Question

human-powered torpedo is being designed for purpose of this competitions. The next year engineering students modeling is to determine the skin friction drag of the torpedo which is acting on L-4 carea (perimeter times the length). The overall length of the prototype torpedo 4.5 m and the diameter D m. Prototype Specification requires the torpedo to travel fully in freshwater water (a lake) at T 20 C with velocity V-2.50m/s. The one-fourth el is used for testing in their universitys wind tunnel (see the figure). A shield the drag balance strut so that the acrodynamic drag of the strut itself does not influence the to ured drag. The air in the wind tunnel is at 20°C and at one standard atmosphere pressure 0.6 (101.325 kPa) showed that the model skin friction drag force = 4.4 N. Fill the missing values in the table between the model and prototype based on the one fourth seale Show all equations, computations and corresponding units. Scale M:P 1:4 Prototype Fluid Water @ 20C Air@20c 118 Density, p (kg/m) 998 Strut Specific weight, Y (N/m)- H(kg/(m.s) 115.758 0.000018 9790 0.001 Shield 1.002E-06 1.52542E-06 Drag balance 4.5 0.6 16.96464 Length Diameter Skin Surface Area Hydrodynamic Parameters Velocity (m/s) 2.5 4.4 Drag force (N) Nondimensional Numbers Reynolds Number - Re- Skin Drag Coefficient C

Answer 1

ans: F_D=C_D*$\rho$*A*V²/2 ............. (1)

Given scale =1:4

velocity of model = 2.5/4 =0.625 m/s

length of model = 4.5/4 = 1.125 m

Diameter of model = 0.6/4 = 0.15 m

skin surface area = 16.96464/4 = 4.24116 m

F_D of prototype = 381.066 kN ...... (from (1))

Re =( $\small \rho$*V*D) / $\small \mu$ => Re for prototype = 1497000

Re for model = 61458

C_D of prototype = 22.026 ............ from (1)

C_D of model = 5.506