The ship in Fig. 8–4 is subjected to a drag FD on its hull, created by water passing under and around its surface. It is anticipated that this force is a function of the density ρ and viscosity μ of the water, and since waves are produced, their weight, defined by gravity g, is important. Also, the

Question

The ship in Fig. 8–4 is subjected to a drag [latex] F_D [/latex] on its hull, created by water passing under and around its surface. It is anticipated that this force is a function of the density ρ and viscosity μ of the water, and since waves are produced, their weight, defined by gravity g, is important. Also, the “characteristic length” of the ship, L, and the velocity of the flow, V, influence the magnitude of the drag. Show how the drag depends on all these variables.

Accepted Answer

Define the Physical Variables. Here the unknown function is [latex] F_D=f( ho, \mu, L, V, g) [/latex], which is then expressed as [latex] h\left(F_D, ho, \mu, L, V, g ight)=0 [/latex], where here n = 6. We will solve the problem using the F-L-T system in Table 8–1. The primary dimensions of the variables are

[latex] \begin{array}{ll} ext { Drag, } F_D & F \ ext { Density, } ho & F T^2 L^{-4} \ ext { Viscosity, } \mu & F T L^{-2} \ ext { Length, } L & L \ ext { Velocity, } V & L T^{-1} \ ext { Gravity, } g & L T^{-2} \end{array} [/latex]

Since the three primary dimensions (F, L, T) are involved in the collection of all these variables, m = 3, and so there will be (n - m) = (6 - 3) = 3 Π terms.

Select the Repeating Variables. Density, length, and velocity will be chosen as the repeating variables, since the collection of their dimensions contains the m = 3 primary dimensions.

[latex] \Pi_1 ext { Term } q=F_D [/latex] and Dimensional Analysis. We will consider [latex] F_D [/latex] as the q variable for the first Π term.

[latex] \begin{aligned} \Pi_1 & = ho^a L^b V^c F_D \ & =\left(F^a T^{2 a} L^{-4 a} ight)\left(L^b ight)\left(L^c T^{-c} ight) F=F^{a+1} L^{-4 a+b+c} T^{2 a-c} \end{aligned} [/latex]

Therefore,

For F: 0 = a + 1
For L: 0 = -4a + b + c
For T: 0 = 2a - c

Solving yields a = -1, b = -2, c = -2, so

[latex] \Pi_1= ho^{-1} L^{-2} V^{-2} F_D=\frac{F_D}{ ho L^2 V^2} [/latex]

[latex] \Pi_2 ext { Term } q=\mu [/latex] and Dimensional Analysis. Here we will consider μ as the q variable, thus creating the second Π term.

[latex] \begin{aligned} \Pi_2 & = ho^d L^e V^h \mu \ & =\left(F^d T^{2 d} L^{-4 d} ight)\left(L^e ight)\left(L^h T^{-h} ight) F T L^{-2}=F^{d+1} L^{-4 d+e+h-2} T^{2 d-h+1} \end{aligned} [/latex]

For F: 0 = d + 1
For L: 0 = -4d + e + h - 2
For T: 0 = 2d - h + 1

Solving yields d = -1, e = -1, h = -1, so

[latex] \Pi_2= ho^{-1} L^{-1} V^{-1} \mu=\frac{\mu}{ ho V L} [/latex]

As in the previous example, we can replace [latex] \Pi_2 ext { by } \Pi_2^{-1} [/latex] since it represents the Reynolds number, Re.

[latex] \Pi_3 ext { Term } q=g [/latex] and Dimensional Analysis. Finally, we consider g as the q variable for the third Π term.

[latex] \begin{aligned} \Pi_3 & = ho^i L^j V^k g \ & =\left(F^i T^{2 i} L^{-4 i} ight)\left(L^j ight)\left(L^k T^{-k} ight)\left(L T^{-2} ight)=F^i L^{-4 i+j+k+1} T^{2 i-k-2} \end{aligned} [/latex]

For F: 0 = i
For L: 0 = -4i + j + k + 1
For T: 0 = 2i - k - 2

Solving yields i = 0, j = 1, k = -2, so

[latex] \Pi_3= ho^0 L^1 V^{-2} g=g L / V^2 [/latex]

Recognizing this term to be equal to the inverse square of the Froude number, for simplicity we will consider using Fr instead of [latex] \Pi_3 [/latex] because both are dimensionless. Thus, the unknown function [latex] f_1 [/latex] between the Π terms takes the form

[latex] f_1\left(\frac{F_D}{ ho L^2 V^2}, \operatorname{Re}, Fr ight)=0 [/latex]

If we solve for [latex] F_D / ho L^2 V^2 [/latex] in this equation, then it can be written
symbolically as a function [latex] f_2 [/latex] of the Reynolds and Froude numbers.

[latex] \frac{F_D}{ ho L^2 V^2}=f_2[ Re , Fr ] [/latex] (1)

[latex] F_D= ho L^2 V^2 f_2[\operatorname{Re}, Fr ] [/latex]

Question 8.3: The ship in Fig. 8–4 is subjected to a drag FD on its hull, ...

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