Respuesta :
Answer:
the volume flow rate per unit depth is:
[tex]\frac{Q}{b} = \frac{2}{3} u_{max} h[/tex]
the ratio is : [tex]\frac{V}{u_{max}}=\frac{2}{3}[/tex]
Explanation:
From the question; the equations of the velocities profile in the system are:
[tex]u = u_{max}(Ay^2+By+C)[/tex] ----- 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 = [tex]\frac{h}{2}[/tex] ; u = [tex]u_{max}[/tex] ------(c)
where ;
A,B and C are constant
h = distance between two plates
u = velocity
[tex]u_{max}[/tex] = maximum velocity
y = measured distance upward from the lower plate
Replacing the boundary condition in (a) into equation (1) ; we have:
[tex]u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*0+B*0+C) \\ \\ 0=u_{max}C \\ \\ C= 0[/tex]
Replacing the boundary condition (b) in equation (1); we have:
[tex]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)[/tex]
Replacing the boundary condition (c) in equation (1); we have:
[tex]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}[/tex]
replacing [tex]A = -\frac{4}{h^2}[/tex] for A in (d); we get:
[tex]B = - ( -\frac{4}{h^2})h[/tex][tex]B = \frac{4}{h}[/tex]
replacing the values of A, B and C into the velocity profile expression; we have:
[tex]u = u_{max}(Ay^2+By+C) \\ \\ u = u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y)[/tex]
To determine the volume flow rate; we have:
[tex]Q = AV \\ \\ Q= \int\limits^h_0 (u.bdy)[/tex]
Replacing [tex]u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y) \ for \ u[/tex]
[tex]\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})[/tex]
[tex]\frac{Q}{b} = u_{max}(\frac{2h}{3}) \\ \\ \frac{Q}{b} = \frac{2}{3} u_{max} h[/tex]
Thus; the volume flow rate per unit depth is:
[tex]\frac{Q}{b} = \frac{2}{3} u_{max} h[/tex]
Consider the discharge ;
Q = VA
where :
A = bh
Q = Vbh
[tex]\frac{Q}{b}= Vh[/tex]
Also; [tex]\frac{Q}{b} = \frac{2}{3} u_{max} h[/tex]
Then;
[tex]\frac{2}{3} u_{max} h = Vh \\ \\ \frac{V}{u_{max}}=\frac{2}{3}[/tex]
Thus; the ratio is : [tex]\frac{V}{u_{max}}=\frac{2}{3}[/tex]
