Answered: The drag force FD acting on a ship…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

The drag force FD acting on a ship depends on the forward speed V , the fluid density ρ and viscosity µ, gravity, its length L and width B and the average roughness height k.

(a) Use dimensional analysis to express this information in terms of a functional dependence on nondimensional groups.

(b) If the ship moved from fresh water to salt water, where the density and viscosity are both increased by 10%, how would its drag force change at fixed speed?

Expert Solution

Step 1

Solution :

$D r a g f o r c e, F_{d} = M L T^{- 2} s p e e d, v = L T^{- 1} D e n s i t y, ρ = M L^{- 3} v i s \cos i t y, μ = M L^{- 1} T^{1} g r a v i t y, g = L T^{- 1} l e n g t h, l = L w i d t h o f s h i p, w = L r o u g h n e s s h e i g h t, k = L$

Step 2

There are 8 variables , so the number of dimensionless numbers are given as:

$l = n - m = 8 - 3 = 5$

Let the reapeatng variables are length, velocity and density.

Step 3

$π_{1} = L^{a} {({LT}^{- 1})}^{b} {({ML}^{- 3})}^{c} \times F_{d} M^{0} L^{0} T^{0} = L^{a} {({LT}^{- 1})}^{b} {({ML}^{- 3})}^{c} \times ({MLT}^{- 2}) equating M : 0 = c + 1 c = - 1 equating L : 0 = a + b - 3 c + 1 0 = a + b + 4 a + b = - 4 equating T : 0 = - b - 2 b = - 2 a = - 2$

$π_{1} = L^{- 2} {(V)}^{- 2} {(ρ)}^{- 1} \times F_{d} = \frac{F_{d}}{L^{2} V^{2} ρ}$

Step 4

$π_{2} = L^{a} {({LT}^{- 1})}^{b} {({ML}^{- 3})}^{c} \times μ M^{0} L^{0} T^{0} = L^{a} {({LT}^{- 1})}^{b} {({ML}^{- 3})}^{c} \times ({ML}^{- 1} T^{- 1}) equating M : 0 = c + 1 c = - 1 equating L : 0 = a + b - 3 c - 1 0 = a + b + 2 a + b = - 2 equating T : 0 = - b - 1 b = - 1 a = - 1$

$π_{2} = L^{- 1} {(V)}^{- 1} {(ρ)}^{- 1} \times μ = \frac{μ}{LVρ}$

Step 5

$π_{3} = L^{a} {({LT}^{- 1})}^{b} {({ML}^{- 3})}^{c} \times g M^{0} L^{0} T^{0} = L^{a} {({LT}^{- 1})}^{b} {({ML}^{- 3})}^{c} \times (L T^{- 2}) equating M : 0 = c c = 0 equating L : 0 = a + b + 1 a + b = - 1 equating T : 0 = - b - 2 b = - 2 a = 1$

$π_{3} = L^{1} {(V)}^{- 2} {(ρ)}^{0} \times g = \frac{g L}{V^{2}}$

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