Respuesta :
Answer:
a) The shear stress is 0.012
b) The shear stress is 0.0082
c) The total friction drag is 0.329 lbf
Explanation:
Given by the problem:
Length y plate = 2 ft
Width y plate = 10 ft
p = density = 1.938 slug/ft³
v = kinematic viscosity = 1.217x10⁻⁵ft²/s
Absolute viscosity = 2.359x10⁻⁵lbfs/ft²
a) The Reynold number is equal to:
[tex]Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar[/tex]
The boundary layer thickness is equal to:
[tex]\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098[/tex] ft
The shear stress is equal to:
[tex]\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012[/tex]
b) If the railing edge is 2 ft, the Reynold number is:
[tex]Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar[/tex]
The boundary layer is equal to:
[tex]\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft[/tex]
The sear stress is equal to:
[tex]\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082[/tex]
c) The drag coefficient is equal to:
[tex]C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019[/tex]
The friction drag is equal to:
[tex]F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf[/tex]
