Question

A small submarine has a triangular stabilizing fin on its stern. The fin is 1 ft tall and 2 ft long. The water temperature where it is traveling is 60°F. Determine the drag on the fin when the submarine is traveling at 2.5 ft/s.

Answer 1

`\mathbf{F_D \approx 1.071 \ lbf}`

Step-by-step 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 =
`1.2075 * 10^(-5) \ ft^2/s`

the density of water ρ = 62.36 lb /ft³

`Re_(max) = (Ux)/(v)`

`Re_(max) = (2.5 \ ft/s * 2 \ ft )/(1.2075 * 10 ^(-5) \ ft^2/s)`

`Re_(max) = 414078.6749`

`Re_(max) = 4.14 * 10^5` 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:

`dF_D = ((0.664 * \rho * U^2 (2-x) dy)/(√(Re_x)))^2`

where;

`(2-x) dy` = strip area

`Re_x = (2.5(2-x))/(1.2075 * 10 ^(-5))`

Therefore;

`dF_D = (\frac{0.664 * 62.36 * 2.5^2 (2-x) dy}{\sqrt{ (2.5(2-x))/(1.2075 * 10 ^(-5))}})`

`dF_D = 1.136 *(2-x)^(1/2) \ dy`

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:

`\int\limits \ dF_D = \int\limits^1_0 \ 1.136 *(2-(y)/(0.5))^(1/2) \ dy`

`\int\limits \ dF_D = \int\limits^1_0 \ 1.136 *(2-2y)^(1/2) \ dy`

Let U = (2-2y)

-2dy = du

dy = -du/2

`F_D = \int\limits^0_2 \ 1.136 *(U)^(1/2) \ (du)/(-2)`

`F_D = - (1.136)/(2) \int\limits^0_2 \ U^(1/2) \ du`

`F_D = -0.568 [ \frac{(1)/(2)U^{ (1)/(2)+1 } }{(1)/(2)+1}]^0__2`

`F_D = -0.568 [ (2)/(3)U^{(3)/(2) } ] ^0__2`

`F_D = -0.568 [0 - (2)/(3)(2)^{(3)/(2) } ]`

`F_D = -0.568 [- (2)/(3) (2.828427125)} ]`

`F_D = 1.071031071 \ lbf`

`\mathbf{F_D \approx 1.071 \ lbf}`