Fluid Mechanics for Mechanical Engineers/Internal Flows

>back to Chapters of Fluid Mechanics for Mechanical Engineers

Internal and External 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(Reynold Number). There might be an analytical solution for laminar flows but not for turbulent flows.

Laminar and Turbulent Flows in Channels and Pipes


At fully developed state the velocity profile becomes parabolic for laminar flow. The average velocity at any cross section is:$$ \bar{U} = U_{0}= V = \frac{1}{A}\int_{Area}{U_{1}dA} $$ 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 $$Re=\frac{U_{0}D}{\nu}$$.



Concept of Fully Developed Flow
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 the fully developed state:

$$\frac{\partial U_{i}}{\partial x_{1}} = 0 \rightarrow \frac{\partial \tau_{ij}}{\partial x_{1}} = 0$$

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.

$$U_{3}= 0\, \ \frac{\partial U_{i}}{\partial x_{3}}= 0\ , \ \frac{\partial \tau_{ij}}{\partial x_{3}}= 0\ , \ \frac{\partial P}{\partial x_{3}}= 0,$$ $$U_{2}= U_{3}= 0$$

Hence,

$$U_{1} = U_{1}(x_{2})\ ,\ \tau_{ij}=\tau_{ij}(x_2).$$

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

$$	\frac{Le_{pipe}}{D} = \left[(0.619)^{1.6} + (0.0567Re)^{1.6}\right]^{\frac{1}{1.6}} $$

$$ \frac{Le_{channel}}{D} = \left[(0.631)^{1.6} + (0.0442Re)^{1.6}\right]^{\frac{1}{1.6}} $$

Fully Developed Laminar Flow Between Infinite Parallel Plates


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

$$ \underbrace{\frac{\partial \rho}{\partial t}}_{= 0} + \frac{\partial \rho U_{i}}{\partial x_{i}} = 0\ \stackrel{\rho = cst.}{\longrightarrow}\ \frac{\partial U_{i}}{\partial x_{i}} = 0 $$

Since $$\frac{\partial U_{1}}{\partial x_{1}} = 0$$, $$\frac{\partial U_{3}}{\partial x_{3}} = 0 \rightarrow \frac{\partial U_{2}}{\partial x_{2}} = 0$$ because it is a fully-developed and two dimensional flow. Hence, $$\ U_2$$ reads $$\ U_2=constant $$ As $$\ U_2$$ is zero on the walls, it should be zero in the whole fully developed region, i.e. $$\ U_2(x_2)=0 $$

Momentum Equation in j-direction 

$$ \underbrace{\rho\frac{\partial U_{j}}{\partial t}}_{= 0\ Steady} + \rho U_{i}\frac{\partial U_{j}}{\partial x_{i}} = -\frac{\partial P}{\partial x_{j}} + \rho g_{j} + \frac{\partial \tau_{ij}}{\partial x_{i}} $$

in $$x_{1}$$ direction ,$$g_{1} = 0$$

$$\underbrace{\rho U_{i}\frac{\partial U_{1}}{\partial x_{i}}}_{A} = -\frac{\partial P}{\partial x_{1}} + \underbrace{\frac{\partial \tau_{i1}}{\partial x_{i}}}_{B}$$

Consider term A:

$$\rho U_{i}\frac{\partial U_{1}}{\partial x_{i}} = \rho \left[U_{1}\underbrace{\frac{\partial U_{1}}{\partial x_{1}}}_{= 0\ Fully-developed} + \underbrace{U_{2}}_{= 0}\frac{\partial U_{1}}{\partial x_{2}} + \underbrace{U_{3}}_{=0\ fully-developed\ 2D }\underbrace{\frac{\partial U_{1}}{\partial x_{3}}}_{= 0\ 2D}\right]=0$$

Consider term B:

$$\frac{\partial \tau_{i1}}{\partial x_{i}} = \underbrace{\frac{\partial \tau_{11}}{\partial x_{1}}}_{= 0\ fully-dev.} + \frac{\partial \tau_{21}}{\partial x_{2}} + \underbrace{\frac{\partial \tau_{31}}{\partial x_{3}}}_{= 0\ 2D} = \frac{\partial \tau_{21}}{\partial x_{2}} = \frac{d\tau_{21}}{dx_{2}}$$

hence $$\tau_{21} = \tau_{21}(x_{2})$$.

Thus the momentum equation in $$x_{1}$$ direction reads:

$$0 = -\frac{\partial P}{\partial x_{1}} + \frac{d\tau_{21}}{dx_{2}}$$

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.

$$ \tau_{ij} = \mu \left(\frac{\partial U_{i}}{\partial x_{j}} + \frac{\partial U_{j}}{\partial x_{i}}\right) - \frac{2}{3}\delta_{ij}\mu\frac{\partial U_{k}}{\partial x_{k}} $$

thus,

$$\tau_{21} = \mu\left(\underbrace{\frac{\partial U_{2}}{\partial x_{1}}}_{= 0} + \frac{\partial U_{1}}{\partial x_{2}}\right) + 0$$

$$\tau_{21} = \mu\frac{\partial U_{1}}{\partial x_{2}}\ \text{since,}\ U_{1} = U_{1}(x_{2})\ \tau_{21} = \mu\frac{dU_{1}}{dx_{2}}$$

Thus the momentum equation reads:

$$ \frac{\partial P}{\partial x_{1}} = \mu\frac{d^{2}U_{1}}{dx_{2}^{2}} $$

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

The momentum equation in $$x_{1}$$ direction,

$$ F_{S1} + \underbrace{F_{B1}}_{=0} = \frac{\partial}{\partial t} \underbrace{\int_{CV}\rho U_{1}dV}_{=0} + \underbrace{\int_{CS}U_{1}\rho U_{i}n_{i}dA}_{=0\ fully-dev.} $$

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

$$\displaystyle F_{S1} = 0$$

That is:

$$ \left(p - \frac{\partial P}{\partial x_{1}}\frac{dx_{1}}{2}\right)dx_{2}dx_{3} - \left(p + \frac{\partial P}{\partial x_{1}}\frac{dx_{1}}{2}\right)dx_{2}dx_{3}$$

$$ + \left(\tau_{21} + \frac{dx_{2}}{2}\frac{\partial \tau_{21}}{\partial x_{2}}\right)dx_{1}dx_{3} - \left(\tau_{21} - \frac{dx_{2}}{2}\frac{\partial \tau_{21}}{\partial x_{2}}\right)dx_{1}dx_{3} = 0 $$

$$-\frac{\partial P}{\partial x_{1}}dV + \frac{\partial \tau_{21}}{\partial x_{2}}dV = 0$$

$$-\frac{\partial P}{\partial x_{1}} + \frac{\partial \tau_{21}}{\partial x_{2}} = 0$$

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

$$ \frac{\partial P}{\partial x_{1}} = \mu\frac{d^{2}U_{1}}{dx_{2}^{2}} $$

with the following boundary conditions:

$$x_{2} = 0 \rightarrow U_{1} = 0$$ and $$x_{2} = D \rightarrow U_{1} = 0$$

Integrating the equation once results in a linear function of $$x_{2}$$:

$$ \mu\frac{dU_{1}}{dx_{2}} = \left(\frac{\partial P}{\partial x_{1}}\right)x_{2} + c_{1} \rightarrow \tau_{21} = \left(\frac{\partial P}{\partial x_{1}}\right)x_{2} + c_{1} $$

The second integration reads:

$$ U_{1} = \frac{1}{2\mu}\left(\frac{\partial P}{\partial x_{1}}\right)x_{2}^{2} + \frac{1}{\mu} c_{1}x_{2} + c_{2} $$

The integration constants is obtained by using the boundary conditions:

$$x_{2} = 0,\ U_{1} = 0 \rightarrow c_{2} = 0$$

$$x_{2} = D,\ U_{1} = 0 \rightarrow c_{1} = -\frac{1}{2}\left(\frac{\partial P}{\partial x_{1}}\right)D$$

Finally, the velocity profile reads:

$$ \begin{array}{lll} U_{1} &=& \frac{1}{2\mu}\left(\frac{\partial P}{\partial x_{1}}x_{2}^{2}\right) - \frac{1}{2\mu}\left(\frac{\partial P}{\partial x_{1}}\right)D{x_{2}}\\ &=&\frac{D^{2}}{2\mu}\left(\frac{\partial P}{\partial x_1}\right) \left[\left(\frac{x_{2}}{D}\right)^{2} - \left(\frac{x_{2}}{D}\right)\right] \end{array} $$

Note that the velocity profile is parabolic!

The shear stress becomes:

$$ \begin{array}{lll} \tau_{21} &=& \left(\frac{\partial P}{\partial x_{1}}\right)x_{2} - \frac{1}{2}\left(\frac{\partial P}{\partial x_{1}}\right)D \\ &=& D\left(\frac{\partial P}{\partial x_{1}}\right)\left[\frac{x_{2}}{D} - \frac{1}{2}\right] \end{array} $$

at the wall i.e. at $$x_{2}$$ = 0 and $$x_{2}$$= D

$$ \tau_{21}(0) = -\frac{1}{2}D\left(\frac{\partial P}{\partial x_{1}}\right) $$

$$ \tau_{21}(D) = \frac{1}{2}D\left(\frac{\partial P}{\partial x_{1}}\right) $$



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,

$$ Q = \int_{A}U_{i}n_{i}dA = \int_{0}^{D}U_{1}w dx_{2} $$

where $$w$$ is the depth of the channel.

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

$$ \frac{Q}{w} = \int_{0}^{D}\frac{1}{2\mu}\left(\frac{\partial P}{\partial x_{1}}\right)\left(x^{2} - D{x_{2}}\right)dx_{2} $$

$$ \frac{Q}{w} = -\frac{1}{12\mu}\left(\frac{\partial P}{\partial x_{1}}\right)D^{3} $$

Note that $$\frac{\partial P}{\partial x_{1}}$$ should be constant for the fully developed flow. Hence, for a channel with a finite length $$L$$:

$$\frac{\partial P}{\partial x_{1}} = \frac{p_{2} - p_{1}}{L} = \frac{-\Delta P}{L}$$

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

$$\frac{Q}{w} = -\frac{1}{12\mu}\left(\frac{-\Delta P}{L}\right)D^{3} = \frac{D^{3}}{12\mu L}\Delta P$$

or the pressure drop can be calculated from:

$$ \Delta P = \frac{Q}{w}\frac{12\mu L}{D^{3}} $$

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

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

$$ \bar{U} = U_{0} = \frac{Q}{A} = \frac{Q}{w D} = -\frac{1}{12\mu}\left(\frac{\partial P}{\partial x_{1}}\right)\frac{D^{3}w}{w D} = -\frac{1}{12\mu}\left(\frac{\partial P}{\partial x_{1}}\right)D^{2} $$

The maximum velocity occurs when:

$$ \frac{dU_{1}}{dx_{2}} = 0 = \frac{D^{2}}{2\mu}\left(\frac{\partial P}{\partial x_{1}}\right)\left(\frac{2x_{2}}{D^{2}} - \frac{1}{D}\right) $$

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

$$U_{1}\left(\frac{D}{2}\right) = U_{1max} = -\frac{1}{8\mu}\left(\frac{\partial P}{\partial x_{1}}\right)D^{2} = \frac{3}{2}\bar{U}$$

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

$$ \begin{array}{lll} U_{1} &=& -4 U_{1max} \left[\left(\frac{x_{2}}{D}\right)^{2} - \left(\frac{x_{2}}{D}\right)\right]\\ &=&-6 \overline{U} \left[\left(\frac{x_{2}}{D}\right)^{2} - \left(\frac{x_{2}}{D}\right)\right] \end{array} $$

Same problem can be solved by using moving plates.

Example


Consider the hydraulic control valve comprising 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 $$\frac{kg}{m\ sec}$$ at 55ºC). The flow can be assumed to be laminar, steady, incompressible, fully-developed flow. $$\left(\frac{L}{a} = 3000\right)$$

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:

$$\frac{Q}{w} = \frac{a^{3}\Delta P}{12\mu L}$$

$$Q = \frac{a^{3}\Delta P}{12\mu L}\pi D = \frac{(0.005\cdot 10^{-3})^{3}}{12*(0.018)*15\cdot10^{-3}}*(20-1)\cdot10^{6}* \pi * 25\cdot10^{-3}$$

$$Q = 57.6\cdot10^{-9} \frac{m^{3}}{s} = 57.6 \frac{mm^{3}}{sec}$$

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

$$\bar{U} = \frac{Q}{A} = \frac{Q}{\pi D\ a} = 0.147 \frac{m}{sec}$$

$$Re = \frac{\rho \bar{U}a}{\mu} = \frac{932*0.147*0.005\cdot10^{-3}}{0.018} = 0.0375$$

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

Layered Channel Flow


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 $$\displaystyle \mu_{A}$$, a density $$\displaystyle \rho_{A}$$ and the mass flow $$\displaystyle \dot{m}_{A}$$. Fluid B, which is located above fluid A, has a viscosity $$\displaystyle \mu_{B}$$, a density $$\displaystyle \rho_{B}$$ and the mass flow $$\displaystyle \dot{m}_{B}$$. The following differential equations correspond to the molecular momentum $$\displaystyle \tau_{21}$$ for each Fluid.

$$ \frac{d\tau_{21}^{A}}{dx_{2}} = \frac{dP}{dx_{1}}$$   and    $$\frac{d\tau_{21}^{B}}{dx_{2}} = \frac{dP}{dx_{1}} $$.

With $$\tau_{21}= \mu \frac{dU_{1}}{dx_{2}}$$ yields the velocity field:

$$\frac{d^{2}U_{1}^{A}}{dx_{2}} = \frac{1}{\mu_{A}}\frac{dP}{dx_{1}} $$   and  $$  \frac{d^{2}U_{1}^{B}}{dx_{2}} = \frac{1}{\mu_{B}}\frac{dP}{dx_{1}}$$

After integration of both equations we obtain:

$$\tau_{21}^{A} = \frac{dP}{dx_{1}}x_{2} + C_{1}^{A}$$ and $$\tau_{21}^{B} = \frac{dP}{dx_{1}}x_{2} + C_{1}^{B}$$

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,

$$C^{A}_{1} = C_{1}^{B} = C_{1}$$

After the integration for the velocity field:

$$ U_{1}^{A} = \frac{1}{2\mu_{A}}\frac{dP}{dx_{1}}x_{2}^{2} + \frac{C_{1}}{\mu_{A}}x_{2} + C_{2}^{A}$$

and

$$ U_{1}^{B} = \frac{1}{2\mu_{B}}\frac{dP}{dx_{1}}x_{2}^{2} + \frac{C_{1}}{\mu_{B}}x_{2} + C_{2}^{B}$$

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

$$ U_{1}^{A}(x_{2} = 0) = U_{1}^{B}(x_{2} = 0)$$, therefore,

$$C_{2}^{A} = C_{2}^{B} = C_{2}$$. The integration constants can be calculated with the following boundary conditions:

At $$ x_{2} = -D \ \rightarrow \ \ U_{1}^{A} = 0$$:

$$ 0 = \frac{dP}{dx_{1}}\frac{1}{2\mu_{A}}D^{2} - \frac{C_{1}D}{\mu_{A}} + C_{2}$$

At $$ x_{2} = +D\ \rightarrow \ \ U_{1}^{B} = 0$$:

$$ 0 = \frac{dP}{dx_{1}}\frac{1}{2\mu_{B}}D^{2} + \frac{C_{1}D}{\mu_{B}} + C_{2}$$

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

$$U_{1}^{A} = -\frac{D^{2}}{2\mu_{A}}\frac{dP}{dx_{1}} \left[+\frac{2\mu_{A}}{(\mu_{A} + \mu_{B})} + \left(\frac{\mu_{A} - \mu_{B}}{\mu_{A}+ \mu_{B}}\right)\left(\frac{x_{2}}{D}\right) - \left(\frac{x_{2}}{D}\right)^{2}\right] \ \ \ \ \ $$

and

$$U_{1}^{B} = -\frac{D^{2}}{2\mu_{B}}\frac{dP}{dx_{1}} \left[+\frac{2\mu_{B}}{(\mu_{A} + \mu_{B})} + \left(\frac{\mu_{A} - \mu_{B}}{\mu_{A}+ \mu_{B}}\right)\left(\frac{x_{2}}{D}\right) - \left(\frac{x_{2}}{D}\right)^{2}\right]$$

For the distribution of the shear stress we get:

$$\tau_{21} = D\frac{dP}{dx_{1}}\left[\left(\frac{x_{2}}{D}\right) - \frac{1}{2} \left(\frac{\mu_{A} - \mu_{B}}{\mu_{A}+ \mu_{B}}\right)\right]$$

If we choose $$\displaystyle\mu_{A} = \mu_{B}$$,

$$U_{1} = \frac{-D^{2}}{2\mu_{A}}\frac{dP}{dx_{1}}\left[1 - \left(\frac{x_{2}}{D}\right)^{2}\right]$$

$$\tau_{21} = D\frac{dP}{dx_{1}}\left(\frac{x_{2}}{D}\right)$$

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 $$\displaystyle\tau_{21}$$, where $$\displaystyle\tau_{21} = 0$$ at the channel's centerline.

If $$\mu_{A} \neq \mu_{B}$$, the position where the maximal velocity occurs can be calculated by introducing $$\displaystyle\tau_{21} = 0$$ on the velocity profile equation:

$$ x_2(U_{1 max}) = \frac{D}{2}\left(\frac{\mu_{A} - \mu_{B}}{\mu_{A}+ \mu_{B}}\right) $$

The shear stress on the upper plate is:

$$ \tau_{W}^{B} = \frac{D}{2}\frac{dP}{dx_{1}}\left[\frac{\mu_{A} + 3\mu_{B}}{\mu_{A} + \mu_{B}}\right]$$

and the shear stress on the lower plate reads:

$$ \tau_{W}^{A} = -\frac{D}{2}\frac{dP}{dx_{1}}\left[\frac{3\mu_{A} + \mu_{B}}{\mu_{A} + \mu_{B}}\right]$$

The average velocities of the fluids A and B are:

$$\tilde{U}_{1}^{A} = -\frac{D^{2}}{12\mu_{A}}\frac{dP}{dx_{1}}\left(\frac{7\mu_{A} + \mu_{B}}{\mu_{A} + \mu_{B}}\right)$$

and

$$\tilde{U}_{1}^{B} = -\frac{D^{2}}{12\mu_{B}}\frac{dP}{dx_{1}}\left(\frac{\mu_{A} + 7\mu_{B}}{\mu_{A} + \mu_{B}}\right)$$

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 Cylindrical Coordinates
A change of variables on the Cartesian equations will yield the following equations of momentum in r, $$\theta$$, and x directions for incompressible and isothermal flows (constant density and viscosity):



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



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



x:\;\;\rho \left(\frac{\partial u_x}{\partial t} + u_r \frac{\partial u_x}{\partial r} + \frac{u_{\theta}}{r} \frac{\partial u_x}{\partial \theta} + u_x \frac{\partial u_x}{\partial x}\right) = -\frac{\partial p}{\partial x} + \mu \left[\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 \theta^2} + \frac{\partial^2 u_x}{\partial x^2}\right] + \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 \theta}\left(\rho u_\theta\right) + \frac{\partial}{\partial x} \left(\rho u_x\right) = 0.$$

Fully Developed Pipe Flow


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.

$$\left(p - \frac{\partial P}{\partial x}\frac{dx}{2}\right)\pi r^{2} - \left(p + \frac{\partial P}{\partial x}\frac{dx}{2}\right)\pi r^{2} + \tau_{r_{x}}2\pi rdx = 0$$

$$-\frac{\partial P}{\partial x}dx \pi r^{2} + \tau_{r_{x}}2\pi rdx = 0$$

Hence, $$ \tau_{r_{x}} = \frac{r}{2}\frac{\partial P}{\partial x} $$which shows that the stress has a negative value and therefore it is in the negative x-direction.

(ii) Infinitesimal thin hollow cylinder at the center
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

$$p2\pi rdr -\left(p + \frac{\partial P}{\partial x}{dx}\right)2\pi rdr\rightarrow -\left(\frac{\partial P}{\partial x}\right) 2\pi rdrdx$$

considering viscous shear stress on the surface of the cylinder $$\left(\tau_{r_{x}}+\frac{\partial \tau_{r_{x}}}{\partial r}dr\right)2\pi(r+dr)dx - \tau_{r_{x}}2\pi rdx$$ $$=\tau_{r_{x}} 2\pi rdx + \tau_{r_{x}} 2\pi drdx + \left(\frac{\partial \tau_{r_{x}}}{\partial r}\right)2\pi rdrdx + \underbrace{\left(\frac{\partial \tau_{r_{x}}}{\partial r}\right)2\pi r{dr}^2dx}_{{dr}^2\approx 0} - \tau_{r_{x}} 2\pi rdx $$ $$ \rightarrow \tau_{r_{x}} 2\pi drdx + \left(\frac{\partial \tau_{r_{x}}}{\partial r}\right)2\pi rdrdx $$

combining both term,we get the balance equation, $$-\left(\frac{\partial P}{\partial x}\right) 2\pi rdrdx + \tau_{r_{x}} 2\pi drdx + \left(\frac{\partial \tau_{r_{x}}}{\partial r}\right)2\pi rdrdx = 0 $$

it could be rewritten as $$-\left(\frac{\partial P}{\partial x}\right) 2\pi rdrdx +(\tau_{r_{x}} + r \frac{\partial \tau_{r_{x}}}{\partial r})2\pi drdx = 0 $$ dividing both side with $$2\pi drdx $$,we get $$-\left(\frac{\partial P}{\partial x}\right)r + \frac{\partial (r \tau_{r_{x}})}{\partial r} = 0$$ or,$$\left(\frac{\partial P}{\partial x}\right)r = \frac{\partial (r \tau_{r_{x}})}{\partial r} $$ integrating both side ,

$$\int \left(\frac{\partial P}{\partial x}\right)r\, dr = \int \frac{\partial (r \tau_{r_{x}})} {\partial r}  dr $$

or,$$\left(\frac{\partial P}{\partial x}\right)\frac{r^2}{2}=r \tau_{r_{x}}$$

or,$$\left(\frac{\partial P}{\partial x}\right)\frac{r}{2}= \tau_{r_{x}}$$

However,in a laminar flow!

$$\mu\frac{dU}{dr} = \frac{1}{2}r\frac{\partial P}{\partial x}$$

integrating,

$$U = \frac{r^{2}}{4\mu}\left(\frac{\partial P}{\partial x}\right) + c_{1}$$

The boundary condition is:

$$U = 0\ at\ r = R$$

Thus $$c_{1}$$ can be calculated from the boundary condition.

$$ c_{1} = -\frac{R^{2}}{4\mu}\left(\frac{\partial P}{\partial x}\right) \Rightarrow U = \frac{1}{4\mu}\left(\frac{\partial P}{\partial x}\right)(r^{2} - R^{2}) $$

or

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

(iii) Using NS in cylindrical co-ordinates:
$$ x:\rho \left(\underbrace{\frac{\partial u_x}{\partial t}}_{steady state} + \underbrace{u_r \frac{\partial u_x}{\partial r}}_{u_r=0} + \underbrace{\frac{u_{\theta}}{r} \frac{\partial u_x}{\partial \theta}}_{u_{\theta}=0}+\underbrace {u_x \frac{\partial u_x}{\partial x}}_{\frac{\partial u_x}{\partial x} =0}\right) = -\frac{\partial p}{\partial x} + \mu \left[\frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial u_x}{\partial r}\right) + \underbrace{\frac{1}{r^2}\frac{\partial^2 u_x}{\partial \theta^2}}_{=0} + \underbrace{\frac{\partial^2 u_x}{\partial x^2}}_{=0}\right] + \underbrace{\rho g_x}_{g_x=0}$$

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

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

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

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

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

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

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

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

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

$$ \tau_{r_{x}} = \mu\frac{dU}{dr} = \frac{r}{2}\left(\frac{\partial P}{\partial x}\right) $$

at r = 0 $$\rightarrow \tau_{r_{x}} = 0$$

at r = R $$\rightarrow \tau_{r_{x}} = \frac{R}{2}\left(\frac{\partial P}{\partial x}\right)$$



The volume flow rate would read

$$\displaystyle Q = \int_{Area} U_{i}n_{i}dA = \int_{Area} U2\pi rdr$$

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

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

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

$$Q = -\frac{\pi R^{4}}{8\mu}\left[\frac{-\Delta P}{L}\right] = \frac{\pi \Delta PR^{4}}{8\mu L} = \frac{\pi \Delta PD^{4}}{128\mu L}$$

$$ \Delta P = \frac{128\mu L}{\pi D^{4}}Q $$

Increase radius to create drastic reduction in the pressure drop.

The mean velocity is:

$$ \bar{U} = \frac{Q}{A} = \frac{Q}{\pi R^{2}} = \frac{-R^{2}}{8\mu}\left(\frac{\partial P}{\partial x}\right) = -\frac{D^{2}}{32\mu}\left(\frac{\partial P}{\partial x}\right) $$

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

$$\frac{dU}{dr} = 0 \rightarrow \frac{dU}{dr} = \frac{1}{2\mu}\left(\frac{\partial P}{\partial x}\right)r$$

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

$$ U_{max} = U(0) = -\frac{R^{2}}{4\mu}\left(\frac{\partial P}{\partial x}\right) = 2\bar{U} $$

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

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

$$ U = \underbrace{-\frac{R^{2}}{4\mu}\left(\frac{\partial P}{\partial x}\right)}_{U_{max}}\left[1 - \left(\frac{r}{R}\right)^{2}\right] $$

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

Again, the velocity profile becomes parabolic.