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MAE 101A - Chapter 5 Sample Solutions

# MAE 101A - Chapter 5 Sample Solutions - 5.12 The Stokes...

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Unformatted text preview: 5.12 The Stokes number. St. used in particle-dynamics studies. is a dimensionless combination of ﬁve variables: acceleration of gravity 3. viscosity In. density p. particle velocity U. and particle diameter D. If St is proportional to p and inversely proportional to g. ﬁnd its form. (b) Show that St is actually the quotient of two more traditional dimensionless groups. Solution: (a) The relevant dimensions are = {LT—2}. {,u} = {h-IL_1T_1}. {p} = {LIL—3}. {U} = {LT—l}. and {D} = To have ,u in the numerator and g in the denominator. we need the combination: _ —1 a a: c_‘M-T_1_i£ai’£b c_ 000 {SQ-{HHS} {P} {U} {D} -{LTHL}1L3} If} {1} -MLT ' luU Solve for a = —1. i: =1. e = —2. or: St = 2 pgD U2.-"l:gD) _ Froude number pUD.-"';r Reynoids number A as. (a) This has the rrm'o form: S: = Ans. (b) 5.17 The pressure drop per unit length in a porous. rotating duct (Really! See Ref. 35:! depends upon average velocity V. density p. viscosity gr. duct height ii. \vall injection velocity mr. and rotation rate (1. Using [p.Vﬁ} as repeating variables. rewrite this relationship in dimensionless form. Solution: The relevant dimensions are = {ML—2T2}. = {LT—1}. {p} = {ML—3}. {a} = {ML—1T1}. {a} = {L}. {vw} = {LT—1}. and {o} = {r1}. wall :1! = 7 and j = 3:. we expect :1 —j = if = 4 pi groups: They are found. as speciﬁed. using V. ii} as repeating variables: M' 1—1 = aV-Enhcg:{ }a 1 a0 L L3 L [M {Eib {L}! 'ILETE } = MULOTO. wire a = —l. b = —2. 9 =1 —1 M L M" 1T2 = pﬂVbiicrfl = {—3}! '[—}b {Sci } =M0L0T0. wire n =1. i) =1. (3 =1 _L [1" _LT M L 1 n3 = paVbl‘tCQ = {L3 }“ {FF {L}r {F} = MOLUTU. solve a = 0. a = —1. 0 =1 M [L L' n = “Vbhcv.={ “-— b L‘{— =M°L°TU. solve a=0.b=—l.c=0 The ﬁnal dimensionless function then is given by: .5 Vi C13? »' . 1T1 =fcn(l_lq_.l_l3.lT4}. or: E jq =fcn '0 j.—. l—“ Aria. ' I. ﬁll“ ,u V V 5.18 Under laminar conditions. the volume ﬂow Q through a small triangular-section pore of side length I: and length L is a function of viscosity ,u. pressure drop per unit length Ap-"L. and b. Using the pi theorem. rewrite this relation in dimensionless form. How does the volume flow change if the pore size I) is doubled:J Solution: Establish the variables and their dimensions: Q =fcn(.ﬂ.p.—"L . In . b} {List} {M.-’L2T2} {Iv-151T} {L} Then I? = 4 andj = 3. hence we expect 11 —j = 4 — 3 = 1 Pi group. found as follows: H. = (Ap.-=Lr<.u)b<brql = {Mr-LEW mm" {L}: {LW = M°L°T° Mra+b=01 L.‘ —2a—b+c+3 =0: Ts—Za—b— 1 =0. solve a=—l.b=+1.c=—4 QF 11 =— 1 (Ap.-l]b4 = constant Ans. lClearly. if b is doubled. the ﬂow rate Q increases by a factor of E. Ana. 5.6? A student needs to measlu‘e the drag on a prototype of characteristic length tip moving at velocity Up in air at sea-level conditions. He constructs a model of characteristic length #131. such that the ratio dps’dm = a factor He then measures the model drag under dynamically similar conditions. in sea-level air. The student claims that the drag force on the prototype will be identical to that of the model. Is this claim correct? Explain. Solution: Assuming no compressibility effects. dynamic similarity requires that —memdm =—'0P rpd‘”. it’hence Ur?” =d—P=f ,um yup DP film Rem = Rap. 0?: Run the tunnel at '3‘“ times the prototype speed. then drag coeﬁcients match: 2 1 F F F U 3' _ + =—Ez. or. —’" ={ I“ m] = = 1 Yes, drags are the some! pmbriidrii PF :39}: FF Dada f 5.72 A one-fifteenth-scale model of a parachute has a drag of 450 lbf when tested at 20 fts’s in a water tunnel. If Reynolds-number effects are negligible. estimate the terminal fall velocity at 5000-ft standard altitude of a parachutist using the prototype if chute and chutist together weigh 160 lbf. Neglect the drag coefficient of the woman. Solution: For water at 20°C. take ,0 = 1.94 leg-"1113. For air at 5000-ft standard altitude (Table A-6) take p = 0.00205 kg—"inB. If Reynolds number is unimportant. then the two cases have the same drag-force coefficient: C _ Fm _ 450 _C _ 160 DE pm-rlei 1.94(20;}1(DP.='15)2 DP 0.00205sﬁnﬁ' solve VP =24.5 f—l Arts. 3 5.?4 A one-tenth-scale model of a supersonic wing tested at 3'00 1115's in air at 20°C and 1 atm shows a pitching moment of 0.25 kN-m. If Reynolds-munber effects are negligible. what will the pitching moment of the prototype wing be ﬂying at the same Mach number at S-kin standard altitude? Solution: If Reynolds number is unimportant. then the dimensimlless moment coefficient MKpVng} must be a function only of the Mach number. Ma = V-‘a. For sea-level air. take ,0 = 1.225 kg-‘in3 and sound speed a = 340 1n-"s. For air at 8000-111 standard altitude (Table A-ﬁ). take ,0: 0.525 kg-‘in3 and sound speed a = 308 ill-"s. Then .r 7 " Mam — Km — ﬂ — 2.06 — Map ——P. solye for VP s..- 634 E am 340 303 a. VET} 0.525” 634 1 10 3 Then L—1P=Mm[%]=025[ j [_] ...33m.v.ml Am pmsmim 1.2.5 .00 1 5.81 An airplane. of overall length 55 ft. is designed to ﬂy at 680 ill-"s at 8000-111 standard altitude. A one-thirtieth-scale model is to be tested in a pressurized helium xviin tunnel at 20°C . 1What is the appropriate tunnel pressure in arm? Even at this {high} pressure. exact dynamic similarity is not achieved. Why"? Solution: For air at 3000-111 standard altitude [Table A-6). take ,0 = 0.525 kgs’mi. Ju = 1.53E—5 leg-"ms. and sound speed a = 308 mis. For helium at 20°C (Table A-4). take gas constant R = 207’? J!(kg-°K). ,u = 1.97E—5 kgfm-s. and a = 1005 ms's. For similarity at this supersonic speed. we must match both the Mach and Reynolds numbers. First convert Lp = 55 ft=16.8 in. Then 680 V 111 Ma =—=2.21=Ma = m . solve for V 22219 — P 303 m 1005 “d” r. r .525 . . Rep 2 ps L ‘p: 0 (680)(16 3) : 39mg : Rem : pHe(2219}(16 3 30} Jr: 1.53E—5 1.97E—5 p pHe Solve for 6.211(2-1113 = — = —. pH“ 9: “ RT (20770093) or pHe :5 3.13 MPa = 31.3 atm Ans. Even with Ma and Re matched. true dynamic similarity is not achieved. because the specific heat ratio of helium. k a: 1.66. is not equal to kajr 9: 1.40. ...
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MAE 101A - Chapter 5 Sample Solutions - 5.12 The Stokes...

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