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Unformatted text preview: 5.12 The Stokes number. St. used in particledynamics 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} = {hIL_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_‘MT_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 = aVEnhcg:{ }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 triangularsection 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} {Iv151T} {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 = {MrLEW 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 sealevel 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 sealevel 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 onefifteenthscale model of a parachute has a drag of 450 lbf when tested at
20 fts’s in a water tunnel. If Reynoldsnumber effects are negligible. estimate the terminal fall
velocity at 5000ft 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 5000ft standard altitude (Table
A6) take p = 0.00205 kg—"inB. If Reynolds number is unimportant. then the two cases have the
same dragforce coefficient: C _ Fm _ 450 _C _ 160
DE pmrlei 1.94(20;}1(DP.='15)2 DP 0.00205sﬁnﬁ'
solve VP =24.5 f—l Arts. 3 5.?4 A onetenthscale 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 kNm. If Reynoldsmunber effects are negligible. what will
the pitching moment of the prototype wing be ﬂying at the same Mach number at Skin 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 sealevel air. take ,0 = 1.225
kg‘in3 and sound speed a = 340 1n"s. For air at 8000111 standard altitude (Table Aﬁ). 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 8000111 standard altitude.
A onethirtiethscale 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 3000111 standard altitude [Table A6). take ,0 = 0.525 kgs’mi. Ju =
1.53E—5 leg"ms. and sound speed a = 308 mis. For helium at 20°C (Table A4). take gas
constant R = 207’? J!(kg°K). ,u = 1.97E—5 kgfms. 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(21113 = — = —.
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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 Spring '08
 Seshadri,K
 Aerodynamics, altitude (Table

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