@Dmitry - please check this link

https://en.wikipedia.org/wiki/Viscosity#/media/File:Laminar_shear_flow.svg

1) the area of the velocity profile is Int[y0,y1] u dy = uav*(y1-y0) that is volumetric flow rate in the section y1-y0.

2) assuming you have the law for the velocity u=u(y) you can get the derivative du/dy. This is only one component of the general tensor mu*Grad_s v. In the case of vanishing normal velocity component, you can write tau=mu*du/dy. Therefore, the stress depends on the velocity derivative and the viscosity of the fluid. Only if you have the Hagen-Poiseulle solution, you can deduce the viscosity from the velocity profile and the pressure loss along the streamwise direction.

In a boundary layer, from the Prandtl equations you see that the normal velocity component is not exactly zero but is of the order of the viscosity

@Filippo

Regarding your first point, it doesn't really satisfy the units for volumetric flow rate (m^3/s), it can be kinematic viscosity or momentum flux with units m^2/s

the volumetric flow rate, for example in a 2D channel, is Q=u_average*H*1 (1 is for the direction normal to the plane), where

u_average = 1/H Int [0,H] u dy

Measuring the effort of stress locally (tangential force in a very small area) to a distance where this graph has been obtained and in this site insurance decides that it is possible to obtain the viscocity on having divided the effort of local stress in the surface between the speed variation with regard to "y"

Tapan, you are right about the asymptotic limit in the BL, it requires the limit y1->+Inf and the normal velocity component is very small but not zero, depending on the viscosity. I am not sure that in academic courses all students are educated to the concept of asymptotic expansion in the BL theory. Sometimes I just observed students considering the BL as a region extending up to the finite value y1 that leads to U=0.99U_inf and considering the normal component vanishing everywhere as weel as the do not consider that the normal pressure gradient is small but not exactly zero. And that is a rude approximation. Also I often observed the introduction of a Reynolds number in a wrong way, since the theory requires a semi-infinite lenght of the plate and this latter cannot be used to define the Reynolds number.

I suspect we are talking about two different things...

I was talking about the measure of the volumetric flow rate at a station x1 as the integral of u(x1,y). This flow rate Q(x1) I was not talking at all equates the value Q(x2), the difference being in the contribution of v between x1 and x2. I simply addressed one face of the control volume. Thè normal velocity magnitude is estimated by the continuity equation and is strictly related to the viscosity of the fluid.

Regards

Tapan, I still don't know if we are focusing the same issue...as I am not considering at all any control volume...

The original question in this post was about the meaning of the area of the curve u(y) and I simply defined it as the quantity

Q(x) =lim y1->+Inf Int [0,y1] u(x,y) dy

that is nothing else that a measure of the general flow rate through one face at a position x that requires only the normal component of the velocity

Q = Int [S] n.v dS

Thus, I simply assume S as the (y1-y0) x 1 face extension in 2D and u(y) is the normal velocity at this face.

As you wrote, if you consider S as the surfaces of a global control volume (a face at x1 and a face at x2), the face x2-x1 at y=0, with a top lid (also extending at y1->+Inf) then Q(x2)-Q(x1) is not zero the difference must be balanced by the integral of the normal velocity (of course for a non-permeable wall).

What are you instead considering from the original question?? Please clarify your issue maybe we are using a different definition of flow rate...

Definitely agree. I was following the original question of Usman about the area underlying u(y) only along the positive semi-infinite plate. I am used (I can say we are used in Italy) to define for any arbitrary surface S (or line in 2d) the integral of the normal velocity component as volumetric flow rate through that face, according to its dimension. Thus, the sum of these volumetric flow rates over all the faces building a control volume must be zero (for divergence-free velocity).

Best regards

Thankyou for your answers, Can the same area be called as momentum diffusivity? which has the units of m^2/s

Usman,

I am not sure at all you can call the integral of u(y) as "momentum diffusivity".

Diffusion of momentum requires to be expresse in terms of the gradient of velocity.