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3 years ago

Define Viscosity. What are the main differences between viscous and inviscid flows?

Engineering

1 answer:

Evgesh-ka [11]3 years ago

5 0

1. Define Viscosity

In physics, Viscosity refers to the level of resistance of a fluid to flow due to internal friction, in other words, viscosity is the result of the magnitude of internal friction in a fluid, as measured by the force per unit area resisting uniform flow. For example, the honey is a fluid with high viscosity while the water has low viscosity.

What are the main differences between viscous and inviscid flows?

Viscous flows are flows that has a thick, sticky consistency between solid and liquid, contain and conduct heat, does not have a rest frame mass density and whose motion at a fixed point always remains constant. Inviscid flows, on the other hand, are flows characterized for having zero viscosity (it does not have a thick, sticky consistency), for not containing or conducting heat, for the lack of steady flow and for having a rest frame mass density

Furthermore, viscous flows are much more common than inviscid flows, while this latter is often considered an idealized model since helium is the only fluid that can become inviscid.

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4 years ago

An incompressible fluid flows between two infinite stationary parallel plates. The velocity profile is given by u=umaxðAy2 + By+

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Answer:

the volume flow rate per unit depth is:

$\frac{Q}{b} = \frac{2}{3} u_{max} h$

the ratio is : $\frac{V}{u_{max}}=\frac{2}{3}$

Explanation:

From the question; the equations of the velocities profile in the system are:

$u = u_{max}(Ay^2+By+C)$ ----- equation (1)

The above boundary condition can now be written as :

At y= 0; u =0 ----- (a)

At y = h; u =0 -----(b)

At y = $\frac{h}{2}$ ; u = $u_{max}$ ------(c)

where ;

A,B and C are constant

h = distance between two plates

u = velocity

$u_{max}$ = maximum velocity

y = measured distance upward from the lower plate

Replacing the boundary condition in (a) into equation (1) ; we have:

$u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*0+B*0+C) \\ \\ 0=u_{max}C \\ \\ C= 0$

Replacing the boundary condition (b) in equation (1); we have:

$u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*h^2+B*h+C) \\ \\ 0 = Ah^2 +Bh + C \\ \\ 0 = Ah^2 +Bh + 0 \\ \\ Bh = - Ah^2 \\ \\ B = - Ah \ \ \ \ \ --- (d)$

Replacing the boundary condition (c) in equation (1); we have:

$u = u_{max}(Ay^2+By+C) \\ \\ u_{max}= u_{max}(A*(\frac{h^2}{2})+B*\frac{h}{2}+C) \\ \\ 1 = \frac{Ah^2}{4} +B \frac{h}{2} + 0 \\ \\ 1 = \frac{Ah^2}{4} + \frac{h}{2}(-Ah) \\ \\ 1= \frac{Ah^2}{4} - \frac{Ah^2}{2} \\ \\ 1 = \frac{Ah^2 - Ah^2}{4} \\ \\ A = -\frac{4}{h^2}$

replacing $A = -\frac{4}{h^2}$ for A in (d); we get:

$B = - ( -\frac{4}{h^2})h$ $B = \frac{4}{h}$

replacing the values of A, B and C into the velocity profile expression; we have:

$u = u_{max}(Ay^2+By+C) \\ \\ u = u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y)$

To determine the volume flow rate; we have:

$Q = AV \\ \\ Q= \int\limits^h_0 (u.bdy)$

Replacing $u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y) \ for \ u$

$\frac{Q}{b} = \int\limits^h_0 u_{max}(-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\ \frac{Q}{b} = u_{max} \int\limits^h_0 (-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\ \frac{Q}{b} = u_{max} (-\frac{-4}{h^2}\frac{y^3}{3} +\frac{4}{h}\frac{y^2}{y})^ ^ h}}__0 }} \\ \\ \frac{Q}{b} =u_{max} (-\frac{-4}{h^2}\frac{h^3}{3} +\frac{4}{h}\frac{h^2}{y})^ ^ h}}__0 }} \\ \\ \frac{Q}{b} = u_{max}(\frac{-4h}{3}+\frac{4h}2} ) \\ \\ \frac{Q}{b} = u_{max}(\frac{-8h+12h}{6}) \\ \\ \frac{Q}{b} =u_{max}(\frac{4h}{6})$

$\frac{Q}{b} = u_{max}(\frac{2h}{3}) \\ \\ \frac{Q}{b} = \frac{2}{3} u_{max} h$

Thus; the volume flow rate per unit depth is:

$\frac{Q}{b} = \frac{2}{3} u_{max} h$

Consider the discharge ;

Q = VA

where :

A = bh

Q = Vbh

$\frac{Q}{b}= Vh$

Also; $\frac{Q}{b} = \frac{2}{3} u_{max} h$

Then;

$\frac{2}{3} u_{max} h = Vh \\ \\ \frac{V}{u_{max}}=\frac{2}{3}$

Thus; the ratio is : $\frac{V}{u_{max}}=\frac{2}{3}$

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