I'm doing the following problem of White's Fluid Mechanics:

Water flows downward in a pipe at $45^0$, as shown in the figure. The pressure drop $p_1 – p_2$ is partly due to gravity and partly due to friction. The mercury manometer reads a 15 cm height difference.

enter image description here

I found the pressure drop between $p_1$ and $p_2$ by using the usual hydrostatic equation $p_1+\gamma _{water}*(1.06+y+0.15)-0.15*\gamma_{Hg}-y*\gamma_{water}=p_2$ and solved for $p_1-p_2$. My question is, which term in the equation corresponds to friction loss? And why?

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    $\begingroup$ I don't believe that equation accounts for friction loss. What does the question want you to find and what we're you given? $\endgroup$ – JMac Jan 24 '17 at 20:40

None of these terms correspond to the friction loss. The equation you wrote down is, as you say, just the hydrostatic pressure equation, which knows nothing about friction. You find the part of the pressure drop that is due to friction loss by comparing the pressure drop you would see without friction (which would be the pressure drop between points 1 and 2, if there was no flow), with the actual pressure drop you just calculated. You should be able to take it from there. Let us know if you need more help.

  • $\begingroup$ Of course! So, comparing both pressure drop equations with and without flow, respectively: $p_1-p_2=\gamma_{Hg}*0.15 - \gamma_{water}*(1.06+0.15)$ and $p_1-p_2= \gamma_{water}*(1.06)$. What I understood is that, considering no flow, I ignore the manometer and assume I have both ends of the pipe closed? If my train of thought is correct, then $\gamma_{Hg}*0.15-\gamma_{water}*0.15$ is the pressure drop due to friction. So the manometer only reads the friction loss I guess? Thank you! $\endgroup$ – RicardoP Jan 24 '17 at 23:29

There are two phenomena you have to investigate here:

  1. Conservation of energy from 1 to 2
  2. Fluid statics in the manometer


Assuming steady, incompressible, inviscid (frictionless) flow and uniform velocity along the tube, you can use the Bernoulli equation (or to be more precise, the extended Bernoulli equation that includes the pressure loss) along a streamline, that is from 1 to 2:

$p_2+\frac{1}{2}\rho_{w}V_2^2+\rho_{w}gz_2 =p_1+\frac{1}{2}\rho_{w}V_1^2+\rho_{w}gz_1 - p_\text{loss}$

Note that by definition $p_\text{loss}$ is always positive and it acts as an energy reducing term to the upstream point. From the conservation of mass, you can immediately notice that $V_1=V_2$ (because $\rho_w A_1 V_1 = \rho_w A_2 V_2$ and density and area are constant), hence the velocity terms cancel out:

$p_2+\rho_{w}gz_2=p_1+\rho_{w}gz_1 - p_\text{loss}$

$p_1-p_2=\rho_{w}g (z_2-z_1) + p_\text{loss}$

$p_1-p_2=\rho_{w}g (-1.061m) + p_\text{loss} \dots \text{Eq. 1}$

From the fluid statics, we know that $p'_1=p'_2$ (see the figure) since we can connect both points through a constant density fluid (mercury) and that fluid is at rest.



Subtracting the equations;

$0 = p_1-p_2+\rho_{w}g(h-h_1)-\rho_{m}g(0.15m)$

$p_1-p_2 = -\rho_{w}g(1.211m)+\rho_{m}g(0.15m) \dots \text{Eq. 2}$

Now equate Eq. 1 and Eq. 2;

$\rho_{w}g (-1.061m) + p_\text{loss} = -\rho_{w}g(1.211m)+\rho_{m}g(0.15m)$

$ p_\text{loss} = -\rho_{w}g(0.15m)+\rho_{m}g(0.15m)$

$ p_\text{loss} = g(0.15m)(\rho_{m}-\rho_{w})$

Using standard values of $\rho_{m}=13593 kg/m^3$, $\rho_{w}=1000kg/m^3$, $g=9.81m/s^2$:

$ p_\text{loss} = 18.53 kPa$

Please check for calculation errors before counting on my solution.


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