Solutions written assignments Poiseuille
Problem 3-15 in White
Since the boundary conditions are independent of \(x\) we may assume a parallel shear solution of the form \(\boldsymbol{u} = (u(y), 0, 0)\), with \(\partial p / \partial x = 0\). The momentum equation reduces to
\[
0 = \mu \nabla^2u + \rho g \sin \theta,
\]
where \(\theta\) is the angle of the plate with the horizontal. The boundary conditions are \(u(0)=0\) and \(du/dx(h)=0\). Integrate twice to obtain
\[
u = -\frac{\rho g \sin \theta y^2}{2 \mu} + Ay + B
\]
The boundary conditions are then used to find A and B. The first, \(u(0)=0\), gives us \(B=0\) and the second
\[\begin{split}
\begin{aligned}
\frac{\mathrm{d} u}{dy}(y=h) = 0 &= -\frac{\rho g \sin \theta h}{\mu} +A \\
A &= \frac{\rho g \sin \theta h}{\mu}
\end{aligned}
\end{split}\]
and thus the final velocity profile
\[
u(y) = \frac{\rho g \sin \theta y}{2 \mu}\left(2h - y\right)
\]
The volumetric flow rate \(Q\) per unit width is found through
\[\begin{split}
\begin{aligned}
Q &= \int_0^h u(y) \mathrm{d}y \\
Q &= \frac{\rho g h^3 \sin \theta}{3 \mu}
\end{aligned}
\end{split}\]
Problem 3-16 in White
Assume negligible pressure gradient in \(z\)-direction since the film is thin and the pressure outside the film is constant atmospheric pressure. The momentum equation in \(z\)-direction reads
\[
\frac{\mu}{r}\frac{\mathrm{d} }{\mathrm{d}r}\left( r \frac{\mathrm{d} u}{\mathrm{d} r} \right) + \rho g = 0.
\]
Integrate twice to obtain
\[
u(r) = -\frac{\rho g r^2}{4 \mu} + A \ln r + B.
\]
The boundary conditions are no-slip at the wall and no shear at the free surface. The second one gives us
\[\begin{split}
\begin{aligned}
\frac{\mathrm{d} u}{\mathrm{d} r} (y=b) &= -\frac{\rho g b}{2 \mu} + \frac{A}{b} = 0 \\
A &= \frac{\rho g b^2}{2 \mu}
\end{aligned}
\end{split}\]
The no-slip boundary condition gives us
\[\begin{split}
\begin{aligned}
u(a) = 0 &= -\frac{\rho g a^2}{4 \mu} + \frac{\rho g b^2}{2 \mu} \ln a + B\\
B &= \frac{\rho g a^2}{4 \mu} - \frac{\rho g b^2}{2 \mu} \ln a,
\end{aligned}
\end{split}\]
which leads to the final expression for the velocity
\[
u(r) = \frac{\rho g}{4 \mu}\left(a^2 - r^2 \right) + \frac{\rho g b^2}{2 \mu} \ln \frac{r}{a}
\]
The volumetric flow rate can be found through
\[\begin{split}
\begin{aligned}
Q &= 2 \pi \int_a^b u(r) r \mathrm{d}r \\
Q &= \ldots
\end{aligned}
\end{split}\]
Problem 3-17 in White
Like in problem 3-15 the velocity is given as
\[\begin{split}
\begin{aligned}
u_1 &= -\frac{\rho_1 g \sin \theta y^2}{2 \mu_1} + A_1y + B_1, \\
u_2 &= -\frac{\rho_2 g \sin \theta y^2}{2 \mu_2} + A_2y + B_2.
\end{aligned}
\end{split}\]
The boundary conditions are now
\[\begin{split}
\begin{aligned}
u_1(0) &= 0, \\
u_1(h_1) &= u_2(h_1), \\
\mu_1 \frac{\mathrm{d} u_1}{\mathrm{d} y}(h_1) &= \mu_2 \frac{\mathrm{d} u_2}{\mathrm{d} y}(h_1) \\
\frac{\mathrm{d} u_2}{\mathrm{d} y}(h_1+h_2) &= 0.
\end{aligned}
\end{split}\]
The first one gives \(B_1=0\). The remaining leads to three equations for three unknowns. They are:
\[\begin{split}
\begin{aligned}
-\frac{\rho_1 g \sin \theta h_1^2}{2 \mu_1} + A_1h_1 &= -\frac{\rho_2 g \sin \theta h_1^2}{2 \mu_2} + A_2h_1 + B_2, \\
-\frac{\rho_1 g \sin \theta h_1}{\mu_1} +A_1 &= -\frac{\rho_2 g \sin \theta h_1}{\mu_2} +A_2 \\
-\frac{\rho_2 g \sin \theta (h_1+h_2)}{\mu_2} + A_2 &= 0
\end{aligned}
\end{split}\]
where \(A_2\) is given by the last equation. Inserting this into the previous gives us \(A_1\)
\[\begin{split}
\begin{aligned}
A_1 &= -\frac{\rho_2 g \sin \theta h_1}{\mu_2} + \frac{\rho_2 g \sin \theta (h_1+h_2)}{\mu_2} + \frac{\rho_1 g \sin \theta h_1}{\mu_1}\\
A_1 &= g \sin \theta \left(\frac{\rho_2 h_2}{\mu_2} + \frac{\rho_1 h_1}{\mu_1} \right)
\end{aligned}
\end{split}\]
And \(A_1\) and \(A_2\) inserted into the first gives us \(B_2\)
\[
B_2 = -\frac{\rho_1 g \sin \theta h_1^2}{2 \mu_1} + A_1 h_1 + \frac{\rho_2 g \sin \theta h_1^2}{2 \mu_2} - A_2 h_1
\]
