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