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Turbulent flow through tubes
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If the Reynolds number exceeds 2000, the flow in a tube always becomes unstable and eventually becomes fully turbulent. As a result, it is no longer possible to use the viscous laminar flow approximation that gives rise to the Hagen-Poiseuille law, and an alternative model is required.
The model that describes a flow where viscosity is irrelevant is the one that gives rise to Bernoulli's equation. However, this model assumes that energy density is conserved. An alternative is to assume that turbulence leads to mixing in such a way that energy density is not conserved but remains constant. In this case, the flow can be modeled using an equation similar to Bernoulli's, but with a correction to account for homogenization due to mixing effects.
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Turbulence modeling
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Laminar flow is described by "sheets" that move at different speeds in a coordinated manner. In turbulent flow, however, these kinds of sheets do not exist. Fluid elements are deflected, can change direction, and participate in circular motions, often in a chaotic manner.
The first consequence of this is that fluid parameters tend to mix, causing the disappearance of velocity differences and the emergence of a type of mean velocity. When averaging the motion, structured patterns emerge, but they no longer originate from individual fluid elements but rather from a temporal mixture. As a result, a relatively constant profile similar to the profiles defined in laminar flow emerges, but in this case, these are mean values with fewer large gradients, making them more uniform.
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Darcy-Weisbach equation
Equation
When modeling flow in a tube while assuming that the density of energy is conserved, we obtain the Bernoulli equation, which in this case describes the flow as:
| $ \Delta p = - \rho \bar{v} \Delta v $ |
where $\Delta p$ is the pressure difference, $\rho$ is the density, and $v$ is the velocity.
In turbulent flow, the mixing process acts as friction, reducing the velocity gradient that exists in laminar flow between the center and the edge of the flow. Assuming that this mixing factor can be empirically modeled with a correction factor, we empirically arrive at the Darcy-Weisbach equation:
 $ \Delta p = \displaystyle\frac{ \Delta L }{ D_H } f_D \displaystyle\frac{1}{2} \rho_w v ^2 $ |
where $f_D$ is the Darcy friction factor, $\Delta L$ is the length, and $D_H$ is the hydraulic diameter of the tube.
The friction factor has been empirically determined for various situations and is expressed as a function of the Reynolds number.
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Hydraulic diameter
Equation
In the case of the Darcy-Weisbach equation, we work with a hydraulic diameter that corresponds to a generalization of the traditional diameter of a circle. This allows us to take a non-circular section and calculate a diameter based on the area of the section, denoted as $S$, and its perimeter, denoted as $P$, using the following formula:
| $ D_H = \displaystyle\frac{ 4 S }{ P }$ |
For the case of a circular section, we obtain the traditional diameter of a circle as follows:
$D_H = \displaystyle\frac{4S}{P} = \displaystyle\frac{4\pi R^2}{2\pi R} = 2R$
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Hydraulic radius
Equation
In the context of the Darcy-Weisbach friction factor, we work with a hydraulic radius that corresponds to a generalization of the traditional radius of a circle. This allows us to calculate a diameter based on the area of the cross-section $S$ and its perimeter in contact with the liquid $P$ using the following formula:
| $ R_H = \displaystyle\frac{ S }{ P_H }$ |
When dealing with a circular section, we obtain the traditional radius of a circle as follows:
$R_H = \displaystyle\frac{S}{P} = \displaystyle\frac{\pi R^2}{2 \pi R} = \displaystyle\frac{1}{2} R$
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Depth of an unfilled tube
Equation
In a cylindrical tube, the depth depends on the flow as follows:
If we integrate the cross-section, we can calculate how the surface area depends on the depth using the following equation:
$S = 2\displaystyle\int_0^h dz \sqrt{2Rz - z^2}=\displaystyle\frac{1}{2}(R-h)\sqrt{2Rh - h^2}+\displaystyle\frac{1}{2}R^2\arcsin\left(\displaystyle\frac{1-h}{R}\right)$
For small flows, where the depth is significantly smaller than the radius, the relationship between the cross-section and the depth simplifies significantly. When solving for the depth, we obtain:
| $ h = \left(\displaystyle\frac{3^2 S^2 }{2^3 R }\right)^{1/3}$ |
The cross-sectional area of the tube containing the liquid can be integrated with respect to height, yielding:
$S = 2\displaystyle\int_0^h dz \sqrt{2Rz - z^2}=\displaystyle\frac{1}{2}(R-h)\sqrt{2Rh - h^2}+\displaystyle\frac{1}{2}R^2\arcsin\left(\displaystyle\frac{1-h}{R}\right)$
If we expand this expression in terms of the factor $h/R$ in the limit $h\ll R$, we obtain, to first order:
$S = \sqrt{\displaystyle\frac{2^3}{3^2} R h ^3}$
If we solve for the depth, we finally obtain:
| $ h = \left(\displaystyle\frac{3^2 S^2 }{2^3 R }\right)^{1/3}$ |
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Hydrodynamic perimeter in unfilled tube
Equation
The hydrodynamic perimeter of a partially filled tube corresponds to the edges of the section in contact with the liquid, namely, the arc touching the tube wall and the surface:
In general, this can be expressed as:
$P_H = 2 R \arccos\left(1-\displaystyle\frac{h}{R}\right) + 2\sqrt{2Rh-h^2}$
For small flows, where the depth is significantly smaller than the radius, this simplifies the relationship between the cross-section and the depth to:
| $ P_H = \sqrt{2^5 R h }$ |
Given that the angle can be determined as:
$\phi =\arccos\left(1-\displaystyle\frac{h}{R}\right)$
The arc is equal to $R\phi$, therefore, the total arc length is:
$2 R \arccos\left(1-\displaystyle\frac{h}{R}\right)$
Similarly, half of the surface can be determined using the Pythagorean theorem, resulting in:
$\sqrt{2Rh - h^2}$
Thus, the hydrodynamic perimeter is expressed as:
$S = 2\displaystyle\int_0^h dz \sqrt{2Rz - z^2}=\displaystyle\frac{1}{2}(R-h)\sqrt{2Rh - h^2}+\displaystyle\frac{1}{2}R^2\arcsin\left(1-\displaystyle\frac{h}{R}\right)$
In the limit of a small height, where $h\ll R$, this expression can be developed, resulting in:
| $ P_H = \sqrt{2^5 R h }$ |
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Moody diagram
Image
In 1944, Lewis Ferry Moody measured the Darcy-Weisbach friction factor as a function of the Reynolds number and the relative roughness of the wall, resulting in the creation of the following diagram:
The relative roughness can be estimated by considering the size of surface irregularities (height of protrusions or depths of grooves) in relation to the hydraulic diameter.
Two distinct behaviors are observed:
• For Reynolds numbers below 2000, the Darcy-Weisbach friction factor depends solely on the Reynolds number, following a relationship of $64/Re$. This corresponds to the laminar flow regime.
• For Reynolds numbers above 2000, a behavior is observed that depends on both the Reynolds number and the relative roughness of the tube's surface.
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Laminar limit
Equation
At low Reynolds numbers, the Moody diagram shows that the Darcy-Weisbach friction factor, $f_D$, is equal to:
| $ f_D = \displaystyle\frac{64}{ Re }$ |
where $Re$ is the Reynolds number. This applies to Reynolds numbers up to 2000. Beyond this value, the wall roughness starts to affect flow destabilization and turbulence formation.
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Flow at the laminar boundary
Description
If we replace the Darcy-Weisbach friction factor in the laminar limit, given by
| $ f_D = \displaystyle\frac{64}{ Re }$ |
in the Darcy-Weisbach equation, represented as
| $ \Delta p = \displaystyle\frac{ \Delta L }{ D_H } f_D \displaystyle\frac{1}{2} \rho_w v ^2 $ |
and employ the definition of the Reynolds number $Re$, we can demonstrate that the flow is governed by
| $ J_V =-\displaystyle\frac{ \pi R ^4}{8 \eta }\displaystyle\frac{ \Delta p }{ \Delta L }$ |
which corresponds to the Hagen-Poiseuille equation.
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Colebrook-White equation in closed tube
Equation
When the pipe is completely closed, meaning the cylinder has no upper opening, and the liquid fills the entire section, the Colebrook-White equation
| $\displaystyle\frac{1}{\sqrt{ f_D }}=-2\log\left(\displaystyle\frac{\epsilon}{3.7 D_H}+\displaystyle\frac{2.51}{ Re \sqrt{ f_D }}\right)$ |
allows for estimating the Darcy-Weisbach friction factor in turbulent flow based on the roughness $\epsilon$, the hydraulic diameter, and the Reynolds number.
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Solution for closed tube
Equation
The Colebrook-White equation for the case of a closed tube:
| $\displaystyle\frac{1}{\sqrt{ f_D }}=-2\log\left(\displaystyle\frac{\epsilon}{3.7 D_H}+\displaystyle\frac{2.51}{ Re \sqrt{ f_D }}\right)$ |
is an implicit equation used to determine the Darcy-Weisbach friction factor ($f_D$). To solve this equation, various approximations have been developed, which vary in complexity and accuracy. One of the most effective approximations that covers a wide range of Reynolds numbers $Re$ is the one proposed by S.E. Haaland:
| $ f_D = \displaystyle\frac{25}{81 \log\left(\left(\displaystyle\frac{ \epsilon }{3.7 D_H }\right)^{1.11} + \displaystyle\frac{6.9}{ Re }\right)^2}$ |
The original solution by S.E. Haaland is as follows:
$\displaystyle\frac{1}{\sqrt{f_D}}=-1.8\log\left(\left(\displaystyle\frac{\epsilon/D_H}{3.7}\right)^{1.11} + \displaystyle\frac{6.9}{Re}\right)$
It can be rearranged to obtain the expression for the Darcy-Weisbach friction factor $f_D$ as follows:
| $ f_D = \displaystyle\frac{25}{81 \log\left(\left(\displaystyle\frac{ \epsilon }{3.7 D_H }\right)^{1.11} + \displaystyle\frac{6.9}{ Re }\right)^2}$ |
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Colebrook-White equation in open tube
Equation
When the pipe is open, meaning the liquid doesn't fill the entire cylinder section, the Colebrook-White equation
| $\displaystyle\frac{1}{\sqrt{ f_D }}=-2\log\left(\displaystyle\frac{\epsilon}{12 R_H}+\displaystyle\frac{2.51}{ Re \sqrt{ f_D }}\right)$ |
allows for estimating the Darcy-Weisbach friction factor in turbulent flow based on the roughness $\epsilon$, the hydraulic diameter, and the Reynolds number.
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Solution for open tube
Equation
The Colebrook-White equation for the case of a closed tube:
| $\displaystyle\frac{1}{\sqrt{ f_D }}=-2\log\left(\displaystyle\frac{\epsilon}{12 R_H}+\displaystyle\frac{2.51}{ Re \sqrt{ f_D }}\right)$ |
is an implicit equation used to determine the Darcy-Weisbach friction factor ($f_D$). To solve this equation, various approximations have been developed, which vary in complexity and accuracy. One of the most effective approximations that covers a wide range of Reynolds numbers $Re$ is the one proposed by S.E. Haaland:
| $ f_D = \displaystyle\frac{25}{81 \log\left(\left(\displaystyle\frac{ \epsilon }{12 R_H }\right)^{1.11} + \displaystyle\frac{6.9}{ Re }\right)^2}$ |
The original solution by S.E. Haaland is as follows:
$\displaystyle\frac{1}{\sqrt{f_D}}=-1.8\log\left(\left(\displaystyle\frac{\epsilon/R_H}{12}\right)^{1.11} + \displaystyle\frac{6.9}{Re}\right)$
It can be rearranged to obtain the expression for the Darcy-Weisbach friction factor $f_D$ as follows:
| $ f_D = \displaystyle\frac{25}{81 \log\left(\left(\displaystyle\frac{ \epsilon }{12 R_H }\right)^{1.11} + \displaystyle\frac{6.9}{ Re }\right)^2}$ |
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Turbulent flow velocity profile in tube
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The velocity profile of turbulent flow through a tube exhibits two distinct zones based on the distance to the surface ($z$), where $\delta$ represents the thickness of the boundary layer. In the region near the surface ($z < \delta$), the flow is essentially laminar, while in the region further away from the surface ($z > \delta$), the flow becomes turbulent.
In the laminar range, the velocity is proportional to the normalized distance:
$u^+ = y^+$
This relationship is akin to a Hagen-Poiseuille velocity profile near the wall and represents a linear approximation.
In the turbulent range, the normalized velocity profile takes on a logarithmic form:
$u^+ = \displaystyle\frac{1}{\kappa} \ln\left(\displaystyle\frac{y^+}{y_0}\right)$
Here, $\kappa$ is the Karman constant (approximately $0.41$), and $y_0\sim 1/8$ is the normalized distance at which the velocity would be zero. However, it's important to note that the width of the laminar layer, as shown in the graph, is approximately 7.072 times larger than $y_0$.
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Shear speed
Equation
Within the laminar boundary layer, the velocity profile is given by
| $ v = v_{max} \left(1-\displaystyle\frac{ r ^2}{ R ^2}\right)$ |
which, in the limit of distances close to the wall ($r \sim R$), allows us to define the dimensionless distance
$y^+=\displaystyle\frac{\rho_w u_{\tau}}{\eta} y$
and dimensionless velocity
$u^+=\displaystyle\frac{u}{u_{\tau}}$
By normalizing these values, we obtain the logarithmic law of turbulent velocity
| $ u_{\tau} =\sqrt{\displaystyle\frac{ R }{2 \rho_w }\displaystyle\frac{ \Delta p }{ \Delta L }}$ |
In laminar flow, according to Hagen-Poiseuille, the velocity profile is given by
| $ v = v_{max} \left(1-\displaystyle\frac{ r ^2}{ R ^2}\right)$ |
with
| $ v_{max} =-\displaystyle\frac{ R ^2}{4 \eta }\displaystyle\frac{ \Delta p }{ \Delta L }$ |
This yields the relationship for velocity as a function of distance from the wall:
$v=v_{max}\displaystyle\frac{2y}{R}=\displaystyle\frac{R}{2\eta}\displaystyle\frac{\Delta p}{\Delta L}y$
which corresponds to the normalized relation
$u^+=y^+$
With the shear velocity $u_{\tau}$, the normalized velocity is
$u^+=\displaystyle\frac{v}{u_{\tau}}$
and the normalized distance from the wall is
$y^+=\displaystyle\frac{\rho_w u_{\tau}}{\eta} y$
Thus, we have
$u^+=\displaystyle\frac{v}{u_{\tau}}=\displaystyle\frac{R\Delta p}{2\eta\Delta L u_{\tau}}y=y^+=\displaystyle\frac{\rho_w u_{\tau}}{\eta}y$
which leads to
| $ u_{\tau} =\sqrt{\displaystyle\frac{ R }{2 \rho_w }\displaystyle\frac{ \Delta p }{ \Delta L }}$ |
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Turbulent flow velocity profile equation
Equation
Empirically, it has been determined that the velocity in turbulent flow within a tube follows the following form:
$u^+ = \displaystyle\frac{1}{\kappa} \ln\left(\displaystyle\frac{y^+}{y_0}\right)$
Using the shear velocity
| $ u_{\tau} =\sqrt{\displaystyle\frac{ R }{2 \rho_w }\displaystyle\frac{ \Delta p }{ \Delta L }}$ |
we can describe the velocity in turbulent flow through a tube as:
| $ v_y = \displaystyle\frac{ u_{\tau} }{ \kappa }\ln\left(\displaystyle\frac{8 \rho_w u_{\tau} y }{ \eta }\right)$ |
Empirically, it has been determined that the velocity in turbulent flow within a tube follows the following form:
$u^+ = \displaystyle\frac{1}{\kappa} \ln\left(\displaystyle\frac{y^+}{y_0}\right)$
Where the normalized velocity is given by:
$u^+=\displaystyle\frac{v}{u_{\tau}}$
And the normalized distance from the wall is defined as:
$y^+=\displaystyle\frac{\rho_w u_{\tau}}{\eta}y$
With the shear velocity provided by:
| $ u_{\tau} =\sqrt{\displaystyle\frac{ R }{2 \rho_w }\displaystyle\frac{ \Delta p }{ \Delta L }}$ |
This leads us to obtain:
| $ v_y = \displaystyle\frac{ u_{\tau} }{ \kappa }\ln\left(\displaystyle\frac{8 \rho_w u_{\tau} y }{ \eta }\right)$ |
where it is assumed that $y_0 \sim 1/8.
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Laminar flow layer thickness
Equation
The thickness of the laminar boundary layer can be defined based on the condition that laminar velocity profiles match the logarithmic profile corresponding to turbulent flow. This occurs when
$y^+=u^+=\displaystyle\frac{1}{\kappa}\ln\left(\displaystyle\frac{y^+}{y_0}\right)$
whose root is $7.072$. Since this is an estimation of thickness, and the normalization includes a factor of $1/8$, the numbers can be simplified at the first order, and the thickness can be estimated as:
| $ \delta = \displaystyle\frac{ \eta }{ \rho_w u_{\tau} }$ |
In the laminar layer, the flow has the form
$u^+ = y^+$
while in the turbulent region, it is
$u^+=\displaystyle\frac{1}{\kappa}\ln\left(\displaystyle\frac{y^+}{y_0}\right)$
By equating both functions to define the boundary, we obtain the equation
$y^+=\displaystyle\frac{1}{\kappa}\ln\left(\displaystyle\frac{y^+}{y_0}\right)$
whose numerical solution is
$y^+=7.072 \text{ and } y_0=\displaystyle\frac{7.072}{8}\sim 1$
where the empirical value for $y_0=1/8$ was used, and since this is an estimation, we assumed that $7/8\sim 1$.
With the definition of normalization
$y^+=\displaystyle\frac{\rho_w u_{\tau}}{\eta} y = 1$
and solving for $y$, we obtain
| $ \delta = \displaystyle\frac{ \eta }{ \rho_w u_{\tau} }$ |
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Shear stress on walls
Equation
The shear speed, $u_{\tau}$, is directly associated with the shear stress, $\tau_w$, generated by the flow on the walls of the tube:
| $ \tau_w = \rho u_{\tau} ^2$ |
This exerts forces on the walls and affects systems that can move, similar to the case where air flows over water.
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