If (u(y),0,0) are the velocity components of an incompressible Newtonian fluid flow due to a pressure gradient in the X-direction, then u(y) is a. (a) linear function of y. (b) quadratic function of y. (c) cubic function of y. (d) constant.

vidamuhae 2022-11-21 Answered
Determining a component of velocity vector of a fluid flow explicitly
If (u(y),0,0) are the velocity components of an incompressible Newtonian fluid flow due to a pressure gradient in the X-direction, then u(y) is a
(a) linear function of y.
(b) quadratic function of y.
(c) cubic function of y.
(d) constant.
I started with Euler dynamical equation as follows:
D q D t = F 1 ρ ( p )
where q = ( u ( y ) , 0 , 0 ) and F is the external force, p being pressure gradient. Now as per question we have along X-direction that
u ( y ) t + u ( y ) u ( y ) x + 0. u ( y ) y + 0. u ( y ) z = 0 1 ρ p x u ( y ) t = 1 ρ p x u ( y ) = ( 1 ρ p x ) d t + C
C being arbitrary constant. But from here I am unable to conclude whether this is linear or quadratic or cubic or constant. Is my thinking correct here?
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Answers (1)

Lena Gomez
Answered 2022-11-22 Author has 14 answers
Step 1
Neglecting body force terms other than gravity which can be absorbed into the pressure field, the x-component of velocity satisfies
(*) u t + u u x + v u y + w u z = 1 ρ p x + μ ρ 2 u ,
where μ is the viscosity.
The flow is steady, fully developed (u depends only on y), and unidirectional. Consequently the partial derivatives of u with respect to t, x, and z vanish and v = w = 0. Thus, (*) reduces to
(**) d 2 u d y 2 = 1 μ p x = G μ
The pressure gradient G is independent of y. This follows from the y-component of the Navier-Stokes equations which reduces to
p y = 0
Step 2
Integrating both sides of (**) twice with respect to y we get the quadratic function
u ( y ) = G 2 μ y 2 + C 1 y + C 2 ,
where C 1 and C 2 are integration constants.
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