Question 3.5: Parallel Flows in Cylindrical Tubes Case A: Consider flow th......

Case A: Of interest is the z-momentum equation, i.e., with the stated postulates (see App. A, Equation Sheet):

$\rm0=-\frac{\partial p}{\partial z} =\mu\left[\frac{1}{r}\frac{\partial}{\partial r}\left\lgroup r\frac{\partial v_z}{\partial r} \right\rgroup \right]$                      (E.3.5.1a)

Thus, with $\rm P\equiv \frac{1}\mu \left\lgroup\frac{-\Delta p}{l} \right\rgroup$

$\rm\frac{d}{dr} \left\lgroup r\frac{dv_z}{dr} \right\rgroup =P\cdot r$                          (E3.5.1b)

and hence after double integration,

$\rm v_z(r)=\frac{P}{4} r^2+C_1\ln r+C_2$                      (E.3.5.2)

Invoking the BCs,

and

yields

$\rm v_z(r)=\frac{PR^2}{4} \left[1-\left\lgroup\frac{r}{R} \right\rgroup^2-\frac{1-a^2}{\ln(1/a)}\ln\left\lgroup\frac{R}{r} \right\rgroup \right]$                          (E.3.5.3)

and

$\rm\tau_{rz}=\mu\frac{dv_z}{dr} :=\frac{\mu \,P}{2} R\left[\left\lgroup\frac{r}{R} \right\rgroup-\frac{1-a^2}{2\ln(1/a)}\left\lgroup\frac{R}{r} \right\rgroup \right]$                                (E.3.5.4a, b)

Notes:
• For Poiseuille flow, i.e., no inner cylinder, the solution is (see Example 3.4):

$\rm v_z(r)=\frac{R^2}{4\mu} \left\lgroup\frac{\Delta p}{l} \right\rgroup \left[1-\left\lgroup\frac{r}{R} \right\rgroup^2 \right]$                    (E.3.5.5)

This solution is not recovered when letting a → 0 because of the prevailing importance of the ln-term near the inner wall.
• The maximum annular velocity is not in the middle of the gap aR ≤ r ≤ R, but closer to the inner cylinder wall, where the velocity gradient is zero and hence

This equation can be solved for b so that $\rm v_z(r=bR)=v_{max}$.

• The average velocity is $\rm v_{av}=\int v_z(r)dA$, where dA = 2πrdr 〈cross-sectional ring of thickness dr〉 , so that

$\rm v_{av}=\frac{R^2 }{8\mu} \left\lgroup\frac{\Delta p}{l} \right\rgroup \left[\frac{1-a^4}{1-a^2}-\frac{1-a^2}{\ln(1/a)} \right]$                        (E.3.5.6)

and hence

• The net force exerted by the fluid on the solid surfaces comes from two wall shear stress contributions:

$\rm F_s=\left(-\tau_{rz}|_{r=aR}\right) (2\pi aRl)+\left(\tau_{rz}|_{r=R}\right) (2\pi Rl)$                (E.3.5.8a)

$\therefore ~~\rm F_s=\pi R^2\Delta p(1-a^2)$                      (E.3.5.8b)

Case B: With $\rm v_r=v_z=0;\;\frac{\partial}{\partial t} =\frac{\partial}{\partial \theta } =0$; and $\rm v_\theta =v_\theta (r)$ only (see Continuity and BCs) we reduce the θ-component of the \ Navier– Sokes equation (see Equation Sheet) to:

$\rm 0=0+\mu\left[\frac{\partial}{\partial r} \left\lgroup\frac{1}{r}\frac{\partial}{\partial r}(rv_\theta ) \right\rgroup \right]$                      (E.3.5.9)

subject to
$\rm v_ θ(r= aR) =ω_0 (aR)~ and~ v_θ (r =R) = ω_1R$.

Again, as in simple Couette flow after start-up, the moving-wall induced frictional effect propagates radially and the forced cylinder rotations balanced by the drag resistance generate an equilibrium velocity profile. Double integration yields:

$\rm v_\theta (r)=C_1r+\frac{C_2}{r}$                (E.3.5.10a)

where

$\rm C_1=\frac{\omega _1R^2-\omega _0(aR)^2}{R^2-(aR)^2} ~~and~~C_2=\frac{a^2R^4(\omega _0-\omega _1)}{R^2-(aR)^2}$              (E.3.5.10b, c)

Notes:
• The r-momentum equation reduces to:

$\rm -\frac{v_\theta ^2}{r} =-\frac{1}{\rho} \frac{\partial p}{\partial r}$                      (E.3.5.11)

Thus, with the solution for $\rm v_θ (r)$  known, Eq. (E.3.5.11) can be used to find ∂p / ∂r and ultimately the load-bearing capacity.

• Applying this solution as a first-order approximation to a journal bearing where the outer tube (or sleeve) is fixed, i.e., $ω_1 ≡0$, we have in dimensionless form:

$\rm\frac{v_\theta (r)}{\omega _0R}=\frac{a^2}{1-a^2} \left\lgroup\frac{R}{r}-\frac{r}{R} \right\rgroup$                  (E.3.5.12)

• The torque necessary to rotate the inner cylinder (or shaft) of length l is

$\rm T=\int(aR)dF:=(aR)\int\limits_0^l\tau_{r\theta} |_{r=aR}dA$                  (E.3.5.13)

where dA = π(aR)dz and $\rm\left.\tau_{r\theta} \right|_{r=aR}=\mu\left[r\frac{d}{dr}\left\lgroup\frac{v_\theta }{r} \right\rgroup \right] _{r=aR}.$

Thus with:

$\rm\tau_{surface}\equiv \tau_{r\theta }|_{r=aR}=2\mu\frac{\omega _oR^2}{R^2-(aR)^2}$                  (E.3.5.14)

$\rm T=\tau_{surf}A_{surf}(aR):=4\pi\mu(aR)^2l\frac{\omega _0}{1-a^2}$                  (E.3.5.15)

Graph:

Comments:
• An electric motor may provide the necessary power, $\rm P = Tω_0$ , which turns into thermal energy which has to be removed to avoid overheating.
• The Graph depicts the nonlinear dependence of T(a) for a given system. As the gap between rotor (or shaft) and stator widens, the wall stress increases (see Eq. (E.3.5.14)) as well as the surface area and hence the necessary torque.