To show that the Froude number (Fr) is dimensionless using dimensional analysis, we need to check if all the terms in the formula have the same dimensions.
Let's consider the dimensions of each term:
1. Acceleration due to gravity (g): [L][T]^-2 (where [L] represents length and [T] represents time).
2. Hydraulic depth (D): [L] (length).
3. Velocity of flow (u): [L][T]^-1 (where [L] represents length and [T] represents time).
Now, let's calculate the dimensions of the Froude number (Fr) using the given formula:
Fr = (g * D) / u
Dimensions of Fr = ( [L][T]^-2 * [L] ) / ( [L][T]^-1 )
Dimensions of Fr = [L^2][T]^-2 * [L] * [T] / [L][T]
Dimensions of Fr = [L^3][T]^-3 * [T] / [L][T]
Dimensions of Fr = [L^3][T]^-3 * T / [L][T]
Now, since [T]^-3 * T = [T]^(-3+1) = [T]^-2 (using the rule [T]^a * [T]^b = [T]^(a+b)), we have:
Dimensions of Fr = [L^3][T]^-2 / [L][T]
Now, [L^3] / [L] = [L^(3-1)] = [L]^2 (using the rule [L]^a / [L]^b = [L]^(a-b)), so we have:
Dimensions of Fr = [L]^2[T]^-2 / [T]
Now, [T]^-2 / [T] = [T]^(-2-1) = [T]^-3 (using the rule [T]^a / [T]^b = [T]^(a-b)), so we have:
Dimensions of Fr = [L]^2[T]^-3
Since there are no dimensions of time in the final expression, the Froude number (Fr) is dimensionless, which confirms that the formula is consistent from a dimensional analysis perspective.