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Chapter 1: Problem 17

1.17 A formula to estimate the volume rate of flow, \(Q\), flowing over a dam of length, \(B\), is given by the equation \\[ Q=3.09 B H^{3 / 2} \\] where \(H\) is the depth of the water above the top of the dam (called the head). This formula gives \(Q\) in \(\mathrm{ft}^{3} / \mathrm{s}\) when \(B\) and \(H\) are in feet. Is the constant, \(3.09,\) dimensionless? Would this equation be valid if units other than feet and seconds were used?

Short Answer

Expert verified
No, the constant 3.09 is not dimensionless. It has the dimensions of square root of Length per Time (\([L]^{1/2}/[T]\)). And, no, the equation would not be valid if units other than feet and seconds were used. The constant value would need to change in accordance with the change in units to maintain the balance of equation.

Step by step solution

01

Identify dimensions of each variable

The first step for dimensional analysis is identifying the dimensionality of each variable in the equation. In our specific case, variables in the given equation are \(Q\) (volume rate of flow, dimensions: \([L^3/T]\)), \(B\) (length, dimensions: \([L]\)) and \(H\) (depth or head, dimensions: \([L]\)), where \(L\) for length and \(T\) for Time.
02

Check the dimensionality on both sides of the equation

The next step is to check if the units or dimensions hold on both sides of the equation. Using the dimensions identified, let's substitute the dimensions of the variables in the given equation. Left-hand side (LHS) = \(Q\) = \([L^3/T]\). Right-hand side (RHS) = \(3.09 * B * H^{3/2}\) = \(3.09 * [L] * [L]^{3/2}\) = \(3.09 * [L]^{5/2}\). As we can see, both sides of the equation do not have the same dimensionality, which indicates that the constant must carry some dimensions.
03

Determine the dimensions of the constant

We can determine the dimensions of the constant by dividing the dimensions of the LHS by the dimensions of the RHS disregarding the constant. The dimension of constant = dimension of LHS/dimension of RHS = \([L^3/T] / [L]^{5/2}\) = \([L]^{1/2}/[T]\). The constant therefore is not dimensionless as it carries the dimension of square root of Length per Time.
04

Evaluating the validity of the equation with different units

The equation would not be valid if units other than feet and seconds were used. This is because the constant, 3.09, was derived specifically considering these units. If we change the units of length and time, the numerical value of the constant would also need to change to compensate for the unit change to maintain the balance of the equation.

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Most popular questions from this chapter

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