How does shape affect fluid flow for a given section area? I'm a designer and I have no background in physics and have almost no understanding of fluid behavior.
I've been asked to change the dimensions of a reflection chamber on a lab spectrometer.
A tube with about 29mmˆ2 section comes into the chamber, the chamber then changes shape to a rectangular section, 2mm by 14.5mm. So far so good.
I've been asked to change the dimensions of the chamber so that it now has 0.5mm thickness (by 4*14.5mm, so that total section areal is still the same). I replied by asking if the restriction of this new shape would be the same, even if the section area is still the same. The person who asked for me to change the shape couldn't provide an assertive answer.
I mean, you can't infinitely vary the thickness of a shape and make it wider and expect it to behave in the same way, right? At some point too thin of a section will be too restrictive irregardless of total section area, right?
If so, is there a way we can calculate, simulate or test this?
The liquids to be analysed are mainly water solutions at ~30degreeC. I don't have exact figures on fluid speed but the chamber is gravity fed through a funnel and a "cup".
 A: Geometry indeed can effect resistance to flow depending how energy losses are incurred and whether they are recoverable or unrecoverable losses. In your case water is an incompressible fluid and so to maintain continuity of mass flow rate through a conduit of changing cross section, the fluid must either speed up or slow down
$$\dot{m}=v_1A_1=v_2A_2 $$
But since energy must also be conserved
$$P_1+\frac{1}{2}\rho {v_1}^2=P_2+\frac{1}{2}\rho {v_2}^2 $$
The $P$'s are the respective pressures at sections 1 and 2. Think of the first term of each side of the the equation as potential fluid energy (pressure) and the second term as kinetic energy (the fluid velocity).
The resistance to flow is
$$R=\frac{P_1-P_2}{v_2A_2}$$
The pressure loss here is said to be recoverable since you can design your conduit to theoretically get back to the original static pressure you started at.
But then there are unrecoverable losses, not due to changes in inertia but rather friction, but in terms of fluids, viscosity. Viscous fluid losses turn kinetic energy into heat. And the heat can either raise the temperature of the fluid (an internal energy) or escape through the fluid conduit boundaries. Viscous losses occur when fluids drag along conduit boundary layers, layers within the fluid itself or when higher dimensional flow occurs (turbulence, vorticity).
Water indeed is a viscous fluid so as you increase surface area in your design you increase the chance for more drag and energy loss which increases resistance to flow, but by a different mechanism (heat production). Borrowing from Robin's post the resistance due to viscous loss would be.
$$R=\frac{K h^3 b}{12 \eta l }$$
Both of these processes can happen concurrently and you might be able to get a rough idea using the two expressions for your problem, but generally speaking simple calculations more often fail to accurately predict. So the best way is to either build a prototype and test it to see if it meets your requirements or enter your design into a Computational Fluid Mechanics software tool.
A: The following is academic knowledge. I'm no pipe flow expert.
For laminar flow through a rectangular pipe, the volumetric flow rate $\dot V$ calculates to
$$
\dot V=\frac{K \cdot h^3 \cdot b}{12 \, \eta \, l} \cdot \Delta p
$$
with $b$ and $h < b$ being the pipe dimensions, $l$ pipe length, $\eta$ viscosity, $\Delta p$ pressure drop along the pipe and $K$ depending on $h/b$. (This is from Wikipedia which I didn't verify, but seems reasonable to me.) For the current geometry, $h/b \approx 0.14$, thus $K \approx 0.9$. For your new geometry, $K \approx 1$.
Assuming constant $\Delta p$, $\eta$ and $l$, the product $K \cdot h^3 \cdot b$ must be the same for two cross-section areas in order for $\dot V$ to remain constant. So almost certainly the volumetric flow rate of the two chambers will not be the same.
Note 1: Since your device is gravity fed, a constant $\Delta p$ seems a reasonable assumption.
Note 2: This is for laminar flow only. For turbulent flow, things are much more complicated. Transition for pipe flow depends on the surface roughness of the pipe. Usually flow is assumed to be laminar as long as $Re < 2300$.
 For this problem, $Re=\frac{v\cdot h \cdot \rho}{\eta}$ with velocity $v$ and density $\rho$. For water at 30°C, $\eta/\rho \approx 0.08 \mathrm{\frac{m^2}{s}}$. Since $h$ is in the millimetre regime, the flow should be laminar.
