Explanation Given information: The vortex shading frequency is f k , the free stream velocity is V , the fluid density is ρ , the fluid dynamic viscosity is μ , the diameter of the cylinder is D and the speed of the sound is c . Expression of function of shading frequency, f k = f ( V , ρ , μ , D , c ) ...... (I) Dimension of density in M L T unit system, ρ = [ M L − 3 ] ...... (II) Here, the dimension of mass is M and the dimension of the length is L . Dimension of the dynamic viscosity, μ = [ M L − 1 T − 1 ] ...... (III) Here, the dimension of time is T . Dimension of free stream velocity, V = [ L T − 1 ] ...... (IV) Dimension of diameter in M L T unit system, D = [ L ] ...... (V) Expression of vortex shading frequency, f k = ( θ t ) ...... (VI) Here, the angle is θ and the time is t . Dimension of time in M L T unit system, t = [ T ] Dimension of angle in M L T unit system, θ = 1 . Substitute T for t and 1 for θ in Equation ( VI). f k = ( 1 T ) = [ T − 1 ] ...... (VII) Dimension of speed of sound in M L T unit system, c = [ L T − 1 ] ...... (VIII) Here, the number of variable is 6 these are f k , V , ρ , μ , D and c and the number of dimension is 3 these are M , L and T . Expression of number of pi-terms, pi-terms = n − j ...... (VIII) Here, the number of variables is n and the number of dimensions is j . Substitute 6 for n and 3 for j in Equation (VIII). pi-terms = 6 − 3 = 3 Here, three pi-terms is present. Here, the basic variable is V , ρ and D . Expression of first pi-terms, ∏ 1 = f k V a ρ b D d ...... (IX) Here, the constants are a , b , and d . Expression of second pi-terms, ∏ 2 = μ V a ρ b D d ...... (X) Expression of third pi-terms, ∏ 3 = c V a ρ b D d ...... (XI) Dimension of first pi-term, ∏ 1 = [ M 0 L 0 T 0 ] Dimension of second pi-term, ∏ 2 = [ M 0 L 0 T 0 ] Dimension of third pi-term, ∏ 3 = [ M 0 L 0 T 0 ] Substitute [ M 0 L 0 T 0 ] for ∏ 1 , [ T − 1 ] for f k , [ L T − 1 ] for V , [ M L − 3 ] for ρ and [ L ] for D in Equation (IX). [ M 0 L 0 T 0 ] = [ T − 1 ] [ L T − 1 ] a [ M L − 3 ] b [ L ] d [ M 0 L 0 T 0 ] = [ M b L a − 3 b + d T − 1 − a ] ...... (XII) Compare the power of M on both sides of Equation (XII). b = 0 Compare the power of T on both sides of Equation (XII). 0 = − 1 − a a = − 1 Compare the power of L on both sides of Equation (XII) and substitute 0 for b and − 1 for a . 0 = − 1 − 3 ( 0 ) + d 0 = − 1 + d d = 1 Substitute − 1 for a , 0 for b and 1 for d in Equation (IX). ∏ 1 = f k V − 1 ρ 0 D 1 = f k D V Substitute [ M 0 L 0 T 0 ] for ∏ 2 , [ M L − 1 T − 1 ] for μ , [ L T − 1 ] for V , [ M L − 3 ] for ρ and [ L ] for D in Equation (X)

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Author: CENGEL

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Chapter 7, Problem 48P

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The dimensionless relationship.

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Chapter 7, Problem 47P

Chapter 7, Problem 49P

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EBK FLUID MECHANICS: FUNDAMENTALS AND A

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The dimensionless relationship.

Solution Summary: The author explains the dimensionless relationship between the vortex shading frequency, free stream velocity, fluid density, and fluid dynamic viscosity.

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The dimensionless relationship.

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