Question

The drag of a sonar transducer is to be predicted, based on wind tunnel test data. The prototype, a 1-ft diameter sphere, is to be towed at 5 knots (i.e. 8.44 ft/s) in seawater at 40°F. The model is 6 in. in diameter and the drag of the model is measured to be 0.6 lbf for wind tunnel air temperature at 70°F 1) (5 pts) Use Buckingham Pi Theorem to prove that the drag coefficient Cd is related to the Reynolds number Rep in F i.e. Cp = f(Rep) 2) (7 pts) Calculate the speed to be used in the wind tunnel model testing, Vm in ft/s. 3) (5 pts) Calculate the drag of the prototype sonar transducer, in lbf. 4) (3 pts) If the sonar probe were to be used in high speed near the free surface of the seawater, can the wind tunnel test be used? Why? FCD). PV2D2 De = 1 ft Dm = 6 in. Vm Fm = 0.60 lbf Ve = 5 knots Water at 40°F Air

Answer 1

(1) To prove: Co=f(Red)- we know FFB.V, ,u) The dimensions of variables are listed below Variable Dimension in МСТ system. F MLT-2 an M2-3 LT-1 L MET-1 So we have s variables & 3 fundamental units. So, we will have 5-3 -2 dimensionless 't' terms. (ire.ñ, fñz). Let S.V&D be repeating variables: Let Ti= F povodo { 2,6,(= constants writing dimensional equation: [MPL"ro] = [Met-2] [ m(3]" (LT-1]" [2] comparing powers of M,TEL on both sides. (M) Now Ita as-1 - 2-b * (T) :- 1b2

0 1 - 3 tbc (1) OF 1 - 3(-) - 2 2. ... A = f gºv D ni F PU²D2 het i X2 = u gavo pa :: CMOLºgo] o C Meigal] [M(-3.2a CLT-iJbCl] (M) o= ita olar.al - 4 - 5 -6 (1) ..lb--| oc-1-3 at btc :.06-1-3(-1)-I + c : FE-1 u - T- 2 PVD T'2 = PVD -Re //

Now we can write : 1, -f (Az) (Am. ). ( for air @ 70 F : vm=1-642 x64 .. F =f(Reo) :lo f(Rep) (2) For prototype (P) jer seawater @ 40°F : V 1.664X1005 fels, Sp = 62.4 ebelft? Vp = 8.44 FEIS Dp = lft For model (m): FE?!). Dm=6in = o.sft Sm=0.0749 ebflft². Fm 0.6 lbf speed of air in wind tunne e um): 1 (12')m= (az! JP e out = ve}. ,, Vmom Vp op Ve Vm xoos 8.44 X 1 i, 1.642x10-t 1.664 X105 Vm=166.57 ftis " (Ans: lo

(31 Dray of prototype (Fp): (Film = (, JP :. Fm Sim V2 DE Fp Spv²p² :. 0.6 0.0799 x 166.572 x(0.512 Fr 62-4%8-44 (1) i Fp=5.133 ebf > Ang. ). (4) No. Near the water surface the ur face tension should also major role which is not consider in current play a model.

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