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Given an orifice rated 1.8 GPM of water at 90 PSI, how do I calculate the flow rate at 30 PSI and what rating I would need to purchase to achieve 1.8 GPM at 30 PSI?

Below added to give some context, based on responses. Sorry, as a programmer I always reduce problems to the simplest form. Apparently, that's a bad thing on this site.

I'm trying to fix the shower in my new apartment in Chicago. The building owner says that as long as warm water comes out she has met her legal requirement. Google seems to agree with her. I've added "check shower, sinks, etc." to my apt inspection checklist for my next move, but that doesn't help me now. I should probably Google "apt inspection checklist" next time to see what else I should have looked at.

I don't know if the problem is in the pipes, values, showerhead, or what. The sink works a lot better now that I removed the 2.2 GPM aerator. Changing the showerhead would be the equivalent for a shower, but simply removing it doesn't let you see if that fixes the problem like removing an aerator does. I'm guessing flow rate is proportional to pressure, but I don't know if it is a major or minor effect.

The showerhead is labeled 1.8GPM, which seems to be fairly normal for a showerhead. I used a Watts gauge with an aerator adapter on the sink to find my water pressure. I spoke with the showerhead manufacturer (Waterpik) and they stated that they rate their showerheads based on 90 PSI. They were unable to tell me how well they would work at 30 PSI. Seems like a scam to me. Who has 90 PSI water pressure?!?!

All the formulas I find on the Internet require all kinds of other information I don't have and are just too complex for me to figure out.

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    $\begingroup$ The mentality here about presenting the problem is fundamentally different. Usually, the devil is in the details, and abstraction does not help. $\endgroup$
    – NMech
    Jun 8 at 16:26
  • $\begingroup$ @Pascal, the original question was perfectly fine IMO, it's just different from your higher level goal. In light of the update I would check the DIY / Home Improvement stack. Consider the possible limits of the building hot water system also. Also you can measure this scale flow easily with a bucket. $\endgroup$
    – Pete W
    Jun 8 at 20:30
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    $\begingroup$ I measured the flow rate from the showerhead using a funnel and gallon jug. It filled up in about 56 seconds, so about 1.07 GPM. Significantly less than the 1.8 GPM it is labeled, but consistent with the calculated value from this question, confirming that the showerhead itself was the problem. I took it apart and figured out how to remove the regulator. That brought the output to about 1.62 GPM. Still not great, but a lot better than it was, and just over the 1.53 GPM maximum from the Energy Policy Act of 1992, so I'm not sure replacing the showerhead would accomplish any more. $\endgroup$
    – Pascal
    Jun 9 at 4:29
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Here is a reference from RoyMech, see section under "Orifice Flow Meter". The setup has the orifice placed in a pipe.


Below is a similar, perhaps simpler, expression, from Jobson 1955, part of the way between equations (4) and (5) of that paper. (The paper is really about incompressible flow, this is just a warmup for the author).

$$\dot{m} = {{\rho}_0}{C_i}{A}{\sqrt{\frac{2(p_0-p)}{{\rho}_0}}}$$

Where $A$ is orifice area, ${\rho}_0$ and $p_0$ are upstream density and pressure, $p$ is downstream pressure, and $C{_i}$ is contraction coefficient. Typical $C_i=\pi/(\pi+2)=0.611$, for a theoretical thin-plate.

The takeaway is that it scales with $\sqrt{{\Delta}p}$

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  • $\begingroup$ @NMech are we now a free homework completion site when no effort is shown? $\endgroup$
    – Solar Mike
    Jun 8 at 5:16
  • $\begingroup$ @SolarMike you are probably right this is an exercise, that's why I stopped from writing the second part of the question, and only did one division and one multiplication. However, because you are posting this on another persons answer - which is essentially identical to mine (apart from said division) (and which btw I upvoted before your comment)- and given our recent exchanges of comments I can't help feeling that I've done something to offend you. If that's the case, I would appreciate if you had some time to explain to me in private what that is. $\endgroup$
    – NMech
    Jun 8 at 5:49
  • $\begingroup$ @NMech posted as relevant to both answers. Is there a change to what we expect? Or to how we respond to these questions? Are you both showing the "new" way? $\endgroup$
    – Solar Mike
    Jun 8 at 5:52
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    $\begingroup$ @solarMike I am more "trigger happy" to some questions because I see them as a way to go out of my comfort zone and learn something new. In some subjects, I intentionally expose myself and my understanding and I expect people to upvode/downvote to show me if I got something right or not. This was one of those cases. $\endgroup$
    – NMech
    Jun 8 at 5:58
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    $\begingroup$ @SolarMike, I'm aware of that possibility, that's a fair criticism and I feel that way about some questions that get answered. But I do think we can make a decision each time if it has some extra value, for future searches or reference or learning, including our own. I also find SE motivates me to actually follow up on things I am curious about. So this time I made a different decision on that. Cheers $\endgroup$
    – Pete W
    Jun 8 at 12:27
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If you know the external diameter and orifice diameter, From Wikipedia the orifice plate equation :

enter image description here

$$q_m =\frac{ C_d}{\sqrt{1-\beta^4}}\epsilon\frac{\pi}{4} d^2 \sqrt{2\Delta p \cdot \rho_1} $$

where:

  • $C_{d}$ = coefficient of discharge, dimensionless, typically between 0.6 and 0.85, depending on the orifice geometry and tappings
  • $\beta$ = diameter ratio of orifice diameter d to pipe diameter D, dimensionless
  • $\epsilon$ = expansibility factor, 1 for incompressible gases and most liquids, and decreasing with pressure ratio across the orifice, dimensionless
  • $d$ = internal orifice diameter under operating conditions, m
  • $\rho _{1}$ = fluid density in plane of upstream tapping, $kg/m^3$
  • $\Delta p$ = differential pressure measured across the orifice, $Pa$

What you'd do is you first solve for $C_d$ for the know flow and pressure and then use that value for the new pressure.


Since you don't mention the Diameter for the Pipe and orifice what you could do is solve for the whole term

$$\frac{ C_d}{\sqrt{1-\beta^4}}\epsilon\frac{\pi}{4} d^2 \sqrt{2\rho_1}$$

in which case the above equation simplifies to:

$$q_m =k \sqrt{\Delta p}$$ where:

  • $k =\frac{ C_d}{\sqrt{1-\beta^4}}\epsilon\frac{\pi}{4} d^2 \sqrt{2\rho_1}$

in the case at hand $k=0.1897 \frac{GPM}{\sqrt{PSI}}$

which results in a flow of 1.0392 GPM at 30 PSI.


All (other) things being equal, you should have an 3 GPM @90psi if you want a 1.8GPM@30psi.

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Flow rate is directly proportional to the square root of pressure. So for any given orifice, all else being equal, $\frac{GPM}{\sqrt{PSI}}$ is constant. Technically PSI here is the pressure difference across the orifice, but with no backpressure only the input pressure matters.

For an orifice rated 1.8 GPM at 90 PSI, $\frac{GPM}{\sqrt{PSI}} = 0.1897$. To calculate GPM for the same orifice, just multiply $0.1897$ by $\sqrt{PSI}$. So for 30 PSI: $0.1897*\sqrt{30}=0.1897*5.477 = 1.039$ GPM.

For an orifice rated 1.8 GPM at 30 PSI, $\frac{1.8}{\sqrt{30}} = 0.3286$. To calculate GPM for the same orifice, just multiply $0.3286$ by $\sqrt{PSI}$. So for 90 PSI: $0.3286*\sqrt{90}=0.3286*9.487 = 3.118$ GPM.

The above is based upon the answers by NMech and Pete W, I just removed all the non-relevant parts and focused on the original question.

Unfortunately, the United States Energy Policy Act of 1992 limited showerheads manufactured after 1993 to 2.5 GPM at 80 PSI. $\frac{2.5}{\sqrt{80}} = 0.2795$ yielding a maximum at 30 PSI of $0.2795*\sqrt{30} = 0.2795*5.477 = 1.531$ GPM.

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  • $\begingroup$ You may want to reduce your significant figures to just 2 or 3. For this scenario, I don't think the accuracy is going to substantially change. $\endgroup$ Jun 11 at 16:10
  • $\begingroup$ Good idea. Thank you. $\endgroup$
    – Pascal
    Jun 12 at 1:27

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