# Shower head for feeding water onto a solar thermal panel

• nyoo
In summary: P1 = p0 + v1 (hole in pipe) v1 = v0 - (p1-p0) (hole in pipe)Bottom figure: The velocity of the water at each hole is divided by the sum of the velocities at all the holes. So if there are n holes in the pipe, the water at each hole has velocity vn/n.
nyoo
"shower head" for feeding water onto a solar thermal panel

## Homework Statement

The question is homework to me, but I haven't actually seen formal physics classwork since college, 41 years ago. Please advise if this question is misplaced.

I'm building a "shower head" to trickle or spray water onto a home-made solar thermal panel. My pump will deliver 180 gallons per hour, unpressurised, to a tee fitting. Pressure at the tee is 2.6 psi. The tee is in the center of an 8' horizontal copper pipe, capped at both ends. All pipes are 3/4" L-type copper. (The wall thickness for L copper is 0.045".)

I will drill evenly-spaced holes into the 8' horizontal pipe, to spray water onto a panel. Think of it as a long skinny shower head, with a single row of holes. I'm looking for an even flow out of the sum of the holes, at greater than 90 GPH. There must be a sufficient number of holes drilled into the 8-foot length to cover the entire panel.

## Homework Equations

I've been staring at the Bernouilli's Principle, which I believe holds the answer to my question. But Bernouilli's Principle talks about gravity, and I'm not sure how to convert that for forced GPH pressure. And Bernouilli's Principle seems to assume a single hole, not a row of holes. I'm not sure how to proceed.

## The Attempt at a Solution

I'm thinking 32 holes of .125" diameter. My naive layman's logic is based on area of the pipe: (3/4)^^2 = .5625, 32 * (1/8)^^2 = .5. But I know this logic does not account for turbulence at the pin-hole openings.

As copper pipe is currently expensive, I'm trying to reduce the number of experimental iterations, by applying science.

The questions are: will the water come out of all holes evenly? Will the holes farthest from the central tee spray hardest? Will the ends not spray at all? Should I drill bigger holes?

Thanks very much for your help.

Welcome to PF

Somebody else in here may be able to calculate what you want, but if not I am thinking an experimental approach would be a good way to figure out the best hole spacing.

Since 8' copper pipe is expensive, maybe you could experiment on cheaper PVC tubes, trying out different hole patterns. They seem to be a few \$ each at Home Depot (if you're in USA).

You could start first with evenly spaced holes. Increase the number of holes until you get the total 90 gph flow you want, then see how evenly distributed that flow is.

You can always add more holes. You can also plug a hole by drilling and tapping it with one of the standard 32 tpi machine screw threads (6-32, 8-32, or 10-32), then a screw and some epoxy or silicone to plug and seal it.

Other than that, I can only offer the (perhaps obvious) observation that your pipe will be somewhere between the following two extremes:
• The pressure hardly varies over the length of the pipe, so evenly spaced holes deliver uniform flow over the length
• The water pressure drops to near zero at the ends of the pipe, so the hole pattern is much more closely spaced at the ends than in the middle.
If I can, I'll try to calculate the relative hole spacing for the 2nd case, so we can see what that imposes as a limit. Will have to think about that some more though.

Meanwhile, perhaps somebody else knows how to do the full calculation?

Thanks for the practical advice. It's a good idea: I will experiment using PVC, and expect uneven spacings.

If only I could get clear in my head some of the principles at work.

Like yourself, I expect the water pressure to reduce near the capped ends, furthest away from the tee. But this is based of a common sense (and probably incorrect to the point of being a cartoon) feeling that, as the water passes each hole, some pressure escapes through the hole.

But those 180 gallons-per-hour of velocity have to go somewhere. My attempts at Bernouilli's Principle say pressure increases with the square of the velocity. But isn't it pressure that's pushing water through the outlier holes?

Thanks for the encouragement.

Post #1 of solution

Okay, I think I see how to solve this thing now. I'm going to break this up into separate posts, and hopefully finish up by the end of today or tonight as I also attend to various errands and other chores at home.

Here's a figure:

Top figure:
Water at pressure p0 enters the middle of the pipe, and has velocity v0 in the middle. At each hole in the pipe, there is a change in pressure and velocity as some of the water is diverted out the hole. After the first hole, the pressure and velocity within the pipe are p1 and v1. u1 is the velocity of the water exiting the hole. After the final Nth hole, the pressure is pN and velocity is zero.

Botom figure:
Closeup of the "ith" hole. The fluid exiting the hole can be thought of as eminating from a cross-section area Bi-1 just before the hole. The water exits the hole at a velocity ui.
If A is the total cross-section, then fluid in an area A-Bi-1 remains inside the pipe as it flows past the hole.

Before going into more details, I'll first outline how to get a solution:
• Use Bernoulli's equation and the continuity equation to solve for all the pressures and velocities: pi, vi, and ui.
• The pressures and velocities don't depend on the hole spacing, but we can use the ui's to figure out the hole spacing. I.e., what hole spacing will give a uniform flow (in say GPH per foot of pipe)

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Post #2 of solution

(By the way, I am ignoring turbulence. Not sure how big a role that would play, to be honest.)

For input parameters, we have the following:

p0, the inlet pressure
A, the cross-sectional area of the pipe
voA, the volume flow rate at the inlet (from which we can calculate v0)
dH and AH, the diameter and area of each hole
ρ, the density of water

Equations for flow at each hole

Assume that we know pi-1 and vi-1, the pressure and velocity of the fluid just before the ith hole in the pipe.

For the water that flows out the ith hole, Bernouli's equation gives
pi-1 + (1/2) ρ vi-12 = (1/2) ρ ui2
from which we can calculate ui.
From the continuity equation,
Bi-1 vi-1 = Ah ui,​
we can calculate Bi-1.

For water that flows past the ith hole (i.e. remains in the pipe), the continuity equation is
(A - Bi-1) vi-1 = A vi.​
We can calculate vi from that.

Finally, from Bernouli's equation
pi-1 + (1/2) ρ vi-12 = pi + (1/2) ρ vi2,​
we can calculate pi.

We now have pi and vi, so the procedure can be repeated for the (i+1)th hole.

The procedure should be repeated until v=0 (or nearly so), since that is the condition at the end of the closed pipe.

The important numbers from this calculation are the ui's, the velocities of the water exiting the different holes. More to the point, uiAH is the volume flow rate of water exiting the ith hole.

We simply need to make the hole spacing proportional to ui, in order to have a constant flow rate per length of pipe.

What you'll need to do is determine the head loss that you have through the piping system. Do a google on pressure loss in piping and a Moody diagram. The pressure loss in the main piping will be something like:
$$\Delta H = f\frac{V^2}{2}\frac{L}{D}$$
Where f is the friction factor obtained from the Moody diagram (which unfortunately is a function of not only pipe roughness, but velocity as well, so it's semi-iterative). However, if you're velocities are high enough, then we can just assume turbulent and grab a decent estimate.

The second point of pressure loss will be the holes themselves. Do another google, and you should be able to find some sort of coefficient of discharge for that type of hole. Multiply it by the number of holes to get the total. Now go back to your Bernoulli's equation with these new head loss values. They get added to the right-hand side, essentially limiting the amount of energy available at the outlet.

From here, you can calculate a new flow. From this new flow, you need to go back and calculate new pressure losses based on this new velocity. Iterate a few times until it converges. Now you have a head that your pump needs to pump against. Try and obtain a pump performance curve for the pump. The maximum flow that you listed is more than likely against no head, or resistance. The curve will be $$\Delta H$$ vs $$Q$$, and it will curve from a maximum pressure output at now flow, to maximum flow at no pressure.

Since you now have what you are "pushing" against, you can calculate what maximum flow your pump can deliver for that system.

Thank you for your immense effort here. Be assured that I shall use this several times, and will given credit when I share it with others.

I assume the idea is to first get rho out of the equations? So we'd work in metric?

ρ, the density of water = 1 kg/m^3
p(0), the inlet pressure = 2.6 psi * 0.0689 = .1792 bars = N/m^2
A, the cross-sectional area of the pipe = (0.0254 * .75)^2 * pi = 0.00114 m^2
v(0)A, the volume flow rate at the inlet = 3 GPM * 4.4 = 13.2 m^3/hr
d(H), the diameter of each hole = .125 inch = 0.003175 m
A(H), the area of each hole = (.125 * .0254)^2 * pi = 3.1669e-5 m^2

v(0) = 13.2 / 0.00114 = 11578 m/hr Hmm? This seems odd.

I understand the pictures. I can carry on in this vain with the calculations, maybe write a macro or small program for the iterations. Please confirm this is the right direction.

I did lose you, though, in your last sentence, "the hole spacing proportional to u(i), in order to have a constant flow". Can you say that another way?

Thanks again.

nyoo said:
Thank you for your immense effort here. Be assured that I shall use this several times, and will given credit when I share it with others.

I assume the idea is to first get rho out of the equations? So we'd work in metric?
Rho can stay, it is a known quantity.

I am fine with working in non-metric, as long as it's a consistent set of units. In my analysis of your system (not posted yet), I used

lengths in inches, therefore volumes in cubic inches
time in seconds
mass in pounds-mass,
density in pounds-mass per in^3

I understand the pictures. I can carry on in this vain with the calculations, maybe write a macro or small program for the iterations. Please confirm this is the right direction.

Yes, in fact I already have done an Excel spreadsheet and will post the results soon. I'll need to double-check it for dumb errors first.

I did lose you, though, in your last sentence, "the hole spacing proportional to u(i), in order to have a constant flow". Can you say that another way?
Okay.

For example, if each hole in the middle emits twice as much water as a hole at the end, then the middle holes should be twice as far apart as the holes at the ends. That way, each foot of pipe will emit the same gph of water.

minger said:
What you'll need to do is determine the head loss that you have through the piping system. Do a google on pressure loss in piping and a Moody diagram. The pressure loss in the main piping will be something like:
$$\Delta H = f\frac{V^2}{2}\frac{L}{D}$$
Where f is the friction factor obtained from the Moody diagram (which unfortunately is a function of not only pipe roughness, but velocity as well, so it's semi-iterative). However, if you're velocities are high enough, then we can just assume turbulent and grab a decent estimate.

Such effects are getting beyond what I'm familiar with, but I'll post my solution (which ignores turbulence) later tonight.

I concur with Minger. You should only really need to model the head loss trough the pipe and out the holes which there are plenty of empirical correlations and models for. You will probably need to write a numerical solver using the Eulerian description to solve the problem. Any fundamental fluid mechanics text will have all the information you require except for actually writing the code to solve the problem.

Hi nyoo. Welcome to the board. I concur with minger also - the proper way to evaluate this is to check permenant pressure drop per the Darcy-Weisbach equation as a first step. Once you do that, you can then design a header system within which the permenant pressure drop through the pipe is small compared to the pressure drop across any given orifice. Then you can use the Bernoulli equation as Redbelly is pointing out. In other words, consider this header to be a big ‘tank’ and all the pressure drop is across the individual restrictions (the holes) in the bottom of your ‘tank’.

So in this case, you want to prove that your header doesn’t produce a large pressure drop in relationship to the individual holes. And if you have only 2.6 psi pressure downstream of your pump to play with, then you want most of that drop (~2.4 psi) across the various holes.

For your header, with a ¾” type L pipe (ID =.785”) and a flow of 90 GPH, the pressure drop is very small. You could even use 1/2” copper (5/8” actual OD) as long as the tube coming out of your pump is ¾” Type L copper (7/8” actual OD) which is no more than 2 feet long, going into a ¾” T and then reducing down to ½” on both sides. If this tube is more than 2 feet long, then you should calculate the dP per the Darcy-Weisbach equation. Note that above tube sizes are nominal except as noted.

So if you have 1/2” Type L copper tube, you can calculate the flow through any given hole from the dP produced. For this case, I’d suggest using a discharge coefficient between 0.6 and 0.8. Let’s use 0.7 for the sake of argument, it won’t be that far off.

For a .125” orifice with 2.5 psid across it, the flow is roughly 0.515 GPM. You can use a calculator found on the internet to do this if you’d like:
http://www.efunda.com/formulae/fluids/calc_orifice_flowmeter.cfm#calc

Since you are only flowing 90 GPH per side (or 1.5 GPM) you only need 3 holes at that size. Needless to say, the .125” holes are too big. Reduce the size of the holes and recalculate using the above calculator.

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Post #3 (final one?) of solution

Ignoring turbulence

Okay, I finished up the solution that neglects turbulence, using the Bernouli and continuity equations as I had outlined earlier.

Not too different than what Q_Goest had, I came up with 4 holes for the stated pressure, flow, tube size, and hole diameter.

If you go to 1/16" diameter holes--1/4 the area of a 1/8" hole--you'll need 4 times as many, or 16, holes. That's a spacing of 6" between holes for the 8' pipe length.

Also, surprisingly to me, the flow velocity out of each hole is identical, so they should be spaced evenly.

The turbulence considerations are well outside my area of expertise, so I will leave things there for now. That PVC prototype may be the way to go at this point, but it does appear that smaller holes are required.

Regards,

Mark

EDIT: I should also mention, I have assumed a horizontal pipe so that gravity (height differences) does not play a role. However, it was never said if this will be the actual situation...

EDIT #2: Just re-read post #1. It IS to be a horizontal pipe after all, good.

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Thank you, all, for your insights.

I now have an answer, and that's marvellous. But, at some quite large effort to yourselves, you have also provided me with the tools so that if the circumstances of my design change, I can accommodate them. (Or if that neighbour down the street, whose back yard does not dip so much, asks for help.)

I shall, of course, do the PVC prototype, as suggested.

You're great. You gave a man a fish, but also taught him to fish for himself.

Um, I knew there was going to be one last thing. There always is.

Q_Goest said:
So if you have 1/2” Type L copper tube, you can calculate the flow through any given hole from the dP produced. For this case, I’d suggest using a discharge coefficient between 0.6 and 0.8. Let’s use 0.7 for the sake of argument, it won’t be that far off.

Could you please show me how you arrived at that 0.6 - 0.8 range for discharge coefficient?

Thanks again.

Hi nyoo. Sorry, but that just comes from experience. The holes are basically a flat plate orifice, but the discharge coefficient changes considerably depending on pressure drop across it. The value of 0.7 is about in the middle of the range of what I've seen and should be close enough for what you're trying to accomplish.

## 1. What is a shower head for feeding water onto a solar thermal panel?

A shower head for feeding water onto a solar thermal panel is a specialized device used in solar water heating systems. It is attached to a water supply and helps to evenly distribute water onto a solar thermal panel to maximize heat absorption.

## 2. How does a shower head for feeding water onto a solar thermal panel work?

The shower head works by channeling water from the water supply onto the solar thermal panel. The water is then heated by the sun's energy and circulated back into the water supply to be used for showering or other purposes.

## 3. What are the benefits of using a shower head for feeding water onto a solar thermal panel?

Using a shower head for feeding water onto a solar thermal panel can help reduce energy costs and carbon footprint. It also allows for more efficient use of solar energy, as the water is evenly distributed onto the panel for maximum heat absorption.

## 4. Can a shower head for feeding water onto a solar thermal panel be used in all types of climates?

Yes, a shower head for feeding water onto a solar thermal panel can be used in various climates. However, it may be more effective in areas with abundant sunlight and warmer temperatures.

## 5. Is it difficult to install a shower head for feeding water onto a solar thermal panel?

The installation process for a shower head feeding water onto a solar thermal panel is similar to that of a regular shower head. It may require some basic plumbing and knowledge of solar water heating systems, but it can be easily installed with the help of a professional or by following installation instructions.

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