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Final Practice Problem 1

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175 views7 pages

Final Practice Problem 1

Practice problem

Uploaded by

Jay
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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7.3 It is desired to determine the wave height when wind blows across a lake. The wave height, H, is assumed to be a function of the wind speed, V, the water density, p, the air density, p,, the water depth, d, the distance from the shore, ¢, and the acceleration of gravity, g, as shown in Fig. P7.7. Use d, V, and p as repeating variables to determine a suitable set of pi terms that could be used to describe this problem. qa ~<~_—"v Hef (6 2d, L, 3) wal velT" paretr pirctr® del hel ger? From The pr Fheorem, 7-3 = 4 pi terms reguired. Use d, V, and Pp as repeating variables, Thus, t= Kasvee® am mayen) er)! (eet) e = FLT So That wun Che ®) f+ are-4e=0 Cr L) —b 420=0 Ger T) It filows that G2-), b=0, 670, and Therefore 1,°$ whith 1s obviously dimensionless. hr T° = Baty? ¢ zea [ectrya)tar rect) = Pur 4 1+#e%0 fer FD —t¢ath —4e =e +) CeLgee ae fr T) Lt follows thet a0, b=0, C=-| So That nme & which is obviously dimension less . For Ty: Pe Tez b dt vps and as fr 7, , 2~l, b=, C20 50 that yes Ror Ty: ie hap Tee 3 d*y> pe (erty Ger) = PUT C20 (for F) / tare — #050 (fr L) —2-68 r2ze20 (fr 7) tt fellows tet @= | be-2, C=O, and There fore - ad Ty = oa Check chimensions ! ~ ATM) 2 pore 4 ir re eR BEE] “Ok 7.10 A liquid spray nozzle is designed to produce a specific size droplet with diameter, d. The droplet size depends on the nozzle diameter, D, nozzle velocity, V, and the liquid properties p, p, - Using the common dimensionless terms found in Table 7.1, deter- mine the functional relationship for the dependent diameter ratio of dD. Given d=f(D,V,e,4,7) $0 that k=6 (there 6 variables) and r=3 (it takes MLT or FLT to describe them). Hence, k-r =6-3=3 which means that 2 pi terms are needed. T, = OCT, %), where Ti, 4 is clearly dimensienless., With the independent variables (i.e, D, V0, 4,7) if is clear that the Reynolds number canbe one of the M terms. Hence, set 7, = eVD/u. 7% must inclvde the surface tension, T, since st does not appear in Ty or 7. Based on the information in Table 7.1 it is seen thal the Weben number, We, can be the other 7 term, Hence, set 1 = eV ’D/r This, A -9( SP, eb) or d= (Re, We) 7.16 Glycerin at 20°C flows with a velocity of 4 mis through a 30-mm-diameter tube. A model of this system is to be developed using standard air as the model fluid. The air velocity is to be 2 ns. What tube diameter is required for the model if dynamic similarity is to be maintained between model and prototype? For dynamic similarity, the Reynolds number must be The same fer model and prototype. Thus, Vn Dm . YD Un Vv so that ote 46 x10 Ny 2, * F é : = NOP) vo som) eerie he ee (119x102) (2) 3 ©.736 x10 ~m = 0,736 mm 4 7.22 A large, rigid, rectangular billboard is supported by an elastic column as shown in Fig. P7.22. There is concern about the deflec- tion, 6, of the top of the structure during a high wind of velocity V. A wind tunnel test is to be conducted with a 1:15 scale model. Assume the pertinent column variables are its length and cross- sectional dimensions, and the modulus of elasticity of the material used for the column. The only important “wind” variables are the air density and velocity. (a) Determine the model design condi- tions and the prediction equation for the deflection. (b) If the same structural materials are used for the model and prototype, and the wind tunnel operates under standard atmospheric conditions, what is the required wind tunnel velocity to match an 80 km/hr wind? Front View Side View Assume S= F(L4 A, VE) where: S~deflecin*l, 2 column lenghsl, 4~oter lengths = L Geb ,-~ ete.) ,. Pa air density FL*T®, Vo wind velocthy 3 LT Ew modulus of elasherty =FL* From the pi theorem, (&4e)-3= 2+6 pr terms. reguired, and a dimensieial analysis ylelds : . 4 (fF) (a) The model design conditions are F S fn faNn » PY sett —-— and The pred’ rs is re or with a lengty scale of [115 Sz Sim (6) From the second

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