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:
![Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar](https://tex.z-dn.net/?f=Re%3D%5Cfrac%7B1%2A3%7D%7B1.217x10%5E%7B-5%7D%20%7D%20%3D246507%2C%20laminar)
The boundary layer thickness is equal to:
ft
The shear stress is equal to:
![\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012](https://tex.z-dn.net/?f=%5Ctau%3D0.332%28%5Cfrac%7B2.359x10%5E%7B-5%7D%2A3%20%7D%7B1%7D%20%20%29%28246507%29%5E%7B0.5%7D%20%3D0.012)
b) If the railing edge is 2 ft, the Reynold number is:
![Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar](https://tex.z-dn.net/?f=Re%3D%5Cfrac%7B2%2A3%7D%7B1.215x10%5E%7B-5%7D%20%7D%20%3D493015.6%2Claminar)
The boundary layer is equal to:
![\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft](https://tex.z-dn.net/?f=%5Cdelta%3D%5Cfrac%7B4.91%2A2%7D%7B493015.6%5E%7B0.5%7D%20%7D%20%3D0.000019ft)
The sear stress is equal to:
![\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082](https://tex.z-dn.net/?f=%5Ctau%3D0.332%28%5Cfrac%7B2.359x10%5E%7B-5%7D%2A3%20%7D%7B2%7D%20%20%29%28493015.6%5E%7B0.5%7D%20%29%3D0.0082)
c) The drag coefficient is equal to:
![C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B1.328%7D%7B%5Csqrt%7BRe%7D%20%7D%20%3D%5Cfrac%7B1.328%7D%7B%5Csqrt%7B493015.6%7D%20%7D%20%3D%3D0.0019)
The friction drag is equal to:
![F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf](https://tex.z-dn.net/?f=F%3DCp%5Cfrac%7Bv%5E%7B2%7D%20%7D%7B2%7D%20wL%3D0.0019%2A1.938%2A%28%5Cfrac%7B3%5E%7B2%7D%20%7D%7B2%7D%20%29%2810%2A2%29%3D0.329lbf)