What determines the dominant pressure-flow relationship for a gas across a flow restriction? If one measures the pressure drop across any gas flow restriction you can generally fit the relationship to
$$\Delta P = K_2Q^2+K_1Q$$
where $\Delta P$ is the pressure drop and $Q$ is the volumetric flow
and what I've observed is that if the restriction is orifice-like, $K_2 >> K_1$ and if the restriction is somewhat more of a complex, tortuous path, $K_1 >> K_2$ and $K_2$ tends towards zero.
I get that the Bernoulli equation will dominate when velocities are large and so the square relationship component. But what's determining the $K_1$ component behavior? Is this due to viscosity effrects becoming dominant? Does the Pouiselle relationship become dominant?
 A: Regimes


*

*1 - low Reynods number:


*

*viscosity (shear stress) dominated

*less likely to present transition to turbulence

*$\Delta P \propto Q,\,\,\,$ Eq.(1): linear ($K_1$)


*2 - high Reynolds number:


*

*inertia dominated

*more likely to present transition to turbulence

*$\Delta P \propto Q^2,\,\,\,$ Eq.(2): quadratic ($K_2$)




Regime 1 is described by equations such as Hagen–Poiseuille equation and Darcy's law.
Original Poster's (OP) expression actually coincides with a known generalization of Darcy' law for higher Reynoulds numbers, the Darcy–Forchheimer law:
$$ \partial P/\partial x  = AQ^2 + BQ $$
And also the Hagen–Poiseuille "fails in the limit of low viscosity, wide and/or short pipe" [Wikipedia] and is then bounded by Eq.(2).

References


*

*Main reference: the Laminar & Turbulent Flow lecture at Jens Ducrée's myFluidix.com, where it's shown, for concrete examples, that:


*

*When shear stress is important, that is, when the Reynolds number is low, viscosity dominates the flow and you have (pg. 26) Regime 1;

*For high Reynolds numbers, Blasius expression for the Fanning friction factor gives (pg.51)
$$\Delta P \propto Q^{7/4},$$
of which Eq.(2) is an approximnation.


*In the sixth chapter [Eq. (6.1.27) and (6.1.13)] of his book Fluid Mechanics for Chemical Engineers, Wilkes detailedly corroborates Scenario 1, i.e., he shows that low-Reynods-number flows (i.e., viscosity dominated) satisfy Eq.(1).

*In Emerson's Differential pressure Engineering Guide it's shown, from Bernoulli's equation$^\mathrm{footnote 1}$, that high-Reynolds-number flows follow Eq.(2) (their Eq. 3.15). Later in the book, a general power-$n$ dependency is found (Eq. 11.1.14) for non-Newtonian fluids.

*This anaesthesiology tutorial (or here) offers an intuitive physical explanation:
"[with turbulence,] flow is less ordered and the eddy currents react with each other, increasing drag or resistance to flow. As a result, a greater energy input is required for a given flow rate when flow is turbulent compared to when flow is laminar. This is best demonstrated by the fact that in turbulent flow, the flow rate is proportional to the square root of the pressure gradient, whereas in laminar flow, flow rate is directly proportional to the pressure gradient."

$^\mathrm{footnote1}$ : Bernoulli's equation is valid as long as shear stress isn't important: high Re away from boundary layers: "outside of the boundary layer, even real, viscous flows can be treated as inviscid [and] you can apply Bernoulli's equation", as argued in this very informative forum thread.

Orifices
Flows through orifices are typically associated to high Reynolds numbers. That can be seen from the variation in speed being likely to be large with abrupt diameter variations or, alternatively, by considering the orifice as a limiting case of a finite tube with increasing diameter (tutorial). And Wikipedia also gives $\Delta P \propto Q^2$ for orifices.
Interestingly, though, there is a highly cited paper, "An orifice flow model for laminar and turbulent conditions", which "provides a linear relation for small pressure differences and the conventional square root law for turbulent conditions".

Navier-Stokes
What follows is a sketch some steps to qualitatively obtain OP's expression from Navier-Stokes (NS) equations. It's more hand waving than mathematical proof, since longer, proper derivations can already be found in the references above.
I consider only one dimension whenever possible, in the fashion that's done in the "Reynolds number" section of Bob McGinty's website. We start with the usual NS equation:
$$\nabla P = \frac{1}{\mathrm{Re}} \nabla^2\mathbf{v} - \frac{D\mathbf{v}}{Dt}.$$


*

*1st: in 1-D, $\nabla$ is a spatial partial derivative ($\nabla P = \partial P/\partial x$, or, for steady flows, simply $dP/dx$) and, since we're interested in the integral drop in pressure along the fixed length $\Delta x = L$ along the obstruction, the term might be further simplified to $\Delta P/L$ (dimensional).

*2nd: $\nabla^2\mathbf{v} = \partial^2v/\partial x^2 + \partial^2v/\partial y^2$ is the viscosity term (it can be understood as a diffusion of moment).

*3rd: $D\mathbf{v}/Dt$ is the material derivative and stands for $\partial \mathbf{v}/ \partial t + \mathbf{v}\cdot\nabla \mathbf{v}$. For a steady flow the first term vanishes, and the second term is $(v \partial v/\partial x + u \partial v/\partial y)$, which is the convective acceleration, i.e., a spatial change of speed, and where $u$ denotes the speed in the direction(s) perpendicular to the (free) flow.
With that, NS becomes:
$$\frac{dP}{dx} = \frac{1}{\mathrm{Re}} \left(\frac{\partial^2v}{\partial x^2} + \frac{\partial^2v}{\partial y^2}\right) - \left(\frac{v\partial v}{\partial x} + u\frac{\partial v}{\partial y}\right) $$


*

*Regime 1 - low Reynods number:


For small $\mathrm{Re}$ values, the viscous term becomes dominant and we can neglect the inertial (material derivative) term. Considering an $x$-independent $v=v(y)$ profile ($\partial^2v/\partial x^2$ then vanishes, and, e.g., $\partial^2v/\partial y^2$ is constant for a parabolic profile), and we can write the NS equation as
$$\Delta P = \frac{1}{\mathrm{Re}}\frac{d^2v}{dy^2} \Rightarrow $$
$$\int \Delta P dy = \frac{1}{\mathrm{Re}} \int \frac{d^2v}{dy^2}dy
                    = \frac{1}{\mathrm{Re}} \int \frac{d}{dy}\left(\frac{dv}{dy}\right)dy\Rightarrow$$
$$\int\int \Delta P dy^2 = \frac{1}{\mathrm{Re}} \int \frac{dv}{dy}dy\Rightarrow
 v \propto \Delta P $$
The flow $Q = \int v\, dA$ is therefore of order
$$ Q \propto \Delta P. \,\,\,\, \mathrm{ Eq.(1)}$$


*

*Regime 2 - high Reynods number:


For large $\mathrm{Re}$ values, the viscous term vanishes from NS equation and, if we consider the flow through the obstruction to be mostly unidirectional (such as through a orifice) we also have $u\approx0$ and can write:
$$ \frac{dP}{dx} = - v\frac{dv}{dx} \Rightarrow$$
$$ \int \frac{dP}{dx}dx = - \int v\frac{dv}{dx} dx = - \int v dv = v_1^2 - v_2^2$$
For an incompressible fluid, $v_1A_1 = v_2A_2$, so $v_1^2-v_2^2 = v_1^2(1-A_1^2/A_2^2)$, thus
$$ \Delta P \propto v^2 \Rightarrow v \propto \sqrt{\Delta P},$$
and the flow $Q = \int v\, dA$ is
$$ Q \propto \sqrt{\Delta P}. \,\,\,\, \mathrm{Eq.(2)}$$

Original Answer
It seems that the unified description you're after is given by the transition between laminar and turbulent flows, quantified by the Reynolds number - with turbulent flow being linked to the quadratic term and laminar flow to the linear one.
One might perhaps be able to glean that much from the nondimensional form of the incompressible Navier–Stokes equations:
$$\nabla p = \frac{1}{\mathrm{Re}}\nabla^2\mathbf{v} - \frac{D\mathbf{v}}{Dt}.$$
But it's the Mathworks Simscape documentation that gives me more confidence about that:

pressure differences [...] proportional to the square of the flow rate [...] is the typical behavior for turbulent flow. However, for laminar flow, the pressure difference becomes linear with respect to flow rate

You can find also some information on the coefficients of the $\Delta P$ equation for the laminar and turbulent cases here.
A: Solutions that depend upon $\Delta P \propto Q^{2}$ are related to the normal aerodynamic drag, which just depends upon the ram pressure, $\rho \ U^{2}/2$, a drag coefficient, $C_{d}$, and the cross-sectional area affected, $A$, or:
$$
F_{d} = \frac{1}{2} \ \rho \ U^{2} \ C_{d} \ A \tag{1}
$$
Solutions that depend upon $\Delta P \propto Q$ are dominated by Stokes drag, where the drag force is given by:
$$
F_{s} = 6 \ \pi \ \eta \ r \ U \tag{2}
$$
where $\eta$ is the dynamic viscosity, $r$ is an effective radius or scale-size of the object, and $U$ is the bulk flow velocity relative to the object.

I get that the Bernoulli equation will dominate when velocities are large and so the square relationship component. But what's determining the $K_{1}$ component behavior? Is this due to viscosity effrects becoming dominant? Does the Pouiselle relationship become dominant?

Yes, it's an effective viscous effect.  When the restriction or obstacle are a single shape and flow is relatively steady, laminar, then Equation 1 above dominates.  If the flow path has multiple turns or the fluid is highly viscous or the flow is turbulent, then Equation 2 above dominates.
Stokes drag (Equation 2) arises from the strain tensor in the Navier-Stokes equations while the aerodynamic drag (Equation 1) arises from an approximation for the pressure tensor from Bernoulli's equation.
The separation between the two is approximated by the Reynolds number given by:
$$
Re = \frac{ \rho \ U \ L }{ \eta } \tag{3}
$$
where $\rho$ is the mass density of the fluid and $L$ is the characteristic scale size.  Turbulent flow onsets for high $Re$, but the onset depends on whether the fluid flows around an obstacle (e.g., $Re > 10^{5}$) or through a pipe (e.g., $Re > 10^{3}$).
There simple limits/examples to consider when Equations 1 and 2 dominate.  If you try to drag a stick/rod through a thick, viscous fluid like honey then obviously Equation 2 is the relevant drag force.  If you drag the same stick/rod through the air as fast as your arm can move, then Equation 1 will dominate.
If you have a fluid that falls in between, then both equations can play significant roles.  For instance, air is generally not tremendously viscous but if you force it through a small pipe with multiple bends at high enough speeds, it can behave like a viscous fluid.  Unfortunately, the Reynolds number is not an exact parameter, in that it does not state that at the value $Re = X$ then the flow is exactly laminar.
