Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat plate 10 ft wi

Question

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat plate 10 ft wide and 2 ft long parallel to the flow, immersed in 60F water (=1.938 slug/ft3, =1.217×10-5ft2/s =2.359×10-5 lbf.s/ft2) flowing at an undisturbed velocity of 3 ft/s. Assume laminar boundary layer over the whole plate. Also (f) find the total friction drag on one side of the plate.

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Yến Oanh 5 years 2021-07-31T06:58:21+00:00 1 Answers 74 views 0

Answers ( )

  1. Euphemia
    0
    2021-07-31T07:00:16+00:00
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    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 = 1.217×10⁻⁵ft²/s

    Absolute viscosity = 2.359×10⁻⁵lbfs/ft²

    a) The Reynold number is equal to:

    Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

    The boundary layer thickness is equal to:

    \delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

    The shear stress is equal to:

    \tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012

    b) If the railing edge is 2 ft, the Reynold number is:

    Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

    The boundary layer is equal to:

    \delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

    The sear stress is equal to:

    \tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082

    c) The drag coefficient is equal to:

    C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

    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

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