A circular section cylinder 0.2 m in diameter is held so that its circular end f

Question

A circular section cylinder 0.2 m in diameter is held so that its circular end faces are normal to a uniform stream of water flowing at 6m/s. Under certain conditions the average values of C_P are +0.7 on the upstream circular face and -0.1 on the down-stream face. If a turbulent boundary layer is assumed to grow on -: the cylindrical surface under zero pressure gradient conditions, calculate the cylinder length to diameter ratio for which the pressure drag will equal the skin friction drag. Assume viscosity mu_0 = 0.002 kg/m-s and density rho_0 = 1000kg/m^3.

Explanation / Answer

solution:

1)here for given flow pressure force is equal to shear force ,hence pressure force is given by

Fp=-(dp/dx)(L.dA)

where shear force is

Fs=t.dA=-mu*(du/dr)*dA

on equating both we get that

Fp=Fs

-(dp/dx)(L.dA)=-mu*(du/dr)*dA

where du/dr=(r/2mu)(dp/dx)

on putting we get that

L=r/2

r=D/2

so we get LD ratio as

L/D=1/4

hence L=.2/4=.05 m

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