Explanation Given information: The viscosity and density of the fluid is constant. The distance between the plate is 2 h The flow of fluid is fully developed with a single velocity component w = w x the pressure gradient ∂ p ∂ x is 0 . The Figure-(1) show two plates having distance 2 h between them, the single velocity component w = w x is acting between the two plates and the fluid is falling under the influence of gravity. Figure-(1) Write the expression for the Navier Stokes equation in z direction ρ g − ∂ p ∂ z + μ ∂ 2 w ∂ x 2 + ∂ 2 w ∂ y 2 + ∂ 2 w ∂ z 2 = ρ d w d t …… (I) Here, the density of the fluid is ρ , the acceleration due to gravity is g , the pressure gradient in z direction is ∂ p ∂ z , the rate of change of velocity with respect to time is d w d t Since there is no pressure gradient in w direction so ∂ p ∂ z = 0 . Since there is no velocity change in y and z -direction so ∂ 2 w ∂ y 2 = 0 and ∂ 2 w ∂ z 2 = 0 . Since w is a function of x only so d w d t = 0 . Calculation: Substitute 0 for ∂ p ∂ z , 0 for ∂ 2 w ∂ y 2 and ∂ 2 w ∂ z 2 , 0 for ∂ w ∂ t in Equation (I). ρ g − 0 + μ ∂ 2 w ∂ x 2 + 0 + 0 = ρ 0 0 = ρ g + μ ∂ 2 w ∂ x 2 ∂ 2 w ∂ x 2 = − ρ g μ ... (II) Integrating equation (II) with respect to x . ∫ ∂ 2 w ∂ x 2 = ∫ − ρ g μ ∂ w ∂ x = − ρ g μ x + c 1 ... (III) Further Integrating Equation (III) with respect to x

8th Edition

ISBN: 9789385965494

Author: Frank White

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 4, Problem 4.37P

To determine

The velocity profile between the plates by solving the Navier stokes equation.

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Chapter 4, Problem 4.36P

Chapter 4, Problem 4.38P

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Chapter 4 Solutions

Fluid Mechanics, 8 Ed

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