Answer:
[tex]\mathbf{F_D \approx 1.071 \ lbf}[/tex]
Explanation:
Given that:
The height of a triangular stabilizing fin on its stern is 1 ft tall
and it length is 2 ft long.
Temperature = 60 °F
The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.
From these information given; we can have a diagrammatic representation describing how the triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.
The diagram can be found in the attached file below.
If we recall ,we know that;
Kinematic viscosity v = [tex]1.2075 \times 10^{-5} \ ft^2/s[/tex]
the density of water ρ = 62.36 lb /ft³
[tex]Re_{max} = \dfrac{Ux}{v}[/tex]
[tex]Re_{max} = \dfrac{2.5 \ ft/s \times 2 \ ft }{1.2075 \times 10 ^{-5} \ ft^2/s}[/tex]
[tex]Re_{max} = 414078.6749[/tex]
[tex]Re_{max} = 4.14 \times 10^5[/tex] which is less than < 5.0 × 10⁵
Now; For laminar flow; the drag on the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:
[tex]dF_D = (\dfrac{0.664 \times \rho \times U^2 (2-x) dy}{\sqrt{Re_x}})^2[/tex]
where;
[tex](2-x) dy[/tex] = strip area
[tex]Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}[/tex]
Therefore;
[tex]dF_D = (\dfrac{0.664 \times 62.36 \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})[/tex]
[tex]dF_D = 1.136 \times(2-x)^{1/2} \ dy[/tex]
Let note that y = 0.5x from what we have in the diagram,
so , x = y/0.5
By applying the rule of integration on both sides, we have:
[tex]\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy[/tex]
[tex]\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-2y)^{1/2} \ dy[/tex]
Let U = (2-2y)
-2dy = du
dy = -du/2
[tex]F_D = \int\limits^0_2 \ 1.136 \times(U)^{1/2} \ \dfrac{du}{-2}[/tex]
[tex]F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du[/tex]
[tex]F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 } }{\frac{1}{2}+1}]^0__2[/tex]
[tex]F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} } ] ^0__2[/tex]
[tex]F_D = -0.568 [0 - \dfrac{2}{3}(2)^{\frac{3}{2} } ][/tex]
[tex]F_D = -0.568 [- \dfrac{2}{3} (2.828427125)} ][/tex]
[tex]F_D = 1.071031071 \ lbf[/tex]
[tex]\mathbf{F_D \approx 1.071 \ lbf}[/tex]
