Fluid Mechanics for MAP/Analytical solutions of internal and external flows

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Internal and External Incompressible Viscous Flows

Flows completely bounded by solid surfaces are called internal flows. External flows are flows over bodies immersed in an unbounded fluid. Internal flows might be laminar or turbulent. The state of the flow regime is dependent on Re. There might be an analytical solution for laminar flows but not for turbulent flows. File:Internal-flow-examples.png Flows through pipes, ducts, nozzles, diffusers, valves and fittings are examples of internal flows.

Flow around an aircraft is classified under external flows. Flow around a car is classified under external flows.

Laminar and Turbulent Channel and Pipe flows

At fully developed state the velocity profile becomes parabolic for laminar flow. The average velocity at any cross section is: Template:Center top$\overline{U}=U_0=V=\frac{1}{A}\underset{Area}{\int }{U}_{1}dA$Template:Center bottom For the same flow value i.e. $U_0$, the fully developed turbulent pipe flow, would have higher velocity close to the wall and lower velocity at the center. The reason is the turbulent eddies, which causes more momentum loss to the wall i.e. higher velocity gradients close to the wall. Note that such a direct comparison is only valid at the same $Re=\frac{U_{0}D}{\nu }$.

Velocity profiles for laminar (upper) and turbulent (lower) states at the same mass flow rate

File:Transition-in-pipe.png Signal measured by a hot-wire at a center of a pipe during transition showing the developement of transitional structures along the pipe at different Reynolds numbers and the change of friction coefficient at laminar, transitional and turbulent regimes.

Development of velocity profile in a pipe with increasing Reynolds number

Concept of Fully Developed Flow

File:IF2 me.png

Flow between two plates

Consider the flow in a channel between two plates having a height of $D$ and an infinite depth in $x_3$ direction. Starting from the entrance, the boundary layers develop due to the no-slip condition on the wall. At a finite distance, the boundary layers merge and the inviscid core (field with no velocity gradient in $x_2$ direction) vanishes. The flow becomes fully viscous. The velocity field in $x_1$ direction adjusts slightly further until $x_1=L_e$ and it no longer changes with $x_1$ direction. This state of the flow is called fully-developed. At that state:

Template:Center top$\frac{\partial U_i}{\partial x_1}=0\to \frac{\partial {\tau }_{ij}}{\partial x_1}=0$Template:Center bottom

Because of the two-dimensional nature of the flow, no gradient of the velocity quantities in $x_3$ direction is expected starting from the entrance.

Template:Center top$U_3=0,\frac{\partial U_i}{\partial x_3}=0,\frac{\partial {\tau }_{ij}}{\partial x_3}=0,\frac{\partial P}{\partial x_3}=0,$Template:Center bottom Template:Center top$U_2=U_3=0$Template:Center bottom

Hence,

Template:Center top$U_1=U_1(x_2),{\tau }_{ij}={\tau }_{ij}(x_2).$Template:Center bottom

The entrance lengths for laminar pipe and channel flows are, respectively:

Template:Center top$\frac{L{e}_{pipe}}{D}={[(0.619{)}^{1.6}+(0.0567Re{)}^{1.6}]}^{\frac{1}{1.6}}$Template:Center bottom

Template:Center top$\frac{L{e}_{channel}}{D}={[(0.631{)}^{1.6}+(0.0442Re{)}^{1.6}]}^{\frac{1}{1.6}}$Template:Center bottom

Fully Developed Laminar Flow Between Infinite Parallel Plates

File:Infinite plate.svg

Fully developed flow in a channel with an infinite depth

Consider the fully developed laminar flow between two infinite plates.

Consider the continuity equation and momentum equation in $x_1$ direction for an incompressible steady flow between two infinite plates as shown.

Continuity Equation

Template:Center top$\underset{=0}{\underbrace{\frac{\partial \rho }{\partial t}}}+\frac{\partial \rho U_i}{\partial x_i}=0\stackrel{\rho =cst.}{⟶}\frac{\partial U_i}{\partial x_i}=0$Template:Center bottom

Since $\frac{\partial U_1}{\partial x_1}=0$ , $\frac{\partial U_3}{\partial x_3}=0\to \frac{\partial U_2}{\partial x_2}=0$ because it is a fully-developed and two dimensional flow. Hence, $U_2$ reads Template:Center top$U_2=constant$Template:Center bottom As $U_2$ is zero on the walls, it should be zero in the whole fully developed region, i.e. Template:Center top$U_2(x_2)=0$Template:Center bottom

Momentum Equation in j-direction

Template:Center top$\underset{=0Steady}{\underbrace{\rho \frac{\partial U_j}{\partial t}}}+\rho U_i\frac{\partial U_j}{\partial x_i}=-\frac{\partial P}{\partial x_j}+\rho g_j+\frac{\partial {\tau }_{ij}}{\partial x_i}$Template:Center bottom

in $x_1$ direction ,$g_1=0$

Template:Center top$\underset{A}{\underbrace{\rho U_i\frac{\partial U_1}{\partial x_i}}}=-\frac{\partial P}{\partial x_1}+\underset{B}{\underbrace{\frac{\partial {\tau }_{i1}}{\partial x_i}}}$Template:Center bottom

Consider term A:

Template:Center top$\rho U_i\frac{\partial U_1}{\partial x_i}=\rho [U_1\underset{=0Fully-developed}{\underbrace{\frac{\partial U_1}{\partial x_1}}}+\underset{=0}{\underbrace{U_2}}\frac{\partial U_1}{\partial x_2}+\underset{=0fully-developed2D}{\underbrace{U_3}}\underset{=02D}{\underbrace{\frac{\partial U_1}{\partial x_3}}}]=0$Template:Center bottom

Consider term B:

Template:Center top$\frac{\partial {\tau }_{i1}}{\partial x_i}=\underset{=0fully-dev.}{\underbrace{\frac{\partial {\tau }_{11}}{\partial x_1}}}+\frac{\partial {\tau }_{21}}{\partial x_2}+\underset{=02D}{\underbrace{\frac{\partial {\tau }_{31}}{\partial x_3}}}=\frac{\partial {\tau }_{{\tau }_{21}}}{\partial x_2}=\frac{d{\tau }_{21}}{d{x}_2}$Template:Center bottom

hence ${\tau }_{21}={\tau }_{21}(x_2)$.

Thus the momentum equation in $x_1$ direction reads:

Template:Center top$0=-\frac{\partial P}{\partial x_1}+\frac{d{\tau }_{21}}{d{x}_2}$Template:Center bottom

This equation should be valid for all $x_1$ and $x_2$. This requires that $\frac{\partial P}{\partial x_1}=\frac{d{\tau }_{21}}{d{x}_2}$ = constant.

Remember ${\tau }_{21}$ is the stress in $x_1$ direction on a face normal to $x_2$ direction.

Template:Center top${\tau }_{ij}=\mu (\frac{\partial U_i}{\partial x_j}+\frac{\partial U_j}{\partial x_i})+\frac{2}{3}{\delta }_{ij}\mu \frac{\partial U_k}{\partial x_k}$Template:Center bottom

thus,

Template:Center top${\tau }_{21}=\mu (\underset{=0}{\underbrace{\frac{\partial U_2}{\partial x_1}}}+\frac{\partial U_1}{\partial x_2})+0$Template:Center bottom

Template:Center top${\tau }_{21}=\mu \frac{\partial U_1}{\partial x_2}\text{since,}{U}_1=U_1(x_2){\tau }_{21}=\mu \frac{d{U}_1}{d{x}_2}$Template:Center bottom

Thus the momentum equation reads:

Template:Center top$\frac{\partial P}{\partial x_1}=\mu \frac{d^2{U}_1}{d{x}_2^2}$Template:Center bottom

This equation can be obtained also by using the Reynold's transport equations for a differential volume.

File:Rtt-treatment-of-channel-flow.png

Treatment of channel flow with the RTT approach

The momentum equation in $x_1$ direction,

Template:Center top$F_{S1}+\underset{=0}{\underbrace{F_{B1}}}=\frac{\partial }{\partial t}\underset{=0}{\underbrace{\underset{CV}{\int }\rho U_{1}dV}}+\underset{=0fully-dev.}{\underbrace{\underset{CS}{\int }{U}_1\rho U_i{n}_{i}dA}}$Template:Center bottom

The flux term becomes zero since for fully-developed flow incoming flux is equal to the outgoing flux. Thus,

Template:Center top$F_{S1}=0$Template:Center bottom

That is:

Template:Center top$(p-\frac{\partial P}{\partial x_1}\frac{d{x}_1}{2})d{x}_{2}d{x}_3-(p+\frac{\partial P}{\partial x_1}\frac{d{x}_1}{2})d{x}_{2}d{x}_3$Template:Center bottom

Template:Center top$+({\tau }_{21}+\frac{d{x}_2}{2}\frac{\partial {\tau }_{21}}{\partial x_2})d{x}_{1}d{x}_3-({\tau }_{21}-\frac{d{x}_2}{2}\frac{\partial {\tau }_{21}}{\partial x_2})d{x}_{1}d{x}_3=0$Template:Center bottom

Template:Center top$-\frac{\partial P}{\partial x_1}dV+\frac{\partial {\tau }_{21}}{\partial x_2}dV=0$Template:Center bottom

Template:Center top$-\frac{\partial P}{\partial x_1}+\frac{\partial {\tau }_{21}}{\partial x_2}=0$Template:Center bottom

Finally, the governing equation of this kind of flow becomes:

Template:Center top$\frac{\partial P}{\partial x_1}=\mu \frac{d^2{U}_1}{d{x}_2^2}$Template:Center bottom

with the following boundary conditions: Template:Center top $x_2=0\to U_1=0$ and $x_2=D\to U_1=0$ Template:Center bottom

Integrating the equation once results in a linear function of $x_2$:

Template:Center top$\mu \frac{d{U}_1}{d{x}_2}=\left(\frac{\partial P}{\partial x_1}\right)x_2+c_1\to {\tau }_{21}=\left(\frac{\partial P}{\partial x_1}\right)x_2+c_1$Template:Center bottom

The second integration reads:

Template:Center top$U_1=\frac{1}{2\mu }\left(\frac{\partial P}{\partial x_1}\right)x_2^2+\frac{1}{\mu }{c}_1{x}_2+c_2$Template:Center bottom

The integration constants is obtained by using the boundary conditions:

Template:Center top$x_2=0,U_1=0\to c_2=0$Template:Center bottom

Template:Center top$x_2=D,U_1=0\to c_1=-\frac{1}{2}\left(\frac{\partial P}{\partial x_1}\right)D$Template:Center bottom

Finally, the velocity profile reads:

Template:Center top$\begin{array}{ccc}{U}_1& =& \frac{1}{2\mu }\left(\frac{\partial P}{\partial x_1}{x}_2^2\right)-\frac{1}{2\mu }\left(\frac{\partial P}{\partial x_1}\right)D{x}_2\\ & =& \frac{D^2}{2\mu }\left(\frac{\partial P}{\partial x_1}\right)[{\left(\frac{x_2}{D}\right)}^2-\left(\frac{x_2}{D}\right)]\end{array}$Template:Center bottom

Note that the velocity profile is parabolic!

The shear stress becomes:

Template:Center top$\begin{array}{ccc}{\tau }_{21}& =& \left(\frac{\partial P}{\partial x_1}\right)x_2-\frac{1}{2}\left(\frac{\partial P}{\partial x_1}\right)D\\ & =& D\left(\frac{\partial P}{\partial x_1}\right)[\frac{x_2}{D}-\frac{1}{2}]\end{array}$Template:Center bottom

at the wall i.e. at $x_2$ = 0 and $x_2$= D

Template:Center top${\tau }_{21}(0)=-\frac{1}{2}D\left(\frac{\partial P}{\partial x_1}\right)$Template:Center bottom

Template:Center top${\tau }_{21}(D)=\frac{1}{2}D\left(\frac{\partial P}{\partial x_1}\right)$Template:Center bottom

File:Fully developed channel-flow-solution.png

Velocity and shear stress profiles in fully developed channel flow

Note that ${\tau }_{21}$ is maximum near the wall, i.e. momentum loss is maximum near the wall. This is due to the maximum velocity gradient $\frac{\partial U_1}{\partial x_2}$ near the wall!

The volume flow rate is,

Template:Center top$Q=\underset{A}{\int }{U}_i{n}_{i}dA=\underset{0}{\overset{D}{\int }}{U}_{1}wd{x}_2$Template:Center bottom

where $w$ is the depth of the channel.

Thus the volume flow rate per depth $w$ is given by:

Template:Center top$\frac{Q}{w}=\underset{0}{\overset{D}{\int }}\frac{1}{2\mu }\left(\frac{\partial P}{\partial x_1}\right)(x^2-D{x}_2)d{x}_2$Template:Center bottom

Template:Center top$\frac{Q}{w}=-\frac{1}{12\mu }\left(\frac{\partial P}{\partial x_1}\right)D^3$Template:Center bottom

Note that $\frac{\partial P}{\partial x_1}$ should be constant for the fully developed flow. Hence, for a channel with a finite length $L$:

Template:Center top$\frac{\partial P}{\partial x_1}=\frac{p_2-p_1}{L}=\frac{-\mathrm{\Delta }P}{L}$Template:Center bottom

Where $\mathrm{\Delta }P$ is the pressure drop along L.

Template:Center top$\frac{Q}{w}=-\frac{1}{12\mu }\left(\frac{-\mathrm{\Delta }P}{L}\right)D^3=\frac{D^3}{12\mu L}\mathrm{\Delta }P$Template:Center bottom

or the pressure drop can be calculated from:

Template:Center top$\mathrm{\Delta }P=\frac{Q}{l}\frac{12\mu L}{D^3}$Template:Center bottom

For the same flow rate, increasing the height of the channel would cause a drastic reduction in the pressure drop.

The average velocity $\overline{U}$ is:

Template:Center top$\overline{U}=U_0=\frac{Q}{A}=\frac{Q}{wD}=-\frac{1}{12\mu }\left(\frac{\partial P}{\partial x_1}\right)\frac{D^{3}w}{wD}=-\frac{1}{12\mu }\left(\frac{\partial P}{\partial x_1}\right)D^2$Template:Center bottom

The maximum velocity occurs when:

Template:Center top$\frac{d{U}_1}{d{x}_2}=0=\frac{D^2}{2\mu }\left(\frac{\partial P}{\partial x_1}\right)(\frac{2x_2}{D^2}-\frac{1}{D})$Template:Center bottom

Hence, at $x_2=\frac{D}{2}$, $U_1=U_{1max}$

Template:Center top$U_1\left(\frac{D}{2}\right)=U_{1max}=-\frac{1}{8\mu }\left(\frac{\partial P}{\partial x_1}\right)D^2=\frac{3}{2}\overline{U}$Template:Center bottom

The velocity profile can be written as functions of bulk velocity $\bar{U}$ or maximum velocity $U_{1max}$ by replacing their value the velocity profile equation:

Template:Center top$\begin{array}{ccc}{U}_1& =& -4U_{1max}[{\left(\frac{x_2}{D}\right)}^2-\left(\frac{x_2}{D}\right)]\\ & =& -6\bar{U}[{\left(\frac{x_2}{D}\right)}^2-\left(\frac{x_2}{D}\right)]\end{array}$Template:Center bottom

Same problem can be solved by using moving plates.

Example

File:IF11.png

Flow through the gap between a cylindrical piston and the surrounding wall

Consider the hydrolic control valve coprising a piston, fitted to a cylinder with a mean radial clearance of 0,005mm. Determine the leakage flow rate. The fluid is SAE low oil ($\rho$ = 932 $\frac{kg}{m^3}$, $\mu$ =0.018 $\frac{kg}{msec}$ at 55ºC). The flow can be assumed to be laminar, steady, incompressible, fully-developed flow. $(\frac{L}{a}=3000)$

Since $\frac{D}{a}=\frac{25}{0,005}$ = 5000 the flow in the clearance can be accepted to be 2-D, with the depth $w=\pi \cdot D$, thus:

Template:Center top$\frac{Q}{w}=\frac{a^3\mathrm{\Delta }P}{12\mu L}$Template:Center bottom

Template:Center top$Q=\frac{a^3\mathrm{\Delta }P}{12\mu L}\pi D=\frac{(0.005\cdot 10^{-3}{)}^3}{12*(0.018)*15\cdot 10^{-3}}*(20-1)\cdot 10^{-6}*\pi *25\cdot 10^{-3}$Template:Center bottom

Template:Center top$Q=57.6\cdot 10^{-9}\frac{m^3}{s}=57.6\frac{m{m}^3}{sec}$Template:Center bottom

Check the Reynolds number to ensure that laminar flow assumption is correct.

Template:Center top$\overline{U}=\frac{Q}{A}=\frac{Q}{\pi Da}=0.147\frac{m}{sec}$Template:Center bottom

Template:Center top$Re=\frac{\rho \overline{U}a}{\mu }=\frac{932*0.147*0.005\cdot 10^{-3}}{0.018}=0.0375$Template:Center bottom

Re $<<1800$, i. e. the flow is laminar.

Layered Channel Flow

File:Layered Channel Flow2.png

This channel flow contains two different and non miscible fluids. Fluids A and B flow at the same time through a channel, which is bounded two flat plates. They both occupy the half height of the channel. The fluid A has a viscosity ${\mu }_A$, a density ${\rho }_A$ and the mass flow ${\dot{m}}_A$. Fluid B, which is located above fluid A, has a viscosity ${\mu }_B$, a density ${\rho }_B$ and the mass flow ${\dot{m}}_B$. The following differential equations correspond to the molecular momentum ${\tau }_{21}$ for each Fluid.

$\frac{d{\tau }_{21}^A}{d{x}_2}=\frac{d\mathrm{\Pi }}{d{x}_1}$ and $\frac{d{\tau }_{21}^B}{d{x}_2}=\frac{d\mathrm{\Pi }}{d{x}_1}$.

With ${\tau }_{21}=\mu \frac{d{U}_1}{d{x}_2}$ yields the velocity field:

$\frac{d^2{U}_1^A}{d{x}_2}=\frac{1}{{\mu }_A}\frac{d\mathrm{\Pi }}{d{x}_1}$ and $\frac{d^2{U}_1^B}{d{x}_2}=\frac{1}{{\mu }_B}\frac{d\mathrm{\Pi }}{d{x}_1}$

After integration of both equations we obtain:

${\tau }_{21}^A=\frac{d\mathrm{\Pi }}{d{x}_1}{x}_2+C_1^A$ and ${\tau }_{21}^B=\frac{d\mathrm{\Pi }}{d{x}_1}{x}_2+C_1^B$

As boundary condition we consider that shear stress on the interface between A and B is the same. Therefore we obtain:

${\tau }_{21}^A(x_2=0)={\tau }_{21}^B(x_2=0)$

Then,

$C_1^A=C_1^B=C_1$

After the integration for the velocity field:

$U_1^A=\frac{1}{2{\mu }_A}\frac{d\mathrm{\Pi }}{d{x}_1}{x}_2^2+\frac{C_1}{{\mu }_A}{x}_2+C_2^A$

and

$U_1^B=\frac{1}{2{\mu }_B}\frac{d\mathrm{\Pi }}{d{x}_1}{x}_2^2+\frac{C_1}{{\mu }_B}{x}_2+C_2^B$

The second boundary condition turns out to be on the interface:

$U_1^A(x_2=0)=U_1^B(x_2=0)$, i.e. $C_2^A\ne C_2^B$

therefore,

$C_2^A=C_2^B=C_2$. The integration constants can be calculated with the following boundary conditions:

At $x_2=-D\to U_1^A=0$:

$0=\frac{d\mathrm{\Pi }}{d{x}_1}\frac{1}{2{\mu }_A}{D}^2-\frac{C_{1}D}{{\mu }_A}+C_2$

At $x_2=+D\to U_1^B=0$:

$0=\frac{d\mathrm{\Pi }}{d{x}_1}\frac{1}{2{\mu }_B}{D}^2+\frac{C_{1}D}{{\mu }_B}+C_2$

Therefore we obtain for the velocity distribution in the fluids A and B:

$U_1^A=-\frac{D^2}{2{\mu }_A}\frac{d\mathrm{\Pi }}{d{x}_1}[+\frac{2{\mu }_A}{({\mu }_A+{\mu }_B)}+\left(\frac{{\mu }_A-{\mu }_B}{{\mu }_A+{\mu }_B}\right)\left(\frac{x_2}{D}\right)-{\left(\frac{x_2}{D}\right)}^2]$

and

$U_1^B=-\frac{D^2}{2{\mu }_B}\frac{d\mathrm{\Pi }}{d{x}_1}[+\frac{2{\mu }_B}{({\mu }_A+{\mu }_B)}+\left(\frac{{\mu }_A-{\mu }_B}{{\mu }_A+{\mu }_B}\right)\left(\frac{x_2}{D}\right)-{\left(\frac{x_2}{D}\right)}^2]$

For the distribution of the shear stress we get:

${\tau }_{21}=D\frac{d\mathrm{\Pi }}{d{x}_1}[\left(\frac{x_2}{D}\right)-\frac{1}{2}\left(\frac{{\mu }_A-{\mu }_B}{{\mu }_A+{\mu }_B}\right)]$

If we choose ${\mu }_A={\mu }_B$,

$U_1=\frac{-D^2}{2{\mu }_A}\frac{d\mathrm{\Pi }}{d{x}_1}[1-{\left(\frac{x_2}{D}\right)}^2]$

${\tau }_{21}=D\frac{d\mathrm{\Pi }}{d{x}_1}\left(\frac{x_2}{D}\right)$

The solution gives that of the channel flow. In other words, velocity has a parabolic profile with the peak in the middle of the channel and a linear shear stress distribution ${\tau }_{21}$, where ${\tau }_{21}=0$ at the channel's centerline.

If ${\mu }_A\ne {\mu }_B$, the position where the maximal velocity occurs can be calculated by introducing ${\tau }_{21}=0$ on the velocity profile equation:

$x_2(U_{1max})=\frac{D}{2}\left(\frac{{\mu }_A-{\mu }_B}{{\mu }_A+{\mu }_B}\right)$

The shear stress on the upper plate is:

${\tau }_W^B=\frac{D}{2}\frac{d\mathrm{\Pi }}{d{x}_1}\left[\frac{{\mu }_A+3{\mu }_B}{{\mu }_A+{\mu }_B}\right]$

and the shear stress on the lower plate reads:

${\tau }_W^A=-\frac{D}{2}\frac{d\mathrm{\Pi }}{d{x}_1}\left[\frac{3{\mu }_A+{\mu }_B}{{\mu }_A+{\mu }_B}\right]$

The average velocities of the fluids A and B are:

${\tilde{U}}_1^A=-\frac{D^2}{12{\mu }_A}\frac{d\mathrm{\Pi }}{d{x}_1}\left(\frac{7{\mu }_A+{\mu }_B}{{\mu }_A+{\mu }_B}\right)$

and

${\tilde{U}}_1^B=-\frac{D^2}{12{\mu }_B}\frac{d\mathrm{\Pi }}{d{x}_1}\left(\frac{{\mu }_A+7{\mu }_B}{{\mu }_A+{\mu }_B}\right)$

Hence the respectively mass flow rates are:

${\dot{m}}_A=BD{\tilde{U}}_1^A$

and

${\dot{m}}_B=BD{\tilde{U}}_1^B$

Navier Stokes Equation in Cyclindrical Coordinates

A change of variables on the Cartesian equations will yield the following equations of momentum in r, $\varphi$, and z directions:

$r:\rho (\frac{\partial u_r}{\partial t}+u_r\frac{\partial u_r}{\partial r}+\frac{u_{\varphi }}{r}\frac{\partial u_r}{\partial \varphi }+u_x\frac{\partial u_r}{\partial x}-\frac{u_{\varphi }^2}{r})=-\frac{\partial p}{\partial r}+\mu [\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_r}{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^2{u}_r}{\partial {\varphi }^2}+\frac{{\partial }^2{u}_r}{\partial x^2}-\frac{u_r}{r^2}-\frac{2}{r^2}\frac{\partial u_{\varphi }}{\partial \varphi }]+\rho g_r$

$\varphi :\rho (\frac{\partial u_{\varphi }}{\partial t}+u_r\frac{\partial u_{\varphi }}{\partial r}+\frac{u_{\varphi }}{r}\frac{\partial u_{\varphi }}{\partial \varphi }+u_x\frac{\partial u_{\varphi }}{\partial x}+\frac{u_r{u}_{\varphi }}{r})=-\frac{1}{r}\frac{\partial p}{\partial \varphi }+\mu [\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_{\varphi }}{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^2{u}_{\varphi }}{\partial {\varphi }^2}+\frac{{\partial }^2{u}_{\varphi }}{\partial z^2}+\frac{2}{r^2}\frac{\partial u_r}{\partial \varphi }-\frac{u_{\varphi }}{r^2}]+\rho g_{\varphi }$

$x:\rho (\frac{\partial u_x}{\partial t}+u_r\frac{\partial u_x}{\partial r}+\frac{u_{\varphi }}{r}\frac{\partial u_x}{\partial \varphi }+u_x\frac{\partial u_x}{\partial x})=-\frac{\partial p}{\partial x}+\mu [\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_x}{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^2{u}_x}{\partial {\varphi }^2}+\frac{{\partial }^2{u}_x}{\partial x^2}]+\rho g_x.$

The continuity equation is:

$\frac{\partial \rho }{\partial t}+\frac{1}{r}\frac{\partial }{\partial r}\left(\rho r{u}_r\right)+\frac{1}{r}\frac{\partial }{\partial \varphi }\left(\rho u_{\varphi }\right)+\frac{\partial }{\partial x}\left(\rho u_x\right)=0.$

Fully Developed Pipe Flow

File:RTT-treatment-of-pipe-flow-with-infinitesimal-cylinder-mdf1.png

Treatment of pipe flow with the RTT approach(with infinitesimal cylinder)

It is possible to use the same mathematical treatment like before to find and understand the velocity profile for fully developed flow inside a pipe with diameter D and infinite length. To show the flexibility, the same solution for this problem will be approached via 3 different ways.

(i) Infinitesimal Cylinder at the center of the pipe

Applying the RTT to the infinitesimal cylindrical CV along the symmetry axis of horizontal pipe, in which the flow is fully developed, the conservation of mass and the transport side of the conservation of momentum equation drops. Only remaining term governing this kind of flow is the balance of the forces on the CV in $x$ direction.

Template:Center top$(p-\frac{\partial P}{\partial x}\frac{dx}{2})\pi r^2-(p+\frac{\partial P}{\partial x}\frac{dx}{2})\pi r^2+{\tau }_{r_x}2\pi rdx=0$Template:Center bottom

Template:Center top$-\frac{\partial P}{\partial x}dx\pi r^2+{\tau }_{r_x}2\pi rdx=0$Template:Center bottom

Template:Center top${\tau }_{r_x}=\frac{r}{2}\frac{\partial P}{\partial x}$Template:Center bottom

(ii) Infinitesimal thin hollow cylinder at the center

File:Rtt-treatment-of-pipe-flow-infinitesimal-tubular-cylindrical-element.png

Treatment of pipe flow with the RTT approach(with infinitesimal hollow cylinder)

This time the pressure and viscous force is considered for a concentric hollow cylinder with radius of R and infinitesimally small thickness dr and length dx(as shown in the image besides) along x-direction. Considering pressure term on the cross-section of cylinder

Template:Center top$p2\pi rdr-(p+\frac{\partial P}{\partial x}dx)2\pi rdr\to -\left(\frac{\partial P}{\partial x}\right)2\pi rdrdx$Template:Center bottom

considering viscous shear stress on the surface of the cylinder Template:Center top$({\tau }_{r_x}+\frac{\partial {\tau }_{r_x}}{\partial r}dr)2\pi (r+dr)dx-{\tau }_{r_x}2\pi rdx$Template:Center bottom Template:Center top$={\tau }_{r_x}2\pi rdx+{\tau }_{r_x}2\pi drdx+\left(\frac{\partial {\tau }_{r_x}}{\partial r}\right)2\pi rdrdx+\underset{{dr}^2\approx 0}{\underbrace{\left(\frac{\partial {\tau }_{r_x}}{\partial r}\right)2\pi r{dr}^{2}dx}}-{\tau }_{r_x}2\pi rdx$Template:Center bottom Template:Center top$\to {\tau }_{r_x}2\pi drdx+\left(\frac{\partial {\tau }_{r_x}}{\partial r}\right)2\pi rdrdx$Template:Center bottom

combining both term,we get the balance equation, Template:Center top$-\left(\frac{\partial P}{\partial x}\right)2\pi rdrdx+{\tau }_{r_x}2\pi drdx+\left(\frac{\partial {\tau }_{r_x}}{\partial r}\right)2\pi rdrdx=0$Template:Center bottom

it could be rewritten as Template:Center top$-\left(\frac{\partial P}{\partial x}\right)2\pi rdrdx+({\tau }_{r_x}+r\frac{\partial {\tau }_{r_x}}{\partial r})2\pi drdx=0$Template:Center bottom dividing both side with $2\pi drdx$,we get Template:Center top$-\left(\frac{\partial P}{\partial x}\right)r+\frac{\partial (r{\tau }_{r_x})}{\partial r}=0$Template:Center bottom or,Template:Center top$\left(\frac{\partial P}{\partial x}\right)r=\frac{\partial (r{\tau }_{r_x})}{\partial r}$Template:Center bottom integrating both side ,

Template:Center top$\int \left(\frac{\partial P}{\partial x}\right)rdr=\int \partial (r{\tau }_{r_x})$Template:Center bottom

or,Template:Center top$\left(\frac{\partial P}{\partial x}\right)\frac{r^2}{2}=r{\tau }_{r_x}$Template:Center bottom

or,Template:Center top$\left(\frac{\partial P}{\partial x}\right)\frac{r}{2}={\tau }_{r_x}$Template:Center bottom

However,in a laminar flow!

Template:Center top$\mu \frac{dU}{dr}=\frac{1}{2}r\frac{\partial P}{\partial x}$Template:Center bottom

integrating,

Template:Center top$U=\frac{r^2}{4\mu }\left(\frac{\partial P}{\partial x}\right)+c_1$Template:Center bottom

The boundary condition is:

Template:Center top$U=0atr=R$Template:Center bottom

Thus $c_1$ can be calculated from the boundary condition.

Template:Center top$c_1=-\frac{R^2}{4\mu }\left(\frac{\partial P}{\partial x}\right)\Rightarrow U=\frac{1}{4\mu }\left(\frac{\partial P}{\partial x}\right)(r^2-R^2)$Template:Center bottom

or

Template:Center top$U=-\frac{R^2}{4\mu }\left(\frac{\partial P}{\partial x}\right)[1-{\left(\frac{r}{R}\right)}^2]$Template:Center bottom

(iii) Using NS in cylindrical co-ordinates:

Template:Center top$x:\rho (\underset{steadystate}{\underbrace{\frac{\partial u_x}{\partial t}}}+\underset{u_r=0}{\underbrace{u_r\frac{\partial u_x}{\partial r}}}+\underset{u_{\varphi }=0}{\underbrace{\frac{u_{\varphi }}{r}\frac{\partial u_x}{\partial \varphi }}}+\underset{\frac{\partial u_x}{\partial x}=0}{\underbrace{u_x\frac{\partial u_x}{\partial x}}})=-\frac{\partial p}{\partial x}+\mu [\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_x}{\partial r}\right)+\underset{=0}{\underbrace{\frac{1}{r^2}\frac{{\partial }^2{u}_x}{\partial {\varphi }^2}}}+\underset{=0}{\underbrace{\frac{{\partial }^2{u}_x}{\partial x^2}}}]+\underset{g_x=0}{\underbrace{\rho g_x}}$Template:Center bottom

thus, Template:Center top$\frac{\partial p}{\partial x}=\mu \left[\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_x}{\partial r}\right)\right]$Template:Center bottom

or,Template:Center top$\frac{\partial p}{\partial x}=\frac{\mu }{r}\left[\frac{\partial }{\partial r}\left(r\frac{\partial u_x}{\partial r}\right)\right]$Template:Center bottom

or,Template:Center top$\left(\frac{\partial p}{\partial x}\right)\frac{r}{\mu }=\left[\frac{\partial }{\partial r}\left(r\frac{\partial u_x}{\partial r}\right)\right]$Template:Center bottom

Now, integrating both side with respect to $r$, we get

Template:Center top$\int \left(\frac{\partial p}{\partial x}\right)\frac{r}{\mu }dr=\int \partial \left(r\frac{\partial u_x}{\partial r}\right)$Template:Center bottom then, Template:Center top$\left(\frac{\partial p}{\partial x}\right)\frac{r^2}{2\mu }+C=\left(r\frac{\partial u_x}{\partial r}\right)$Template:Center bottom dividing by r in both side , we get then Template:Center top$\left(\frac{\partial p}{\partial x}\right)\frac{r}{2\mu }+\frac{C}{r}=\left(\frac{\partial u_x}{\partial r}\right)$Template:Center bottom integrating again with respect to $r$ gives , Template:Center top$\left(\frac{\partial p}{\partial x}\right)\frac{r^2}{4\mu }+C\ln r+D=U_x(r)$Template:Center bottom

Consequently, when $r=0$ then $U_x=U_{max}$ and $\ln r=\mathrm{\infty }$, as a result C=0.

On the other hand, when $r=R$ then $U_x=0$ , so Template:Center top$\left(\frac{\partial p}{\partial x}\right)\frac{R^2}{4\mu }+D=0$Template:Center bottom or, Template:Center top$D=-\left(\frac{\partial p}{\partial x}\right)\frac{R^2}{4\mu }$Template:Center bottom

By putting D to the primitive equation, we get, Template:Center top$U=-\frac{R^2}{4\mu }\left(\frac{\partial P}{\partial x}\right)[1-{\left(\frac{r}{R}\right)}^2]$Template:Center bottom

Knowing the velocity profile we can evaluate relevant quantities. The shear stress profile will look like:

Template:Center top${\tau }_{r_x}=\mu \frac{dU}{dr}=\frac{r}{2}\left(\frac{\partial P}{\partial x}\right)$Template:Center bottom

at r = 0 $\to {\tau }_{r_x}=0$

at r = R $\to {\tau }_{r_x}=\frac{R}{2}\left(\frac{\partial P}{\partial x}\right)$

File:IF13.png

Shear stress distribution in the fully developed laminar pipe flow.

The volume flow rate would read

Template:Center top$Q=\underset{Area}{\int }{U}_i{n}_{i}dA=\underset{Area}{\int }U2\pi rdr$Template:Center bottom

Template:Center top$Q=\int \frac{1}{4\mu }\left(\frac{\partial P}{\partial x}\right)(r^2-R^2)2\pi rdr$Template:Center bottom

Template:Center top$Q=-\frac{\pi R^4}{8\mu }\left(\frac{\partial P}{\partial x}\right)$Template:Center bottom

When we approximate $\frac{\partial P}{\partial x}=-\frac{\mathrm{\Delta }P}{L}$

Template:Center top$Q=-\frac{\pi R^4}{8\mu }\left[\frac{-\mathrm{\Delta }P}{L}\right]=\frac{\pi \mathrm{\Delta }P{R}^4}{8\mu L}=\frac{\pi \mathrm{\Delta }P{D}^4}{128\mu L}$Template:Center bottom

Template:Center top$\mathrm{\Delta }P=\frac{128\mu L}{\pi D^4}Q$Template:Center bottom

Increase radius to create drastic reduction in the pressure drop.

The mean velocity is:

Template:Center top$\overline{U}=\frac{Q}{A}=\frac{Q}{\pi R^2}=\frac{-R^2}{8\mu }\left(\frac{\partial P}{\partial x}\right)=-\frac{D^2}{32\mu }\left(\frac{\partial P}{\partial x}\right)$Template:Center bottom

The location where maximum velocity occurs can be found be setting:

Template:Center top$\frac{dU}{dr}=0\to \frac{dU}{dr}=\frac{1}{2\mu }\left(\frac{\partial P}{\partial x}\right)r$Template:Center bottom

at r = 0 $\to$ U = $U_{max}$.

Template:Center top$U_{max}=U(0)=-\frac{R^2}{4\mu }\left(\frac{\partial P}{\partial x}\right)=2\overline{U}$Template:Center bottom

Note that in a channel was $U_{max}=\frac{3}{2}\overline{U}$.

$U$ can be written as a function of $U_{max}$ i.e.

Template:Center top$U=\underset{U_{max}}{\underbrace{-\frac{R^2}{4\mu }\left(\frac{\partial P}{\partial x}\right)}}[1-{\left(\frac{r}{R}\right)}^2]$Template:Center bottom

Template:Center top$\frac{U}{U_{max}}=[1-{\left(\frac{r}{R}\right)}^2]$Template:Center bottom

Again, the velocity profile becomes parabolic.