chpt_5_suggested_questions

chpt_5_suggested_questions - Chapter 5 Dimensional Analysis...

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Chapter 5 Dimensional Analysis and Similarity 5.1 Solution: For kerosene at 20 ° C, take ρ = 804 kg/m 3 and μ = 0.00192 kg/m s. The only unknown in the transition Reynolds number is the fluid velocity: tr tr Vd (804)V(0.05) Re 2300 , solve for V 0.11 m/s 0.00192 ≈== = 3 2 m Then Q VA (0.11) (0.05) 2.16E 4 3600 4s Ans. π == = × m 0.78 hr 3 5.4 Solution: For water at 20 ° C take 998 kg/m 3 and 0.001 kg/m s. For sea-level standard air take 1.2255 kg/m 3 and 1.78E 5 kg/m s. The balloon velocity follows from dynamic similarity , which requires identical Reynolds numbers: balloon model model proto 1.2255V (1.5) VD 998(2.0)(0.08) Re 1.6E5 Re 0.001 1.78E 5 = = = | ρ or Vballoon 1.55 m/s . Ans . Then the two spheres will have identical drag coefficients: balloon D,model D,proto 22 2 2 2 2 F F5 N C 0.196 C V D 998(2.0) (0.08) 1.2255(1.55) (1.5) = = = Solve for . Ans F 1.3 N balloon 5.8 Solution: The relevant dimensions are { g } = {LT 2 }, { } = {ML 1 T 1 }, { } = {ML 3 }, and { Y } = {MT 2 }. To have g in the numerator, we need the combination: 00 0 23 2 { } {} {}{}{Y } abc LM M M ab c M og M L T LT TL T μρ ⎧⎫ = ⎨⎬ ⎩⎭
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Solve for 4, 1, 3, or: . a b c Ans == = g Mo μ ρ 3 Y = 4 5.10 Solution: Note that { u/ x} {U/L}, { p dA} {pA}, = = etc. The results are: (a) (b) (c) (d) . Ans ⎧⎫ ⎨⎬ ⎩⎩ ⎭⎭ 22 2 32 2 MM LMM L ;; ; LT T T P5.11 Solution : The proposed function is R = f ( E, , t ). There are four variables ( n = 4) and three primary dimensions (MLT, or j = 3), thus we expect n-j = 4-3 = 1 pi group. List the dimensions: T} { } { ; } M/L { } { ; } T / ML { } { ; } L { } { 3 2 2 = = = = t E R Assume arbitrary exponents and make the group dimensionless: 5 2 ; 5 1 ; 5 1 Solve ; 0 2 ; 0 3 2 1 ; 0 whence , T L M T) ( ) M/L ( ) T / ML ( L) ( 0 0 0 c 3 2 2 1 1 = + = = = + = + = + = = c b a c a b a b a t E R b a c b a The single pi group is 1/5 2/5 1 1/5 2/5 constant, thus . wave R R tA Et Π= = n s
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5.12 Solution: (a) The relevant dimensions are { g } = {LT 2 }, { μ = {ML 1 T 1 ρ = 3 U } = {LT 1 }, and { D } = {L}. To have in the numerator and g in the denominator, we need the combination: 2 0 0 = 10 3 {} {} {}{}{}{} {} abc c ab MT M L St g U D L M L T LT L T L μρ ⎧⎫ == ⎨⎬ ⎩⎩ ⎭⎭ 2 Solve for 1, 1, 2, or: (a) U a b c St Ans. gD =− = = 2 /( ) This has the ratio form: (b) / Ug D Froude number Ans. UD Reynolds number ρμ St 5.13 Solution: The “function” of , λ , and Y must have velocity units. Thus ac b 32 LM M {C} {f( , ,Y)}, or C const Y , or: {L} T LT ρλ ⎫⎧⎫ ⎧ ⎫ = ⎬⎨⎬ ⎨ ⎬ ⎩⎭⎩ ⎭ ⎩ ⎭ Solve for a b 1/2 and c 1/2, or: Ans. = + Y Cc o n s t = Thus, for constant and , if Y is doubled, C increases as 2, or . Ans. + 41% 5.15 Soln: {LT 1 }, { } = {ML 3 }, and {d p /d x } = {ML 2 T 2 }. With n = 6 and j = 3, we expect n j = k = 3 pi groups: 00 0 1 , 1 ,
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chpt_5_suggested_questions - Chapter 5 Dimensional Analysis...

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