d du dδ m = ( m + m ) = ρuδ = ρ δ + U bc b cd m bc e. Give the expression for the x-momentum flux crossing in through the surfce bc, in terms of µ, ρ, U, du/, δ, nd dδ/ (or subset of these terms). Key physicl insight: this x-momentum flux equls Um bc becuse where the B.L. meets the free strem, the velocity is simply the free strem velocity U nd the momentum flux integrl simplifies s: uρv ida = U ρv ida = U ρv i da = Um. bc bc bc Inserting the previous result for the mss flow rte gives the required nswer. bc du = + dδ Um ρ U δ U bc