p0218

# p0218 - ^ 1 ρ dp dx ± = L 3 M M L 2 T 2 = L T 2 the...

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2.18. CHAPTER 2, PROBLEM 18 127 2.18 Chapter 2, Problem 18 2.18(a): The dimensions of Q and D are [ Q ]= L 3 T , [ D ]= L Also, the dimensions of dp/dx are ^ dp dx ± = FL 2 L = MLT 2 L 3 = M L 2 T 2 So, the dimensions of the coefficient 61.9 follow from substituting dimensions into the Hazen-Williams formula, viz., [61 . 9] = ^ Q D 2 . 63 ( dp/dx ) 0 . 54 ± = L 3 T 1 L 2 . 63 L 1 . 08 T 1 . 08 M 0 . 54 = L 1 . 45 T 0 . 08 M 0 . 54 Therefore, the dimensions of the constant 61.9 in the Hazen-Williams formula are [61 . 9] = M 0 . 54 L 1 . 54 T 0 . 08 2.18(b): Now, if we rewrite the formula in terms of a new coefficient C =61 . 9 ρ 0 . 54 ,the equation becomes Q = CD 2 . 63 X 1 ρ dp dx ~ 0 . 54 Hence, noting that
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Unformatted text preview: ^ 1 ρ dp dx ± = L 3 M M L 2 T 2 = L T 2 the dimensions of C become [ C ] = Q D 2 . 63 p 1 ρ dp dx Q . 54 = L 3 T 1 L 2 . 63 T 1 . 08 L . 54 = L − . 17 T . 08 Thus, since these powers of L and T are so small, C is very nearly a dimensionless coefficient. It would be dimensionless, for example, if the exponents 2.63 and 0.54 were replaced by 5/2 and 1/2, respectively....
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