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If I know the pressure on one side of the straw (atmospheric) and the pressure inside the mouth, can I compute the volume flow rate? This is first disregarding the transient rise and fall and only consider the steady-state suck.

My current list of assumptions includes:

  • Newtonian fluid (water)
  • Laminar flow ($\text{Re} < 2000$), estimated assuming $D = 7\text{ mm}$ and an assumed upper fluid speed of $0.25\text{ m/s}$ Of course this will have to be re-confirmed once a flow rate has been calculated.
  • $T_{\text{atm}} = 20\text{ C}$
  • $P_{\text{atm}}$ at $20\text{ C} = 101325\text{ Pa}$

This model and calculator seemed relevant, but my inlet and outlet diameters are the same and my attempts to compute the same volumetric flow rate are around $50\%$ too small:

$$Q = C_f \times A_o \times \sqrt{2 \times \frac{\delta_p}{\rho}}$$

  • $C_f = 0.9$
  • $A_o = 3.84\times10^{-5} \text{ m^2} (D = 7\text{ mm})$
  • $\delta_p = 2000 \text{ Pa}$
  • $\rho = 999.97 \text{ kg/m}^3$

If the flow is laminar and fully-developed, I'm considering using the following taken from here and setting $\theta = 90$ degrees:

$$Q = \frac{\pi D^4 (\delta_p - \rho g L \sin \theta)}{128 \mu L}$$

Example geometry:

$D = 7E^-3 m$

$L = 20E^-2m$

$\theta = 0degrees$

Constants:

$\rho = 999.97 kg / m^3$

$g = 9.81 m / s^2$

$\mu = 1.002E^-3 Pa * s$

$\nu = 1.004E^-6 Pa * s$

Computing $Q(\delta_p = 50 Pa)$:

$Q => 2.51E^-05m^3/s => 25.1mL/s$

$V_{avg} = Q / A_c => 2.51E^-05m^3/s / 5.03E-05 => 0.5m / s$

Re-checking laminar assumption:

$Re = V_{avg} * D / \nu \Longrightarrow 10K > Re_{cr}$

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    $\begingroup$ I would suggest to use the Hagen–Poiseuille equation in combination with the pressure of the liquid column. If the straw diameter is small, you would also have to take the capillary pressure into account. $\endgroup$
    – CuriousOne
    Dec 15, 2014 at 0:05
  • $\begingroup$ Is fully developed laminar flow in a horizontal pipe relevant here? Forgive my formatting: Q = pi * delta_p * D^4 / (128 * u * L) $\endgroup$
    – tarabyte
    Dec 15, 2014 at 0:24
  • $\begingroup$ If we are using the conventional definition of "straw" as a straight pipe that is much longer than its diameter, yes, I would say so. Now, you can create non-laminar flow in a straw by sucking really hard and you can even create cavitation, in which case the formula does not apply, but I would leave those cases out of the discussion. $\endgroup$
    – CuriousOne
    Dec 15, 2014 at 0:28
  • $\begingroup$ Possibly related: physics.stackexchange.com/questions/61673/… $\endgroup$
    – Kyle Kanos
    Dec 15, 2014 at 17:42
  • $\begingroup$ Why do you say that you're not getting laminar flow? Looks to me like you'd need a mean flow velocity greater than 5 m/s. to get above $Re=2300$. $\endgroup$ Dec 16, 2014 at 4:59

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