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babunello [35]
3 years ago
5

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

CÞ, where A, B, and C are constants and y is measured upward from the lower plate. The total gap width is h units. Use appropriate boundary conditions to express the magnitude and units of the constants in terms of h. Develop an expression for volume flow rate per unit depth and evaluate the ratio V=umax
Engineering
1 answer:
nexus9112 [7]3 years ago
5 0

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})hB = \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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