@mfb and
@russ_watters,
In order to try to more clearly illustrate the point I've been trying to make, I have solved the full set of equations (iteratively) that are usually used to design these sorts of systems. Here is what I did. I started with the Bernoulli equation modified to incorporate viscous losses through the system:
\dfrac{p_1}{\rho g} + \dfrac{v_1^2}{2g} = \dfrac{p_2}{\rho g} + \dfrac{v_2^2}{2 g} + h_{L,\mathrm{major}} + h_{L,\mathrm{minor}}.
Here, ##h_{L,\mathrm{major}}## is the major head loss due to the length of the hose, and ##h_{L,\mathrm{minor}}## is the minor loss associated with the sudden contraction. The definition of the minor loss is
h_{L,\mathrm{minor}} = K\dfrac{v_2^2}{2g},
where ##K## is an experimentally determined coefficient (see
https://www.amazon.com/dp/0070621780/?tag=pfamazon01-20). When ##A_2/A_1## is zero, which is the worst case scenario, then ##K = 0.5##, so I used that here to be conservative with this. The definition of the major loss is obtained from the Darcy-Weisebach equation,
h_{L,\mathrm{major}} = f\dfrac{\ell}{D}\dfrac{v_2^2}{2g},
where ##f## is the Colebrook friction factor, ##\ell## is the length of the hose (100 ft in this example), and ##D## is the diameter (5/8" for most hoses in the US). ##f## can be determined implicitely from the Colebrook formula,
\dfrac{1}{\sqrt{f}} = -2\log_{10}\left( \dfrac{\epsilon}{3.7 D} + \dfrac{2.51}{Re\sqrt{f}} \right),
where ##\epsilon## is the absolute surface roughness (determined as 1.35 microns according to my previous source's mass flow plot, which is in line with some Googling to check that value) and ##Re## is the Reynolds number.
So, at the end of the day, if you define the volumetric flow rate ##Q = v_1 A_1##, you can use all of these previous equations and combine them to get
Q = A_1\sqrt{\dfrac{2(p_{\mathrm{tot}} - p_2)}{\rho\left[ \left( \dfrac{A_1}{A_2} \right)^2\left( 1 + K \right) + f\dfrac{\ell}{D} \right]}}.
Using this plus the Colebrook equation in a predictor-corrector scheme, you can plot the flow rate as a function of the area ratio. This relationship is plotted below:
As you can see, for the most conservative case in the contraction, closing off half of the orifice implies that the resulting mass flow through the hose is about 95% that of the uncapped case with the end of the hose completely open. You have to block about 65% of the hose to bring that down to 90%, and almost 90% of the hose to bring that down to 50% of the original mass flow.
The moral of the story is that my original post where ##A_2/A_1 = 0.5## that assumed constant mass flow was only in error by about 5%. I think that makes it a reasonably valid assumption, particularly for illustrative purposes.
russ_watters said:
Approach 1: Boneh3ad: Assume large orifice, all losses are in the pipe, ignore exit velocity/pressure.
I'd also like to point out that this is never what I did. I did assume an arbitrary orifice size for my example, but I made no assumptions about all losses being in the pipe. In fact, I was assuming the losses in the pipe would be nearly identical in the two situations and using that to only calculate the losses due to the contraction at the end. I also didn't ignore the dynamics pressure (or velocity pressure if you prefer that nomenclature) as I previously addressed.
russ_watters said:
Not a huge issue, since 75 psi is pretty big and pretty arbitrary.
It wasn't arbitrary. That was the nominal water pressure supplied to my house that I lived in when I was in graduate school.
