Rotierender Zylinder < mehrere Veränderl. < reell < Analysis < Hochschule < Mathe < Vorhilfe
|
| Aufgabe | An open top can in the shape of a circular cylinder of equal radius and height, $h = r = a$, is half full of water of constant density [mm] $\rho$.
[/mm]
(i) It is made to rotate steadily with velocity $u = [mm] R\omega \boldsymbol{\hat \theta}$, [/mm] where $(R, [mm] \theta, [/mm] z)$ are cylindrical polar coordinates based on the cylinder axis and [mm] $\omega$ [/mm] is constant. Find the vorticity, and hence write down the R and z components of the steady rotational Bernoulli equation
[mm] $$\rho(\mathbf{u} \times \zeta) [/mm] = [mm] \nalba(p [/mm] + [mm] \frac{1}{2}\mathbf{u}^2 [/mm] + [mm] \rho [/mm] gz)$$
If the pressure on the free top surface is constant, find the equation of the top surface. Show that if [mm] $\omega^2 [/mm] = 2g/a$ the water will reach the top of the can at $R = a$.
(ii) If the flow is now [mm] $\mathbf{u} [/mm] = [mm] \frac{k}{R} \boldsymbol{\hat \theta}$, [/mm] as in the line-vortex flow, show that if the water reaches z = a at R = a there is no water in the region r < b, where b is given by
[mm] $$b=\frac{ka}{\sqrt{k^2 + 2ga^3}}$$
[/mm]
and k satisfies
[mm] $$\log \frac {k^2+2ga^3}{k^2} [/mm] = [mm] \frac{ga^3}{k^2}$$ [/mm] |
ganz ehrlich - ich verstehe absolut nicht, wie ich bei der frage 6b vorgehen soll... wäre daher über ansätze / tipps SEHR dankbar 
frage 6a habe ich denke ich richtig beantwortet, hier nur der vollständigkeit halber:
Vorticity [mm] $\boldsymbol{\zeta}$
[/mm]
[mm] $\mathbf{u} [/mm] = [mm] R\omega\boldsymbol{\hat \theta} [/mm] = [mm] -\omega [/mm] y [mm] \mathbf{i} [/mm] + [mm] \omega [/mm] x [mm] \mathbf{j} [/mm] $
[mm] $\boldsymbol{\zeta} [/mm] = [mm] \nabla \times \mathbf{u} [/mm] = [mm] \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -\omega y & \omega x & 0 \end{vmatrix} [/mm] = [mm] (\omega [/mm] + [mm] \omega) \mathbf{k} [/mm] = [mm] 2\omega \mathbf{k} [/mm] = 2 [mm] \omega \mathbf{\hat z} [/mm] $
Steady Rotational Bernoulli Equation:
[mm] $$\rho(\mathbf{u} \times \boldsymbol{\zeta}) [/mm] = [mm] \nabla [/mm] (p + [mm] \frac{1}{2} \rho \mathbf{u}^2 [/mm] + [mm] \rho [/mm] gz)$$
[mm] $$\mathbf{u} \times \boldsymbol{\zeta} [/mm] = 2 [mm] \omega^2 [/mm] R [mm] \mathbf{\hat R} [/mm] = [mm] \nabla (\omega^2 R^2)$$
[/mm]
[mm] $\Rightarrow \rho(\mathbf{u} \times \boldsymbol{\zeta}) [/mm] = [mm] \rho \nabla (\frac{p}{\rho} [/mm] + [mm] \frac{1}{2} \mathbf{u}^2 [/mm] + gz) [mm] $\\
[/mm]
[mm] $\Rightarrow \nabla (\omega^2 R^2) [/mm] = [mm] \nabla (\frac{p}{\rho} [/mm] + [mm] \frac{1}{2} \mathbf{u}^2 [/mm] + gz) [mm] $\\
[/mm]
[mm] $\Rightarrow \nabla (\frac{p}{\rho} [/mm] - [mm] \frac{1}{2} \mathbf{u}^2 [/mm] + gz) = 0 [mm] $\\
[/mm]
[mm] $\Rightarrow \frac{p}{\rho} [/mm] - [mm] \frac{1}{2} \mathbf{u}^2 [/mm] + gz = C = const. [mm] $\\
[/mm]
At the surface, [mm] $p=p_0$ [/mm] atmospheric pressure:
[mm] $\frac{p_0}{\rho} [/mm] - [mm] \frac{1}{2} \mathbf{u}^2 [/mm] + gz = C [mm] $\\
[/mm]
[mm] $\Rightarrow [/mm] z = [mm] \frac{1}{g}(C [/mm] - [mm] \frac{p_0}{\rho} [/mm] + [mm] \frac{1}{2} \omega^2R^2) [/mm] = [mm] z_0 [/mm] + [mm] \frac{\omega^2 R^2}{2g} \quad (z_0 [/mm] = [mm] \text{all constants})$\\
[/mm]
[mm] $\Rightarrow R^2 [/mm] = [mm] \frac{2}{\omega^2} (\frac{p_0}{\rho} [/mm] + gz -C) = [mm] R_0 [/mm] + [mm] \frac{2gz}{\omega^2} \quad (R_0 [/mm] = [mm] \text{all constants})$\\
[/mm]
[mm] $\quad \Rightarrow [/mm] R = [mm] \sqrt{R_0 + \frac{2gz}{\omega^2}}$
[/mm]
Equation of the top surface
$V = [mm] \frac{1}{2}\pi a^3 \quad \text{Cylinder is half-full} $\\
[/mm]
$V = [mm] \int_0^a 2\pi [/mm] R [mm] \cdot [/mm] z(R)~dR [mm] \quad \text{Volume of rotating fluid} $\\
[/mm]
$= [mm] \int_0^a 2\pi [/mm] R [mm] \cdot (z_0 [/mm] + [mm] \frac{\omega^2 R^2}{2g})~dR [/mm] = [mm] \int_0^a \Big(2\pi Rz_0 [/mm] + [mm] \frac{\pi \omega^2 R^3}{g}\Big)~dR$\\
[/mm]
$= [mm] \bigg[2\pi z_0 \frac{R^2}{2} \bigg]_0^a [/mm] + [mm] \bigg[\frac{\pi \omega^2 R^4}{4g}\bigg]_0^a [/mm] = [mm] \pi z_0 a^2 [/mm] + [mm] \frac{\pi \omega^2 a^4}{4g} [/mm] = [mm] \frac{1}{2}\pi a^3$\\
[/mm]
[mm] $\Rightarrow z_0 [/mm] = [mm] \frac{a}{2} [/mm] - [mm] \frac{\omega^2 a^2}{4g}$\\
[/mm]
[mm] $\Rightarrow [/mm] z(R) = [mm] \frac{a}{2} [/mm] - [mm] \frac{\omega^2 a^2}{4g} [/mm] + [mm] \frac{\omega^2 R^2}{2g} [/mm] $
Show: If [mm] $\omega^2 [/mm] = [mm] \frac{2g}{a}$, [/mm] water will reach top of can at $R = a$
$z(a) = [mm] \frac{a}{2} [/mm] - [mm] \frac{2ga^2}{4ga} [/mm] + [mm] \frac{2ga^2}{2ga} [/mm] = [mm] \frac{a}{2} [/mm] - [mm] \frac{a}{2} [/mm] + a = a $
|
|
| |
|
| Status: |
(Antwort) fertig  | | Datum: | 15:11 Mo 29.03.2010 | | Autor: | rainerS |
Hallo!
Was spricht denn dagegen, Teil (ii) genauso anzugehen wie Teil (i) und die Oberfläche zu berechnen?
Viele Grüße
Rainer
|
|
|
| |
|
ok, hab das jetzt mal versucht, aber irgendwie verstehe ich die frage wohl immer noch nicht richtig.
hier soweit ich gekommen bin:
Vorticity
[mm] $\boldsymbol{\zeta} [/mm] = [mm] \nabla \times \mathbf{u} [/mm] = [mm] \frac{1}{R} \frac{\partial (R\cdot F_{\theta})}{\partial R} \mathbf{\hat z} [/mm] = 0 $
Bernoulli equation
[mm] $\nabla (\frac{p}{\rho} [/mm] + [mm] \frac{1}{2} \mathbf{u}^2 [/mm] + gz) = 0 [mm] $\\
[/mm]
[mm] $\Rightarrow \frac{p}{\rho} [/mm] + [mm] \frac{1}{2} \mathbf{u}^2 [/mm] + gz = C = const. [mm] $\\
[/mm]
[mm] $\Rightarrow [/mm] z = [mm] z_0 [/mm] - [mm] \frac{\omega^2 R^2}{2g} \quad (z_0 [/mm] = [mm] \text{all constants})$
[/mm]
Equation of the top surface
$ z(R) = [mm] \frac{a}{2} [/mm] + [mm] \frac{\omega^2 a^2}{4g} [/mm] + [mm] \frac{\omega^2 R^2}{2g} [/mm] $
Water reaches $z=a$ at [mm] $R=a$\\
[/mm]
$z(a) = [mm] \frac{a}{2} [/mm] + [mm] \frac{\omega^2 a^2}{4g} [/mm] + [mm] \frac{\omega^2 a^2}{2g} [/mm] = [mm] \frac{a}{2} [/mm] + [mm] \frac{3 \omega^2 a^2}{4g} [/mm] = a$
erste verständnisfrage: soll ich zeigen, dass "innen" quasi ein paraboloid "frei" ist, oder geht es darum, dass ein teil der grundfläche des zylinders wasserfrei ist?
und was hat das mit diesem b und k auf sich? ich sehe überhaupt nicht, was ich damit anfangen soll...... ![[cry01]](/images/smileys/cry01.gif "[cry01]")
danke schonmal.
|
|
|
| |
|
| Status: |
(Antwort) fertig | | Datum: | 15:23 Do 01.04.2010 | | Autor: | rainerS |
Hallo!
> ok, hab das jetzt mal versucht, aber irgendwie verstehe ich
> die frage wohl immer noch nicht richtig.
>
> hier soweit ich gekommen bin:
> Vorticity
> [mm]\boldsymbol{\zeta} = \nabla \times \mathbf{u} = \frac{1}{R} \frac{\partial (R\cdot F_{\theta})}{\partial R} \mathbf{\hat z} = 0[/mm]
>
> Bernoulli equation
> [mm]\nabla (\frac{p}{\rho} + \frac{1}{2} \mathbf{u}^2 + gz) = 0[/mm][mm] \\[/mm]
>
> [mm]\Rightarrow \frac{p}{\rho} + \frac{1}{2} \mathbf{u}^2 + gz = C = const.[/mm][mm] \\[/mm]
>
> [mm]\Rightarrow z = z_0 - \frac{\omega^2 R^2}{2g} \quad (z_0 = \text{all constants})[/mm]
Wo kommt das [mm] $\omega$ [/mm] her? Du hast hier einen anderen Flow mit dem Parameter $k$:
[mm]\Rightarrow z = z_0 - \frac{k^2 }{2gR^2} \quad (z_0 = \text{all constants})[/mm]
[...]
> erste verständnisfrage: soll ich zeigen, dass "innen"
> quasi ein paraboloid "frei" ist, oder geht es darum, dass
> ein teil der grundfläche des zylinders wasserfrei ist?
Du sollst zeigen, dass die Oberfläche außerhalb des Zylinders mit Radius b liegt.
Viele Grüße
Rainer
|
|
|
|
