Explanation Write the expression of function of static deflection. δ = f ( l , d , ρ , μ , V , E ) (I) Here the length of the wire is l , the diameter of the wire is d , the density of the fluid is ρ , the dynamic viscosity of fluid is μ , the velocity of fluid is V and the modulus of elasticity is E . Write the dimension of deflection in M L T unit system. δ = [ L ] (II) Here, the dimension of length is L . Write the dimension of length of wire in M L T unit system. l = [ L ] (III) Write the dimension of diameter of wire in M L T unit system. d = [ L ] (IV) Write the dimension of density of fluid in M L T unit system. ρ = [ M L − 3 ] (V) Here, the dimension of mass is M and the dimension of length is L . Write the dimension of velocity of fluid in M L T unit system. V = [ L T − 1 ] (VI) Here, the dimension of length is L and the dimension of time is T . Write the dimension of dynamic viscosity in M L T unit system. μ = [ M L − 1 T − 1 ] (VII) Write the expression of modulus of elasity. E = σ ε (VIII) Here the stress is σ and the strain is ε . Write the expression of stress. σ = F A (IX) Here, the force is F and the area is A . Write the expression of strain. ε = Δ l l (X) Here the change in length is Δ l and the original length is l . Write the dimension of force in M L T unit system. F = [ M L T − 2 ] (XI) Write the dimension of area in M L T unit system. A = [ L 2 ] (XII) Write the dimension of change in length in M L T unit system. Δ l = [ L ] (XIII) Substitute [ M L T − 2 ] for F and [ L 2 ] for A in Equation (IX). σ = [ M L T − 2 ] [ L 2 ] = [ M L − 1 T − 2 ] Substitute L for Δ l and L for l in Equation (X). ε = L L = 1 Substitute [ M L − 1 T − 2 ] for σ and 1 for ε in Equation (VIII). E = [ M L − 1 T − 2 ] 1 = [ M L − 1 T − 2 ] (XIV) Here, the number of variables is 7 these are δ , l , d , ρ , μ , V and E and the number of dimension is 3 these are M , L and T . Write the expression of pi-term. pi-term = n − r (XV) Here the number of variables is n and number of dimension is r . Substitute 7 for n and 3 for r in Equation (XV). pi-term = 7 − 3 = 4 Consider d , V and E is the independent variable. Write the expression of first pi-term. ∏ 1 = δ ( d ) a ( V ) b ( E ) c (XVI) Here, the static deflection is δ , the diameter is d , the velocity of fluid is V , the modulus of elasticity is E and the constants are a , b and c . Write the expression of second pi-term. ∏ 2 = l ( d ) a ( V ) b ( E ) c (XVII) Here, the length is l . Write the expression of third pi-term. ∏ 3 = ρ ( d ) a ( V ) b ( E ) c (XVIII) Here, the fluid density is ρ . Write the expression of fourth pi-term. ∏ 4 = μ ( d ) a ( V ) b ( E ) c (XIX) Here, the dynamic viscosity is μ . Write the dimension of first pi-term in M L T unit system. ∏ 1 = [ M 0 L 0 T 0 ] (XX) Write the dimension of second pi-term in M L T unit system. ∏ 2 = [ M 0 L 0 T 0 ] (XXI) Write the dimension of third pi-term in M L T unit system. ∏ 3 = [ M 0 L 0 T 0 ] (XXII) Write the dimension of fourth pi-term in M L T unit system. ∏ 4 = [ M 0 L 0 T 0 ] (XXIII) Substitute [ M 0 L 0 T 0 ] for ∏ 1 , [ L ] for δ , [ L ] for d , [ L T − 1 ] for V and [ M L − 1 T − 2 ] for E in Equation (XVI)

8th Edition

ISBN: 9781119547990

Author: HOCHSTEIN

Publisher: JOHN WILEY+SONS INC.

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Chapter 7.3, Problem 13P

To determine

The suitable set of pi terms.

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EBK MUNSON, YOUNG AND OKIISHI'S FUNDAME

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