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Q1. A viscous, incompressible fluid flows between the two infinite, vertical, parallel plates of the Fig. Determine, by use of the Navier-Stokes equations, an expression for the pressure gradient in the direction of flow. Express your answer in terms of the mean velocity. Assume that the flow is laminar, steady, and uniform. with the coordinate system shown u= 0, w=o and from the continuity equation ou = "=0. Thus, from the y-component of the Navier-Stokes equations (Eg. 6.1276), with gy = -2, d²u -35-pg + M dx² (1) 0 = Since the pressure is not a function of x, Eq. (1) can be written as d²= P dx² pe (where P = 2 +pg) and integrated to obtain P du = x + C₁ From symmetry du = of Eg. (2) yields The flowrate at x=0 =0 P4²³ + C₂ Since at x=±h, v=0 it follows that and therefore P مار it follows that V= 2h 30 That per P = f*dx = √² + (x²³h²³) dx = -² 2£³ 2μ h h Direction of flow Thus, with V (mean velocity ) given by the equation. 1 PR² 3μ = v= (x²-1²) unit width in the Z-direction can be expressed as C₁=0. Integration (₂= - 12/12 (4²) 38=-344 - pg ap ay h (2)

Q1. A viscous, incompressible fluid flows between the two infinite, vertical, parallel plates of the Fig. Determine, by use of the Navier-Stokes equations, an expression for the pressure gradient in the direction of flow. Express your answer in terms of the mean velocity. Assume that the flow is laminar, steady, and uniform. with the coordinate system shown u= 0, w=o and from the continuity equation ou = "=0. Thus, from the y-component of the Navier-Stokes equations (Eg. 6.1276), with gy = -2, d²u -35-pg + M dx² (1) 0 = Since the pressure is not a function of x, Eq. (1) can be written as d²= P dx² pe (where P = 2 +pg) and integrated to obtain P du = x + C₁ From symmetry du = of Eg. (2) yields The flowrate at x=0 =0 P4²³ + C₂ Since at x=±h, v=0 it follows that and therefore P مار it follows that V= 2h 30 That per P = f*dx = √² + (x²³h²³) dx = -² 2£³ 2μ h h Direction of flow Thus, with V (mean velocity ) given by the equation. 1 PR² 3μ = v= (x²-1²) unit width in the Z-direction can be expressed as C₁=0. Integration (₂= - 12/12 (4²) 38=-344 - pg ap ay h (2)

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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answer is given but can you please explain everything clearly and why P is not a function of x, and how they got what.

Thank you

Q1. A viscous, incompressible fluid flows between the two infinite, vertical, parallel plates of the Fig. Determine, by use of the Navier-Stokes equations, an expression for the pressure gradient in the direction of flow. Express your answer in terms of the mean velocity. Assume that the flow is laminar, steady, and uniform. with the coordinate system shown u= 0, w=o and from the continuity equation ou = "=0. Thus, from the y-component of the Navier-Stokes equations (Eg. 6.1276), with gy = -2, d²u -35-pg + M dx² (1) 0 = Since the pressure is not a function of x, Eq. (1) can be written as d²= P dx² pe (where P = 2 +pg) and integrated to obtain P du = x + C₁ From symmetry du = of Eg. (2) yields The flowrate at x=0 =0 P4²³ + C₂ Since at x=±h, v=0 it follows that and therefore P مار it follows that V= 2h 30 That per P = f*dx = √² + (x²³h²³) dx = -² 2£³ 2μ h h Direction of flow Thus, with V (mean velocity ) given by the equation. 1 PR² 3μ = v= (x²-1²) unit width in the Z-direction can be expressed as C₁=0. Integration (₂= - 12/12 (4²) 38=-344 - pg ap ay h (2)
Transcribed Image Text:Q1. A viscous, incompressible fluid flows between the two infinite, vertical, parallel plates of the Fig. Determine, by use of the Navier-Stokes equations, an expression for the pressure gradient in the direction of flow. Express your answer in terms of the mean velocity. Assume that the flow is laminar, steady, and uniform. with the coordinate system shown u= 0, w=o and from the continuity equation ou = "=0. Thus, from the y-component of the Navier-Stokes equations (Eg. 6.1276), with gy = -2, d²u -35-pg + M dx² (1) 0 = Since the pressure is not a function of x, Eq. (1) can be written as d²= P dx² pe (where P = 2 +pg) and integrated to obtain P du = x + C₁ From symmetry du = of Eg. (2) yields The flowrate at x=0 =0 P4²³ + C₂ Since at x=±h, v=0 it follows that and therefore P مار it follows that V= 2h 30 That per P = f*dx = √² + (x²³h²³) dx = -² 2£³ 2μ h h Direction of flow Thus, with V (mean velocity ) given by the equation. 1 PR² 3μ = v= (x²-1²) unit width in the Z-direction can be expressed as C₁=0. Integration (₂= - 12/12 (4²) 38=-344 - pg ap ay h (2)
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