Under certain conditions, wind blowing past a rectangular speed limit Sign can cause the sing to oscillate with a frequency omega. Assume that omega is a function of the Sign width, b, Sign height, h, wind velocity, V, air density p, and an elastic constant, K, for the supporting pole. Hint: The constant, k, has dimensions of [force x length]. (i) How many (non-dimensional) Pi-groups are there? (ii) Find these non-dimensional groups. (iii) Can one of the pi-groups be considered a Reynolds number?

Explanation:

Given that,

Omega is a function of the following

ω = f(b, h, v, ρ, k)

Where, all unit have a dimension of

ω = T^-1

b = L

h = L

V = LT^-1

ρ = FL^-4T²

k = FL

Then,

From the pie theorem

The required pi term is 6—3 = 3 terms,

So, we use V, p and b as a repeating term.

For first pi

π1 = ω•b^a•v^b•ρ^c.

Since

ω= T^-1, b = L, v = LT^-1 and

ρ= FL^-4T²

Since π is dimensionless then,

π = F^0•L^0•T^0

(T^-1)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0

Rearranging

F^0•L^0•T^0 = F^c•T^(2c-b-1)• L^(a+b-4c)

Comparing coefficient

c = 0

2c - b - 1 = 0

b = 2c - 1 = 0 - 1 = -1

a + b - 4c = 0

a = 4c - b = 0 - -1 = 0+1

a = 1

Then, a = 1, b = -1 and c = 0

So, π1 = ω•b^a•v^b•ρ^c.

π1 = ω•b^1•v^-1•ρ^0

π1 = ω•b / v

Check dimensions

ωb/v = (T^-1)L / LT^-1 = L^0•T^0 = 1

Then, π1 is dimensionless

For second pi

π2 = h•b^a•v^b•ρ^c.

Since

h = L, b = L, v = LT^-1 and ρ= FL^-4T²

Since π is dimensionless then,

π = F^0•L^0•T^0

(L)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0

Rearranging

F^0•L^0•T^0 = F^c•T^(2c-b)• L^(1+a+b-4c)

Comparing coefficient

c = 0

2c - b = 0

b = 2c = 0

1 + a + b - 4c = 0

a = 4c - b - 1 = 0 -0 - 1  = -1

a = -1

Then, a = -1, b = 0 and c = 0

So, π2 = h•b^a•v^b•ρ^c.

π2 = h•b^-1•v^0•ρ^0

π2 = h / b

Check dimensions

h / b = L / L = 1

Then, π2 is dimensionless

For third pi

π3 = k•b^a•v^b•ρ^c.

Since

k= FL, b = L, v = LT^-1 and ρ=FL^-4T²

Since π is dimensionless then,

π = F^0•L^0•T^0

(FL)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0

Rearranging

F^0•L^0•T^0 = F^(c+1)•T^(2c-b)• L^(1+a+b-4c)

Comparing coefficient

c + 1= 0

Then, c = -1

2c - b = 0

b = 2c = -2

1 + a + b - 4c = 0

a = 4c - b - 1 = -4 +2 - 1  = -3

a = -3

Then, a = -3, b = -2 and c = -1

So, π3 = k•b^a•v^b•ρ^c.

π3 = k•b^-3•v^-2•ρ^-1

Therefore,

π3 = k / b³•v²•ρ

Let check for dimension

π3 = FL / (L³• L²T^-2 • FL^-4T²)

π3 = FL / (L^(3+2-4) • T^(-2+2) •F)

π3 = FL / (L• T^(0) •F)

π3 = FL / LF = 1

π3 is also dimensions less

So.

I. There are three none dimensional pi

II. The none dimensional group are

π1 = ω•b / v

π2 = h / b

π3 = k / b³•v²•ρ

III. Reynolds Number. The Reynolds number is the ratio of inertial forces to viscous forces and it is dimensionless

So, the π3 can be considered as a Reynolds number

energy can cross the boundaries of a closed system in the form of heat and mass

explanation:

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