For our design project we are working with a client in Wisconsin to help them remove snow from their roof-mounted solar panels in the winter. We are designing a system to spray water a vertical distance of 40-50 ft.

• Flow rate coming from the hose in the house measured at 4.9 $\frac{gal}{min}$
• The pressure is being pumped to the house from a series wells at 60 psi

We want to increase this flow rate to 10 $\frac{gal}{min}$ to help melt the snow off the panels more effectively. Our idea was to use a 100 gallon storage tank, we would fill this with the garden hose, and have water leave the tank to get our desired flow rate. From the tank the hose would approximately be 50 ft. in length, and at the end of the hose we were planning to use nozzle designed to give us a solid stream.
If from the tank the water leaves at 10 $\frac{gal}{min}$, using the inner diameter of the hose (0.62992 in.) we calculated the velocity of the water to be 3.1375 $\frac{m}{s}$. Will this velocity stay constant until the end of the hose and be the Vin for the nozzle? Then if we had a nozzle diameter of 0.2 in, using $A_{in}V_{in}=A_{out}V_{out}$ we calculated the water leaving the nozzle to be at a velocity of 31.09 $\frac{m}{s}$. Using the trajectory equations
$$R= Vi^2sin\frac{(2 θ)}{g}$$ and $$h=Vi^2*sin^2\frac{θ}{2g}$$
we found our stream would travel R=92.58 m and h=62.55 m Could we put a valve at the bottom of the tank, and get that flow rate and velocity to shoot the water that distance, or will we have a problem with pressure with this system? If we use a pump to pump out the water at 10 gal/min would the water still be able to shoot that far, or would we have to model our system similarly to a pressure washer and have a pump attached closer to where we would be spraying the water from?

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Hi @Jay, and welcome to Physics Stack Exchange! I edited your question to make it more generally useful –  Vineet Menon Nov 9 '11 at 5:04
@Jay: I am pretty sure this does not work as easily as you hoped. There are several reasons for that, first the water will not be a solid stream over a distance of almost 100 m, just look at a firetruck hose. The smaller the nozzle diameter the faster the stream will spread out. The flow rate might not be the best parameter to look at this problem. Calculate the necessary pressure, it will be quite high to reach such distances. –  Alexander Nov 9 '11 at 13:05
It's a great practical question. I looked it up in the engineer's handbook and found nothing. Your theorizing is good, but it's going to need some experimentation. –  Mike Dunlavey Nov 9 '11 at 22:06

Sending a stream of water through the air involves lots of losses, since the stream breaks up into small droplets and these have a lot of drag (as you ask for higher and higher flowrates, or higher heights, the scaling is going to work against you nastily). You might want to consider an alternate solution (for example, install a hose up to the panels and then route a slotted hose across the tops of the panels; instead of spraying the panels, "irrigate" them).

FWIW I don't know if your approach is practical but I would guess that it might not be. You can try to find tables of water-in-air losses, but I would be tempted to ask a fireman what practical distances are for delivering water streams into the air. They probably are able to deliver water 50' into the air, the important thing to find out is what pressure they need to do so (and then you can compare this against the no-drag calculation you have made, and therefore know a real-world fudge-factor). You are dealing with a smaller scale system than a firehose so your losses will be greater so you'll need to make your own fudge factor a bit bigger.

W.r.t. your fluid questions, you have some existing data; from this you can figure out the actual real world losses in your system:

• knowing the flow rate exiting from your hose, and its diameter, you can figure out the water pressure at the end of the hose. 60psi is the pressure you get at the exit when there is no flow; when your 4.9gpm is flowing, nearly all of this 60psi will be dissipated in losses along your hose, the small bit left over will be used to deliver the kinetic energy of the 3.13m/s discharge stream. Using p=rho*v^2/2 you can figure out mow much pressure is used to produce the jet; the rest of the 60psi is flow resistance.

• from this you know how much flow resistance your system has (e.g. 59psi of pressure drop at 4.9gpm flow). You can make some assumptions about the nature of the flow regime and the shape of the pressure-drop-vs-flowrate curve (is it linear, quadratic, cubic, etc?).

To deliver a 10gpm jet at 31m/s you have two things to accomplish:

• firstly, what does it take to deliver 10gpm to your nozzle? (you might need bigger delivery piping and hoses, or a higher pressure well pump)
• at your nozzle, how much pressure is needed to accelerate that 10gpm to 31m/s? (use p=0.5*rho*v^2 again)
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Not to forget the heat loss during "flight"! What keeps that storage tank from freezing? (Or is it planned to be indoor?). But even that irrigation system with a hose up to the roof will have problems from frosty conditions. –  Georg Nov 9 '11 at 17:21
@georg excellent point about heat loss (I completely overlooked it). Spraying small droplets through the air is a great way to chill them down to 0C, leaving virtually no sensible heat left to melt the ice/snow. –  Daniel Chisholm Nov 9 '11 at 17:48
FWIW: That's a method we used to use to melt ice dams on the roof, but we used hot water, and the scale was much smaller. –  Mike Dunlavey Nov 9 '11 at 22:09

I calculated that for an area of 159.25 sqft and 1 inch of snow we would need 100.2 gallons of water to melt all the snow, I chose to use a factor of safety of 1.5 to account for stream separation and temperature loss in the water stream. So for 1 inch of snow with that factor of safety we would need about 150.31 gallons of water. Which seems like a lot.

I assumed the temp of snow to be -10 C, the temp of the water to be 10 C, the snow and water mixing to 1 C, used 6.24 lb/ft^3 for the density of snow fall on a roof, and used for pure water C=4181.3 J/kg*K

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