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I made this problem for more understanding of pressure and pressure loss in vertical flow.

Consider the following steady system, where a fluid enters a tank and exits through a vertical pipe of length $L$ and diameter $D=2R$. Height of fluid in the tank is constant and equals to $H$. Density and viscosity of liquid are $\rho$ and $\mu$, respectively. If flow is laminar find $Q$.

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Now if I write Bernoulli's equation for the free surface of tank and exiting point of pipe, then I get$$\frac{P_{atm}}{\gamma}+\frac{v_0^2}{2g}+z_0=\frac{P_{atm}}{\gamma}+\frac{v^2}{2g}+z+h_l,$$where $h_L$ is the friction loss head of exiting pipe and $v=Q/(\pi R^2)$ and $\gamma= \rho g$. We know that $v_0 \approx 0$, thus$$H+L=\frac{v^2}{2g}+h_L$$ Now we need to find another relation between $v$ and $h_L$. Can we use Darcy–Weisbach equation? I think we can not due to vertical flow! I'm interested in writing momentum balance and derive the relation between friction loss and velocity (like Hagen–Poiseuille equation), but I don't know how to treat pressure terms! Is there a pressure distribution along the exiting pipe?

Edit1: Momentum balance for the laminar flow in the pipe gives the velocity as$$v_z(r)=\frac{R^2}{4 \mu}\left(-\frac{dp}{dz}+ \rho g \right) \left(1-\frac{r^2}{R^2} \right) $$And integrating over cross section of pipe for the flow rate gives$$Q=\pi r^2 v=\int_{0}^{R} 2 \pi r v_z(r) \ dr=\frac{\pi R^4}{8 \mu} \left(-\frac{dp}{dz}+ \rho g \right)$$And finally$$-\frac{dp}{dz}+ \rho g=\frac{32 \mu v}{D^2}$$Now which one of the following is right and why?

1) $h_L=L(-dp/dz)/ \gamma=\frac{32 \mu v L}{\gamma D^2}-L$

2) $h_L=L(-dp/dz+ \rho g)/ \gamma=\frac{32 \mu v L}{\gamma D^2}$

I don't have a sense of $p$ here! Can you give a physical sense of pressure within the exiting pipe?

Edit2: The answers and discussions in this question may solve the following similar questions:

Q1, Q2, Q3.

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    $\begingroup$ Your problem would be solved by explicitly calculating the pressure loss due to friction (which you express as "friction loss head" $h_l$). That depends on whether your flow is fast enough to be turbulent or remains laminar (and hence on its Reynolds number in the pipe). Laminar (slow or viscous) flow is sufficiently easy that if you're pressed, you could get a solution by integration, using the fluid's viscosity; the result will be a force proportional to velocity. For turbulent flows, there are only phenomenological approximations; the force becomes quadratic in velocity. $\endgroup$ – pyramids Jul 31 '16 at 14:50
  • $\begingroup$ @pyramids I want to derive formulas for laminar flow, if its possible. $\endgroup$ – Ghartal Jul 31 '16 at 14:53
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    $\begingroup$ In your differential equation, you are measuring z downward. So, at z = L, the pressure is atmospheric. If you integrate your differential equation subject to this boundary condition, you obtain $$p(z)=p_{atm}-\rho g(L-z)+\frac{32\mu v}{D^2}(L-z)$$The weight of the fluid causes the pressure to increase with z, and the viscous drag causes the pressure to decrease with z. $\endgroup$ – Chet Miller Aug 1 '16 at 12:55
  • $\begingroup$ Oh, this makes sense :) I really appreciate the time you put on answering my question. $\endgroup$ – Ghartal Aug 1 '16 at 16:33
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    $\begingroup$ In Q1 and Q2, for an inviscid fluid, if the pressure is atmospheric at both ends, the fluid will not be able to maintain contact with the walls of the tube. The fluid cross section will become smaller as the fluid moves downward, and its velocity will be increasing. Otherwise, if it is able to maintain contact, then the hydrostatic equation will be satisfied in the vertical tube, and the pressure at the inlet will be less than atmospheric. $\endgroup$ – Chet Miller Aug 2 '16 at 12:06
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You can use the Darcy-Weisbach equation, but you have to modify it a little for vertical flow. In vertical flow, a differential force balance on the flow gives:

$$(P(z+\Delta z)-P(z))\frac{\pi D^2}{4}+\rho g \frac{\pi D^2}{4}\Delta z=\tau_w\Delta z \pi D$$where z is the elevation above the bottom of the tube and $\tau_w$ is the shear stress at the wall. So, $$\frac{d(P+\rho gz)}{dz}=\frac{4}{D}\tau_w$$ For laminar flow,$$\tau_w=\frac{f}{4}\frac{\rho v^2}{2}$$where f is the Darcy-Weisbach friction factor. So, combining the two equations, you get: $$\frac{d(P+\rho gz)}{dz}=\frac{f}{D}\frac{\rho v^2}{2}$$ For a horizontal tube, you would have just: $$\frac{dP}{dz}=\frac{f}{D}\frac{\rho v^2}{2}$$ So, for vertical flow, you simply replace the P in the horizontal flow equation by $P+\rho gz$.

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  • $\begingroup$ Thanks sir. Please take a look at my new edit to the question. $\endgroup$ – Ghartal Aug 1 '16 at 8:44
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This is not a complete answer, just an outline of how to go about to derive a formula for the velocity profile and hence the pressure loss across the cross section of a pipe. I'll model the pipe as infinitely long and horizontal to simplify the problem---the result should still be applicable, but needs to be used with a proper water pressure due to gravity in the case of a vertical pipe.

The viscous force is parametrized by a fluid's viscosity $\eta$. Expressed as pressure perpendicular to a velocity gradient in the $x$-direction, it is $\frac{\partial p}{\partial y} = \eta \frac{\partial v}{\partial x}$ where $v$ is the local velocity (essentially, $\dot{y}$). This expression needs to be transformed into cylindrical coordinates (for the cylindrical pipe). Together with the usual relations (and perhaps a continuity condition), that should allow to derive a flow profile $v(r)$, which is parabolic; the solution is giveb e.g. at Hyperphysics.

Having a flow profile $v(r)$ allows to calculate the viscous forces and to get the total pressure drop (per unit length of pipe) by integrating over the pipe's cross section. That would be the (partial) answer I set out to sketch.

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  • $\begingroup$ Actually my main problem is the difference between horizontal and vertical flows. In horizontal pipe flow you have an applied pressure (for example using a pump) and you have a pressure distribution along the pipe. But in my case, what will be the pressure? Can we calculate pressure in an arbitrary point along the exiting pipe? $\endgroup$ – Ghartal Jul 31 '16 at 16:14
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    $\begingroup$ Due to continuity and symmetry, your flow profile is the same at any vertical location. Hence the pressure loss per unit length is the same everywhere, except that in your case of a vertical pipe, you need to add the effect of gravity (which I believe you have separately accounted for already). So it really doesn't make a difference that your pipe is vertical, unless it is so long that the pressure somewhere drops to zero and voids are created, violating continuity. $\endgroup$ – pyramids Jul 31 '16 at 16:18
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Water is flowing in the pipe driven by a constant pressure gradient equal to $\rho g$. So you can apply D-W equation provided flow is laminar.

Response to question edit:

In writing $-\frac{dp}{dz}+\rho g$ you separated out contribution due to gravity to pressure gradient acting on the fluid. Therefore $-\frac{dp}{dz}$ is any pressure gradient applied by means other than due to gravity (for e.g. using pump), which in your case is zero.

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  • $\begingroup$ This means that $v=\gamma D^2 /32 \mu$, which makes velocity independent of $L$. $\endgroup$ – Ghartal Aug 1 '16 at 11:17
  • $\begingroup$ You have assumed fully developed flow, so $v$ depends only on radial distance and not on axial distance. $\endgroup$ – Deep Aug 1 '16 at 11:39
  • $\begingroup$ Then friction loss in the pipe does not have an effect on exiting velocity. Also it gives $h_l=L$ and reduces Bernoulli's equation to $v^2=2gH$, but I thought that this problem must be different from Torricelli's law. $\endgroup$ – Ghartal Aug 1 '16 at 11:58

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